Another Peek Beyond the Point - NCERT Solutions Class 7 Maths (Ganita Prakash)

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Another Peek Beyond the Point – NCERT Solutions Class 7 Maths (Ganita Prakash) includes all the questions with solutions given in the NCERT Class 7 Maths (Ganita Prakash).

NCERT Solutions Class 7

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Another Peek Beyond the Point – NCERT Solutions

Q.1: Recall that a tenth is 0.1, a hundredth is 0.01, and so on. Find the following products in tenths, hundredths and so on:

{tex} 6 \times 4 \text { tenths }=24 \text { tenths } {/tex}
{tex} 7 \times 0.3 {/tex}
{tex} 9 \times 5 \text { hundredths } {/tex}

Solution:

Already given:
{tex} 6 \times 4 \text { tenths }=24 \text { tenths } {/tex}
Since 1 tenth {tex}=0.1{/tex},
{tex} 24 \text { tenths }=24 \times 0.1=2.4 {/tex}
24 tenths {tex}=2.4{/tex}
{tex} 7 \times 0.3=7 \times 3 \text { tenths }=21 \text { tenths } {/tex}
{tex} 21 \text { tenths }=21 \times 0.1=2.1 {/tex}
{tex}21 \text { tenths = } 2.1 {/tex}
{tex} 9 \times 5 \text { hundredths }=45 \text { hundredths } {/tex}
{tex} 45 \text { hundredths }=45 \times 0.01=0.45 {/tex}
{tex}45 \text { hundredths = } 0.45 {/tex}

Q.2: Find the product: {tex}27.34 \times 6{/tex}

Solution:

{tex} 27.34 \times 6=(27 \times 6)+(0.34 \times 6) {/tex}
{tex} =162+2.04=164.04 {/tex}
{tex} 27.34 \times 6=164.04 {/tex}

Q.3: Find the product: {tex} 4.23 \times 3.7 {/tex}

Solution:

Ignore decimals first: {tex}423 \times 37=423 \times(30+7){/tex}
{tex} 423 \times 30=12690, 423 \times 7=2961 {/tex}
{tex} 12690+2961=15651 {/tex}
Now count decimal places {tex}\rightarrow(2{/tex} in {tex}4.23+1{/tex} in 3.7{tex})=3{/tex} decimal places
{tex} 15651 \div 1000=15.651 {/tex}
{tex}4.23 \times 3.7=15.651{/tex}

Q.4: Find the product: {tex}0.432 \times 0.23{/tex}

Solution:

Ignore decimals first: {tex}432 \times 23=432 \times(20+3){/tex}
{tex} 432 \times 20=8640, 432 \times 3=1296 {/tex}
{tex} 8640+1296=9936 {/tex}
Total decimal places {tex}\rightarrow(3{/tex} in {tex}0.432+2{/tex} in 0.23{tex})=5{/tex} decimal places
{tex} 9936 \div 100000=0.09936 {/tex}
Answer: {tex}0.432 \times 0.23=0.09936{/tex}

Q.5: Thejus needs 1.65 m of cloth for a shirt. How many metres of cloth are needed for 3 shirts?

Solution:

Each shirt needs 1.65 m of cloth.
For 3 shirts:
{tex} 1.65 \times 3=4.95 {/tex}
Thejus needs 4.95 metres of cloth for 3 shirts.

Q.6: Meenu bought 4 notebooks and 3 erasers. The cost of each book was ₹ 15.50 and each eraser was ₹ 2.75. How much did she spend in all?

Solution:

Cost of 1 notebook {tex}=₹ 15.50{/tex}
Number of notebooks {tex}=4{/tex}
{tex} 4 \times 15.50=62.00 {/tex}
Cost of 1 eraser {tex}=₹ 2.75{/tex}
Number of erasers = 3
{tex} 3 \times 2.75=8.25 {/tex}
Total cost {tex}=62.00+8.25=70.25{/tex}
Meenu spent ₹ 70.25 in all.

Q.7: The thickness of a rupee coin is 1.45 mm. What is the total height of the cylinder formed by placing 36 rupee coins one over the other? Write the answer in centimeters.

Solution:

Thickness of one rupee coin {tex}=1.45 \mathrm{~mm}{/tex}
Number of coins {tex}=36{/tex}
Total height in millimeters:
{tex} 1.45 \times 36=52.2 {~mm} {/tex}
Now, convert millimeters to centimeters:
{tex} 1 {~cm}=10 {~mm} {/tex}
{tex} 52.2 {~mm}=52.2 \div 10=5.22 {~cm} {/tex}
The total height of the cylinder is 5.22 cm.

Q.8: The price of 1 kg of oranges is ₹ 56.50. What is the price of 2.250 kg of oranges? Can we write 56.50 as 56.5 and 2.250 as 2.25 and multiply? Will we get the same product? Why?

Solution:

Given: Price of 1 kg oranges {tex}=₹ 56.50{/tex}
Weight {tex}=2.250 \mathrm{~kg}{/tex}
Step 1: {tex} 56.50 \times 2.250=127.125 {/tex}
Total price = ₹ 127.13 (rounded to two decimal places)
Step 2: {tex} 56.50=56.5 \text { and } 2.250=2.25 {/tex}
Adding or removing zeros after the decimal point does not change the value of a decimal number.
If we multiply:
{tex} 56.5 \times 2.25=127.125 {/tex}
It gives the same product.
Final Answer:
Price of 2.250 kg oranges {tex}=₹ 127.13{/tex}
Yes, we can write 56.50 as 56.5 and 2.250 as 2.25 because trailing zeros after the decimal point do not change the value, and the product remains the same.

Q.9: Dwarakanath purchases notebooks at a wholesale price of ₹ 23.6 per piece and sells each notebook at ₹ 30/-. How much profit does he make if he sells 50 books in a week?

Solution:

Cost price (C.P.) of 1 notebook {tex}=₹ 23.60{/tex}
Selling price (S.P.) of 1 notebook {tex}=₹ 30.00{/tex}
Step 1: Profit on 1 notebook
Profit per notebook {tex}=30-23.6=6.4{/tex}
Step 2: Profit on 50 notebooks
{tex} 6.4 \times 50=320 {/tex}
Hence, Dwarakanath makes a profit of ₹320 by selling 50 notebooks in a week.

Q.10: Given that {tex}18 \times 12=216{/tex}, find the products:

{tex}18 \times 1.2{/tex}
{tex}18 \times 0.12{/tex}
{tex}1.8 \times 1.2{/tex}
{tex}0.18 \times 0.12{/tex}
{tex}0.018 \times 0.012{/tex}
{tex}1.8 \times 12{/tex}

In which of the cases above is the product less than 1?

Solution:

{tex} \text { Since } 1.2=12 \div 10_x {/tex}
{tex} 18 \times 1.2=\frac{18 \times 12}{10}=\frac{216}{10}=21.6 {/tex}
{tex} \text { Since } 0.12=12 \div 100 \text {, } {/tex}
{tex} 18 \times 0.12=\frac{18 \times 12}{100}=\frac{216}{100}=2.16 {/tex}
{tex} \text { Here } 1.8=18 \div 10 \text { and } 1.2=12 \div 10 \text {, } {/tex}
{tex} 1.8 \times 1.2=\frac{18 \times 12}{100}=\frac{216}{100}=2.16 {/tex}
{tex} \text { Here } 0.18=18 \div 100 \text { and } 0.12=12 \div 100 {/tex}
{tex} 0.18 \times 0.12=\frac{18 \times 12}{10000}=\frac{216}{10000}=0.0216 {/tex}
{tex} \text { Here } 0.018=18 \div 1000 \text { and } 0.012=12 \div 1000 \text {, } {/tex}
{tex} 0.018 \times 0.012=\frac{18 \times 12}{1000000}=\frac{216}{1000000}=0.000216 {/tex}
{tex} \text { Here } 1.8=18 \div 10 \text {, } {/tex}
{tex} 1.8 \times 12=\frac{18 \times 12}{10}=\frac{216}{10}=21.6 {/tex}

Products less than 1 are:
{tex} 0.18 \times 0.12=0.0216 {/tex}
{tex} 0.018 \times 0.012=0.000216 {/tex}

Q.11: In which of the following multiplications is the product less than 1? Can you find the answer without actually doing the multiplications?

{tex}7 \times 0.6{/tex}
{tex}0.7 \times 0.6{/tex}
{tex}0.7 \times 6{/tex}
{tex}0.07 \times 0.06{/tex}

Solution:

Here, {tex}7>1{/tex} and {tex}0.6<1{/tex}
The large number dominates.
{tex} 7 \times 0.6=4.2>1 {/tex}
Product is greater than 1
Both numbers are less than 1
Product of two numbers less than 1 is less than 1
{tex} 0.7 \times 0.6=0.42<1 {/tex}
Product is less than 1
{tex} 6>1 \text { and } 0.7<1 {/tex}
{tex} 0.7 \times 6=4.2>1 {/tex}
{tex} \text { Product is greater than } 1 {/tex}
Both are less than 1
{tex} 0.07 \times 0.06=0.0042<1 {/tex}
Product is less than 1

Q.12: Multiplying the following numbers by 10, 100 and 1000 to complete the table.

	{tex}\times 10{/tex}	{tex}\times 100{/tex}	{tex}\times 1000{/tex}
5.7
23.02
0.92
0.306
24.67

Solution:

Number	×10	×100	×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

{tex}0.306 \times 100=30.6{/tex} (decimal moves two places right)
{tex}23.02 \times 1000=23020{/tex} (decimal moves three places right)

Q.13: Find the quotient by converting the denominator into 1, 10, 100 or 1000 and verify the solution by the long division method (division by place value).

{tex}\frac{18}{5}{/tex}
{tex}\frac{415}{4}{/tex}
{tex}\frac{1217}{2}{/tex}
{tex}\frac{4827}{8}{/tex}

Solution:

To make 5 into 10, multiply numerator and denominator by 2:
{tex} \frac{18}{5}=\frac{18 \times 2}{5 \times 2}=\frac{36}{10}=3.6 {/tex}
Quotient {tex}=3.6{/tex}
Verification (long division):
5 goes into {tex}18 \rightarrow 3{/tex} times (remainder 3). Bring down decimal {tex}\rightarrow 30 \div 5=6 \rightarrow 3.6{/tex}
{tex} 4 \times 100=400, \text { close to } 415 {/tex}
Long division directly:
{tex} 4 \mid 415 {/tex}
{tex}\rightarrow 4 \times 1=4 \rightarrow{/tex} remainder {tex}0 \rightarrow{/tex} bring down {tex}1 \rightarrow{/tex} remainder {tex}1 \rightarrow{/tex} bring down {tex}5 \rightarrow 15 \div 4=3{/tex} remainder 3
Add decimal {tex}\rightarrow 30 \div 4=7.5{/tex}
{tex} \text { Quotient = } 103.75 {/tex}
Divide normally:
{tex} 2 \text { | } 1217 {/tex}
{tex}\rightarrow 2 \times 6=12 \rightarrow 1{/tex} remainder {tex}\rightarrow{/tex} bring down {tex}1 \rightarrow 1 \times 2=0 \rightarrow{/tex} remainder {tex}1 \rightarrow{/tex} bring down {tex}7 \rightarrow 17{/tex}
{tex}\div 2=8{/tex} remainder 1
Quotient {tex}=608.5{/tex}
Divide:
{tex}8 \times 603=4824 \rightarrow{/tex} remainder 3
So,
{tex} \frac{4827}{8}=603+\frac{3}{8}=603.375 {/tex}
Quotient {tex}=603.375{/tex}

Q.14: Choose the correct answer:
{tex}\frac{1526}{4}={/tex}

Options:
(1) 38.15
(2) 380.15
(3) 381.5 ✅
(4) 381.05

Explanation:

{tex}4 \times 3=12 \rightarrow{/tex} remainder 1
Bring down {tex}2 \rightarrow 12 \div 4=3 \rightarrow{/tex} remainder 0
Bring down {tex}6 \rightarrow 6 \div 4=1{/tex} remainder 2
Add a decimal point {tex}\rightarrow{/tex} bring down {tex}0 \rightarrow 20 \div 4=5{/tex}
Quotient = 381.5

Q.15: Choose the correct answer: {tex} \frac{3567}{8}= {/tex}

Options:
(1) 4458.75
(2) 44.5875
(3) 445.875 ✅
(4) 4458.7

Explanation: 445.875

Q.16: What is the quotient?
{tex}132 \div 4={/tex}

Solution: {tex}132 \div 4=33{/tex}

Q.17: What is the quotient?
{tex}13.2 \div 4={/tex}

Solution: {tex}13.2 \div 4=3.3{/tex}

Q.18: What is the quotient?
{tex}1.32 \div 4={/tex}

Solution: {tex}1.32 \div 4=0.33{/tex}

Q.19: What is the quotient?
{tex}0.132 \div 4={/tex}

Solution: {tex}0.132 \div 4=0.033{/tex}

Q.20: What is the quotient?
{tex}126 \div 8={/tex}

Solution: {tex} 8 \times 15=120 \rightarrow \text { remainder } 6 \rightarrow 6 \div 8=0.75 {/tex}
{tex} \text { Quotient }=15.75 {/tex}

Q.21: What is the quotient?
{tex} 1.26 \div 8 {/tex}

Solution:

{tex}8 \times 0=0 \rightarrow{/tex} remainder {tex}1.26 \rightarrow 12 \div 8=1{/tex} remainder {tex}4 \rightarrow 46 \div 8=5{/tex} remainder {tex}6 \rightarrow 60 \div 8 =7.5{/tex}
Quotient = 0.1575

Q.22: What is the quotient?
{tex}0.126 \div 8{/tex}

Solution:

{tex}8 \times 0=0 \rightarrow{/tex} remainder {tex}0.126 \rightarrow 12 \div 8=1{/tex} remainder {tex}4 \rightarrow 46 \div 8=5{/tex} remainder {tex}6 \rightarrow 60 \div 8=7{/tex} remainder {tex}4 \rightarrow 40 \div 8=5{/tex}
Quotient {tex}=0.01575{/tex}

Q.23: What is the quotient?
{tex} 0.0126 \div 8= {/tex}

Solution:

Q.24: Express the fraction in decimal form: {tex}\frac{2}{5}{/tex}

Solution:

To make the denominator 10:
{tex} \frac{2}{5}=\frac{2 \times 2}{5 \times 2}=\frac{4}{10}=0.4 {/tex}
Answer: 0.4

Q.25: Express the fraction in decimal form: {tex}\frac{13}{4}{/tex}

Solution:

{tex} \text { Divide } 13 \text { by } 4 \text { : } {/tex}
{tex} 4 \times 3=12, \text { remainder } 1 \rightarrow 1 \div 4=0.25 {/tex}
{tex} \therefore \frac{13}{4}=3.25 {/tex}

Q.26: Express the fraction in decimal form: {tex} \frac{4}{50} {/tex}

Solution:

To make denominator 100:
{tex} \frac{4}{50}=\frac{4 \times 2}{50 \times 2}=\frac{8}{100}=0.08 {/tex}
Answer: 0.08

Q.27: Express the fraction in decimal form: {tex}\frac{5}{8}{/tex}

Solution:

This is found by dividing 5 by 8:
{tex} \frac{5}{8}=0.625 {/tex}

Q.28: Find the quotient: {tex}24.86 \div 1.2{/tex}

Solution:

To make the divisor a whole number, multiply both numbers by 10:
{tex} \frac{24.86}{1.2}=\frac{248.6}{12} {/tex}
Now divide:
{tex} 12 \times 20=240 \rightarrow \text { remainder }=8.6 {/tex}
{tex} 8.6 \div 12=0.7166 \ldots {/tex}
Quotient {tex}=20.7166 \approx 20.72{/tex}

Q.29: Find the quotients: 5.728 {tex}\div{/tex} 1.52

Solution:

Multiply both by 100 to remove decimals:
{tex} \frac{5.728}{1.52}=\frac{572.8}{152} {/tex}
Now divide:
{tex} 152 \times 3=456 \rightarrow \text { remainder }=116.8 {/tex}
{tex} 1168 \div 152=7.68 \rightarrow \text { approximately } {/tex}
Quotient {tex}=3.77{/tex}

Q.30: Evaluate the following using the information {tex}156 \times 12=1872{/tex}.

{tex}15.6 \times 1.2={/tex} ________
{tex}187.2 \div 1.2={/tex} ________

Solution:

To make the divisor a whole number, multiply both numbers by 10:
{tex} \frac{24.86}{1.2}=\frac{248.6}{12} {/tex}
Now divide:
{tex} 12 \times 20=240 \rightarrow \text { remainder }=8.6 {/tex}
{tex} 8.6 \div 12=0.7166 \ldots {/tex}
Quotient {tex}=20.7166 \approx 20.72{/tex}
Multiply both by 100 to remove decimals:
{tex} \frac{5.728}{1.52}=\frac{572.8}{152} {/tex}
Now divide:
{tex} 152 \times 3=456 \rightarrow \text { remainder }=116.8 {/tex}
{tex} 1168 \div 152=7.68 \rightarrow \text { approximately } {/tex}
Quotient {tex}=3.77{/tex}

Q.31: Evaluate the following using the information {tex}156 \times 12=1872{/tex}.

{tex}18.72 \div 15.6={/tex} ________
{tex}0.156 \times 0.12={/tex} ________

Solution:

Convert both numbers to remove decimals:
{tex} \frac{18.72}{15.6}=\frac{1872}{156} {/tex}
Now use the given information:
{tex} \frac{1872}{156}=12 {/tex}
Quotient {tex}=12{/tex}
Using {tex}156 \times 12=1872{/tex}, shift decimals {tex}3+2=5{/tex} places to the left:
{tex} 0.156 \times 0.12=0.001872 {/tex}
Product {tex}=0.001872{/tex}

Q.32: Evaluate the following:

{tex}25\ \div{/tex} ________ = 0.025
{tex}25\ \div {/tex} ________ = 250

Solution:

{tex}25 \div {/tex} __ = 0.025
{tex} \rightarrow25 \div 1000=0.025 {/tex}
Answer: 1000
{tex}25\ \div {/tex} ___ = 250
{tex} \rightarrow 25 \div 0.1=250 {/tex}
{tex} \text { Answer: } 0.1 {/tex}

Q.33: Evaluate the following:

{tex} 25\ \div{/tex} ________ = 2.5
{tex}25 \div 10=25 \times{/tex} ________

Solution:

{tex}25 \div 10=2.5{/tex}
Answer: 10
{tex} 25 \div 10=25 \times \frac{1}{10} {/tex}
Answer: 1/10

Q.34: Evaluate the following:

{tex} 25 \div 0.10=25 \times {/tex} ________
{tex}25 \div 0.01=25 \times{/tex} ________

Solution:

{tex} 25 \div 0.10=25 \times 10 {/tex}
Answer: 10
{tex} 25 \div 0.01=25 \times 100 {/tex}
Answer: 100

Q.35: Find the quotient:

{tex} 2.46 \div 1.5= {/tex}
{tex}2.46 \div 0.15= {/tex}
{tex} 2.46 \div 0.015= {/tex}

Is the quotient obtained in {tex}24.6 \div 1.5{/tex} the same as the quotient obtained in {tex}2.46 \div 0.15 ?{/tex}

Solution:

Multiply both by 10 to remove decimals: {tex}24.6 \div 15=1.64{/tex}
Quotient {tex}=1.64{/tex}
Multiply both by {tex}100: 246 \div 15=16.4{/tex}
Quotient = 16.4
Multiply both by {tex}1000: 2460 \div 15=164{/tex}
Quotient {tex}=164{/tex}

{tex} 24.6 \div 1.5=16.4 \text { and } 2.46 \div 0.15=16.4 {/tex}
{tex} \text { Yes, the quotients are the same. } {/tex}

Q.36: A 4 m long wooden block has to be cut into 5 pieces of equal length. What is the length of each piece?

Solution:

Total length of wooden block {tex}=4 \mathrm{~m}{/tex}
Number of pieces {tex}=5{/tex}
Length of each piece
{tex} =\frac{4}{5}=0.8 {~m} {/tex}
Each piece is 0.8 m long.

Q.37: If the perimeter of a regular polygon with 12 sides is 208.8 cm, what is the length of its side?

Solution:

Perimeter of polygon {tex}=208.8 \mathrm{~cm}{/tex}
Number of sides {tex}=12{/tex}
Length of each side
{tex} =\frac{208.8}{12}=17.4 {~cm} {/tex}
Each side of the polygon is 17.4 cm long.

Q.38: 3 litres of watermelon juice is shared among 8 friends equally. How much watermelon juice will each get? Express the quantity of juice in millilitres.

Solution:

Total juice = 3 litres
Number of friends {tex}=8{/tex}
Juice per friend
{tex} =\frac{3}{8}=0.375 \text { litres } {/tex}
Now, convert litres to millilitres:
{tex} 0.375 \times 1000=375 {~mL} {/tex}
Each friend will get 375 mL of watermelon juice.

Q.39: A car covers 234.45 km using 12.6 litres of petrol. What is the distance travelled per litre?

Solution:

Total distance = 234.45 km
Petrol used = 12.6 litres
{tex} \text { Distance per litre }=\frac{234.45}{12.6}=18.615 {~km} {/tex}
Distance travelled per litre {tex}=18.615 \mathrm{~km}(\approx 18.62 \mathrm{~km}){/tex}

Q.40: 13.5 kg of flour (aata) was distributed equally among 15 students. How much flour did each student receive?

Solution:

Total flour {tex}=13.5 \mathrm{~kg}{/tex}
Number of students {tex}=15{/tex}
{tex} \text { Flour per student }=\frac{13.5}{15}=0.9 {~kg} {/tex}
Each student received 0.9 kg of flour.

Q.41: A 210 gram packet of peanut chikki costs ₹ 70.5, while a 110 gram packet of potato chips costs ₹ 33.25. Which is cheaper?

Solution:

To find which is cheaper, compare cost per gram of each item.
For peanut chikki:
{tex}\text {Cost per gram }=\frac{70.5}{210}=0.3357 {/tex} {tex}\text { (approx ₹0.34 per gram) } {/tex}
For potato chips:
{tex}\text {Cost per gram }=\frac{33.25}{110}=0.3023{/tex} {tex} \text { (approx ₹ } 0.30 \text { per gram) } {/tex}
Potato chips are cheaper ({tex}₹ 0.30{/tex} per gram < ₹ 0.34 per gram).

Q.42: Write the decimal number at the arrow mark:

Solution:

First arrow: 3.15
Second arrow: 2.158

Q.43: Shyamala bought 3 kg bananas at ₹ {tex}30 /{/tex} – per kg. She counted 35 bananas in all. She sells each banana for ₹ 5/-. How much profit does she make selling all the bananas?

Solution:

Let’s calculate Shyamala’s profit step by step:
Step 1: Find the total cost.
She bought 3 kg bananas at ₹ 30 per kg:
{tex} \text { Total Cost }=3 \times 30=₹ 90 {/tex}
Step 2: Find the total selling price.
She sells 35 bananas at ₹5 each:
{tex} \text { Total Selling Price }=35 \times 5=₹ 175 {/tex}
Step 3: Calculate profit.
{tex} \text { Profit }=\text { Total Selling Price }- \text { Total Cost }=175-90=\text { ₹ } 85 {/tex}
Final Answer:
Shyamala makes a profit of ₹ 85 by selling all the bananas.

Q.44: A teacher placed textbooks that are 2.5 cm thick on a bookshelf. The teacher wanted to place 80 textbooks on the shelf. The bookshelf is 160 cm long. How many books could be placed on the shelf? Was there any space left? If yes, how much?

Solution:

A teacher wants to arrange 80 textbooks, each 2.5 cm thick, on a shelf that is 160 cm long. She wonders how many books can actually fit and if there will be any extra space left.
First, the total thickness for 80 books is {tex}80 \times 2.5=200 \mathrm{~cm}{/tex}.
The shelf is only 160 cm long, so 200 cm of books will not fit.
To find how many books the shelf can hold, divide the length of the shelf by the thickness of each book:
{tex} 160 \div 2.5=64 {/tex}
This means only 64 books will fit on the shelf.
After placing 64 books, the shelf will be completely full because {tex}64 \times 2.5=160 \mathrm{~cm}{/tex}, which is exactly the length of the shelf.
Therefore, the teacher can fit 64 textbooks on the shelf and there will be no space left over.

Q.45: Fill in the following blanks appropriately:

Solution:

{tex} 5.5 {~km}=5500 {~m} {/tex}
{tex} 35 {~cm}=0.35 {~m} {/tex}
{tex} 14.5 {~cm}=145 {~mm} {/tex}
{tex} 68 {~g}=0.068 {~kg} {/tex}
{tex} 9.02 {~m}=9020 {~mm} {/tex}
{tex} 125.5 {ml}=0.1255 {l} {/tex}

Q.46: The following problem was set by Sridharacharya in his book, Patiganita. ” {tex}6 \frac{1}{4}{/tex} is divided by {tex}2 \frac{1}{2}{/tex}, and {tex}60 \frac{1}{4}{/tex} is divided by {tex}3 \frac{1}{2}{/tex}. Tell the quotients separately.” Can you try to solve it by converting the fractions into decimals?

Solution:

{tex}6 \frac{1}{4} \div 2 \frac{1}{2}{/tex}
{tex}6 \frac{1}{4}=6+0.25=6.25{/tex}
{tex}2 \frac{1}{2}=2+0.5=2.5{/tex}
Quotient: {tex}\frac{6.25}{2.5}=2.5{/tex}
{tex}60 \frac{1}{4} \div 3 \frac{1}{2}{/tex}
{tex}60 \frac{1}{4}=60+0.25=60.25{/tex}
{tex}3 \frac{1}{2}=3+0.5=3.5{/tex}
Quotient: {tex}\frac{60.25}{3.5}=17.2142857{/tex}

Q.47: Fill the box in at least 2 different ways:

Solution:

{tex} 1.2 \times 2=2.4 {/tex}
{tex} 0.6 \times 4=2.4 {/tex}

Q.48: Fill the box in at least 2 different ways:

Solution:

{tex} 2.9 \times 5=14.5 {/tex}
{tex} 1 \times 14.5=14.5 {/tex}

Q.49: {tex} \text { Find the following quotients given that } 756 \div 36=21 \text { : } {/tex}

{tex}75.6 \div 3.6{/tex}
{tex}7.56 \div 0.36{/tex}

Solution:

Move decimal one place right (both numbers): {tex}756 \div 36=21{/tex} So, {tex}75.6 \div 3.6=21{/tex}
Move decimal two places right: {tex}756 \div 36=21{/tex}
So, {tex}7.56 \div 0.36=21{/tex}

Q.50: Find the following quotients given that {tex}756 \div 36=21{/tex}:

{tex}756 \div 0.36{/tex}
{tex}75.6 \div 360{/tex}

Solution:

Move decimal two places right in divisor: {tex}75600 \div 36=2100{/tex}
{tex} 75.6 \div 360=(756 \div 3600){/tex} 0 {tex}=(756 \div 36) \div 100{/tex} {tex}=21 \div 100=0.21 {/tex}

Q.51: {tex} \text { Find the following quotients given that } 756 \div 36=21 \text { : } {/tex}

{tex} 7560 \div 3.6 {/tex}
{tex}7.56 \div 0.36 {/tex}

Solution:

{tex}7560 \div 3.6=2100{/tex}, since {tex}7560 \div 3.6=(7560 \times 10) \div(3.6 \times 10){/tex} {tex}=75600 \div 36=2100{/tex}
using {tex}756 \div 36=21{/tex}.
{tex} 7.56 \div 0.36=21, \text { since } 7.56 \div 0.36{/tex} {tex}=(7.56 \times 100) \div(0.36 \times 100){/tex} {tex}=756 \div 36=21 {/tex}

Q.52: {tex} \text { Find the missing cells if each cell represents } {a} \div {b} \text { : } {/tex}

Solution:

b {tex}\downarrow{/tex} a {tex}\rightarrow{/tex}	1517	151.7	15.17	1.517	15170
37	41	4.1	0.41	0.041	410
41	37	3.7	0.37	0.037	370
3.7	410	41	4.1	0.41	4100
4.1	370	37	3.7	0.37	3700
0.37	4100	410	41	4.1	41000
4100	0.37	0.037	0.0037	0.00037	3.7
0.037	41000	4100	410	41	410000
370	4.1	0.41	0.041	0.0041	41

Q.53: Using the digits {tex}2,4,5,8{/tex}, and 0 fill the boxes {tex}\square{/tex}{tex}\square{/tex}.{tex}\square{/tex} {tex}\times{/tex} {tex}\square{/tex}.{tex}\square{/tex} to get the: