NCERT Solutions for Class 6 Maths Chapter 12 (Ex 12.3)
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Exercise:  Exercise  12.3 
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Access NCERT solutions for Class 6 Chapter 12 Ratio and Proportion
Exercise 12.3
1. If the cost of 7 m of cloth is ₹ 294, find the cost of 5 m of cloth.
Ans: The cost of 7 m of cloth is₹ 294.
Therefore, the cost of 1 m of cloth can be obtained by division operation as follows.
${\text{Cost of }}1{\text{ m cloth}} = \dfrac{{{\text{Total cost}}}}{{{\text{Total length of cloth}}}}$
${\text{Cost of }}1{\text{ m cloth}} = \dfrac{{294}}{7}$
${\text{Cost of }}1{\text{ m cloth}} = 42$
Hence, the cost of 5 m of cloth can be obtained by multiplication operation as follows.
${\text{Cost of }}5{\text{ m cloth}} = {\text{Cost of }}1{\text{ m cloth}} \times 5$
${\text{Cost of }}5{\text{ m cloth}} = 42 \times 5$
${\text{Cost of }}5{\text{ m cloth}} = 210$
Therefore, the obtained cost of 5 m of cloth is ₹ 210.
2. Ekta earns ₹ 1500 in 10 days. How much will she earn in 30 days?
Ans: The earning of Ekta is ₹ 1500 in 10 days.
Therefore, the earning of 1 day can be obtained by division operation as follows.
${\text{Earning of }}1{\text{ day}} = \dfrac{{{\text{Total Earning}}}}{{{\text{Number of days}}}}$
${\text{Earning of }}1{\text{ day}} = \dfrac{{{\text{1500}}}}{{{\text{10}}}}$
${\text{Earning of }}1{\text{ day}} = {\text{150}}$
Hence, the earning of 30 days can be obtained by multiplication operation as follows.
${\text{Earning of }}30{\text{ days}} = {\text{Earning of 1 day}} \times 30$
${\text{Earning of }}30{\text{ days}} = 150 \times 30$
${\text{Earning of }}30{\text{ days}} = 4500$
Therefore, the obtained earning of 30 days is ₹4500.
3. If it has rained 276 mm in the last 3 days, how many cm of rain will fall in one full week (7 days)? Assume that the rain continues to fall at the same rate.
Ans: The measure of rain in the last 3 days is 276 mm.
Therefore, the measure of rain in 1 day can be obtained by division operation as follows.
${\text{Measure of rain in }}1{\text{ day}} = \dfrac{{{\text{Total rain}}}}{{{\text{Total days}}}}$
${\text{Measure of rain in }}1{\text{ day}} = \dfrac{{{\text{276}}}}{{\text{3}}}$
${\text{Measure of rain in }}1{\text{ day}} = {\text{92}}$
Hence, the measure of rain in 7 days can be obtained by multiplication operation as follows.
${\text{Measure of rain in }}7{\text{ days}} = {\text{Measure of rain in 1 day}} \times 7$
${\text{Measure of rain in }}7{\text{ days}} = {\text{92}} \times 7$
${\text{Measure of rain in }}7{\text{ days}} = 644{\text{ mm}}$
Since $1{\text{ mm}} = 0.1{\text{ cm}}$, then 644 mm can be written as follows.
$644{\text{ mm}} = 644 \times 0.1{\text{ cm}}$
$644{\text{ mm}} = 64.4{\text{ cm}}$
Therefore, the obtained measure of 7 days is $64.4$ cm.
4. Cost of 5 kg of wheat is ₹$30.50$.
(a) What will be the cost of 8 kg of wheat?
Ans: The cost of 5 kg wheat is ₹ $30.50$.
Therefore, the cost of 1 kg wheat can be obtained by division operation as follows.
${\text{Cost }}1{\text{ kg wheat}} = \dfrac{{{\text{Total cost}}}}{{{\text{Total weight}}}}$
${\text{Cost }}1{\text{ kg wheat}} = \dfrac{{30.50}}{5}$
${\text{Cost }}1{\text{ kg wheat}} = 6.10$
Hence, the cost of 8 kg wheat can be obtained by multiplication operation as follows.
${\text{Cost }}8{\text{ kg wheat}} = {\text{Cost }}1{\text{ kg wheat}} \times 8$
${\text{Cost }}8{\text{ kg wheat}} = 6.10 \times 8$
${\text{Cost }}8{\text{ kg wheat}} = 48.80$
Therefore, the obtained cost of 8 kg of wheat is ₹$48.80$.
(b) What quantity of wheat can be purchased in ₹ 61?
Ans: The cost of 5 kg wheat is ₹ $30.50$.
Therefore, the quantity of wheat in ₹ 1 can be obtained by division operation as follows.
${\text{Quantity of wheat in }}1{\text{ rupees}} = \dfrac{{{\text{Total quantity}}}}{{{\text{Total cost}}}}$
${\text{Quantity of wheat in }}1{\text{ rupees}} = \dfrac{5}{{30.50}}$
Hence, the quantity of wheat in ₹ 61 can be obtained by multiplication operation as follows.
${\text{Quantity of wheat in }}61{\text{ rupees}} = {\text{Quantity of wheat in }}1{\text{ rupees}} \times 61$
${\text{Quantity of wheat in }}61{\text{ rupees}} = \dfrac{5}{{30.50}} \times 61$
${\text{Quantity of wheat in }}61{\text{ rupees}} = \dfrac{{500}}{{3050}} \times 61$
${\text{Quantity of wheat in }}61{\text{ rupees}} = \dfrac{{10}}{{61}} \times 61$
${\text{Quantity of wheat in }}61{\text{ rupees}} = 10{\text{ kg}}$
Therefore, the obtained quantity of wheat in ₹ 61 is 10 kg.
5. The temperature dropped 15 degree Celsius in the last 30 days. If the rate of temperature drop remains the same, how many degrees will the temperature drop in the next ten days?
Ans: The temperature drop in the last 30 days is 15 degree Celsius.
Therefore, the temperature drop in 1 day can be obtained by division operation as follows.
${\text{Temperature drop in }}1{\text{ day}} = \dfrac{{{\text{Total temperature drop}}}}{{{\text{Total days}}}}$
${\text{Temperature drop in }}1{\text{ day}} = \dfrac{{15}}{{30}}$
${\text{Temperature drop in }}1{\text{ day}} = \dfrac{1}{2}$
Hence, the temperature drop in 10 days can be obtained by multiplication operation as follows.
${\text{Temperature drop in }}10{\text{ days}} = {\text{Temperature drop in }}1{\text{ day}} \times 10$
${\text{Temperature drop in }}10{\text{ days}} = \dfrac{1}{2} \times 10$
${\text{Temperature drop in }}10{\text{ days}} = 5$
Therefore, the obtained temperature drop in 10 days is 5 degree Celsius.
6. Shains pays ₹ 7500 as rent for 3 months. How much does she have to pay for a whole year, if the rent per month remains the same?
Ans: The rent for 3 months is ₹ 7500.
Therefore, the rent of 1 month can be obtained by division operation as follows.
${\text{Rent of }}1{\text{ month}} = \dfrac{{{\text{Total rent}}}}{{{\text{Total number of months}}}}$
${\text{Rent of }}1{\text{ month}} = \dfrac{{{\text{7500}}}}{{\text{3}}}$
${\text{Rent of }}1{\text{ month}} = {\text{2500}}$
1 year can be written as 12 months.
Hence, the rent of 12 month can be obtained by multiplication operation as follows.
${\text{Rent of }}12{\text{ months}} = {\text{Rent of }}1{\text{ month}} \times 12$
${\text{Rent of }}12{\text{ months}} = 2500 \times 12$
${\text{Rent of }}12{\text{ months}} = 30000$
Therefore, the obtained rent of 12 month is ₹ 30,000.
7. Cost of 4 dozen bananas is ₹ 60. How many bananas can be purchased for ₹$12.50$?
Ans: 1 dozen can be expressed as 12.
Therefore,
$4{\text{ dozens}} = 4 \times 12$
$4{\text{ dozens}} = 48$
The cost of 48 bananas is ₹ 60.
Therefore, the quantity of banana in ₹ 1 can be obtained by division operation as follows.
${\text{Quantity of banana in }}1{\text{ rupees}} = \dfrac{{{\text{Total quantity}}}}{{{\text{Total cost}}}}$
${\text{Quantity of banana in }}1{\text{ rupees}} = \dfrac{{48}}{{60}}$
${\text{Quantity of banana in }}1{\text{ rupees}} = \dfrac{4}{5}$
Hence, the quantity of bananas in ₹$12.50$ can be obtained by multiplication operation as follows.
${\text{Quantity of banana in }}12.50{\text{ rupees}} = {\text{Quantity of banana in }}1{\text{ rupees}} \times 12.50$
${\text{Quantity of banana in }}12.50{\text{ rupees}} = \dfrac{4}{5} \times 12.50$
${\text{Quantity of banana in }}12.50{\text{ rupees}} = 4 \times 2.5$
${\text{Quantity of banana in }}12.50{\text{ rupees}} = 10$
Therefore, the obtained quantity of bananas in ₹$12.50$ is 10.
8. The weight of 72 books is 9 kg. What is the weight of 40 such books?
Ans: The weight of 72 books is 9 kg.
Therefore, the weight of 1 book can be obtained by division operation as follows.
\[{\text{Weight of }}1{\text{ book}} = \dfrac{{{\text{Total weight}}}}{{{\text{Total number of books}}}}\]
\[{\text{Weight of }}1{\text{ book}} = \dfrac{9}{{{\text{72}}}}\]
\[{\text{Weight of }}1{\text{ book}} = \dfrac{1}{8}\]
Hence, the weight of 40 books can be obtained by multiplication operation as follows.
\[{\text{Weight of }}40{\text{ books}} = {\text{Weight of }}1{\text{ book}} \times 40\]
\[{\text{Weight of }}40{\text{ books}} = \dfrac{1}{8} \times 40\]
\[{\text{Weight of }}40{\text{ books}} = 5{\text{ kg}}\]
Therefore, the obtained weight of 40 books is 5 kg.
9. A truck requires 108 litres of diesel for covering a distance of 594 km. How much diesel will be required by the truck to cover a distance of 1650 km?
Ans: For covering 594 km distance requirement of diesel is 108 litres.
Therefore, the requirement of diesel for covering 1 km can be obtained by division operation as follows.
${\text{Required diesel for }}1{\text{ km}} = \dfrac{{{\text{Total diesel}}}}{{{\text{Total distance}}}}$
${\text{Required diesel for }}1{\text{ km}} = \dfrac{{{\text{108}}}}{{{\text{594}}}}$
${\text{Required diesel for }}1{\text{ km}} = \dfrac{2}{{11}}$
Hence, the requirement of diesel for covering 1650 km can be obtained by multiplication operation as follows.
${\text{Required diesel for }}1650{\text{ km}} = {\text{Required diesel for }}1{\text{ km}} \times 1650$
${\text{Required diesel for }}1650{\text{ km}} = \dfrac{2}{{11}} \times 1650$
${\text{Required diesel for }}1650{\text{ km}} = 2 \times 150$
${\text{Required diesel for }}1650{\text{ km}} = 300{\text{ litres}}$
Therefore, the obtained requirement of diesel for covering 1650 km is 300 litres.
10. Raju purchases 10 pens for ₹150 and Manish buys 7 pens for ₹84. Can you say who got the pen cheaper?
Ans: The cost of 10 pens of Raju is ₹150.
Therefore, the cost of 1 pens of Raju can be obtained by division operation as follows.
${\text{Cost of }}1{\text{ pen}} = \dfrac{{{\text{Total cost}}}}{{{\text{Total number of pens}}}}$
${\text{Cost of }}1{\text{ pen}} = \dfrac{{150}}{{{\text{10}}}}$
${\text{Cost of }}1{\text{ pen}} = 15$
The cost of 7 pens of Manish is ₹84.
Therefore, the cost of 1 pen of Manish can be obtained by division operation as follows.
${\text{Cost of }}1{\text{ pen}} = \dfrac{{{\text{Total cost}}}}{{{\text{Total number of pens}}}}$
${\text{Cost of }}1{\text{ pen}} = \dfrac{{84}}{7}$
${\text{Cost of }}1{\text{ pen}} = 12$
Since $12 < 15$, therefore Manish got the pens at a cheaper rate.
11. Anish made 42 runs in 6 overs and Anup made 63 runs in 7 overs. Who made more runs per over?
Ans: The number of runs made by Anish in 6 overs is 42.
Therefore, the number of runs made by Anish in 1 overs can be obtained by division operation as follows.
${\text{Run made in }}1{\text{ over}} = \dfrac{{{\text{Total run}}}}{{{\text{Total number of overs}}}}$
${\text{Run made in }}1{\text{ over}} = \dfrac{{{\text{42}}}}{{\text{6}}}$
${\text{Run made in }}1{\text{ over}} = {\text{7}}$
The number of runs made by Anup in 7 overs is 63.
Therefore, the number of runs made by Anup in 1 over can be obtained by division operation as follows.
${\text{Run made in }}1{\text{ over}} = \dfrac{{{\text{Total run}}}}{{{\text{Total number of overs}}}}$
${\text{Run made in }}1{\text{ over}} = \dfrac{{63}}{7}$
${\text{Run made in }}1{\text{ over}} = 9$
Since $9 > 7$, Anup scored more runs in each over.
NCERT Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Exercise 12.3
Opting for the NCERT solutions for Ex 12.3 Class 6 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 12.3 Class 6 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.
Vedantu inhouse subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 6 students who are thorough with all the concepts from the Maths textbook and quite wellversed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 6 Maths Chapter 12 Exercise 12.3 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.
Besides these NCERT solutions for Class 6 Maths Chapter 12 Exercise 12.3, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our inhouse subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.
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