Rs Aggarwal 2020 2021 for Class 7 Maths Chapter 10 - Percentage

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Page No 142:

Answer:

We have the following:

(i) $\frac{47}{100} = (\frac{47}{100} \times 100) % = 47 %$

(ii) $\frac{9}{20} = (\frac{9}{20} \times 100) % = (9 \times 5) % = 45 %$

(iii) $\frac{3}{8} = (\frac{3}{8} \times 100) % = (\frac{3 \times 25}{2}) % = (\frac{75}{2}) % = 37 \frac{1}{2} %$

(iv) $\frac{8}{125} = (\frac{8}{125} \times 100) % = (\frac{8 \times 4}{5}) % = (\frac{32}{5}) % = 6.4 %$

(v) $\frac{19}{500} = (\frac{19}{500} \times 100) % = (\frac{19}{5}) % = 3.8 %$

(vi) $\frac{4}{15} = (\frac{4}{15} \times 100) % = (\frac{4 \times 20}{3}) % = (\frac{80}{3}) % = 26 \frac{2}{3} %$

(vii) $\frac{2}{3} = (\frac{2}{3} \times 100) % = (\frac{200}{3}) % = 66 \frac{2}{3} %$

(viii) $1 \frac{3}{5} = \frac{8}{5} = (\frac{8}{5} \times 100) % = (8 \times 20) % = 160 %$

Page No 142:

Answer:

We have the following:

(i) $32 % = (\frac{32}{100}) = \frac{8}{25}$

(ii) $6 \frac{1}{4} % = (\frac{25}{4}) % = (\frac{25}{4} \times \frac{1}{100}) = \frac{1}{16}$

(iii) $26 \frac{2}{3} % = (\frac{80}{3}) % = (\frac{80}{3} \times \frac{1}{100}) = (\frac{4 \times 1}{3 \times 5}) = \frac{4}{15}$

(iv) $120 % = (\frac{120}{100}) = \frac{6}{5} = 1 \frac{1}{5}$

(v) $6.25 % = (\frac{6.25}{100}) = (\frac{625}{100 \times 100}) = (\frac{25}{400}) = \frac{1}{16}$

(vi) $0.8 % = (\frac{0.8}{100}) = (\frac{8}{10 \times 100}) = (\frac{8}{1000}) = \frac{1}{125}$

(vii) $0.06 % = (\frac{0.06}{100}) = (\frac{6}{100 \times 100}) = (\frac{6}{10000}) = \frac{3}{5000}$

(viii) $22.75 % = (\frac{22.75}{100}) = (\frac{2275}{100 \times 100}) = \frac{91}{400}$

Page No 142:

Answer:

We have:

(i) $43 % = \frac{43}{100} = 43 : 100$

(ii) $36 % = \frac{36}{100} = \frac{9}{25} = 9 : 25$

(iii) $7.5 % = (\frac{7.5}{100}) = (\frac{75}{10 \times 100}) = \frac{3}{40} = 3 : 40$

(iv) $125 % = \frac{125}{100} = \frac{5}{4} = 5 : 4$

Page No 142:

Answer:

We have the following:

(i) 37 : 100 = $\frac{37}{100} = (\frac{37}{100} \times 100) % = 37 %$

(ii) 16 : 25 = $\frac{16}{25} = (\frac{16}{25} \times 100) % = (16 \times 4) % = 64 %$

(iii) 3 : 5 = $\frac{3}{5} = (\frac{3}{5} \times 100) % = (3 \times 20) % = 60 %$

(iv) 5 : 4 = $\frac{5}{4} = (\frac{5}{4} \times 100) % = (5 \times 25) % = 125 %$

Page No 142:

Answer:

We have the following:

(i) 45% = $(\frac{45}{100}) = 0.45$

(ii) 127% = $(\frac{127}{100}) = 1.27$

(iii) 3.6% = $(\frac{3.6}{100}) = (\frac{36}{10 \times 100}) = \frac{36}{1000} = 0.036$

(iv) 0.23% = $(\frac{0.23}{100}) = (\frac{23}{100 \times 100}) = \frac{23}{10000} = 0.0023$

Page No 142:

Answer:

We have:

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

Page No 142:

Answer:

We have:
(i) 32% of 425 = $(\frac{32}{100} \times 425) = (\frac{32 \times 17}{4}) = (8 \times 17) = 136$

(ii) $16 \frac{2}{3} %$ of 16 = $\frac{50}{3} %$ of 16 = $(\frac{50}{3 \times 100} \times 16) = (\frac{1}{6} \times 16) = \frac{8}{3} = 2 \frac{2}{3}$

(iii) 6.5% of 400 = $(\frac{6.5}{100} \times 400) = (\frac{65}{10 \times 100} \times 400) = (\frac{65 \times 4}{10}) = \frac{260}{10} = 26$

(iv) 136% of 70 = $(\frac{136}{100} \times 70) = (\frac{136 \times 7}{10}) = (\frac{952}{10}) = 95.2$

(v) 2.8% of 35 = $(\frac{2.8}{100} \times 35) = (\frac{28}{10 \times 100} \times 35) = (\frac{14 \times 7}{100}) = \frac{98}{100} = 0.98$

(vi) 0.6% of 45 = $(\frac{0.6}{100} \times 45) = (\frac{6}{10 \times 100} \times 45) = (\frac{3 \times 45}{5 \times 100}) = (\frac{3 \times 9}{100}) = \frac{27}{100} = 0.27$

Page No 142:

Answer:

We have the following:

(i) 25% of Rs 76 = Rs $(76 \times \frac{25}{100})$ = Rs $(76 \times \frac{1}{4})$ = Rs 19

(ii) 20% of Rs 132 = Rs $(132 \times \frac{20}{100})$ = Rs $(132 \times \frac{1}{5})$ = Rs 26.4

(iii) 7.5% of 600 m = $(600 \times \frac{7.5}{100}) m = (6 \times 7.5) m = 45 m$

(iv) $3 \frac{1}{3} %$ of 90 km = $\frac{10}{3} %$ of 90 km = $(90 \times \frac{10}{3 \times 100}) km = (90 \times \frac{1}{30}) km = 3 km$

(v) 8.5% of 5 kg = $(5 \times \frac{8.5}{100}) kg = (5 \times \frac{85}{1000}) kg = 0.425 kg = 425 g$ [∵1 kg = 1000 g]

(vi) 20% of 12 L = $(12 \times \frac{20}{100}) L = (12 \times \frac{1}{5}) L = 2.4 L$

Page No 142:

Answer:

Let x be the required number.

Then, 13% of x = 65
⇒ $(\frac{13}{100} \times x) = 65$
⇒ x = $(65 \times \frac{100}{13}) = 500$
Hence, the required number is 500.

Page No 142:

Answer:

Let x be the required number.
Then, $6 \frac{1}{4} %$ of x = 2
⇒ $(6 \frac{1}{4} % \times x) = 2$
⇒ $(\frac{25}{400} \times x) = 2$
⇒ x = $(2 \times \frac{400}{25}) = 32$
Hence, the required number is 32.

Page No 143:

Answer:

10% of Rs 90 = Rs $(\frac{10}{100} \times 90)$ = 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:

Answer:

20% of Rs 60 = Rs $(60 \times \frac{20}{100})$ = 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:

Answer:

3% of x = 9
⇒ $(\frac{3}{100} \times x) = 9$
⇒ x = $(9 \times \frac{100}{3}) = 300$
Hence, the value of x is 300.

Page No 143:

Answer:

12.5% of x = 6
⇒ $(\frac{12.5}{100} \times x) = 6$
⇒ x = $(6 \times \frac{100}{12.5}) = (6 \times 8) = 48$
Hence, the value of x is 48.

Page No 143:

Answer:

Let x% of 84 be 14.
Then, $(\frac{x}{100} \times 84) = 14$
⇒ $\frac{21 x}{25} = 14$
⇒ x = $(14 \times \frac{25}{21}) = (\frac{2 \times 25}{3}) = \frac{50}{3} = 16 \frac{2}{3} %$
Hence, $16 \frac{2}{3} %$ of 84 is 14.

Page No 143:

Answer:

(i) Let x% of Rs 120 be Rs 15.
Then, Rs $(\frac{x}{100} \times 120)$ = Rs 15
    ⇒ $(\frac{6 x}{5})$ = 15
    ∴ x = $(\frac{15 \times 5}{6}) %$ = $(\frac{25}{2}) %$ = 12.5%
Hence, 12.5% of Rs 120 is Rs 15.

(ii) Let x% of 2 h be 36 min.
Then, $(\frac{x}{100} \times 2 \times 60)$ min = 36 min
      ⇒ $(\frac{120 x}{100})$ = 36
     ∴ x = $(\frac{36 \times 100}{120}) %$ = 30%
Hence, 30% of 2 h is 36 min.

(iii) Let x% of 2 days be 8 h.
Then, $(\frac{x}{100} \times 2 \times 24)$ h = 8 h
     ⇒ $(\frac{48 x}{100})$ = 8
     ∴ x = $(\frac{8 \times 100}{48}) %$ = $16 \frac{2}{3} %$
     Hence, $16 \frac{2}{3} %$ of 2 days is 8 h.

(iv) Let x% of 4 km be 160 m.
Then, $(\frac{x}{100} \times 4 \times 1000)$ m = 160 m
     ⇒ 40x = 160
     ∴ x = $(\frac{160}{40}) %$ = 4%
Hence, 4% of 4 km is 160 m.

(v) Let x% of 1 L be 175 mL.
Then, $(\frac{x}{100} \times 1 \times 1000)$ mL = 175 mL
     ⇒ 10x = 175
     ∴ x = $(\frac{175}{10}) %$ = 17.5%
Hence, 17.5% of 1 L is 175 mL.

(vi) Let x% of Rs 4 be 25 paise.
Then, $(\frac{x}{100} \times 4 \times 100)$ paise = 25 paise
     ⇒ 4x = 25
     ∴ x = $(\frac{25}{4}) %$ = $6 \frac{1}{4} %$
     Hence, $6 \frac{1}{4} %$ of Rs 4 is 25 paise.

Page No 147:

Answer:

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

Hence, Rupesh scored 66% in the examination.

Page No 147:

Answer:

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

Page No 147:

Answer:

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

Page No 147:

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

Page No 147:

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 = $(\frac{100}{30} \times 504)$ = 1680
∴ Number of boys = (1680 − 504) = 1176

Hence, there are 1176 boys in the school.

Page No 147:

Answer:

Let x kg be the amount of the required ore.

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

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

Page No 147:

Answer:

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

Page No 147:

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

Page No 147:

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 = $(\frac{100}{28} \times 392)$ = 1400
Hence, the total number of examinees is 1400.

Page No 147:

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 $(x - \frac{5 x}{100})$ = Rs $(\frac{100 x - 5 x}{100})$ = Rs $(\frac{95 x}{100})$
Now, price of the moped after deducting the commission = Rs 15200
Then, Rs $(\frac{95 x}{100})$ = Rs 15200
∴ x = Rs $(\frac{15200 \times 100}{95})$ = Rs (160 $\times$ 100) = Rs 16000
Hence, the gross value of the moped is Rs 16000.

Page No 147:

Answer:

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

Quantity of sulphur in it = 10% of 8000 g
                                         = $(\frac{10}{100} \times 8000)$ 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:

Answer:

Total quantity of chalk = 1 kg = 1000 g

Now, we have the following:

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

Page No 148:

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

Page No 148:

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
⇒ $(\frac{3}{100} \times x)$ = 42660
∴ x = $(\frac{42660 \times 100}{3}) = 1422000$
Hence, the total value of the property is Rs 14,22,000.

Page No 148:

Answer:

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

Hence, candidate B recieved 19,200 votes.

Page No 148:

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 $(\frac{12}{100} \times x)$ = Rs $(\frac{12 x}{100})$
Value of the shirt after discount = Rs $(x - \frac{12 x}{100})$
                                                      = Rs $(\frac{100 x - 12 x}{100})$ = Rs $(\frac{88 x}{100})$
Present price of the shirt = Rs 1188
Then, Rs $(\frac{88 x}{100})$ = Rs 1188
     ⇒ 88x = (1188 $\times$ 100)
     ⇒ 88x = 118800
∴ x = $(\frac{118800}{88})$ = 1350

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

Page No 148:

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 $(\frac{8}{100} \times x)$ = Rs $(\frac{2 x}{25})$
Increased price of the sweater = Rs $(x + \frac{2 x}{25})$
= Rs $(\frac{25 x + 2 x}{25})$ = Rs $(\frac{27 x}{25})$
However, increased price of the sweater = Rs 1566
Then, Rs $(\frac{27 x}{25})$ = Rs 1566
∴ x = $(\frac{1566 \times 25}{27})$ = 1450
Hence, the original price of the sweater is Rs 1450

Page No 148:

Answer:

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

Page No 148:

Answer:

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

Page No 148:

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 = $(\frac{20}{120} \times 100)$ = $(\frac{100}{6})$ = $16 \frac{2}{3}$
Hence, the required reduction on the new salary is $16 \frac{2}{3} %$ .

Page No 148:

Answer:

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

Commission on the next Rs 200000 = 1% of Rs 200000
                                                           = $(\frac{1}{100} \times 200000)$ = Rs 2000
Remaining amount = Rs (540000 − 400000) = Rs 140000
∴ Commission on Rs 140000 = 0.5% of Rs 140000
                                                  = Rs $(\frac{0.5}{100} \times 140000)$
                                                  = Rs $(\frac{5}{1000} \times 140000)$ = 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:

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 $(\frac{100}{80} \times 100)$ = 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:

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

Page No 148:

Answer:

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

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

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

Page No 148:

Answer:

The present value of the car = Rs 450000
The decrease in its value after the first year = 20% of Rs 450000
= Rs $(\frac{20}{100} \times 450000)$ = 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 $(\frac{20}{100} \times 360000)$ = 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:

Answer:

Present population of the town = 60000
Increase in population of the town after the 1 year = 10% of 60000
= $(\frac{10}{100} \times 60000)$ = 6000
Thus, population of the town after 1 year = 60000 + 6000 = 66000
Increase in population after 2 years = 10% of 66000
= $(\frac{10}{100} \times 66000)$ = 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:

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 $(\frac{1}{125} \times 100)$ unit = $(\frac{4}{5})$ unit of sugar.
Reduction in consumption = $(1 - \frac{4}{5})$ = $(\frac{1}{5})$ unit
∴ Reduction percent in consumption = $(\frac{1}{5} \times \frac{1}{1} \times 100)$ %= $(\frac{100}{5})$ %= 20%

Page No 148:

Answer:

(b) 75%

$\frac{3}{4}$ = $(\frac{3}{4} \times 100) %$ = 75%

Page No 148:

Answer:

Page No 148:

Answer:

(c) $\frac{1}{12}$

$8 \frac{1}{3} %$ = $\frac{25}{3} %$ = $(\frac{25}{3} \times \frac{1}{100}) = (\frac{1}{3 \times 4}) = \frac{1}{12}$

Page No 149:

Answer:

(c) 12
We have x% of 75 = 9
⇒ $(\frac{x}{100} \times 75) = 9$
∴ x = $(\frac{9 \times 100}{75}) = 12$
Hence, the value of x is 12

Page No 149:

Answer:

(d) 10%

Let x be the required percent.
Then, x % of $\frac{2}{7}$ = $\frac{1}{35}$
⇒ $(\frac{x}{100} \times \frac{2}{7}) = \frac{1}{35}$
∴ x = $(\frac{100 \times 7}{35 \times 2})$ = 10
Hence, 10% of $\frac{2}{7}$ is $\frac{1}{35}$

Page No 149:

Answer:

(b) 2.5%

Let x % of 1 day be 36 min.
Then, $(\frac{x}{100} \times 1 \times 24 \times 60)$ min = 36 min
∴ x = $(\frac{36 \times 100}{24 \times 60})$ = $(\frac{3 \times 5}{2 \times 3}) % = (\frac{5}{2}) % = 2.5 %$

Hence, 2.5% of 1 day is 36 min.

Page No 149:

Answer:

(a) 35
Let the required number be x.
Then, x + 20% of x = 42
⇒ $(x + \frac{20 x}{100}) = 42$
⇒ $(x + \frac{x}{5}) = 42$
⇒ $(\frac{5 x + x}{5}) = 42$ [∵ LCM of 1 and 5 = 5]
⇒ $(\frac{6 x}{5}) = 42$
∴ x = $(\frac{42 \times 5}{6}) = 35$
Hence, the required number is 35.

Page No 149:

Answer:

(b) 75
Let the required number be x.
Then, x − 8% of x = 69
⇒ $(x - \frac{8 x}{100})$ = 69
⇒ $(x - \frac{2 x}{25})$ = 69
⇒ $(\frac{25 x - 2 x}{25})$ = 69 [Since L.C.M. of 1 and 25 = 25]
⇒ $(\frac{23 x}{25}) = 69$
∴ x = $(\frac{69 \times 25}{23})$ = 75
Hence, the required number is 75

Page No 149:

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]
⇒ $(\frac{5}{100} \times x) = 0.4$
⇒ x = $(\frac{0.4 \times 100}{5})$ = 8
Hence, 8 kg of ore is required to obtain 400 g of copper.

Page No 149:

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 $(x - \frac{10}{100} x)$ = Rs $(\frac{100 x - 10 x}{100})$ = Rs $(\frac{9 x}{10})$
However, price of the TV after deducting the commission = Rs 18000
Then, Rs $(\frac{9 x}{10})$ = Rs 18000
∴ x = $(\frac{18000 \times 10}{9})$ = Rs (2000 $\times$ 10) = Rs 20000
Hence, the gross value of the TV is Rs 20,000

Page No 149:

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 $(\frac{25}{100} \times x)$ = Rs $(\frac{x}{4})$
Salary after increment= Rs $(x + \frac{x}{4})$ = Rs $(\frac{5 x}{4})$
However, increased salary = Rs 20000
Then, Rs $(\frac{5 x}{4})$ = Rs 20000
∴ x = Rs $(\frac{20000 \times 4}{5})$ = Rs 16000
Hence, the original salary of the man is Rs 16,000

Page No 149:

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 = $(\frac{100}{5} \times 28) = (20 \times 28) = 560$
Hence, there were 560 examinees.

Page No 149:

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 = $(\frac{100}{60} \times 420)$ = 700
Hence, the fruit seller originally had 700 apples with him.

Page No 149:

Answer:

(c) Rs. 25250

Present value of the machine = Rs 25000
Decrease in its value after 1 year = 10% of Rs 25000
= Rs $(\frac{10}{100} \times 25000)$ = 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:

Answer:

(c) 75

Let the required number be x. Then, we have:
8% of x = 6
⇒ $(\frac{8}{100} \times x) = 6$
∴ x = $(\frac{6 \times 100}{8}) = 75$
Hence, the required number is 75

Page No 149:

Answer:

Page No 149:

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 $(\frac{6}{100} \times x)$ = Rs $(\frac{3 x}{50})$
Reduced price of the chair = Rs $(x - \frac{3 x}{50})$
= Rs $(\frac{50 x - 3 x}{50})$ = Rs $(\frac{47 x}{50})$
However, present price of the chair = Rs 658
Then, Rs $(\frac{47 x}{50})$ = Rs 658
⇒ Rs $(\frac{47 x}{50})$ = Rs 658
⇒ x = Rs $(\frac{658 \times 50}{47})$ = Rs $(14 \times 50) = 700$
Hence, the original price of the chair is Rs 700

Page No 149:

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 = $(\frac{100}{30} \times 240) = 800$
∴ Number of boys = (800 − 240) = 560

Hence, there are 560 boys in the school.

Page No 149:

Answer:

(c) 450

Let x be the number.
(11% of x) − (7% of x) = 18
⇒ $(\frac{11 x}{100} - \frac{7 x}{100}) = 18$
⇒ $\frac{4 x}{100} = 18$
∴ x = $(\frac{18 \times 100}{4}) = (18 \times 25) = 450$
Hence, the required number is 450

Page No 149:

Answer:

(a) 60
Let x be the number.
According to question, we have:
(35% of x ) + 39 = x
⇒ $(\frac{35}{100} \times x) + 39 = x$
⇒ $(\frac{7 x}{20}) + 39 = x$
⇒ $(x - \frac{7 x}{20}) = 39$
⇒ $(\frac{20 x - 7 x}{20}) = 39$
⇒ $(\frac{13 x}{20}) = 39$
∴ x = $(\frac{39 \times 20}{13})$ = 60
Hence, the required number is 60

Page No 149:

Answer:

(c) 500
Let x be the maximum marks.
Pass marks = (145 + 35) = 180
∴ 36% of x = 180
⇒ $(\frac{36}{100} \times x) = 180$
⇒ x = $(\frac{180 \times 100}{36}) = (5 \times 100) = 500$
Hence, maximum marks = 500

Page No 150:

Answer:

(d) 225
Let x be the number.
According to question, we have:
x − 40% of x = 135
⇒ $(x - \frac{40 x}{100}) = 135$
⇒ $(\frac{100 x - 40 x}{100}) = 135$
⇒ $(\frac{60 x}{100}) = 135$
⇒ x = $(\frac{135 \times 100}{60})$ = 225
Hence, the required number is 225

Page No 151:

Answer:

We have:

(i) $\frac{4}{5}$ = $(\frac{4}{5} \times 100) % = (4 \times 20) % = 80 %$

(ii) $\frac{7}{4}$ = $(\frac{7}{4} \times 100) % = (7 \times 25) % = 175 %$

(iii) 45% = $(\frac{45}{100}) = (\frac{9}{20})$

(iv) 105% = $(\frac{105}{100}) = (\frac{21}{20})$

(v) 15% = $\frac{15}{100} = \frac{3}{20}$ = 3 : 20

(vi) 12 : 25 = $\frac{12}{25} = (\frac{12}{25} \times 100) % = (12 \times 4) % = 48 %$

Page No 151:

Answer:

(i) Let x% of 1 kg be 125g.
Then, $(\frac{x}{100} \times 1 \times 1000) g = 125 g$
     ⇒ 10x = 125
     ⇒ x = $(\frac{125}{10}) % = 12 \frac{1}{2} %$
Hence, $12 \frac{1}{2} %$ of 1 kg is 125 g.

(ii) Let x% of 80 m be 24 m.
Then, $(\frac{x}{100} \times 80) m = 24 m$
     ⇒ $(\frac{4 x}{5})$ = 24
     ⇒ x = $(24 \times \frac{5}{4}) % = 30 %$
Hence, $30 %$ of 80 m is 24 m.

Page No 151:

Answer:

(i) $16 \frac{2}{3} %$ of 30 = $\frac{50}{3} %$ of 30
                          = $(\frac{50}{3 \times 100} \times 30)$
                          = 5

(ii) 15% of Rs 140 = Rs $(\frac{15}{100} \times 140)$
                              = Rs (3 $\times$ 7)
                              = Rs 21

Page No 151:

Answer:

(i) Let x be the required number.
Then, $6 \frac{1}{4} %$ of x = 5
     ⇒ $\frac{25}{4} %$ of x = 5
     ⇒ $(\frac{25}{4 \times 100} \times x) = 25$
     ⇒ $(\frac{x}{16}) = 5$
∴ x = (5 $\times$ 16) = 80
Hence, the required number is 80.

(i) 0.8% of 45 = $(\frac{0.8}{100} \times 45)$
                         = $(\frac{8}{10 \times 100} \times 45)$
                         = $(\frac{72}{200}) = (\frac{36}{100}) = 0.36$
Hence, 0.8% of 45 is 0.36.

Page No 151:

Answer:

Let x be the number.
The number is increased by 10%.
∴ Increased number = 110% of x = $(x \times \frac{110}{100}) = (\frac{11 x}{10})$
The number is, then, decreased by 10%.

∴ Decreased number = 90% of $(\frac{11 x}{10})$ = $(\frac{11 x}{10} \times \frac{90}{100}) = (\frac{99 x}{100})$

Net decrease = $(x - \frac{99 x}{100}) = \frac{(100 x - 99 x)}{100} = \frac{x}{100}$
Net decrease percentage = $(\frac{x}{100} \times \frac{1}{x} \times 100) = 1$

Page No 151:

Answer:

The present value of the machine = Rs 10000
The decrease in its value after the 1^st year = 10% of Rs 10000
= Rs $(\frac{10}{100} \times 10000)$ = Rs 1000
The depreciated value of the machine after the 1^st year = Rs (10000 − 1000) =Rs 9000
The decrease in its value after the 2^nd year = 10% of Rs 9000
= Rs $(\frac{10}{100} \times 9000)$ = Rs 900
The depreciated value of the machine after the 2^nd year = Rs (9000 − 900) = Rs 8100

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

Page No 151:

Answer:

The present population of the town = 16000
Increase in population after 1 year = 5% of 16000
= $(\frac{5}{100} \times 16000)$ = 800
Thus, population after one year = 16000 + 800 = 16800
Increase in population after 2 years = 5% of 16800
= $(\frac{5}{100} \times 16800)$ = 840
Increased population after two years = 16800 + 840 = 17640

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

Page No 151:

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. $(\frac{5}{100} \times x)$ = Rs. $(\frac{x}{20})$
Then, increased price of the tea set = Rs. $(x + \frac{x}{20})$
= Rs. $(\frac{20 x + x}{20})$ = Rs. $(\frac{21 x}{20})$
However, increased price = Rs. 441
Then, Rs. $(\frac{21 x}{20})$ = Rs. 441
∴ x = $(\frac{441 \times 20}{21})$ = 420
Hence, the original price of the tea set is Rs 420

Page No 151:

Answer:

(b) $\frac{1}{16}$

$6 \frac{1}{4} %$ = $\frac{25}{4} %$ = $(\frac{25}{4 \times 100}) = \frac{1}{16}$

Page No 151:

Answer:

(c) 12

Given that x% of 75 = 12
Then, $(\frac{x}{100} \times 75) = 12$
⇒ x = $(\frac{12 \times 100}{75})$ =16
Hence, the value of x is 16

Page No 151:

Answer:

(c) 25

Let the number be x. Then, we have:
120% of x = increased number
⇒ 30 = $(x \times \frac{120}{100})$
⇒ 30 = $(\frac{6 x}{5})$
⇒ x = $(30 \times \frac{5}{6}) = 25$
Hence, the required number is 25

Page No 151:

Answer:

(d) 180

Let the required number be x. Then, we have:
5% of x = 9
⇒ $(\frac{5}{100} \times x) = 9$
⇒ x = $(9 \times \frac{100}{5}) = (9 \times 20) = 180$

Page No 151:

Answer:

(a) 60
Let the number be x.
According to question, we have:
(35% of x ) + 39 = x
⇒ $(\frac{35}{100} \times x) + 39 = x$
⇒ $(\frac{7 x}{20}) + 39 = x$
⇒ $(x - \frac{7 x}{20}) = 39$
⇒ $(\frac{20 x - 7 x}{20}) = 39$
⇒ $(\frac{13 x}{20}) = 39$
∴ x = $(\frac{39 \times 20}{13})$ = 60
Hence, the required number is 60.

Page No 151:

Answer:

(c) 500
Let x be the maximum marks.
Pass marks = (160 + 20) = 180
∴ 36% of x = 180
⇒ $(\frac{36}{100} \times x) = 180$
⇒ x = $(\frac{180 \times 100}{36}) = (5 \times 100) = 500$
Hence, maximum marks = 500

Page No 151:

Answer:

We have the following:

(i) 3 : 4 = (75)%
     Explanation: 3 : 4 = $\frac{3}{4}$ = $(\frac{3}{4} \times 100) % = (3 \times 25) % = 75 %$

(ii) 0.75 = (75)%
      Explanation: ( 0.75 $\times$ 100)% = 75%

(iii) 6% = 0.06 (expressed in decimals)
       Explanation: 6% = $\frac{6}{100} = 0.06$

(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
       ⇒ $(x - \frac{40 x}{100}) = 135$
       ⇒ $(\frac{100 x - 40 x}{100}) = 135$
       ⇒ $(\frac{60 x}{100}) = 135$
        ⇒ x = $(\frac{135 \times 100}{60})$ = 225

(v) (11% of x) − (7% of x) = 18
⇒ x = 450
      Explanation:
       (11% of x) − (7% of x) = 18
        ⇒ $(\frac{11 x}{100} - \frac{7 x}{100}) = 18$
        ⇒ $\frac{4 x}{100} = 18$
         ∴ x = $(\frac{18 \times 100}{4}) = (18 \times 25) = 450$

Page No 151:

Answer:

(i) True (T)
     Justification: $(\frac{3}{4} \times 100) %$ = 75%

(ii) True (T)
     Justification: $12 \frac{1}{2} % = \frac{25}{2} %$ = $(\frac{25}{2 \times 100}) = \frac{1}{8}$

(iii) False (F)
       Justification: $\frac{2}{5}$ = $(\frac{2}{5} \times 100)$ % = $(2 \times 20) %$ = 40%

(iv) True (T)
       Justification: 80% of 450 = $(\frac{80}{100} \times 450) = (\frac{80 \times 9}{2}) = (40 \times 9) = 360$

(v) True (T)
      Justification: 20% of 1 L = 20% of 1000 mL
                                                     = $(\frac{20}{100} \times 1000)$ mL = 200 mL

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