# GATE Questions

Q:

5 men and 4 women are to be seated in a row so that the women occupy the even places . How many such arrangements are possible?

 A) 2880 B) 1440 C) 720 D) 2020

Explanation:

There are total 9 places out of which 4 are even and rest 5 places are odd.

4 women can be arranged at 4 even places in 4! ways.

and 5 men can be placed in remaining 5 places in 5! ways.

Hence, the required number of permutations  = 4! x 5! = 24 x 120 = 2880

3 4556
Q:

A Certain sum of money an amounts to Rs 2500 in a span Of 5 years and further to Rs.3000 in a span of 7 years at simple interest The sum is ?

 A) Rs. 1800 B) Rs. 2000 C) Rs. 1400 D) Rs. 1250

Answer & Explanation Answer: D) Rs. 1250

Explanation:

2500 in 5th year and 3000 in 7th year
So in between 2 years Rs. 500 is increased => for a year 500/2 = 250
So, per year it is increasing Rs.250 then in 5 years => 250 x 5 = 1250
Hence, the initial amount must be 2500 - 1250 = Rs. 1250

8 4539
Q:

The calendar for the year 1988 is same as which upcoming year ?

 A) 2012 B) 2014 C) 2016 D) 2010

Explanation:

We already know that the calendar after a leap year repeats again after 28 years.

Here year 1988 is a Leap year, then the same calendar will be in the year = 1988 + 28 = 2016.

42 4506
Q:

If the numerator of a fraction is increased by 150% and the denominator of the fraction is increased by 350%, the resultant fraction is 25/51. What is the original fraction ?

 A) 31/25 B) 15/17 C) 14/25 D) 11/16

Explanation:

The original fraction is    = 15/17.

10 4331
Q:

Artists are generally whimsical. Some of them are frustrated. Frustrated people are prone to be drug addicts.

Based on these statements which of the following conclusions is true?

 A) All frustrated people are drug addicts B) Frustrated people are whimsical C) All drug addicts are artists D) Some artists may be drug addicts

Answer & Explanation Answer: D) Some artists may be drug addicts

Explanation:

5 4128
Q:

A vessel contains 20 liters of a mixture of milk and water in the ratio 3:2. 10 liters of the mixture are removed and replaced with an equal quantity of pure milk. If the process is repeated once more, find the ratio of milk and water in the final mixture obtained ?

 A) 5:3 B) 1:4 C) 4:1 D) 9:1

Explanation:

Milk = 3/5 x 20 = 12 liters, water = 8 liters

If 10 liters of mixture are removed, amount of milk removed = 6 liters and amount of water removed = 4 liters.

Remaining milk = 12 - 6 = 6 liters

Remaining water = 8 - 4 = 4 liters

10 liters of pure milk are added, therefore total milk = (6 + 10) = 16 liters.

The ratio of milk and water in the new mixture = 16:4 = 4:1

If the process is repeated one more time and 10 liters of the mixture are removed,
then amount of milk removed = 4/5 x 10 = 8 liters.

Amount of water removed = 2 liters.

Remaining milk = (16 - 8) = 8 liters.

Remaining water = (4 -2) = 2 liters.

Now 10 lts milk is added => total milk = 18 lts

The required ratio of milk and water in the final mixture obtained

= (8 + 10):2 = 18:2 = 9:1.

7 4105
Q:

Equal quantities of three mixtures of milk and water are mixed in the ratio 1:2, 2:3 and 3:4. The ratio of water and milk in the mixture is ?

 A) 193 : 122 B) 97 : 102 C) 115 : 201 D) 147 : 185

Answer & Explanation Answer: A) 193 : 122

Explanation:

Given the three mixtures ratio as (1:2),(2:3),(3:4)

(1+2),(2+3),(3+4)

Total content = 3,5,7

Given equal quantities of the three mixtures are mixed, then LCM of 3, 5, 7 = 105

105/3 = 35 , 105/5 = 21 , 105/7 = 15

Now, the individual equal quantity ratios are (35x1, 35x2), (21x2, 21x3), (15x3, 15x4)

i.e (35,70), (42,63), (45,60)

So overall mixture ratio of milk and water is

35+42+45 : 70+63+60

122:193

But in the question asked the ratio of water to milk = 193 : 122

7 3990
Q:

How many different words can be made using the letters of the word ' HALLUCINATION ' if all constants are together?

 A) 129780 B) 1587600 C) 35600 D) None of these

Explanation:

H   L   C   N    T    A   U   I   O

L        N          A        I

There are total 131 letters out of which 7 are consonants and 6 are vowels. Also ther are 2L's , 2N's, 2A's and 2I's.

If all the consonants  are together then the numberof arrangements = $\frac{\mathbf{7}\mathbf{!}}{\mathbf{2}\mathbf{!}}$x 1/2! .

But the 7 consonants  can be arranged themselves in  $\frac{\mathbf{7}\mathbf{!}}{\mathbf{2}\mathbf{!}}$ x 1/2! ways.

Hence the required number of ways = ${\left(\frac{7!}{2!2!}\right)}^{2}$ = 1587600