Mix reasoning practice set in english ,SSC Reasoning Practice Set PDF Download
Mix reasoning practice set in English, SSC Previous Year Reasoning Questions PDF, Most important Reasoning questions with answers which includes circle problems, box problems, triangles problems for competitive exams like SSC CGL , SSC CHSL ,CPO ,Bank exams and RRB NTPC etc have been explained in this post.
Ten Important Reasoning questions with answers for competitive exams
Problem # 1
This box problem consist of four rows and four columns . And we have to find the value of question mark in 4th row after studying the pattern of all the numbers in this box. This question mark is in 4th row of 2nd column.
Formula :- (R2 - R3) × R1 = R4
Calculation
Column 1
(15 -14 ) × 10 = 10
Column 3
(8 - 6 ) × 7 = 14
Column 4
(25 -12 ) × 5 = 65
Column 2
(7 - 3 ) × 4 = 16
Option (A)16 is correct option.
Problem # 2
This problem figure consists of four rows and every rows from top to bottom have one more row in comparison to previous row.
If we take the product of all the numbers in each row and compare the result so obtained ,then we have same result in every row.
Product of all the numbers in every row is 120.
2 × 5 × 4 × 3 = 120 ( Last Row )
4 × 15 × 2 = 120 (2nd Last Row )
10 × 12 = 120 (2nd Row )
Hence the value of "?" must be 20
Option (4)120 is correct option.
Problem # 3
This circle consists of four quadrants and every quadrant consists of three numbers . And every quadrant have two numbers in outer part and one number in the inner part . To find the value of question mark "?" , we shall use two numbers which are in the outer part to calculate the value of the number which is in the inner part of every quadrant .
Since the question mark is in 1st quadrant ,so we have to start the calculation from 2nd quadrant.
2nd Quadrant
Step 1. Take sum of both the numbers (16 and 20 ) in outer part of this quadrant.
Step 2. Take the square root of the number obtained in step 1.
Calculation
√( 20 +16) = √(36) =6
3rd Quadrant
Step 1. Take sum of both the numbers (5 and 4 ) in outer part of this quadrant.
Step 2. Take the square root of the number obtained in step 1.
Calculation
√( 5 + 4 ) = √( 9) =3
4th Quadrant
Step 1. Take sum of both the numbers (5 and 4 ) in outer part of this quadrant.
Step 2. Take the square root of the number obtained in step 1.
Calculation
√(15 +10) = √(25) =5
1st Quadrant
Step 1. Take sum of both the numbers (5 and 4 ) in outer part of this quadrant.
Step 2. Take the square root of the number obtained in step 1.
Calculation
√( 30 +19) = √(49) = 7
Option (1)7 is correct option.
Problem # 4
Writing the 1st Number as it is and then place its cube on the right side of it. Now moving clockwise pick the next odd number for next position.
1³ = 1 ⇒ 11
3³ = 27 ⇒327
5³ = 125 ⇒5125
7³ = 343 ⇒7343
9³ = 729 ⇒9729
Option (3)9729 is correct answer.
Problem # 5
This box problem consists of three rows and three columns. To find the value of question marks in 3rd column of 3rd row, we have to add 1st ,2nd and 3th columns since since in every column is same in every row.
Formula :- .
F + 4 = J , J + 4 = N
M + 4 = Q , Q + 4 = U
O + 4 = S , S + 4 = W
Option (B)W is correct option.
Problem # 6
This box problem consists of three rows and three columns. To find the value of question marks in 3rd column of 3rd row, we have to add 1st ,2nd and 3th columns since since in every column is same in every row.
Formula :- The sum of 1st three column in every row is same .
B + 1 = D , D + 1 = F
N + 1 = P , P + 1 = R
H + 1 = J , J + 1 = L = ?
And 3 + 3 = 6
7 + 9 = 16
5 + 6 = 11
Hence L11
Option (B)L11 is correct option.
Problem # 7
4th Quadrant
9 × 8 = 72 ⟺ 27 (After reversing the orders of digits)1st Quadrant
8 × 3 = 24 ⟺ 42 (After reversing the orders of digits)2nd Quadrant
5 × 7 = 35 ⟺ 53 (After reversing the orders of digits)3rd Quadrant
7 × 7 = 49 ⟺ 94 (After reversing the orders of digits) (The value of question mark).Hence option (2)94 is correct.
Problem # 8
Splitting all the numbers in the given circle into two series
1st Series
28, 32 , 36, 40 with a difference of 4
Any Term = Preceding Term + 4
1st term = 28
2nd term = 1st term + 4
2nd term = 28 + 4 = 32
2nd term = 32
3rd term = 2nd term + 4
3rd term = 32 + 4 = 36
3rd term = 36
4th term = 3rd term + 4
4th term = 36 + 4 = 40
4th term = 40
2nd Series
? , 31 , 33, 35 with a difference of 2 , So the 1st term of this series must be 2 less than the 2nd term.
Any Term = Preceding Term - 2
4th term = 5th term - 2
4th term = 37 - 2
4th term = 35
3rd term = 4th term - 2
3rd term = 35 - 2
3rd term = 33
2nd term = 3rd term - 2
2nd term = 33 - 2
2nd term = 31
Therefore 1st term = 2nd term - 2
1st term = 31 - 2
1st term = 29
Option (3)29 is correct Answer
Problem # 9
Every term can be expressed with the help of preceding term.
(2 × 1) + 2 = 2 + 2 = 4 ( 2nd number in series)
(4 × 2) - 3 = 8 - 3 = 5 ( 3rd number in series)
(5 × 3) + 4 = 15 + 4 = 19 ( 4th number in series)
(19 × 4) - 5 = 76 - 5 = 71 ( 5th number in series)
(71 × 5) - 6 = 355 - 6 = 349 ( 6th number in series)
Option (3)349 is correct Answer
Problem # 10
5 = 1st term
(5 × 3) + 1 = 15 + 1 =16 = 2nd term(16 × 3) + 2 = 48 + 2 =50 = 3rd term
(50 × 3) + 3 = 150 + 3 =153 = 4th term = value of question mark
(153 × 3) + 4 = 459 + 4 =463 = 5th term
(463 × 3) + 5 = 1389 + 5 = 1394 = 2nd term
Option (2)153 is correct Answer
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