Chapter 1 Number System (Class 6 - Maths NCERT Exemplar Solutions)

The given expression is a sum of place values multiplied by digits.

Let's evaluate each term in the expression:

$3 \times 10000 = 30000$

$0 \times 1000 = 0$

$8 \times 100 = 800$

$0 \times 10 = 0$

$7 \times 1 = 7$

Now, we sum these values:

$30000 + 0 + 800 + 0 + 7 = 30807$

The value of the expression $3 \times 10000 + 0 \times 1000 + 8 \times 100 $$ \ + 0 \times 10 + 7 \times 1$ is 30807.

Comparing this value with the given options, we find that it matches option (B).

The correct option is (B).

We need to determine which of the given options is equivalent to 1 billion.

Let's define the values of a billion, a million, and a lakh according to the international and Indian number systems:

1 billion = $1,000,000,000$ (One thousand million) in the international system.

1 million = $1,000,000$ (Ten lakh) in the international and Indian systems respectively.

1 lakh = $100,000$ in the Indian system.

Now let's evaluate each option:

(A) 100 millions

$100 \times 1,000,000 = 100,000,000$

This is equal to 100 million, which is $0.1$ billion.

(B) 10 millions

$10 \times 1,000,000 = 10,000,000$

This is equal to 10 million, which is $0.01$ billion.

(C) 1000 lakhs

$1000 \times 100,000 = 100,000,000$

This is equal to 100 million, which is $0.1$ billion.

(D) 10000 lakhs

$10000 \times 100,000 = 1,000,000,000$

This is equal to 1 billion.

Comparing the results, we find that 1 billion is equal to 10000 lakhs.

The correct option is (D).

Solution:

Let's evaluate the rules for each Roman numeral provided:

(A) LXII: $L + X + I + I = 50 + 10 + 1 + 1 = 62$. This is correct.

(B) XCI: $C - X + I = 100 - 10 + 1 = 91$. This is correct as $X$ can be subtracted from $C$.

(C) LC: Here, $L$ (50) is placed before $C$ (100) to imply subtraction ($100 - 50$). However, according to Roman numeral rules, the symbols V, L, and D are never subtracted. To represent 50, we simply use '$L$'. Therefore, $LC$ is incorrect.

(D) XLIV: $(L - X) + (V - I) = (50 - 10) + (5 - 1) = 40 + 4 = 44$. This is correct.

The incorrect Roman numeral is $LC$.

The correct option is (C).

Solution:

Let's check the result of each operation:

(A) $5 + 0 = 5$ (Defined)

(B) $5 - 0 = 5$ (Defined)

(C) $5 \times 0 = 0$ (Defined)

(D) $5 \div 0$

In mathematics, division of any number by zero is undefined. This is because there is no number which, when multiplied by $0$, gives $5$.

The correct option is (D).

Given:

A non-zero whole number and its successor.

To Find:

The number which always divides the product of these two numbers.

Solution:

A whole number starts from $0, 1, 2, 3, ...$ and so on. Since the question mentions a non-zero whole number, we can start taking numbers from $1$.

The successor of a number is the very next number, which we get by adding $1$.

When we multiply a number by its successor, we are multiplying two consecutive (one after another) numbers.

In any two consecutive numbers, one number is always even and the other is always odd.

$\text{Even number} \times \text{Odd number} = \text{Even number}$

Since the product of an even number and an odd number is always an even number, the final result will always be divisible by $2$.

Verification with Examples:

Let us check the product for different whole numbers to see which option is always correct:

From the table, we can see that the product is always divisible by $2$. It is not always divisible by $3, 4,$ or $5$.

Alternate Solution:

We can think of this using the property of even numbers. Any group of two consecutive numbers must contain one multiple of $2$.

For example, in the pair $(1, 2)$, the number $2$ is a multiple of $2$.

In the pair $(2, 3)$, the number $2$ is a multiple of $2$.

In the pair $(3, 4)$, the number $4$ is a multiple of $2$.

Because there is always a multiple of $2$ in the pair, their product will always be divisible by $2$.

Therefore, the correct option is (A).

We need to find the number of factors of 36.

A factor of 36 is a number that divides 36 completely without leaving any remainder.

We can list the factors of 36 by checking which whole numbers divide 36 evenly, starting from 1:

• $36 \div 1 = 36$. So, 1 and 36 are factors.
• $36 \div 2 = 18$. So, 2 and 18 are factors.
• $36 \div 3 = 12$. So, 3 and 12 are factors.
• $36 \div 4 = 9$. So, 4 and 9 are factors.
• $36 \div 5$ is not a whole number.
• $36 \div 6 = 6$. So, 6 is a factor (we only list it once).

Once we reach a factor where the quotient is less than or equal to the divisor (like 6 and 6), we have found all pairs of factors.

The factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, and 36.

Let's count the number of factors:

There are 9 factors.

Alternatively, using prime factorization:

First, find the prime factorization of 36.

$\begin{array}{c|cc} 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

So, $36 = 2^2 \times 3^2$.

To find the number of factors, we add 1 to each exponent in the prime factorization and multiply the results.

Number of factors $= (2+1) \times (2+1) = 3 \times 3 = 9$.

Both methods show that the number of factors of 36 is 9.

The correct option is (D).

We need to find the sum of the first three common multiples of 3, 4, and 9.

First, we find the Least Common Multiple (LCM) of 3, 4, and 9. The common multiples of these numbers are the multiples of their LCM.

Let's find the prime factorization of each number:

Prime factorization of 3 is 3.

Prime factorization of 4 is $2 \times 2 = 2^2$.

Prime factorization of 9 is $3 \times 3 = 3^2$.

To find the LCM, we take the highest power of all prime factors involved in the factorizations:

LCM$(3, 4, 9) = 2^2 \times 3^2 = 4 \times 9 = 36$.

The common multiples of 3, 4, and 9 are the multiples of 36.

The multiples of 36 are $36 \times 1$, $36 \times 2$, $36 \times 3$, $36 \times 4$, ...

These are 36, 72, 108, 144, ...

The first three common multiples of 3, 4, and 9 are 36, 72, and 108.

Now, we find the sum of these three common multiples:

Sum $= 36 + 72 + 108$

Sum $= 108 + 108$

Sum $= 216$

The sum of the first three common multiples of 3, 4, and 9 is 216.

The correct option is (D).

In the Indian System of Numeration, the place value chart is as follows:

Ones, Tens, Hundreds, Thousands, Ten Thousands, Lakhs, Ten Lakhs, Crores, Ten Crores, ...

Commas are placed as follows:

• The first comma comes after the Hundreds place (3 digits from the right).
• Subsequent commas come after every two digits, marking the Lakhs and Crores periods.

Let's apply this to the number 61711682:

Starting from the rightmost digit (2):

• The first comma is placed after 3 digits: 61711,682
• The second comma is placed after the next 2 digits: 617,11,682
• The third comma is placed after the next 2 digits: 6,17,11,682

So, the number 61711682 is written as 6,17,11,682 in the Indian System of Numeration.

The complete statement is: In Indian System of Numeration, the number 61711682 is written, using commas, as 6,17,11,682.

Solution:

To form the smallest 4-digit number, we must use the smallest available digits: $0, 1, 2, 3$.

1. The digit at the thousands place cannot be $0$ (otherwise it becomes a 3-digit number). So, the smallest non-zero digit is $1$.

2. For the hundreds place, we can now use $0$.

3. For the tens place, the next smallest available digit is $2$.

4. For the ones place, the next smallest available digit is $3$.

Thus, the number is $1023$.

The complete statement is: The smallest 4 digit number with different digits is 1023.

A factor of a number is any number that divides it evenly, leaving no remainder.

Let's consider the number of factors for different types of whole numbers:

• Numbers with exactly two factors (1 and the number itself) are called prime numbers (e.g., 2, 3, 5, 7).
• The number 1 has only one factor (1) and is neither prime nor composite.
• Numbers with more than two factors are called composite numbers (e.g., 4 (factors 1, 2, 4), 6 (factors 1, 2, 3, 6), 8 (factors 1, 2, 4, 8), 9 (factors 1, 3, 9)).

The question asks for the name given to numbers having more than two factors.

Based on the definitions, numbers with more than two factors are called composite numbers.

The complete statement is: Numbers having more than two factors are called composite numbers.

To round a number to the nearest hundred, we look at the digit in the tens place.

In the number $58963$, the digit in the tens place is $6$.

Since the tens digit ($6$) is $5$ or greater, we round up the hundreds digit.

The hundreds digit is $9$. Rounding $9$ up requires carrying over $1$ to the thousands place.

$58963$ rounded to the nearest hundred becomes $59000$.

The given statement says the number $58963$ rounded off to nearest hundred is $58900$.

This statement is False.

We need to convert the Roman numerals to their Hindu-Arabic equivalents.

The Roman numeral $L$ represents $50$.

The Roman numeral $X$ represents $10$.

The Roman numeral $V$ represents $5$.

The Roman numeral $I$ represents $1$.

For $LXXV$:

$LXXV = L + X + X + V = 50 + 10 + 10 + 5 = 75$.

For $LXXIV$:

In $IV$, $I$ is placed before $V$, which means $I$ is subtracted from $V$.

$IV = V - I = 5 - 1 = 4$.

So, $LXXIV = L + X + X + IV = 50 + 10 + 10 + 4 = 74$.

Now we compare the values: $75$ and $74$.

Clearly, $75$ is greater than $74$.

Therefore, $LXXV$ is greater than $LXXIV$.

The given statement is True.

Solution:

The first part of the statement is true because 2 and 3 are coprime ($HCF = 1$). If a number is divisible by two coprime numbers, it is divisible by their product.

However, 2 and 4 are not coprime ($HCF = 2$). Therefore, a number divisible by both 2 and 4 is not necessarily divisible by their product ($2 \times 4 = 8$).

Counter-example:

Consider the number $12$.

$12$ is divisible by $2$.

$12$ is divisible by $4$.

But $12$ is not divisible by $8$.

The given statement is False.

Given:

Population of Agra district in the year 2001 = $36,20,436$.

Population of Aligarh district in the year 2001 = $29,92,286$.

To Find:

Total population of the two districts in that year.

Solution:

The total population of the two districts is the sum of the individual populations.

Total Population = Population of Agra + Population of Aligarh

Total Population = $36,20,436 + 29,92,286$

Let's perform the addition:

$\begin{array}{ccccccc} & 3 & 6 & 2 & 0 & 4 & 3 & 6 \\ + & 2 & 9 & 9 & 2 & 2 & 8 & 6 \\ \hline & 6 & 6 & 1 & 2 & 7 & 2 & 2 \\ \hline \end{array}$

So, the total population is $66,12,722$.

The total population of Agra and Aligarh districts in the year 2001 was $66,12,722$.

(i) Rounding off to nearest tens:

$5981$ rounded to nearest ten = $5980$

$4428$ rounded to nearest ten = $4430$

Estimated Product $= 5980 \times 4430$

The estimated product is $2,64,91,400$.

(ii) Rounding off to nearest hundreds:

$5981$ rounded to nearest hundred = $6000$

$4428$ rounded to nearest hundred = $4400$

Estimated Product $= 6000 \times 4400$

Estimated Product $= 2,64,00,000$

Solution:

Using the Distributive Property of Multiplication over Addition:

$a \times (b + c) = (a \times b) + (a \times c)$

Here, we can write $102$ as $(100 + 2)$.

$8739 \times 102 = 8739 \times (100 + 2)$

$8739 \times 102 = (8739 \times 100) + (8739 \times 2)$

Now, calculating the parts:

$8739 \times 100 = 873900$

$8739 \times 2 = 17478$

Adding the results:

$\begin{array}{ccccccc} & 8 & 7 & 3 & 9 & 0 & 0 \\ + & & 1 & 7 & 4 & 7 & 8 \\ \hline & 8 & 9 & 1 & 3 & 7 & 8 \\ \hline \end{array}$

The product is $8,91,378$.

Given:

Length of the floor = $4.5\text{ m} = 450\text{ cm}$

Breadth of the floor = $3\text{ m} = 300\text{ cm}$

To Find: Minimum number of square slabs.

Solution:

To minimize the number of slabs, each slab must be of the maximum possible size. The side of such a square slab must be the HCF of the floor's dimensions.

Finding HCF of $450$ and $300$:

$\begin{array}{c|cc} 2 & 450 \;, & 300 \\ \hline 5 & 225 \;, & 150 \\ \hline 5 & 45 \;, & 30 \\ \hline 3 & 9 \;, & 6 \\ \hline & 3 \;, & 2 \end{array}$

$\text{HCF} = 2 \times 5 \times 5 \times 3 = 150\text{ cm}$.

So, the side of the square slab is $150\text{ cm}$.

$\text{Area of the floor} = 450 \times 300\text{ cm}^2$

$\text{Area of one square slab} = 150 \times 150\text{ cm}^2$

$\text{Number of slabs} = \frac{\text{Area of floor}}{\text{Area of one slab}}$

$\text{Number of slabs} = \frac{\cancel{450}^3 \times \cancel{300}^2}{\cancel{150} \times \cancel{150}}$

$\text{Number of slabs} = 3 \times 2 = 6$.

The minimum number of square slabs required is 6.

The given number is $428721$.

The digit $2$ appears at two places in the number.

The first $2$ is in the ten thousands place.

Its place value is $2 \times 10000 = 20000$.

The second $2$ is in the tens place.

Its place value is $2 \times 10 = 20$.

The product of the place values of the two 2’s is the product of $20000$ and $20$.

Product = $20000 \times 20 = 400000$.

The correct answer is (C) 400000.

We need to evaluate the given expression: $3 \times 10000 + 7 \times 1000 + 9 \times 100 + 0 \times 10 + 4$.

Let's calculate each term:

$3 \times 10000 = 30000$

$7 \times 1000 = 7000$

$9 \times 100 = 900$

$0 \times 10 = 0$

$4 = 4$

Now, we add these values together:

$30000 + 7000 + 900 + 0 + 4$

$\begin{array}{cccccc} & 3 & 0 & 0 & 0 & 0 \\ & & 7 & 0 & 0 & 0 \\ & & & 9 & 0 & 0 \\ & & & & 0 & 0 \\ + & & & & & 4 \\ \hline & 3 & 7 & 9 & 0 & 4 \\ \hline \end{array}$

The sum is $37904$.

The correct answer is (C) 37904.

The greatest 7-digit number is the number with $7$ nines.

Greatest 7-digit number = $9,999,999$.

We are asked to find the number obtained by adding $1$ to the greatest 7-digit number.

$9,999,999 + 1$.

Let's perform the addition:

$\begin{array}{ccccccc} & 9 & 9 & 9 & 9 & 9 & 9 & 9 \\ + & & & & & & & 1 \\ \hline 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline \end{array}$

The result is $10,000,000$.

The correct answer is (D) 1 crore.

The given number is $9578$.

To write the expanded form, we represent the number as the sum of the place values of its digits.

The digit $9$ is in the thousands place, so its place value is $9 \times 1000$.

The digit $5$ is in the hundreds place, so its place value is $5 \times 100$.

The digit $7$ is in the tens place, so its place value is $7 \times 10$.

The digit $8$ is in the units place, so its place value is $8 \times 1$.

So, the expanded form of $9578$ is $9 \times 1000 + 5 \times 100 + 7 \times 10 + 8 \times 1$.

The standard expanded form expressing each digit's place value is given in option (B).

The correct answer is (B) 9 × 1000 + 5 × 100 + 7 × 10 + 8 × 1.

To round a number to the nearest thousands, we look at the digit in the hundreds place.

The given number is $85642$.

The digit in the hundreds place is $6$.

Since the hundreds digit ($6$) is $5$ or greater, we round up the thousands digit.

The thousands digit is $5$. Rounding up $5$ makes it $6$.

All digits to the right of the thousands place become $0$.

So, $85642$ rounded off to the nearest thousands is $86000$.

The correct answer is (D) 86000.

Given:

Digits: $5, 9, 2, 6$.

Solution:

To form the largest 4-digit number using one digit twice:

• We must choose the largest available digit to repeat, which is $9$.
• This gives us the set of digits: $9, 9, 6, 5$ (discarding the smallest digit 2 to maintain a 4-digit requirement with one repetition).

To maximize the number, we arrange these digits in descending order:

$9, 9, 6, 5 \rightarrow 9,965$

The largest 4-digit number formed is $9,965$.

The correct answer is (D) 9965.

Solution:

In the Indian System of Numeration, commas are placed to separate the periods of ones, thousands, lakhs, and crores.

• The first comma is placed after the hundreds place (3 digits from the right).
• Subsequent commas are placed after every two digits.

Given number: $58695376$

1. First comma: $58695,376$
2. Second comma: $586,95,376$
3. Third comma: $5,86,95,376$

The number is written as $5,86,95,376$ (Five crore eighty-six lakh ninety-five thousand three hundred seventy-six).

The correct answer is (C) 5, 86, 95, 376.

Solution:

According to the International System of Numeration:

$1 \text{ million} = 1,000,000$

In the Indian System of Numeration, we place commas as $10,00,000$.

Comparing this to the place value chart:

$10,00,000 = \text{Ten Lakhs}$

Therefore, $1$ million is equal to $10$ lakh.

The correct answer is (B) 10 lakh.

Solution:

A number rounds to $5,000$ (nearest thousand) if it lies in the range $4,500$ to $5,499$.

• If the number is $\geq 5,500$, it rounds to $6,000$.
• If the number is $\geq 4,500$ and $< 5,500$, it rounds to $5,000$.

Let's check the options:

• $5,001$ rounds to $5,000$.
• $5,559$ rounds to $6,000$.
• $5,999$ rounds to $6,000$.
• $5,499$ rounds to $5,000$.

Comparing $5,001$ and $5,499$, the greatest number is $5,499$.

The correct answer is (D) 5499.

Given:

Original number: $63,50,947$.

Fixed digit: $6$ at the first place.

Solution:

The digits available for rearrangement are: $3, 5, 0, 9, 4, 7$.

To obtain the smallest number, we must arrange these digits in ascending order:

$0, 3, 4, 5, 7, 9$

Now, we place these digits after the fixed digit $6$:

Number $= 6,0,3,4,5,7,9 \rightarrow 60,34,579$

The correct answer is (C) 6034579.

We need to check the validity of each given Roman numeral.

Recall the basic Roman numeral symbols and their values:

$I = 1$

$V = 5$

$X = 10$

$L = 50$

$C = 100$

$D = 500$

$M = 1000$

Also, recall the rules for forming Roman numerals, particularly regarding repetition and subtraction. Symbols $V, L,$ and $D$ are never repeated.

The correct answer is (D) LLX.

Solution:

To form the largest 5-digit number with three different digits:

• We should use the three largest digits: $9, 8, 7$.
• To make the number largest, the largest digit ($9$) should be placed in the highest place values as much as possible.

We use $9$ for the Ten-Thousands, Thousands, and Hundreds places, then $8$ for Tens and $7$ for Ones.

Number $= 99987$

Checking the options:

• (A) $98978$
• (B) $99897$
• (C) $99987$
• (D) $98799$

Clearly, $99987$ is the largest.

The correct answer is (C) 99987.

Solution:

To form the smallest 4-digit number with three different digits:

• We should use the three smallest available digits: $0, 1, 2$.
• The first digit (Thousands place) cannot be $0$. So, we use $1$.
• To minimize the number, we should use the smallest digit ($0$) for the next available place values.

Placing $1$ at Thousands, $0$ at Hundreds, $0$ at Tens, and $2$ at Ones:

Number $= 1002$

This uses exactly three different digits: $\{0, 1, 2\}$.

The correct answer is (D) 1002.

Solution:

The numbers "between" two given numbers $a$ and $b$ do not include $a$ and $b$ themselves.

The formula to find the count of whole numbers between $a$ and $b$ is:

$\text{Count} = (b - a) - 1$

Given, $a = 38$ and $b = 68$.

$\text{Count} = (68 - 38) - 1$

$\text{Count} = 30 - 1 = 29$

Thus, there are $29$ whole numbers between $38$ and $68$.

The correct answer is (C) 29.

The given number is $999$.

The successor of a number is the number that comes just after it.

Successor of $999 = 999 + 1 = 1000$.

The predecessor of a number is the number that comes just before it.

Predecessor of $999 = 999 - 1 = 998$.

We need to find the product of the successor and the predecessor of $999$.

Product = Successor $\times$ Predecessor

Product = $1000 \times 998$.

$1000 \times 998 = 998000$.

The correct answer is (B) 998000.

Solution:

Let the non-zero whole number be $n$. Its successor is $n + 1$.

The product is $n(n + 1)$.

Out of any two consecutive numbers ($n$ and $n + 1$), one must be even and the other must be odd.

Example:

• If $n = 4$ (even), successor $= 5$ (odd). Product $= 4 \times 5 = 20$ (Even).
• If $n = 5$ (odd), successor $= 6$ (even). Product $= 5 \times 6 = 30$ (Even).

Thus, the product is always an even number.

The correct answer is (A) an even number.

Solution:

Let the whole number be $x$.

According to the question, the two resulting numbers are:

1. Result 1: $25 + x$
2. Result 2: $25 - x$

We need to find the sum of these results:

$\text{Sum} = (25 + x) + (25 - x)$

$\text{Sum} = 25 + x + 25 - x$

$\text{Sum} = 25 + 25 + (x - x)$

$\text{Sum} = 50 + 0 = 50$

The sum is always $50$.

The correct answer is (C) 50.

Solution:

Let's check each option:

• (A) $(7 + 8) + 9 = 7 + (8 + 9) \rightarrow$ This is the Associative Property of Addition. (True)
• (B) $(7 \times 8) \times 9 = 7 \times (8 \times 9) \rightarrow$ This is the Associative Property of Multiplication. (True)
• (D) $7 \times (8 + 9) = (7 \times 8) + (7 \times 9) \rightarrow$ This is the Distributive Property of Multiplication over Addition. (True)

Now, let's evaluate (C):

$\text{L.H.S.} = 7 + 8 \times 9 = 7 + 72 = 79$

$\text{R.H.S.} = (7 + 8) \times (7 + 9) = 15 \times 16 = 240$

Thus, option (C) is not true.

The correct answer is (C) 7 + 8 × 9 = (7 + 8) × (7 + 9).

Solution:

• Line: Every whole number greater than 1 can be arranged in a line.
• Triangle: Numbers like $3, 6, 10, 15...$ (Triangular Numbers) can form a triangle.
• Rectangle: Composite numbers (excluding 1) can be arranged as a rectangle.

Checking the number $10$:

1. Line: $1 \times 10$ dots. (Possible)
2. Triangle: $1 + 2 + 3 + 4 = 10$. (Possible)
3. Rectangle: $2 \times 5$ dots. (Possible)

Other options:

• $9$ is not triangular.
• $11$ is prime (only a line).
• $12$ is not triangular.

The correct answer is (B) 10.

Solution:

Let's check the properties of whole numbers:

• (A) Associativity holds for both addition and multiplication. (True)
• (C) Commutativity holds for both addition and multiplication. (True)
• (D) Distributive property of multiplication over addition is valid. (True)

Evaluating (B):

The multiplicative identity is a number that, when multiplied by any number $n$, gives $n$.

However, $n \times 0 = 0$. So, zero is not the identity for multiplication.

Zero is the additive identity ($n + 0 = n$).

The correct answer is (B) Zero is the identity for muliplication of whole numbers.

Solution:

Let's evaluate the mathematical operations with zero:

(A) $0 + 0 = 0$. (True)

(B) $0 - 0 = 0$. (True)

(C) $0 \times 0 = 0$. (True)

(D) $0 \div 0$: Division by zero is not defined in mathematics. The expression $0 \div 0$ is indeterminate/undefined.

Thus, statement (D) is not true.

The correct answer is (D) 0 ÷ 0 = 0.

Solution:

$1 \text{ lakh} = 1,00,000$

Predecessor is obtained by subtracting $1$ from the number.

$\text{Predecessor} = 1,00,000 - 1$

$\begin{array}{ccccccc} & 1 & 0 & 0 & 0 & 0 & 0 \\ - & & & & & & 1 \\ \hline & & 9 & 9 & 9 & 9 & 9 \\ \hline \end{array}$

The predecessor of 1 lakh is 99,999.

The correct answer is (B) 99999.

Solution:

$1 \text{ million} = 1,000,000$

Successor is obtained by adding $1$ to the number.

$\text{Successor} = 1,000,000 + 1$

$\text{Successor} = 1,000,001$

The correct answer is (B) 1000001.

Solution:

The even numbers "between" $58$ and $80$ are:

$60, 62, 64, 66, 68, 70, 72, 74, 76, 78$

Counting these numbers:

1. $60$
2. $62$
3. $64$
4. $66$
5. $68$
6. $70$
7. $72$
8. $74$
9. $76$
10. $78$

There are total 10 even numbers.

The correct answer is (A) 10.

Solution:

Primes between 16 and 80 are:

$17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79$

Primes between 90 and 100 is:

$97$

Sum of the number of primes $= 16 + 1 = 17$.

The correct answer is (C) 17.

Solution:

• (A) Prime numbers have no common factors other than 1. (True)
• (B) By definition, coprime numbers have HCF 1. (True)
• (C) Consecutive even numbers (e.g., 8 and 10) always have 2 as their greatest common factor. (True)

Evaluating (D):

An even number has 2 as a factor. An odd number does not have 2 as a factor.

For a number to be an HCF, it must divide both numbers. Since the odd number is not divisible by 2, no even number can divide it.

Thus, the HCF of an even and an odd number must be odd.

The correct answer is (D) The HCF of an even and an odd number is even.

Solution:

The largest 4-digit number is $9,999$.

Performing prime factorization of $9,999$:

$\begin{array}{c|cc} 3 & 9999 \\ \hline 3 & 3333 \\ \hline 11 & 1111 \\ \hline 101 & 101 \\ \hline & 1 \end{array}$

The prime factorization is: $9,999 = 3 \times 3 \times 11 \times 101 = 3^2 \times 11 \times 101$.

The distinct prime factors are 3, 11, and 101.

The count of distinct prime factors is 3.

The correct answer is (B) 3.

Solution:

The smallest 5-digit number is $10,000$.

Performing prime factorization of $10,000$:

$\begin{array}{c|cc} 2 & 10000 \\ \hline 2 & 5000 \\ \hline 2 & 2500 \\ \hline 2 & 1250 \\ \hline 5 & 625 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

The prime factorization is: $10,000 = 2^4 \times 5^4$.

The distinct prime factors are 2 and 5.

The count of distinct prime factors is 2.

The correct answer is (A) 2.

Solution:

A number is divisible by $22$ if it is divisible by both 2 and 11.

• Divisibility by 2: The number $7254*98$ ends in $8$, so it is divisible by 2.

Divisibility by 11:

The difference between the sum of digits at odd places and even places must be $0$ or a multiple of $11$.

Let the missing digit be $x$. The number is $7254x98$.

Sum of digits at odd places (from right) $= 8 + x + 5 + 7 = 20 + x$

Sum of digits at even places (from right) $= 9 + 4 + 2 = 15$

Difference $= (20 + x) - 15 = 5 + x$

For $5 + x$ to be a multiple of $11$, $x$ must be $6$ (since $5 + 6 = 11$).

The digit at * is 6.

The correct answer is (C) 6.

Solution:

Let the two consecutive odd numbers be $2n - 1$ and $2n + 1$.

Their sum is:

$\text{Sum} = (2n - 1) + (2n + 1)$

$\text{Sum} = 4n$

Since the sum is always a multiple of 4, it is always divisible by 4.

Examples:

• $1 + 3 = 4$ (Divisible by 4)
• $3 + 5 = 8$ (Divisible by 4)
• $5 + 7 = 12$ (Divisible by 4)

The correct answer is (B) 4.

Solution:

If a number is divisible by both $5$ and $6$, it must be divisible by their LCM.

$\text{LCM}(5, 6) = 30$

Any multiple of $30$ is also divisible by the factors of $30$, which are $1, 2, 3, 5, 6, 10, 15, 30$.

Thus, the number must be divisible by $10, 15,$ and $30$.

However, it is not necessarily divisible by $60$.

Example: $30$ is divisible by $5$ and $6$, but $30$ is not divisible by $60$.

The correct answer is (D) 60.

Solution:

We find the prime factors of $1729$ (known as the Hardy-Ramanujan number):

1. $1729 \div 7 = 247$
2. $247 \div 13 = 19$
3. $19 \div 19 = 1$

The prime factors are 7, 13, and 19.

Sum of these factors $= 7 + 13 + 19 = 39$.

The correct answer is (D) 39.

Given:

An odd natural number other than $1$. We need to find the product of its predecessor (the number before it) and its successor (the number after it).

To Find:

The greatest number that always divides this product.

Solution:

Natural numbers are $1, 2, 3, 4, 5, ...$ and so on. Odd natural numbers other than $1$ are $3, 5, 7, 9, ...$

For any odd number, the number just before it (predecessor) and the number just after it (successor) are both even numbers.

Since both the predecessor and the successor are even, they are both divisible by $2$.

Because each even number has at least $2$ as a factor, their product will have $2 \times 2 = 4$ as a factor.

Verification with Examples:

Let us take different odd natural numbers and calculate the product of their predecessor and successor:

Now, let us check the divisibility of these products ($8, 24, 48, 80$) by the given options:

The products are all divisible by $4$. While they are also divisible by $8$ in these cases, according to the options and general property of consecutive even numbers, the number $4$ is a consistent divisor.

Conclusion:

The product of the predecessor and successor of an odd natural number is always a multiple of $4$.

The correct option is (B) 4.

Solution:

Let's find the prime factorization of each number:

• $75 = 3 \times 5 \times 5 = 3^1 \times 5^2$
• $60 = 2 \times 2 \times 3 \times 5 = 2^2 \times 3^1 \times 5^1$
• $105 = 3 \times 5 \times 7 = 3^1 \times 5^1 \times 7^1$

The common prime factors present in all three numbers are:

1. 3
2. 5

The number of common prime factors is 2.

The correct answer is (A) 2.

Solution:

Coprime numbers are pairs of numbers whose HCF is 1.

Checking the options:

• (B) 11, 12: HCF is 1. (Coprime)
• (C) 1, 3: HCF is 1. (Coprime)
• (D) 31, 33: HCF is 1. (Coprime)

Evaluating (A) 8, 10:

Factors of 8: $1, 2, 4, 8$

Factors of 10: $1, 2, 5, 10$

Common Factor: 2. Since the HCF is $2$ (not 1), they are not coprime.

The correct answer is (A) 8, 10.

Solution:

A number is divisible by $11$ if the difference between the sum of digits at odd places and even places is $0$ or a multiple of $11$.

Evaluating (C) 22222222:

• Sum of digits at odd places: $2 + 2 + 2 + 2 = 8$
• Sum of digits at even places: $2 + 2 + 2 + 2 = 8$

Difference $= 8 - 8 = 0$.

Since the difference is $0$, the number is divisible by 11. Other options have an odd number of digits with the same repeating digit, resulting in a non-zero difference.

The correct answer is (C) 22222222.

Solution:

To find the Least Common Multiple (LCM), we use the common division method:

$\begin{array}{c|cc} 2 & 10 \;, & 15 \;, & 20 \\ \hline 2 & 5 \; , & 15 \; , & 10 \\ \hline 3 & 5 \; , & 15 \; , & 5 \\ \hline 5 & 5 \; , & 5 \; , & 5 \\ \hline & 1 \; , & 1 \; , & 1 \end{array}$

$\text{LCM} = 2 \times 2 \times 3 \times 5 = 60$

The LCM of 10, 15, and 20 is 60.

The correct answer is (B) 60.

Solution:

A key property of numbers is that the HCF must always be a factor of the LCM.

Given $\text{LCM} = 180$. We need to check which option is not a factor of $180$.

1. $180 \div 45 = 4$ (Factor)
2. $180 \div 60 = 3$ (Factor)
3. $180 \div 90 = 2$ (Factor)

Evaluating $180 \div 75$:

$\frac{180}{75} = \frac{\cancel{36}^{12}}{\cancel{15}_{5}} = 2.4$

Since $75$ is not a factor of $180$, it cannot be the HCF.

The correct answer is (C) 75.

The statement is True.

In Roman numeration, the symbols I, X, C, and M can be repeated up to three times consecutively to denote a sum (e.g., III = 3, XX = 20, CCC = 300, MMM = 3000). Symbols V, L, and D are never repeated.

The statement is False.

In Roman numeration, when a symbol is repeated, its value is added as many times as it occurs, not multiplied.

For example:

II means $1 + 1 = 2$ (not $1 \times 1 = 1$)

XX means $10 + 10 = 20$ (not $10 \times 10 = 100$)

The statement is True.

The expression $5 \times 1000 + 5 \times 100 + 5 \times 10 + 5 \times 1$ is the expanded form of the number 5555 based on its place values.

Calculation:

$5 \times 1000 = 5000$

$5 \times 100 = 500$

$5 \times 10 = 50$

$5 \times 1 = 5$

Adding these values:

$5000 + 500 + 50 + 5 = 5555$

So, $5555 = 5555$, which is correct.

The statement is True.

The given expression is the expanded form of the number 39746 based on its place values.

Let's calculate the value of the right side:

$3 \times 10000 = 30000$

$9 \times 1000 = 9000$

$7 \times 100 = 700$

$4 \times 10 = 40$

$6$

Summing these values:

$30000 + 9000 + 700 + 40 + 6 = 39746$

Since $39746 = 39746$, the statement is correct.

The statement is False.

The expanded form of 82546 should be based on the place value of each digit.

The place values are:

• 8 is in the ten thousands place ($10000$)
• 2 is in the thousands place ($1000$)
• 5 is in the hundreds place ($100$)
• 4 is in the tens place ($10$)
• 6 is in the ones place ($1$)

So, the correct expanded form is:

$8 \times 10000 + 2 \times 1000 + 5 \times 100 + 4 \times 10 + 6 \times 1$

Let's calculate the value of the expression given in the statement:

$8 \times 1000 = 8000$

$2 \times 1000 = 2000$

$5 \times 100 = 500$

$4 \times 10 = 40$

$6$

Adding these values:

$8000 + 2000 + 500 + 40 + 6 = 10546$

Since $82546 \neq 10546$, the given statement is incorrect.

The statement is True.

The given expression is the expanded form of the number 532235 based on the place value of each digit.

Let's calculate the value of the right side:

$5 \times 100000 = 500000$

$3 \times 10000 = 30000$

$2 \times 1000 = 2000$

$2 \times 100 = 200$

$3 \times 10 = 30$

$5 \times 1 = 5$

Summing these values:

$500000 + 30000 + 2000 + 200 + 30 + 5 = 532235$

Since $532235 = 532235$, the statement is correct.

The statement is False.

In Roman numeration:

• XX represents $10 + 10 = 20$.
• IX represents $10 - 1 = 9$ (since the symbol I, representing a smaller value, is placed before X, representing a larger value).

Combining these values:

XXIX = XX + IX = $20 + 9 = 29$.

Therefore, XXIX is equal to 29, not 31.

The statement is True.

In Roman numeration:

• L represents 50.
• XX represents $10 + 10 = 20$.
• IV represents $5 - 1 = 4$ (due to the subtraction rule, where a smaller value I is placed before a larger value V).

Combining these values:

LXXIV = L + XX + IV = $50 + 20 + 4 = 74$.

Therefore, LXXIV is indeed equal to 74.

The statement is False.

Let's convert the Roman numerals to Hindu-Arabic numerals:

• LIV: L represents 50, IV represents $5 - 1 = 4$. So, LIV = $50 + 4 = 54$.
• LVI: L represents 50, V represents 5, I represents 1. So, LVI = $50 + 5 + 1 = 56$.

Comparing the values:

LIV = 54

LVI = 56

Since $54 < 56$, LIV is not greater than LVI. LVI is greater than LIV.

The statement is False.

Descending order means arranging numbers from the largest to the smallest.

The given numbers are: 4578, 4587, 5478, 5487.

Let's compare them:

• Comparing 4578 and 4587: 4587 > 4578
• Comparing 5478 and 5487: 5487 > 5478
• Comparing 4587 and 5478: 5478 > 4587 (comparing the thousands digit)

Arranging the numbers in descending order:

5487, 5478, 4587, 4578

The given order is 4578, 4587, 5478, 5487, which is actually in ascending order (from smallest to largest).

The statement is False.

To round off 85764 to the nearest hundreds, we look at the tens digit, which is 6.

Since the tens digit (6) is 5 or greater, we round up the hundreds digit.

The hundreds digit is 7. Rounding up 7 gives 8.

The digits to the right of the hundreds place become zeros.

So, 85764 rounded off to the nearest hundreds is 85800.

The statement is False.

First, let's round off each number to the nearest hundreds:

• 7826: The tens digit is 2. Since $2 < 5$, we round down the hundreds digit. 7826 rounded to the nearest hundreds is 7800.
• 12469: The tens digit is 6. Since $6 \geq 5$, we round up the hundreds digit. 12469 rounded to the nearest hundreds is 12500.

Now, let's find the estimated sum by adding the rounded numbers:

$7800 + 12500 = 20300$

The estimated sum rounded off to hundreds is 20300, not 20000.

The statement is False.

To form the largest possible six-digit number using the digits 5, 3, 4, 7, 0, and 8 exactly once, we must arrange the digits in descending order from left to right (from the largest place value to the smallest).

The given digits are 8, 7, 5, 4, 3, 0.

Arranging them in descending order gives: 8, 7, 5, 4, 3, 0.

The largest number formed by these digits is 875430.

The given number is 875403.

Comparing the two numbers, $875430 > 875403$.

Therefore, the largest six-digit number is 875430, not 875403.

Given:

Number: $81652318$

Provided reading: "eighty one crore six lakh fifty two thousand three hundred eighteen"

Solution:

In the Indian System of Numeration, commas are used to separate periods. The first comma is placed after the hundreds place (three digits from the right), and subsequent commas are placed after every two digits.

Let's place commas in the given number $81652318$:

Now, let's identify the periods for each group of digits:

• 8 is in the Crores period.
• 16 is in the Lakhs period.
• 52 is in the Thousands period.
• 318 is in the Ones period.

Based on these periods, the correct way to read $8,16,52,318$ is:

Eight crore sixteen lakh fifty-two thousand three hundred eighteen.

Comparing this with the reading given in the question:

Given reading: "Eighty-one crore six lakh fifty-two thousand three hundred eighteen."

The given reading is incorrect because it treats the number as having 9 digits (beginning with eighty-one crore), whereas the number $81652318$ has only 8 digits and starts with eight crore.

Therefore, the statement is False.

The statement is True.

To form the largest 4-digit number using the digits 6, 7, 0, and 9 exactly once, we need to arrange the digits in descending order.

The given digits are 9, 7, 6, 0.

Arranging them in descending order gives: 9, 7, 6, 0.

Placing these digits in the thousands, hundreds, tens, and ones places respectively, we get the number 9760.

This is the largest 4-digit number that can be formed using these digits.

The statement is False.

These are prefixes used in the metric system, representing powers of 10.

Let's look at their values relative to the base unit (e.g., meter, gram, liter):

• Kilo (k) means $10^3 = 1000$ times the base unit.
• Centi (c) means $10^{-2} = \frac{1}{100} = 0.01$ times the base unit.
• Milli (m) means $10^{-3} = \frac{1}{1000} = 0.001$ times the base unit.

Comparing the values:

• Kilo is the largest (1000).
• Centi is smaller (0.01).
• Milli is the smallest (0.001).

Therefore, among kilo, milli, and centi, the smallest is milli.

The statement is False.

The successor of a number is the number that comes immediately after it, which is obtained by adding 1 to the number.

One-digit numbers are $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$.

Let's find the successor of the largest one-digit number, which is 9.

Successor of 9 = $9 + 1 = 10$.

10 is a two-digit number.

Since the successor of 9 is 10, which is not a one-digit number, the statement is false.

The statement is False.

The successor of a number is obtained by adding 1 to it.

Consider the largest 3-digit number, which is 999.

The successor of 999 is $999 + 1 = 1000$.

1000 is a 4-digit number.

Since the successor of the largest 3-digit number (999) is a 4-digit number (1000), the statement is false.

The statement is False.

The predecessor of a number is the number that comes immediately before it, which is obtained by subtracting 1 from the number.

Two-digit numbers range from 10 to 99.

Let's find the predecessor of the smallest two-digit number, which is 10.

Predecessor of 10 = $10 - 1 = 9$.

9 is a one-digit number.

Since the predecessor of 10 is 9, which is not a two-digit number, the statement is false.

The statement is True.

Whole numbers are $0, 1, 2, 3, 4, \dots$ (non-negative integers).

The successor of any whole number $n$ is $n + 1$. Since the set of whole numbers is infinite and for any whole number $n$, $n+1$ is also a whole number, every whole number has a unique successor which is also a whole number.

There is no largest whole number.

The statement is False.

Whole numbers are $0, 1, 2, 3, 4, \dots$

The predecessor of a number is obtained by subtracting 1 from it.

Let's find the predecessor of the smallest whole number, which is 0.

Predecessor of 0 = $0 - 1 = -1$.

-1 is an integer, but it is not a whole number (since whole numbers are non-negative).

Therefore, the whole number 0 does not have a predecessor within the set of whole numbers.

The statement is False.

Natural numbers are $1, 2, 3, 4, \dots$ (positive integers).

Consider two consecutive natural numbers, for example, 1 and 2.

There is no natural number between 1 and 2.

If we take two non-consecutive natural numbers, say 3 and 5, there is one natural number (4) between them. But this is not true for *any* two natural numbers.

For the statement to be true, there must always be at least one natural number between any two given natural numbers. As shown with 1 and 2, this is not the case.

The statement is True.

The largest 3-digit number is 999.

The smallest 4-digit number is 1000.

The successor of the largest 3-digit number (999) is obtained by adding 1 to it:

Successor of 999 = $999 + 1 = 1000$.

Since the successor of the largest 3-digit number (1000) is equal to the smallest 4-digit number (1000), the statement is true.

The statement is True.

This is a fundamental rule when comparing natural numbers (or positive integers).

A number with more digits will always have a value greater than any number with fewer digits, regardless of the value of the digits themselves.

For example:

• The smallest 2-digit number is 10. The largest 1-digit number is 9. $10 > 9$.
• The smallest 3-digit number is 100. The largest 2-digit number is 99. $100 > 99$.
• Any 5-digit number (like 10000) is greater than any 4-digit number (like 9999).

This is because the leftmost digit in the number with more digits represents a much larger place value than any digit in the number with fewer digits.

The statement is True.

A set of numbers is closed under an operation if performing that operation on any two numbers in the set always results in a number that is also in the same set.

Natural numbers are $1, 2, 3, 4, \dots$.

Addition is the operation.

If we take any two natural numbers, say $a$ and $b$, and add them ($a + b$), the result will always be a natural number.

For example:

• $1 + 1 = 2$ (2 is a natural number)
• $5 + 8 = 13$ (13 is a natural number)
• $100 + 200 = 300$ (300 is a natural number)

There is no pair of natural numbers whose sum is not a natural number.

Therefore, natural numbers are closed under addition.

The statement is False.

Natural numbers are $1, 2, 3, 4, \dots$.

A set is closed under an operation if the result of the operation on any two elements of the set is also in the set.

Multiplication is the operation.

If we take any two natural numbers, say $a$ and $b$, and multiply them ($a \times b$), the result will always be a natural number.

For example:

• $1 \times 1 = 1$ (1 is a natural number)
• $2 \times 3 = 6$ (6 is a natural number)
• $10 \times 5 = 50$ (50 is a natural number)

There is no pair of natural numbers whose product is not a natural number.

Therefore, natural numbers are closed under multiplication.

The statement is False.

Natural numbers are $1, 2, 3, 4, \dots$.

A set is closed under an operation if the result of the operation on any two elements of the set is also in the set.

Subtraction is the operation.

If we take two natural numbers, say $a$ and $b$, and subtract them ($a - b$), the result is not always a natural number.

For example:

• $2 - 3 = -1$. -1 is an integer but not a natural number.
• $5 - 5 = 0$. 0 is a whole number but not a natural number (if natural numbers start from 1).
• $10 - 12 = -2$. -2 is not a natural number.

Since the result of subtracting two natural numbers can be an integer that is not a natural number, the set of natural numbers is not closed under subtraction.

The statement is True.

An operation is commutative for a set of numbers if the order of the operands does not affect the result.

For addition, this means that for any two natural numbers $a$ and $b$, $a + b = b + a$.

Natural numbers are $1, 2, 3, 4, \dots$.

Let's test with examples:

• $2 + 3 = 5$ and $3 + 2 = 5$. So, $2 + 3 = 3 + 2$.
• $10 + 7 = 17$ and $7 + 10 = 17$. So, $10 + 7 = 7 + 10$.
• For any natural numbers $a$ and $b$, their sum is the same regardless of the order.

Therefore, addition is commutative for natural numbers.

The statement is False.

An additive identity is a number such that when it is added to any number in a set, the result is the original number.

For whole numbers (0, 1, 2, 3, ...), the additive identity is 0.

This is because for any whole number $a$, $a + 0 = a$ and $0 + a = a$.

For example:

• $5 + 0 = 5$
• $0 + 12 = 12$

If we use 1 as the identity:

$5 + 1 = 6$, which is not equal to 5.

Therefore, 1 is not the additive identity for whole numbers. 0 is the additive identity.

The statement is True.

A multiplicative identity is a number such that when it is multiplied by any number in a set, the result is the original number.

For whole numbers (0, 1, 2, 3, ...), the multiplicative identity is 1.

This is because for any whole number $a$, $a \times 1 = a$ and $1 \times a = a$.

For example:

• $5 \times 1 = 5$
• $1 \times 12 = 12$
• $0 \times 1 = 0$

Therefore, 1 is the multiplicative identity for whole numbers.

The statement is True.

The whole number described in the statement is the additive identity for whole numbers.

The whole number is 0.

When 0 is added to any whole number $a$, the result is $a$ itself ($a + 0 = a$).

For example:

• $5 + 0 = 5$
• $0 + 0 = 0$
• $100 + 0 = 100$

So, there exists such a whole number, which is 0.

The statement is False.

We are looking for a natural number, let's call it $e$, such that for any natural number $a$, $a + e = a$.

If $a + e = a$, then subtracting $a$ from both sides gives $e = a - a = 0$.

The number required for this property is 0.

However, natural numbers are $1, 2, 3, \dots$. The number 0 is not a natural number.

Therefore, there is no natural number which, when added to a natural number, gives the number itself.

Given:

A whole number is divided by another whole number which is greater than the first one.

Solution:

Let the first whole number be $a$ and the second whole number be $b$. According to the condition, $b > a$.

Whole numbers are $\{0, 1, 2, 3, \dots\}$.

Consider the case where the first whole number $a = 0$. Since $b > a$, let's take $b = 5$.

$\text{Quotient} = 0 \div 5 = 0$

In this case, the quotient is equal to zero. The statement claims the quotient is not equal to zero, which is contradicted when the dividend is $0$.

Therefore, the statement is False.

Given:

A non-zero whole number divided by itself.

Solution:

Let the non-zero whole number be $x$, where $x \neq 0$.

According to the property of division, when any number (except zero) is divided by itself, the result is always $1$.

For example, $15 \div 15 = 1$ and $7 \div 7 = 1$.

The statement is True.

Solution:

Whole numbers are $\{0, 1, 2, 3, \dots\}$. The set of whole numbers is closed under multiplication.

This means that if we multiply any two whole numbers $a$ and $b$, the product $a \times b$ will always be a whole number.

Example:

$5 \times 4 = 20$ (A whole number)

$0 \times 10 = 0$ (A whole number)

Since the product is always a whole number, the statement "need not be a whole number" is incorrect.

The statement is False.

Given:

A whole number is divided by another whole number which is greater than 1.

To Find:

Whether the quotient can ever be equal to the first whole number.

Solution:

When we divide a non-zero whole number by a number larger than $1$, the value always decreases. This means the result (quotient) will always be smaller than the original number.

Let us take an example of the whole number $6$ and divide it by $2$ (which is greater than $1$):

In this case, the quotient $3$ is not equal to the original number $6$.

Let us try another example with the whole number $10$ divided by $5$:

Here also, the quotient $2$ is not equal to the original number $10$.

In the Indian school curriculum, the property of division for whole numbers states that for any non-zero whole number, dividing by a number greater than $1$ reduces its value.

Therefore, the quotient will never be equal to the original number.

The statement is True.

Given:

Multiples of a number.

To Find:

Whether every multiple is either larger than or equal to the number itself.

Solution:

In mathematics, we find the multiples of a number by multiplying it by $1, 2, 3, 4,$ and so on.

Let us look at the multiples of $5$:

Observation:

As we can see from the examples above:

• The smallest multiple of any number is the number itself (so it is equal to the number).
• All other multiples are obtained by multiplying with numbers larger than $1$, so they are always greater than the number.

Since no multiple of a natural number can be smaller than the number itself, the statement is correct.

The statement is True.

Solution:

Multiples of a number $n$ are $n \times 1, n \times 2, n \times 3, \dots$ and so on.

Since we can multiply a number by infinitely many natural numbers, the list of multiples never ends.

For example, multiples of $2$ are $2, 4, 6, 8, \dots$, which is an infinite set.

Therefore, the number of multiples is infinite, not finite.

The statement is False.

Solution:

A multiple of a number $n$ is any number that can be expressed as $n \times k$, where $k$ is a natural number.

For any number $n$, we can write:

$n = n \times 1$

Since $n$ is $1$ times itself, it satisfies the definition of being a multiple.

The statement is True.

Solution:

Let two consecutive odd numbers be $2n + 1$ and $2n + 3$.

Sum $= (2n + 1) + (2n + 3) = 4n + 4$

$\text{Sum} = 4(n + 1)$

Since the sum is a multiple of $4$, it is always divisible by $4$.

Example: $3 + 5 = 8$ (divisible by 4); $7 + 9 = 16$ (divisible by 4).

The statement is True.

Solution:

Let a number $n$ divide $a, b,$ and $c$ exactly. This means $a = nx, b = ny,$ and $c = nz$ for some integers $x, y, z$.

$\text{Sum} = a + b + c = nx + ny + nz$

Since the sum is a multiple of $n$, it is exactly divisible by $n$.

The statement is True.

Solution:

Consider the sum $10 = 2 + 3 + 5$.

The number $5$ exactly divides the sum $10$ ($10 \div 5 = 2$).

However, $5$ does not divide $2$ or $3$ separately.

Therefore, divisibility of the sum does not imply divisibility of the individual addends.

The statement is False.

Solution:

If a number is divisible by $2$ and $3$, it must be divisible by their LCM, which is $6$.

It is not necessary for the number to be divisible by $12$.

Example:

The number $6$ is divisible by both $2$ and $3$, but $6$ is not divisible by $12$.

The statement is False.

Solution:

The divisibility rule for $6$ is that the number must be divisible by both $2$ and $3$.

Checking only the last two digits is not a valid rule for $6$.

Example:

In the number $212$, the last two digits are $12$, which is divisible by $6$.

However, the sum of digits of $212$ is $2 + 1 + 2 = 5$, which is not divisible by $3$. Thus, $212$ is not divisible by $6$.

The statement is False.

Solution:

This is the standard divisibility rule for 8.

A number is divisible by $8$ if the number formed by its hundreds, tens, and ones digits is divisible by $8$.

Example:

In $7120$, the last three digits are $120$. Since $120 \div 8 = 15$, the entire number $7120$ is divisible by $8$.

The statement is True.

Solution:

If the sum of digits is divisible by $3$, the number is divisible by $3$.

For a number to be divisible by $9$, the sum of its digits must be divisible by $9$.

Example:

The sum of digits of $12$ is $1 + 2 = 3$. This is divisible by $3$, so $12$ is divisible by $3$.

However, $3$ is not divisible by $9$, so $12$ is not divisible by $9$.

The statement is False.

Solution:

Multiples of $4$: $4, 8, 12, 16, 20, \dots$

Multiples of $8$: $8, 16, 24, \dots$

We can see that numbers like $4, 12,$ and $20$ are divisible by $4$ but not by $8$.

Thus, it is correct to say they "may not" be divisible by $8$.

The statement is True.

Solution:

The HCF of numbers is always less than or equal to the smallest of the numbers.

The LCM of numbers is always greater than or equal to the largest of the numbers.

Example:

For $6$ and $8$:

$\text{HCF} = 2$

$\text{LCM} = 24$

Clearly, $2 < 24$. The HCF can never be greater than the LCM.

The statement is False.

Solution:

HCF is a factor of the numbers, and the numbers are factors of the LCM.

Therefore, the HCF is always a factor of the LCM.

Example:

For $12$ and $18$:

$\text{HCF} = 6$

$\text{LCM} = 36$

$36 \div 6 = 6$ (exactly divisible).

The statement is True.

Solution:

As established in the previous question, the LCM must be divisible by the HCF.

Let's check for the given values:

$\text{LCM} \div \text{HCF} = 28 \div 8$

$28$ is not exactly divisible by $8$ ($28 = 8 \times 3 + 4$).

Since the condition is not met, no such pair of numbers can exist.

The statement is False.

Solution:

This happens when one number is a multiple of the other(s).

Example:

For $5$ and $10$:

$\text{LCM} = 10$

Here, the LCM ($10$) is one of the given numbers.

The statement is True.

Solution:

This happens when one number is a factor of the other(s).

Example:

For $5$ and $15$:

$\text{HCF} = 5$

Here, the HCF ($5$) is one of the given numbers.

The statement is True.

Solution:

Whole numbers are $\{0, 1, 2, 3, \dots\}$.

The successor of $n$ is $n + 1$. This means any number $x$ is a successor if it can be written as $n + 1$.

Consider the whole number $0$. If $0$ is a successor, then $n + 1 = 0$, which gives $n = -1$.

But $-1$ is not a whole number.

Therefore, $0$ is the only whole number that is not the successor of any other whole number.

The statement is False.

Solution:

Let's check with small whole numbers:

If $a = 0$ and $b = 5$:

$\text{Sum} = 0 + 5 = 5$

$\text{Product} = 0 \times 5 = 0$

Here, $5 > 0$. (The sum is greater than the product).

If $a = 1$ and $b = 2$:

$\text{Sum} = 1 + 2 = 3$

$\text{Product} = 1 \times 2 = 2$

Here, $3 > 2$. (The sum is greater than the product).

The statement is False.

Solution:

If the sum of two numbers is odd, then one number must be even and the other must be odd.

Let $a = \text{Even}$ and $b = \text{Odd}$.

$\text{Sum} = \text{Even} + \text{Odd} = \text{Odd}$

$\text{Difference} = \text{Even} - \text{Odd} = \text{Odd}$ (or $\text{Odd} - \text{Even} = \text{Odd}$)

Example:

Numbers $4$ and $7$.

$\text{Sum} = 4 + 7 = 11$ (Odd)

$\text{Difference} = 7 - 4 = 3$ (Odd)

The statement is True.

Solution:

Consecutive numbers are of the form $n$ and $n + 1$.

Their only common factor is $1$.

Example:

• $4$ and $5$: HCF is $1$.
• $15$ and $16$: HCF is $1$.
• $20$ and $21$: HCF is $1$.

Numbers with HCF $1$ are called coprime numbers.

The statement is True.

Solution:

We know the relation:

Let the numbers be $a$ and $b$. Suppose $\text{HCF} = a$.

$a \times \text{LCM} = a \times b$

$\text{LCM} = b$

So, the LCM is indeed the other number.

The statement is True.

Solution:

The HCF of two numbers is always less than or equal to the smaller number.

It "may be" equal, so it is not always strictly smaller.

Example:

For $4$ and $8$, the smaller number is $4$.

$\text{HCF} = 4$.

Here, the HCF ($4$) is equal to the smaller number ($4$), not smaller than it.

The statement is False.

Solution:

The LCM of two numbers is always greater than or equal to the larger number.

It "may be" equal, so it is not always strictly greater.

Example:

For $5$ and $10$, the larger number is $10$.

$\text{LCM} = 10$.

Here, the LCM ($10$) is equal to the larger number ($10$), not greater than it.

The statement is False.

Solution:

For coprime numbers, the HCF is $1$.

Using the formula:

$\text{HCF} \times \text{LCM} = \text{Product}$

$1 \times \text{LCM} = \text{Product}$

$\text{LCM} = \text{Product}$

The statement is True.

(a) We know that 1 million is equal to 10 lakh, and 1 crore is equal to 100 lakh.

Therefore, 10 million = $10 \times 10$ lakh = 100 lakh.

Since 100 lakh = 1 crore, 10 million = 1 crore.

(b) We know that 1 million is equal to 10 lakh.

Therefore, 10 lakh = 1 million.

(a) We know that 1 metre = 100 centimetres.

Also, 1 centimetre = 10 millimetres.

Therefore, 1 metre = $100 \times 10$ millimetres = 1000 millimetres.

So, 1 metre = 1000 millimetres.

(b) We know that 1 centimetre = 10 millimetres.

So, 1 centimetre = 10 millimetres.

(c) We know that 1 kilometre = 1000 metres.

Also, 1 metre = 1000 millimetres.

Therefore, 1 kilometre = $1000 \times 1000$ millimetres = 10,00,000 millimetres.

So, 1 kilometre = 10,00,000 millimetres.

(a) We know that 1 gram is equal to 1000 milligrams.

So, 1 gram = 1000 milligrams.

(b) We know that 1 litre is equal to 1000 millilitres.

So, 1 litre = 1000 millilitres.

(c) We know that 1 kilogram = 1000 grams.

Also, 1 gram = 1000 milligrams.

Therefore, 1 kilogram = $1000 \times 1000$ milligrams = 10,00,000 milligrams.

So, 1 kilogram = 1,000,000 milligrams.

We know that 1 lakh is equal to 100 thousands.

Therefore, 100 thousands = 1 lakh.

Given height = 1m 65cm.

We need to convert this height into millimetres.

We know that 1 metre = 1000 millimetres.

So, 1m = 1000 mm.

We also know that 1 centimetre = 10 millimetres.

So, 65cm = $65 \times 10$ millimetres = 650 millimetres.

Total height in millimetres = Height in metres (in mm) + Height in centimetres (in mm).

Total height = $1000 \text{ mm} + 650 \text{ mm} = 1650 \text{ mm}$.

His height in millimetres is 1650.

Given length of river Narmada = 1290 km.

We need to convert this length into metres.

We know that 1 kilometre = 1000 metres.

Therefore, 1290 km = $1290 \times 1000$ metres = 12,90,000 metres.

Its length in metres is 12,90,000.

Given distance between Srinagar and Leh = 422 km.

We need to convert this distance into metres.

We know that 1 kilometre = 1000 metres.

Therefore, 422 km = $422 \times 1000$ metres = 4,22,000 metres.

The same distance in metres is 4,22,000.

The greatest number made by five different non-zero digits uses the digits 9, 8, 7, 6, and 5 arranged in descending order.

The greatest number is 98765.

Reversing the order of digits of 98765 gives 56789.

This new number, 56789, is formed by the same five digits (5, 6, 7, 8, 9) but arranged in ascending order.

Arranging digits in ascending order forms the smallest number possible using those specific digits.

Therefore, the new number is the smallest number of five digits formed by these digits.

The new number is the smallest number of five digits (using the same digits).

We know that ten lakh is written as 10,00,000.

The number that is 1 less than 10,00,000 is $10,00,000 - 1 = 9,99,999$.

The number 9,99,999 has 6 digits and is the greatest 6-digit number.

So, by adding 1 to the greatest 6-digit number (9,99,999), we get ten lakh (10,00,000).

By adding 1 to the greatest six digit number, we get ten lakh.

The number given in words is five crore twenty three lakh seventy eight thousand four hundred one.

We can break this down by place value in the Indian System:

Five crore: 5,00,00,000

Twenty three lakh: 23,00,000

Seventy eight thousand: 78,000

Four hundred one: 401

Adding these values together:

$5,00,00,000 + 23,00,000 + 78,000 + 401 = 5,23,78,401$.

Using commas in the Indian System of Numeration, the number is written as 5,23,78,401.

The number can be written, using commas, in the Indian System of Numeration as 5,23,78,401.

Given:

The symbol $X$ in Roman Numeration and the rule for its subtraction.

To Find:

The missing symbol in the statement.

Solution:

The rules for subtracting Roman numerals are very specific. The symbol $X$ (which represents 10) can only be subtracted from symbols that have a value of 50 or 100.

The symbols for these values are:

• $L = 50$
• $C = 100$

Rules Summary Table:

By comparing the standard rules with the question, we find that the missing symbol is L.

In Roman Numeration, the symbol $X$ can be subtracted from L, $M$ and $C$ only.

To write the number 66 in Roman numerals, we decompose it based on the standard Roman numeral values:

$66 = 50 + 10 + 5 + 1$

The Roman numeral for 50 is L.

The Roman numeral for 10 is X.

The Roman numeral for 5 is V.

The Roman numeral for 1 is I.

Combining these symbols in descending order of their values:

$50 \to \text{L}$

$10 \to \text{X}$

$5 \to \text{V}$

$1 \to \text{I}$

Putting them together, we get LXVI.

The number 66 in Roman numerals is LXVI.

Given population is 2,538,473.

We need to round this number to the nearest thousands.

The thousands digit is 8.

The digit to the right of the thousands digit (in the hundreds place) is 4.

Since the digit 4 is less than 5, we keep the thousands digit as it is (8) and replace all digits to its right with zeros.

The digits to the right are 4, 7, and 3.

Replacing these digits with zeros, we get 2,538,000.

Rounded off to nearest thousands, the population was 2,538,000.

Whole numbers are the set of non-negative integers. They start from 0 and go upwards infinitely (0, 1, 2, 3, ...).

The smallest number in this set is the first number listed.

The smallest whole number is 0.

The successor of a number is the number that comes immediately after it. It is obtained by adding 1 to the given number.

Successor of 106159 = $106159 + 1$.

$106159 + 1 = 106160$.

Successor of 106159 is 106160.

The predecessor of a number is the number that comes immediately before it. It is obtained by subtracting 1 from the given number.

Predecessor of 100000 = $100000 - 1$.

$100000 - 1 = 99999$.

Predecessor of 100000 is 99999.

If 400 is the predecessor of a number, it means that number is 1 more than 400.

The number = $400 + 1 = 401$.

400 is the predecessor of 401.

The largest 3-digit number is 999.

The successor of 999 is the number immediately after it, which is $999 + 1 = 1000$.

1000 is the smallest 4-digit number.

1000 is the successor of the largest 3 digit number.

Let $a$ be a whole number.

Subtracting 0 from $a$ means performing the operation $a - 0$.

Any number minus 0 is the number itself.

$a - 0 = a$

So, the result is the whole number itself.

If 0 is subtracted from a whole number, then the result is the number itself .

Given:

A natural number having 6 digits that must end with the digit 5.

To Find:

The smallest such 6-digit natural number.

Solution:

In the number system, the smallest 6-digit number is formed by placing 1 at the highest place value (Lakhs place in the Indian system) and 0 in all other places.

$\text{Smallest 6-digit number} = 1,00,000$

We are looking for a number that ends in 5. This means the digit at the ones place must be 5.

To keep the number as small as possible, we should not change the digits at the higher place values (Lakhs, Ten-thousands, Thousands, Hundreds, and Tens).

Therefore, we keep the first five digits as $1, 0, 0, 0, 0$ and replace the last digit $0$ with $5$.

Alternate Solution:

We can find the answer by listing the smallest 6-digit numbers in increasing order and identifying the first one that ends in 5:

$1,00,000$ (Ends in 0)

$1,00,001$ (Ends in 1)

$1,00,002$ (Ends in 2)

$1,00,003$ (Ends in 3)

$1,00,004$ (Ends in 4)

$1,00,005$ (Ends in 5)

Since $1,00,005$ is the first number in this sequence that satisfies the condition, it is the smallest.

Conclusion:

The smallest 6-digit natural number ending in 5 is 1,00,005.

Given:

The property of "closure" for whole numbers under different mathematical operations.

To Find:

The two operations under which whole numbers are always closed.

Solution:

Whole numbers are $0, 1, 2, 3, 4, ...$ and so on. A set of numbers is said to be closed under an operation if, after performing that operation on any two whole numbers, the result is always a whole number.

Let us check the four basic operations:

1. Addition: When we add any two whole numbers, the sum is always a whole number.

So, whole numbers are closed under addition.

2. Multiplication: When we multiply any two whole numbers, the product is always a whole number.

So, whole numbers are closed under multiplication.

Verification Table:

Let us see why subtraction and division are not included:

Alternate Solution:

We can remember this using a simple rule: if you can find even one example where the result is not a whole number, the operation is not closed. Since subtracting a larger number from a smaller number or dividing numbers that don't divide perfectly results in non-whole numbers, only addition and multiplication satisfy the closure property for all cases.

Conclusion:

Whole numbers are closed under addition and under multiplication.

Given:

The property of "closure" for natural numbers under basic arithmetic operations.

To Find:

The two operations under which natural numbers are closed.

Solution:

Natural numbers are the counting numbers $1, 2, 3, 4, ...$ and so on. A set of numbers is closed under an operation if, when we use any two natural numbers, the result is also a natural number.

Let us check the operations:

1. Addition: If we add any two natural numbers, the result is always a natural number.

Since the sum is always a natural number, natural numbers are closed under addition.

2. Multiplication: If we multiply any two natural numbers, the product is always a natural number.

Since the product is always a natural number, natural numbers are closed under multiplication.

Verification Table:

We can see why subtraction and division do not follow this property:

Alternate Solution:

We define natural numbers as numbers starting from $1$. For any two natural numbers $a$ and $b$:

Subtraction is not closed because $3 - 3 = 0$, and $0$ is not a natural number. Similarly, $2 - 5 = -3$, which is also not a natural number.

Conclusion:

Natural numbers are closed under addition and under multiplication.

In mathematics, division by zero is an undefined operation.

If we try to divide a non-zero number by zero, the result is undefined.

If we try to divide zero by zero, the result is also undefined (it is considered an indeterminate form).

Since 0 is a whole number, the division of any whole number (including 0) by 0 is not defined.

Division of a whole number by zero is not defined.

Given:

The distributive property of whole numbers.

To Find:

The operation over which multiplication is distributive for whole numbers.

Solution:

The distributive property of multiplication allows us to "distribute" a multiplier to each number inside a bracket. This is most commonly applied over addition.

The general rule for any three whole numbers $a, b,$ and $c$ is:

Example:

Let us take the whole numbers $3, 10,$ and $5$ to verify this property:

Since both methods give the same result ($45$), we say multiplication is distributive over addition.

Verification for Subtraction:

Multiplication is also distributive over subtraction for whole numbers, provided the result inside the bracket remains a whole number.

For example, if we take $4 \times (10 - 2)$:

$4 \times 8 = 32$

$(4 \times 10) - (4 \times 2) = 40 - 8 = 32$

Conclusion:

The most common operation referred to in this property is addition.

Multiplication is distributive over addition for whole numbers.

Given:

An incomplete equation: $2395 \times \text{_____} = 6195 \times 2395$

To Find:

The missing number that makes both sides of the equation equal.

Solution:

This question is based on a very important property of whole numbers called the Commutative Property of Multiplication.

According to this property, we can multiply two whole numbers in any order, and the answer (product) will remain the same.

$\text{First Number} \times \text{Second Number} = \text{Second Number} \times \text{First Number}$

Now, let us compare the two sides of our equation:

$\text{Left Side} = 2395 \times \text{_____}$

$\text{Right Side} = 6195 \times 2395$

On the right side, we have two numbers: $6195$ and $2395$. On the left side, we only have $2395$.

To make the left side equal to the right side, we must have the same two numbers being multiplied, just in a different order.

Therefore, the missing number is 6195.

Verification:

Let us take a smaller example to understand this better. Suppose we have:

$2 \times 3 = 6$

$3 \times 2 = 6$

Just like $2 \times 3$ is the same as $3 \times 2$, the product $2395 \times 6195$ will be exactly the same as $6195 \times 2395$.

Conclusion:

The blank should be filled with $6195$.

The complete statement is: $2395 \times \mathbf{6195} = 6195 \times 2395$.

The given equation is $1001 \times 2002 = 1001 \times (1001 + \text{_____})$.

This equation demonstrates the distributive property of multiplication over addition, which states $a \times (b + c) = a \times b + a \times c$.

In the given equation, the left side is $1001 \times 2002$.

The right side has the factor 1001 multiplied by an expression in parentheses, $(1001 + \text{_____})$.

For the equality to hold, the expression inside the parentheses must be equal to 2002.

So, we have the equation:

To find the value of the blank, we subtract 1001 from 2002.

Thus, the number in the blank is 1001.

1001 × 2002 = 1001 × (1001 + 1001 ).

The given expression is $10001 \times 0$.

According to the property of multiplication by zero, any number multiplied by zero is equal to zero.

$a \times 0 = 0$ for any number $a$.

In this case, $a = 10001$.

So, $10001 \times 0 = 0$.

10001 × 0 = 0.

The given equation is $2916 \times \text{_____} = 0$.

We know that for the product of two numbers to be zero, at least one of the numbers must be zero.

In this equation, one factor is 2916, which is not zero.

Therefore, the other factor (the number in the blank) must be zero for the product to be zero.

$2916 \times 0 = 0$.

2916 × 0 = 0.

The given equation is $9128 \times \text{_____} = 9128$.

This equation demonstrates the property of the multiplicative identity for whole numbers.

The multiplicative identity for whole numbers is 1, because any whole number multiplied by 1 results in the same whole number.

$a \times 1 = a$ for any whole number $a$.

In this case, $a = 9128$.

For the equation $9128 \times \text{_____} = 9128$ to be true, the number in the blank must be 1.

$9128 \times 1 = 9128$.

9128 × 1 = 9128.

Property:

This equation is based on the Associative Property of Addition. It states that the way in which numbers are grouped while adding does not change the final sum.

Solution:

Let us compare the given equation with the property:

Left Side: $125 + (68 + 17)$

Right Side: $(125 + \text{_____}) + 17$

By comparing both sides, we identify the three numbers as $125$, $68$, and $17$.

Final Answer:

The number in the blank is 68.

The complete statement is: $125 + (68 + 17) = (125 + \mathbf{68} ) + 17$.

The given expression is $8925 \times 1$.

This demonstrates the property of the multiplicative identity.

According to the multiplicative identity property, when any number is multiplied by 1, the result is the number itself.

$a \times 1 = a$ for any number $a$.

In this case, $a = 8925$.

So, $8925 \times 1 = 8925$.

8925 × 1 = 8925.

Given:

An incomplete equation: $19 \times 12 + 19 = 19 \times (12 + \text{_____})$

To Find:

The missing number in the blank.

Solution:

Let us look at the left side of the equation: $19 \times 12 + 19$.

We know that any number multiplied by $1$ remains the same value. This is called the multiplicative identity. So, we can write $19$ as $19 \times 1$.

Now, we can rewrite the left side of the equation:

$19 \times 12 + 19 \times 1$

This expression is now in the form of the Distributive Property of Multiplication over Addition. The property states:

By comparing our expression $(19 \times 12 + 19 \times 1)$ with the property:

• The common number (multiplier) is $19$.
• The first number inside the bracket is $12$.
• The second number inside the bracket must be $1$.

Conclusion:

By comparing the result with the right side of the given question, we find that the missing number is 1.

The complete statement is: $19 \times 12 + 19 = 19 \times (12 + \mathbf{1})$.

Given:

The equation: $24 \times 35 = 24 \times 18 + 24 \times \text{_____}$

To Find:

The missing number in the blank.

Solution:

This problem is based on the Distributive Property of Multiplication over Addition. This property states that multiplying a number by a sum is the same as multiplying the number by each part of the sum separately and then adding them together.

In our question, the number being multiplied is $24$. On the left side, it is multiplied by $35$. On the right side, it is distributed over $18$ and the missing number.

This means that the sum of $18$ and the missing number must be equal to $35$.

To find the blank, we subtract $18$ from $35$:

Conclusion:

The missing number is 17.

The complete statement is: $24 \times 35 = 24 \times 18 + 24 \times \mathbf{17}$.

Given:

An incomplete equation: $32 \times (27 \times 19) = (32 \times \text{_____} ) \times 19$

To Find:

The missing whole number in the blank.

Solution:

This equation is based on the Associative Property of Multiplication for whole numbers. This property states that when multiplying three or more numbers, the way they are grouped (using brackets) does not change the final product.

Let us identify the three numbers from the left side (LHS) of the equation:

• The first number is $32$.
• The second number is $27$.
• The third number is $19$.

Now, let us look at the right side (RHS) of the equation:

By comparing both sides, we can see that $32$ and $19$ are already present. To maintain the equality, the missing number in the bracket must be the second number, which is 27.

Verification:

If we use $27$ in the blank:

Conclusion:

The number that completes the equation is 27.

The complete statement is: $32 \times (27 \times 19) = (32 \times \mathbf{27} ) \times 19$.

The given expression is $786 \times 3 + 786 \times 7$.

We can use the distributive property of multiplication over addition, which states that $a \times b + a \times c = a \times (b + c)$.

In this expression, we have $a = 786$, $b = 3$, and $c = 7$.

Applying the distributive property:

Now, we perform the addition inside the parentheses:

Substitute this back into the expression:

Perform the multiplication:

So, the value of the expression is 7860.

786 × 3 + 786 × 7 = 7860.

The given equation is $24 \times 25 = 24 \times \text{_____}$.

For the equality to be true, the expression on the left side must be equal to the expression on the right side.

The left side is $24 \times 25$.

The right side is $24 \times \text{_____}$.

Since 24 is a common factor on both sides, the number in the blank must be equal to 25 for the equality to hold.

Thus, the blank should be filled with 25.

24 × 25 = 24 × 25.

Given:

A statement describing the relationship between a number and its factors.

To Find:

the missing word that correctly relates a number to its factors.

Solution:

In the Indian school curriculum, we learn that factors and multiples are two sides of the same coin. Let us understand this with an example.

Consider the number $12$. The factors of $12$ are those numbers that divide $12$ exactly without leaving any remainder.

Now, let us see how $12$ is related to these factors through multiplication:

A multiple of a number is what we get after multiplying that number by a whole number ($1, 2, 3,$ etc.).

From the table above, we can see that $12$ is obtained by multiplying each of its factors by another number. This means $12$ is a multiple of $1$, a multiple of $2$, a multiple of $3$, and so on.

Conclusion:

Since the original number is always found in the multiplication table of its factors, it is always a multiple of those factors.

A number is a multiple of each of its factor.

A factor of a number is a number that divides the given number exactly, with no remainder.

Let's consider a few numbers and their factors:

Factors of 5 are 1, 5.

Factors of 10 are 1, 2, 5, 10.

Factors of 12 are 1, 2, 3, 4, 6, 12.

Factors of any positive integer $n$ include 1, because $n \div 1 = n$, which is a whole number with a remainder of 0. So, 1 divides every positive integer exactly.

For the number 0, any non-zero number is a factor of 0 because $0 \div a = 0$ for any $a \neq 0$. However, 1 is also a factor of 0 because $0 \div 1 = 0$.

Thus, 1 is a factor of every whole number.

1 is a factor of every number.

A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.

Let's consider a few prime numbers and find their factors:

Prime number 2: The only positive integers that divide 2 exactly are 1 and 2. Factors are {1, 2}. Number of factors = 2.

Prime number 3: The only positive integers that divide 3 exactly are 1 and 3. Factors are {1, 3}. Number of factors = 2.

Prime number 5: The only positive integers that divide 5 exactly are 1 and 5. Factors are {1, 5}. Number of factors = 2.

Prime number 7: The only positive integers that divide 7 exactly are 1 and 7. Factors are {1, 7}. Number of factors = 2.

By definition, a prime number $p$ has exactly two distinct positive divisors: 1 and $p$ itself.

The number of factors of a prime number is two.

Let's consider an example to understand this property.

Consider the number 6.

The factors of 6 are the numbers that divide 6 exactly: 1, 2, 3, and 6.

The sum of the factors of 6 is $1 + 2 + 3 + 6 = 12$.

Twice the number 6 is $2 \times 6 = 12$.

Since the sum of the factors (12) is equal to twice the number (12), 6 satisfies the given condition.

Consider another example, the number 28.

The factors of 28 are 1, 2, 4, 7, 14, and 28.

The sum of the factors of 28 is $1 + 2 + 4 + 7 + 14 + 28 = 56$.

Twice the number 28 is $2 \times 28 = 56$.

Since the sum of the factors (56) is equal to twice the number (56), 28 also satisfies the given condition.

Numbers that have this property (the sum of all their factors is equal to twice the number) are called perfect numbers.

A number for which the sum of all its factors is equal to twice the number is called a perfect number.

We know that a prime number is a natural number greater than 1 that has exactly two factors: 1 and itself.

Natural numbers greater than 1 that are not prime are called composite numbers.

Let's consider a few examples of numbers greater than 1:

Number 2: Factors are {1, 2}. Number of factors = 2 (Prime).

Number 3: Factors are {1, 3}. Number of factors = 2 (Prime).

Number 4: Factors are {1, 2, 4}. Number of factors = 3 (More than two factors).

Number 5: Factors are {1, 5}. Number of factors = 2 (Prime).

Number 6: Factors are {1, 2, 3, 6}. Number of factors = 4 (More than two factors).

Numbers with more than two factors must have factors other than 1 and themselves. This is the definition of a composite number.

The numbers having more than two factors are called composite numbers.

We categorize natural numbers based on their factors. A prime number is a number that has exactly two factors: 1 and the number itself.

Let us look at the first few prime numbers:

An even number is a number that is divisible by 2. All even numbers except 2 are composite numbers because they have 2 as a factor in addition to 1 and themselves.

Conclusion:

The number 2 is the only even number that has only two factors (1 and 2). All other even numbers have more than two factors.

Therefore, 2 is the only prime number which is even.

When we find the factors of two numbers and the only factor they share is 1, these numbers have a special relationship.

Let's consider an example:

Factors of 4 are {1, 2, 4}.

Factors of 9 are {1, 3, 9}.

The common factors of 4 and 9 are the numbers that appear in both lists. The only common factor is 1.

Numbers like 4 and 9, which have only 1 as their common factor, are called coprime numbers or relatively prime numbers.

Two numbers having only 1 as a common factor are called coprime numbers.

To find the number of prime numbers between 1 and 100, we can list the prime numbers in this range.

Remember, a prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.

Let's list them:

Primes between 1 and 10: 2, 3, 5, 7 (4 primes)

Primes between 10 and 20: 11, 13, 17, 19 (4 primes)

Primes between 20 and 30: 23, 29 (2 primes)

Primes between 30 and 40: 31, 37 (2 primes)

Primes between 40 and 50: 41, 43, 47 (3 primes)

Primes between 50 and 60: 53, 59 (2 primes)

Primes between 60 and 70: 61, 67 (2 primes)

Primes between 70 and 80: 71, 73, 79 (3 primes)

Primes between 80 and 90: 83, 89 (2 primes)

Primes between 90 and 100: 97 (1 prime)

Total number of primes = $4 + 4 + 2 + 2 + 3 + 2 + 2 + 3 + 2 + 1 = 25$.

The prime numbers between 1 and 100 are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.

Counting these, there are 25 prime numbers.

Number of primes between 1 to 100 is 25.

A number is divisible by 10 if and only if its last digit is 0.

The last digit of a number is the digit in the ones place.

For example:

The number 50 is divisible by 10 because its ones digit is 0 ($50 \div 10 = 5$).

The number 120 is divisible by 10 because its ones digit is 0 ($120 \div 10 = 12$).

The number 345 is not divisible by 10 because its ones digit is 5.

Thus, for a number to be divisible by 10, the digit in the ones place must be 0.

If a number has 0 in ones place, then it is divisible by 10.

A number is divisible by 5 if and only if its last digit is either 0 or 5.

The last digit of a number is the digit in its ones place.

Examples:

The number 20 is divisible by 5 ($20 \div 5 = 4$). Its ones digit is 0.

The number 35 is divisible by 5 ($35 \div 5 = 7$). Its ones digit is 5.

The number 105 is divisible by 5 ($105 \div 5 = 21$). Its ones digit is 5.

The number 230 is divisible by 5 ($230 \div 5 = 46$). Its ones digit is 0.

The number 123 is not divisible by 5. Its ones digit is 3.

Thus, for a number to be divisible by 5, the digit in its ones place must be either 0 or 5.

A number is divisible by 5, if it has 0 or 5 in its ones place.

Given:

A number with any of the digits $0, 2, 4, 6,$ or $8$ in its ones place.

Solution:

According to the Divisibility Rule of 2, a number is divisible by $2$ if it is an even number. Even numbers always end with an even digit.

The even digits are $0, 2, 4, 6,$ and $8$.

Conclusion:

Since the numbers ending in these specific digits are all even, they are always divisible by $2$.

A number is divisible by 2 if it has any of the digits $0, 2, 4, 6,$ or $8$ in its ones place.

This is the divisibility rule for 3.

Let's consider some examples:

Consider the number 123.

Sum of digits = $1 + 2 + 3 = 6$.

6 is divisible by 3 ($6 \div 3 = 2$).

Is 123 divisible by 3? Yes, $123 \div 3 = 41$.

Consider the number 549.

Sum of digits = $5 + 4 + 9 = 18$.

18 is divisible by 3 ($18 \div 3 = 6$).

Is 549 divisible by 3? Yes, $549 \div 3 = 183$.

Consider the number 14.

Sum of digits = $1 + 4 = 5$.

5 is not divisible by 3.

Is 14 divisible by 3? No, $14 \div 3 = 4$ with a remainder of 2.

The divisibility rule for 3 states that a number is divisible by 3 if the sum of its digits is divisible by 3.

Another way to say "divisible by 3" is "a multiple of 3".

If the sum of the digits in a number is a multiple of 3, then the number is divisible by 3.

Given:

A rule to check the divisibility of a number by $11$ based on the position of its digits.

To Find:

The number that completes the divisibility rule for $11$.

Solution:

We use the Divisibility Test for 11 to find out if a large number can be divided by $11$ without doing actual long division.

To apply this rule, we follow these steps:

1. Find the sum of the digits at the odd places (1st, 3rd, 5th... digits) starting from the right side.

2. Find the sum of the digits at the even places (2nd, 4th, 6th... digits) starting from the right side.

3. Calculate the difference between these two sums.

Example Verification:

Let us check the number $1331$:

Since the difference is $0$, the number $1331$ is divisible by $11$.

Now, let us check the number $616$:

Since the difference is $11$, and $11$ is divisible by $11$, the number $616$ is also divisible by $11$.

Conclusion:

The rule states that for the number to be divisible by $11$, this difference must be either $0$ or a multiple of $11$.

Therefore, the missing number is 11.

The complete statement is: If the difference between the sum of digits at odd places (from the right) and the sum of digits at even places (from the right) of a number is either 0 or divisible by 11, then the number is divisible by 11.

LCM stands for Least Common Multiple.

The LCM of two or more numbers is defined as the smallest positive integer that is a multiple of all the given numbers.

Let's consider an example with two numbers, say 4 and 6.

Multiples of 4 are: 4, 8, 12, 16, 20, 24, 28, 32, 36, ...

Multiples of 6 are: 6, 12, 18, 24, 30, 36, 42, ...

The common multiples of 4 and 6 are the numbers that appear in both lists: 12, 24, 36, ...

The lowest (smallest) of these common multiples is 12.

So, the LCM of 4 and 6 is 12.

The definition explicitly states that the LCM is the lowest of the common "multiples".

The LCM of two or more given numbers is the lowest of their common multiples.

HCF stands for Highest Common Factor.

The HCF of two or more numbers is defined as the largest positive integer that divides each of the given numbers without leaving a remainder.

Let's consider an example with two numbers, say 12 and 18.

Factors of 12 are the numbers that divide 12 exactly: 1, 2, 3, 4, 6, 12.

Factors of 18 are the numbers that divide 18 exactly: 1, 2, 3, 6, 9, 18.

The common factors of 12 and 18 are the numbers that appear in both lists: 1, 2, 3, and 6.

The highest (largest) of these common factors is 6.

So, the HCF of 12 and 18 is 6.

The definition explicitly states that the HCF is the highest of the common "factors".

The HCF of two or more given numbers is the highest of their common factors.

Given:

Two columns with mathematical descriptions and their corresponding results.

To Match:

Each item in Column I with its correct value or property in Column II.

Solution:

(i) The difference of two consecutive whole numbers

Whole numbers are $0, 1, 2, 3, \dots$. Consecutive numbers follow one after another, such as $4$ and $5$. The difference is always calculated by subtracting the smaller number from the larger one.

This matches with (d) 1.

(ii) The product of two non-zero consecutive whole numbers

In any two consecutive numbers, one must be even and the other must be odd. When we multiply an even number by any other number, the result is always even.

This matches with (f) even.

(iii) Quotient when zero is divided by another non-zero whole number

When $0$ is divided by any number (except $0$), the result (quotient) is always $0$.

This matches with (b) 0.

(iv) 2 added three times, to the smallest whole number

The smallest whole number in the Indian number system is $0$. Adding $2$ to it three times means:

This matches with (e) 6.

(v) Smallest odd prime number

Prime numbers are $2, 3, 5, 7, \dots$. Among these, $2$ is the smallest prime number, but it is even. The next prime number is $3$, which is odd.

This matches with (c) 3.

Final Match Table:

Conclusion:

The final mapping is: (i)-d, (ii)-f, (iii)-b, (iv)-e, (v)-c.

To Find:

Arrange the given numbers from the greatest to the smallest (descending order).

Solution:

The given numbers are $8435$, $4835$, $13584$, $5348$, and $25843$.

First, we compare the number of digits in each number:

• 4-digit numbers: $8435, 4835, 5348$
• 5-digit numbers: $13584, 25843$

Since 5-digit numbers are greater than 4-digit numbers, we compare the 5-digit numbers first.

Comparing 5-digit numbers:

In $25843$ and $13584$, the digits at the ten-thousands place are $2$ and $1$ respectively.

Therefore, $25843 > 13584$.

Comparing 4-digit numbers:

In $8435, 4835,$ and $5348$, the digits at the thousands place are $8, 4,$ and $5$.

Therefore, $8435 > 5348 > 4835$.

Final Arrangement:

Arranging all the numbers in descending order (greatest to smallest):

25843, 13584, 8435, 5348, 4835.

Given:

Numbers: $3,80,51,425$; $3,00,40,700$; $6,72,05,602$

Solution:

All the given numbers have 8 digits. We compare them by looking at their digits from the leftmost place (highest place value).

The digits at the Crores place are:

• $3$ (in 38051425)
• $3$ (in 30040700)
• $6$ (in 67205602)

Therefore, 67205602 is the greatest number.

Now, we compare the remaining two numbers starting with 3: $38051425$ and $30040700$.

We look at the next digit to the right (Ten Lakhs place):

• $8$ (in 38051425)
• $0$ (in 30040700)

Therefore, 30040700 is the smallest number.

Conclusion:

Greatest number: 67205602

Smallest number: 30040700

Solution:

To write the expanded form, we multiply each digit by its respective place value.

(a) 74836

$74836 = 7 \times 10000 + 4 \times 1000 + 8 \times 100 + 3 \times 10 + 6 \times 1$

(b) 574021

$574021 = 5 \times 100000 + 7 \times 10000 + 4 \times 1000 $$ \ + 0 \times 100 + 2 \times 10 + 1 \times 1$

(c) 8907010

$8907010 = 8 \times 1000000 + 9 \times 100000 + 0 \times 10000 + 7 \times 1000 $$ \ + 0 \times 100 + 1 \times 10 + 0 \times 1$

Solution:

The populations are as follows:

• Andhra Pradesh (AP): $7,62,10,007$ (8 digits)
• Bihar (BR): $8,29,98,509$ (8 digits)
• Maharashtra (MH): $9,6,878,627$ (8 digits)
• Uttar Pradesh (UP): $16,61,97,921$ (9 digits)

Ascending Order (Smallest to Greatest):

Comparing 8-digit numbers, we look at the leftmost digit: $7 < 8 < 9$. The 9-digit number is the greatest.

Andhra Pradesh < Bihar < Maharashtra < Uttar Pradesh

Descending Order (Greatest to Smallest):

Uttar Pradesh > Maharashtra > Bihar > Andhra Pradesh

Solution:

In the International System of Numeration, commas are placed after every three digits starting from the right (ones, thousands, millions, etc.).

Given number: $142800000$

Starting from the right, group the digits in threes: $142, 800, 000$.

Diameter $= \mathbf{142,800,000}$ metres.

In words: One hundred forty-two million eight hundred thousand metres.

To Find:

Total increase in population and its representation in the Indian System.

Solution:

Population in 1961 $= 439 \text{ million}$

Population in 2001 $= 1028 \text{ million}$

$\text{Increase} = 1028 - 439$

Increase in population $= 589 \text{ million}$.

Now, convert million to the Indian System:

$1 \text{ million} = 10,00,000$

$589 \text{ million} = 589 \times 10,00,000 = 58,90,00,000$

In the Indian System, the increase is 58,90,00,000 (Fifty-eight crore ninety lakh).

Given:

Radius of Earth $= 6400 \text{ km}$

Radius of Mars $= 4300000 \text{ m}$

Solution:

To compare, let's convert the Earth's radius to metres ($1 \text{ km} = 1000 \text{ m}$):

$\text{Radius of Earth} = 6400 \times 1000 = 64,00,000 \text{ m}$

Comparing the values:

Thus, the radius of Earth is bigger.

To find the difference:

$\begin{array}{ccccccc} & 6 & 4 & 0 & 0 & 0 & 0 & 0 \\ - & 4 & 3 & 0 & 0 & 0 & 0 & 0 \\ \hline & 2 & 1 & 0 & 0 & 0 & 0 & 0 \\ \hline \end{array}$

Difference $= \mathbf{21,00,000}$ metres (or 2100 km).

Solution:

Population of Tripura (3,199,203):

Words: Three million, one hundred ninety-nine thousand, two hundred three.

Population of Meghalaya (2,318,822):

Words: Two million, three hundred eighteen thousand, eight hundred twenty-two.

Solution:

Children in March $= 2,12,583$

Children in April $= 2,16,813$

To find the difference:

$\begin{array}{cccccc} & 2 & 1 & 6 & 8 & 1 & 3 \\ - & 2 & 1 & 2 & 5 & 8 & 3 \\ \hline & & & 4 & 2 & 3 & 0 \\ \hline \end{array}$

The difference is 4230.

Given:

Total money $= \textsf{₹} \ 10,00,000$

Expenditure on T.V. $= \textsf{₹} \ 16,580$

Expenditure on motorcycle $= \textsf{₹} \ 45,890$

Expenditure on flat $= \textsf{₹} \ 8,70,000$

Solution:

Total money spent:

$\begin{array}{cccccc} & & 1 & 6 & 5 & 8 & 0 \\ & & 4 & 5 & 8 & 9 & 0 \\ + & 8 & 7 & 0 & 0 & 0 & 0 \\ \hline & 9 & 3 & 2 & 4 & 7 & 0 \\ \hline \end{array}$

Total spent $= \textsf{₹} \ 9,32,470$.

Money left:

$\begin{array}{ccccccc} & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ - & & 9 & 3 & 2 & 4 & 7 & 0 \\ \hline & & & 6 & 7 & 5 & 3 & 0 \\ \hline \end{array}$

The money left is ₹ 67,530.

Solution:

Total tablets $= 1,80,000$

Distributed tablets $= 18,734$

Remaining tablets:

$\begin{array}{cccccc} & 1 & 8 & 0 & 0 & 0 & 0 \\ - & & 1 & 8 & 7 & 3 & 4 \\ \hline & 1 & 6 & 1 & 2 & 6 & 6 \\ \hline \end{array}$

The number of remaining tablets is 1,61,266.

Solution:

Total money spent:

$\begin{array}{cccccc} & & 8 & 7 & 5 & 0 & 0 \\ & 1 & 2 & 6 & 3 & 8 & 0 \\ + & 3 & 5 & 0 & 0 & 0 & 0 \\ \hline & 5 & 6 & 3 & 8 & 8 & 0 \\ \hline \end{array}$

Money left with Chinmay:

$\begin{array}{cccccc} & 6 & 1 & 0 & 0 & 0 & 0 \\ - & 5 & 6 & 3 & 8 & 8 & 0 \\ \hline & & 4 & 6 & 1 & 2 & 0 \\ \hline \end{array}$

Money left is ₹ 46,120.

Solution:

Largest 7-digit number $= 99,99,999$

Smallest 8-digit number $= 1,00,00,000$

Difference:

$\begin{array}{cccccccc} & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ - & & 9 & 9 & 9 & 9 & 9 & 9 & 9 \\ \hline & & & & & & & & 1 \\ \hline \end{array}$

The difference is 1.

Given:

Mobile number structure: $9 \ 9 \ 8 \ 7 \ \_ \ \_ \ \_ \ 3 \ 5 \ 5$

Solution:

To make the number the greatest possible, we should use the largest available distinct digits for the middle positions.

The digits already used in the number are: $9, 8, 7, 3, 5$.

The available distinct digits from the set $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$ that have not been used are:

$\{6, 4, 2, 1, 0\}$

To maximize the number, we choose the three largest from this set and arrange them in descending order: 6, 4, and 2.

The mobile number becomes $9987\mathbf{642}355$.

The digits are 6, 4, 2.

Given:

First four digits: $9, 9, 7, 9$.

Remaining 6 digits to be filled from $\{8, 3, 5, 6, 0\}$ with one digit used twice.

Solution:

To make the smallest number, we pick the smallest digit to repeat, which is $0$.

The digits to use are: $\{0, 0, 3, 5, 6, 8\}$.

Arranging them in ascending order to keep the number smallest:

Remaining digits: $0, 0, 3, 5, 6, 8$.

Smallest mobile number: $9979\mathbf{003568}$.

The digits used are 0, 0, 3, 5, 6, 8.

Solution:

Let's determine the digits place by place:

• Tens place digit $= 4$
• Units place digit $= \frac{1}{4} \times 4 = 1$
• Hundreds place digit $= 0$
• Thousands place digit $= 5 \times 1 = 5$
• Ten-thousands place digit $= 2 \times 4 = 8$

Arranging the digits: 8 5 0 4 1.

The number is 85041.

Given:

Digits: $2, 0, 4, 7, 6, 5$

Solution:

Greatest 6-digit number (Descending order): $7,65,420$

Least 6-digit number (Ascending order, first digit non-zero): $2,04,567$

Sum of these numbers:

$\begin{array}{ccccccc} & 7 & 6 & 5 & 4 & 2 & 0 \\ + & 2 & 0 & 4 & 5 & 6 & 7 \\ \hline & 9 & 6 & 9 & 9 & 8 & 7 \\ \hline \end{array}$

The sum is 9,69,987.

Given:

Total volume $= 35,874 \text{ L}$

Capacity of bottle $= 200 \text{ ml}$

Solution:

First, convert litres to millilitres:

$1 \text{ L} = 1000 \text{ ml}$

$35874 \text{ L} = 35874 \times 1000 = 3,58,74,000 \text{ ml}$

Number of bottles $=$ Total volume $\div$ capacity of one bottle

$\text{Number of bottles} = \frac{3,58,74,000}{200}$

$\text{Number of bottles} = \frac{3,58,740}{2}$

$\text{Number of bottles} = 1,79,370$

The number of bottles is 1,79,370.

Solution:

Total population $= 4,50,772$

Fraction of illiterate persons $= \frac{1}{14}$

Number of illiterate persons $= 4,50,772 \div 14$

$\begin{array}{r} 32198 \phantom{)} \\ 14{\overline{\smash{\big)}\, 450772 \phantom{)}}} \\ \underline{- 42 \phantom{)}} \\ 30 \phantom{)} \\ \underline{- 28 \phantom{)}} \\ 27 \phantom{)} \\ \underline{- 14 \phantom{)}} \\ 137 \phantom{)} \\ \underline{- 126 \phantom{)}} \\ 112 \phantom{)} \\ \underline{- 112 \phantom{)}} \\ 0 \phantom{)} \end{array}$

Total illiterate persons $= \mathbf{32,198}$.

Solution:

We use the common division method to find the LCM:

$\begin{array}{c|cc} 2 & 80, & 96, & 125, & 160 \\ \hline 2 & 40, & 48, & 125, & 80 \\ \hline 2 & 20, & 24, & 125, & 40 \\ \hline 2 & 10, & 12, & 125, & 20 \\ \hline 2 & 5, & 6, & 125, & 10 \\ \hline 5 & 5, & 3, & 125, & 5 \\ \hline 5 & 1, & 3, & 25, & 1 \\ \hline 5 & 1, & 3, & 5, & 1 \\ \hline 3 & 1, & 3, & 1, & 1 \\ \hline & 1, & 1, & 1, & 1 \end{array}$

$\text{LCM} = 2 \times 2 \times 2 \times 2 \times 2 \times 5 \times 5 \times 5 \times 3$

$\text{LCM} = 32 \times 125 \times 3$

$\text{LCM} = 4000 \times 3 = 12,000$

The LCM is 12,000.

Solution:

Number structure: $\_ \ \_ \ \_ \ \mathbf{5} \ \_$

Greatest 5-digit number:

Use the largest digits: $9, 8, 7$ and for the units place, the next largest available digit after $5$ is $6$.

Greatest number $= 98,756$

Smallest 5-digit number:

Use the smallest digits: $1, 0, 2$ and for the units place, the next smallest available digit is $3$.

Smallest number $= 10,253$

Solution:

Initial weight $= 2 \text{ kg } 300 \text{ g} = 2300 \text{ g}$

Final weight $= 5 \text{ kg } 68 \text{ g} = 5068 \text{ g}$

Weight to be added:

$\begin{array}{cccc} & 5 & 0 & 6 & 8 \\ - & 2 & 3 & 0 & 0 \\ \hline & 2 & 7 & 6 & 8 \\ \hline \end{array}$

The amount to be added is 2,768 grams.

Solution:

Weight of one box $= 50 \times 120 \text{ g} = 6,000 \text{ g}$

Weight of one box in kg $= \frac{6000}{1000} = 6 \text{ kg}$

Maximum capacity $= 900 \text{ kg}$

Number of boxes $= 900 \div 6$

$\text{Number of boxes} = 150$

Total 150 boxes can be loaded.

Solution:

$1 \text{ billion} = 10,000 \text{ lakhs}$

$5 \text{ billions} = 5 \times 10,000 = 50,000 \text{ lakhs}$

There are 50,000 lakhs in five billions.

Solution:

$1 \text{ crore} = 10 \text{ millions}$

$3 \text{ crores} = 3 \times 10 = 30 \text{ millions}$

There are 30 millions in 3 crores.

Solution:

(a) $874 + 478 \approx 900 + 500 = \mathbf{1400}$

(b) $793 + 397 \approx 800 + 400 = \mathbf{1200}$

(c) $11244 + 3507 \approx 11200 + 3500 = \mathbf{14700}$

(d) $17677 + 13589 \approx 17700 + 13600 = \mathbf{31300}$

Solution:

(a) $11963 - 9369 \approx 11960 - 9370 = \mathbf{2590}$

(b) $76877 - 7783 \approx 76880 - 7780 = \mathbf{69100}$

(c) $10732 - 4354 \approx 10730 - 4350 = \mathbf{6380}$

(d) $78203 - 16407 \approx 78200 - 16410 = \mathbf{61790}$

Solution:

(a) $87 \times 32 \approx 90 \times 30 = \mathbf{2700}$

(b) $311 \times 113 \approx 310 \times 110 = \mathbf{34100}$

(c) $3239 \times 28 \approx 3240 \times 30 = \mathbf{97200}$

(d) $1385 \times 789 \approx 1390 \times 790 = \mathbf{1098100}$

Solution:

Rounding to hundreds:

$78,787 \approx 78,800$

$95,833 \approx 95,800$

Estimated increase $= 95,800 - 78,800 = 17,000$

Estimated increase is 17,000.

Solution:

General rule: Round to the highest place value.

$758 \approx 800$ (hundreds place)

$6784 \approx 7000$ (thousands place)

Product $= 800 \times 7000 = \mathbf{56,00,000}$

Given:

Number of shirts produced = $2,16,315$

Number of trousers produced = $1,82,736$

Number of jackets produced = $58,704$

To Find:

The total production of all the three items in the year.

Solution:

To calculate the total production, we need to find the sum of the quantities of shirts, trousers, and jackets produced by the factory.

$\text{Total Production} = \text{Number of shirts} + \text{Number of trousers} $$ \ + \text{Number of jackets}$

Let's perform the addition using the column method:

$\begin{array}{ccccccc} & 2 & 1 & 6 & 3 & 1 & 5 \\ & 1 & 8 & 2 & 7 & 3 & 6 \\ + & & 5 & 8 & 7 & 0 & 4 \\ \hline & 4 & 5 & 7 & 7 & 5 & 5 \\ \hline \end{array}$

The sum of the three quantities is $4,57,755$.

Final Answer: The total production of all the three items in that year was 4,57,755.

Solution:

We use the division method:

$\begin{array}{c|cc} 2 & 160, & 170, & 90 \\ \hline 5 & 80, & 85, & 45 \\ \hline 2 & 16, & 17, & 9 \\ \hline 2 & 8, & 17, & 9 \\ \hline 2 & 4, & 17, & 9 \\ \hline 2 & 2, & 17, & 9 \\ \hline 3 & 1, & 17, & 9 \\ \hline 3 & 1, & 17, & 3 \\ \hline 17 & 1, & 17, & 1 \\ \hline & 1, & 1, & 1 \end{array}$

$\text{LCM} = 2 \times 5 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 17$

$\text{LCM} = 2^5 \times 3^2 \times 5 \times 17 = 24,480$

The LCM is 24,480.

Solution:

Total volume $= 13 \text{ L } 200 \text{ ml} = 13,200 \text{ ml}$

Glass capacity $= 60 \text{ ml}$

Number of glasses $= 13200 \div 60$

$\text{Number of glasses} = 220$

It can be filled in 220 glasses.

Solution:

• Successor of $32 = 33$
• Predecessor of $49 = 48$
• Predecessor of predecessor of $56 = 56 - 1 - 1 = 54$
• Successor of successor of $67 = 67 + 1 + 1 = 69$

Sum $= 33 + 48 + 54 + 69 = \mathbf{204}$

Solution:

Total boxes $= 482 + 518 = 1000$

Weight per box $= 15 \text{ kg}$

Total weight $= 1000 \times 15 = \mathbf{15,000 \text{ kg}}$

Given:

The individual amounts spent by Leela are as follows:

• Amount spent on food and decoration = $\textsf{₹} \ 2,16,766$
• Amount spent on jewellery = $\textsf{₹} \ 1,22,322$
• Amount spent on furniture = $\textsf{₹} \ 88,234$
• Amount spent on kitchen items = $\textsf{₹} \ 26,780$

To Find:

The total amount spent by Leela on all the above items.

Solution:

To find the total expenditure, we need to calculate the sum of the amounts spent on food and decoration, jewellery, furniture, and kitchen items.

$\text{Total Amount Spent} = \text{Amount (Food)} + \text{Amount (Jewellery)} $$ \ + \text{Amount (Furniture)} + \text{Amount (Kitchen)}$

Let's arrange the numbers in columns according to their place values and perform addition:

$\begin{array}{ccccccc} & 2 & 1 & 6 & 7 & 6 & 6 \\ & 1 & 2 & 2 & 3 & 2 & 2 \\ & & 8 & 8 & 2 & 3 & 4 \\ + & & 2 & 6 & 7 & 8 & 0 \\ \hline & 4 & 5 & 4 & 1 & 0 & 2 \\ \hline \end{array}$

The sum of the digits from the units place to the lakhs place gives a total of $4,54,102$.

Comparing the result, we find that the total amount spent by Leela is $\textsf{₹} \ 4,54,102$.

Final Answer: The total amount spent by Leela on her daughter's marriage items is ₹ 4,54,102.

Solution:

Capsules in one box $= 5 \times 12 = 60$

Medicine in one box $= 60 \times 500 \text{ mg} = 30,000 \text{ mg}$

Medicine in 32 boxes $= 32 \times 30,000 = 9,60,000 \text{ mg}$

Convert to grams ($1000 \text{ mg} = 1 \text{ g}$):

Weight $= 960,000 \div 1000 = 960 \text{ g}$

Total weight is 960 grams.

Solution:

First, find the LCM of $3, 4,$ and $5$:

$\text{LCM}(3, 4, 5) = 3 \times 4 \times 5 = 60$

Required number $= \text{LCM} + \text{remainder}$

Number $= 60 + 2 = 62$

The least number is 62.

Solution:

To find the greatest capacity, we find the HCF of 120, 180, and 240.

Prime factorizations:

• $120 = 2^3 \times 3 \times 5$
• $180 = 2^2 \times 3^2 \times 5$
• $240 = 2^4 \times 3 \times 5$

$\text{HCF} = 2^2 \times 3 \times 5 = 60$

The greatest capacity is 60 litres.

To find the required 4-digit odd number using digits $1, 2, 4,$ and $5$, we follow these simple steps:

1. Condition for Odd Number: The number must end with an odd digit. So, the last digit must be either $1$ or $5$. Let's choose 1 as the last digit.

2. Condition for Divisibility by 4: When the first and last digits are interchanged, the new number's last two digits must be divisible by $4$.

Let the original number be $ABCD$. If we fix $D = 1$, the interchanged number is $1BCA$. Here, the two-digit number $CA$ must be divisible by $4$.

The remaining digits for $A, B,$ and $C$ are $\{2, 4, 5\}$. From these, the possible two-digit numbers ending in $A$ that are divisible by $4$ are:

• $52$ (where $C = 5, A = 2$)
• $24$ (where $C = 2, A = 4$)

Taking $A = 2$ and $C = 5$, the remaining digit for $B$ is $4$.

Original Number: 2451

Verification:

Original number $2451$ is odd. After interchanging the first ($2$) and last ($1$) digits, we get $1452$. Since $52$ is divisible by $4$ ($13 \times 4 = 52$), the condition is satisfied.

Final Answer: The required 4-digit number is 2451.

Solution:

Divisibility by 4 means the last two digits must be a multiple of 4.

Multiples from $\{1, 2, 3, 4\}$ are: $12, 24, 32$.

To make the number smallest, we want the thousands digit to be 1.

• If last two digits are $24$, remaining digits are $1, 3$. Smallest: $1324$.
• If last two digits are $32$, remaining digits are $1, 4$. Smallest: $1432$.

Comparing $1324$ and $1432$, the smallest is 1324.

Solution:

To find the greatest denomination, we calculate the HCF of 20, 28, and 36.

$20 = 2^2 \times 5$

$28 = 2^2 \times 7$

$36 = 2^2 \times 3^2$

$\text{HCF} = 2^2 = 4$.

The greatest denomination is ₹ 4.

Solution:

Find the LCM of 12, 15, and 21:

$\text{LCM}(12, 15, 21) = 420$.

Number of packets for Brand A $= 420 \div 12 = \mathbf{35}$

Number of packets for Brand B $= 420 \div 15 = \mathbf{28}$

Number of packets for Brand C $= 420 \div 21 = \mathbf{20}$

Solution:

Dimensions in cm: $896 \text{ cm}$ and $672 \text{ cm}$.

Tile size $= \text{HCF}(896, 672) = 224 \text{ cm}$.

Number of tiles $=$ Area of floor $\div$ Area of tile

$\text{Number of tiles} = \frac{896 \times 672}{224 \times 224}$

$\text{Number of tiles} = 4 \times 3 = 12$

The minimum number of tiles is 12.

Solution:

To have the same number of books of each subject per shelf, the number of shelves must divide both 780 and 364.

Max shelves $= \text{HCF}(780, 364) = 52$.

English per shelf $= 780 \div 52 = 15$

Science per shelf $= 364 \div 52 = 7$

Total books per shelf $= 15 + 7 = 22$

The minimum number of books per shelf is 22.

Solution:

Common stops are multiples of $\text{LCM}(6, 10)$.

$\text{LCM}(6, 10) = 30$.

Multiples of 30 within 1-100 are: 30, 60, and 90.

Both will stop at blocks 30, 60, and 90.

Solution:

(a) 5335: $(5+3) - (3+5) = 8 - 8 = 0$. (Divisible)

(b) 9020814: $(4+8+2+9) - (1+0+0) = 23 - 1 = 22$. ($22 \div 11 = 2$). (Divisible)

Solution:

(a) 4096: Last two digits 96. $96 \div 4 = 24$. (Yes)

(b) 21084: Last two digits 84. $84 \div 4 = 21$. (Yes)

(c) 31795012: Last two digits 12. $12 \div 4 = 3$. (Yes)

Solution:

(a) 672: Sum $= 6+7+2 = 15$. (Not divisible by 9)

(b) 5652: Sum $= 5+6+5+2 = 18$. (Divisible by 9)