**SSC CGL Maths Repeated Questions PDF:**

Download SSC CGL Maths Repeated questions with answers PDF based on previous papers very useful for SSC CGL exams. 25 Very important Maths Repeated objective questions for SSC exams.

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**Question 1: **If the cost prices of articles A and B are in the ratio 3:4 and the selling prices are in the ratio 5:6 and the profit obtained on selling both of them is the same. What is the profit percentage on selling article B?

a) 66.67%

b) 33.33%

c) 50%

d) 25%

**Question 2: **If a trader sold two articles each for Rs.3600/- with no profit or loss. If the first article is sold at 20 %profit, at what loss is the second one sold?

a) 20%

b) 14.29%

c) 16.52%

d) 21.32%

**Question 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

a) 10% increase

b) 10% decrease

c) 12% increase

d) 12% decrease

**Question 4: **A shopkeeper bought 30 kg of rice at the rate of Rs. 70 per kg and 20 kg of rice at the rate of Rs. 70.75 per kg. If he mixed the two brands of rice and sold the mixture at Rs. 80.50 per kg, his gain is

a) Rs. 450

b) Rs. 510

c) Rs. 525

d) Rs. 485

**Question 5: **Two numbers are in the ratio 3:4. Their L.C.M. is 84. The greater number is

a) 21

b) 24

c) 28

d) 84

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**Question 6: **The sum of two numbers is 36 and their H.C.F and L.C.M. are 3 and 105 respectively. The sum of the reciprocals of two numbers is

a) 2/35

b) 3/25

c) 4/35

d) 2/25

**Question 7: **What is the HCF (highest common factor) of 57 and 513?

a) 10

b) 57

c) 3

d) 27

**Question 8: **The amount received at 10% per annum Compound interest after 3 yrs is Rs 5324. What was the principal (in Rs)?

a) 4100

b) 4200

c) 4000

d) 4300

**Question 9: **At what rate of compound interest (in %) per annum will a sum of Rs. 15,000 become Rs. 18,150 in 2 years?

a) 11

b) 10

c) 9

d) 12

**Question 10: **If the amount received at the end of 2nd and 3rd year at Compound Interest on a certain Principal is Rs 1,800, and Rs 1,926 respectively, what is the rate of interest?

a) 7.5%

b) 7%

c) 6%

d) 6.5%

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**Question 11: **A sum of ₹ 3000 yields an interest of ₹ 1080 at 12% per annum simple interest in how many years ?

a) 4 Years

b) 3 Years

c) 5 years

d) 2½ Years

**Question 12: **The simplest form of 3774/2958 is

a) 43/19

b) 37/29

c) 31/13

d) 31/23

**Question 13: **Two fractions are such that their product is 4 and sum is 68/15. Find the two fractions.

a) 6/15, 10/3

b) 6/5, 10/3

c) 7/2, 8/7

d) 10/7, 14/5

**Question 14: **The simplest form of 3565/1495 is

a) 31/13

b) 43/19

c) 23/13

d) 31/23

**Question 15: **The reciprocal of the sum of the reciprocals of 8/7 and 5/6 is:

a) 83/40

b) 42/83

c) 83/42

d) 40/83

**Question 16: **The average age of a husband and is wife was 23 years at the time of their marriage. After five years they have a one year old child. The average age of the family now is

a) 29.3 years

b) 19 years

c) 23 years

d) 28.5 years

**Question 17: **The average weight of 8 persons increases by 2.5 kg when a new persons comes in place of one of them weighing 65 kg. The weight of the new person is

a) 84 kg

b) 85 kg

c) 76 kg

d) 76.5 kg

**Question 18: **The cost price of an article is Rs.100. A discount series of 5%, 10% successively reduces the price of a article by

a) Rs 4.5

b) Rs 14.5

c) Rs 24.5

d) None of the above

**Question 19: **A container containing 400 litres of oil lost 8% by leakage. Oil left in the container is

a) 320 litres

b) 368 litres

c) 332 litres

d) 32 litres

**Question 20: **The ﬁrst and last terms of an arithmetic progression are 29 and -49. If the sum of the series is -140, then it has how many terms?

a) 13

b) 14

c) 12

d) 11

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**Question 21: **The first and last terms of an arithmetic progression are -23 and 42. What is the sum of the series if it has 14 terms?

a) 91

b) 133

c) 93

d) -133

**Question 22: **The first and last terms of an arithmetic progression are 33 and -57. What is the sum of the series if it has 16 terms?

a) -135

b) -192

c) -207

d) -165

**Question 23: **The first and last terms of an arithmetic progression are 25 and -52. What is the sum of the series if it has 12 terms?

a) -162

b) -110

c) 162

d) 110

**Question 24: **The volume of the largest right circular cone that can be cut out of a cube of edge 7cm ? $(Use \pi = \frac{22}{7})$.

a) $13.6 cm^3$

b) $121 cm^3$

c) $147.68 cm^3$

d) $89.8 cm^3$

**Question 25: **The sum of two numbers is 75 and their difference is 25. The product of the two numbers is:

a) 1350

b) 1250

c) 1000

d) 125

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**Answers & Solutions:**

**1) Answer (C)**

Let the cost price of A and B be 3x and 4x respectively.

Let the selling price of A and B be 5y and 6y respectively.

5y-3x = 6y-4x

x= y

Hence profit on selling B = 6y-4x = 2x

Profit % = $\frac{2x}{4x} \times 100$ = 50%

**2) Answer (B)**

Given that the first article is sold at Rs.3600/- with 20% profit.

Let x,y be the cost prices of first and second articles respectively.

1.2x = 3600

x = 3000

Profit on first article = Loss on second article

3600-3000 =y-3600

y = Rs.4200/-

Loss % = $\frac{600*100}{4200}$ = $14.2857$%

Hence, option B is the correct answer.

**3) Answer (D)**

Let us assume the price of the article to be Rs. 100. Let the number of articles sold be 100.

Total sales = 100*100 = 10,000

Now, the price increases by 10% => New cost = Rs. 110

Sales decrease by 20% => New sales = 80 units.

Now, new sales value = 80*110 = Rs. 8800

%age change =(10000 – 8800) / 10000 = 1200/10000 = 12% decrease.

Option D is the right answer.

**4) Answer (B)**

The shopkeeper bought 30 kg rice at Rs. 70/kg

C.P. = 70*30 = Rs. 2100

Similarly, C.P. for second type of rice = 70.75*20 = Rs. 1415

=> Total C.P. = 2100+1415 = Rs. 3515

He sold these two brands at Rs. 80.50/kg

=> Total S.P. = 80.50*50 = Rs. 4025

Profit = 4025-3515 = Rs. 510

**5) Answer (C)**

Let the numbers be 3x, 4x

LCM of 3x and 4x is = 12x

So the number 84 is divisible by 12

$\frac{84}{12}$ = 7

The numbers are 7×3 = 21 , 7x 4 = 28

The greatest number is 28

**6) Answer (C)**

let’s say numbers are $x$ and $y$

hence sum of the reciprocals will be $\frac{1}{x} + \frac{1}{y}$

or $\frac{x +y}{xy}$

as $x+y$ = 36 (given)

and $xy$ = $HCF \times LCM$

= $3 \times 105 = 315$

after putting the values we will get summation of reciprocals equals to $\frac{4}{35}$

**7) Answer (B)**

Factors of 57 = 1 , 3 , 19 , 57

Factors of 513 = 1 , 3 , 9 , 19 , 27 , 57 , 171 , 513

The common factors are = 1 , 3 , 19 , 57

=> Highest common factor = 57

=> Ans – (B)

**8) Answer (C)**

Total time = $P \times (1+\frac{R}{100})^{n}$

$\Rightarrow 5324 = P \times (1+\frac{10}{100})^{3}$

$\Rightarrow 5324 = P \times (1+\frac{1}{10})^{3}$

$\Rightarrow 5324 = P \times (\frac{11}{10})^{3}$

$\Rightarrow 5324 = P \times \frac{11}{10} \times \frac{11}{10} \times \frac{11}{10}$

$\Rightarrow P = 4000$

So the answer is option C.

**9) Answer (B)**

Amount received at the end of 2nd yr = P$(1+\frac{R}{100})^{2}$

18150 = 15000$(1+\frac{R}{100})^{2}$

1.21 = $(1+\frac{R}{100})^{2}$

1.1 = $1+\frac{R}{100}$

0.1 =$\frac{R}{100}$

$R = 10$%

So the answer is option B.

**10) Answer (B)**

Amount received at the end of 2nd yr = P$(1+\frac{R}{100})^{2}$ = 1800 ——-(1)

Amount received at the end of 3rd yr = P$(1+\frac{R}{100})^{3}$ = 1926 ——-(2)

divide (2) with (1)

$\Rightarrow\frac{P(1+\frac{R}{100})^{3}}{P(1+\frac{R}{100})^{2}} = \frac{1926}{1800}$

$\Rightarrow 1+\frac{R}{100}$ = $\frac{1926}{1800}$

$\Rightarrow \frac{R}{100}$ = $\frac{1926}{1800}-1$

$\Rightarrow \frac{R}{100}$ = $\frac{1926-1800}{1800}$

$\Rightarrow \frac{R}{100}$ = $\frac{126}{1800}$

$\Rightarrow \frac{R}{100}$ = $\frac{7}{100}$

$\Rightarrow R = 7$%

So the answer is option B.

**11) Answer (B)**

Principal sum = P = Rs. 3000

Let time period = $t$ years and rate of interest = 12%

Simple interest = $\frac{P \times r \times t}{100}$

=> $\frac{3000 \times 12 \times t}{100}=1080$

=> $360t=1080$

=> $t=\frac{1080}{360}=3$ years

=> Ans – (B)

**12) Answer (B)**

Expression : $\frac{3774}{2958}$

Both are multiples of 2, thus dividing numerator and denominator by 2

= $\frac{1887}{1479}$

Now, dividing by 51, = $\frac{37}{29}$

=> Ans – (B)

**13) Answer (B)**

Let the two numbers be x and y

=> $x + y = \frac{68}{15}$ and $x.y = 4$

=> $x(\frac{68}{15} – x) = 4$

=> $x(\frac{68 – 15x}{15}) = 4$

=> $68x – 15x^2 = 60$

=> $15x^2 – 68x + 60 = 0$

=> $15x^2 – 50x – 18x + 60 = 0$

=> $5x(3x – 10) – 6(3x – 10) = 0$

=> $(5x – 6) (3x – 10) = 0$

=> $x = \frac{6}{5} , \frac{10}{3}$

**14) Answer (A)**

Expression : $\frac{3565}{1495}$

Both are multiples of 5, thus dividing numerator and denominator by 5

= $\frac{713}{299}$

Now, dividing by 23, = $\frac{31}{13}$

=> Ans – (A)

**15) Answer (D)**

Sum of the reciprocals of 8/7 and 5/6

= $\frac{7}{8} + \frac{6}{5}$

= $\frac{7(5)+6(8)}{40} = \frac{35+48}{40}$

= $\frac{83}{40}$

=> Reciprocal of 83/40 = $\frac{40}{83}$

=> Ans – (D)

**16) Answer (B)**

Sum of ages of husband and wife at the time of their marriage = $23\times2=46$ years

Sum of the family after 5 years = 5 years of husband + 5 years of wife + 1 year of child

=> Total age = $46+5+5+1=57$ years

=> Required average = $\frac{57}{3}=19$ years

=> Ans – (B)

**17) Answer (B)**

Let average weight of 8 persons = $x$ kg and weight of new person = $y$ kg

=> Sum of weights of persons = $8x$ kg

According to ques,

=> $\frac{8x-65+y}{8}=x+2.5$

=> $8x-65+y=8x+20$

=> $y=20+65=85$

$\therefore$ The weight of the new person = **85 kg**

=> Ans – (B)

**18) Answer (B)**

Cost price = Rs. 100

Selling price after first discount of 5% = $100-(\frac{5}{100}\times100)$

= $100-5=Rs.$ $95$

Similarly, selling price after second discount of 10% = $95-(\frac{10}{100}\times95)$

= $95-9.5=Rs.$ $85.5$

$\therefore$ Amount is reduced by = $100-85.5=Rs.$ $14.5$

=> Ans – (B)

**19) Answer (B)**

Quantity of oil originally in the container = 400 litres

Quantity of oil left = $400-(\frac{8}{100}\times400)$

= $400-32=368$ litres

=> Ans – (B)

**20) Answer (B)**

In an arithmetic progression with first term, $a = 29$ , last term, $l = -49$

Let number of terms = $n$

$\therefore$ Sum of A.P. = $\frac{n}{2} (a + l) = -140$

=> $\frac{n}{2} (29 – 49) = -140$

=> $\frac{-20n}{2} = -140$

=> $n = \frac{(-140) \times 2}{-20} = 7 \times 2$

=> $n=14$

=> Ans – (B)

**21) Answer (B)**

In an arithmetic progression with first term, $a = -23$ , last term, $l = 42$

Number of terms = $n = 14$

$\therefore$ Sum of A.P. = $\frac{n}{2} (a + l)$

= $\frac{14}{2} (-23 + 42)$

= $7 \times 19 = 133$

=> Ans – (B)

**22) Answer (B)**

In an arithmetic progression with first term, $a = 33$ , last term, $l = -57$

Number of terms = $n = 16$

$\therefore$ Sum of A.P. = $\frac{n}{2} (a + l)$

= $\frac{16}{2} (33 – 57)$

= $8 \times (-24) = -192$

=> Ans – (B)

**23) Answer (A)**

In an arithmetic progression with first term, $a = 25$ , last term, $l = -52$

Number of terms = $n = 12$

$\therefore$ Sum of A.P. = $\frac{n}{2} (a + l)$

= $\frac{12}{2} (25 – 52)$

= $6 \times (-27) = -162$

=> Ans – (A)

**24) Answer (D)**

Height of largest circular cone = $7$ cm and radius = $\frac{7}{2}=3.5$ cm

Volume of cone = $\frac{1}{3}\pi r^2h$

= $\frac{1}{3}\times\frac{22}{7}\times(3.5)^2\times7$

= $\frac{1}{3}\times22\times12.25$

= $\frac{269.5}{3}=89.8$ $cm^3$

=> Ans – (D)

**25) Answer (B)**

Let the numbers be $x$ and $y$

=> Sum = $x+y=75$ ————(i)

and difference = $x-y=25$ ————(ii)

Adding both equations, we get : $2x=75+25=100$

=> $x=\frac{100}{2}=50$

Substituting it in equation (i), => $y=75-50=25$

$\therefore$ Product = $50\times25=1250$

=> Ans – (B)