GATE Questions

Q:

How many zeros are there from 1 to 10000 ?

A) 2893 B) 4528
C) 6587 D) 4875
 
Answer & Explanation Answer: A) 2893

Explanation:

For solving this problem first we would break the whole range in 5 sections


1) From 1 to 9
Total number of zero in this range = 0


2) From 10 to 99
Total possibilities = 9*1 = 9 ( here 9 is used for the possibilities of a non zero integer)


3) From 100 to 999 - three type of numbers are there in this range
a) x00 b) x0x c) xx0 (here x represents a non zero number)
Total possibilities
for x00 = 9*1*1 = 9, hence total zeros = 9*2 = 18
for x0x = 9*1*9 = 81, hence total zeros = 81
similarly for xx0 = 81
total zeros in three digit numbers = 18 + 81 +81 = 180


4) From 1000 to 9999 - seven type of numbers are there in this range
a)x000 b)xx00 c)x0x0 d)x00x e)xxx0 f)xx0x g)x0xx
Total possibilities
for x000 = 9*1*1*1 = 9, hence total zeros = 9*3 = 27
for xx00 = 9*9*1*1 = 81, hence total zeros = 81*2 = 162
for x0x0 = 9*1*9*1 = 81, hence total zeros = 81*2 = 162
for x00x = 9*1*1*9 = 81, hence total zeros = 81*2 = 162
for xxx0 = 9*9*9*1 = 729, hence total zeros = 729*1 = 729
for xx0x = 9*9*1*9 = 729, hence total zeros = 729*1 = 729
for x0xx = 9*1*9*9 = 729, hence total zeros = 729*1 = 729
total zeros in four digit numbers = 27 + 3*162 + 3*729 = 2700
thus total zeros will be 0+9+180+2700+4 (last 4 is for 4 zeros of 10000)
= 2893

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Q:

A man pointing to a lady says, "Her brother is the father of my only son's sister."  How is that lady related to the man ?

A) sister of father-in-law B) daughter-in-law
C) Sister D) daughter
 
Answer & Explanation Answer: C) Sister

Explanation:
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14 8239
Q:

A train moving at 2/3 rd of its normal speed reaches its destination 20 minutes late. Find the normal time taken ?

A) 4/3 hrs B) 2/3 hrs
C) 3/2 hrs D) 1/4 hrs
 
Answer & Explanation Answer: B) 2/3 hrs

Explanation:

Let the original speed and time is S and T
then distance = S x T
Now the speed changes to 2/3S and T is T+20
As the distance is same
S x T = 2/3Sx(T+20)
solving this we get t = 40 minutes
=40/60 = 2/3 hour

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Q:

14 persons are seated around a circular table. Find the probability that 3 particular persons always seated together.

A) 11/379 B) 21/628
C) 24/625 D) 26/247
 
Answer & Explanation Answer: C) 24/625

Explanation:

Total no of ways = (14 – 1)! = 13!

Number of favorable ways = (12 – 1)! = 11!

 

So, required probability = 11!×3!13! = 39916800×66227020800 = 24625

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22 8234
Q:

How many meshes are there in 1 squrare meter of wire gauge if each mesh is 8mm long and 5mm wide ?

A) 2500 B) 25000
C) 12500 D) 15400
 
Answer & Explanation Answer: B) 25000

Explanation:

We know that,

 

One square meter is equal to 10,00,000 sq mm.
Given, each mesh is 8mm length and 5mm wide
Therefore for each mesh : 8 x 5 = 40 sq mm.
So, number of meshes = 100000040 = 25000.

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6 8213
Q:

When the price of whear was increased by 32%, a family reduced its consumption in such a way that the expenditure on wheat was only 10% more than before. If 30 kg per month were consumed before, find the new monthly consumption of family?

A) 27 kgs B) 25 kgs
C) 24 kgs D) 21 kgs
 
Answer & Explanation Answer: B) 25 kgs

Explanation:

Initial wheat per kg cost be Rs. 100

Then, monthly expenditure of the family = 30 x 100 = 3000

After increase in cost,

Let P be new monthly consumption of family.

P x 132 = 110/100 x 3000

=> P = 25 kg.

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7 8206
Q:

A batsman scores 26 runs and increases his average from 14 to 15. Find the runs to be made if he wants top increasing the average to 19 in the same match ?

A) 74 B) 79
C) 72 D) 60
 
Answer & Explanation Answer: A) 74

Explanation:

Number of runs scored more to increse the ratio by 1 is 26 - 14 = 12
To raise the average by one (from 14 to 15), he scored 12 more than the existing average.
Therefore, to raise the average by five (from 14 to 19), he should score 12 x 5 = 60 more than the existing average. Thus he should score 14 + 60 = 74.

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Q:

19 C in a CB

Answer

19 C in a CB means 19 Coins in a Carrom Board.


We have 9 white coins, 9 black coins and one red coin in a carrom board.

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