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Page No 142:

Question 1:

Convert each of the following fraction into a percentage:

(i) 47100
(ii)  920
(iii) 38
(iv) 8125
(v) 19500
(vi) 415
(vii) 23
(viii) 135

Answer:

We have the following:

(i) 47100=47100×100% = 47%

(ii) 920=920×100% =9×5%= 45%

(iii) 38=38×100% =3×252%=752%=3712%

(iv) 8125=8125×100%=8×45%=325%=6.4%

(v) 19500=19500×100%=195%=3.8%

(vi) 415=415×100%=4×203%=803%=2623%

(vii) 23=23×100%=2003%=6623%

(viii) 135=85=85×100%=(8×20)%=160%

Page No 142:

Question 2:

Convert each of the following into a fraction:

(i) 32%
(ii) 614%
(iii) 2623%
(iv) 120%
(v) 6.25%
(vi) 0.8%
(vii) 0.06%
(viii) 22.75%

Answer:

We have the following:

(i) 32%=32100=825

(ii) 614%=254%=254×1100=116

(iii) 2623%=803%=803×1100=4×13×5=415

(iv) 120%=120100=65=115

(v) 6.25%=6.25100=625100×100=25400=116

(vi) 0.8%=0.8100=810×100=81000=1125

(vii) 0.06%=0.06100=6100×100=610000=35000

(viii)  22.75%=22.75100=2275100×100=91400

Page No 142:

Question 3:

Express each of the following as a ratio:

(i) 43%
(ii) 36%
(iii) 7.5%
(iv) 125%

Answer:

We have:

(i) 43%=43100=43 : 100

(ii) 36%=36100=925=9 : 25

(iii) 7.5%=7.5100=7510×100=340=3 : 40

(iv) 125%=125100=54=5 : 4

Page No 142:

Question 4:

Convert each of the following ratios into a percentage:

(i) 37 : 100
(ii) 16 : 25
(iii) 3 : 5
(iv) 5 : 4

Answer:

We have the following:

(i) 37 : 100 = 37100=37100×100%=37%

(ii) 16 : 25 = 1625=1625×100%=16×4%=64%

(iii) 3 : 5 =35=35×100%=3×20%=60%

(iv) 5 : 4 = 54=54×100%=5×25%=125%

Page No 142:

Question 5:

Convert each of the following into decimal form:

(i) 45%
(ii) 127%
(iii) 3.6%
(iv) 0.23%

Answer:

We have the following:

(i) 45% = 45100=0.45

(ii) 127% = 127100=1.27

(iii) 3.6% =3.6100=3610×100=361000=0.036

(iv) 0.23% =0.23100=23100×100=2310000=0.0023

Page No 142:

Question 6:

Convert each of the following decimals into a percentage:

(i) 0.6
(ii) 0.42
(iii) 0.07
(iv) 0.005

Answer:

We have:

(i) 0.6 = (0.6 × 100)% = 60%
(ii) 0.42 = (0.42 × 100)% = 42%
(iii) 0.07 = (0.07 × 100)% = 7%
(iv) 0.005 = (0.005 × 100)% = 0.5%

Page No 142:

Question 7:

Find:

(i) 32% of 425
(ii) 1623% of 16
(iii) 6.5% of 400
(iv) 136% of 70
(v) 2.8% of 35
(vi) 0.6% of 45

Answer:

We have:
 (i) 32% of 425 = 32100×425=32×174=8×17=136

(ii) 1623% of 16 = 503% of 16 = 503×100×16=16×16=83=223

(iii) 6.5% of 400 = 6.5100×400=6510×100×400=65×410=26010=26

(iv) 136% of 70 = 136100×70=136×710=95210=95.2

(v) 2.8% of 35 = 2.8100×35=2810×100×35=14×7100=98100=0.98

(vi) 0.6% of 45  = 0.6100×45=610×100×45=3×455×100=3×9100=27100=0.27

Page No 142:

Question 8:

Find:

(i) 25% of Rs 76
(ii) 20% of Rs 132
(iii) 7.5% of 600 m
(iv) 313% of 90 km
(v) 8.5% of 5 kg
(vi) 20% of 12 litres

Answer:

We have the following:

(i) 25% of Rs 76 = Rs 76×25100 = Rs 76×14 = Rs 19

(ii) 20% of Rs 132 = Rs 132×20100= Rs 132×15 = Rs 26.4

(iii) 7.5% of 600 m = 600×7.5100 m=6×7.5 m=45 m

(iv) 313% of 90 km = 103%of 90 km =90×103×100 km=90×130 km=3 km

(v) 8.5% of 5 kg = 5×8.5100 kg=5×851000 kg=0.425 kg=425 g         [∵1 kg = 1000 g]

(vi) 20% of 12 L = 12×20100 L=12×15 L=2.4 L

Page No 142:

Question 9:

Find the number whose 13% is 65.

Answer:

Let x be the required number.

Then, 13% of x = 65  
13100×x=65
x = 65×10013=500
Hence, the required number is 500.

Page No 142:

Question 10:

Find the number whose 614% is 2.

Answer:

Let x be the required number.
Then, 614% of x = 2
614%×x=2
25400×x=2
⇒  x = 2×40025=32
Hence, the required number is 32.



Page No 143:

Question 11:

What amount is 10% more than Rs 90?

Answer:

10% of Rs 90 = Rs 10100×90 = Rs 9
∴ Amount that is 10% more than Rs 90 = Rs (90 + 9) = Rs 99

Hence, the required amount is Rs 99.

Page No 143:

Question 12:

What amount is 20% less than Rs 60?

Answer:

20% of Rs 60 = Rs 60×20100= Rs 12
∴ Amount that is 20% less than Rs 60 = Rs (60 − 12) = Rs 48

Hence, the required amount is Rs 48.

Page No 143:

Question 13:

If 3% of x is 9, find the value of x.

Answer:

3% of x = 9
3100×x=9
x = 9×1003=300
Hence, the value of x is 300.

Page No 143:

Question 14:

If 12.5% of x is 6, find the value of x.

Answer:

12.5% of x = 6
12.5100×x=6
x = 6×10012.5=6×8=48
Hence, the value of x is 48.

Page No 143:

Question 15:

What per cent of 84 is 14?

Answer:

Let x% of 84 be 14.
Then, x100×84=14
21x25=14
x = 14×2521=2×253=503=1623%
Hence, 1623% of 84 is 14.

Page No 143:

Question 16:

What percentage is

(i) Rs 15 of Rs 120?
(ii) 36 minutes of 2 hours?
(iii) 8 hours of 2 days
(iv) 160 metres of 4 km?
(v) 175 mL of 1 litre?
(vi) 25 paise of Rs 4?

Answer:

(i) Let x% of Rs 120 be Rs 15.
Then, Rs x100×120 = Rs 15  
    ⇒ 6x5 = 15
    ∴ x = 15×56% = 252% = 12.5%
Hence, 12.5% of Rs 120 is Rs 15.

(ii) Let x% of 2 h be 36 min.
Then, x100×2×60 min = 36 min
      ⇒ 120x100 = 36
     ∴ x = 36×100120% = 30%
Hence, 30% of 2 h is 36 min.

(iii) Let x% of 2 days be 8 h.
Then, x100×2×24 h = 8 h
     ⇒ 48x100 = 8
     ∴ x = 8×10048% = 1623%
     Hence, 1623% of 2 days is 8 h.

(iv) Let x% of 4 km be 160 m.
Then, x100×4×1000 m = 160 m
     ⇒ 40x = 160
     ∴ x = 16040% = 4%
Hence, 4% of 4 km is 160 m.

(v) Let x% of 1 L be 175 mL.
Then, x100×1×1000 mL = 175 mL
     ⇒ 10x  = 175
     ∴ x = 17510% = 17.5%
Hence, 17.5% of 1 L is 175 mL.

(vi) Let x% of Rs 4 be 25 paise.
Then, x100×4×100 paise = 25 paise
     ⇒ 4x  = 25
     ∴ x = 254% = 614%
     Hence, 614% of Rs 4 is 25 paise.



Page No 147:

Question 1:

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

Answer:

Maximum marks of the examination = 750
Marks secured by Rupesh = 495
Percentage of marks secured = 495750×100% = 66%

Hence, Rupesh scored 66% in the examination.

Page No 147:

Question 2:

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

Answer:

Total monthly salary = Rs 15625
Increase percentage = 12%
∴ Amount increase = 12% of Rs 15625
                                 = Rs 15625×12100 = Rs 1875
∴ New salary = Rs 15625 + Rs 1875
                       = Rs 17500
Hence, the new salary of the typist is Rs 17,500.

Page No 147:

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:

Original excise duty on the item = Rs 950
Amount reduced on excise duty = Rs (950 − 760) = Rs 190
∴ Reduction percent = Reduction amountOriginal value×100
                                  = 190950×100 = 20
Hence, the excise duty on that item is reduced by 20%.

Page No 147:

Question 4:

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

Answer:

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
96100×x = 10464
x = 10464×10096 =  10900
Hence, the total cost of the TV set is Rs 10900.

Page No 147:

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:

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 = 10030×504 = 1680
∴ Number of boys = (1680 − 504) = 1176

Hence, there are 1176 boys in the school.

Page No 147:

Question 6:

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

Answer:

Let x kg be the amount of the required ore.

Then, 12% of x kg = 69 kg
12100×x kg = 69 kg
x = 69×10012 kg = 575 kg

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

Page No 147:

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:

Let x be the maximum marks.
Pass marks = (123 + 39) = 162
Then, 36% of x = 162
36100×x=162
x162×10036 = 450
∴ Maximum marks = 450

Page No 147:

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:

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 = 10060
Initial amount of apples if 420 of them are remaining = 10060×420 = 700
Hence, the fruit seller originally had 700 apples.

Page No 147:

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:

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 = 10028×392 = 1400
Hence, the total number of examinees is 1400.

Page No 147:

Question 10:

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

Answer:

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 x-5x100 = Rs 100x-5x100 = Rs 95x100
Now, price of the moped after deducting the commission = Rs 15200
Then, Rs 95x100= Rs 15200
x  = Rs 15200×10095 = Rs (160 × 100) = Rs 16000
Hence, the gross value of the moped is Rs 16000.

Page No 147:

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:

Total quantity of gunpowder = 8 kg = 8000 g                         (1 kg = 1000 g)
Quantity of nitre in it = 75% of 8000 g
                                   = 75100×8000 g = 6000 g = 6 kg

Quantity of sulphur in it = 10% of 8000 g
                                         = 10100×8000 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.

Page No 147:

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:

Total quantity of chalk = 1 kg = 1000 g

Now, we have the following:

Quantity of carbon in it = 3% of 1000 g
                                       =3100×1000 = 30 g
Quantity of calcium in it = 10% of 1000 g
                                          = 10100×1000 g = 100 g
Quantity of oxygen in it = 12% of 1000 g
                                        = 12100×1000 g = 120 g



Page No 148:

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:

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
75100×x=219
x = 219×10075=292 days
Hence, the school was open for a total of 292 days.

Page No 148:

Question 14:

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

Answer:

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
3100×x = 42660
x = 42660×1003=1422000
Hence, the total value of the property is Rs 14,22,000.

Page No 148:

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:

Total number of eligible voters = 60000
Number of voters who gave their votes = 80% of 60000
                                                                 = 80100×60000 = 48000
Number of votes in favour of candidate A = 60% of 48000
                                                                    = 60100×48000 = 28800
∴ Number of votes received by candidate B = (48000 − 28800) = 19200

Hence, candidate B recieved 19,200 votes.

Page No 148:

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:

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 12100×x = Rs 12x100
Value of the shirt after discount = Rs x-12x100
                                                      = Rs 100x-12x100 = Rs 88x100
Present price of the shirt = Rs 1188
Then, Rs 88x100 = Rs 1188
     ⇒ 88x = (1188 × 100)
     ⇒ 88x = 118800
∴  x = 11880088 = 1350

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

Page No 148:

Question 17:

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

Answer:

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 8100×x = Rs 2x25
Increased price of the sweater = Rs x+2x25
                                                  = Rs 25x+2x25 = Rs 27x25
However, increased price of the sweater = Rs 1566
Then, Rs 27x25 = Rs 1566
∴  x1566×2527 = 1450
Hence, the original price of the sweater is Rs 1450

Page No 148:

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:

Let the income of the man be Rs x.
Then, income spent = 80% of Rs. x
                                =
Rs 80100×x = Rs 80x100 = Rs 4x5
Amount left after all the expenditure = Rs x-4x5 = Rs 5x-4x5 = Rs x5
Amount given to the charity = 10% of Rs x5
                                       = Rs 10100×x5 = Rs 10x500= Rs x50
Amount left after the charity = Rs x5-x50
                                                = Rs 10x-x50 = Rs 9x50
Now, we have:
Rs 9x50 = Rs 46260
x = Rs 46260×509 = Rs 257000
Hence, the income of the man is Rs 2,57,000.

Page No 148:

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:

Let the number be 100.
Increase in the number = 20%
Increased number = (100 + 20) =120
Now, decrease in the number = (20% of 120)
                                                = 20100×120=24
New number = (120 − 24) = 96
Net decrease = (100 − 96) = 4
Net decrease percentage = 4100×100 = 4
Hence, the net decrease is 4%.

Page No 148:

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:

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 = 20120×100 = 1006 = 1623
Hence, the required reduction on the new salary is 1623%.

Page No 148:

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:

Total cost of the property = Rs 540000
Commission on the first Rs 200000 = 2% of Rs 200000
                                                            = 2100×200000 = Rs 4000

Commission on the next Rs 200000 = 1% of Rs 200000
                                                           = 1100×200000 = Rs 2000
Remaining amount = Rs (540000 − 400000) = Rs 140000
∴ Commission on Rs 140000 = 0.5% of Rs 140000
                                                  = Rs 0.5100×140000
                                                  = Rs 51000×140000 = 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.

Page No 148:

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:

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 10080×100 = 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%.

Page No 148:

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:

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 100120×100 = Rs 8313
Hence, Mr Thomas' income is less than that of John's by 1623%.

Page No 148:

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:

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

Then, its present value = 90% of Rs x
                                    
= Rs 90100×x = Rs 9x10
It is given that present value of the machine = Rs 387000
x = Rs387000×109 = Rs 43000×10 = Rs 430000

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

Page No 148:

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:

The present value of  the car = Rs 450000
The decrease in its value after the first year = 20% of Rs 450000
                                                                         = Rs 20100×450000= 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 20100×360000 = 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.

Page No 148:

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:

Present population of the town = 60000
Increase in population of the town after the 1 year = 10% of 60000
                                                                              = 10100×60000 = 6000
Thus, population of the town after 1 year = 60000 + 6000 = 66000
Increase in population after 2 years = 10% of 66000
                                                           = 10100×66000 = 6600
Thus, population after the second year = 66000 + 6600 = 72600
Hence, the population of the town after 2 years will be 72600.

Page No 148:

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:

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 1125×100 unit = 45 unit of sugar.
Reduction in consumption = 1-45 = 15 unit
∴ Reduction percent in consumption = 15×11×100 %= 1005 %= 20%

Page No 148:

Question 1:

Mark (✓) against the correct answer
34 as rate per cent is

(a) 7.5%
(b) 75%
(c) 0.75%
(d) none of these

Answer:

(b) 75%

34 = 34×100% = 75%

Page No 148:

Question 2:

Mark (✓) against the correct answer
The ratio 2 : 5 as rate per cent is

(a) 4%
(b) 0.4%
(c) 40%
(d) 14%

Answer:

(c) 40%

2 : 5 = 25 = 25×100% = 40%

Page No 148:

Question 3:

Mark (✓) against the correct answer
813% expressed as a fraction, is

(a) 253
(b) 325
(c) 112
(d) 14

Answer:

(c) 112

813% = 253% = 253×1100=13×4=112



Page No 149:

Question 4:

Mark (✓) against the correct answer
If x% of 75 = 9, then the value of x is

(a) 16
(b) 14
(c) 12
(d) 8

Answer:

(c) 12
We have x% of 75 = 9
 x100×75=9
x = 9×10075=12
Hence, the value of x is 12

Page No 149:

Question 5:

Mark (✓) against the correct answer
What per cent of 27 is 135?

(a) 25%
(b) 20%
(c) 15%
(d) 10%

Answer:

(d) 10%

Let x be the required percent.
Then, x % of 27 = 135
x100×27=135
x = 100×735×2 = 10
Hence, 10% of 27 is 135

Page No 149:

Question 6:

Mark (✓) against the correct answer
What per cent of 1 day is 36 minutes?

(a) 25%
(b) 2.5%
(c) 3.6%
(d) 0.25%

Answer:

(b) 2.5%

Let x % of 1 day be 36 min.
Then, x100×1×24×60 min = 36 min
x = 36×10024×60 = 3×52×3%=52%=2.5% 

Hence, 2.5% of 1 day is 36 min.

Page No 149:

Question 7:

Mark (✓) against the correct answer
A number increased by 20% gives 42. The number is

(a) 35
(b) 28
(c) 36
(d) 30

Answer:

(a) 35
Let the required number be x.
Then, x + 20% of x = 42
x+20x100=42
x+x5=42
5x+x5=42        [∵ LCM of 1 and 5 = 5]
6x5=42
x = 42×56=35
Hence, the required number is 35.

Page No 149:

Question 8:

Mark (✓) against the correct answer
A number decreased by 8% gives 69. The number is

(a) 80
(b) 75
(c) 85
(d) none of these

Answer:

(b) 75
Let the required number be x.
Then, x − 8% of x = 69
x-8x100 = 69
x-2x25 = 69
⇒ 25x-2x25 = 69            [Since L.C.M. of 1 and 25 = 25]
23x25=69
x = 69×2523 = 75
Hence, the required number is 75

Page No 149:

Question 9:

Mark (✓) against the correct answer
An ore contains 5% copper. How much ore is required to obtain 400 g of copper?

(a) 2 kg
(b) 4 kg
(c) 6 kg
(d) 8 kg

Answer:

(d) 8 kg
Let x kg be the required amount of ore.
Then, 5% of x kg = 400 g = 0.4 kg        [∵ 1 kg = 1000 g]
5100×x=0.4
x = 0.4×1005 = 8
Hence, 8 kg of ore is required to obtain 400 g of copper.

Page No 149:

Question 10:

Mark (✓) against the correct answer
After deducting a commission of 10% a TV costs Rs 18000. What is its gross value?

(a) Rs 18800
(b) Rs 20000
(c) Rs 19800
(d) none of these

Answer:

(b) Rs. 20000
Suppose that the gross value of the TV is Rs x.
Commission on the TV = 10%
Price of the TV after deducting the commission = Rs (x − 10% of x)
                                                                  = Rs x-10100x = Rs 100x-10x100 = Rs 9x10
However, price of the TV after deducting the commission = Rs 18000
Then, Rs 9x10 = Rs 18000
x = 18000×109 = Rs (2000 × 10) = Rs 20000
Hence, the gross value of the TV is Rs 20,000

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Question 11:

Mark (✓) against the correct answer
On increasing the salary of a man by 25%, it becomes Rs 20000. What was his original salary?

(a) Rs 15000
(b) Rs 16000
(c) Rs 18000
(d) Rs 25000

Answer:

(b) Rs. 16000
Let us assume that the original salary of the man is Rs x.
Increase in it = 25%
Value increased in the salary = 25% of Rs. x
                                               = Rs 25100×x = Rs x4
Salary after increment= Rs x+x4 = Rs 5x4
However, increased salary = Rs 20000
Then, Rs 5x4 = Rs 20000
∴  x = Rs 20000×45 = Rs 16000
Hence, the original salary of the man is Rs 16,000

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Question 12:

Mark (✓) against the correct answer
In an examination, 95% of the total examinees passed. If the number of failures is 28, how many examinees were there?

(a) 600
(b) 480
(c) 560
(d) 840

Answer:

(c) 560
Suppose that the number of examinees is 100.
Number of passed examinees = 95
Number of failed examinees = (100 − 95) = 5

Total number of examinees if 5 of them failed = 100
Total number of examinees if 28 of them failed = 1005×28=20×28=560
Hence, there were 560 examinees.

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Question 13:

Mark (✓) against the correct answer
A fruit-seller had some apples. He sells 40% of them and still has 420 apples. How many apples had he in all?

(a) 588
(b) 600
(c) 700
(d) 725

Answer:

(c) 700
Suppose that the fruit seller initially had 100 apples.
Number of apples sold = 40
∴ Number of remaining apples = (100 − 40) = 60

Initial number of apples if 60 of them are remaining = 100
Initial number of apples if 420 of them are remaining = 10060×420 = 700
Hence, the fruit seller originally had 700 apples with him.

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Question 14:

Mark (✓) against the correct answer
The value of a machine depreciated 10% annually. If its present value is Rs 25000, what will be its value after 1 year?

(a) Rs 27500
(b) Rs 22500
(c) Rs 25250
(d) none of these

Answer:

(c) Rs. 25250

Present value of the machine = Rs 25000
Decrease in its value after 1 year = 10% of Rs 25000
                                                      = Rs 10100×25000 = Rs 2500
Depreciated value after 1 year = Rs (25000 − 2500) = Rs 22500

Hence, the value of the machine after 1 year will be Rs 22500

Page No 149:

Question 15:

Mark (✓) against the correct answer
8% of a number is 6. What is the number?

(a) 48
(b) 96
(c) 75
(d) 60

Answer:

(c) 75

Let the required number be x. Then, we have:
8% of x = 6
8100×x=6
x = 6×1008=75
Hence, the required number is 75

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Question 16:

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60% of 450 = ?

(a) 180
(b) 210
(c) 270
(d) none of these

Answer:

(c) 270
60% of 450 = 60100×450
                    = 35×450 = (3 × 90) = 270

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Question 17:

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On reducing the value of a chair by 6% it becomes Rs 658. The original value of the chair is

(a) Rs 750
(b) Rs 720
(c) Rs 500
(d) Rs 700

Answer:

(d) Rs. 700
Let us assume that the original price of the chair is Rs x.
Reduce percentage on the chair = 6%
So, value of reduction on the chair = 6% of Rs. x
                                                        = Rs 6100×x = Rs 3x50
Reduced price of the chair = Rs x-3x50
                                            = Rs 50x-3x50 = Rs 47x50
However, present price of the chair = Rs 658
Then, Rs 47x50 = Rs 658
⇒ Rs 47x50 = Rs 658
x = Rs 658×5047 = Rs 14×50=700
Hence, the original price of the chair is Rs 700

Page No 149:

Question 18:

Mark (✓) against the correct answer
70% of students in a school are boys. If the number of girls is 240, how many boys are there in the school?

(a) 420
(b) 560
(c) 630
(d) 480

 

Answer:

(b) 560

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

Now, total number of students if there are 30 girls = 100
Total number of students if there are 240 girls = 10030×240=800
∴ Number of boys = (800 − 240) = 560

Hence, there are 560 boys in the school.

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Question 19:

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If 11% of a number exceeds 7% of the number by 18, the number is

(a) 72
(b) 360
(c) 450
(d) 720

Answer:

(c) 450

Let x be the number.
(11% of x) − (7% of x) = 18
11x100-7x100=18
4x100=18
x = 18×1004=18×25=450
Hence, the required number is 450

Page No 149:

Question 20:

Mark (✓) against the correct answer
If 35% of a number added to 39 is the number itself, the number is

(a) 60
(b) 65
(c) 75
(d) 105

Answer:

(a) 60
Let x be the number.
According to question, we have:
(35% of x ) + 39 = x
35100×x+39=x
7x20+39=x
x-7x20=39
20x-7x20=39
13x20=39
x = 39×2013 = 60
Hence, the required number is 60

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Question 21:

Mark (✓) against the correct answer
In an examination it is required to get 36% to pass. A student gets 145 marks and fails by 35 marks. The maximum marks are

(a) 400
(b) 450
(c) 500
(d) 600

Answer:

(c) 500
Let x be the maximum marks.
Pass marks = (145 + 35) = 180
∴ 36% of x = 180
36100×x=180
x = 180×10036=5×100=500
Hence, maximum marks = 500



Page No 150:

Question 22:

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A number decreased by 40% gives 135. The number is

(a) 175
(b) 200
(c) 250
(d) 225

Answer:

(d) 225
Let x be the number.
According to question, we have:
x − 40% of x = 135
x-40x100=135
100x-40x100=135
60x100=135
x = 135×10060 = 225
Hence, the required number is 225



Page No 151:

Question 1:

Convert:

(i) 45 into a percentage
(ii) 74 into a percentage
(iii) 45% into a percentage
(iv) 105% into a percentage
(v) 15% into a percentage
(vi) 12 : 25 into a percentage

Answer:

We have:

(i) 45= 45×100%=4×20%=80%

(ii) 74=74×100%=7×25%=175%

(iii) 45% = 45100=920

(iv) 105% =105100=2120

(v) 15% =15100=320=  3 : 20

(vi) 12 : 25 = 1225=1225×100%=12×4%=48%

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Question 2:

(i) What per cent of 1 kg is 125 g?
(ii) What per cent of 80 m is 24 m?

Answer:

(i) Let x% of 1 kg be 125g.
Then, x100×1×1000 g=125 g
     ⇒ 10x = 125
     ⇒ x =12510%=1212%
Hence, 1212% of 1 kg is 125 g.

(ii) Let x% of 80 m be 24 m.
Then, x100×80 m=24 m
     ⇒ 4x5 = 24
     ⇒ x =24×54%=30%
Hence, 30% of 80 m is 24 m.

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Question 3:

(i) Find 1623% of 30.
(ii) Find 15% of Rs 140

Answer:

(i) 1623% of 30 = 503% of 30
                          = 503×100×30
                          = 5

(ii) 15% of Rs 140 = Rs 15100×140
                              = Rs (3 × 7)
                              = Rs 21

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Question 4:

(i) Find the number whose 614% is 5.
(ii) Find 0.8% of 45.

Answer:

(i) Let x be the required number.
Then, 614% of x = 5
     ⇒ 254% of x = 5
     ⇒ 254×100×x=25
     ⇒x16=5
x = (5 × 16) = 80
Hence, the required number is 80.

(i) 0.8% of 45 =0.8100×45
                         =810×100×45
                         = 72200=36100=0.36
Hence, 0.8% of 45 is 0.36.

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Question 5:

A number is increased by 10% and the increased number is decreased by 10%. Show that the net decrease is 1%.

Answer:

Let x be the number.
The number is increased by 10%.
∴ Increased number = 110% of x = x×110100=11x10
The number is, then, decreased by 10%.

∴ Decreased number = 90% of 11x10 = 11x10×90100=99x100

Net decrease = x-99x100=100x-99x100=x100
Net decrease percentage = x100×1x×100=1

Page No 151:

Question 6:

The value of a machine depriciates at the rate of 10% per annum. If its present value is Rs 10000, what will be its value after 2 years?

Answer:

The present value of the machine = Rs 10000
The decrease in its value after the 1st year = 10% of Rs 10000
                                                            = Rs 10100×10000 = Rs 1000
The depreciated value of the machine after the 1st year = Rs (10000 − 1000) =Rs 9000
The decrease in its value after the 2nd year = 10% of Rs 9000
                                                               = Rs 10100×9000 = Rs 900
The depreciated value of the machine after the 2nd year = Rs (9000 − 900) = Rs 8100

Hence, the value of the machine after two years will be Rs 8100.

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Question 7:

The population of a town increases at 5% per annum. Its present population is 16000. What will be its population after 2 years?

Answer:

The present population of the town = 16000
Increase in population after 1 year = 5% of 16000
                                                          = 5100×16000 = 800
Thus, population after one year = 16000 + 800 = 16800
Increase in population after 2 years = 5% of 16800
                                                           = 5100×16800 = 840
Increased population after two years = 16800 + 840 = 17640

Hence, the population of the town after two years will be 17,640.

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Question 8:

The price of a teaset is increased by 5%. If the increased price is Rs 441, what is its original price?

Answer:

Let us assume that the original price of the tea set is Rs. x
Increase in it = 5%
So, value increased on the tea set = 5% of Rs. x
                                                      = Rs. 5100×x = Rs. x20
Then, increased price of the tea set = Rs. x+x20
                                                          = Rs. 20x+x20 = Rs. 21x20
However, increased price = Rs. 441
Then, Rs. 21x20 = Rs. 441
∴  x = 441×2021 = 420
Hence, the original price of the tea set is Rs 420

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Question 9:

Mark (✓) against the correct answer
614% expressed as a fraction is

(a) 18
(b) 116
(c) 425
(d) 125

Answer:

(b) 116

614% = 254% =254×100=116

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Question 10:

Mark (✓) against the correct answer
If x% of 75 = 12, then the value of x is

(a) 8
(b) 10
(c) 12
(d) 16

Answer:

(c) 12

Given that x% of 75 = 12
Then, x100×75=12
x = 12×10075 =16
Hence, the value of x is 16

Page No 151:

Question 11:

Mark (✓) against the correct answer
A number increased by 20% gives 30. The number is

(a) 150
(b) 6
(c) 25
(d) 60

Answer:

(c) 25

Let the number be x. Then, we have:
120% of x  = increased number
⇒ 30 = x×120100
⇒ 30 = 6x5
x = 30×56=25
Hence, the required number is 25

Page No 151:

Question 12:

Mark (✓) against the correct answer
5% of a number is 9. The number is

(a) 120
(b) 140
(c) 160
(d) 180

Answer:

(d) 180

Let the required number be x. Then, we have:
5% of x = 9
5100×x=9
x = 9×1005=9×20=180

Page No 151:

Question 13:

Mark (✓) against the correct answer
If 35% of a number added to 39 is the number itself, the number is

(a) 60
(b) 65
(c) 75
(d) 70

Answer:

(a) 60
Let the number be x.
According to question, we have:
(35% of x ) + 39 = x
35100×x+39=x
7x20+39=x
x-7x20=39
20x-7x20=39
13x20=39
x = 39×2013 = 60
Hence, the required number is 60.

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Question 14:

Mark (✓) against the correct answer
In an examination it is required to get 36% to pass. A student gets 160 marks and fails by 20 marks. The maximum marks are

(a) 400
(b) 450
(c) 500
(d) 600

Answer:

(c) 500
Let x be the maximum marks.
Pass marks = (160 + 20) = 180
∴ 36% of x = 180
36100×x=180
x = 180×10036=5×100=500
Hence, maximum marks = 500

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Question 15:

Fill in the blanks.

(i) 3 : 4 = (......)%
(ii) 0.75 = (......)%
(iii) 6% = ...... (express in decimals)
(iv) If x decreased by 40% gives 135, then x = ...... .
(v) (11% of x) − (7% of x) = 18 ⇨ x = ......

Answer:

We have the following:

(i) 3 : 4 = (75)%
     Explanation: 3 : 4 = 34 = 34×100%=3×25%=75%

(ii) 0.75 = (75)%
      Explanation: ( 0.75 × 100)% = 75%


(iii) 6% = 0.06 (expressed in decimals)
       Explanation: 6% = 6100=0.06

(iv) If x decreased by 40% gives 135, then x = 225
       Explanation:
       Let the number be x.
       According to question, we have:
       x − 40% of x = 135
       ⇒ x-40x100=135
       ⇒ 100x-40x100=135
       ⇒ 60x100=135
        ⇒ x = 135×10060 = 225

(v) (11% of x) − (7% of x) = 18
      ⇒ x = 450
      Explanation:
      
(11% of x) − (7% of x) = 18
        ⇒ 11x100-7x100=18
        ⇒ 4x100=18
         ∴ x = 18×1004=18×25=450

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Question 16:

Write 'T' for true and 'F' for false

(i) 34 as rate per cent is 75%
(ii) 1212% expressed as a fraction is 18.
(iii) 2 : 5 = 25%
(iv) 80 % of 450 = 360.
(v) 20% of 1 litre = 200 mL.

Answer:

(i) True (T)
     Justification: 34×100% = 75%

(ii) True (T)
     Justification: 1212%=252% = 252×100=18

(iii) False (F)
       Justification: 25 = 25×100% = 2×20% = 40%

(iv) True (T)
       Justification: 80% of 450 = 80100×450=80×92=40×9=360

(v) True (T)
      Justification: 20% of 1 L = 20% of 1000 mL
                                                     = 20100×1000 mL = 200 mL



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