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1. Three numbers are such that the first number

is 30% of third number and second number is

40% of the third number. First number is

what percent of the second number?

a) 133.33% b) 25% c) 33.33% d) 75%

Sol: Let the third number be x. Then,

3First number = 30% of x = x

10

4and second number = 40% of x = x

10

First number∴Percentage value = × 100

Second number

3 x 10 3

= × 100 = × 100 = 75%. 4 4

x 10 Ans: d

2. What is the percentage increase in the area of

a rectangle if its length is increased by 20%?

2a) 44% b) 40% c) 20% d) 16 %

3

Sol: Area of rectangle = L × B

New length = 1.2 L; New area = 1.2 L × B

Increase in area = 20%. Ans: c

3. After spending 20% on clothes, 10% on

books, 9% on purchasing gift for girl friend

and 7% on others, Chandra has a balance of

Rs. 2,700. How much money was there with

him initially?

a) Rs.5,000 b) Rs.5,400

c) Rs.2,500 d) Rs.2,700

Sol: Let the initial money that Chandra had be x.

Balance left with Chandra = x − 20% of

x − 10% of x − 9% of x − 7% of x = 2,700

⇒ 0.54 x = 2700; ⇒ x = Rs. 5,000.

Ans: a

4. In an election contested by two, the loser

loses by a margin of 20% of the total votes

polled, which is equivalent to 20,000 votes. If

only 50% of the total eligible people cast their

votes, then how many total people were

eligible for casting their votes?

a) 1,00,000 b) 50,000

c) 4,00,000 d) 2,00,000

Sol: 20% of total votes = 20,000 votes

⇒ Total votes = 2,00,000. Ans: d

5. A batsman scored 110 runs which included 3

boundaries and 8 sixes. What percent of his

total score did he make by running between

the wickets?

6 5a) 50% b) 54 % c) 45 % d) 48%

11 11

Sol: Runs by boundaries and sixes

= (3 × 4) + (8 × 6) = 60

Total number of runs scored = 110

∴ Runs scored by running between the wickets

= 110 − 60 = 50

Percentage of runs scored by running

50 5 between the wickets × 100 = 45 %.

110 11

Ans: c

6. If the numerator of a fraction be increased by

15% and its denominator be diminished by

158%, then the new value of the fraction is .

16

Find the original fraction.

4 10 6 3a) b) c) d)

5 13 7 4

xSol: Let the fraction be .

y

x + 15% of x 15Then, =

y − 8% of y 16

1.15x 15 x 15 0.92 3⇒ = ⇒ = × = .

0.92 16 y 16 1.15 4

Ans: d

7. In an examination consisting of 4 subjects,

the marks obtained by Shyam in 3 of them

are 90%, 95% and 95% respectively. Each

subject is of equal marks. Under the given

circumstances his average percentage

marks for the examination cannot be

a) 94% b) 90% c) 93% d) 96%

Sol: Let 100 be the maximum marks for each

subject. Then, from option (d),

90 + 95 + 95 + x = 96 ⇒ x = 104

4

which is notpossible.

Hence, his average % marks cannot be 96%.

Ans: d

8. If X = 37.5% of 20% of 48 and Y = 14.28% of

27.27% of 77, then

a) X >Y b) X = Y

c) X < Y d) X − Y = 1.4

Sol: X = 37.5% of 20% of 48

3 1= × × 48 = 3.6

8 5

Y = 14.28% of 27.27% of 77

1 3= × × 77 = 3 ∴ X > Y. Ans: a

7 11

9. A candidate who gets 20% marks fails by 10

marks but another candidate who gets 42%

marks gets 12 marks more than the passing

marks. Find the maximum marks.

a) 50 b) 100 c) 150 d) None

Sol: Let maximum marks be x.

20 42Then, x + 10 = x − 12

100 100

⇒ x = 100. Ans: b

10. In a town of population 1,20,000, 55% are

males and rest are females. If 48% of males

and 60% of females can vote, then what is

the total number of voters in the town?

a) 64,320 b) 64,080

c) 70,000 d) None of these

Sol: Male population = 1,20,000 × 0.55 = 66,000

∴ Female population = 1,20,000 − 66,000

= 54,000.

Number of male voters = 66,000 × 0.48

= 31,680

Number of female voters = 54,000 × 0.60

= 32,400

∴ Total number of voters = 32,400 + 31,680

= 64,080. Ans: b

11. In an election, there were three candidates.

Out of total 1200 cast votes, Ram received

30%, Balu received 720 votes and Kapil

received the rest of the votes. Find out per-

cent of votes which the winner got in com-

parison to his closest rival?

a) 100% b) 200% c) 180% d) 90%

Sol: Total votes = 1200

Ram received = 0.30 × 1200 = 360

Balu received = 720

Kapil received = 1200 − (360 + 720) = 120

Percentages of votes which the winner got in

comparison to his closest rival is given by

720= × 100 = 200%. Ans: b

360

12. The incomes of X, Y and Z are in the ratio

2 : 3 : 5 respectively. If the income of Y is

Rs.9,000, then by what percent is income of

Z more than that of X?

a) 50% b) 60% c) 150% d) 25%

Sol: Let X = 2x, Y = 3x and Z = 5x. Then,

3x = 9000 ⇒ x = Rs.3,000

∴ X = Rs.6,000 and Z = Rs.15,000

9000∴ Percentage value = × 100 = 150%.

6000

Ans: c

13. The income of a property dealer remains

unchanged though the rate of commission

is increased from 8% to 10%. The percent-

age change in the value of the business is

a) 2% b) 20% c) 28% d) 15%

Sol: Take initial value of business = Rs.100

So, commission of dealer = 8% of 100 = Rs.8

For next case also, commission for dealer

= Rs.8

Take final value of business = a

a × 10% = 8 ⇒ a = Rs.80

Percentage change in value of business

20= × 100 = 20%. Ans: b

100

14. The value of a machine depreciates at the

rate of 10% per annum. If its present value

is Rs.1,62,000, then what was the value of

the machine 2 years ago?

a) Rs.1,00,000 b) Rs.2,00,000

c) Rs.2,50,000 d) Rs.1,80,000

Sol: Let the value of the machine two years ago

be x.

Then, value of machine after one year

= x − 10% of x = 0.9x

Further value of machine after two years,

i.e., present value = 0.9x − 10% of 0.9x

= 0.81x

∴ Present value = 0.81x = 1,62,000

⇒ x = Rs.2,00,000. Ans: b

15. The radius of a sphere is 14 cm. The cost of

painting the surface of sphere is Rs.25 per

square cm. If the radius of sphere is

increased by 10%, then the cost of painting

is increased by 20%. What is the percent-

age increase in the total cost of painting per

square cm?

a) 54.27% b) 20.3% c) 2.58% d) 45.2%

Sol: Total cost of painting = Rate × Surface

area of sphere

As the radius increases by 10%.

Surface area will change by 21%

a + b + (a × b)[using formula ]100

Total change in cost of painting

20 × 21= 21 + 20 + = 45.2%.

100

Ans: d

16. A student multiplies a number by 5 instead

of dividing it by 5. What is the percentage

change in the result due to this mistake?

a) 2500% b) 98% c) 100% d) 2400%

Sol: Let the number be x.

xCorrect answer = .; Wrong answer = 5x

5

x(5x − )5Percentage error = × 100 = 2400%.

x 5

Ans: d

17. A candidate who gets 30% of the total

marks fails by 14 marks but another candi-

date who gets 45% of the total marks gets

16 marks more than the passing marks.

Find the passing marks.

a) 200 b) 74 c) 60 d) 84

Sol: Let maximum marks be x.

30 45 x + 14 = x − 16 ⇒ x = 200100 100

30Hence, passing marks = × 200 + 14 = 74

100

Ans: b

18. Sandeep saves 30% of his salary and

spends remaining. Out of his total savings,

he invests 40% in LIC policy, 35% in HDFC

and the remaining on other. If the difference

between the amount invested in LIC and

others is Rs.135. What is his salary?

a) Rs.3,000 b) Rs.2,000

c) Rs.2,800 d) Rs.1,500

Sol: Let x be the salary of Sandeep.

Then, savings = 30% of x

Expenses = 70% of x

LIC = 40% of 30% of x

Others = 25% of 30% of x

Difference = 40% of 30% of x − 25% of 30%

of x

⇒ 135 = 15% of 30% of x ⇒ x = Rs.3,000.

Ans: a

19. Number of students who passed in a class

is 20% greater than those who failed. Find

by what percent, failure are lesser than

those who passed?

a) 54.27% b) 20.3% c) 2.58% d) 16.66%

Sol: Let number of students failed = 100

Number of students who passed = 120

Percentages by which number of students

failed is less than who have passed is given by

100= (1 − ) × 100 = 16.66%.

120 Ans: d

Percentage

SSC ExamsSpecial

Quantitative Aptitude

Number of voters in the town are...

These model questions were prepared by

Subject experts of Career Launcher,

Hyderabad

ÎCî¦ô¢Ù áì÷J 24, 2021 n e-mail: [email protected]

17

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32.- Ú¨ÙC Ô ÓûÂâËµj-îÂªÚÛª >ÙÚÂ ú£ï£°-Ú¥-ô¢-ÚÛÙÞ¥sÚÁðÆ§u-ÚÛdôÂ) ÑÙåªÙC?

1) óŸ«ú‡èÂ ðÆ§ú£pÄ-çË¶âËÀ 2) Îõ\-öËµjûÂ ðÆ§ú£pÄ-çË¶âËÀ3) Óîµªi-öË¶âËÀ 4) öËµj›íâËÀ

33.- Ú¨ÙC-î¦-æ¨öËº óŸ«Ùæ©-Î-Ú¨q-èËµÙæÀ Nå-NªìªxÔN?

i.- D1 ii. A iii. C iv.- E

1) i, ii, iii 2) ii, iii, iv

3) i, iii, iv 4) i, ii, iii, iv

N-å-Nª-ûÂõª n- õ÷-é°õª

áì-ô¢öËÀ šújûÂq n ñóŸ«õ@

ÎôÂÎôÂH, ÏêŸô¢ ð¼æ©̈í£K¤Ûõ ví£ê¶uÚÛÙ

ô¢àŸô³êŸÚ•ô¦xÙ þ§ô³î�µÙÚÛçË¶ùÃ

Íú‡šúdÙæÀ vð»šíÆú£ôÂ

ú£-÷«-ëÅ¯-û¦õª1-4

2-3

3-1

4-2

5-3

6-3

7-4

8-4

9-2

10-2

11-4

12-1

13-1

14-1

15-3

16-1

17-3

18-3

19-2

20-4

21-3

22-4

23-2

24-3

25-3

26-1

27-2

28-1

29-2

30-3

31-4

32-1

33-2

34-2

35-4

36-4

37-2

38-1

39-2

40-4

34.- Nå-Nª-ûÂõª ú£JÞ¥ @ô¢gÙ Ú¥ÚÛ ÚÛé-â°-ö°öËºx›íô¢ª-ÚÛª-ð¼-÷è[Ù ÷õx ú£ÙòÅ¡-NÙà¶ ú‡–A?

1) šïj°ð¼ Nå-Nª-ûÁ-ú‡úÃ 2) šïj°í£ôÂ Nå-Nª-ûÁ-ú‡úÃ3) šïj°í£ôÂ óŸ«Ú¨dîËÂ Íîµªi-öË¶-ù£ûÂ 4) šïj°í£-î¶ª-è…-óŸªîÂª

35.-- Ú¨ÙC-î¦-æ¨öËº ú£·ôjì î¦Ú¥uõª ÔN?i) Nå-Nª-ûÂõª øŒÚ¨hE ÑêŸpAh à¶ú£«h

ë¶ï£° vë]÷u-ô¦-P šíÙð»Ù-ë]-è¯-EÚ¨ ú£ï£°-ÚÛ-J-þ§hô³.-

ii) õ÷-é°õª ·ÚöËº-J-íÆ‡ÚÂ Nõª-÷ìª ÚÛLTÑÙèË¶ ÚÛô¢(ì ú£î¶ªt-üŒ-û¦õª

iii) Nå-Nª-ûÂõª ÍCÅ-ÚÛÙÞ¥ Bú£ª-ÚÛª-ìo-í£±è[ªÍN øŒK-ô¢ÙöËº Ú•÷±y ô¢«í£ÙöËº EõyÑÙæ°ô³.-

1) i, ii 2) ii, iii

3) i, ii, iii 4) iii ÷«vêŸî¶ª36.--- Ú¨ÙC-î¦-æ¨öËº þ¼è…óŸªÙ NëÅ]ªõª ÔNªæ¨?

i) ÚÛÙè[ô¢ ú£ÙÚÁàŸÙii) øŒK-ô¢ÙöËº sÓõ-vÚÁd-öËµjæÀ) õ÷é

ú£÷ª-ê½-õuêŸiii) ô¢ÚÛh-íˆ-è[ì EóŸªÙ-vêŸé1) i, ii 2) ii, iii 3) i, iii 4) i, ii, iii

37.-- ›úyà¦aÄ vð§A-í£-C-ÚÛõª svíÆˆô¦-è…-ÚÛ-öËÀÀqzìªêŸT_Ùà¶ ÓûÂâËµj-îÂª-õìª v›í¸ô-í‡Ùà¶ ú£«¤Ût÷´õÚÛÙ ÔC?

1) îµªUo-ù‡óŸªÙ 2) ÷«ÙÞœ-FúÃ 3) >ÙÚÂ 4) ÷«L-G“ìÙ

38.--- Ú¨ÙC-î¦-æ¨öËº ú£·ôjì î¦Ú¥uõª ÔN?i) ðÆ§ú£pÄ-ô¢úÃ, Ú¥L{-óŸªÙêÁ ÚÛLú‡ Ó÷³-ÚÛõª,

ë]Ùê¦õ ë]”èÅ[-ê¦y-EÚ¨ Ú¥ô¢-é-÷ª-÷±-êŸªÙC.-ii) ðÆ§ú£pÄ-ô¢úÃ, Lí‡-èÂ-õêÁ ÚÛLú‡ ðÆ§þ¼pÄ-L-í‡-èÂ-

õìª Ôô¦påª à¶ú‡ ÚÛé-êŸyàŸÙ ú£÷ª-vÞœ-êŸÚÛª Ú¥ô¢-é-÷ª-÷±-êŸªÙC.-

iii) ðÆ§þ§pÄ-J-öË¶-ù£ûÂ àŸô¢u ë¯yô¦ ÓûÂâËµj-îÂªõàŸô¢u-Q-õ-êŸìª ðÆ§ú£pÄ-ô¢úÃ ví£òÅ°-NêŸÙà¶ú£ªhÙC.-

1) i, ii, iii 2) i, ii 3) ii, iii 4) i, iii

39.--- Ú¨ÙC-î¦-æ¨öËº Ô Nå-NªûÂ öËºí£Ù ÷õxÚÛÙè[-ô¦õ ñõ-ïˆ°-ìêŸ, ÷ÙëÅ]uêŸyÙ, Þœô¢(Ä-vþ§÷Ù ÚÛõª-Þœªê¦ô³?

1) Nå-NªûÂ K 2) Nå-NªûÂ E

3) Nå-NªûÂ B 4) Nå-NªûÂ C

40.--- ú‡våúÃ íÆ£ö°õª, vë¯¤Û, â°÷ª, òË¹ð§pô³ö°Ùæ¨ î¦æ¨öËº ÍCÅ-ÚÛÙÞ¥ õGÅÙà¶ Nå-NªûÂ ÔC?

1) ·ôæ¨-û¦öËÀ 2) ëÇ]óŸ«-NªûÂ 3) ·ôjòËº-ðÆ§xNûÂ 4) Îþ§\-J(ÚÂ óŸ«ú‡èÂ

íÆ£ÙÚÂ

1. If base of a triangle increases by 14.28%,

then what should be change in height if the

area remains constant?

a) − 5.5% b) − 7.14% c) − 9.09% d) − 12.5%

2. In an election there were only two candi-

dates A and B, B got 50% of the votes that

A got. Had A got 200 votes less, there

would have been a tie. How many people

cast their votes in all? (All votes were

valid.)

a) 800 b) 1000 c) 1200 d) 1600

Key: 1- d 2-c

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