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# SSC CGL Arithmetic questions

Download Top-20 Arithmetic questions for SSC CGL exam. Most important arithmetic questions based on asked questions in previous exam papers for SSC CGL.

Question 1: If X = 0.3 $\times$ 0.3, the value of X is

a) 0.009

b) 0.03

c) 0.09

d) 0.08

Question 2: An equation of the form ax + by + c = 0. Where, a ≠ 0, b ≠ 0 and c = 0 represents a straight line which passes through

a) (2, 4)

b) (0, 0)

c) (3, 2)

d) None of these

Question 3: The fifth term of the sequence for which $t_{1}=1$, $t_{2}=2$ and $t_{n+2}$ = $t_{n}+t_{n+1}$, is

a) 5

b) 10

c) 6

d) 8

Question 4: Reduce 3596 / 4292 to lowest terms.

a) 29/37

b) 17/43

c) 31/37

d) 19/23

Question 5: Reduce 2530/1430 to lowest terms.

a) 47/17

b) 23/13

c) 47/19

d) 29/17

Question 6: The first and last terms of an arithmetic progression are -32 and ­43. If the sum of the series is ­88, then it has how many terms?

a) 16

b) 15

c) 17

d) 14

Question 7: 29 is 0.8% of?

a) 3625

b) 1450

c) 7250

d) 10875

Question 8: 5*[-0.6 (2.8 + 1.2)] of 0.3 is equal to

a) -1.44

b) -1.08

c) -1.2

d) -3.6

Question 9: Find the value of p if 3x + p, x – 10 and -x + 16 are in arithmetic progression.

a) 16

b) 36

c) -16

d) -36

Question 10: If 9/4th of 7/2 of a number is 126, then 7/2th of that number is …………..

a) 56

b) 284

c) 72

d) 26

Question 11: The 4th term of an arithmetic progression is 15, 15th term is -29, ﬁnd the 10th term?

a) -5

b) -13

c) -17

d) -9

Question 12: (91 + 92 + 93 + ……… +110) is equal to

a) 4020

b) 2010

c) 6030

d) 8040

Question 13: 40.36 – (9.347 – x ) – 29.02 = 3.68. Find x.

a) -56.353

b) 1.687

c) -17.007

d) 82.407

Question 14: What is the value of (81 + 82 + 83 + ……… +130)?

a) 5275

b) 10550

c) 15825

d) 21100

Question 15: In an arithmetic progression if 13 is the 3rd term, ­47 is the 13th term, then ­30 is which term?

a) 9

b) 10

c) 7

d) 8

Question 16: The first and last terms of an arithmetic progression are 37 and ­-18. If the sum of the series is 114, then it has how many terms?

a) 13

b) 12

c) 14

d) 15

Question 17: If 4/5th of 6/7th of a number is 216, then 8/9th of that number will be

a) 179

b) 280

c) 160

d) 269

Question 18: 199994 x 200006 = ?

a) 39999799964

b) 39999999864

c) 39999999954

d) 39999999964

Question 19: In an arithmetic progression, if 17 is the 3rd term, -25 is the 17th term, then -1 is which term?

a) 10

b) 11

c) 9

d) 12

Question 20: In an arithmetic progression, if 9 is the 5th term, -26 is the 12th term, then -6 is which term?

a) 11

b) 8

c) 10

d) 7

Expression : $X=0.3\times0.3$

=> $X=0.09$

=> Ans – (C)

As c=0, and substituting the point (0,0) in the equation, we get ax+by+c = 0 at the point (0,0).
Hence, the line passes through origin.

$t_{1}=1$, $t_{2}=2$

$t_{n+2}$ = $t_{n}+t_{n+1}$

put n=3, then  $t_{5}$ = $t_{3}+t_{4}$

$t_{3}$ = $t_{1}+t_{2}$ = 1+2 = 3

$t_{4}$ = $t_{2}+t_{3}$ = 2+3 = 5

$t_{5}$ = $t_{3}+t_{4}$ = 3+5 = 8

so the answer is option D.

Expression : $\frac{3596}{4292}$

Dividing both numerator and denominator by 4, = $\frac{899}{1073}$

Similarly, dividing by 29, we get :

= $\frac{31}{37}$

=> Ans – (C)

Expression : $\frac{2530}{1430}$

Dividing both numerator and denominator by 10, we get = $\frac{253}{143}$

Similarly, dividing by 11, we get :

= $\frac{23}{13}$

=> Ans – (B)

First term of AP, $a=-32$ and last term, $l=43$

Let there be $n$ terms

Sum of AP = $\frac{n}{2}(a+l) = 88$

=> $\frac{n}{2}(-32+43)=88$

=> $\frac{11n}{2}=88$

=> $n=88 \times \frac{2}{11}$

=> $n=8 \times 2=16$

=> Ans – (A)

Let the number be $x$

According to ques, 0.8% of $x$ = 29

=> $\frac{0.8}{100} \times x = 29$

=> $\frac{x}{125} = 29$

=> $x = 29 \times 125 = 3625$

=> Ans – (A)

Expression : 5*[-0.6 (2.8 + 1.2)] of 0.3

= $5 [(-0.6) \times (4)] \times 0.3$

= $5 \times (-2.4) \times 0.3$

= $(-12) \times 0.3 = -3.6$

=> Ans – (D)

Terms in arithmetic progression : $(3x + p) , (x – 10) , (-x + 16)$

=> Difference between first two terms is equal to the difference between last two terms

=> $(x – 10) – (3x + p) = (-x + 16) – (x – 10)$

=> $-2x -10 – p = -2x + 16 + 10$

=> $-p = 26 + 10 = 36$

=> $p = -36$

=> Ans – (D)

Let the number be $x$

According to ques,

=> $\frac{9}{4} \times \frac{7}{2} \times x = 126$

=> $\frac{63}{8} x = 126$

=> $x = \frac{126}{63} \times 8$

=> $x = 2 \times 8 = 16$

$\therefore (\frac{7}{2})^{th}$ of the number = $\frac{7}{2} \times 16$

= $7 \times 8 = 56$

=> Ans – (A)

The $n^{th}$ term of an A.P. = $a + (n – 1) d$, where ‘a’ is the first term , ‘n’ is the number of terms and ‘d’ is the common difference.

4th term, $A_4 = a + (4 – 1) d = 15$

=> $a + 3d = 15$ —————–(i)

Similarly, 15th term, $A_{15} = a + 14d = -29$ ——————(ii)

Subtracting equation (i) from (ii), we get :

=> $(14d – 3d) = -29 – 15$

=> $d = \frac{-44}{11} = -4$

Substituting it in equation (i), => $a – 12 = 15$

=> $a = 15 + 12 = 27$

$\therefore$ 10th term, $A_{10} = a + (10 – 1)d$

= $27 + (9 \times -4) = 27 – 36 = -9$

=> Ans – (D)

Expression : (91 + 92 + 93 + ……… +110)

This is an arithmetic progression with first term, $a = 91$ , last term, $l = 110$ and common difference, $d = 1$

Let number of terms = $n$

Last term in an A.P. = $a + (n – 1)d = 110$

=> $91 + (n – 1)(1) = 110$

=> $n – 1 = 110 – 91 = 19$

=> $n = 19 + 1 = 20$

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

= $\frac{20}{2} (91 + 110)$

= $10 \times 201 = 2010$

Expression : 40.36 – (9.347 – x ) – 29.02 = 3.68

=> 40.36 – 9.347 + x = 3.68 + 29.02

=> 31.013 + x = 32.7

=> x = 32.7 – 31.013

=> x = 1.687

=> Ans – (B)

Expression : (81 + 82 + 83 + ……… +130)

This is an arithmetic progression with first term, $a = 81$ , last term, $l = 130$ and common difference, $d = 1$

Let number of terms = $n$

Last term in an A.P. = $a + (n – 1)d = 130$

=> $81 + (n – 1)(1) = 130$

=> $n – 1 = 130 – 81 = 49$

=> $n = 49 + 1 = 50$

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

= $\frac{50}{2} (81 + 130)$

= $25 \times 211 = 5275$

The $n^{th}$ term of an A.P. = $a + (n – 1) d$, where ‘a’ is the first term , ‘n’ is the number of terms and ‘d’ is the common difference.

3rd term, $A_3 = a + (3 – 1) d = 13$

=> $a + 2d = 13$ —————–(i)

Similarly, 13th term, $A_{13} = a + 12d = 47$ ——————(ii)

Subtracting equation (i) from (ii), we get :

=> $(12d – 2d) = 47 – 13 = 34$

=> $d = \frac{34}{10} = 3.4$

Substituting it in equation (i), => $a + 2 \times 3.4 = 13$

=> $a = 13 – 6.8 = 6.2$

Let $n^{th}$ term = 30

=> $a + (n – 1) d = 30$

=> $6.2 + (n – 1) (3.4) = 30$

=> $(n – 1) (3.4) = 30 – 6.2 = 23.8$

=> $(n – 1) = \frac{23.8}{3.4} = 7$

=> $n = 7 + 1 = 8$

In an arithmetic progression with first term, $a = 37$ , last term, $l = -18$

Let number of terms = $n$

$\therefore$ Sum of A.P. = $\frac{n}{2} (a + l) = 114$

=> $\frac{n}{2} (37 – 18) = 114$

=> $19n = 114 \times 2 = 228$

=> $n = \frac{228}{19} = 12$

=> Ans – (B)

Let the number be $x$

According to ques,

=> $\frac{4}{5} \times \frac{6}{7} \times x = 216$

=> $x = 216 \times \frac{35}{24} = 9 \times 35$

$\therefore$ 8/9th of the number = $\frac{8}{9} \times (35 \times 9)$

= $8 \times 35 = 280$

=> Ans – (B)

Expression :  199994 x 200006

= (200000 – 6) x (200000 + 6)

= $(200000)^2 – (6)^2$

= 40000000000 – 36 = 39999999964

=> Ans – (D)

The $n^{th}$ term of an A.P. = $a + (n – 1) d$, where ‘a’ is the first term , ‘n’ is the number of terms and ‘d’ is the common difference.

3rd term, $A_3 = a + (3 – 1) d = 17$

=> $a + 2d = 17$ —————–(i)

Similarly, 17th term, $A_{17} = a + 16d = -25$ ——————(ii)

Subtracting equation (i) from (ii), we get :

=> $(16d – 2d) = -25 – 17$

=> $d = \frac{-42}{14} = -3$

Substituting it in equation (i), => $a – 6 = 17$

=> $a = 17 + 6 = 23$

Let $n^{th}$ term = -1

=> $a + (n – 1) d = -1$

=> $23 + (n – 1) (-3) = -1$

=> $(n – 1) (-3) = -1 – 23 = -24$

=> $(n – 1) = \frac{-24}{-3} = 8$

=> $n = 8 + 1 = 9$

The $n^{th}$ term of an A.P. = $a + (n – 1) d$, where ‘a’ is the first term , ‘n’ is the number of terms and ‘d’ is the common difference.

5th term, $A_5 = a + (5 – 1) d = 9$

=> $a + 4d = 9$ —————–(i)

Similarly, 12th term, $A_{12} = a + 11d = -26$ ——————(ii)

Subtracting equation (i) from (ii), we get :

=> $(11d – 4d) = -26 – 9$

=> $d = \frac{-35}{7} = -5$

Substituting it in equation (i), => $a – 20 = 9$

=> $a = 9 + 20 = 29$

Let $n^{th}$ term = -6

=> $a + (n – 1) d = -6$

=> $29 + (n – 1) (-5) = -6$

=> $(n – 1) (-5) = -6 – 29 = -35$

=> $(n – 1) = \frac{-35}{-5} = 7$

=> $n = 7 + 1 = 8$