Chapter 2 Fractions & Decimals (Class 7 - Maths NCERT Exemplar Solutions)

Given:

Total weight of sweets = $1\frac{3}{4}$ kg

Number of friends = $7$

To Find:

Weight of sweets received by each friend.

Solution:

First, we convert the mixed fraction into an improper fraction:

To find the share of each friend, we divide the total weight by the number of friends:

Therefore, each friend receives $\frac{1}{4}$ kg of sweets.

Correct Option: (b)

To Find:

The unknown number.

Solution:

Let the required number be $x$.

According to the question:

The number is 16.

Correct Option: (b)

Solution:

To find the product of two fractions, we use the following rule:

Applying this rule to $\frac{2}{7}$ and $\frac{5}{9}$:

The product is $\frac{10}{63}$, which is represented by option (d).

Correct Option: (d)

Solution:

A fraction is smallest when its numerator is the smallest possible and its denominator is the largest possible.

Given $0 < p < q < r < s$, we observe the following:

1. The smallest possible sum of two distinct integers from the set $\{p, q, r, s\}$ is $(p + q)$.

2. The largest possible sum of two distinct integers from the set is $(r + s)$.

Comparing the options:

(a) $\frac{p + q}{r + s}$ has the smallest numerator and the largest denominator among all combinations of these four integers.

(d) $\frac{r + s}{p + q}$ would be the largest as it has the largest numerator and smallest denominator.

Thus, $\frac{p + q}{r + s}$ is the smallest value.

Correct Option: (a)

Solution:

Let us observe the relation between the consecutive terms of the pattern:

The rule of the pattern is that each subsequent term is half of the previous term.

Therefore, the next term will be:

Correct Option: (d)

Given:

Amount = $8$ rupees and $8$ paise.

Solution:

We know that in the Indian currency system:

Now, converting $8$ paise into rupees:

Total amount in rupees:

Correct Option: (b)

Given:

Length of each side of the regular hexagon $= 3.5\text{ cm}$.

To Find:

The perimeter of the regular hexagon.

Solution:

A regular hexagon is a polygon with six equal sides.

Therefore, the perimeter of the regular hexagon is $21\text{ cm}$.

Correct Option: (b)

Solution:

To divide a decimal number by $1000$, we shift the decimal point to the left by three places (since there are three zeros in $1000$).

Given expression: $2.5 \div 1000$

Starting with $2.5$:

1. Shift one place left: $0.25$

2. Shift two places left: $0.025$

3. Shift three places left: $0.0025$

Thus, $2.5 \div 1000 = 0.0025$.

Correct Option: (b)

Solution:

Let us evaluate the value of each option to compare them:

(a) $0.0002$

(b) $\frac{2}{1000} = 0.002$

(c) $\frac{(0.2)^2}{2} = \frac{0.04}{2} = 0.02$

(d) $\frac{2}{100} \div 0.01 = 0.02 \div 0.01 = \frac{0.02}{0.01} = 2$

Comparing the values: $0.0002, 0.002, 0.02,$ and $2$.

We can see that $0.0002$ is the smallest value.

Correct Option: (a)

Solution:

Let us evaluate each expression:

(a) $\frac{32}{0.05} = \frac{32 \times 100}{5} = \frac{3200}{5} = 640$

(b) $\frac{0.320}{50} = \frac{0.32}{50} = 0.0064$

(c) $\frac{3.2}{0.05} = \frac{3.2 \times 100}{5} = \frac{320}{5} = 64$

(d) $\frac{3.2}{50} = 0.064$

Comparing the results: $640, 0.0064, 64,$ and $0.064$.

The largest value is $640$, which corresponds to option (a).

Correct Option: (a)

Solution:

Evaluating the numerical values of each option:

(a) $0.0001$

(b) $\frac{1}{1000} = 0.001$

(c) $(0.100)^2 = 0.1 \times 0.1 = 0.01$

(d) $\frac{1}{10} \div 0.1 = 0.1 \div 0.1 = 1$

Comparing the values: $0.0001, 0.001, 0.01,$ and $1$.

The largest value is $1$.

Correct Option: (d)

Solution:

In mathematics, especially when dealing with word problems involving portions, a fraction often acts as an operator represented by the word 'of'.

For example: $\frac{1}{2}$ 'of' $10$ means $\frac{1}{2} \times 10$.

Answer: of

Solution:

The reciprocal of a fraction $\frac{a}{b}$ is obtained by interchanging the numerator and the denominator, i.e., $\frac{b}{a}$.

Given fraction is $\frac{2}{3}$.

Its reciprocal is $\frac{3}{2}$.

Answer: $\frac{3}{2}$

Solution:

Let us consider a proper fraction (numerator < denominator) and an improper fraction (numerator > denominator).

Example: Let proper fraction $= \frac{1}{2}$ and improper fraction $= \frac{3}{2}$.

Product $= \frac{1}{2} \times \frac{3}{2} = \frac{3}{4}$

Comparing the product ($\frac{3}{4} = 0.75$) with the improper fraction ($\frac{3}{2} = 1.5$):

$0.75 < 1.5$

Since the product of any number and a fraction less than $1$ (proper fraction) is always smaller than the original number, the product is less than the improper fraction.

Answer: less than

Solution:

By definition, if the product of two fractions is $1$, then each fraction is the multiplicative inverse or reciprocal of the other.

Example: $\frac{4}{5} \times \frac{5}{4} = 1$

Answer: reciprocals (or multiplicative inverse)

Given:

Amount = $5$ rupees and $5$ paise.

Solution:

We know that $100$ paise $= \textsf{₹} 1$.

Therefore, $1$ paisa $= \textsf{₹} \frac{1}{100}$.

Converting $5$ paise to rupees:

$5 \text{ paise} = \textsf{₹} \frac{5}{100} = \textsf{₹} 0.05$

Total amount in rupees:

Total $= \textsf{₹} 5 + \textsf{₹} 0.05 = \textsf{₹} 5.05$

Answer: $5.05$

Solution:

We use the standard metric conversion factors:

$10 \text{ mm} = 1 \text{ cm}$

$100 \text{ cm} = 1 \text{ m}$

Therefore, $1000 \text{ mm} = 1 \text{ m}$.

To convert mm to m, we divide the value by $1000$:

$45 \text{ mm} = \frac{45}{1000} \text{ m}$

$45 \text{ mm} = 0.045 \text{ m}$

Answer: $0.045$

Solution:

When we multiply a decimal number by $1000$, we shift the decimal point to the right by three places.

Starting with $2.4$:

1st shift: $24$

2nd shift: $240$

3rd shift: $2400$

Answer: $2400$

Solution:

The rule for dividing a decimal number by powers of 10 ($10, 100, 1000$, etc.) states that we must move the decimal point to the left.

Since $100$ has two zeros, the decimal point is shifted by two places to the left.

Answer: left, two

Solution:

An improper fraction is a fraction where the numerator is greater than or equal to the denominator (e.g., $\frac{3}{2}$).

The reciprocal is obtained by interchanging the numerator and the denominator.

Example: Reciprocal of $\frac{3}{2}$ is $\frac{2}{3}$.

Since $\frac{2}{3}$ is a proper fraction (numerator < denominator), the statement is false.

Answer: False

Solution:

Let us evaluate the LHS (Left Hand Side) by converting mixed fractions into improper fractions:

Now, perform the division:

Since $\frac{12}{11} \neq 2$, the statement is false.

Answer: False

Solution:

Let us evaluate the division:

To simplify, multiply both numerator and denominator by $100$:

Since the result matches the RHS, the statement is true.

Answer: True

Solution:

To multiply decimals, we first multiply them as whole numbers and then place the decimal point.

Product of whole numbers: $2 \times 3 = 6$.

Total decimal places in the factors: $1$ (in $0.2$) $+ 1$ (in $0.3$) $= 2$ decimal places.

Therefore, the product must have two decimal places:

$0.2 \times 0.3 = 0.06$

Since $0.06 \neq 0.6$, the statement is false.

Answer: False

To Find: The value of $\frac{2}{3}$ of $6$.

Solution:

$\frac{2}{3}$ of $6$ means we divide $6$ items into $3$ equal groups and take $2$ of those groups.

Visual Representation:

Looking at the provided figure, we have $6$ circles. Each circle is divided into $3$ equal parts. If we shade $2$ parts out of $3$ in each of the $6$ circles, we get:

$6 \times \frac{2}{3} = \frac{12}{3} = 4$ whole circles.

Therefore, 4 circles will be shaded in total if we combine the parts.

To Find: The sum of the given complex fractions.

Solution:

Step 1: Convert the mixed fractions in the denominators into improper fractions.

Step 2: Express the reciprocals.

Step 3: Find the sum of the fractions.

The expression becomes: $\frac{7}{30} + \frac{13}{50} + \frac{9}{5}$

To add these, we find the LCM of denominators $30, 50, \text{ and } 5$.

$\begin{array}{c|cc} 2 & 30 \;, & 50 \;, & 5 \\ \hline 3 & 15 \; , & 25 \; , & 5 \\ \hline 5 & 5 \; , & 25 \; , & 5 \\ \hline 5 & 1 \; , & 5 \; , & 1 \\ \hline & 1 \; , & 1 \; , & 1 \end{array}$

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

Step 4: Equivalent fractions with denominator 150.

$\frac{7 \times 5}{30 \times 5} + \frac{13 \times 3}{50 \times 3} + \frac{9 \times 30}{5 \times 30}$

$= \frac{35}{150} + \frac{39}{150} + \frac{270}{150}$

$= \frac{35 + 39 + 270}{150} = \frac{344}{150}$

Step 5: Simplify the fraction.

$\frac{\cancel{344}^{172}}{\cancel{150}_{75}} = \frac{172}{75} = 2\frac{22}{75}$

The final value is $2\frac{22}{75}$.

Given:

Total number of small $1 \times 1 \times 1$ cubes in a $3 \times 3 \times 3$ cube $= 3 \times 3 \times 3 = 27$.

The cube is "tunneled" by removing a column of cubes from the center of each face.

To Find: Fractions as specified in the question.

Solution:

Step 1: Calculate cubes removed

1. Center of front-to-back: $3$ cubes removed.

2. Center of left-to-right: $3$ cubes, but the center-most cube is already removed in step 1. So, $2$ new cubes removed.

3. Center of top-to-bottom: $3$ cubes, but the center-most cube is already removed in step 1. So, $2$ new cubes removed.

Total small cubes removed $= 3 + 2 + 2 = 7$.

Total small cubes left $= 27 - 7 = 20$.

(i) Fraction of cubes removed to cubes left:

$\text{Fraction} = \frac{\text{Cubes removed}}{\text{Cubes left}} = \frac{7}{20}$

(ii) Fraction of cubes removed to total cubes:

$\text{Fraction} = \frac{\text{Cubes removed}}{\text{Total cubes}} = \frac{7}{27}$

(iii) What part is (ii) of (i)?

Let part be $x$.

$(ii) = x \times (i)$

$\frac{7}{27} = x \times \frac{7}{20}$

$x = \frac{7}{27} \div \frac{7}{20}$

$x = \frac{7}{27} \times \frac{20}{7}$

$x = \frac{20}{27}$

The required part is $\frac{20}{27}$.

Given:

Work finished in $1$ hour $= \frac{1}{3}$ part.

Total time $= 2\frac{1}{5}$ hours.

Solution:

To find the total work finished, we multiply the rate of work by the total time.

Total time in improper fraction $= 2\frac{1}{5} = \frac{11}{5}$ hours.

$\text{Work finished} = \text{Work in 1 hour} \times \text{Total hours}$

$\text{Work finished} = \frac{1}{3} \times \frac{11}{5}$

$\text{Work finished} = \frac{1 \times 11}{3 \times 5}$

$\text{Work finished} = \frac{11}{15}$

Ramu will finish $\frac{11}{15}$ part of the work.

Given:

Total weight of the cake $= 4$ kg.

Weight of each piece $= \frac{2}{3}$ kg.

Solution:

To find the number of pieces, we divide the total weight by the weight of one piece.

$\text{Number of pieces} = 4 \div \frac{2}{3}$

$\text{Number of pieces} = 4 \times \frac{3}{2}$

$\text{Number of pieces} = \frac{\cancel{4}^2 \times 3}{\cancel{2}_1}$

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

Therefore, 6 pieces can be cut from the cake.

Given:

Weight of potatoes $= 3.5$ kg.

Cost per kg $= \textsf{₹} 13.75$.

Solution:

Total money to be paid $= \text{Weight} \times \text{Rate per kg}$

$\text{Total cost} = 3.5 \times 13.75$

Performing multiplication:

$\begin{array}{cc}& & 1 & 3 & 7 & 5 \\ \times & & & & 3 & 5 \\ \hline && 6 & 8 & 7 & 5 \\ & 4 & 1 & 2 & 5 & \times \\ \hline & 4 & 8 & 1 & 2 & 5 \\ \hline \end{array}$

Since there are $2 + 1 = 3$ decimal places, the product is $48.125$.

$\text{Total cost} = \textsf{₹} 48.125$

Rounding to nearest Rupee:

Since the decimal part $.125$ is less than $.50$, we round down.

$\textsf{₹} 48.125 \approx \textsf{₹} 48$.

Harmeet should pay $\textsf{₹} 48$.

Given:

Total length of the rope = $9.5\text{ m}$

Length of each small piece = $1.9\text{ m}$

To Find:

Number of small pieces Kavita will get.

Solution:

The number of pieces is obtained by dividing the total length by the length of one piece.

$\text{Number of pieces} = 9.5 \div 1.9$

$\text{Number of pieces} = \frac{9.5}{1.9}$

Multiplying the numerator and denominator by 10 to remove the decimals:

$\text{Number of pieces} = \frac{95}{19}$

$\text{Number of pieces} = \frac{\cancel{95}^5}{\cancel{19}_1}$

$\text{Number of pieces} = 5$

Therefore, Kavita will get 5 pieces of the rope.

Given:

Total amount earned by three boys = $\textsf{₹} 235.50$

Number of boys = $3$

To Find:

Average amount earned per boy.

Solution:

The average amount is calculated by dividing the total earnings by the number of boys.

$\text{Average amount} = \frac{\text{Total amount}}{\text{Number of boys}}$

$\text{Average amount} = \frac{\textsf{₹} 235.50}{3}$

$\begin{array}{r} 78.50 \phantom{3)} \\ 3{\overline{\smash{\big)}\,235.50 \phantom{)}}} \\ \underline{-~ 21 \phantom{(00)}} \\ 25 \phantom{(0)} \\ \underline{-~ 24 \phantom{(0)}} \\ 1.5 \phantom{)} \\ \underline{-~ 1.5 \phantom{)}} \\ 0.0 \phantom{)} \end{array}$

$\text{Average amount} = \textsf{₹} 78.50$

Therefore, the average amount earned per boy was $\textsf{₹} 78.50$.

Solution (i):

$\frac{1}{2} \times \frac{5}{8} = \frac{1 \times 5}{2 \times 8} = \frac{5}{16}$

Solution (ii):

$\frac{1}{3} \times \frac{7}{5} = \frac{1 \times 7}{3 \times 5} = \frac{7}{15}$

Solution (iii):

$\frac{4}{3} \times \frac{5}{2} = \frac{\cancel{4}^2 \times 5}{3 \times \cancel{2}_1} = \frac{2 \times 5}{3 \times 1} = \frac{10}{3} = 3\frac{1}{3}$

Solution:

(i) Effect of interchanging fractions:

No, interchanging the fractions does not affect the answer. This is because multiplication of fractions follows the Commutative Property.

$\frac{1}{12} \times \frac{5}{8} = \frac{5}{96}$

$\frac{5}{8} \times \frac{1}{12} = \frac{5}{96}$

(ii) Comparing product value:

Case (i) and (ii) of Example 32 involve proper fractions. When we multiply two proper fractions, the product is less than each of the fractions.

In Case (iii), we multiply an improper fraction ($\frac{4}{3}$) and an improper fraction ($\frac{5}{2}$). In this case, the product is greater than each of the fractions.

In general, if we multiply a proper fraction by another fraction, the value of the product is less than the other fraction.

Given:

Cloth required for one shirt = $\frac{3}{4}\text{ m}$

Total length of cloth available = $2\frac{1}{4}\text{ m}$

To Find:

Number of shirts Reshma can make.

Solution:

First, convert the mixed fraction into an improper fraction:

Total cloth = $2\frac{1}{4} = \frac{(4 \times 2) + 1}{4} = \frac{9}{4}\text{ m}$

Number of shirts = $\text{Total cloth} \div \text{Cloth per shirt}$

Number of shirts = $\frac{9}{4} \div \frac{3}{4}$

Number of shirts = $\frac{9}{4} \times \frac{4}{3}$

Number of shirts = $\frac{\cancel{9}^3}{\cancel{4}_1} \times \frac{\cancel{4}^1}{\cancel{3}_1} = 3$

Therefore, Reshma can make 3 shirts.

Given:

Frequency of note D = $297$

Frequency of note B = $495$

To Find:

The fraction of the frequency of note D to note B and check if they are harmonious.

Solution:

The fraction of frequencies of D and B is given by:

$\text{Fraction} = \frac{\text{Frequency of D}}{\text{Frequency of B}}$

$\text{Fraction} = \frac{297}{495}$

Now, we simplify the fraction by finding common factors:

Dividing both numerator and denominator by 9:

$\frac{297 \div 9}{495 \div 9} = \frac{33}{55}$

Dividing by 11:

$\frac{33 \div 11}{55 \div 11} = \frac{3}{5}$

Since there was a common factor (9 and 11), the notes are harmonious. The resulting fraction is $\frac{3}{5}$.

Given:

Distance covered = $120\text{ km}$

Fraction of the way covered = $\frac{2}{3}$

To Find:

The remaining distance to the camp site.

Solution:

Let the total distance to the camp site be $x\text{ km}$.

According to the problem:

$\frac{2}{3}$ of $x = 120$

$\frac{2}{3} \times x = 120$

$x = \frac{120 \times 3}{2}$

$x = \frac{\cancel{120}^{60} \times 3}{\cancel{2}_1}$

$x = 180\text{ km}$

Now, the distance remaining is:

$\text{Remaining distance} = \text{Total distance} - \text{Distance covered}$

$\text{Remaining distance} = 180 - 120 = 60\text{ km}$

Alternate Solution:

If $\frac{2}{3}$ of the way is covered, the remaining part is $1 - \frac{2}{3} = \frac{1}{3}$.

If 2 parts (out of 3) = $120\text{ km}$

Then 1 part (out of 3) = $\frac{120}{2} = 60\text{ km}$

Therefore, they have to go $60\text{ km}$ farther.

Given:

The expression is $\frac{2}{5} \times 5\frac{1}{5}$.

Solution:

First, convert the mixed fraction into an improper fraction:

Now, perform the multiplication:

Comparing the result with the options, it matches (b).

Correct Option: (b)

Solution:

Convert the mixed fraction into an improper fraction:

Now, perform the division by multiplying with the reciprocal of the divisor:

Correct Option: (c)

Given:

Total length of ribbon = $5\frac{1}{4}$ m

Length of each piece = $\frac{3}{4}$ m

To Find:

The total number of pieces.

Solution:

Total length in improper fraction:

Number of pieces = Total length $\div$ length of one piece

Correct Option: (c)

Solution:

To arrange the fractions in ascending order, we must first make their denominators equal by finding the LCM of 3, 7, and 21.

Now, convert each fraction to an equivalent fraction with denominator 21:

1. $\frac{2}{3} = \frac{2 \times 7}{3 \times 7} = \frac{14}{21}$

2. $\frac{6}{7} = \frac{6 \times 3}{7 \times 3} = \frac{18}{21}$

3. $\frac{13}{21}$ is already $\frac{13}{21}$

Comparing the numerators: $13 < 14 < 18$.

Therefore, $\frac{13}{21} < \frac{14}{21} < \frac{18}{21}$.

The original fractions in ascending order are: $\frac{13}{21}, \frac{2}{3}, \frac{6}{7}$.

Correct Option: (b)

Solution:

The reciprocal of a fraction is obtained by interchanging the numerator and the denominator.

Therefore, the reciprocal of $\frac{2}{3}$ is $\frac{3}{2}$.

Correct Option: (d)

Solution:

To find the product, multiply the numerator by the integer:

Now, convert the improper fraction into a mixed fraction:

$\begin{array}{r} 3 \phantom{13)} \\ 13{\overline{\smash{\big)}\,44 \phantom{)}}} \\ \underline{-~ 39 \phantom{x}} \\ 5 \phantom{)} \end{array}$

Thus, $\frac{44}{13} = 3\frac{5}{13}$.

Correct Option: (a)

Solution:

First, convert the mixed fraction into an improper fraction:

Now, multiply by 3:

Now, convert the result back into a mixed fraction:

$\begin{array}{r} 13 \phantom{5)} \\ 5{\overline{\smash{\big)}\,66 \phantom{)}}} \\ \underline{-~ 5 \phantom{xx}} \\ 16 \phantom{(} \\ \underline{-~ 15 \phantom{x}} \\ 1 \phantom{)} \end{array}$

Thus, $\frac{66}{5} = 13\frac{1}{5}$.

Correct Option: (c)

Solution:

The expression $3 \times \frac{2}{3}$ represents three groups of $\frac{2}{3}$.

1. The fraction $\frac{2}{3}$ means that out of 3 equal parts of a whole, 2 parts are shaded.

2. Therefore, $3 \times \frac{2}{3}$ will show three circles, where each circle has 2 out of 3 parts shaded.

Looking at the given options:

In option (b), there are three circles and each has exactly 2 parts shaded out of 3 total parts.

Correct Option: (b)

Solution:

To divide one fraction by another, we multiply the first fraction by the reciprocal of the second fraction.

Now, we cancel the common factor $5$ from the numerator and denominator:

Correct Option: (d)

Solution:

To multiply decimals, we first multiply the numbers as whole numbers and then place the decimal point based on the total number of decimal places in the factors.

1. Multiply whole numbers: $3 \times 9 = 27$

2. Count decimal places:

In $0.03$, there are 2 decimal places.

In $0.9$, there is 1 decimal place.

Total decimal places $= 2 + 1 = 3$.

3. Place the decimal point in $27$ such that there are 3 decimal places:

Correct Option: (c)

Solution:

The division of a fraction by a whole number is done by multiplying the fraction by the reciprocal of the whole number.

Given expression: $\frac{5}{7} \div 6$

The reciprocal of $6$ is $\frac{1}{6}$.

Correct Option: (b)

Solution:

First, convert the mixed fraction into an improper fraction.

Now, perform the division:

Cancel the common factor $2$ from the numerator and denominator:

Correct Option: (d)

Solution:

In the context of fractions, the word "of" acts as an operator for multiplication.

Therefore, $\frac{1}{3}$ of $\frac{1}{6}$ is mathematically represented as:

Correct Option: (c)

Solution:

Using the rule that "of" represents multiplication:

Multiply the numerators and denominators:

Correct Option: (d)

Given:

Flour required for 1 packet = $2\frac{1}{2}$ cups

Sugar required for 1 packet = $1\frac{2}{3}$ cups

Number of packets = $10$

To Find:

The estimated total quantity of both ingredients for 10 packets.

Solution:

First, we find the total quantity of ingredients for one packet:

Quantity for 1 packet = $2\frac{1}{2} + 1\frac{2}{3}$

$= \frac{5}{2} + \frac{5}{3}$

Taking LCM of 2 and 3, which is 6:

$= \frac{5 \times 3 + 5 \times 2}{6} = \frac{15 + 10}{6} = \frac{25}{6}$ cups

Now, calculate the quantity for 10 packets:

Total quantity $= 10 \times \frac{25}{6}$

$= \frac{\cancel{10}^5 \times 25}{\cancel{6}_3}$

$= \frac{125}{3}$

Converting to a mixed fraction:

$= 41\frac{2}{3}$ cups

The value $41\frac{2}{3}$ lies between 40 cups and 50 cups.

Correct Option: (c)

Solution:

Convert the mixed fraction into an improper fraction:

$6\frac{3}{4} = \frac{(4 \times 6) + 3}{4} = \frac{24 + 3}{4} = \frac{27}{4}$

Now, find the product with 7:

Product $= 7 \times \frac{27}{4}$

$= \frac{7 \times 27}{4} = \frac{189}{4}$

Convert the improper fraction back into a mixed fraction:

$\begin{array}{r} 47 \phantom{4)} \\ 4{\overline{\smash{\big)}\,189 \phantom{)}}} \\ \underline{-~ 16 \phantom{xx}} \\ 29 \phantom{(} \\ \underline{-~ 28 \phantom{x}} \\ 1 \phantom{)} \end{array}$

So, the result is $47\frac{1}{4}$.

Correct Option: (b)

Solution:

To divide a whole number by a fraction, we multiply the whole number by the reciprocal of the fraction.

The reciprocal of $\frac{2}{5}$ is $\frac{5}{2}$.

Correct Option: (d)

Solution:

Convert the mixed fraction into an improper fraction:

$2\frac{2}{3} = \frac{(3 \times 2) + 2}{3} = \frac{8}{3}$

Now, divide by 5 by multiplying with its reciprocal $\frac{1}{5}$:

Correct Option: (a)

Given:

Total weight of apples $= 5$ kg

Apples used on Monday $= \frac{4}{5}$ of total

Apples used on Tuesday $= \frac{1}{3}$ of the remainder

To Find:

Weight of apples left at the end.

Solution:

1. Apples used on Monday:

Usage $= \frac{4}{5} \times 5 = 4$ kg

Remaining apples $= 5 - 4 = 1$ kg

2. Apples used on Tuesday:

Usage $= \frac{1}{3}$ of remaining $= \frac{1}{3} \times 1 = \frac{1}{3}$ kg

3. Apples left finally:

Left $= 1 - \frac{1}{3} = \frac{3 - 1}{3}$

Left $= \frac{2}{3}$ kg

Correct Option: (c)

Solution:

Looking at the pictorial representation:

1. There are three identical circles on the left side of the equal sign.

2. Each circle is divided into 4 equal parts, and one part is shaded. This represents the fraction $\frac{1}{4}$.

3. On the right side, there is one circle where three parts are shaded, representing the fraction $\frac{3}{4}$.

So, the interpretation is the addition of $\frac{1}{4}$ three times:

$\frac{1}{4} + \frac{1}{4} + \frac{1}{4} = 3 \times \frac{1}{4} = \frac{3}{4}$

Correct Option: (b)

Given:

Part of cake eaten by Rani = $\frac{2}{7}$

Part of cake eaten by Ravi = $\frac{4}{5}$ of the remaining cake.

Solution:

Let the total cake be $1$.

Remaining cake after Rani ate = $1 - \frac{2}{7}$

$= \frac{7 - 2}{7} = \frac{5}{7}$

Part of cake eaten by Ravi = $\frac{4}{5}$ of $\frac{5}{7}$

$= \frac{4}{\cancel{5}_1} \times \frac{\cancel{5}^1}{7} = \frac{4}{7}$

Total part of cake eaten = $\frac{2}{7} + \frac{4}{7} = \frac{6}{7}$

Part of the cake left = $1 - \frac{6}{7} = \frac{1}{7}$

Answer: $\frac{1}{7}$

Solution:

The reciprocal of a fraction is obtained by inverting it (swapping the numerator and denominator).

Reciprocal of $\frac{3}{7} = \frac{7}{3}$

Answer: $\frac{7}{3}$

Solution:

$\frac{2}{3}$ of $27 = \frac{2}{3} \times 27$

$= 2 \times \frac{\cancel{27}^9}{\cancel{3}_1}$

$= 2 \times 9 = 18$

Answer: $18$

Solution:

$\frac{4}{5}$ of $45 = \frac{4}{5} \times 45$

$= 4 \times \frac{\cancel{45}^9}{\cancel{5}_1}$

$= 4 \times 9 = 36$

Answer: $36$

Solution:

First, convert the mixed fraction into an improper fraction:

$6\frac{1}{3} = \frac{(6 \times 3) + 1}{3} = \frac{19}{3}$

Now, multiply by 4:

$4 \times \frac{19}{3} = \frac{76}{3}$

Converting to mixed fraction:

$= 25\frac{1}{3}$

Answer: $25\frac{1}{3}$ (or $\frac{76}{3}$)

Solution:

First, convert the mixed fraction into an improper fraction:

$4\frac{2}{7} = \frac{(4 \times 7) + 2}{7} = \frac{30}{7}$

Now, find $\frac{1}{2}$ of it:

$\frac{1}{2} \times \frac{30}{7} = \frac{1}{\cancel{2}_1} \times \frac{\cancel{30}^{15}}{7} = \frac{15}{7}$

Converting to mixed fraction:

$= 2\frac{1}{7}$

Answer: $2\frac{1}{7}$ (or $\frac{15}{7}$)

Solution:

$\frac{1}{9} \times \frac{6}{5} = \frac{1 \times \cancel{6}^2}{\cancel{9}_3 \times 5}$

$= \frac{2}{15}$

Answer: $\frac{2}{15}$

Solution:

First, convert the mixed fraction into an improper fraction:

$2\frac{3}{7} = \frac{(2 \times 7) + 3}{7} = \frac{17}{7}$

Now, multiply by $\frac{7}{9}$:

$\frac{17}{7} \times \frac{7}{9} = \frac{17}{\cancel{7}_1} \times \frac{\cancel{7}^1}{9} = \frac{17}{9}$

Converting to mixed fraction:

$= 1\frac{8}{9}$

Answer: $1\frac{8}{9}$ (or $\frac{17}{9}$)

Solution:

To divide a fraction by a whole number, we multiply the fraction by the reciprocal of that whole number.

$\frac{4}{5} \div 4 = \frac{4}{5} \times \frac{1}{4}$

Using the cancellation method:

$\frac{\cancel{4}^1}{5} \times \frac{1}{\cancel{4}_1} = \frac{1}{5}$

Answer: $\frac{1}{5}$

Solution:

The operator "of" represents multiplication.

$\frac{2}{5} \times 25 = 2 \times \frac{\cancel{25}^5}{\cancel{5}_1}$

$= 2 \times 5 = 10$

Answer: 10

Solution:

When we divide a fraction by another fraction, it is equivalent to multiplying the first fraction by the reciprocal of the second fraction.

Reciprocal of $\frac{5}{6}$ is $\frac{6}{5}$.

Thus, $\frac{1}{5} \div \frac{5}{6} = \frac{1}{5} \times \frac{6}{5}$

Answer: $\times$ (multiplication sign)

Solution:

When a decimal number is multiplied by 10, the decimal point shifts one place to the right.

$3.2 \times 10 = 32.0 = 32$

Answer: 32

Solution:

When a decimal number is multiplied by 1000, the decimal point shifts three places to the right.

$25.4 \times 1000 = 25400.0 = 25400$

Answer: 25400

Solution:

When a decimal number is multiplied by 100, the decimal point shifts two places to the right.

$93.5 \times 100 = 9350.0 = 9350$

Answer: 9350

Solution:

When a decimal number is divided by 10, the decimal point shifts one place to the left.

$4.7 \div 10 = 0.47$

Answer: 0.47

Solution:

When a decimal number is divided by 100, the decimal point shifts two places to the left.

$4.7 \div 100 = 0.047$

Answer: 0.047

Solution:

When dividing a decimal number by $1000$, the decimal point shifts three places to the left.

$4.7 \div 1000 = 0.0047$

Answer: $0.0047$

Solution:

A proper fraction has a value less than $1$. When we multiply two numbers that are both less than $1$, the resulting product is always smaller than both original numbers.

Example: $\frac{1}{2} \times \frac{1}{3} = \frac{1}{6}$. Here, $\frac{1}{6} < \frac{1}{2}$ and $\frac{1}{6} < \frac{1}{3}$.

Answer: less (or smaller)

Solution:

According to the rule of division for fractions, we keep the first fraction as it is and change the division sign to multiplication, then take the reciprocal of the second fraction.

Answer: multiply, reciprocal

To Find: The missing value.

Solution:

Let the missing value be $x$.

Multiplying numerator and denominator by $10$:

$x = \frac{84}{21} = 4$

Answer: $4$

Solution:

Observe the shift in the decimal point. The decimal moved from after the digit $2$ to before the digit $5$. This is a shift of two places to the left.

A shift of two places to the left occurs when we divide by $100$.

Answer: $100$

Solution:

If we multiply $5$ and $7$, we get $35$. Since there are two decimal places in total ($0.5$ and $0.7$), the product is $0.35$.

Answer: $\times$ (multiplication sign)

Solution:

To get $\frac{10}{3}$, we multiply the whole number $2$ by the numerator $5$.

$2 \times 5 = 10$.

Answer: $\times$ (multiplication sign) or 'of'

Solution:

To solve this, we can remove the decimals by multiplying both numbers by $1000$ (as both have three decimal places).

Now, perform the division:

$\frac{\cancel{2001}^{667}}{\cancel{3}_{1}} = 667$

Answer: $667$

Solution:

A proper fraction is a fraction where the numerator is less than the denominator ($a < b$). Its reciprocal is $\frac{b}{a}$, where the numerator is now greater than the denominator.

For example, the reciprocal of $\frac{2}{3}$ is $\frac{3}{2}$. Since $\frac{3}{2}$ is an improper fraction, the statement is false.

Answer: False

Solution:

An improper fraction is a fraction where the numerator is greater than the denominator ($a > b$). Its reciprocal is $\frac{b}{a}$, where the numerator is now less than the denominator.

For example, the reciprocal of $\frac{5}{4}$ is $\frac{4}{5}$. Since $\frac{4}{5}$ is a proper fraction, the statement is false.

Answer: False

Solution:

According to the rules of fraction multiplication:

Since the given formula is inverted, the statement is false.

Answer: False

Solution:

Improper fractions are always greater than or equal to 1. When two numbers greater than 1 are multiplied, the product is always greater than both original numbers.

Example: $\frac{3}{2} \times \frac{4}{3} = \frac{12}{6} = 2$. Here, $2$ is greater than both $\frac{3}{2}$ ($1.5$) and $\frac{4}{3}$ ($1.33$).

Answer: False

Solution:

To find the reciprocal of a fraction $\frac{a}{b}$, we interchange the numerator and denominator to get $\frac{b}{a}$. This is commonly described as inverting the fraction.

Answer: True

Solution:

When multiplying by a power of 10, the decimal point shifts to the right. Since $1000$ has three zeros, the decimal point moves exactly three places to the right.

Example: $1.2345 \times 1000 = 1234.5$.

Answer: True

Solution:

When a decimal number is divided by a power of 10, the decimal point shifts to the left. The number of places it shifts is equal to the number of zeros in the divisor.

Since 100 has two zeros, the decimal point moves two places to the left.

Answer: True

Solution:

The reciprocal of a number $x$ is $\frac{1}{x}$. A number is its own reciprocal if $x = \frac{1}{x}$, which implies $x^2 = 1$.

The solutions to $x^2 = 1$ are $1$ and $-1$.

1. Reciprocal of $1 = \frac{1}{1} = 1$

2. Reciprocal of $-1 = \frac{1}{-1} = -1$

Since both $1$ and $-1$ are their own reciprocals, the statement that $1$ is the only such number is false.

Answer: False

Solution:

Let us evaluate both expressions:

LHS: $\frac{2}{3}$ of $8$

RHS: $\frac{2}{3} \div 8$

Since $\frac{16}{3} \neq \frac{1}{12}$, the statement is false.

Answer: False

Solution:

The reciprocal of a fraction $\frac{a}{b}$ is obtained by interchanging the numerator and the denominator, resulting in $\frac{b}{a}$.

Reciprocal of $\frac{4}{7} = \frac{7}{4}$

Since $\frac{7}{4} \neq \frac{4}{7}$, the statement is false.

Answer: False

Given:

Original fraction $= \frac{5}{9}$

Value added to numerator and denominator $= 5$

Solution:

New numerator $= 5 + 5 = 10$

New denominator $= 9 + 5 = 14$

To compare $\frac{5}{9}$ and $\frac{5}{7}$, we can find their decimal values or use cross-multiplication.

Original: $\frac{5}{9} \approx 0.555$

New: $\frac{5}{7} \approx 0.714$

Since $0.714 > 0.555$, the value changes and it increases.

Solution:

A fraction is represented as $\frac{\text{Numerator}}{\text{Denominator}}$. The denominator indicates the number of parts a whole is divided into.

When the denominator decreases, the size of each part becomes larger while the number of parts taken (numerator) remains the same.

Therefore, the value of the fraction increases.

Solution:

The letters from A to J are: A, B, C, D, E, F, G, H, I, J.

The total number of letters in this sequence is $10$.

To find $\frac{2}{5}$ of the way, we calculate:

The 4th letter in the sequence is D.

To Find: The value of $1.75$ times the number.

Solution:

Let the number be $x$.

Now, we need to find $1.75$ times of $15$:

The required value is $26.25$.

Given:

Fraction of students liking rice only $= \frac{1}{5}$

Fraction of students liking chapati only $= \frac{2}{5}$

Solution:

Let the total fraction of students be $1$.

Fraction of students liking only rice or only chapati:

Fraction of students liking both:

The fraction of students who like to eat both is $\frac{2}{5}$.

Given:

Part of homework completed in $2$ hours $= \frac{2}{3}$

Time given $= 1\frac{1}{4}$ hours $= \frac{5}{4}$ hours

Solution:

First, find the part of homework completed in 1 hour:

Now, calculate the part completed in $\frac{5}{4}$ hours:

Renu completed $\frac{5}{12}$ part of her homework.

Given:

Initial part of the book read = $\frac{1}{5}$

Additional pages read = $40$

Final part of the book read = $\frac{7}{10}$

To Find:

Number of pages left to be read.

Solution:

Let the total number of pages in the book be $x$.

According to the question:

Total pages in the book = $80$.

Pages already read = $\frac{7}{10} \times 80 = 56$ pages.

Pages left to be read = $80 - 56 = 24$ pages.

Answer: 24 pages

Solution:

Let the missing fraction be $\frac{a}{b}$.

Using cancellation:

Answer: $\frac{5}{14}$

Solution:

First, convert the mixed fractions into improper fractions:

$7\frac{1}{6} = \frac{43}{6}$ and $3\frac{2}{3} = \frac{11}{3}$

Now, perform the division:

Since $1.95 > 1.5$, the quotient is greater than 1.5.

Solution:

Method 1: Converting fraction to decimal

We divide $13$ by $17$ to get a decimal value.

$13 \div 17 \approx 0.764$

Comparing $0.764$ and $0.82$, we see that $0.764 < 0.82$.

Method 2: Converting decimal to fraction

$0.82 = \frac{82}{100} = \frac{41}{50}$

Now compare $\frac{13}{17}$ and $\frac{41}{50}$ by cross-multiplication:

$13 \times 50 = 650$

$17 \times 41 = 697$

Since $650 < 697$, $\frac{13}{17} < 0.82$.

Conclusion: Method 1 is generally easier because long division quickly gives a comparative decimal value without needing to multiply large numbers.

(a) Grams received by a 72 kg adult:

A 72 kg adult falls in the "60 kg and over" category, so they take 4 tablets.

(b) Recommended dose for a 46 kg adult:

A 46 kg adult falls in the "40 to 50 kg" category, so they take $2\frac{1}{2} = \frac{5}{2}$ tablets.

Given:

1. An adult dog with weight = $18$ kg.

2. A cat with weight = $6$ kg.

3. A pregnant dog with weight = $18$ kg.

To Find:

The required vitamin dosage for each animal based on the provided guidelines.

Solution:

(a) For an $18$ kg adult dog:

The guideline states $\frac{1}{2}$ tsp is required for every $9$ kg of body weight.

Number of $9$ kg units in $18$ kg is calculated as:

$\text{Units} = \frac{18}{9} = 2$

Therefore, the total dosage is:

$\text{Dosage} = 2 \times \frac{1}{2}$ tsp

$\text{Dosage} = 1$ tsp

(b) For a $6$ kg cat:

The guideline states $\frac{1}{4}$ tsp is required for every $1$ kg of body weight.

Number of $1$ kg units in $6$ kg is $6$.

Therefore, the total dosage is:

$\text{Dosage} = 6 \times \frac{1}{4}$ tsp

$\text{Dosage} = \frac{\cancel{6}^3}{\cancel{4}_2}$ tsp

$\text{Dosage} = \frac{3}{2}$ tsp = $1.5$ tsp

(c) For an $18$ kg pregnant dog:

The guideline states $\frac{1}{2}$ tsp is required for every $4.5$ kg of body weight for pregnant dogs.

Number of $4.5$ kg units in $18$ kg is calculated as:

$\text{Units} = \frac{18}{4.5} = \frac{180}{45} = 4$

Therefore, the total dosage is:

$\text{Dosage} = 4 \times \frac{1}{2}$ tsp

$\text{Dosage} = 2$ tsp

Summary of Dosages:

(a) Adult dog ($18$ kg): $1$ tsp

(b) Cat ($6$ kg): $1.5$ tsp (or $1 \frac{1}{2}$ tsp)

(c) Pregnant dog ($18$ kg): $2$ tsp

Given:

Total chocolates = $1\frac{1}{2} = \frac{3}{2}$ kg

Box capacity = $\frac{1}{16}$ kg

Solution:

Answer: 24 boxes

Given:

Length of one bookmark (from Fig. 2.6) = $10 \frac{1}{2}$ cm

Total length of the ribbon = $15$ m

To Find:

Number of bookmarks Anvi can make.

Solution:

First, we need to ensure both lengths are in the same unit. We know that $1$ m = $100$ cm.

Now, let's convert the mixed fraction of the bookmark length into an improper fraction:

To find the number of bookmarks, we divide the total length of the ribbon by the length of one bookmark:

$\text{Number of bookmarks} = \text{Total length} \div \text{Length of one bookmark}$

$\text{Number of bookmarks} = 1500 \div \frac{21}{2}$

$\text{Number of bookmarks} = 1500 \times \frac{2}{21}$

$\text{Number of bookmarks} = \frac{3000}{21}$

On dividing $3000$ by $21$:

$\text{Number of bookmarks} = 142.85...$

Since the number of bookmarks must be a whole number, Anvi can make $142$ bookmarks.

Given Rule:

Solution (a):

Length of a side = $8.3$ cm

$\text{Length of diagonal} = 8.3 \times 1.414$

$\text{Length of diagonal} = 11.7362$ cm

Solution (b):

Length of a side = $7.875$ cm

$\text{Length of diagonal} = 7.875 \times 1.414$

$\text{Length of diagonal} = 11.13525$ cm

Given Rule:

Solution (a):

Diameter = $14.35$ cm

$\text{Length of side} = 0.707 \times 14.35$

$\text{Length of side} = 10.14545$ cm

Solution (b):

Diameter = $8.63$ cm

$\text{Length of side} = 0.707 \times 8.63$

$\text{Length of side} = 6.10141$ cm

Given Rule:

Solution (a):

Diameter = $18.7$ cm

$\text{Distance around disc} = 18.7 \times 3.14$

$\text{Distance} = 58.718$ cm

Solution (b):

Radius = $6.45$ cm

We know that Diameter = $2 \times$ Radius

$\text{Diameter} = 2 \times 6.45 = 12.9$ cm

$\text{Distance around disc} = 12.9 \times 3.14$

$\text{Distance} = 40.506$ cm

Given:

Total length of cloth = $27.5$ m

Cost of cloth per metre = $\textsf{₹} 53.50$

To Find:

Total cost of the cloth.

Solution:

$\text{Total Cost} = \text{Length of cloth} \times \text{Cost per metre}$

$\text{Total Cost} = 27.5 \times 53.50$

To calculate the product:

$\begin{array}{cc} & & & 5 & 3 & 5 & 0 \\ \times & & & & 2 & 7 & 5 \\ \hline && 2 & 6 & 7 & 5 & 0 \\ & 3 & 7 & 4 & 5 & 0 & \times \\ 1 & 0 & 7 & 0 & 0 & \times & \times \\ \hline 1 & 4 & 7 & 1 & 2 & 5 & 0 \\ \hline \end{array}$

After adjusting the decimal places (one place in 27.5 and two places in 53.50, total three places):

$\text{Total Cost} = \textsf{₹} 1471.25$

The total cost of $27.5$ m of cloth is $\textsf{₹} 1471.25$.

Given:

As per the given figure and information, the race is divided into equal segments by hurdles A, B, C, D, and E.

Since Hurdle B represents the second segment out of six, we can conclude that each hurdle represents a successive sixth of the total race distance:

Hurdle A = $\frac{1}{6}$, Hurdle B = $\frac{2}{6}$, Hurdle C = $\frac{3}{6}$, Hurdle D = $\frac{4}{6}$, Hurdle E = $\frac{5}{6}$.

Solution:

(a) Where will Nidhi be at $\frac{4}{6}$ of the way?

Based on the pattern established above, the position $\frac{4}{6}$ corresponds to Hurdle D.

(b) Where will Nidhi be at $\frac{5}{6}$ of the way?

Based on the pattern established above, the position $\frac{5}{6}$ corresponds to Hurdle E.

(c) Two fractions for the part of the race finished at Hurdle C:

At Hurdle C, Nidhi is at the third marker out of six segments.

First fraction: $\frac{3}{6}$

To find the second fraction, we simplify the first one:

$\frac{\cancel{3}^1}{\cancel{6}_2} = \frac{1}{2}$

So, the two fractions are $\frac{3}{6}$ and $\frac{1}{2}$.

Given:

Diameter of Earth = $12756000$ m

Ratio of the new planet's diameter to Earth's diameter = $\frac{5}{86}$

To Find:

Diameter of the new planet in km.

Solution:

First, let's convert the diameter of Earth from metres to kilometres.

$\text{Diameter of Earth} = 12756$ km

Now, calculate the diameter of the new planet:

$\text{Diameter of new planet} = \frac{5}{86} \times 12756$

$\text{Diameter of new planet} = 5 \times \frac{12756}{86}$

$\text{Diameter of new planet} = 5 \times 148.325...$

$\text{Diameter of new planet} \approx 741.63$ km

The diameter of the new planet is approximately $741.63$ km.

Given:

The fraction is $\frac{5}{129}$.

Solution:

The reciprocal of a fraction $\frac{a}{b}$ is $\frac{b}{a}$.

$\text{Reciprocal of } \frac{5}{129} = \frac{129}{5}$

Now, find the product:

$\text{Product} = \frac{5}{129} \times \frac{129}{5}$

$\text{Product} = \frac{\cancel{5}^1}{\cancel{129}_1} \times \frac{\cancel{129}^1}{\cancel{5}_1}$

$\text{Product} = 1$

The product of any non-zero fraction and its reciprocal is always $1$.

Given Expression:

$\frac{2\frac{1}{2}+\frac{1}{5}}{2\frac{1}{2}\div\frac{1}{5}}$

Solution:

First, convert the mixed fraction $2\frac{1}{2}$ to an improper fraction:

$2\frac{1}{2} = \frac{(2 \times 2) + 1}{2} = \frac{5}{2}$

Now, simplify the Numerator:

$\text{Numerator} = \frac{5}{2} + \frac{1}{5}$

Taking LCM of $2$ and $5$, which is $10$:

$\text{Numerator} = \frac{5 \times 5}{10} + \frac{1 \times 2}{10} = \frac{25 + 2}{10} = \frac{27}{10}$

Now, simplify the Denominator:

$\text{Denominator} = \frac{5}{2} \div \frac{1}{5}$

$\text{Denominator} = \frac{5}{2} \times \frac{5}{1} = \frac{25}{2}$

Finally, divide the simplified numerator by the simplified denominator:

$\text{Result} = \frac{27}{10} \div \frac{25}{2}$

$\text{Result} = \frac{27}{10} \times \frac{2}{25}$

$\text{Result} = \frac{27 \times \cancel{2}^1}{\cancel{10}_5 \times 25}$

$\text{Result} = \frac{27}{125}$

The simplified value is $\frac{27}{125}$.

Given Expression:

$\frac{\frac{1}{4}+\frac{1}{5}}{1-\frac{3}{8}\times\frac{1}{5}}$

Solution:

Simplify the Numerator:

$\text{Numerator} = \frac{1}{4} + \frac{1}{5}$

Taking LCM of $4$ and $5$, which is $20$:

$\text{Numerator} = \frac{5 + 4}{20} = \frac{9}{20}$

Simplify the Denominator:

$\text{Denominator} = 1 - \left( \frac{3}{8} \times \frac{1}{5} \right)$

$\text{Denominator} = 1 - \frac{3}{40}$

$\text{Denominator} = \frac{40 - 3}{40} = \frac{37}{40}$

Now, divide the numerator by the denominator:

$\text{Result} = \frac{9}{20} \div \frac{37}{40}$

$\text{Result} = \frac{9}{20} \times \frac{40}{37}$

$\text{Result} = \frac{9 \times \cancel{40}^2}{\cancel{20}_1 \times 37}$

$\text{Result} = \frac{18}{37}$

The simplified value is $\frac{18}{37}$.

To Find:

$\frac{3}{10} \div \left( \frac{1}{4} \text{ of } \frac{3}{5} \right)$

Solution:

First, solve the part within the brackets using the 'of' operator (multiplication):

$\left( \frac{1}{4} \text{ of } \frac{3}{5} \right) = \frac{1}{4} \times \frac{3}{5} = \frac{3}{20}$

Now, perform the division:

$\text{Result} = \frac{3}{10} \div \frac{3}{20}$

To divide by a fraction, we multiply by its reciprocal:

$\text{Result} = \frac{3}{10} \times \frac{20}{3}$

$\text{Result} = \frac{\cancel{3}^1}{\cancel{10}_1} \times \frac{\cancel{20}^2}{\cancel{3}_1}$

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

The final answer is $2$.

Given:

One-eighth of a number is equal to the quotient of $\frac{2}{5}$ divided by $\frac{1}{20}$.

To Find:

The unknown number.

Solution:

Let the required number be $x$.

According to the question, the mathematical expression is:

$\frac{1}{8} \text{ of } x = \frac{2}{5} \div \frac{1}{20}$

First, let's solve the Right Hand Side (RHS):

$\frac{2}{5} \div \frac{1}{20} = \frac{2}{5} \times \frac{20}{1}$

$\text{RHS} = \frac{2 \times \cancel{20}^4}{\cancel{5}_1}$

$\text{RHS} = 2 \times 4 = 8$

Now, substitute this value back into the equation:

$\frac{1}{8} \times x = 8$

Multiply both sides by $8$ to find $x$:

$x = 8 \times 8$

$x = 64$

The required number is $64$.

Given:

Amount paid by Heena's father = $\textsf{₹} 500.00$

Amount of the electric bill = $\textsf{₹} 385.70$

To Find:

The change received by him.

Solution:

To find the change, we subtract the bill amount from the total amount paid:

$\text{Change} = 500.00 - 385.70$

$\begin{array}{cc} & 5 & 0 & 0 & . & 0 & 0 \\ - & 3 & 8 & 5 & . & 7 & 0 \\ \hline & 1 & 1 & 4 & . & 3 & 0 \\ \hline \end{array}$

Heena's father should have received $\textsf{₹} 114.30$ as change.

Given:

Normal body temperature = $98.6^\circ F$

Temperature during illness = $103.1^\circ F$

To Find:

The temperature difference (rise in temperature).

Solution:

To find how many degrees above normal the temperature was, we subtract the normal temperature from the measured temperature:

$\text{Temperature Rise} = 103.1^\circ F - 98.6^\circ F$

$\begin{array}{cc} & 1 & 0 & 3 & . & 1 \\ - & & 9 & 8 & . & 6 \\ \hline & & & 4 & . & 5 \\ \hline \end{array}$

The temperature was $4.5^\circ F$ above normal.

Given:

Variation from base temperature for different years. Let's convert all variations to decimal form for easy comparison:

Solution (a):

To order the years from coldest to warmest, we arrange the temperature differences in ascending order (lowest value to highest value):

Comparing the values: $-0.17 < -0.10 < 0.02 < 0.10 < 0.54$

The order of years from coldest to warmest is: 1964, 1965, 1978, 1958, 2002.

Solution (b):

For the year 1946, the variation is $-0.03^\circ C$.

We need to place $-0.03$ in our sorted list of values:

$-0.10 < -0.03 < 0.02$

Since $-0.10$ corresponds to 1965 and $0.02$ corresponds to 1978, the year 1946 should fall between 1965 and 1978.

Given:

Atomic weight of Helium (He) = $4.0030$

Atomic weight of Hydrogen (H) = $1.0080$

Atomic weight of Oxygen (O) = $16.0000$

Solution:

(a) Difference between Oxygen and Hydrogen:

$\begin{array}{cc} & 1 & 6 & . & 0 & 0 & 0 & 0 \\ - & & 1 & . & 0 & 0 & 8 & 0 \\ \hline & 1 & 4 & . & 9 & 9 & 2 & 0 \\ \hline \end{array}$

The difference is $14.9920$.

(b) Difference between Oxygen and Helium:

$\begin{array}{cc} & 1 & 6 & . & 0 & 0 & 0 & 0 \\ - & & 4 & . & 0 & 0 & 3 & 0 \\ \hline & 1 & 1 & . & 9 & 9 & 7 & 0 \\ \hline \end{array}$

The difference is $11.9970$.

(c) Difference between Helium and Hydrogen:

$\begin{array}{cc} & 4 & . & 0 & 0 & 3 & 0 \\ - & 1 & . & 0 & 0 & 8 & 0 \\ \hline & 2 & . & 9 & 9 & 5 & 0 \\ \hline \end{array}$

The difference is $2.9950$.

Given:

Correct length of the iron rod = $19.33$ cm

Ravi's measurement = $19.34$ cm

Kamal's measurement = $19.25$ cm

Tabish's measurement = $19.27$ cm

Solution:

The error is the difference between the measured length and the correct length.

Error by Ravi:

$|19.34 - 19.33| = 0.01$ cm

Error by Kamal:

$|19.33 - 19.25|$

$\begin{array}{cc} & 1 & 9 & . & 3 & 3 \\ - & 1 & 9 & . & 2 & 5 \\ \hline & & 0 & . & 0 & 8 \\ \hline \end{array}$

The error is $0.08$ cm.

Error by Tabish:

$|19.33 - 19.27|$

$\begin{array}{cc} & 1 & 9 & . & 3 & 3 \\ - & 1 & 9 & . & 2 & 7 \\ \hline & & 0 & . & 0 & 6 \\ \hline \end{array}$

The error is $0.06$ cm.

Summary:

The errors made are: Ravi ($0.01$ cm), Kamal ($0.08$ cm), and Tabish ($0.06$ cm).

To Find:

The quotient of $0.02964 \div 0.004$.

Solution:

First, we can express the division as a fraction:

$\frac{0.02964}{0.004}$

To eliminate the decimals in the denominator, we multiply both numerator and denominator by $1000$ (since there are three decimal places in the denominator):

$\frac{0.02964 \times 1000}{0.004 \times 1000} = \frac{29.64}{4}$

Now, perform the division:

$\begin{array}{r} 7.41\phantom{)} \\ 4{\overline{\smash{\big)}\,29.64\phantom{)}}} \\ \underline{-~28\phantom{.00}} \\ 1.6\phantom{4)} \\ \underline{-~1.6\phantom{4)}} \\ 0.04\phantom{)} \\ \underline{-~0.04\phantom{)}} \\ 0\phantom{)} \end{array}$

The quotient is $7.41$.

Given:

We need to find a number $x$ such that:

Solution:

First, let's calculate the quotient of $85 \div 0.625$:

$\frac{85}{0.625} = \frac{85 \times 1000}{625}$

$\frac{85000}{625} = 136$

Now, substitute this value back into equation (i):

$\frac{x}{520} = 136$

$x = 136 \times 520$

To find the product of $136$ and $520$:

$\begin{array}{cc}& & & 5 & 2 & 0 \\ \times & & & 1 & 3 & 6 \\ \hline && 3 & 1 & 2 & 0 \\ & 1 & 5 & 6 & 0 & \times \\ & 5 & 2 & 0 & \times & \times \\ \hline & 7 & 0 & 7 & 2 & 0 \\ \hline \end{array}$

The required number is $70720$.

Given:

Length of the floor = $4.5$ m = $450$ cm

Width of the floor = $3.6$ m = $360$ cm

Side of the square tile = $6$ cm

Cost of one tile = $\textsf{₹} 23.25$

To Find:

The total cost to cover the floor with tiles.

Solution:

First, we calculate the area of the floor:

$\text{Area of floor} = 450 \times 360 = 162000 \text{ cm}^2$

Next, we calculate the area of one square tile:

$\text{Area of one tile} = \text{side} \times \text{side} = 6 \times 6 = 36 \text{ cm}^2$

Now, we find the number of tiles required:

$\text{Number of tiles} = \frac{\text{Area of floor}}{\text{Area of one tile}}$

$\text{Number of tiles} = \frac{162000}{36} = 4500$

Finally, we calculate the total cost:

$\text{Total Cost} = \text{Number of tiles} \times \text{Cost per tile}$

$\text{Total Cost} = 4500 \times 23.25$

$\text{Total Cost} = 45 \times 2325$

$\begin{array}{cc}& & 2 & 3 & 2 & 5 \\ \times & & & & 4 & 5 \\ \hline & 1 & 1 & 6 & 2 & 5 \\ & 9 & 3 & 0 & 0 & \times \\ \hline 1 & 0 & 4 & 6 & 2 & 5 \\ \hline \end{array}$

The total cost to cover the floor is $\textsf{₹} 104625$.

Given:

Total cloth Sunita has = $\frac{3}{4}$ m

Fraction of cloth given to Rehana = $\frac{1}{3}$ of Sunita's cloth

Solution:

Amount of cloth Rehana has:

$\text{Cloth with Rehana} = \frac{1}{3} \text{ of } \frac{3}{4}$

$\text{Cloth with Rehana} = \frac{1}{\cancel{3}} \times \frac{\cancel{3}}{4}$

$\text{Cloth with Rehana} = \frac{1}{4}$ m

Rehana had $\frac{1}{4}$ m of cloth (or $0.25$ m).

Given:

Length of the garden side = $22.50$ m

Length of one brick = $0.25$ m

Solution:

To find the number of bricks needed, we divide the total length of the garden side by the length of one brick:

$\text{Number of bricks} = \frac{22.50}{0.25}$

To simplify, multiply both numerator and denominator by $100$:

$\text{Number of bricks} = \frac{2250}{25}$

$\text{Number of bricks} = \frac{\cancel{2250}^{90}}{\cancel{25}_{1}}$

Sheela will need $90$ bricks.

Given:

Cloth required for one shirt = $2 \frac{1}{4}$ m

Waste per shirt = $\frac{1}{8}$ m

Number of shirts = $6$

Solution:

First, calculate the total cloth needed for one shirt including waste:

$\text{Total per shirt} = 2 \frac{1}{4} + \frac{1}{8}$

$\text{Total per shirt} = \frac{9}{4} + \frac{1}{8}$

Taking LCM of $4$ and $8$, which is $8$:

$\text{Total per shirt} = \frac{18}{8} + \frac{1}{8} = \frac{19}{8}$ m

Now, calculate the cloth for $6$ shirts:

$\text{Total cloth} = 6 \times \frac{19}{8}$

$\text{Total cloth} = \frac{\cancel{6}^3 \times 19}{\cancel{8}_4}$

$\text{Total cloth} = \frac{57}{4}$ m

Converting to decimal or mixed fraction:

$\text{Total cloth} = 14 \frac{1}{4}$ m or $14.25$ m

The total cloth used will be $14.25$ m.

Given:

Total capacity of the hall = $820$ seats

Actual ticket sales = $648$

Usher 1 guess = $\frac{3}{4}$ full

Usher 2 guess = $\frac{2}{3}$ full

Solution:

First, calculate the actual number of people for each guess:

Usher 1's guess:

$\text{Guess 1} = \frac{3}{4} \times 820 = 3 \times 205 = 615$ persons

Usher 2's guess:

$\text{Guess 2} = \frac{2}{3} \times 820 = \frac{1640}{3} \approx 546.67$ persons

Now, find the error (difference from actual sales) for each:

$\text{Error 1} = |648 - 615| = 33$

$\text{Error 2} = |648 - 546.67| = 101.33$

Since the first usher's error ($33$) is smaller than the second usher's error ($101.33$), the first usher made the better guess.

Given:

Cost of sweets = $\textsf{₹} 740.25$

Cost of cold drinks = $\textsf{₹} 70.00$

Number of students = $35$

To Find:

Contribution per student.

Solution:

First, find the total expenditure:

$\text{Total Expense} = 740.25 + 70.00$

$\begin{array}{cc} & 7 & 4 & 0 & . & 2 & 5 \\ + & & 7 & 0 & . & 0 & 0 \\ \hline & 8 & 1 & 0 & . & 2 & 5 \\ \hline \end{array}$

Now, divide the total expense by the number of students:

$\text{Contribution per student} = \frac{810.25}{35}$

$\text{Contribution per student} = 23.15$

Each student contributed $\textsf{₹} 23.15$.

Given:

Times taken in 5 races (in minutes): $3.20,\ 3.37,\ 3.29,\ 3.17,\ 3.32$

To Find:

Average time taken by Rohan.

Solution:

Average is calculated by dividing the sum of all observations by the total number of observations.

First, we find the sum of the times:

$\begin{array}{cc} & 3 & . & 2 & 0 \\ & 3 & . & 3 & 7 \\ & 3 & . & 2 & 9 \\ & 3 & . & 1 & 7 \\ + & 3 & . & 3 & 2 \\ \hline 1 & 6 & . & 3 & 5 \\ \hline \end{array}$

Now, substituting the sum in equation (i):

$\text{Average Time} = \frac{16.35}{5}$

$\text{Average Time} = 3.27$ minutes

The average time taken by Rohan in the races is $3.27$ minutes.

Given:

Total length of the sewer line = $80\frac{1}{4}$ m

Work completed in one day = $7.5$ m

To Find:

Total number of days to complete the project.

Solution:

First, convert the mixed fraction and decimal into improper fractions:

To find the total time, we divide the total length by the daily work rate:

$\text{Number of days} = \text{Total Length} \div \text{Daily Work}$

$\text{Number of days} = \frac{321}{4} \div \frac{15}{2}$

$\text{Number of days} = \frac{321}{4} \times \frac{2}{15}$

$\text{Number of days} = \frac{321 \times \cancel{2}^1}{\cancel{4}_2 \times 15}$

$\text{Number of days} = \frac{\cancel{321}^{107}}{2 \times \cancel{15}_5}$

$\text{Number of days} = \frac{107}{10} = 10.7$ days

The project will take $10.7$ (round off 11) days to complete.

Given:

Weight of the object on Earth = $5\frac{3}{5}$ kg

Solution:

Convert the mixed fraction to an improper fraction:

$\text{Weight on Earth} = 5\frac{3}{5} = \frac{(5 \times 5) + 3}{5} = \frac{28}{5}$ kg

Now, calculate the weight on the moon:

$\text{Weight on Moon} = \frac{1}{6} \times \frac{28}{5}$

$\text{Weight on Moon} = \frac{1 \times \cancel{28}^{14}}{\cancel{6}_3 \times 5}$

$\text{Weight on Moon} = \frac{14}{15}$ kg

The object would weigh $\frac{14}{15}$ kg on the moon.

Given:

Total number of students surveyed = $200$

Fractions from the graph:

• Radio = $\frac{9}{20}$

• Music Video Channel = $\frac{2}{25}$

• Friend/Relative = $\frac{3}{20}$

• Heard/Saw in Shop = $\frac{1}{10}$

• Live Performance = $\frac{7}{100}$

• Heard/Saw in Shop = $\frac{3}{20}$

Solution (a):

Number of students influenced by Radio:

$\text{Students} = \frac{9}{20} \times 200$

$\text{Students} = 9 \times 10 = 90$

$90$ students said radio influenced them most.

Solution (b):

First, find the number of students influenced by Music Video Channel:

$\text{Students} = \frac{2}{25} \times 200$

$\text{Students} = 2 \times 8 = 16$

Now, find the difference between Radio and Music Video Channel:

$\text{Difference} = 90 - 16 = 74$

$74$ more students were influenced by radio than by a music video channel.

Solution (c):

Find the number of students for both categories:

1. Friend or Relative = $\frac{3}{20} \times 200 = 3 \times 10 = 30$ students

2. Heard/Saw in Shop = $\frac{1}{10} \times 200 = 20$ students

Total for both = $30 + 20 = 50$ students

$50$ students said a friend/relative influenced them or they heard it in a shop.

Given:

Total milk initially = $5\frac{1}{2}$ L = $\frac{11}{2}$ L

Milk sold to Renu, Kamla, and Renuka = $3 \times \frac{3}{4}$ L

Milk sold to Shadma = $\frac{7}{8}$ L

Milk sold to Jassi = $1\frac{1}{2}$ L = $\frac{3}{2}$ L

To Find:

The remaining quantity of milk in the can.

Solution:

First, calculate the Total Milk Sold:

$\text{Total Sold} = \left( 3 \times \frac{3}{4} \right) + \frac{7}{8} + \frac{3}{2}$

$\text{Total Sold} = \frac{9}{4} + \frac{7}{8} + \frac{3}{2}$

Taking the LCM of $4, 8,$ and $2$, which is $8$:

$\text{Total Sold} = \frac{9 \times 2}{8} + \frac{7}{8} + \frac{3 \times 4}{8}$

$\text{Total Sold} = \frac{18 + 7 + 12}{8} = \frac{37}{8}$ L

Now, find the Milk Left by subtracting the sold quantity from the initial quantity:

$\text{Milk Left} = \text{Total Milk} - \text{Total Sold}$

$\text{Milk Left} = \frac{11}{2} - \frac{37}{8}$

Taking LCM of $2$ and $8$, which is $8$:

$\text{Milk Left} = \frac{11 \times 4}{8} - \frac{37}{8}$

$\text{Milk Left} = \frac{44 - 37}{8}$

$\text{Milk Left} = \frac{7}{8}$ L

The quantity of milk left in the can is $\frac{7}{8}$ L.

Given:

Time taken by Anuradha to complete the whole work = $6$ hours.

To Find:

Part of the work done in $1$ hour, $5$ hours, and $6$ hours.

Solution:

The rate of work done per hour is the reciprocal of the total time taken to finish the work.

1. Part of work done in $1$ hour:

Since the whole work takes $6$ hours, in $1$ hour, she will complete $\frac{1}{6}$ of the work.

2. Part of work done in $5$ hours:

$\text{Work done in } 5 \text{ hours} = 5 \times (\text{Work done in } 1 \text{ hour})$

$\text{Work done} = 5 \times \frac{1}{6} = \frac{5}{6}$

3. Part of work done in $6$ hours:

$\text{Work done in } 6 \text{ hours} = 6 \times \frac{1}{6}$

$\text{Work done} = \frac{\cancel{6}^1}{\cancel{6}_1} = 1$ (which represents the whole work).

Final Answer:

Anuradha can do $\frac{1}{6}$ of the work in $1$ hour, $\frac{5}{6}$ in $5$ hours, and $1$ full work in $6$ hours.

Given:

Total time required to paint the whole saree = $5$ hours.

To Find:

Portion of saree painted in $1$ hour, $4\frac{3}{5}$ hours, and $3\frac{1}{2}$ hours.

Solution:

1. Portion painted in $1$ hour:

$\text{Portion in } 1 \text{ hour} = \frac{1}{\text{Total time}} = \frac{1}{5}$

2. Portion painted in $4\frac{3}{5}$ hours:

First, convert $4\frac{3}{5}$ to an improper fraction: $4\frac{3}{5} = \frac{(4 \times 5) + 3}{5} = \frac{23}{5}$ hours.

$\text{Portion painted} = \text{Time} \times \text{Work per hour}$

$\text{Portion painted} = \frac{23}{5} \times \frac{1}{5} = \frac{23}{25}$

3. Portion painted in $3\frac{1}{2}$ hours:

First, convert $3\frac{1}{2}$ to an improper fraction: $3\frac{1}{2} = \frac{(3 \times 2) + 1}{2} = \frac{7}{2}$ hours.

$\text{Portion painted} = \frac{7}{2} \times \frac{1}{5} = \frac{7}{10}$

Final Answer:

Portion painted in $1$ hour is $\frac{1}{5}$, in $4\frac{3}{5}$ hours is $\frac{23}{25}$, and in $3\frac{1}{2}$ hours is $\frac{7}{10}$.

Given:

Total weight of cotton wool = $6 \frac{1}{4}$ kg

Cotton wool required for one pillow = $1 \frac{1}{4}$ kg

To Find:

Number of pillows Rama can make.

Solution:

First, let us convert the mixed fractions into improper fractions:

$\text{Total cotton} = 6 \frac{1}{4} = \frac{(6 \times 4) + 1}{4} = \frac{25}{4}$ kg

$\text{Cotton per pillow} = 1 \frac{1}{4} = \frac{(1 \times 4) + 1}{4} = \frac{5}{4}$ kg

To find the number of pillows, divide the total cotton by cotton per pillow:

$\text{Number of pillows} = \frac{25}{4} \div \frac{5}{4}$

$\text{Number of pillows} = \frac{25}{4} \times \frac{4}{5}$

$\text{Number of pillows} = \frac{\cancel{25}^5}{\cancel{4}_1} \times \frac{\cancel{4}^1}{\cancel{5}_1} = 5$

Rama can make $5$ pillows.

Given:

Total length of cloth = $9 \frac{1}{3}$ m

Cloth required for one shirt = $2 \frac{1}{3}$ m

To Find:

Number of shirts Radhika can make.

Solution:

Converting mixed fractions into improper fractions:

$\text{Total length} = 9 \frac{1}{3} = \frac{28}{3}$ m

$\text{Cloth per shirt} = 2 \frac{1}{3} = \frac{7}{3}$ m

Number of shirts = $\text{Total length} \div \text{Cloth per shirt}$

$\text{Number of shirts} = \frac{28}{3} \div \frac{7}{3}$

$\text{Number of shirts} = \frac{28}{3} \times \frac{3}{7}$

$\text{Number of shirts} = \frac{\cancel{28}^4}{\cancel{3}_1} \times \frac{\cancel{3}^1}{\cancel{7}_1} = 4$

Radhika can make $4$ shirts.

Given:

Speed of Ravi = $3 \frac{1}{3}$ km/h

Distance to office = $10$ km

To Find:

Time taken to reach the office.

Solution:

Convert speed into an improper fraction:

$\text{Speed} = 3 \frac{1}{3} = \frac{10}{3}$ km/h

We know the formula for time:

$\text{Time} = 10 \div \frac{10}{3}$

$\text{Time} = 10 \times \frac{3}{10}$

$\text{Time} = \frac{\cancel{10}^1 \times 3}{\cancel{10}_1} = 3$ hours

It will take Ravi $3$ hours to reach his office.

Given:

Distance covered with $\frac{3}{5}$ tank of petrol = $360$ km.

To Find:

Distance covered with a full tank (i.e., $1$ full tank).

Solution:

Let the distance covered with a full tank be $D$.

According to the question:

$\frac{3}{5} \times D = 360$

To find $D$, we multiply $360$ by the reciprocal of $\frac{3}{5}$:

$D = 360 \div \frac{3}{5}$

$D = 360 \times \frac{5}{3}$

$D = \frac{\cancel{360}^{120} \times 5}{\cancel{3}_1}$

$D = 120 \times 5 = 600$ km

Raj would travel $600$ km with a full tank of petrol.

Given:

Amount Kajol has = $\textsf{₹} 75$

This amount is $\frac{3}{8}$ of her total earnings.

To Find:

The total amount she earned.

Solution:

Let the total amount earned by Kajol be $\textsf{₹} x$.

According to the question:

$\frac{3}{8} \times x = 75$

$x = \frac{75 \times 8}{3}$

$x = \frac{\cancel{75}^{25} \times 8}{\cancel{3}_{1}}$

$x = 25 \times 8$

$x = 200$

Kajol earned a total of $\textsf{₹} 200$.

Given:

Number of trees required for $1$ tonne of paper = $17$

Total number of trees in the forest = $221$

Solution (i):

First, find the number of trees required for the given tonnes of paper:

(a) For 5 tonnes of paper:

$\text{Trees needed} = 5 \times 17 = 85$

$\text{Fraction of forest used} = \frac{\text{Trees used}}{\text{Total trees}} = \frac{85}{221}$

Simplifying by $17$:

$\text{Fraction} = \frac{\cancel{85}^{5}}{\cancel{221}_{13}} = \frac{5}{13}$

(b) For 10 tonnes of paper:

$\text{Trees needed} = 10 \times 17 = 170$

$\text{Fraction of forest used} = \frac{170}{221}$

Simplifying by $17$:

$\text{Fraction} = \frac{\cancel{170}^{10}}{\cancel{221}_{13}} = \frac{10}{13}$

Solution (ii):

To find the amount of paper to save, we first find the total paper capacity of the forest:

$\text{Total paper potential} = \frac{\text{Total trees}}{\text{Trees per tonne}} = \frac{221}{17} = 13$ tonnes.

We need to save $\frac{7}{13}$ part of the forest. This corresponds to the same fraction of the total paper potential.

$\text{Paper to be saved} = \frac{7}{13} \times 13 \text{ tonnes}$

$\text{Paper to be saved} = 7$ tonnes.

We have to save $7$ tonnes of paper.

Given Expression:

$\left(1 \div \frac{2}{9} \right) + \left(1 \div 3\frac{1}{5} \right) + \left(1 \div 2\frac{2}{3} \right)$

Solution:

First, convert the mixed fractions into improper fractions:

$3\frac{1}{5} = \frac{16}{5}$

$2\frac{2}{3} = \frac{8}{3}$

Now, simplify each term:

1. $\left(1 \div \frac{2}{9}\right) = 1 \times \frac{9}{2} = \frac{9}{2} = 4.5$

2. $\left(1 \div \frac{16}{5}\right) = 1 \times \frac{5}{16} = \frac{5}{16} = 0.3125$

3. $\left(1 \div \frac{8}{3}\right) = 1 \times \frac{3}{8} = \frac{3}{8} = 0.375$

Finally, add the decimal results:

$\text{Total} = 4.5 + 0.3125 + 0.375$

$\begin{array}{cc} & 4 & . & 5 & 0 & 0 & 0 \\ & 0 & . & 3 & 1 & 2 & 5 \\ + & 0 & . & 3 & 7 & 5 & 0 \\ \hline & 5 & . & 1 & 8 & 7 & 5 \\ \hline \end{array}$

The result in decimal form is $5.1875$.

By observing the shaded parts in each figure:

(a) There are 3 squares, each divided into 9 small squares. In each, 2 parts are shaded. This represents $3 \times \frac{2}{9}$.

(b) There are 4 squares, each divided into 4 parts. In each, 1 part is shaded. This represents $\frac{1}{4} \times 4$.

(c) There are 2 circles, each divided into 3 parts. In each, 1 part is shaded. This represents $2 \times \frac{1}{3}$.

(d) There are 2 circles, each divided into 4 parts. In each, 1 part is shaded. This represents $2 \times \frac{1}{4}$.

(e) There are 3 triangles, each divided into 4 smaller triangles. In each, 1 part is shaded. This represents $\frac{1}{4} \times 3$.

(f) There are 2 rectangles, each divided into 7 columns. In each, 3 columns are shaded. This represents $2 \times \frac{3}{7}$.

To Evaluate:

$(0.3) \times (0.3) - (0.2) \times (0.2)$

Solution:

First, find the product of each term:

$(0.3) \times (0.3) = 0.09$

$(0.2) \times (0.2) = 0.04$

Now, perform the subtraction:

$\text{Result} = 0.09 - 0.04$

$\begin{array}{cc} & 0 & . & 0 & 9 \\ - & 0 & . & 0 & 4 \\ \hline & 0 & . & 0 & 5 \\ \hline \end{array}$

The final value is $0.05$.

To Evaluate:

$\frac{0.6}{0.3} + \frac{0.16}{0.4}$

Solution:

First, simplify the first term:

$\frac{0.6}{0.3} = \frac{6}{3} = 2$

Next, simplify the second term:

$\frac{0.16}{0.4} = \frac{1.6}{4} = 0.4$

Now, add the two results:

$\text{Result} = 2 + 0.4 = 2.4$

The final evaluated value is $2.4$.

Given Expression:

$\frac{(0.2 \times 0.14) + (0.5 \times 0.91)}{(0.1 \times 0.2)}$

Solution:

First, calculate the products in the numerator:

$0.2 \times 0.14 = 0.028$

$0.5 \times 0.91 = 0.455$

Add the results in the numerator:

$\text{Numerator Sum} = 0.028 + 0.455 = 0.483$

Next, calculate the product in the denominator:

$0.1 \times 0.2 = 0.02$

Finally, divide the numerator by the denominator:

$\text{Result} = \frac{0.483}{0.02}$

$\text{Result} = \frac{48.3}{2} = 24.15$

The final value is $24.15$.

Given:

Side of the equilateral triangle ($ABC$) = $\frac{4}{3}$ cm

The square ($BCED$) and the triangle share a common side $BC$.

To Find:

Perimeter of the entire figure $ABDEC$.

Solution:

The perimeter of the combined figure consists of the outer boundary only. The common side $BC$ is inside the figure and is not counted in the perimeter.

The outer sides are: $AB, AC, CE, ED,$ and $DB$.

Since the triangle is equilateral, all its sides are equal:

$AB = AC = BC = \frac{4}{3}$ cm

Since the square has all sides equal and shares $BC$ with the triangle:

$BC = CE = ED = DB = \frac{4}{3}$ cm

Total number of outer sides = $2$ (from triangle) + $3$ (from square) = $5$ equal sides.

$\text{Perimeter} = 5 \times \text{side}$

$\text{Perimeter} = 5 \times \frac{4}{3} = \frac{20}{3}$ cm

Converting to a mixed fraction:

$\text{Perimeter} = 6 \frac{2}{3}$ cm

The perimeter of the figure is $6 \frac{2}{3}$ cm.

Given:

Dimensions of carpet = $4$ m $\times$ $6 \frac{2}{3}$ m

Dimensions of room = $3 \frac{1}{3}$ m $\times$ $5 \frac{1}{3}$ m

Solution:

First, calculate the Area of the carpet:

$\text{Area of carpet} = 4 \times \frac{20}{3} = \frac{80}{3} \text{ m}^2$

Next, calculate the Area of the room:

$\text{Area of room} = \frac{10}{3} \times \frac{16}{3} = \frac{160}{9} \text{ m}^2$

The area to be cut off is the difference between the carpet area and the room area:

$\text{Area to be cut} = \frac{80}{3} - \frac{160}{9}$

Taking LCM as $9$:

$\text{Area to be cut} = \frac{240 - 160}{9} = \frac{80}{9} \text{ m}^2$

Now, we find the fraction of the carpet area that needs to be cut off:

$\text{Fraction} = \frac{\text{Area to be cut}}{\text{Total carpet area}}$

$\text{Fraction} = \frac{80}{9} \div \frac{80}{3} = \frac{80}{9} \times \frac{3}{80}$

$\text{Fraction} = \frac{\cancel{80}^1}{\cancel{9}_3} \times \frac{\cancel{3}^1}{\cancel{80}_1} = \frac{1}{3}$

Therefore, $\frac{1}{3}$ of the area should be cut off.

Given:

Length of photograph ($l$) = $14 \frac{2}{5} = \frac{72}{5}$ cm

Breadth of photograph ($b$) = $10 \frac{2}{5} = \frac{52}{5}$ cm

Width of border ($w$) = $2 \frac{3}{5} = \frac{13}{5}$ cm

Solution:

The border is uniform on all sides. So, the total length and breadth of the framed photograph will increase by twice the width of the border.

$\text{New Length} (L) = \frac{72}{5} + 2 \times \left( \frac{13}{5} \right) = \frac{72}{5} + \frac{26}{5} = \frac{98}{5}$ cm

$\text{New Breadth} (B) = \frac{52}{5} + 2 \times \left( \frac{13}{5} \right) = \frac{52}{5} + \frac{26}{5} = \frac{78}{5}$ cm

Now, calculate the area of the framed photograph:

$\text{Area} = L \times B$

$\text{Area} = \frac{98}{5} \times \frac{78}{5} = \frac{7644}{25}$ cm$^2$

Converting to decimal form:

$\text{Area} = 305.76$ cm$^2$

The area of the framed photograph is $305.76$ cm$^2$.

Given:

Cost of one burger = $\textsf{₹} 20 \frac{3}{4} = \textsf{₹} \frac{83}{4}$

Cost of one Macpuff = $\textsf{₹} 15 \frac{1}{2} = \textsf{₹} \frac{31}{2}$

Solution:

1. Cost of 4 burgers:

$\text{Cost} = 4 \times \textsf{₹} \frac{83}{4}$

$\text{Cost} = \textsf{₹} \cancel{4} \times \frac{83}{\cancel{4}} = \textsf{₹} 83$

2. Cost of 14 Macpuffs:

$\text{Cost} = 14 \times \textsf{₹} \frac{31}{2}$

$\text{Cost} = \textsf{₹} \cancel{14}^7 \times \frac{31}{\cancel{2}_1}$

$\text{Cost} = 7 \times 31 = \textsf{₹} 217$

Total Cost:

$\text{Total} = 83 + 217 = 300$

The total cost of 4 burgers and 14 Macpuffs is $\textsf{₹} 300$.

Given:

Total height of the hill = $101\frac{1}{3}$ m

Fraction of height under water = $\frac{1}{4}$ of the total height

To Find:

Height of the hill visible above the water.

Solution:

First, convert the mixed fraction into an improper fraction:

$\text{Total height} = 101\frac{1}{3} = \frac{(101 \times 3) + 1}{3} = \frac{304}{3}$ m

If $\frac{1}{4}$ of the height is under water, then the fraction of the hill visible above water is:

Now, calculate the visible height:

$\text{Visible height} = \frac{3}{4} \times \frac{304}{3}$

$\text{Visible height} = \frac{\cancel{3}^1}{4} \times \frac{304}{\cancel{3}_1}$

$\text{Visible height} = \frac{304}{4}$

$\text{Visible height} = 76$ m

The height of the hill visible above the water is $76$ m.

Given:

Total race time = $11.03$ seconds

Reaction time = $0.214$ seconds

To Find:

Actual running time (excluding reaction time).

Solution:

To find the actual running time, we subtract the reaction time from the total race time:

$\text{Actual running time} = 11.03 - 0.214$

Aligning the decimals for subtraction:

$\begin{array}{cc} & 1 & 1 & . & 0 & 3 & 0 \\ - & 0 & 0 & . & 2 & 1 & 4 \\ \hline & 1 & 0 & . & 8 & 1 & 6 \\ \hline \end{array}$

It took her $10.816$ seconds to run the actual distance.

To determine if the result is greater than or less than 1, we solve each expression:

1. $\frac{2}{3} \div \frac{1}{2} = \frac{2}{3} \times 2 = \frac{4}{3} \approx 1.33$ (Greater than 1)

2. $\frac{2}{3} \div \frac{2}{1} = \frac{2}{3} \times \frac{1}{2} = \frac{1}{3} \approx 0.33$ (Less than 1)

3. $6 \div \frac{1}{4} = 6 \times 4 = 24$ (Greater than 1)

4. $\frac{1}{5} \div \frac{1}{2} = \frac{1}{5} \times 2 = \frac{2}{5} = 0.4$ (Less than 1)

5. $4 \frac{1}{3} \div 3 \frac{1}{2} = \frac{13}{3} \div \frac{7}{2} = \frac{13}{3} \times \frac{2}{7} = \frac{26}{21} \approx 1.24$ (Greater than 1)

6. $\frac{2}{3} \times 8 \frac{1}{2} = \frac{2}{3} \times \frac{17}{2} = \frac{17}{3} \approx 5.67$ (Greater than 1)

The completed table is as follows:

Given:

The containers are arranged in ascending order of height. The smallest container is labeled IV and the largest is labeled I.

Height of the smallest container (IV) = $10.5$ cm

Equation relating the height of the smallest to the largest ($x$):

To Find:

The height of the largest container, denoted by $x$.

Solution:

Using equation (i), we solve for $x$:

$x = 10.5 \div \frac{7}{25}$

$x = 10.5 \times \frac{25}{7}$

First, divide $10.5$ by $7$:

$10.5 \div 7 = 1.5$

Now, multiply by $25$:

$x = 1.5 \times 25$

$x = 37.5$ cm

The height of the largest container is $37.5$ cm.

Given:

A sequence of fractions: $\frac{7}{8}, \frac{7}{24}, \frac{7}{72}, \frac{7}{216}$.

To Find:

The value at the vertex marked with ‘?’.

Solution:

Let us observe the relationship between the denominators of the fractions:

The pattern is that each term is obtained by multiplying the previous term by $\frac{1}{3}$ (or multiplying the denominator by $3$).

To find the missing value (?):

$? = \frac{7}{216} \times \frac{1}{3}$

$? = \frac{7}{216 \times 3}$

$? = \frac{7}{648}$

The value of ‘?’ is $\frac{7}{648}$.

Given:

A pentagonal flow chart starting from the top vertex and moving clockwise through the values: $\frac{3}{32}$, $\frac{3}{16}$, $\frac{3}{8}$, and $\frac{3}{4}$.

To Find:

The value at the vertex marked with ‘?’.

Solution:

Let us analyze the relationship between consecutive values in the sequence:

The pattern is that each term is obtained by multiplying the previous term by $2$.

To find the missing value (?):

$? = \frac{3}{4} \times 2$

$? = \frac{3 \times \cancel{2}^1}{\cancel{4}_2}$

$? = \frac{3}{2}$

The value of ‘?’ is $\frac{3}{2}$ (which is $1 \frac{1}{2}$ or $1.5$).

Given:

A sequence of decimal numbers: $0.05, 0.5, 5, 50$.

To Find:

The value at the vertex marked with ‘?’.

Solution:

Let us observe the multiplier used between terms:

The pattern is that each term is obtained by multiplying the previous term by $10$.

To find the missing value (?):

$? = 50 \times 10$

$? = 500$

The value of ‘?’ is $500$.

Given:

A sequence of decimal numbers: $0.1, 0.01, 0.001, 0.0001$.

To Find:

The value at the vertex marked with ‘?’.

Solution:

Let us analyze the decimal movement in the sequence:

The pattern shows that each term is multiplied by $0.1$ (or divided by $10$).

To find the missing value (?):

$? = 0.0001 \times 0.1$

$? = 0.00001$

The value of ‘?’ is $0.00001$.

Given:

The student compared two negative values: $-\frac{1}{4}$ and $-0.3$.

Calculation: $-\frac{1}{4} = -0.25$

Student's logic: Since $0.3 > 0.25$, then $-0.3 > -0.25$.

To Find:

The error in the student's reasoning.

Solution:

The student made an error in comparing negative numbers. While it is true that for positive numbers $0.3 > 0.25$, the rule for negative numbers is the exact opposite.

On a number line, a number to the right is always greater than a number to the left. Since $-0.25$ is closer to zero than $-0.3$, it lies to the right of $-0.3$.

In negative numbers, the number with the smaller absolute value is actually the greater number.

Therefore, the student's error was assuming that the inequality sign remains the same when comparing negative magnitudes as it does for positive magnitudes.

Given:

The multiplication: $2\frac{4}{7} \times 3\frac{1}{4} = 6\frac{1}{7}$

To Find:

The procedural error committed by the student.

Solution:

By observing the student's result ($6\frac{1}{7}$), we can identify the following incorrect steps taken:

1. The student multiplied the whole numbers separately: $2 \times 3 = 6$.

2. The student multiplied the fractional parts separately: $\frac{4}{7} \times \frac{1}{4} = \frac{1}{7}$.

3. They then combined these to get $6\frac{1}{7}$.

The Error: Mixed fractions cannot be multiplied by treating whole numbers and fractions as independent entities. They must first be converted into improper fractions.

Correct Method (Alternate Solution):

First, convert to improper fractions:

Now, multiply the improper fractions:

$\text{Product} = \frac{18}{7} \times \frac{13}{4}$

$\text{Product} = \frac{\cancel{18}^9}{7} \times \frac{13}{\cancel{4}_2} = \frac{117}{14}$

$\text{Product} = 8\frac{5}{14}$

The student's result $6\frac{1}{7}$ is incorrect because they failed to convert the mixed fractions first.

Given:

The sequence of addition is $\frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \dots$

To Find:

The specific fraction in the sequence that makes the cumulative sum cross the value of $1$ for the first time.

Solution:

Let us calculate the sum step-by-step by adding one fraction at a time:

Step 1: Sum of first two fractions

$\text{Sum}_2 = \frac{1}{3} + \frac{1}{4} = \frac{4 + 3}{12} = \frac{7}{12} \approx 0.583$

Step 2: Add the third fraction ($\frac{1}{5}$)

$\text{Sum}_3 = \frac{7}{12} + \frac{1}{5} = \frac{35 + 12}{60} = \frac{47}{60} \approx 0.783$

Step 3: Add the fourth fraction ($\frac{1}{6}$)

$\text{Sum}_4 = \frac{47}{60} + \frac{1}{6} = \frac{47 + 10}{60} = \frac{57}{60}$

$\text{Sum}_4 = \frac{19}{20} = 0.95$

Step 4: Add the fifth fraction ($\frac{1}{7}$)

$\text{Sum}_5 = \frac{19}{20} + \frac{1}{7} = \frac{133 + 20}{140} = \frac{153}{140}$

$\text{Sum}_5 \approx 1.093$

Explanation:

We see that the cumulative sum after adding $\frac{1}{6}$ was $0.95$, which is less than $1$. Upon adding the next fraction in the pattern, which is $\frac{1}{7}$, the sum becomes approximately $1.093$.

Therefore, the fraction $\frac{1}{7}$ makes the sum greater than $1$ for the first time.