Chapter 4 Another Peek Beyond the Point (Class 7 - Latest Maths NCERT (Ganita Prakash II) Solutions)

(a) $6 \times 4$ tenths

$6 \times 4 \text{ tenths} = 24 \text{ tenths}$

As a decimal: $\mathbf{2.4}$

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(b) $7 \times 0.3$

$0.3$ can be expressed as $3$ tenths.

$7 \times 3 \text{ tenths} = 21 \text{ tenths}$

As a decimal: $\mathbf{2.1}$

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(c) $9 \times 5$ hundredths

$9 \times 5 \text{ hundredths} = 45 \text{ hundredths}$

As a decimal: $\mathbf{0.45}$

(a) $27.34 \times 6$

Multiplying the numbers ignoring the decimal: $2734 \times 6 = 16404$

Since there are $2$ decimal places in $27.34$, we place the decimal point $2$ places from the right in the product.

Result: $164.04$

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(b) $4.23 \times 3.7$

Multiplying the numbers ignoring the decimal: $423 \times 37 = 15651$

Total decimal places in the factors $= 2$ (in $4.23$) $+ 1$ (in $3.7$) $= 3$.

Result: $15.651$

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(c) $0.432 \times 0.23$

Multiplying the numbers ignoring the decimal: $432 \times 23 = 9936$

Total decimal places in the factors $= 3$ (in $0.432$) $+ 2$ (in $0.23$) $= 5$.

Result: $0.09936$

Given:

Cloth needed for $1$ shirt $= 1.65\text{ m}$

Number of shirts $= 3$

To Find:

Total cloth needed for $3$ shirts.

Solution:

$\text{Total cloth} = 1.65 \times 3$

$\text{Total cloth} = 4.95\text{ m}$

Therefore, $4.95\text{ m}$ of cloth is needed for $3$ shirts.

Given:

Number of notebooks $= 4$

Cost of each notebook $= \textsf{₹} 15.50$

Number of erasers $= 3$

Cost of each eraser $= \textsf{₹} 2.75$

To Find:

Total money spent by Meenu.

Solution:

$\text{Total cost of notebooks} = 4 \times 15.50 = \textsf{₹} 62.00$

$\text{Total cost of erasers} = 3 \times 2.75 = \textsf{₹} 8.25$

$\text{Total money spent} = \textsf{₹} 62.00 + \textsf{₹} 8.25 = \textsf{₹} 70.25$

Therefore, she spent $\textsf{₹} 70.25$ in all.

Given:

Thickness of one rupee coin $= 1.45\text{ mm}$

Number of coins $= 36$

To Find:

Total height of the cylinder in centimeters.

Solution:

$\text{Total height in mm} = 1.45 \times 36$

$\text{Total height} = 52.2\text{ mm}$

Now, converting the height from millimetres to centimetres (as $10\text{ mm} = 1\text{ cm}$):

$\text{Height in cm} = \frac{52.2}{10} = 5.22\text{ cm}$

Therefore, the total height of the cylinder is $5.22\text{ cm}$.

Given:

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To Find:

The total price of $2.250\text{ kg}$ oranges and check if removing trailing zeros affects the result.

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Solution:

To find the total price, we multiply the price per kg by the weight.

$\text{Total Price} = 56.50 \times 2.250$

Yes, we can write $56.50$ as $56.5$ and $2.250$ as $2.25$. We will get the same product.

$\text{Total Price} = 56.5 \times 2.25$

Multiplying $565 \times 225$ and placing the decimal point $3$ places from the right:

Therefore, the price of $2.250\text{ kg}$ of oranges is $\textsf{₹} 127.125$.

Given:

Cost Price (CP) of one notebook $= \textsf{₹} 23.6$

Selling Price (SP) of one notebook $= \textsf{₹} 30$

Number of notebooks sold $= 50$

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To Find:

Total profit made in a week.

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Solution:

First, let us calculate the profit made on a single notebook.

$\text{Profit per book} = \text{Selling Price} - \text{Cost Price}$

$\text{Profit per book} = 30 - 23.6 = \textsf{₹} 6.4$

Now, calculate the total profit for $50$ notebooks.

$\text{Total Profit} = \text{Profit per book} \times \text{Total books}$

$\text{Total Profit} = 6.4 \times 50$

$\text{Total Profit} = 64 \times 5 = \textsf{₹} 320$

Dwarakanath makes a total profit of $\textsf{₹} 320$.

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Alternate Solution:

$\text{Total Selling Price} = 30 \times 50 = \textsf{₹} 1500$

$\text{Total Cost Price} = 23.6 \times 50 = \textsf{₹} 1180$

$\text{Total Profit} = 1500 - 1180 = \textsf{₹} 320$

Given:

The product of two whole numbers:

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To Find:

1. The products for cases (a) to (f) using the given information.

2. Identify the cases where the final product is less than $1$.

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Solution:

When multiplying decimal numbers, we first multiply them as whole numbers and then place the decimal point in the product. The number of decimal places in the product is equal to the sum of the decimal places in the numbers being multiplied.

We use the fact that the digits in the product will always be $216$.

(a) $18 \times 1.2$

(b) $18 \times 0.12$

(c) $1.8 \times 1.2$

(d) $0.18 \times 0.12$

(e) $0.018 \times 0.012$

(f) $1.8 \times 12$

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Final Conclusion:

A number is less than $1$ if its integral part (the part before the decimal point) is $0$.

Looking at the results above:

Case (d): $0.0216 < 1$

Case (e): $0.000216 < 1$

Therefore, the products in cases (d) and (e) are less than $1$.

Given:

Multiplication expressions: (a) $7 \times 0.6$, (b) $0.7 \times 0.6$, (c) $0.7 \times 6$, and (d) $0.07 \times 0.06$.

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To Find:

Identify which products are less than $1$ without performing the full calculation.

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Solution:

We can identify if a product is less than $1$ by using the following mathematical principles:

1. Rule of Proper Fractions: If both numbers being multiplied are less than $1$ (i.e., proper decimals), their product will always be less than $1$.

2. Estimation: If one number is significantly larger than $1$, we check if the other decimal factor is small enough to pull the product below $1$.

Let us analyze the options:

(a) $7 \times 0.6$

Here, $7$ is a large whole number. $0.6$ is more than half ($\frac{1}{2}$). Since half of $7$ is $3.5$, the product must be greater than $1$.

(b) $0.7 \times 0.6$

Both factors, $0.7$ and $0.6$, are less than $1$. According to the rule, their product must be less than $1$.

(c) $0.7 \times 6$

Here, $6$ is a whole number greater than $1$. Since $0.7$ is more than half, the product (roughly $4.2$) is clearly greater than $1$.

(d) $0.07 \times 0.06$

Both factors, $0.07$ and $0.06$, are less than $1$. Therefore, their product must be less than $1$.

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Final Conclusion:

The products in cases (b) and (d) are less than $1$. We found this by observing that in these cases, both factors are decimals smaller than $1$.

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Verification through Calculation:

Given:

A set of decimal numbers: $5.7$, $23.02$, $0.92$, $0.306$, and $24.67$.

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To Find:

The product of these numbers when multiplied by $10$, $100$, and $1000$.

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Solution:

When multiplying a decimal number by powers of $10$ (like $10, 100, 1000$), we shift the decimal point to the right by as many places as there are zeros in the multiplier.

1. Multiplying by $10$: Shift decimal point one place to the right.

2. Multiplying by $100$: Shift decimal point two places to the right.

3. Multiplying by $1000$: Shift decimal point three places to the right.

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Completed Table:

Number	$\times 10$	$\times 100$	$\times 1000$
$5.7$	$57$	$570$	$5700$
$23.02$	$230.2$	$2302$	$23020$
$0.92$	$9.2$	$92$	$920$
$0.306$	$3.06$	$30.6$	$306$
$24.67$	$246.7$	$2467$	$24670$

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Exemplary Calculation:

For the number $23.02$:

Given:

The following fractions are to be converted into decimals:

(a) $\frac{18}{5}$   (b) $\frac{415}{4}$   (c) $\frac{1217}{2}$   (d) $\frac{4827}{8}$

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To Find:

The quotient for each case by converting the denominator into a power of $10$ and verifying through long division.

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Solution:

(a) Solving $\frac{18}{5}$

To convert the denominator $5$ into $10$, we multiply both the numerator and the denominator by $2$.

Verification: Upon dividing $18$ by $5$, we get $5 \times 3 = 15$. The remainder is $3$. Adding a decimal point and a zero makes it $30$. Since $5 \times 6 = 30$, the quotient is $3.6$.

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(b) Solving $\frac{415}{4}$

To convert the denominator $4$ into $100$, we multiply both the numerator and the denominator by $25$.

Verification: Dividing $415$ by $4$ gives $103$ as the whole number part with a remainder of $3$. Extending the division with decimals, $3.0 \div 4 = 0.7$ with remainder $2$, and $0.20 \div 4 = 0.05$. The total quotient is $103.75$.

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(c) Solving $\frac{1217}{2}$

To convert the denominator $2$ into $10$, we multiply both the numerator and the denominator by $5$.

Verification: Dividing $1217$ by $2$ gives $608$ with a remainder of $1$. Adding a decimal zero makes it $10$, which divided by $2$ is $5$. Thus, the quotient is $608.5$.

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(d) Solving $\frac{4827}{8}$

To convert the denominator $8$ into $1000$, we multiply both the numerator and the denominator by $125$.

Verification: Dividing $4827$ by $8$ gives $603$ with a remainder of $3$. Continuing the division into decimals: $30 \div 8 = 3$ (rem $6$), $60 \div 8 = 7$ (rem $4$), and $40 \div 8 = 5$. The final quotient is $603.375$.

(a) Solution:

$\frac{1526}{4} = \frac{1526 \times 25}{4 \times 25} = \frac{38150}{100} = 381.5$

The correct answer is (iii) $381.5$.

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(b) Solution:

$\frac{3567}{8} = \frac{3567 \times 125}{8 \times 125} = \frac{445875}{1000} = 445.875$

The correct answer is (iii) $445.875$.

Given:

A series of division problems where the divisor is $4$ and the dividends are powers of $10$ variations of $132$.

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To Find:

The quotients for (a) $132 \div 4$, (b) $13.2 \div 4$, (c) $1.32 \div 4$, and (d) $0.132 \div 4$.

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Solution:

To solve these efficiently, we first calculate the quotient of the whole numbers and then apply the rules for decimal division.

Step 1: Divide the whole numbers

Step 2: Apply decimal rules

When dividing a decimal by a whole number, we perform the division as if there were no decimal point, and then place the decimal point in the quotient directly above the decimal point in the dividend. Effectively, the quotient has the same number of decimal places as the dividend.

(a) $132 \div 4$

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(b) $13.2 \div 4$

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(c) $1.32 \div 4$

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(d) $0.132 \div 4$

Given:

Dividends starting from $126$ decreasing by powers of $10$, all divided by the divisor $8$.

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To Find:

The quotients for parts (a) through (e).

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Solution:

Step 1: Find the basic quotient for $126 \div 8$

We can simplify the fraction $\frac{126}{8}$ by dividing both the numerator and the denominator by their common factor $2$.

Now, dividing $63$ by $4$:

Step 2: Determine quotients by shifting the decimal point

For every power of $10$ that the dividend is reduced, the decimal point in the quotient shifts one place to the left.

(a) $126 \div 8$

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(b) $12.6 \div 8$

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(c) $1.26 \div 8$

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(d) $0.126 \div 8$

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(e) $0.0126 \div 8$

(a) $\frac{2}{5}$

To convert $\frac{2}{5}$ into decimal form, we multiply both the numerator and the denominator by $2$ to make the denominator $10$.

$\frac{2 \times 2}{5 \times 2} = \frac{4}{10} = \mathbf{0.4}$

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(b) $\frac{13}{4}$

To convert $\frac{13}{4}$ into decimal form, we multiply both the numerator and the denominator by $25$ to make the denominator $100$.

$\frac{13 \times 25}{4 \times 25} = \frac{325}{100} = \mathbf{3.25}$

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(c) $\frac{4}{50}$

To convert $\frac{4}{50}$ into decimal form, we multiply both the numerator and the denominator by $2$ to make the denominator $100$.

$\frac{4 \times 2}{50 \times 2} = \frac{8}{100} = \mathbf{0.08}$

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(d) $\frac{5}{8}$

To convert $\frac{5}{8}$ into decimal form, we multiply both the numerator and the denominator by $125$ to make the denominator $1000$.

$\frac{5 \times 125}{8 \times 125} = \frac{625}{1000} = \mathbf{0.625}$

Given:

Two decimal division problems:

(a) $24.86 \div 1.2$

(b) $5.728 \div 1.52$

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To Find:

The quotients by converting divisors into whole numbers and performing long division with proper alignment.

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Solution:

(a) Solving $24.86 \div 1.2$

First, we convert the divisor $1.2$ into a whole number by multiplying both the dividend and the divisor by $10$.

Now, we perform the long division for $248.6 \div 12$:

$\begin{array}{r} 20.716...\phantom{)} \\ 12{\overline{\smash{\big)}\,248.600\phantom{)}}} \\ \underline{-~24\phantom{.600)}} \\ 08\phantom{.600)} \\ \underline{-~~0\phantom{.600)}} \\ 86\phantom{.00)} \\ \underline{-~84\phantom{.00)}} \\ 20\phantom{.0)} \\ \underline{-~12\phantom{.0)}} \\ 80\phantom{)} \\ \underline{-~72\phantom{)}} \\ 8\phantom{)} \end{array}$

The division is non-terminating. Rounding the quotient to two decimal places, we get:

Quotient $\approx 20.72$

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(b) Solving $5.728 \div 1.52$

We convert the divisor $1.52$ into a whole number by multiplying both the dividend and the divisor by $100$.

Now, we perform the long division for $572.8 \div 152$:

$\begin{array}{r} 3.768...\phantom{)} \\ 152{\overline{\smash{\big)}\,572.800\phantom{)}}} \\ \underline{-~456\phantom{.800)}} \\ 1168\phantom{.00)} \\ \underline{-~1064\phantom{.00)}} \\ 1040\phantom{.0)} \\ \underline{-~~912\phantom{.0)}} \\ 1280\phantom{)} \\ \underline{-~1216\phantom{)}} \\ 64\phantom{)} \end{array}$

Rounding the quotient to two decimal places, we get:

Quotient $\approx 3.77$

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Final Answer:

(a) The quotient is approximately $20.72$.

(b) The quotient is approximately $3.77$.

Given:

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To Find:

Evaluate (a) $15.6 \times 1.2$, (b) $187.2 \div 1.2$, (c) $18.72 \div 15.6$ and (d) $0.156 \times 0.12$.

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Solution:

(a) Evaluating $15.6 \times 1.2$

When multiplying decimal numbers, we multiply them as whole numbers first and then place the decimal point. The number of decimal places in the product is the sum of decimal places in the factors.

Number of decimal places in $15.6 = 1$

Number of decimal places in $1.2 = 1$

Total decimal places in the product = $1 + 1 = 2$

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(b) Evaluating $187.2 \div 1.2$

To divide, we convert the divisor into a whole number by shifting the decimal point.

$\frac{187.2}{1.2} = \frac{1872}{12}$

From the given information $156 \times 12 = 1872$, we can derive the division fact:

Therefore,

$187.2 \div 1.2 = 156$

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(c) Evaluating $18.72 \div 15.6$

Similarly, we convert the divisor into a whole number by multiplying both terms by $10$.

$\frac{18.72}{15.6} = \frac{187.2}{156}$

From the given information, $1872 \div 156 = 12$. Since we are dividing $187.2$ (which is $1872 \div 10$), the result will be $12 \div 10$.

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(d) Evaluating $0.156 \times 0.12$

Number of decimal places in $0.156 = 3$

Number of decimal places in $0.12 = 2$

Total decimal places in the product = $3 + 2 = 5$

We take the base product $1872$ and move the decimal $5$ places to the left. We add a zero as a placeholder.

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Final Answers:

(a) $18.72$

(b) $156$

(c) $1.2$

(d) $0.01872$

(a) $25 \div \text{______} = 0.025$

To get $0.025$ from $25$, the decimal point has shifted three places to the left. This happens when we divide by $1000$.

$25 \div \mathbf{1000} = 0.025$

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(b) $25 \div \text{______} = 250$

To increase the value from $25$ to $250$ through division, we must divide by a decimal number less than $1$. Here, the value is multiplied by $10$, which is the same as dividing by $0.1$.

$25 \div \mathbf{0.1} = 250$

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(c) $25 \div \text{______} = 2.5$

The decimal point has shifted one place to the left. This happens when we divide by $10$.

$25 \div \mathbf{10} = 2.5$

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(d) $25 \div 10 = 25 \times \text{_____}$

Division by a number is the same as multiplication by its reciprocal. The reciprocal of $10$ is $\frac{1}{10}$, which is $0.1$.

$25 \div 10 = 25 \times \mathbf{0.1}$

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(e) $25 \div 0.10 = 25 \times \text{______}$

The reciprocal of $0.10$ (which is $\frac{1}{10}$) is $10$.

$25 \div 0.10 = 25 \times \mathbf{10}$

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(f) $25 \div 0.01 = 25 \times \text{______}$

The reciprocal of $0.01$ (which is $\frac{1}{100}$) is $100$.

$25 \div 0.01 = 25 \times \mathbf{100}$

Given:

Division expressions: (a) $2.46 \div 1.5$, (b) $2.46 \div 0.15$, and (c) $2.46 \div 0.015$.

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To Find:

The quotients for the given expressions and a comparison between the quotients of $24.6 \div 1.5$ and $2.46 \div 0.15$.

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Solution:

To divide by a decimal, we first convert the divisor into a whole number by multiplying both the dividend and the divisor by an appropriate power of $10$.

(b) Solving $2.46 \div 0.15$ (Base Calculation)

The divisor $0.15$ has two decimal places. Multiplying both terms by $100$ gives:

Performing long division for $246 \div 15$:

$\begin{array}{r} 16.4\phantom{6.0)} \\ 15{\overline{\smash{\big)}\,246.0\phantom{)}}} \\ \underline{-~15\phantom{.00)}} \\ 096\phantom{.00)} \\ \underline{-~90\phantom{.00)}} \\ 06.0\phantom{)} \\ \underline{-~6.0\phantom{)}} \\ 0\phantom{)} \end{array}$

Therefore, the quotient for (b) is $16.4$.

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(a) Solving $2.46 \div 1.5$

Multiplying both terms by $10$ to make the divisor a whole number:

Since we know from equation (i) that $246 \div 15 = 16.4$, then $24.6 \div 15$ will have the decimal shifted one place to the left.

Therefore, the quotient for (a) is $1.64$.

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(c) Solving $2.46 \div 0.015$

Multiplying both terms by $1000$ to make the divisor a whole number:

Since $246 \div 15 = 16.4$, then $2460 \div 15$ will be $16.4 \times 10$.

Therefore, the quotient for (c) is $164$.

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Comparison:

To compare $24.6 \div 1.5$ and $2.46 \div 0.15$, we convert both into fractions with whole number denominators:

For $24.6 \div 1.5$:

For $2.46 \div 0.15$:

Conclusion: Yes, the quotient obtained in $24.6 \div 1.5$ is the same as the quotient obtained in $2.46 \div 0.15$. Both are equal to $16.4$.

Given:

Total length of the wooden block = $4\text{ m}$

Number of equal pieces = $5$

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To Find:

The length of each piece.

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Solution:

To find the length of each piece, we divide the total length by the number of pieces.

$\text{Length of each piece} = \text{Total length} \div \text{Number of pieces}$

$\text{Length of each piece} = 4 \div 5$

Since $4$ is smaller than $5$, we add a decimal point and a zero to make it $40$.

$\text{Length of each piece} = 0.8\text{ m}$

Therefore, the length of each piece is $0.8\text{ m}$ (or $80\text{ cm}$).

Given:

Perimeter of the regular polygon = $208.8\text{ cm}$

Number of equal sides = $12$

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To Find:

The length of one side of the polygon.

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Solution:

For a regular polygon, all sides are equal. The perimeter is the sum of all sides.

$\text{Side length} = \text{Perimeter} \div \text{Number of sides}$

$\text{Side length} = 208.8 \div 12$

On dividing $208.8$ by $12$:

$\text{Side length} = 17.4\text{ cm}$

Therefore, the length of each side is $17.4\text{ cm}$.

Given:

Total watermelon juice = $3\text{ L}$

Number of friends = $8$

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To Find:

Quantity of juice each friend gets in millilitres (ml).

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Solution:

First, we convert the total juice from litres to millilitres.

$\text{Total juice in ml} = 3 \times 1000 = 3000\text{ ml}$

Now, divide the total quantity by the number of friends:

$\text{Share of each friend} = 3000 \div 8$

$\text{Share of each friend} = 375\text{ ml}$

Therefore, each friend will get $375\text{ ml}$ of watermelon juice.

Given:

Total distance covered = $234.45\text{ km}$

Petrol used = $12.6\text{ L}$

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To Find:

The distance travelled per litre of petrol.

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Solution:

Distance per litre is found by dividing the total distance by the quantity of petrol.

$\text{Distance per litre} = 234.45 \div 12.6$

To simplify, we shift the decimal point one place to the right in both numbers:

$\text{Distance per litre} = 2344.5 \div 126$

On performing the division:

$\text{Distance per litre} = 18.607\dots\text{ km}$

Rounding to one decimal place, the distance travelled per litre is approximately $18.6\text{ km}$.

Given:

Total quantity of flour = $13.5\text{ kg}$

Number of students = $15$

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To Find:

Amount of flour received by each student.

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Solution:

$\text{Flour per student} = \text{Total flour} \div \text{Number of students}$

$\text{Flour per student} = 13.5 \div 15$

Since $13$ is smaller than $15$, the quotient starts with $0$. Treating $135$ as the number and dividing by $15$ gives $9$.

$\text{Flour per student} = 0.9\text{ kg}$

Therefore, each student received $0.9\text{ kg}$ (or $900\text{ g}$) of flour.

Given:

Weight of peanut chikki = $210\text{ g}$

Cost of peanut chikki = $\textsf{₹} 70.5$

Weight of potato chips = $110\text{ g}$

Cost of potato chips = $\textsf{₹} 33.25$

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To Find:

Which item is cheaper (by comparing the cost per gram).

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Solution:

To compare the prices, we find the cost of $1\text{ gram}$ for each item.

For Peanut Chikki:

$\text{Cost per gram} = \frac{70.5}{210}$

$\text{Cost per gram} \approx \textsf{₹} 0.3357$

For Potato Chips:

$\text{Cost per gram} = \frac{33.25}{110}$

$\text{Cost per gram} \approx \textsf{₹} 0.3022$

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Comparison:

Since $\textsf{₹} 0.3022 < \textsf{₹} 0.3357$, the potato chips cost less per gram than the chikki.

Therefore, the potato chips are cheaper.

Solution for the First Number Line:

The number line starts at $3.1$ and ends at $3.2$. There are $10$ equal divisions between them.

The value of each division is $\frac{3.2 - 3.1}{10} = \frac{0.1}{10} = 0.01$.

The arrow is pointing to the $6^{\text{th}}$ mark after $3.1$.

$\text{Value} = 3.1 + (6 \times 0.01) = 3.1 + 0.06 = 3.16$

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Solution for the Second Number Line:

The number line starts at $2.15$ and ends at $2.17$. There are $10$ equal divisions between them.

The value of each division is $\frac{2.17 - 2.15}{10} = \frac{0.02}{10} = 0.002$.

The arrow is pointing to the $6^{\text{th}}$ mark after $2.15$.

$\text{Value} = 2.15 + (6 \times 0.002) = 2.15 + 0.012 = 2.162$

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Final Answers:

The first arrow indicates $3.16$ and the second arrow indicates $2.162$.

Given:

Quantity of bananas bought = $3\text{ kg}$

Cost price per kg = $\textsf{₹} 30$

Total number of bananas = $35$

Selling price per banana = $\textsf{₹} 5$

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To Find:

Total profit made.

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Solution:

First, we calculate the Total Cost Price (CP):

$\text{Total CP} = 3 \times 30 = \textsf{₹} 90$

Next, we calculate the Total Selling Price (SP):

$\text{Total SP} = 35 \times 5 = \textsf{₹} 175$

Finally, we calculate the Profit:

$\text{Profit} = \text{Total SP} - \text{Total CP}$

$\text{Profit} = 175 - 90 = 85$

Therefore, Shyamala makes a profit of $\textsf{₹} 85$.

Given:

Thickness of one textbook = $2.5\text{ cm}$

Length of the bookshelf = $160\text{ cm}$

Number of books the teacher wanted to place = $80$

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To Find:

1. Number of books that can actually fit.

2. Space left on the shelf.

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Solution:

First, let us find how many books can fit on the shelf:

$\text{Number of books} = \frac{\text{Total Length}}{\text{Thickness per book}}$

$\text{Number of books} = \frac{160}{2.5}$

To simplify, multiply numerator and denominator by 10:

$\text{Number of books} = \frac{1600}{25} = 64$

The shelf can only hold $64$ books. Since the teacher wanted to place $80$ books, they will not all fit.

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Checking for Space Left:

Since $64$ books occupy exactly $64 \times 2.5 = 160\text{ cm}$, the entire shelf is filled.

If the teacher places the maximum possible ($64$ books), there is no space left ($0\text{ cm}$).

Using the conversion factors provided:

• $5.5\text{ km} = 5.5 \times 1000\text{ m} = \mathbf{5500\text{ m}}$

• $35\text{ cm} = \frac{35}{100}\text{ m} = \mathbf{0.35\text{ m}}$

• $14.5\text{ cm} = 14.5 \times 10\text{ mm} = \mathbf{145\text{ mm}}$

• $68\text{ g} = \frac{68}{1000}\text{ kg} = \mathbf{0.068\text{ kg}}$

• $9.02\text{ m} = 9.02 \times 1000\text{ mm} = \mathbf{9020\text{ mm}}$

• $125.5\text{ ml} = \frac{125.5}{1000}\text{ l} = \mathbf{0.1255\text{ l}}$

First Case: $6 \frac{1}{4}$ divided by $2 \frac{1}{2}$

Converting to decimals: $6 \frac{1}{4} = 6.25$ and $2 \frac{1}{2} = 2.5$

Calculation: $6.25 \div 2.5 = 62.5 \div 25$

$\text{Quotient} = \mathbf{2.5}$

Second Case: $60 \frac{1}{4}$ divided by $3 \frac{1}{2}$

Converting to decimals: $60 \frac{1}{4} = 60.25$ and $3 \frac{1}{2} = 3.5$

Calculation: $60.25 \div 3.5 = 602.5 \div 35$

$\text{Quotient} = \mathbf{17.214...}$ (or $17 \frac{3}{14}$ in fraction form)

(a) $\text{_____} \times \text{_____} = 2.4$

1. $2 \times 1.2 = 2.4$

2. $3 \times 0.8 = 2.4$

(b) $\text{_____} \times \text{_____} = 14.5$

1. $2 \times 7.25 = 14.5$

2. $5 \times 2.9 = 14.5$

Given:

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To Find:

The quotients for parts (a) to (f) based on the given base calculation.

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Solution:

To solve these, we use the principle that the quotient remains unchanged if both the dividend and the divisor are multiplied or divided by the same non-zero number. If only one is changed, or they are changed by different powers of $10$, we shift the decimal point accordingly.

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(a) $75.6 \div 3.6$

Both the dividend and divisor have one decimal place. Multiplying both by $10$:

From the given fact (i), the quotient is $21$.

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(b) $7.56 \div 0.36$

Both the dividend and divisor have two decimal places. Multiplying both by $100$:

From the given fact (i), the quotient is $21$.

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(c) $756 \div 0.36$

The divisor has two decimal places. Multiplying both terms by $100$ to make the divisor a whole number:

Since $756 \div 36 = 21$, then $75600 \div 36$ will be $21$ followed by two zeros.

Therefore, the quotient is $2100$.

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(d) $75.6 \div 360$

Converting the expression to a fraction:

We know $\frac{756}{36} = 21$. So, $\frac{756}{36 \times 100} = \frac{21}{100}$.

Therefore, the quotient is $0.21$.

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(e) $7560 \div 3.6$

Multiplying both dividend and divisor by $10$ to make the divisor a whole number:

As calculated in part (c), $75600 \div 36 = 2100$.

Therefore, the quotient is $2100$.

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(f) $7.56 \div 0.36$

This is the same as part (b). Multiplying both by $100$:

Therefore, the quotient is $21$.

Given: The base division is $1517 \div 37 = 41$.

Solution: In decimal division, if the dividend ($a$) shifts its decimal point to the left, the quotient shifts its decimal to the left. If the divisor ($b$) shifts its decimal point to the left, the quotient shifts its decimal to the right.

Using this logic, we complete the table:

$b \setminus a$	$1517$	$151.7$	$15.17$	$1.517$	$15170$
$37$	$41$	$4.1$	$0.41$	$0.041$	$410$
$3.7$	$410$	$41$	$4.1$	$0.41$	$4100$
$0.37$	$4100$	$410$	$41$	$4.1$	$41000$
$0.037$	$41000$	$4100$	$410$	$41$	$410000$
$370$	$4.1$	$0.41$	$0.041$	$0.0041$	$41$

Given:

Available digits: $\{2, 4, 5, 8, 0\}$

Expression structure: $\square \square . \square \times \square . \square$ (A 3-digit decimal multiplied by a 2-digit decimal).

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To Find:

Combinations of the given digits that result in the maximum, minimum, and specified range or proximity products.

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Solution:

To solve these, we place the largest digits in the most significant place values (tens and units) to maximize the product, and the smallest digits in those positions to minimize it.

(a) Maximum product

To get the maximum product, we use the largest available digits ($8, 5, 4$) in the leading positions.

Another combination giving the same result is $54.0 \times 8.2 = 442.8$.

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(b) Minimum product

To get the minimum product, we place $0$ in the units place of the second number (making it a pure decimal less than $1$) and use the smallest remaining digits for the leading tens place.

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(c) Product greater than $150$

We can use digits that ensure a product significantly above $150$.

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(d) Product nearest to $100$

We look for numbers whose leading digits multiply to approximately $100$ (like $20 \times 5$ or $25 \times 4$).

Another close option is $25.8 \times 4.0 = 103.2$ (Difference = $3.2$). Thus, $20.5 \times 4.8$ is the nearest.

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(e) Product nearest to $5$

Using the smallest possible configuration from the available digits:

Other combinations like $25.8 \times 0.4 = 10.32$ or $28.5 \times 0.4 = 11.4$ result in products further away from $5$.

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Final Summary:

(a) Maximum product: $82.0 \times 5.4 = 442.8$

(b) Minimum product: $45.8 \times 0.2 = 9.16$

(c) Product greater than $150$: $85.4 \times 2.0 = 170.8$

(d) Product nearest to $100$: $20.5 \times 4.8 = 98.4$

(e) Product nearest to $5$: $45.8 \times 0.2 = 9.16$

Given:

The following expressions and values:

(a) $245.05 \times 0.942368$    (b) $245.05 \times 7.9682$    (c) $245.05 \div 7.9682$

(d) $245.05 \div 0.942368$    (e) $245.05$    (f) $7.9682$

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To Find:

The correct arrangement of these expressions in increasing order (from smallest to largest) without performing the full multiplications or divisions.

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Solution:

To sort these expressions, let us analyze the effect of multiplying and dividing by numbers greater than or less than $1$ relative to the base value $245.05$.

Let $X = 245.05$. Note that $X$ is significantly larger than the other factors ($0.942368$ and $7.9682$).

1. Smallest values:

The value in (f) is $7.9682$, which is a very small number compared to $245.05$.

The value in (c) is $245.05 \div 7.9682$. Dividing a number by a factor greater than $1$ reduces its value. Since $245.05$ is divided by roughly $8$, the result will be around $30.75$.

Comparing these two: $7.9682 < (245.05 \div 7.9682)$. Thus, (f) is the smallest, followed by (c).

2. Mid-range values (Comparing with $245.05$):

In (a), we have $245.05 \times 0.942368$. Multiplying a positive number by a fraction less than $1$ results in a value smaller than the original number.

In (e), we have the base value $245.05$.

In (d), we have $245.05 \div 0.942368$. Dividing by a number less than $1$ results in a value larger than the original number.

Therefore: $(245.05 \times 0.942368) < 245.05 < (245.05 \div 0.942368)$. This orders (a), (e), and (d).

3. Largest value:

In (b), we have $245.05 \times 7.9682$. Multiplying a large number by another number much greater than $1$ results in a very large product (roughly $245 \times 8 \approx 1960$). This is clearly the largest value.

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Final Arrangement (Increasing Order):

Based on the analysis above, the order is:

$7.9682 < (245.05 \div 7.9682) < (245.05 \times 0.942368) < 245.05 $$ < (245.05 \div 0.942368) < (245.05 \times 7.9682)$

Using the option letters:

(f) < (c) < (a) < (e) < (d) < (b)