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#### Page No 142:

We have the following:

(i)

(ii)

(iii)

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

(v) $\frac{19}{500}=\left(\frac{19}{500}×100\right)%=\left(\frac{19}{5}\right)%=3.8%$

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

(vii) $\frac{2}{3}=\left(\frac{2}{3}×100\right)%=\left(\frac{200}{3}\right)%=66\frac{2}{3}%$

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

#### Page No 142:

We have the following:

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

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

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

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

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

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

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

(viii)  $22.75%=\left(\frac{22.75}{100}\right)=\left(\frac{2275}{100×100}\right)=\frac{91}{400}$

We have:

(i)

(ii)

(iii)

(iv)

#### Page No 142:

We have the following:

(i) 37 : 100 = $\frac{37}{100}=\left(\frac{37}{100}×100\right)%=37%$

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

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

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

#### Page No 142:

We have the following:

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

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

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

(iv) 0.23% =$\left(\frac{0.23}{100}\right)=\left(\frac{23}{100×100}\right)=\frac{23}{10000}=0.0023$

#### Page No 142:

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:

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

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

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

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

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

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

#### Page No 142:

We have the following:

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

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

(iii) 7.5% of 600 m =

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

(v) 8.5% of 5 kg =          [∵1 kg = 1000 g]

(vi) 20% of 12 L =

#### Page No 142:

Let x be the required number.

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

#### Page No 142:

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

#### Page No 143:

10% of Rs 90 = Rs $\left(\frac{10}{100}×90\right)$ = 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:

20% of Rs 60 = Rs $\left(60×\frac{20}{100}\right)$= 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:

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

#### Page No 143:

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

#### Page No 143:

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

#### Page No 143:

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

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

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

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

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

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

#### Page No 147:

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

Hence, Rupesh scored 66% in the examination.

#### Page No 147:

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

#### Page No 147:

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

#### Page No 147:

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

#### Page No 147:

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

Hence, there are 1176 boys in the school.

#### Page No 147:

Let x kg be the amount of the required ore.

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

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

#### Page No 147:

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

#### Page No 147:

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 = $\left(\frac{100}{60}×420\right)$ = 700
Hence, the fruit seller originally had 700 apples.

#### Page No 147:

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

#### Page No 147:

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 $\left(\frac{15200×100}{95}\right)$ = Rs (160 $×$ 100) = Rs 16000
Hence, the gross value of the moped is Rs 16000.

#### Page No 147:

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

Quantity of sulphur in it = 10% of 8000 g
= $\left(\frac{10}{100}×8000\right)$ 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.

#### Page No 147:

Total quantity of chalk = 1 kg = 1000 g

Now, we have the following:

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

#### Page No 148:

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

#### Page No 148:

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

#### Page No 148:

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

Hence, candidate B recieved 19,200 votes.

#### Page No 148:

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 $\left(\frac{12}{100}×x\right)$ = 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 $×$ 100)
⇒ 88x = 118800
∴  x = $\left(\frac{118800}{88}\right)$ = 1350

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

#### Page No 148:

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 $\left(\frac{8}{100}×x\right)$ = 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$\left(\frac{1566×25}{27}\right)$ = 1450
Hence, the original price of the sweater is Rs 1450

#### Page No 148:

Let the income of the man be Rs x.
Then, income spent = 80% of Rs. x
=
Rs $\left(\frac{80}{100}×x\right)$ = 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 $\left(\frac{10}{100}×\frac{x}{5}\right)$ = 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 $\left(\frac{46260×50}{9}\right)$ = Rs 257000
Hence, the income of the man is Rs 2,57,000.

#### Page No 148:

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

#### Page No 148:

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

#### Page No 148:

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

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

#### Page No 148:

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 $\left(\frac{100}{80}×100\right)$ = 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%.

#### Page No 148:

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

#### Page No 148:

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

Then, its present value = 90% of Rs x

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

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

#### Page No 148:

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

#### Page No 148:

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

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 $\left(\frac{1}{125}×100\right)$ 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 = $\left(\frac{1}{5}×\frac{1}{1}×100\right)$ %= $\left(\frac{100}{5}\right)$ %= 20%

#### Page No 148:

(b) 75%

$\frac{3}{4}$ = $\left(\frac{3}{4}×100\right)%$ = 75%

#### Page No 148:

(c) 40%

2 : 5 = $\frac{2}{5}$ = $\left(\frac{2}{5}×100\right)%$ = 40%

#### Page No 148:

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

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

#### Page No 149:

(c) 12
We have x% of 75 = 9
$\left(\frac{x}{100}×75\right)=9$
x = $\left(\frac{9×100}{75}\right)=12$
Hence, the value of x is 12

#### Page No 149:

(d) 10%

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

#### Page No 149:

(b) 2.5%

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

Hence, 2.5% of 1 day is 36 min.

#### Page No 149:

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

#### Page No 149:

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

#### Page No 149:

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

#### Page No 149:

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

#### Page No 149:

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

#### Page No 149:

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

#### Page No 149:

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

#### Page No 149:

(c) Rs. 25250

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

(c) 75

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

#### Page No 149:

(c) 270
60% of 450 = $\left(\frac{60}{100}×450\right)$
= $\left(\frac{3}{5}×450\right)$ = (3 $×$ 90) = 270

#### Page No 149:

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

Hence, there are 560 boys in the school.

#### Page No 149:

(c) 450

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

#### Page No 149:

(a) 60
Let x be the number.
According to question, we have:
(35% of x ) + 39 = x
$\left(\frac{35}{100}×x\right)+39=x$
$\left(\frac{7x}{20}\right)+39=x$
$\left(x-\frac{7x}{20}\right)=39$
$\left(\frac{20x-7x}{20}\right)=39$
$\left(\frac{13x}{20}\right)=39\phantom{\rule{0ex}{0ex}}$
x = $\left(\frac{39×20}{13}\right)$ = 60
Hence, the required number is 60

#### Page No 149:

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

#### Page No 149:

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

#### Page No 150:

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

#### Page No 151:

We have:

(i) $\frac{4}{5}$= $\left(\frac{4}{5}×100\right)%=\left(4×20\right)%=80%$

(ii) $\frac{7}{4}$=$\left(\frac{7}{4}×100\right)%=\left(7×25\right)%=175%$

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

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

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

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

#### Page No 151:

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

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

#### Page No 151:

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

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

#### Page No 151:

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

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

#### Page No 151:

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

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

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

#### Page No 151:

The present value of the machine = Rs 10000
The decrease in its value after the 1st year = 10% of Rs 10000
= Rs $\left(\frac{10}{100}×10000\right)$ = 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 $\left(\frac{10}{100}×9000\right)$ = 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.

#### Page No 151:

The present population of the town = 16000
Increase in population after 1 year = 5% of 16000
= $\left(\frac{5}{100}×16000\right)$ = 800
Thus, population after one year = 16000 + 800 = 16800
Increase in population after 2 years = 5% of 16800
= $\left(\frac{5}{100}×16800\right)$ = 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:

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

#### Page No 151:

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

$6\frac{1}{4}%$ = $\frac{25}{4}%$ =$\left(\frac{25}{4×100}\right)=\frac{1}{16}$

#### Page No 151:

(c) 12

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

#### Page No 151:

(c) 25

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

#### Page No 151:

(d) 180

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

#### Page No 151:

(a) 60
Let the number be x.
According to question, we have:
(35% of x ) + 39 = x
$\left(\frac{35}{100}×x\right)+39=x$
$\left(\frac{7x}{20}\right)+39=x$
$\left(x-\frac{7x}{20}\right)=39$
$\left(\frac{20x-7x}{20}\right)=39$
$\left(\frac{13x}{20}\right)=39\phantom{\rule{0ex}{0ex}}$
x = $\left(\frac{39×20}{13}\right)$ = 60
Hence, the required number is 60.

#### Page No 151:

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

#### Page No 151:

We have the following:

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

(ii) 0.75 = (75)%
Explanation: ( 0.75 $×$ 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
⇒ $\left(x-\frac{40x}{100}\right)=135$
⇒ $\left(\frac{100x-40x}{100}\right)=135$
⇒ $\left(\frac{60x}{100}\right)=135$
⇒ x = $\left(\frac{135×100}{60}\right)$ = 225

(v) (11% of x) − (7% of x) = 18
⇒ x = 450
Explanation:

(11% of x) − (7% of x) = 18
⇒ $\left(\frac{11x}{100}-\frac{7x}{100}\right)=18$
⇒ $\frac{4x}{100}=18$
∴ x = $\left(\frac{18×100}{4}\right)=\left(18×25\right)=450$

#### Page No 151:

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

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

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

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

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

View NCERT Solutions for all chapters of Class 7