RS Aggarwal 2021-2022 for Class 6 Maths Solutions Chapter 10- Ratio, Proportion And Unitary Method
RS Aggarwal Class 6 Math Solution Chapter 10- Ratio, Proportion And Unitary Method, Exercise 10C is available here. RS Aggarwal Class 6 Math Solutions are solved by expert teachers in step by step, which help the students to understand easily. RS Aggarwal textbooks are responsible for a strong foundation in Maths. These textbook solutions help students in exams as well as their daily homework routine. The solutions included are easy to understand, and each step in the solution is described to match the students’ understanding.
Rs Aggarwal Class 6 Math Solution Chapter 10- Ratio, Proportion And Unitary Method
Exercise 10C
1. If the cost of 14 m of cloth is Rs. 1890, find the cost of 6 m of cloth.
Solution:
Cost of 14 m of cloth = Rs 1890
Cost of 1 m of cloth = `frac\{cancel 1890^135}{cancel 14}` = Rs 135
Cost of 6 m of cloth = 6 × 135 = Rs 810
2. If the cost of a dozen soaps is Rs. 285.60, what will be the cost of 15 such soaps?
Solution:
Cost of dozen soaps = Rs 285.60
Cost of 1 soap = `frac\{285.60}{12}`
Cost of 15 soaps = 15 × `frac\{285.60}{12}`
= `frac\{cancel 4284^357}{cancel 12}` = Rs 357
3. If 9 kg of rice costs Rs. 327.60, what will be the cost of 50 kg of rice?
Solution:
Cost of 9 kg of rice = Rs 327.60
Cost of 1 kg of rice = `frac\{327.60}{9}`
Cost of 50 kg of rice = 50 × `frac\{327.60}{9}`
= `frac\{cancel 16380^1820}{cancel 9}` = Rs 1820
Hence, the cost of 50 kg of rice is Rs 1820.
4. If 22.5 m of a uniform iron rod weighs 85.5 kg, what will be the weight of 5 m of the same rod?
Solution:
Weight of 22.5 m of iron rod = 85.5 kg
Weight of 1 m of iron rod = `frac\{85.5}{22.5}` kg
Weight of 5 m of iron rod = 5 × `frac\{85.5}{22.5}`
= `frac\{427.5}{22.5}` = 19 kg
Thus, the weight of 5 m of iron rod is 19 kg.
5. If 15 tins of the same size contain 234 kg of oil, how much oil will there be in 10 such tins?
Solution:
Oil contained by 15 tins = 234 kg
Oil contained by 1 tin = `frac\{234}{15}` kg
Oil contained by 10 tins = 10 × `frac\{ 234}{15}`
= `frac\{cancel 2340^156}{cancel 15}` = 156 kg
6. If 12 L of diesel is consumed by a car in covering a distance of 222 km. how many kilometres will it go in 22 L of diesel?
Solution:
Distance covered by a car in 12 L diesel = 222 km
Distance covered by it in 1 L diesel = `frac\{222}{12}` km
Distance covered by it in 22 L diesel = 22 × `frac\{222}{12}`
= `frac\{cancel 4884^407}{cancel 12}` = 407 km
7. A transport company charges Rs. 540 to carry 25 tonnes of weight. What will it charge to carry 35 tonnes?
Solution:
Cost of transporting 25 tonnes of weight = Rs 540
Cost of transporting 1 tone of weight = `frac\{540}{25}`
Cost of transporting 35 tonnes of weight = 35 × `frac\{540}{25}`
= `frac\{cancel 18900^756}{cancel 25}` = Rs 756
8. 4.5 g of an alloy of copper and zinc contains 3.5 g of copper. What weight of copper will there be in 18.9 g of the alloy?
Solution:
Let the weight of copper be x g.
Then, 4.5 : 3.5 : : 18.9 : x
Product of extremes = Product of means
4.5 × x = 3.5 × 18.9
⇒ x = `frac\{66.15}{4.5}` = 14.7
So, the weight of copper is 14.7 g.
9. 35 inland letters cost Rs. 87.50. How many such letters can we buy for Rs. 315?
Solution:
Number of inland letters in Rs 87.50 = 35
Number of inland letters in Re 1 = `frac\{35}{87.50}`
Number of inland letters in Rs 315 = 315 × `frac\{35}{87.50}`
= `frac\{11025}{87.50}` = 126
Hence, we can buy 126 inland letters for Rs 315.
10. Cost of 4 dozen bananas is Rs. 104. How many bananas can be purchased for Rs. 6.50?
Solution:
Number of bananas in Rs 104 = 48 (4 dozen)
Number of bananas in Re 1 = `frac\{48}{104}`
Number of bananas in Rs 6.50 = 6.50 × `frac\{48}{104}`
= `frac\{cancel 312^3}{cancel 104}`
= 3 bananas
Hence, 3 bananas can be purchased for Rs 6.50.
11. The cost of 18 chairs is Rs. 22770. How many such chairs can be bought for Rs. 10120?
Solution:
Number of chairs bought in Rs 22770 = 18
Number of chairs bought in Re 1 = `frac\{18}{22770}`
Number of chairs bought in Rs 10120 = 10120 × `frac\{18}{22770}`
= `frac\{cancel 182160^8}{cancel 22770}`
= 8 chairs
12. A car travels 195 km in 3 hours.
(i) How long will it take to travel 520 km?
(ii) How far will it travel in 7 hours with the same speed?
Solution:
(i) Time taken to travel 195 km = 3 hours
Time taken to travel 1 km = `frac\{3}{195}` hours
Time taken to travel 520 km = 520 × `frac\{3}{195}`
= `frac\{cancel 1560^8}{cancel 195}`
= 8 hours
(ii) Distance covered by the car in 3 hours = 195 km
Distance covered in 1 hour = `frac\{cancel 195^65}{cancel 3}` = 65 km
Distance covered in 7 hours = 7 × 65 = 455 km
13. A labourer earns Rs.1980 in 12 days.
(i) How much does he earn in 7 days?
(ii) In how many days will he earn Rs. 2640?
Solution:
(i) A labour earns in 12 days = Rs 1980
A labour earns in 1 day = `frac\{cancel 1980^165}{cancel 12}`
= Rs 165
A labour earns in 7 days = 7 × 165 = Rs 1155
(ii) The labour earn Rs 1980 in = 12 days
The labour earn Re 1 in = `frac\{12}{1980}` days
The labour earn Rs 2640 in = 2640 × `frac\{12}{1980}`
= `frac\{cancel 31680^16}{cancel 1980}`
= 16 days
14. The weight of 65 books is 13 kg.
(i) What is the weight of 80 such books?
(ii) How many such books weigh 6.4 kg?
Solution:
(i) Weight of 65 books = 13 kg
Weight of 1 book = `frac\{13}{65}` kg
Weight of 80 books = 80 × `frac\{13}{65}`
= `frac\{cancel 1040^16}{cancel 65}` = 16 kg
(ii) Number of books weighing 13 kg = 65
Number of books weighing 1 kg = `frac\{65}{13}` = 5
Number of books weighing 6.4 kg = 6.4 × 5 = 32
15. If 48 boxes contain 6000 pens, how many such boxes will be needed for 1875 pens?
Solution:
Boxes required for 6000 pens = 48
Boxes required for 1 pen = `frac\{48}{6000}`
Boxes required for 1875 pens = 1875 × `frac\{48}{6000}`
= `frac\{90000}{6000}`
= 15 boxes
15 boxes are needed for 1875 pens.
16. 24 workers can build a wall in 15 days. How many days will 9 workers take to build a similar wall?
Solution:
24 workers build a wall in = 15 days
1 worker build the wall in = 15 × 24 = 360 days
(less worker means more days)
9 workers build the wall in = `frac\{cancel 360^40}{cancel 9}` = 40 days
17. 40 men can finish a piece of work in 26 days. How many men will be needed to finish it in 16 days?
Solution:
Men needed to complete the work in 26 days = 40
Men needed to complete the work in 1 day = 40 × 26 = 1040 men
(less men more days)
Men needed to complete the work in 16 days = `frac\{cancel 1040^65}{cancel 16}`
= 65 men
18. In an army camp, there were provisions for 550 men for 28 days. But, 700 men attended the camp. How long did the provisions last?
Solution:
The provisions will last for 550 men = 28 days
The provisions will last for 1 man = 28 × 550 = 15400 days
(less men means more days)
The provisions will last for 700 men = `frac\{15400}{700}`
= 22 days
The provision will last for 22 days.
19. A given quantity of rice is sufficient for 60 persons for 3 days. How many days would the rice last for 18 persons?
Solution:
The given quantity of rice is sufficient for 60 persons = 3 days
The given quantity of rice is sufficient for 1 person = 3 × 60 = 180 days
(less men means more days )
The given quantity of rice is sufficient for 18 persons = `frac\{cancel 180^10}{cancel 18}`
= 10 days
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