Logical Reasoning questions and answers for competitive Exams
Problem # 1
All the letters in each option are written from alphabet either in order or in reverse order.
From 1st option MNOP are written in order starting from M in alphabet.
From 2nd option TSRQ are written in reverse order starting from Q in alphabet. i.e.
QRST ⇒ QRST ⟺ TSRQ (After reversing the orders of letters).
From 3rd option EDCB are written in reverse order starting from B in alphabet. i.e.
BCDE ⇒ BCDE ⟺ EDCB (After reversing the orders of letters).
From 4th option XWVU are written in reverse order starting from U in alphabet. i.e.
UVWX ⇒ UVWX ⟺ XWVU ( After reversing the orders of letters).
Since 2nd, 3rd and 4th options are following the same pattern because all the letters in these options are written in order. But letters in 1st option are written in reverse order. And this option is odd one.
Hence Option (1)MNOP is correct option.
Problem # 2
All the letters are written alternatively from alphabet either in order or in reverse order.
From 1st option MKIG are written in reverse order starting from G with a gap of one letter in alphabet. i.e.
G H I J K L M ⇒ G H I J K L M ( Omitting blue colored letters ) ⇒ G I K M ⟺ MKIG (After reversing the orders of letters).
From 2nd option VTRP are written in reverse order starting from P with a gap of one letter in alphabet. i.e.
PQRSTUV ⇒ PQRSTUV (Omitting blue colored letters) ⇒ PRTV ⟺ VTRP (After reversing the orders of letters).
From 3rd option EGIK are NOT written in reverse order starting from K with a gap of one letter in alphabet.
From 4th option YWUS are written in reverse order starting from S with a gap of one letter in alphabet. i.e.
STUVWXY ⇒ STUVWXY (Omitting blue colored letters) ⇒ SUWY ⟺ YWUP (After reversing the orders of letters).
Since 1st , 2nd and 4th options are following the same pattern because all the letters in these options are written in reverse order with a gap of one letter. But letter in 3rd option are not written in reverse order.
Hence Option (3)EGIK is correct option.
Problem # 3
In this reasoning problem 1st number (5) is associated to 100 with the help of any rule , in the same rule we have to associate 7 to a number out of four given options.
Look carefully the given numbers consists of three digits. These three digits can be utilised with the help a formula given below.
Formula :- 4 × (1st number)² = 2nd Number
4 × (3rd number)² = 4th Number
⇒ {5} = 4 × 5² = 4 × 25 = 100 (2nd number)
⇒{7} = 4 × 7² = 4 × 49 = 196 (4th number)
Problem # 4
In this problem of reasoning we have to combine all the three given numbers in such a way that after applying any mathematical operation/operations we could get the number in the right hand side of all the three problems given above. Because in these types of reasoning problems we can change mathematical sign according to our requirements.
Replacing - sign with + , + sign with × in above conditions . Now place 1st two numbers in bracket . Then apply the BODMAS rule.(7 + 2) × 4 = 9 × 4 = 36
(10 + 5) × 2 = 15 × 2 = 30
Similarly
(20 + 5) × 6 = 25 × 6 = 150
Option (4)150 is correct option.
Problem # 5
In this reasoning problem 1st number (392) is associated to 28 with the help of any rule , in the same way we have to associate 722 to a number out of four given options.
Formula :- 2 × √(1st Number ÷ 2) = 2nd Number
2 × √(392 ÷ 2) = 2 × √(196) = 2 × 14 = 28 .
Similarly
2 × √(722 ÷ 2) = 2 × √(361) = 2 × 19 = 38
Option (3)38 is correct option.
Problem # 6
These words are used when a book is prepared, published and read by Reader. To publish a book , first of all it has to be written by the Author . After this book will be with the Editor for any change or editing. Then that book will be published by Publisher. After publication it will be sold by Bookseller. And at last it will be purchased by read by Reader.
The journey of the book is as follow
- Author
- Editor
- Publisher
- Bookseller
- Reader
Problem # 7
Divide the sum of both the numbers with 10 to get the number in right hand side.
( 15 + 55 ) ÷ 10 = 70 ÷ 10 = 7
( 25 + 35 ) ÷ 10 = 60 ÷ 10 = 6
( 35 + 45 ) ÷ 10 = 80 ÷ 10 = 8
Similarly ( 23 + 17 ) ÷ 10 = 40 ÷ 10 = 4
Hence option (3)4 is correct.
Problem # 8
In this reasoning problem three out of four options have been calculated by Multiplying the 1st number with its next number(successive number) to get 2nd number. This rule is applicable to only three out of four option. And one which do not follow this rule will be the correct option.
(1) 9 × ( 9 + 1 ) = 9 × 10 = 90(2) 13 × ( 13 + 1 ) = 13 × 14 = 182
(3) 8 × ( 8 + 1 ) = 8 × 9 = 72
(4) 12 × ( 12 + 1 ) = 12 × 13 = 156, but this is given as 144
Hence option (4)12 - 144 is odd one.
Also Read these articles
Problem # 9
Take the difference of two successive numbers, this difference is square of continuous natural numbers.
6 - 5 = 1 = 1²
10 - 6 = 4 = 2²
19 - 10 = 9 = 3²
35 - 19 = 16 = 4²
Similarly the difference of last number and 2nd last number can be calculated. This difference will be the square of some natural number.
? - 35 = 25 = 5²
? = 25 + 35 = 60
Hence option (2)60 is correct.
Problem # 10
Take the difference of two successive numbers, this difference is cube of continuous natural numbers.
3085 - 1357 = 1728 = 12³
5282 - 3085 = 2197 = 13³
8026 - 5282 = 2744 = 14³
Similarly the difference of last number and 2nd last number can be calculated. This difference will be the cube of some natural number.
? - 8026 = 3375 = 15³
? = 3375 + 8026 = 11401
Hence option (4)11401 is correct.
Conclusion
So these were the ten problems regarding the post Ten Logical Reasoning questions and answers pdf for competitive exams like SSC CGL ,SSC CHSL ,CPO ,Bank exams and RRB NTPC etc which were explained in this post. Feel free to comment your valuable suggestions.
Thanks for the reasoning questions
ReplyDelete