Aptitude and Reasoning Questions

AS HARD is coded as "1357" and SOFT is coded as "2468"
So H is coded as 1, A as 3, R as 5, D as 7
and S as 2, O as 4, F as 6, T as 8

Now in 21448, go number by number
2 is the code for S
1 is the code for H
4 is the code for O
and
8 is the code for T

hence, 21448 will stand for SHOOT

Let the total earnings be Rs.x 

Then 412% of x + 14% of (x-12000) =2400

 ⇒92×x100+14×x-12000100=2400

=>  x=50,589.47

Total age seven persons = (26 x 7)years
Total age of the first three persons and the last three persons are (19 x 3) years and (32 x 3) years respectively.
Age of the person sitting in the middle of the row = 26 x 7 - 19 x 3 - 32 x 3 = 182 - 57 - 96 = 29 years.

The toys are different; The boxes are identical
If none of the boxes is to remain empty, then we can pack the toys in one of the following ways
a. 2,2,1
b. 3,1,1

Case a. Number of ways of achieving the first option 2−2–1

Two toys out of the 5 can be selected in5C2 5 ways. Another 2 out of the remaining 3 can be selected in3C2 ways and the last toy can be selected in 1C1  way.

However, as the boxes are identical, the two different ways of selecting which box holds the first two toys and which one holds the second set of two toys will look the same. Hence, we need to divide the result by 2.

Therefore, total number of ways of achieving the 2−2–1 option is:

5C2*3C22=10*32=15 ways.

Case b. Number of ways of achieving the second option 3−1–1

Three toys out of the 5 can be selected in 5C3  ways. As the boxes are identical, the remaining two toys can go into the two identical looking boxes in only one way.
Therefore, total number of ways of getting the 3−1–1 option is5C3=10 ways.

Total ways in which the 5 toys can be packed in 3 identical boxes
=number of ways of achieving Case a + number of ways of achieving Case b
=15 + 10 = 25 ways

Ankit's daughter's mother---Ankit's wife; Ankit's wife's father---Ankit's father-in-law;
Father-in-law's daughter--- Ankit's sister-in-law. So, Sanvi is sister-in-law of Ankit.

Relative speed = 120 + 80 = 200 km/hr.
= 200 x 5/18 = 500/9 m/sec.

Let the length of the other train be L mts.
Then, (L + 260)/9 = 500/9 => L = 240 mts.

Let the numbers be x, x+1, x+2 

Then x + x+1 + x+2 = 42

           =>  3x + 3 = 42

           => 3x = 39 

          =>  x = 13

The numbers would be 13, 14, 15 

Therefore, Sum of the digits of middle number = 1+4 = 5