GATE Questions

Q:

A Committee of 5 persons is to be formed from a group of 6 gentlemen and 4 ladies. In how many ways can this be done if the committee is to be included atleast one lady?

 A) 123 B) 113 C) 246 D) 945

Explanation:

A Committee of 5 persons is to be formed from 6 gentlemen and 4 ladies by taking.

(i) 1 lady out of 4 and 4 gentlemen out of 6

(ii) 2 ladies out of 4 and 3 gentlemen out of 6

(iii) 3 ladies out of 4 and 2 gentlemen out of 6

(iv) 4 ladies out of 4 and 1 gentlemen out of 6

In case I the number of ways = $C_{1}^{4}×C_{4}^{6}$ = 4 x 15 = 60

In case II the number of ways = $C_{2}^{4}×C_{3}^{6}$ = 6 x 20 = 120

In case III the number of ways = $C_{3}^{4}×C_{2}^{6}$ = 4 x 15 = 60

In case IV the number of ways = $C_{4}^{4}×C_{1}^{6}$ = 1 x 6 = 6

Hence, the required number of ways = 60 + 120 + 60 + 6 = 246

5 4866
Q:

Ram sell his goods 25% cheaper than Shyam and 25% dearer than Bram. How much % is Bram's good cheaper than Shyam ?

 A) 60% B) 40% C) 50% D) 30%

Explanation:

Lets say shyam sells at 100,

Since Ram sells 25% cheaper than Shyam,

Therefor Ram sells at less than 25% of100 or 75 rs.

Ram sells 25% dearer than Bram or 125% of Ram =100% of Bram or

125% of x (say x price of bram )=75rs.
or 100% of x =60rs.
hence price of bram is 60rs.

now Bram's good is cheaper than Shyam's as (100-60)x100/100% or 40%.

Hence Brams Good is 40% cheaper than that of Shyam's good.

13 4642
Q:

A number is increased by 20% and then decreased by 20%, the final value of the number is ?

 A) increase by 2% B) decrease by 3% C) decrease by 4% D) increase by 5%

Explanation:

Here, x = 20 and y = - 20
Therefore, the net % change in value
= %
%  or  - 4%

Since the sign is negative, there is a decrease in value by 4%.

9 4640
Q:

is related to 81 in the same way as 64 is related to ____

 A) 225 B) 425 C) 525 D) 625

Explanation:

The relationship is

Now ,

So, required number = ${\left(4+1\right)}^{\left(3+1\right)}$ = ${5}^{4}$ = 625

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88 4461
Q:

There are three rooms in a Hotel: one single, one double and one for four persons. How many ways are there to house seven persons in these rooms ?

 A) 105 B) 7! x 6! C) 7!/5! D) 420

Explanation:

Choose 1 person for the single room & from the remaining choose 2 for the double room & from the remaining choose 4 people for the four person room,

Then, 7C1 x 6C2 x 4C

= 7 x 15 x 1 = 105

9 4427
Q:

8 couples (husband and wife) attend a dance show "Nach Baliye' in a popular TV channel ; A lucky draw in which 4 persons picked up for a prize is held, then the probability that there is atleast one couple will be selected is :

 A) 8/39 B) 15/39 C) 12/13 D) None of these

Explanation:

P( selecting atleast one couple) = 1 - P(selecting none of the couples for the prize)

=

44 4303
Q:

A letter is takenout at random from 'ASSISTANT'  and another is taken out from 'STATISTICS'. The probability that they are the same letter is :

 A) 35/96 B) 19/90 C) 19/96 D) None of these

Explanation:

$ASSISTANT\to AAINSSSTT$

$STATISTICS\to ACIISSSTTT$

Here N and C are not common and same letters can be A, I, S, T. Therefore

Probability of choosing A =  $\frac{{2}_{{C}_{1}}}{{9}_{{C}_{1}}}×\frac{{1}_{{C}_{1}}}{{10}_{{C}_{1}}}$ = 1/45

Probability of choosing I = $\frac{1}{{9}_{{C}_{1}}}×\frac{{2}_{{C}_{1}}}{{10}_{{C}_{1}}}$ = 1/45

Probability of choosing S = $\frac{{3}_{{C}_{1}}}{{9}_{{C}_{1}}}×\frac{{3}_{{C}_{1}}}{{10}_{{C}_{1}}}$ = 1/10

Probability of choosing T = $\frac{{2}_{{C}_{1}}}{{9}_{{C}_{1}}}×\frac{{3}_{{C}_{1}}}{{10}_{{C}_{1}}}$ = 1/15

Hence, Required probability =

55 4219
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