RS Aggarwal 2019,2020 solution class 7 chapter 10 Percentage Exercise 10B

Exercise 10B

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Question 1:

Rupesh secures 495 marks out of 750 in his annual examination. Find the percentage of marks obtained by him.

Answer 1:

Maximum marks of the examination = 750
Marks secured by Rupesh = 495
Percentage of marks secured = $\left(\frac{495}{750}\times 100\right)\%$ = 66%

Hence, Rupesh scored 66% in the examination.

Question 2:

The monthly salary of a typist is Rs 15625. If he gets an increase of 12%, find his new salary.

Answer 2:

Total monthly salary = Rs 15625
Increase percentage = 12%
∴ Amount increase = 12% of Rs 15625
                                 = Rs $\left(15625\times \frac{12}{100}\right)$ = Rs 1875
∴ New salary = Rs 15625 + Rs 1875
                       = Rs 17500
Hence, the new salary of the typist is Rs 17,500.

Question 3:

The excise duty on a certain item has been reduced to Rs 760 from Rs 950. Find the reduction per cent in the excise duy on that item.

Answer 3:

Original excise duty on the item = Rs 950
Amount reduced on excise duty = Rs (950 − 760) = Rs 190
∴ Reduction percent = $\left(\frac{\mathrm{Reduction} \mathrm{amount}}{\mathrm{Original} \mathrm{value}}\times 100\right)$
                                  = $\left(\frac{190}{950}\times 100\right)$ = 20
Hence, the excise duty on that item is reduced by 20%.

Question 4:

96% of the cost of a TV is Rs 10464. What is its total cost?

Answer 4:

Let Rs x be the total cost of the TV set.

Now, 96% of the total cost of TV = Rs 10464
⇒ 96% of Rs x = Rs 10464
⇒ $\left(\frac{96}{100}\times x\right)$ = 10464
∴ x = $\left(\frac{10464\times 100}{96}\right)$ =  10900
Hence, the total cost of the TV set is Rs 10900.

Question 5:

70% of the students in a school are boys and the number of girls is 504. Find the number of boys in the school.

Answer 5:

Let the total number of students be 100.
Then, number of boys = 70
∴ Number of girls = (100 − 70) = 30

Now, total number of students when the number of girls is 30 = 100
Then, total number of students when the number of girls is 504 = $\left(\frac{100}{30}\times 504\right)$ = 1680
∴ Number of boys = (1680 − 504) = 1176

Hence, there are 1176 boys in the school.

Question 6:

An ore contains 12% copper. How many kilograms of the ore are required to get 69 kg of copper?

Answer 6:

Let x kg be the amount of the required ore.

Then, 12% of x kg = 69 kg
⇒ $\left(\frac{12}{100}\times x\right)$ kg = 69 kg
⇒ x = $\left(\frac{69\times 100}{12}\right)$ kg = 575 kg

Hence, 575 kg of ore is required to get 69 kg of copper.

Question 7:

36% of the maximum marks are required to pass a test. A student gets 123 marks and is declared failed by 39 marks. Find the maximum marks.

Answer 7:

Let x be the maximum marks.
Pass marks = (123 + 39) = 162
Then, 36% of x = 162
⇒ $\left(\frac{36}{100}\times x\right)=162$
⇒ x = $\left(\frac{162\times 100}{36}\right)$ = 450
∴ Maximum marks = 450

Question 8:

A fruit-seller had some apples. He sells 40% of them and still has 420 apples. Find the number of apples he had originally.

Answer 8:

Suppose that the fruit seller initially had 100 apples.
Apples sold = 40
∴ Remaining apples = (100 − 40) = 60

Initial amount of apples if 60 of them are remaining = 100
Initial amount of apples if 1 of them is remaining = $\left(\frac{100}{60}\right)$
Initial amount of apples if 420 of them are remaining = $\left(\frac{100}{60}\times 420\right)$ = 700
Hence, the fruit seller originally had 700 apples.

Question 9:

In an examination, 72% of the total examinees passed. If the number of failures is 392, find the total number of examinees.

Answer 9:

Suppose that 100 candidates took the examination.
Number of passed candidates = 72
Number of failed candidates = (100 − 72) = 28

Total number of candidates if 28 of them failed = 100
Total number of candidates if 392 of them failed = $\left(\frac{100}{28}\times 392\right)$ = 1400
Hence, the total number of examinees is 1400.

Question 10:

After decuting a commission of 5%, a moped costs Rs 15200. What is its gross value?

Answer 10:

Suppose that the gross value of the moped is Rs x.
Commission on the moped = 5%
Price of moped after deducting the commission = Rs ( x − 5% of x)
                                                                              = Rs $\left(x-\frac{5x}{100}\right)$ = Rs $\left(\frac{100x-5x}{100}\right)$ = Rs $\left(\frac{95x}{100}\right)$
Now, price of the moped after deducting the commission = Rs 15200
Then, Rs $\left(\frac{95x}{100}\right)$= Rs 15200
∴ x  = Rs $\left(\frac{15200\times 100}{95}\right)$ = Rs (160 $\times$ 100) = Rs 16000
Hence, the gross value of the moped is Rs 16000.

Question 11:

Gunpowder contains 75% of nitre and 10% of sulphur, and the rest of it is charcoal. Find the amount of charcoal in 8 kg of gunpowder.

Answer 11:

Total quantity of gunpowder = 8 kg = 8000 g                         (1 kg = 1000 g)
Quantity of nitre in it = 75% of 8000 g
                                   = $\left(\frac{75}{100}\times 8000\right)$ g = 6000 g = 6 kg

Quantity of sulphur in it = 10% of 8000 g
                                         = $\left(\frac{10}{100}\times 8000\right)$ g = 800 g = 0.8 kg
∴ Quantity of charcoal in it = {8000 − (6000 + 800)} g
                                              = (8000 − 6800) g
                                              = 1200 g = 1.2 kg

Hence, the amount of charcoal in 8 kg of gunpowder is 1.2 kg.

Question 12:

Chalk contains 3% of carbon, 10% of calcium and 12% of oxygen. Find the amount in grams of each of these substances in 1 kg of chalk.

Answer 12:

Total quantity of chalk = 1 kg = 1000 g

Now, we have the following:

Quantity of carbon in it = 3% of 1000 g
                                       =$\left(\frac{3}{100}\times 1000\right)$ = 30 g
Quantity of calcium in it = 10% of 1000 g
                                          = $\left(\frac{10}{100}\times 1000\right)$ g = 100 g
Quantity of oxygen in it = 12% of 1000 g
                                        = $\left(\frac{12}{100}\times 1000\right)$ g = 120 g

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Question 13:

Sonal went to school for 219 days in a full year. If her attendance is 75%, find the number of days on which the school was open.

Answer 13:

Let x be the total number of days on which the school was open.
Number of days when Sonal went to school = 219
Percentage of attendance = 75

Thus, 75% of x = 219
⇒ $\left(\frac{75}{100}\times x\right)=219$
∴ x = $\left(\frac{219\times 100}{75}\right)=292$ days
Hence, the school was open for a total of 292 days.

Question 14:

3% commission on the sale of a property amounts to Rs 42660. What is the total value of the property?

Answer 14:

Let the total value of the property be Rs x.
Percentage of commission = 3
Amount of commission  = Rs 42660
Thus, 3% of Rs x = Rs 42660
⇒ $\left(\frac{3}{100}\times x\right)$ = 42660
∴ x = $\left(\frac{42660\times 100}{3}\right)=1422000$
Hence, the total value of the property is Rs 14,22,000.

Question 15:

In an election, there were two candidates A and B. The total number of voters in this constituency was 60000 and 80% of the total votes were polied. If 60% of the polied votes were cast in favour of A, how many votes were received by B?

Answer 15:

Total number of eligible voters = 60000
Number of voters who gave their votes = 80% of 60000
                                                                 = $\left(\frac{80}{100}\times 60000\right)$ = 48000
Number of votes in favour of candidate A = 60% of 48000
                                                                    = $\left(\frac{60}{100}\times 48000\right)$ = 28800
∴ Number of votes received by candidate B = (48000 − 28800) = 19200

Hence, candidate B recieved 19,200 votes.

Question 16:

The price of a shirt is reduced by 12% in a discount sale. If its present price is Rs 1188, find its original price.

Answer 16:

Let us assume that the original price of the shirt is Rs x.
Discount on the shirt = 12%
So, value of discount on the shirt = 12% of Rs x
                                                      = Rs $\left(\frac{12}{100}\times x\right)$ = Rs $\left(\frac{12x}{100}\right)$
Value of the shirt after discount = Rs $\left(x-\frac{12x}{100}\right)$
                                                      = Rs $\left(\frac{100x-12x}{100}\right)$ = Rs $\left(\frac{88x}{100}\right)$
Present price of the shirt = Rs 1188
Then, Rs $\left(\frac{88x}{100}\right)$ = Rs 1188
     ⇒ 88x = (1188 $\times$ 100)
     ⇒ 88x = 118800
∴  x = $\left(\frac{118800}{88}\right)$ = 1350

Hence, the original price of the shirt is Rs 1350.

Question 17:

The price of a sweater is increased by 8%. If its increased price is Rs 1566, find the original price.

Answer 17:

Let us assume that the original price of the sweater is Rs. x
Increased percentage = 8%
So, value of increase on the sweater = 8% of Rs x
                                                             = Rs $\left(\frac{8}{100}\times x\right)$ = Rs $\left(\frac{2x}{25}\right)$
Increased price of the sweater = Rs $\left(x+\frac{2x}{25}\right)$
                                                  = Rs $\left(\frac{25x+2x}{25}\right)$ = Rs $\left(\frac{27x}{25}\right)$
However, increased price of the sweater = Rs 1566
Then, Rs $\left(\frac{27x}{25}\right)$ = Rs 1566
∴  x =  $\left(\frac{1566\times 25}{27}\right)$ = 1450
Hence, the original price of the sweater is Rs 1450

Question 18:

After spending 80% of his income and giving 10% of the remainder in a charity, a man has Rs 46260 left with him. Find his income.

Answer 18:

Let the income of the man be Rs x.
Then, income spent = 80% of Rs. x
                                = Rs $\left(\frac{80}{100}\times x\right)$ = Rs $\left(\frac{80x}{100}\right)$ = Rs $\left(\frac{4x}{5}\right)$
Amount left after all the expenditure = Rs $\left(x-\frac{4x}{5}\right)$ = Rs $\left(\frac{5x-4x}{5}\right)$ = Rs $\left(\frac{x}{5}\right)$
Amount given to the charity = 10% of Rs $\left(\frac{x}{5}\right)$
                                       = Rs $\left(\frac{10}{100}\times \frac{x}{5}\right)$ = Rs $\left(\frac{10x}{500}\right)$= Rs $\left(\frac{x}{50}\right)$
Amount left after the charity = Rs $\left(\frac{x}{5}-\frac{x}{50}\right)$
                                                = Rs $\left(\frac{10x-x}{50}\right)$ = Rs $\left(\frac{9x}{50}\right)$
Now, we have:
Rs $\left(\frac{9x}{50}\right)$ = Rs 46260
∴ x = Rs $\left(\frac{46260\times 50}{9}\right)$ = Rs 257000
Hence, the income of the man is Rs 2,57,000.

Question 19:

A number is increased by 20% and the increased number is decreased by 20%. Find the net increase of decrease per cent.

Answer 19:

Let the number be 100.
Increase in the number = 20%
Increased number = (100 + 20) =120
Now, decrease in the number = (20% of 120)
                                                = $\left(\frac{20}{100}\times 120\right)=24$
New number = (120 − 24) = 96
Net decrease = (100 − 96) = 4
Net decrease percentage = $\left(\frac{4}{100}\times 100\right)$ = 4
Hence, the net decrease is 4%.

Question 20:

The salary of an officer is increased by 20%. By what percentage should the new salary be reduced to restore the original salary?

Answer 20:

Let the original salary be Rs 100.
Increase in it = 20%
Salary after increment = Rs (100 + 20) = Rs 120
To restore the original salary, reduction required = Rs (120 − 100) = Rs 20
Reduction on Rs 120 = Rs 20
∴ Reduction percentage = $\left(\frac{20}{120}\times 100\right)$ = $\left(\frac{100}{6}\right)$ = $16\frac{2}{3}$
Hence, the required reduction on the new salary is $16\frac{2}{3}\%$.

Question 21:

A property dealer charges commission at the rate of 2% on the first Rs 200000, 1% on the next Rs 200000 and 0.5% on the remaining price. Find his commission on the property that has been sold for Rs 540000.

Answer 21:

Total cost of the property = Rs 540000
Commission on the first Rs 200000 = 2% of Rs 200000
                                                            = $\left(\frac{2}{100}\times 200000\right)$ = Rs 4000

Commission on the next Rs 200000 = 1% of Rs 200000
                                                           = $\left(\frac{1}{100}\times 200000\right)$ = Rs 2000
Remaining amount = Rs (540000 − 400000) = Rs 140000
∴ Commission on Rs 140000 = 0.5% of Rs 140000
                                                  = Rs $\left(\frac{0.5}{100}\times 140000\right)$
                                                  = Rs $\left(\frac{5}{1000}\times 140000\right)$ = Rs 700
Thus, total commission on the property worth Rs 540000 = Rs (4000 + 2000 + 700)
                                                                                             = Rs 6700
Hence, the commission of the property dealer on the property that has been sold for Rs 540000 is Rs 6700.

Question 22:

Nikhil's income is 20% less than that of Akhil. How much per cent is Akhil's income more than that of Nikhil's?

Answer 22:

Let Akhil's income be Rs 100.
∴ Nikhil's income = Rs 80
Akhil's income when Nikhil's income is Rs 80 = Rs 100
Akhil's income when Nikhil's income is Rs 100 = Rs $\left(\frac{100}{80}\times 100\right)$ = Rs 125
i.e., if Nikhil's income is Rs.100, then Akhil's income is Rs 125.
Hence, Akhil's income is more than that of Nikhil's by 25%.

Question 23:

Jhon's income is 20% more than that of Mr Thomas. How much per cent is the income of Mr Thomas less than that of John?

Answer 23:

Let Rs 100 be the income of Mr. Thomas.
∴ John's income = Rs 120
Mr. Thomas' income when John's income is Rs 120 = Rs 100
Mr. Thomas' income when John's income is Rs 100 = Rs $\left(\frac{100}{120}\times 100\right)$ = Rs $83\frac{1}{3}$
Hence, Mr Thomas' income is less than that of John's by $16\frac{2}{3}\%$.

Question 24:

The value of a machine depreciated 10% every year. If its present value is Rs 387000, what was its value 1 year ago?

Answer 24:

Let Rs x be the value of the machine one year ago.

Then, its present value = 90% of Rs x
                                     = Rs $\left(\frac{90}{100}\times x\right)$ = Rs $\left(\frac{9x}{10}\right)$
It is given that present value of the machine = Rs 387000
⇒ x = Rs$\left(\frac{387000\times 10}{9}\right)$ = Rs $\left(43000\times 10\right)$ = Rs 430000

Hence, the value of the machine a year ago was Rs 430000.

Question 25:

The value of a car decreases annually by 20%. If the present value of the car be Rs 450000, what will be its value after 2 years?

Answer 25:

The present value of  the car = Rs 450000
The decrease in its value after the first year = 20% of Rs 450000
                                                                         = Rs $\left(\frac{20}{100}\times 450000\right)$= Rs 90000
The depreciated value of the car after the first year = Rs (450000 − 90000) = Rs 360000
The decrease in its value after the second year = 20% of Rs 360000
                                                                            = Rs $\left(\frac{20}{100}\times 360000\right)$ = Rs 72000
The depreciated value of the car after the second year = Rs (360000 − 72000) = Rs 288000

Hence, the value of the car after two years will be Rs 288000.

Question 26:

The population of a town increases 10% annually. If its present population is 60000, what will be its population after 2 years?

Answer 26:

Present population of the town = 60000
Increase in population of the town after the 1 year = 10% of 60000
                                                                              = $\left(\frac{10}{100}\times 60000\right)$ = 6000
Thus, population of the town after 1 year = 60000 + 6000 = 66000
Increase in population after 2 years = 10% of 66000
                                                           = $\left(\frac{10}{100}\times 66000\right)$ = 6600
Thus, population after the second year = 66000 + 6600 = 72600
Hence, the population of the town after 2 years will be 72600.

Question 27:

Due to an increase in the price of sugar by 25%, by how much per cent must a householder decrease the consumption of sugar so that there is no increase in the expenditure on sugar?

Answer 27:

Let the consumption of sugar originally be 1 unit and let its cost be Rs 100
New cost of 1 unit of sugar = Rs 125
Now, Rs 125 yield 1 unit of sugar.
∴ Rs 100 will yield $\left(\frac{1}{125}\times 100\right)$ unit = $\left(\frac{4}{5}\right)$ unit of sugar.
Reduction in consumption = $\left(1-\frac{4}{5}\right)$ = $\left(\frac{1}{5}\right)$ unit
∴ Reduction percent in consumption = $\left(\frac{1}{5}\times \frac{1}{1}\times 100\right)$ %= $\left(\frac{100}{5}\right)$ %= 20%