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SSC CGL Reasoning Questions

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

Question 1: Which of the following interchanges in signs and numbers will make the given equation correct?
8 x 9 + 2 = 74

a) x and =, 2 and 9

b) x and +, 8 and 2

c) + and x, 8 and 9

d) + and =, 74 and 2

Question 2: Five students P, Q, R, S and T are sitting on a bench. Q is to the left of P and right of T. S is at the extreme right end and R is to the immediate left of S. Who is sitting third from the left?

a) P

b) Q

c) R

d) T

Question 3: Which of the following interchanges in signs and numbers will make the given equation correct?
8 x 6 + 4 = 52

a) x and +, 8 and 52

b) + and x, 6 and 4

c) x and +, 8 and 4

d) + and x, 6 and 8

Question 4: A, B, C,D and E are standing in a row. D is the immediate neighbour of A and E. B is at the immediate right of E and C is in the extreme right. Who is fourth to the left of C?

a) B

b) E

c) C

d) A

Question 5: B is older than C but not as old as D. E is not as old as B. Who is the oldest of all?

a) B

b) E

c) C

d) D

Question 6: Five planes, A, B, P, Q and R were saluting on Independence Day in such a way that R was second to the left of Q but to the immediate right of A. There was one plane between B and P. B was not at any of the ends. P and Q were neighbours. Who was at the left end?

a) A

b) B

c) R

d) Q

Question 7: Vishu,Pooja,Vishakha,Rani and Ram are sitting in a line. Pooja is third to the extreme right end. Vishu is second to the left of Pooja. Vishakha is to the right of Pooja. Rani is third to the right of Ram, who is the immediate neighbour of Vishu. Who is sitting in the middle?

a) Pooja

b) Ram

c) Visakha

d) Rani

Question 8: In a company all Mondays and Sundays are offs. If a month starts with a Monday and has 31 days then how many offs will be there in that month?

a) 7

b) 8

c) 9

d) 5

Question 9: Rajan and Manu went to a market. Rajan bought 3 erasers and 5 pens for Rs 105 and Manu bought 4 erasers and 6 pens for Rs 130. What is the price of one eraser?

a) Rs 25

b) Rs 20

c) Rs 10

d) Rs 19

Question 10: Rahul’s age is three times the age of Sumit. If Rahul is 15 years old presently, determine the age (in years) of Sumit after 12 years.

a) 16

b) 15

c) 10

d) 17

Question 11: I hire a taxi from my home to go to my workplace. The fare system in the city is such that for the ﬁrst kilometre, I am charged Rs 25, and after that, I am charged Rs 6 per kilometre. If my workplace is 10 km far from my home, what amount do I have to pay if I go by taxi?

a) Rs. 64

b) Rs. 89

c) Rs. 90

d) Rs. 79

Question 12: P, Q, R, S and T are sitting together. T is at one extreme end. P is the neighbour of T and is third to the left of Q. Who is fourth to the right of T?

a) P

b) T

c) Q

d) S

Question 13: If 19 (36) 13 and 37 (81) 28, then what is the value of ‘A’ in 43 (A) 38?

a) 49

b) 25

c) 34

d) 64

Question 14: In the following question, by using which mathematical operator will the expression become correct?
18 ? 6 ? 9 ? 27

a) x, ÷ and =

b) ÷, x and =

c) x, + and =

d) +, – and =

Question 15: If 18 (9) 3 and 36 (30) 5, then what is the value of A in 19 (A) 18?

a) 33

b) 57

c) 75

d) 96

Question 16: In the following question, correct the given equation by interchanging two numbers.
8 x 3 ÷ 4 + 9 – 5 = 16

a) 3 and 4

b) 4 and 8

c) 5 and 3

d) 5 and 9

Question 17: In the following question, by using which mathematical operators will the expression become correct?
30 ? 6 ? 4 ? 5 ? 4

a) +, =, x and –

b) -, =, x and +

c) =, x, + and –

d) -, +, = and x

Question 18: If 13 # 9 = 94 and 18 # 7 = 100, then 24 # 6 = ?

a) 121

b) 113

c) 148

d) 115

Question 19: If 6 @ 4 @ 7 = 101 and 2 @ 5 @ 11 = 150, then what is the value of A in A @ 8 @ 9 = 289?

a) 5

b) 8

c) 12

d) 17

Question 20: In the following question, by using which mathematical operators will the expression become correct?
15 _ 3 _ 4 _ 20

a) x, ÷ and >

b) ÷, x and <

c) ÷, x and =

d) +, x and =

Expression : 8 x 9 + 2 = 74

(A) : x and =, 2 and 9

$\equiv8=2+9\times74$

R.H.S. $=666+2=668\neq$ L.H.S.

(B) : x and +, 8 and 2

$\equiv2+9\times8=74$

L.H.S. $=2+72=74=$ R.H.S.

=> Ans – (B)

S is at the extreme right end and R is to the left of S.

=> R sits second from right end.

Q is to the left of P and right of T.

=> T sits at extreme left end.

Thus, arrangement :

$\therefore$ P is sitting third from the left.

=> Ans – (A)

Expression : 8 x 6 + 4 = 52

(A) : x and +, 8 and 52

$\equiv52+6\times4=8$

L.H.S. $=52+24=76\neq$ R.H.S.

(B) : + and x, 6 and 4

$\equiv8+4\times6=52$

L.H.S. $=8+24=32\neq$ R.H.S.

(C) : x and +, 8 and 4

$\equiv4+6\times8=52$

L.H.S. $=4+48=52=$ R.H.S.

=> Ans – (C)

It is given that C is in the extreme right.

D is the immediate neighbour of A and E. B is at the immediate right of E.

=> D sits to the immediate left of E, and A sits second to the left of E.

Thus, Arrangement :

$\therefore$ A is fourth to the left of C.

=> Ans – (D)

B is older than C but not as old as D, => D > B > C.

E is not as old as B, => B > E

Thus, combining above equations, we get : D > B > E, C

Thus, D is the oldest of all.

=> Ans – (D)

R was second to the left of Q but to the immediate right of A, => A R _ Q

There was one plane between B and P, => the last plane must be to the immediate right of Q.

Also, B was not at any of the ends, => B was between R and Q.

Arrangement :

$\therefore$ A was at the left end.

=> Ans – (A)

Pooja is third to the extreme right end. Vishu is second to the left of Pooja.

=> Vishu sits at extreme left end.

Vishakha is to the right of Pooja. Rani is third to the right of Ram, who is the immediate neighbour of Vishu.

=> Rani sits at extreme right end, and Ram sits between Vishu and Pooja.

Thus, arrangement :

$\therefore$ Pooja is sitting in the middle.

=> Ans – (A)

It is given that all Mondays and Sundays are offs.

If a month starts with a Monday and has 31 days, then number of Mondays = 5

But number of Sundays = 4

Thus, total off days = 4+5 = 9

=> Ans – (C)

Let price of 1 eraser = Rs. $x$ and price of 1 pen = Rs. $y$

According to ques, => $3x+5y=105$ —————(i)

and $4x+6y=130$

$\equiv2x+3y=65$ ————(ii)

Now, multiplying equation (i) by 2 and equation (ii) by 3 and subtracting them : $2(i)-3(ii)$

=> $10y-9y=210-195$

=> $y=15$

Substituting it in equation (ii), we get : $2x+3(15)=65$

=> $2x=65-45=20$

=> $x=\frac{20}{2}=10$

Thus, price of 1 eraser is Rs. 10

=> Ans – (C)

Rahul’s present age = 15 years

Also, Rahul’s age is three times the age of Sumit, => Sumit’s present age = $\frac{15}{3}=5$ years

$\therefore$ Age (in years) of Sumit after 12 years = $12+5=17$ years

=> Ans – (D)

Fare for 1st km = Rs. 25 and fare or rest of the journey = Rs. 6/km

Distance travelled = 10 km

=> Amount to be paid = $(1\times25)+(9\times6)$

= $25+54=Rs.$ $79$

=> Ans – (D)

T is at one extreme end. P is the neighbour of T and is third to the left of Q.

=> T sits at extreme left end, and P sits to the immediate right of T. Also, Q sits at extreme right end.

Thus, arrangement :

$\therefore$ is fourth to the right of T.

=> Ans – (C)

The pattern is that the number in the middle is obtained by taking square of the difference of the numbers on each sides.

Eg : $(19-13)^2=6^2=36$

and $(37-28)^2=9^2=81$

Similarly, $A=(43-38)^2=5^2=25$

=> Ans – (B)

Expression : 18 ? 6 ? 9 ? 27

(A) : x, ÷ and =

=> $18\times6\div9=27$

L.H.S. = $2\times6=12\neq$ R.H.S.

(B) : ÷, x and =

=> $18\div6\times9=27$

L.H.S. = $3\times9=27=$ R.H.S.

(C) : x, + and =

=> $18\times6+9=27$

L.H.S. = $108+9=117\neq$ R.H.S.

(D) : +, – and =

=> $18+6-9=27$

L.H.S. = $24-9=15\neq$ R.H.S.

=> Ans – (B)

The pattern is that the number in the middle is obtained by multiplying the numbers on each sides and then dividing the result by ‘6’.

Eg : $\frac{18\times3}{6}=3\times3=9$

and $\frac{36\times5}{6}=6\times5=30$

Similarly, $\frac{18\times19}{6}=3\times19=57$

=> Ans – (B)

Expression : 8 x 3 ÷ 4 + 9 – 5 = 16

(A) : $8\times4\div3+9-5=16$

L.H.S. = $10.67+4=14.67\neq$ R.H.S.

(B) : $4\times3\div8+9-5=16$

L.H.S. = $1.5+4=5.5\neq$ R.H.S.

(C) : $8\times5\div4+9-3=16$

L.H.S. = $10+6=16=$ R.H.S.

(D) : $8\times3\div4+5-9=16$

L.H.S. = $6-4=2\neq$ R.H.S.

=> Ans – (C)

Expression : 30 ? 6 ? 4 ? 5 ? 4

(A) : $30+6=4\times5-4$

L.H.S. = $36$

R.H.S. = $20-4=16\neq$ L.H.S.

(B) : $30-6=4\times5+4$

L.H.S. = $24$

R.H.S. = $20+4=24=$ L.H.S.

=> Ans – (B)

The pattern followed is that both the numbers are multiplied and then one more than the sum of numbers is subtracted from the result.

Eg : $(13\times9)-(13+9+1)=117-23=94$

and $(18\times7)-(18+7+1)=126-26=100$

Similarly, $(24\times6)-(24+6+1)=144-31=113$

=> Ans – (B)

The pattern followed is that sum of squares of the number is written on the right.

Eg = $6^2+4^2+7^2=36+16+49=101$

and $2^2+5^2+11^2=4+25+121=150$

Similarly, $A^2+8^2+9^2=289$

=> $A^2+64+81=289$

=> $A^2=289-145=144$

=> $A=\sqrt{144}=12$

=> Ans – (C)

Expression : 15 _ 3 _ 4 _ 20

(A) : $15\times3\div4>20$

L.H.S. = $\frac{45}{4}=11.25$ is not greater than R.H.S.

(B) : $15\div3\times4<20$

L.H.S. = $5\times4=20$ is not less than R.H.S.

(C) : $15\div3\times4=20$

L.H.S. = $5\times4=20=$ R.H.S.

=> Ans – (C)