29
Q:

# Ravi's brother is 3 years senior to him. His father was 28 years of age when his sister was born while his mother was 26 years of age when he was born. If his sister was 4 years of age when his brother was born, what were the ages of Ravi's father and mother respectively when his brother was born ?

 A) 32 years, 23 years B) 32 years, 29 years C) 35 years, 29 years D) 35 years, 33 years

Answer:   A) 32 years, 23 years

Explanation:

When Ravi's brother was born, let Ravi's father's age = x years and mother's age = y years.

Then, sister's age = (x - 28) years. So, x - 28 = 4 or x = 32.

Ravi's age = (y - 26) years. Age of Ravi's brother = (y - 26 + 3) years = (y - 23) years.

Now, when Ravi's brother was born, his age = 0 i.e. y - 23 = 0 or y = 23.

Q:

How many number of weights do you need to weight 31 kg of Rice if you have one kg of weight stone ?

 A) 3 B) 5 C) 7 D) 9

Explanation:

There are total 5 ways.
we have only one kg of weight so,
2kg = 1kg + 1kg
4kg = 1 + 2 + 1
8kg = 1 + 2 + 4 + 1
16kg = 1 + 2 + 4 + 8 + 1
31kg = 1 + 2 + 4 + 8 + 16
So 5 weights are needed.

1 21
Q:

If the positions of the first and the fifth digits of the number 14278359 are interchanged, similarly the positions of the second and the sixth digits are interchanged and so on, then which of the following will be the second digit from the right end after the rearrangement ?

 A) 5 B) 2 C) 7 D) 4

Explanation:

Given number is 14278359
As given 1st and 5th digits, 2nd & 6th, 3rd & 7th and 4th & 8th digits are to to be interchanged.
Therfore, the rearranged number is 83591427
Then, the second digit from the right end is 2.

4 40
Q:

A Child has 1200 five rupee coins. He wants to place them in such a way that the number of rows and the number of columns remains the same. What is the minimum number of coins that he needs more for this purpose?

 A) 125 B) 96 C) 35 D) 25

Explanation:

We know that, to be in the form that the number of rows and the number of columns to be equal the number should be a perfect square ($a^{2}$).
Given number is 1200
The perfect number which is above and near to 1200 is 1225 which is 35 $\times$ 35.
$\therefore$ The minimum number of coins he need is 25.

10 112
Q:

In a note collection, there is one new Rs.2000 note for every three old Rs.500 notes. 10 more new Rs.2000 notes are added to the collection and the ratio of new Rs.2000 notes to old Rs.500 notes would be 1: 2. Based on the information, the total number of notes in the collection now becomes?

 A) 60 B) 30 C) 70 D) 90

Explanation:

Let 'x' be the number of new Rs.2000 notes.
Given for one new 2000 note there are three old 500 notes.
$\fn_jvn&space;\small&space;\Rightarrow$ 3x
Given 10 more new Rs.2000 notes are added to the collection
and the ratio of new Rs.2000 notes to old Rs.500 notes
$\fn_jvn&space;\small&space;\Rightarrow$ $\inline \fn_jvn \small \frac{x+10}{3x}=\frac{1}{2} \Rightarrow x=20$
Therefore, number of new Rs.2000 notes $\fn_jvn&space;\small&space;\Rightarrow$  x+10 = 30.
old Rs.500 notes $\fn_jvn&space;\small&space;\Rightarrow$ 3x = 60.
Thus, the total number of notes in the collection $\fn_jvn&space;\small&space;\Rightarrow$  30 + 60 = 90.

8 119
Q:

In a certain office, $\inline \frac{1}{3}$ of workers are women, $\inline \frac{1}{2}$of the women are married and $\inline \frac{1}{3}$ of the married women have children. If $\inline \frac{3}{4}$ of the men are married and $\inline \frac{2}{3}$ of the married men have children, what part of the workers are without children ?

 A) 5/18 B) 4/9 C) 11/18 D) 17/36

Explanation:

Let the total number of workers be x. Then,

Number of women = $\inline \frac{x}{3}$ and number of men = $\inline \left ( x-\frac{x}{3} \right )$ = $\inline \frac{2x}{3}$

Number of women having children = $\inline \fn_jvn \frac{1}{3}\;of\; \frac{1}{2}\; of\; \frac{x}{3}=\frac{x}{18}$

Number of men having children = $\inline \fn_jvn \frac{2}{3}\;of\; \frac{3}{4}\; of\; \frac{2x}{3}=\frac{x}{3}$

Number of workers having children = $\inline \fn_jvn \left ( \frac{x}{18}+\frac{x}{3} \right )=\frac{7x}{18}$

$\inline \fn_jvn \therefore$ workers having no children = $\inline \fn_jvn \left ( x-\frac{7x}{18} \right )=\frac{11x}{18}=\frac{11}{18}$ of all workers