SSC CGL Previous Year Maths Questions PDF

0
2529
ssc cgl previous year maths questions pdf
ssc cgl previous year maths questions pdf

SSC CGL Previous Year Maths Questions PDF

Download SSC CGL Maths Previous Year questions with answers PDF based on previous papers very useful for SSC CGL exams. 20 Very important Maths objective questions (MCQ’s) for SSC exams.

Download SSC CGL Previous Year Maths Questions PDF

25 SSC CGL Mocks – Just Rs. 149

FREE SSC EXAM YOUTUBE VIDEOS

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

SSC CGL Previous Papers Download PDF

SSC CHSL PREVIOUS PAPERS DOWNLOAD

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

18000+ Questions – Free SSC Study Material

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: If 6/7th of 8/5th of a number is 192, then 3/4th of that number is .———–

a) 105

b) 77

c) 36

d) 80

Question 14: 40.36 – (9.347 – x ) – 29.02 = 3.68. Find x.

a) -56.353

b) 1.687

c) -17.007

d) 82.407

Question 15: What is the value of (81 + 82 + 83 + ……… +130)?

a) 5275

b) 10550

c) 15825

d) 21100

SSC CGL Free Mock Test

SSC CHSL Free Mock Test

Question 16: 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 17: 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 18: 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 19: 199994 x 200006 = ?

a) 39999799964

b) 39999999864

c) 39999999954

d) 39999999964

Question 20: 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

1500+ Free SSC Questions & Answers

Answers & Solutions:

1) Answer (C)

Expression : $X=0.3\times0.3$

=> $X=0.09$

=> Ans – (C)

2) Answer (B)

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.

3) Answer (D)

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

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

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.

4) Answer (C)

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)

5) Answer (B)

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)

100+ Free GK Tests for SSC Exams

Download Free GK PDF

6) Answer (A)

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)

7) Answer (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)

8) Answer (D)

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)

9) Answer (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)

10) Answer (A)

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)

11) Answer (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.

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)

12) Answer (B)

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. = $\frac{n}{2} (a + l)$

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

= $10 \times 201 = 2010$

15000 Questions – Free SSC Study Material

13) Answer (A)

Let the number be $x$

According to ques,

=> $\frac{6}{7} \times \frac{8}{5} \times x = 192$

=> $\frac{48}{35} x = 192$

=> $x = \frac{192}{48} \times 35$

=> $x = 4 \times 35 = 140$

$\therefore (\frac{3}{4})^{th}$ of the number = $\frac{3}{4} \times 140$

= $3 \times 35 = 105$

=> Ans – (A)

14) Answer (B)

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)

15) Answer (A)

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. = $\frac{n}{2} (a + l)$

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

= $25 \times 211 = 5275$

16) Answer (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, $A_3 = a + (3 – 1) d = 13$

=> $a + 2d = 13$ —————–(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

=> $a + (n – 1) d = 30$

=> $6.2 + (n – 1) (3.4) = 30$

=> $(n – 1) (3.4) = 30 – 6.2 = 23.8$

=> $(n – 1) = \frac{23.8}{3.4} = 7$

=> $n = 7 + 1 = 8$

17) Answer (B)

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)

18) Answer (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)

19) Answer (D)

Expression :  199994 x 200006

= (200000 – 6) x (200000 + 6)

= $(200000)^2 – (6)^2$

= 40000000000 – 36 = 39999999964

=> Ans – (D)

20) Answer (C)

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

=> $a + (n – 1) d = -1$

=> $23 + (n – 1) (-3) = -1$

=> $(n – 1) (-3) = -1 – 23 = -24$

=> $(n – 1) = \frac{-24}{-3} = 8$

=> $n = 8 + 1 = 9$

Free SSC Online Coaching

DOWNLOAD APP FOR SSC FREE MOCKS

We hope this Maths Previous Year questions for SSC Exam will be highly useful for your preparation.

LEAVE A REPLY

Please enter your comment!
Please enter your name here