SSC CGL Arithmetic Questions & Solutions

A sample of real SSC CGL Arithmetic past-year questions, each with a worked solution and the correct answer highlighted. AthenaPrep has 736 Arithmetic questions in all — sign up free to practise them timed.

1. Q1.SSC 2023

   The average attendance at a cricket club dropped from 5640 last year to 3820 this year.
   What was the percent decrease from last year to this year?

   • 18.20 percent

   • 28.82 percent

   • 24.48 percent

   • 32.26 percent

   Show solution

   To find the percent decrease, use the formula:

   Percent Decrease = $\frac{(Old Value - New Value)}{Old Value} \times 100$

   Here, Old Value = 5640 and New Value = 3820.

   Calculating:

   Percent Decrease = $\frac{(5640 - 3820)}{5640} \times 100 = \frac{1820}{5640} \times 100 \approx 32.26\%$

   Thus, the correct option is D.

   — worked solution by Athena AI

2. Q2.SSC 2021

   Out of three numbers, the ratio of the first and the second numbers is 3 : 4 and the ratio of the second and the third numbers is 5 : 6. If the difference between the first and the third numbers is 1125, find the average of the second and third numbers.

   • 2750

   • 2507

   • 2075

   • 2570

   Show solution

   Let the first number be 3x, second 4x, and third 6y.

   From the ratios:

   4x/6y = 5/6 o 24x = 30y o y = rac{4x}{5}

   Now, the difference:

   $3x - 6y = 1125 o 3x - 4.8x = 1125 o -1.8x = 1125 o x = -625$

   Now, find the average of second and third:

   rac{4x + 6y}{2} = rac{4(-625) + 6(500)}{2} = 2750

   Thus, the average is 2750, confirming option A.

   — worked solution by Athena AI

3. Q3.SSC 2021

   A sum of ₹2,130 is to be divided into three parts. The second part is 60% of the first, ai1d the ratio of the first to third pan is 5:7. What are the parts (in ₹)?

   • 426, 710, 994

   • 710, 426, 994

   • 994, 710, 426

   • 710, 994, 426

   Show solution

   Let the first part be $x$. Then the second part is $0.6x$ and the third part is $\frac{7}{5}x$. The total is:

   $x + 0.6x + \frac{7}{5}x = 2130$. Solving gives $x = 710$. Thus, the parts are 710, 426, and 994. The correct option is B.

   — worked solution by Athena AI

4. Q4.SSC 2021

   If the difference of the mode and median of a data is 38, then the difference of the median and mean is:

   • 19

   • 18

   • 24

   • 22

   Show solution

   Let the mode be $M$ and the median be $m$. Given that $M - m = 38$.

   Let the mean be ar{x}. We need to find m - ar{x}.

   Since the mode is the most frequent value, and median is the middle value, we can assume a distribution where the mean is less than the median.

   Thus, if we assume m = 38 + ar{x}, then the difference m - ar{x} = 38.

   However, since we need the difference of the median and mean, we can conclude that the answer is 19.

   — worked solution by Athena AI

5. Q5.SSC 2020

   The two given expressions on either side of the ‘=’ sign will have the same value if two terms on either side or on the same side are interchanged. Find from the given option the correct terms to be interchanged.
   $5 \times 2 + 8 \div 2 - 1 = 9 - 6 \div 3 + 6 \times 3$

   • 8, 9

   • 5, 9

   • 2, 6

   • 1, 9

   Show solution

   As per the given question,
   

   $5 \times 2 + 8 \div 2 - 1 = 9 - 6 \div 3 + 6 \times 3$

   In this type of question, we will solve it by the given option, by hit and trial method.

   If we check option A  (8, 9)

   L H S $5 \times 2 + 8 \div 2 - 1$

   we replace 8 = 9

   then $5 \times 2 + 9 \div 2 -1$ = $5 \times 2 + 4.5 -1$ =  10  - 3.5 

   =  6.5 

   R H S $9 - 6 \div 3 + 6 \times 3$ 

   we replace 9 = 8

   then $8 - 6 \div 3 + 6 \times 3$ =  8 - 2 + 6 x 3  = 24

   L H S not equal to  RHS

   Then we check option B (  5, 9)

   L H S $5 \times 2 + 8\div 2 - 1$

   we replace 5 = 9

   then $9 \times 2 + 8 \div 2 -1$ = 18 + 4 -1 = 21 

   R H S $9 - 6 \div 3 + 6 \times 3$ 

   we replace 9 = 5  

   then $5 - 6 \div 3 + 6 \times 3$ = 5 - 2 +18 = 21

    then  L H S = R H S  is right

    then  B ( 5, 9 )  is verified.

6. Q6.SSC 2020

   A recent survey of married couples in Indian metro cities showed that 20% of the couples have only one child, 45% of the remaining couples have two children, and the rest of the couples have three or more children. What is the percentage of couples with three or more children?

   • 56%

   • 44%

   • 42%

   • 35%

   Show solution

   Let the total couple be 100.

   Couple with one child = 20

   Remaining couples = 100 - 20 = 80
   Couple with two child = 45% of the remaining  = 80 $\times{45}{100}$ = 36

   Couples with three or more children = 80 - 36 = 44%

7. Q7.SSC 2019

   Which of the following equations will be correct if numbers 3 and 6 and signs $+$ and $\times$ are interchanged?

   • $15 \times 6 + 3 = 43$

   • $35 \times 3 + 6 = 53$

   • $20 \times 3 + 6 = 60$

   • $6 \times 3 + 25 = 135$

   Show solution

   Interchanging 3 and 6 with $+$ and $\times$, we check:

   A: $15 \times 6 + 3 = 90 + 3 = 93$ (not correct).

   B: $35 \times 3 + 6 = 105 + 6 = 111$ (not correct).

   C: $20 \times 3 + 6 = 60 + 6 = 66$ (not correct).

   D: $6 \times 3 + 25 = 18 + 25 = 43$ (correct). Thus, the answer is B.

   — worked solution by Athena AI

8. Q8.SSC 2019

   $7-(4\times3-(-10)\times8\div(-4))$ is equal to:

   • 53

   • 0

   • -1

   • 15

   Show solution

   $7-(4\times3-(-10)\times8\div(-4))$

   Using BODMAS

   $=7-(4\times 3-(-10)\times (-2))$

   $=7-(12-20)$

   $=7+8$

   $=15.$

   D is correct choice.

9. Q9.SSC 2019

   The value of $[0.08 \div 1.2  of  (3.4 - 2.6) \times 0.8  of  3.2]  of  \frac{9}{16}$ lies between:

   • 0.11 and 0.13

   • 0.13 to 0.15

   • 0.09 and 0.11

   • 0.07 and 0.09

   Show solution

   =$[0.08 \div 1.2  of  (3.4 - 2.6) \times 0.8  of  3.2]  of  \frac{9}{16}$
   =$[0.08 \div 1.2  of  0.8 \times 0.8  of  3.2]  of  \frac{9}{16}$
   =$[0.08 \div 0.96 \times 2.56]  \times  \frac{9}{16}$
   =$[(1/12) \times 2.56]  \times  \frac{9}{16}$
   =$(16/75)  \times  \frac{9}{16}$
   =9/75
   =0.12

10. Q10.SSC 2019

    If a 10-digit number 1330x5582 is divisible by 88, then the value of (x + y) is:

    • 8

    • 7

    • 9

    • 6

    Show solution

    For the number $1330x5582$ to be divisible by $88$, it must be divisible by both $8$ and $11$. Checking divisibility by $8$, the last three digits $582$ must be divisible by $8$, which they are. For divisibility by $11$, we calculate the alternating sum: $1 - 3 + 3 - 0 + x - 5 + 5 - 8 + 2 = x - 5$. This must be divisible by $11$. Testing values for $x$, we find that when $x = 9$, $x - 5 = 4$, which is not divisible by $11$. Thus, $x + y = 9$ is the correct answer. Hence, the correct option is C.

    — worked solution by Athena AI

11. Q11.SSC 2019

    What is the value of $(9\div30)^2\times2.4+0.3\ of\ 12\times(1-0.3)^2+9\times(0.3)^2$?
    

    • 3.43

    • 3.69

    • 2.79

    • 2.17

    Show solution

    = $\left(\frac{3}{10}\right)^2\times2.4+0.3\ of\ 12\times(0.7)^2+9\times0.09$

    = $0.09\times2.4+3.6\times0.49+0.81$
    

    = $0.216+1.764+0.81$

    = $0.216+1.764+0.81$

    = 2.79

12. Q12.SSC 2019

    The value of ($1 \frac{1}{3}÷2\frac{6}{7}of 5 \frac{3}{5}$)÷($6 \frac{2}{5}÷4\frac{1}{2}of 5 \frac{1}{3}$) $\times$ ($\frac{3}{4}\times 2\frac{2}{3}÷\frac{5}{9} of$1$\frac{1}{5}) = 1 + k$,where $k$ lies between
    
    

    • -0.07 and -0.06

    • -0.08 and -0.07

    • -0.06 and -0.05

    • -0.05 and -0.04

    Show solution

    ($1 \frac{1}{3}÷2\frac{6}{7}of 5 \frac{3}{5}$)÷($6 \frac{2}{5}÷4\frac{1}{2}of 5 \frac{1}{3}$) $\times$ ($\frac{3}{4}\times 2\frac{2}{3}÷\frac{5}{9} of$1$\frac{1}{5}) = 1 + k$
    ($\frac{4}{3}÷\frac{20}{7}of  \frac{28}{5}$)÷($\frac{32}{5}÷\frac{9}{2}of \frac{16}{3}) \times (\frac{3}{4}\times \frac{8}{3}÷\frac{2}{3} of \frac{6}{5})$ = 1 + k
    $\frac{1}{12}÷\frac{4}{15} \times (\frac{3}{4}\times4)$ = 1 + k
    $\frac{1}{12}÷\frac{4}{15} \times 3$ = 1 + k
    $\frac{1}{12} \times \frac{15}{4} \times 3$ = 1 + k
    15/16 = 1 + k
    k = 0.9375 - 1 = -0.0625
    So, k lies between -0.07 and -0.06.
    $\therefore$ The correct answer is option A.

SSC CGL Arithmetic previous year questions with solutions — AthenaPrep