P                               Q
|_________________|
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A
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B

When B starts from city P towards city Q, the distance which A would already have covered = 54 x 3 =162 km
A 9 a.m. , the train B is separated from train A by a distance of 162 km.
Train B overtakes train A after a time of
162/(72-54) = 162/18 = 9 hrs
In 9 hrs; the distance traveled by B = 72 x 9 = 648 km.
Distance from city Q, when they meet
=(1440 - 648) = 792 km

Let the usual time taken by the man to reach his office be t. The speed is 4/3rd the normal speed.
Hence time is 4/3rd, 4/3t -t =1/3t = 20
t = 60 minutes.
Had the person walked at 4/3rd of his usual speed. time taken by him = 3t/4
=3/4 (60) = 45 minutes

Speed = 54 x 5/18 = 15 m/sec.

Length of the train = (15 x 20)m = 300 m.

Let the length of the platform be x metres.

Then (x+300)/36 = 15

x + 300 = 540

or, x = 240 m.

Speed upstream = 7.5 kmph.

Speed downstream = 10.5 kmph.

  Total time taken = (105/7.5 + 105/10.5) hours = 24 hours

Let us say x kg of A was mixed with x kg of B to form 2x kg of the third alloy.
Weight of copper in the third alloy = 5/16(x) +3/8(x)=11/16(x)
Weight of zinc in the third alloy = 2x - 11/16(x) = 21/16(x)
The required ratio = 11:21

Let the amount of alcohol initially present be x litres. After two successive dilutions,
(x-6)/x = 9/ (9+16)=(3/5)² ; x=15 litres

Let the sum be P. Let the rate of interest be R% p.a.
(P) (R/100)4 = P
P(R/100)5 = 12500
P=10000
Required interest = P(R/100)3 = 7500

Let CP and SP denote the cost price and selling price of apples respectively.
40SP = 40CP + 10SP
30SP = 40CP
SP = 4/3CP
Profit = 33.33%

πr²h/2πrh = 924/264
r= (924/264 x 2) = 7 m
And, 2πrh = 264
or h= 264 x 7/22 x 1/2 x 1/7 = 6 m
Required ratio = 2r/h = 14/6 = 7:3

Let the number of lead shots be N.
(N) (Volume of the lead shot) = Increase in the volume of water
π (40)²[(8.5-6)10]=N x 4/3 π (2)³
(all measurements are converted to mm)
N= 1600 x 2.5 x 10 x 3/4 x 1/8= 3750

3^{x - y} = 27 = 3³       x - y = 3 ....(i)

3^{x + y} = 243 = 3⁵      x + y = 5 ....(ii)

On solving (i) and (ii), we get x = 4.

Let the number of sweets received by Ram, Shyam and Tarun be r,s and t respectively.
r/s=5/4
s/t=2/3=4/6
r:s:t = 5:4:6
or t = 6/15(60) = 24

(6m +8w)10=(8m+22w)5
(6m+8w)2=8m+22w
12m+16w=8m+22w
4m=6w
2m=3w
work = (6m+8w)10
=(6x3/2w+8w)10 = 170 w days
new work = 170 x 3 days
34 women will do it in (170 x 3)/34 = 15 days

Let x boys be required to work,
Then x(1/8)+1/2+1/4=1
(x+6)/8=1,
x=2

4^log₄52=5^2
x=25

if log₂(1-1/2^x)=x-2
(1-1/2^x)=2^x-2=2^x/4
if 2^x=a, then (a-1)/a=a/4
or a²-4a+4=(a-2)²=0
or a=2
or 2^x=2
x=1

log₂₅125-log₁₂₅25
= (3/2)log₅5-(2/3)log₅5
=3/2-2/3
=5/6

The probabilities and the corresponding number of ways in these possibilities can be realised or tabulated below
 

Boys	Girls	No of Ways
0	4	1
1	3	4
2	2	6
3	1	4
		15

There are 23(26-3) sets of 4 consecutive letters in the alphabet
Total number of favourable probability = 17
Required probability = 17/23

We may have(1 black and 2 non-black) or (2 black and 1 non-black) or (3 black).
So, Required number of ways = (³C₁ x ⁶C₂) + (³C₂ x ⁶C₁) + (³C₃)
=64

Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it.

The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.

The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.

So, Required number of numbers = (1 x 5 x 4) = 20.

The word QUESTION has letters of which 4 are vowels and 4 are consonents. There are 4 even places and the vowels can be arranged in these 4 places in 4! ways while the consonants can be arranged in the remaining 4 places in  4! ways.
CVCVCVCV
Hence total arrangements are 4! x 4! = 24 x 24 = 576

We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).
So, Required number of ways = (7C3 x 6C2) + (7C4 x 6C1) + (7C5)=756

The father of the boy's uncle → the grandfather of the boy and daughter of the grandfather → sister of father.

Let the age of Prashant is A and Dishant is B.
A+B = 2(A+B-14)
or, A+B = 28
A² + B² =400or (A+B)² - 2AB = 400
AB = 192

24700 = 247 (100) = (19)(13)(12²)(5 ²)
Number of ways in which a number having n prime factors can be expressed as a product of two co-prime factors = 2^n-1
Required number of ways = 2^4-1 = 8