Staff Selection Commission Mathematics - Percentage (1999-2017)

TYPE–XI (Contd.)

Q.1

A reduction of 21% in the price of an item enables a person to buy 3 kg more for 100. The reduced price of item per kg is

5.50

7.50

10.50

7.00

Ans .

7.00

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

Q.2

A reduction of 20% in the price of sugar enables a purchaser to obtain 8 kg more for Rs. 160. Then the price per kg before reduction was

Rs. 5

Rs. 6

Rs. 10

Rs. 4

Ans .

Rs. 4

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

Q.3

The price of an item was increased by 10%. This reduced the monthly total sales by 20%. The overall effect on the value of monthly sales is a

10% increase

10% decrease

12% increase

12% decrease

Ans .

12% decrease

Explanation :
Required percentage change = 10 - 20 + \(\frac{10 \times -20}{100}\) % = -12%

Negative sign shows decrease.

Q.4

The price of rice has increased by 60%. In order to restore the original price, the new price must be reduced by

\( 33 \frac{1}{3}\)%

\( 37 \frac{1}{2}\)%

40%

45%

Ans .

\( 37 \frac{1}{2}\)%

Explanation :
Required per cent = \(\frac{x}{100 + x} \times 100\)

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

Q.5

If the price of sugar increases by 20%, one can buy 2 kg less for Rs. 50. What is the amount of sugar that could be bought before price hike?

10 kg.

12 kg.

14 kg.

16 kg.

Ans .

12 kg.

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.

Q.6

. The price of an article is decreased by 10%. To restore it to its former value, the new price must be increased by :

\( 9 \frac{1}{11}\)%

10%

11%

\( 11 \frac{1}{9}\)%

Ans .

\( 11 \frac{1}{9}\)%

Explanation :
Required per cent = \(\frac{Decrease%}{100-Decrease%} \times 100\)

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

Q.7

. The salary of a person is reduced by 20%. To restore the previous salary, his present salary is to be increased by

20%

25%

17.5%

22.5%

Ans .

25%

Explanation :
Required percentage increase = \(\frac{x}{100-x} \times 100\)

= \(\frac{20}{100-20} \times 100\) = 25%

Q.8

. Due to a price hike of 20%, 4 kg. less sugar is available for Rs. 120. What is the initial price per kg of sugar ?

Rs. 5 per kg

Rs. 4 per kg.

Rs. 6 per kg

Rs. 5.5 per kg

Ans .

Rs. 5 per kg

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.

Q.9

In 2001, the price of a building was 80% of its original price. In 2002, the price was 60% of its original price. By what percent did the price decrease ?

15%

20%

25%

30%

Ans .

25%

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

Q.10

In a school 70% of the students are girls. The number of boys are 510. Then the total number of students in the school is :

850

1700

1830

1900

Ans .

1700

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

Q.11

. If 60% of the students in a school are boys and the number of girls is 972, how many boys are there in the school ?

1258

1458

1324

1624

Ans .

1458

Explanation :
40% of students = 972

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

Q.12

. If 70% of the students in a school are boys and the number of girls be 504, the number of boys is :

1176

1008

1208

3024

Ans .

1176

Explanation :
Number of boys = \(\frac{70}{30} \times 504\) = 1176

Q.13

If the sales tax on a television set increases from to 8%, \( 7 \frac{1}{2}\)% what more amount will have to be paid for the television whose price (excluding sales taxes) is 19000 ?

190

180

Ans .

Explanation :
Required sum = 0.5% of 19000 = \(19000 \times \frac{0.5}{100}\) = 95

Q.14

. A spider climbed \( 62 \frac{1}{2}\)% of the height of the pole in one hour and in the next hour it covered \( 12 \frac{1}{2}\)% of the remaining height. If pole’s height is 192 m, then distance climbed in second hour is

3 m

5 m

7 m

9 m

Ans .

9 m

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

Q.15

. Fresh fruit contains 68% water and dry fruit contains 20% water. How much dry fruit can be obtained from 100 kgs of fresh fruits ?

32 kgs

40 kgs.

52 kgs.

80 kgs

Ans .

52 kgs.

Explanation :
Water in 100 kg fresh fruit = 68%

Water in dry fruit = 20%

Decrease = 48%

∴ Dry fruit obtained = 100 – 48 = 52 kg

Q.16

An individual pays 30% income tax. On this tax he has to pay a surcharge of 10%. Thus, the net tax rate, he has to pay, is

45%

40%

33%

27%

Ans .

33%

Explanation :
The net tax rate = 30 + \(30 \times \frac{10}{100}\) % = 33%

Q.17

X has twice as much money as that of Y and Y has 50% more money than that of Z. If the average money of all of them is 110, then the money, which X has, is

180

Ans .

180

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

Q.18

p is six times as large as q. The per cent that q is less than p, is

\( 83 \frac{1}{3}\)%

70%

\( 63 \frac{1}{3}\)%

50%

Ans .

\( 83 \frac{1}{3}\)%

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 = \(\frac{500}{100+500} \times 100\) = \( 83 \frac{1}{3}\)%

Q.19

. In the expression x\(y^2\), the values of both variables x and y are decreased by 20%. By this, the value of the expression is decreased by

40%

80%

48.8%

51.2%

Ans .

48.8%

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%

Q.20

Two numbers are respectively 10% and 25% more than a third number. What per cent is the first of the second ?

88%

65%

75%

80%

Ans .

88%

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%

Q.21

A boy who was asked to find \( 3 \frac{1}{2}\)% of a sum of money misread the question and found \( 5 \frac{1}{2}\)% of it. His answer was 220. What would have been the correct answer ?

120

140

160

150

Ans .

140

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

Q.22

In a factory 60% of the workers are above 30 years and of these 75% are males and the rest are females. If there are 1350 male workers above 30 years, the total number of workers in the factory is

3000

2000

1800

1500

Ans .

3000

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

Q.23

First and second numbers are less than a third number by 30% and 37% respectively. The second number is less than the first by

10%

Ans .

10%

Explanation :
Let the third number = 100.

∴ First number = 70

Second number = 63

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

Q.24

Rani’s weight is 25% that of Meena’s and 40% that of Tara’s. What percentage of Tara’s weight is equal to Meena’s weight?

160%

140%

120%

100%

Ans .

160%

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%

Q.25

Out of 2500 people, only 60% have the saving habit. If 30% save with bank, 32% with post office and the rest with shares, the number of shareholders are

450

570

950

1250

Ans .

570

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

Q.26

The value of a commodity depreciates 10% annually. If it was purchased 3 years ago and its present value is 5,832, what was its purchase price ?

7200

7862

8000

8500

Ans .

8000

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

Q.27

A and B are two fixed points 5 cm apart and C is a point on AB such that AC is 3 cm. If the length of AC is increased by 6%, the length of CB is decreased by

Ans .

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%

Q.28

If 24-carat gold is considered to be hundred per cent pure gold, then the percentage of pure gold in 22-carat gold is :

\( 91 \frac{3}{4}\)%

\( 91 \frac{2}{3}\)%

\( 91 \frac{1}{3}\)%

\( 90 \frac{2}{3}\)%

Ans .

\( 91 \frac{2}{3}\)%

Explanation :
∵ 24= 100% ∴22 = \(\frac{100}{24} \times 22\) = \( 91 \frac{2}{3}\)%

Q.29

In a class, the average score of girls in an examination is 73 and that of boys is 71. The average score for the whole class is 71.8. Find the percentage of girls.

40%

50%

55%

60%

Ans .

40%

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%

Q.30

. Shelf A has \(\frac{4}{5}\) of the number of books that shelf B has. If 25% of the books in A are transferred to B and then 25% of the books from B are transferred to A, then the percentage of the total number of books that A will have is

25%

50%

75%

100%

Ans .

50%

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%

Q.31

Tickets for all but 100 seats in a 10,000 seat stadium were sold. Of the tickets sold, 20% were sold at half price and the remaining tickets were sold at the full price of 20. The total revenue from the ticket sales, in was

158400

178200

180000

198000

Ans .

178200

Explanation :
Total revenue earned = \(9900 \times \frac{20}{100} \times 10 \) + \(9900 \times \frac{80}{100} \times 20 \)

= (19800 + 158400)

= 178200

Q.32

Neha’s weight is 140% of Tina’s weight. Mina’s weight is 90% of Lina’s weight. Lina weighs twice as much as Tina. If Neha’s weight is x% of Mina’s weight, then x is equal to :

\( 64 \frac{2}{9}\)

\( 77 \frac{7}{9}\)

\( 128 \frac{4}{7}\)

Ans .

\( 77 \frac{7}{9}\)

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 = \( 77 \frac{7}{9}\)

Q.33

. The number of seats in a cinema hall is increased by 25%. The cost of each ticket is also increased by 10%. The effect of these changes on the revenue collection will be an increase of

37.5%

45.5%

47.5%

49.5%

Ans .

37.5%

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%

Q.34

. A man had a certain amount with him. He spent 20% of that to buy an article and 5% of the remaining on transport. Then he gifted 120. If he is left with 1,400, the amount he spent on transport is

Ans .

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

Q.35

. 31% of employees pay tax in the year 2008. Non-tax paying employees are 20,700. The total number of employees are :

31,160

64,750

30,000

66,775

Ans .

30,000

Explanation :
Let the total number of employees be x.

\(x \times \frac{69}{100}\) = 20700

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

Q.36

. A fruit seller had some apples. He sells 40% apples and still has 420 apples. Originally, he had :

588 apples

600 apples

672 apples

700 apples

Ans .

700 apples

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

Q.37

Two numbers are more than the third number by 20% and 50% respectively. First number is what per cent of the second number ?

100%

150%

80%

120%

Ans .

80%

Explanation :
Let third number = 100

First number = 120

Second number = 150

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

Q.38

A batsman scored 110 runs which included 3 boundaries and 8 sixes. What per cent of his total score, did he make by running between the wickets ?

45%

\( 45 \frac{5}{11}\)%

\( 54 \frac{6}{11}\)%

55%

Ans .

\( 45 \frac{5}{11}\)%

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

Then, Runs scored by running

= 110 – 60 = 50

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

Q.39

If a number x is 10% less than another number y and y is 10% more than 125, then x is equal to

150

143

140.55

123.75

Ans .

123.75

Explanation :
y = \(\frac{110}{100} \times 125\) = 137 5

∴ x = 90% of y

= \(\frac{90 \times 137.5}{100}\) = 123.75

Q.40

An interval of 3 hours 40 minutes is wrongly estimated as 3 hours 45.5 minutes. The error percentage is

5.5%

5.2%

2.5%

Ans .

2.5%

Explanation :
Error = 5.5 minutes

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

Q.41

In a village, each of the 60% of families has a cow; each of the 30% of families has a buffalo and each of the 15% of families has both a cow and buffalo. In all there are 96 families in the village. How many families do not have a cow or a buffalo ?

Ans .

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

Q.42

A man invests a part of 10,000 at 5% and the remainder at 6%. The 5% investment yields annually 76.50 more than the 6% investment . The amount invested at 6% is

3,600

3,550

3,850

4,000

Ans .

3,850

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

Q.43

For every set of 19 kites sold, a vendor gives 1 kite extra, free of cost. In order to give a discount of 10%, the number of extra kites he should give in a sale of 27 kites to the nearest integer is

Ans .

Explanation :
Kites of 20 are available for 19.

Hence, discount = 5%

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

∴ Required answer

20 kites \(→\) 2 kites

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

Q.44

A number is divided into two parts in such a way that 80% of 1st part is 3 more than 60% of 2nd part and 80% of 2nd part is 6 more than 90% of the 1st part. Then the number is

125

130

135

145

Ans .

135

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

Q.45

A man invested 27,000 in \( 12 \frac{1}{2}\)% stock at 108, then his yield percentage is

\( 18 \frac{3}{4}\)%

\( 11 \frac{31}{54}\)%

15%

\( 8 \frac{1}{2}\)%

Ans .

\( 11 \frac{31}{54}\)%

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 = \(\frac{3125}{27000} \times 100\) = \( 11 \frac{31}{54}\)%

Q.46

The percentage of metals in a mine of lead ore is 60%. Now the percentage of silver is \(\frac{3}{4}\)% of metals and the rest is lead. If the mass of ore extracted from this mine is 8000 kg, the mass (in kg.) of lead is :

4763

4762

4764

4761

Ans .

4764

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.

Q.47

The sum of two positive numbers is 20% of the sum of their squares and 25% of the difference of their squares. If the numbers are x and y then, \(\frac{x+y}{x^2}\) is equal to

\(\frac{1}{4}\)

\(\frac{3}{8}\)

\(\frac{1}{3}\)

\(\frac{2}{9}\)

Ans .

\(\frac{2}{9}\)

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

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

Q.48

A man bought some eggs of which 10% are rotten. He gives 80% of the remainder to his neighbours. Now he is left out with 36 eggs. How many eggs he bought ?

100

200

Ans .

200

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

Q.49

A man gives 50% of his money to his son and 30% to his daughter. 80% of the rest is donated to a trust. If he is left with Rs. 16,000 now, how much money did he have in the beginning?

Rs. 40,000

Rs. 8,00,000

Rs. 80,000

Rs. 4,00,000

Ans .

Rs. 4,00,000

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

Q.50

. A businessman’s earning increases by 25% in one year but decreases by 4% in the next year. Going by this pattern, after 5 years, his total earnings would be Rs.72000. What is his present earning ?

Rs. 10000

Rs. 80000

Rs. 40000

Rs. 54000

Ans .

Rs. 40000

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

Q.51

The red blood cells in a blood sample grows by 10% per hour in first two hours, decreases by 10% in next one hour, remains constant in next one hour and again increases by 5% per hour in next two hours. If the original count of the red blood cells in the sample is 40000, find the approximate red blood cell count at the end of 6 hours.

40000

45025

48025

50025

Ans .

48025

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

Q.52

A factory’s yearly budget for the purchase of employees’ protection shoes increased by 60% this year over last year. If the price of the shoes increased by 20% this year, then the number of shoes it can purchase this year is what percent greater than the number of shoes it purchased last year ?

40%

\( 33 \frac{1}{3}\)%

42%

48%

Ans .

\( 33 \frac{1}{3}\)%

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 = \( 33 \frac{1}{33}\)%

Q.53

A Set A consists of integers 27, 28, 30, 32, and 33. If integer k is included in the average of set A will increase by 30%. What is the value of integer K ?

Ans .

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

Q.54

. A person loses 75% of his money in the first bet, 75% of the remaining in the second and 75% of the remaining in the third bet and returns home with Rs. 2 only. His initial money was

Rs. 64

Rs. 128

Rs. 256

Rs. 512

Ans .

Rs. 128

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

Q.55

An army lost 10% of its men in war, 10% of the remaining died due to disease and 10% of the rest were declared disabled. Thus the strength of the army was reduced to 7,29,000 active men. The original strength of the army was

1500000

1000000

1200000

1100000

Ans .

1000000

Explanation :
Initial number of soldiers in the army = x

According to the question

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

\(→\) x = = 1000000

Q.56

If the value of a company stock drops from Rs. 25 per share to Rs. 21 per share, the percentage decrease per share is :

Ans .

Explanation :
Required percentage decrease = \(\frac{25-21}{25} \times 100\) = \(\frac{400}{25}\) = 16%

Q.57

Starting with 8000 workers, the company increases the number of workers by 5%, 10% and 20% at the end of first, second and third year respectively. The number of workers in the fourth year was

10188

11088

11008

11808

Ans .

= 11088

Explanation :
Required number of workers = \(8000 \times \frac{105}{100} \times \frac{110}{100} \times \frac{120}{100}\)

= 11088

Q.58

. If “basis points” are defined so that 1 per cent is equal to 100 basis points, then by how many basis points is 82.5 per cent greater than 62.5 per cent ?

0.2

200

2000

Ans .

2000

Explanation :
1% = 100 basis points

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

∴ Required difference

= 8250 – 6250

= 2000 basis points

Q.59

In the last financial year, a car company sold 41,800 cars. In this year, the target is to sell 51,300 cars. By what per cent must the sale be increased ?

\( 11 \frac{9}{22}\)%

\( 8 \frac{9}{22}\)%

\( 8 \frac{11}{23}\)%

\( 22 \frac{8}{11}\)%

Ans .

\( 22 \frac{8}{11}\)%

Explanation :
Percentage increase in sales = \(\frac{51300 - 41800}{41800} \times 100\)

= \(\frac{9500}{418}\) = \( 22 \frac{8}{11}\)%

Q.60

In a motor of 120 machine parts, 5% parts were defective. In another motor of 80 machine parts, 10% parts were defective. For the two motors considered together, the percentage of defective machine parts was

6.5

7.5

Ans .

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%

Q.61

A line of length 1.5 metres was measured as 1.55 metres by mistake. What will be the value of error per cent ?

0.05%

\( 3 \frac{7}{31}\)%

\( 3 \frac{1}{3}\)%

0.5%

Ans .

\( 3 \frac{1}{3}\)%

Explanation :
Error = (1.55 – 1.5) metre = 0.55 metre

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

Q.62

A businessman imported Laptops, worth Rs. 210000, Mobile phones worth Rs. 100000 and Television sets worth Rs. 150000. He had to pay 10% duty on laptops, 8% on Phones and 5% on Television sets as a special case. How much total duty (in Rupees) he had to pay on all items as per above details?

36500

37000

37250

37500

Ans .

36500

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

Q.63

A man spends \( 7 \frac{1}{2}\)% of his money and after spending 75% of the remaining he had Rs. 370 left. How much money did he have?

Rs. 1200

Rs. 1600

Rs. 1500

Rs. 1400

Ans .

Rs. 1600

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

Q.64

. On a certain date, Pakistan has a success rate of 60% against India in all the ODIs played between the two countries. They lost the next 30 ODIs in a row to India and their success rate comes down to 30%. The total number of ODIs played between the two countries is

Ans .

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

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