Logical Reasoning (Set-IV)

Question:

Rahul is the father of Anupreet’s husband. Priya and Ankit are the only two children of Rahul. Priya is the sister-in-law of Anupreet. How is Ankit related to Anupreet?

Code Language Problem Solution

Problem: In a certain code language:

• ‘-‘ represents ‘×’ (multiplication)
• ‘÷’ represents ‘+’ (addition)
• ‘+’ represents ‘÷’ (division)
• ‘x’ represents ‘-‘ (subtraction)

Find the answer to the following expression:

Expression: 12 – 6 + 4 x 10 ÷ 16

Solution:

Substituting the symbols as per the given code language, we get:

12 × 6 ÷ 4 – 10 + 16

Step-by-Step Calculation:

1. Step 1: Multiplication: \( 12 \times 6 = 72 \)
2. Step 2: Division: \( 72 \div 4 = 18 \)
3. Step 3: Subtraction: \( 18 – 10 = 8 \)
4. Step 4: Addition: \( 8 + 16 = 24 \)

Answer:

24

Solution:

6 boys P, Q, R, S, T, and U are standing in a row facing north. P and Q cannot be either at the 1st or 2nd place. R and S will always be together, and R must be at one of the ends. R doesn’t have S to his right. U is standing between T and P.

Who is standing on the west end?

Code Language Problem

In a certain code language, “INSTANT” is written as “IOUWESZ”. We need to find how “FORGET” is written in that code language.

Question:

Pointing towards a boy, a lady said, “That boy is the brother of my husband’s only daughter’s daughter.” How is that lady’s son related to that boy?

Age Ratio Problem Solution

The ratio of present ages of Mohan and Sohan is 4 : 6, and the difference between their ages is 12 years.

Let the present ages of Mohan and Sohan be 4x and 6x respectively.

According to the problem, the difference between their ages is 12 years:

6x – 4x = 12

Simplifying, we get:

2x = 12

x = 6

Now, the present age of Sohan is 6x = 6 × 6 = 36 years.

Answer:

(c) 36

Weight Comparison Problem

We are given the following relationships about the weights of Alok, Karan, Roma, Tipu, and Shivam:

• Weight of Alok is twice the weight of Karan.
• Weight of Karan is one-fourth of the weight of Roma.
• Weight of Roma is twice the weight of Tipu.
• Weight of Tipu is twice the weight of Shivam.

Kishankant’s Walk Problem

Kishankant walks 10 km towards North. From there, he walks 6 km towards South. Then he walks 3 km towards East.

Question: How far and in which direction is he with reference to his starting point?

Solution:

1. First Step: He walks 10 km North.
2. Second Step: He then walks 6 km South.

This results in a net distance of \(10 – 6 = 4\) km North.

3. Third Step: He then walks 3 km East.

Now, to find his distance from the starting point, we have a right-angled triangle where:

• The vertical side (North-South distance) = 4 km
• The horizontal side (East distance) = 3 km

Using the Pythagorean theorem:

\[
\text{Distance} = \sqrt{(4^2 + 3^2)} = \sqrt{16 + 9} = \sqrt{25} = 5 \text{ km}
\]

Direction: Since he is 4 km North and 3 km East of the starting point, the direction is North-East.

Answer:

(b) 5 km North-East

Solution:

1. Mayank’s position from the bottom is 8.

2. Pramod is 12 ranks above Mayank, so Pramod’s position from the bottom is:

8 + 12 = 20

This means Pramod is 20th from the bottom.

3. Pramod’s position from the top is given as 9.

4. The total number of people in the row can be found by adding Pramod’s positions from the top and bottom, then subtracting 1 (since we are counting him in both positions):

Total people = 9 + 20 – 1 = 28

Answer:

(c) 28

Alphabet Series Problem Solution

Problem: Find the eighth letter to the right of ‘O’ in the alphabet series given in reverse order.

Solution:

Alphabet series in reverse order:

Z, Y, X, W, V, U, T, S, R, Q, P, O, N, M, L, K, J, I, H, G, F, E, D, C, B, A.

Step 1: Find the position of ‘O’ in the reverse alphabet series:

‘O’ is in the 12th position from Z.

Step 2: Move 8 letters to the right from ‘O’.

Starting from ‘O’ (12th position), the eighth letter to the right is at the 20th position.

Step 3: The 20th letter is ‘H’.

Answer:

(a) H