- Staff Selection Commission Mathematics - Percentage (1999-2017) Part 5

Staff Selection Commission Mathematics - Percentage (1999-2017)

TYPE–XI (Contd.)

Ans .

7.00

1. Explanation :

Original price of article = x per kg.

New price = $$\frac{79x}{100}$$per kg

∴ $$\frac{100}{\frac{79x}{100}}$$ - $$\frac{79x}{100}$$ = 3

$$→$$ 79x = 700 $$→$$ x = $$\frac{700}{79}$$

∴ New price = $$\frac{79x}{100}$$ = $$\frac{79}{100} \times \frac{700}{79}$$

= 7 per kg

Ans .

Rs. 4

1. Explanation :

Let the original price of sugar be Rs. x per kg.

Reduced price = Rs. $$\frac{80x}{100}$$ = Rs. $$\frac{4x}{5}$$per kg.

According to the question,

$$\frac{160}{\frac{4x}{5}}$$ - $$\frac{4x}{5}$$ = 8

$$→$$ 8x = 40

$$→$$ x = $$\frac{40}{8}$$ = 5 per kg.

Reduced Price = Rs. $$\frac{4 \times 5}{5}$$ = Rs. 4 per kg

Ans .

12% decrease

1. Explanation :

Required percentage change = 10 - 20 + $$\frac{10 \times -20}{100}$$ % = -12%

Negative sign shows decrease.

Ans .

$$37 \frac{1}{2}$$%

1. Explanation :

Required per cent = $$\frac{x}{100 + x} \times 100$$

where x = 60% = $$\frac{60}{160} \times 100$$ = $$\frac{75}{2}$$ = $$37 \frac{1}{2}$$%

Ans .

12 kg.

1. Explanation :

Let original price of sugar be Rs. x per kg.

New price = Rs.$$\frac{120x}{100}$$ = Rs.$$\frac{6x}{5}$$per kg

According to the question,

$$\frac{50}{x}$$ - $$\frac{50}{\frac{6x}{5}}$$ = 2

$$→$$ 6x = 25

$$→$$ x = Rs $$\frac{25}{6}$$kg

∴ Required quantity of sugar = $$\frac{50}{x}$$kg

= $$\frac{50}{\frac{25}{6}}$$kg = $$\frac{50 \times 6}{25}$$ = 12kg.

Ans .

$$11 \frac{1}{9}$$%

1. Explanation :

Required per cent = $$\frac{Decrease%}{100-Decrease%} \times 100$$

= $$\frac{10}{100-10} \times 100$$ = $$11 \frac{1}{9}$$%

Ans .

25%

1. Explanation :

Required percentage increase = $$\frac{x}{100-x} \times 100$$

= $$\frac{20}{100-20} \times 100$$ = 25%

Ans .

Rs. 5 per kg

1. Explanation :

Original price of sugar = Rs. x/kg. (let)

New price = Rs. $$\frac{120x}{100}$$ per kg.

= Rs. $$\frac{6x}{5}$$ per kg.

According to the question,

$$\frac{120}{x}$$ - $$\frac{120}{\frac{6x}{5}}$$ = 4

$$→$$ x = Rs. 5 per kg.

Ans .

25%

1. Explanation :

Original price of building = Rs. 100 (let)

∴ Its price in 2001 = Rs. 80

Its price in 2002 = Rs. 60

Required percentage decrease = $$\frac{80-60}{80} \times 100$$ = 25%

TYPE–XII

Ans .

1700

1. Explanation :

Percentage of boys = 100% – 70% = 30%

Let total no. of students be x

\ According to question,

30% of x = 510

x = $$\frac{510}{30} \times 100$$ = 1700

Ans .

1458

1. Explanation :

40% of students = 972

∴ 60% of students = $$\frac{972}{40} \times 60$$ = 1458

Ans .

1176

1. Explanation :

Number of boys = $$\frac{70}{30} \times 504$$ = 1176

Ans .

95

1. Explanation :

Required sum = 0.5% of 19000 = $$19000 \times \frac{0.5}{100}$$ = 95

Ans .

9 m

1. Explanation :

Remaining height = 192 - $$\frac{125}{2}$$% of 192

= 192 – 120 = 72 m

∴ Required distance (distance covered in second hour) then,

= $$\frac{25}{2}$$% of 72

= $$\frac{25 \times 72}{2 \times 100}$$ = 9m

Ans .

52 kgs.

1. Explanation :

Water in 100 kg fresh fruit = 68%

Water in dry fruit = 20%

Decrease = 48%

∴ Dry fruit obtained = 100 – 48 = 52 kg

Ans .

33%

1. Explanation :

The net tax rate = 30 + $$30 \times \frac{10}{100}$$ % = 33%

Ans .

180

1. Explanation :

Let z have x

∴ Money with Y = $$\frac{3}{2}$$x and Money with X = 3x

∴ 3x + $$\frac{3x}{2}$$ + x = $$3 \times 110$$

$$→$$ $$\frac{6x+3x+2x}{2}$$ = 330

$$→$$ 11 x = 2 × 330

x = $$\frac{2 \times 330}{11}$$ = 60

∴ Money with X = 3x = (3×60) = 180

Ans .

$$83 \frac{1}{3}$$%

1. Explanation :

If a number is x% more than other, then the other number is less than the first number by

$$\frac{x}{100+x} \times 100$$%

\ Required answer = $$\frac{500}{100+500} \times 100$$ = $$83 \frac{1}{3}$$%

Ans .

48.8%

1. Explanation :

Let x = 10 and y = 10

∴ x$$y^2$$ = 10 × 10 × 10=1000 units

Decreasing values of x and y by 20%,

Expression = x$$y^2$$

= 8 × 8 × 8

= 512

Decrease= 1000–512 = 488 units

Percentage decrease = $$\frac{488}{1000} \times 100$$ = 48.8%

Ans .

88%

1. Explanation :

If two numbers are respectively x% and y% more than a third number, the first as a per cent of

second is = $$\frac{100+x}{100+y} \times 100$$ = $$\frac{110}{125} \times 100$$ = 88%

Ans .

140

1. Explanation :

Let sum of money be x.

$$\frac{11}{2}$$% of x = 220

$$→$$ x = $$\frac{220 \times 200}{11}$$ = 4000

$$3 \frac{1}{2}$$% of 4000 = $$\frac{7}{2} \times \frac{4000}{100}$$ = 140

Ans .

3000

1. Explanation :

Let the total number of workers in the factory be x

$$x \times \frac{60}{100} \times \frac{75}{100}$$ = 1350

$$→$$ x = $$\frac{1350 \times 100^2}{60 \times 75}$$ = 3000

Ans .

10%

1. Explanation :

Let the third number = 100.

∴ First number = 70

Second number = 63

∴ Required per cent = $$\frac{70-63}{70} \times 100$$ = 10%

Ans .

160%

1. Explanation :

Let Rani’s weight be x kg.

∴ Meena’s weight = 4x kg.

Tara’s weight = $$\frac{5x}{4}$$kg

∴ Required percentage = $$\frac{4x}{\frac{5x}{2}} \times 100$$ = 160%

Ans .

570

1. Explanation :

Number of people who have the saving habit = $$\frac{2500 \times 60}{100}$$

∴ Number of shareholders = (100 – 62)% of 1500

= $$\frac{1500 \times 38}{100}$$ = 570

Ans .

8000

1. Explanation :

Let the original price of the article be x.

According to the question,

5832 = $$x(1-\frac{1}{100})^3$$

5832 = $$x(\frac{9}{10})^3$$

x = 8000

Ans .

9%

1. Explanation :

Increase in AC = 6%

∴ Increased AC = $$\frac{106}{100} \times 3$$ = 3.18 cm

∴ Decreased CB = 5 – 3.18 = 1.82 cm.

∴ Decrease = 2 – 1.82 = 0.18 cm

∴ Percentage decrease = $$\frac{0.18}{2} \times 100$$ = 9%

Ans .

$$91 \frac{2}{3}$$%

1. Explanation :

∵ 24= 100% ∴22 = $$\frac{100}{24} \times 22$$ = $$91 \frac{2}{3}$$%

Ans .

40%

1. Explanation :

Let the number of boys be x and that of girls be y.

Then, 71x + 73y = 71.8 (x + y)

$$→$$ 71.8x – 71x = 73y – 71.8y

$$→$$ 0.8x = 1.2y

$$\frac{x}{y}$$ = $$\frac{1.2}{0.8}$$ = $$\frac{12}{8}$$ = $$\frac{3}{2}$$

∴ $$\frac{x}{y}$$ + 1 = $$\frac{3}{2}$$ + 1

$$\frac{x + y}{y}$$ = $$\frac{5}{2}$$

∴ Percentage of girls = $$\frac{y}{x+y} \times 100$$ = $$\frac{2}{5} \times 100$$ = 40%

Ans .

50%

1. Explanation :

Let the number of books in shelf B be 100.

∴ Number of books in shelf A = 80

On transferring 25% i.e $$\frac{1}{4}$$ of books of shelf A to shelf B.

B = 100 + 20 = 120

Again, on transferring $$\frac{1}{4}$$ of books of shelf B to shelf A.

A = 60 + $$\frac{120}{4}$$ = 90

∴ Required percentage = $$\frac{90}{180} \times 100$$ = 50%

Ans .

178200

1. Explanation :

Total revenue earned = $$9900 \times \frac{20}{100} \times 10$$ + $$9900 \times \frac{80}{100} \times 20$$

= (19800 + 158400)

= 178200

Ans .

$$77 \frac{7}{9}$$

1. Explanation :

Let Tina’s weight = 1 kg

Lina’s weight = 2 kg

Neha’s weight = 1.4kg

Mina’s weight = 1.8 kg.

$$\frac{1.8x}{100}$$ = 1.4

$$→$$ x = $$77 \frac{7}{9}$$

Ans .

37.5%

1. Explanation :

Let the number of seats initially in the cinema hall be 100 and the cost of each ticket be

100.

∴ Total revenue = 100 × 100 = 10000

In second condition,

Number of seats = 125

Cost of each ticket = 110

∴ New revenue = 125 × 110 = 13750

Increase in revenue collection = . (13750 – 10000) = 3750

∴ Percentage increase = $$\frac{3750}{10000} \times 100$$ = 37.5%

Ans .

80

1. Explanation :

Total amount = x

∴ x - $$\frac{x}{5}$$ - $$\frac{4x}{5} \times \frac{5}{100}$$ - 120 = 1400

$$\frac{19x}{25}$$ = 1520

$$→$$ x = 2000

∴ Expenditure on transport = $$\frac{1}{25} \times 2000$$ = 80

Ans .

30,000

1. Explanation :

Let the total number of employees be x.

$$x \times \frac{69}{100}$$ = 20700

$$→$$ x = $$\frac{20700 \times 100}{69}$$ = 30,000

Ans .

700 apples

1. Explanation :

Let the fruit seller had originally x apples.

According to the question;

x – 40% of x = 420

x - $$\frac{40}{100} \times x$$ = 420

$$\frac{3x}{5}$$ = 420

x = 700

Ans .

80%

1. Explanation :

Let third number = 100

First number = 120

Second number = 150

Required percentage = $$\frac{120}{150} \times 100$$ = 80%

Ans .

$$45 \frac{5}{11}$$%

1. Explanation :

The batsman scored 3 × 4 + 8 × 6 = 60 runs by boundariesand sixes respectively.

Then, Runs scored by running

= 110 – 60 = 50

\ Required percentage = $$\frac{50}{110} \times 100$$ = $$\frac{500}{11}$$ =$$45 \frac{5}{11}$$%

Ans .

123.75

1. Explanation :

y = $$\frac{110}{100} \times 125$$ = 137 5

∴ x = 90% of y

= $$\frac{90 \times 137.5}{100}$$ = 123.75

Ans .

2.5%

1. Explanation :

Error = 5.5 minutes

∴ Error per cent = $$\frac{5.5}{3 \times 60 + 40} \times 100$$ = 2.5 per cent

Ans .

24

1. Explanation :

Per cent of families having either a cow or a buffalo or both = 60 + 30 – 15 = 75

It means 25 per cent of families do not have a cow or a buffalo.

∴ Required number of families = 25% of 96 =$$96 \times \frac{25}{100}$$ = 24

Ans .

3,850

1. Explanation :

Let the amount invested at 6% = x

∴ Amount invested at 5% = (10000 – x )

According to the question

$$\frac{(10000 - x) \times 5}{100}$$ - $$\frac{x \times 6}{100}$$ = 76.50

$$→$$ 50000 – 5x – 6x = 7650

$$→$$ 50000 – 11x = 7650

$$→$$ 11x = 50000 – 7650 = 42350

x = 3850

Ans .

3

1. Explanation :

Kites of 20 are available for 19.

Hence, discount = 5%

If one gets kites of 20 for 18, discount = 10%

20 kites $$→$$ 2 kites

27 kites $$→$$ $$\frac{2}{20} \times 27$$ = 3

Ans .

135

1. Explanation :

First part = x and second part = y.

According to question

$$\frac{x \times 80}{100}$$ = $$\frac{y \times 60}{100}$$ + 3

$$→$$ 4x – 3y = 15 ...(i)

Again

$$→$$ 8y = 9x + 60

$$→$$ 8y – 9x = 60 ...(ii)

By equation (i) × 8 + (ii) × 3,

32x – 24y = 120

24y – 27x = 180

5x = 300 $$→$$ x = 60

From equation (i)

4 × 60– 3y = 15

$$→$$ 3y = 240 – 15 = 225

$$→$$ y = $$\frac{225}{3}$$ = 75

∴ x + y = 60 + 75 = 135

Ans .

$$11 \frac{31}{54}$$%

1. Explanation :

Value of 100 stock = 108

∵ Income on investing 108 = $$\frac{25}{2}$$

∴ Income on investment of 27000

= $$\frac{25}{2 \times 108} \times 27000$$

= 3125

∴ Gain per cent = $$\frac{3125}{27000} \times 100$$ = $$11 \frac{31}{54}$$%

Ans .

4764

1. Explanation :

Required mass of lead = $$8000 \times \frac{60}{100} \times (1-\frac{3}{400})$$

= $$8000 \times \frac{60}{100} \times (\frac{397}{400})$$ = 4764 kg.

Ans .

$$\frac{2}{9}$$

1. Explanation :

According to the question $$x + y = (x^2 + y^2) \times \frac{1}{5}$$

Again,

$$x + y = (x^2 - y^2) \times \frac{1}{4}$$

$$\frac{x^2 + y^2}{5} = \frac{x^2 - y^2}{4}$$

$$→$$$$5x^2 -4x^2 = 5y^2 + 4y^2$$

$$→$$$$x^2 = 9y^2$$

$$→$$ x = 3y

$$\frac{x + y}{x^2} = \frac{x^2 + y^2}{5x^2} = \frac{2}{9}$$

Ans .

200

1. Explanation :

Let the total number of eggs bought be x.

10% of eggs are rotten.

\ Remaining eggs = $$\frac{90x}{100}$$ = $$\frac{9x}{10}$$

After giving 80% of eggs to the neighbour,

Remaining eggs = $$\frac{9x}{50}$$

According to the question

$$\frac{9x}{50}$$ = 36

$$→$$ 9x = 36 × 50

$$→$$ x = 200

Ans .

Rs. 4,00,000

1. Explanation :

Amount with man in the

beginning = Rs. x (let).

Amount given to son and daughter = 80%.

Remaining amount = 20% of x = Rs.$$\frac{x}{5}$$

Remaining amount after donations to trust = $$\frac{x}{5} \times \frac{20}{100}$$ = Rs. $$\frac{x}{25}$$

∴ $$\frac{x}{25}$$ = 16000

$$→$$ x = 16000 × 25 = Rs. 400000

Ans .

Rs. 40000

1. Explanation :

Let the business man’s present earning be Rs. x.

According to the question,

$$x \times \frac{125}{100} \times \frac{96}{100} \times \frac{125}{100} \times \frac{96}{100} \times \frac{125}{100}$$ = 72000

$$x \times \frac{5}{4} \times \frac{24}{25} \times \frac{5}{4} \times \frac{24}{25} \times \frac{5}{4}$$ = 72000

$$x \times \frac{9}{5}$$ = 100

x = Rs. 40000

Ans .

48025

1. Explanation :

Number of blood cells in first 6 hours = $$40000 (1+\frac{10}{100})^2(1-\frac{10}{100})(1+\frac{5}{100})^2$$ = 48024.9 = 48025

Ans .

$$33 \frac{1}{3}$$%

1. Explanation :

Let 100 pairs of shoes be bought for Rs. 100.

New budget = Rs. 160

New price = Rs. 1.20 pair of shoes

∴ Number of shoes bought = $$\frac{160}{1.2}$$ = $$133 \frac{1}{3}$$

∴ Percentage increase = $$33 \frac{1}{33}$$%

Ans .

84

1. Explanation :

Average of set A = $$\frac{27 + 28 + 30 + 32 + 33}{5}$$ = $$\frac{150}{5}$$ = 30

Case II,

New average = $$\frac{30 \times 130}{100}$$ = 39

∴ 150 + k = 39 × 6 = 234

$$→$$ k = 234 – 150 = 84

Ans .

Rs. 128

1. Explanation :

Initial amount with the man = Rs. x (let).

Remaining amount after first bet = Rs. $$\frac{x}{4}$$

Remaining amount after second bet = Rs.$$\frac{x}{16}$$

Remaining amount after third bet = Rs.$$\frac{x}{64}$$

∴ $$\frac{x}{64}$$ = 2

x = 128

Ans .

1000000

1. Explanation :

Initial number of soldiers in the army = x

According to the question

$$x \times (\frac{90}{100})^3$$ = 729000

$$→$$ x = = 1000000

Ans .

16

1. Explanation :

Required percentage decrease = $$\frac{25-21}{25} \times 100$$ = $$\frac{400}{25}$$ = 16%

Ans .

= 11088

1. Explanation :

Required number of workers = $$8000 \times \frac{105}{100} \times \frac{110}{100} \times \frac{120}{100}$$

= 11088

Ans .

2000

1. Explanation :

1% = 100 basis points

∴ 82.5% = 8250 basis points and 62.5% = 6250 basis points

∴ Required difference

= 8250 – 6250

= 2000 basis points

Ans .

$$22 \frac{8}{11}$$%

1. Explanation :

Percentage increase in sales = $$\frac{51300 - 41800}{41800} \times 100$$

= $$\frac{9500}{418}$$ = $$22 \frac{8}{11}$$%

Ans .

7

1. Explanation :

Defective parts of 120 machine parts = $$\frac{120 \times 5}{100}$$ = 6

Defective parts of 80 machine parts = $$\frac{80 \times 10}{100}$$ = 8

Total defective parts = 6 + 8 = 14

∴ Required percent = $$\frac{14}{200} \times 100$$ = 7%

Ans .

$$3 \frac{1}{3}$$%

1. Explanation :

Error = (1.55 – 1.5) metre = 0.55 metre

∴ Error per cent = $$\frac{0.05}{15} \times 100$$ = $$\frac{50}{15}$$ = $$3 \frac{1}{3}$$%

Ans .

36500

1. Explanation :

Duty payment :

Laptop $$→$$ Rs $$\frac{210000 \times 10}{100}$$ = Rs. 21000

Mobile phone $$→$$ Rs $$\frac{100000 \times 8}{100}$$ = Rs. 8000

Television set $$→$$ Rs $$\frac{150000 \times 5}{100}$$ = Rs. 7500

Total Duty Payment

= Rs. (21000 + 8000 + 7500)

= Rs. 36500

Ans .

Rs. 1600

1. Explanation :

Initial amount with the person = Rs. x (let)

After an expense of $$\frac{15}{2}$$%

Remaining amount = $$(100 - \frac{15}{2})% of x.$$

= $$(\frac{185}{2})% of x.$$ = Rs. $$\frac{185x}{200}$$

After an expense of 75% of it,

Remaining amount = $$\frac{37x}{40 \times 4}$$ = $$\frac{37x}{160}$$

According to the question,

$$\frac{37x}{160}$$ = 370

$$→$$ 37x = 370 × 160

$$→$$ x = Rs. 1600

Ans .

60

1. Explanation :

Let the number of matches played between India and Pakistan in the first case be x.

Number of wins by Pakistan = $$\frac{60x}{100}$$ = $$\frac{3x}{5}$$

$$\frac{\frac{3x}{5}}{x+30}$$ = $$\frac{30}{100}$$

$$→$$ 2x = x + 30

$$→$$ x = 30

∴ Total number of matches = 30 + 30 = 60