Quantitative Aptitude - Arithmetic Ability Questions

Let distance = x km and usual rate = y kmph.
Then, x/y - x/(y+3) = 40/60 --> 2y (y+3) = 9x ----- (i)
Also, x/(y-2) - x/y = 40/60 --> y(y-2) = 3x -------- (ii)
On dividing (i) by (ii), we get:

x = 40 km.

We have: l = 20 ft and lb = 680 sq. ft.So, b = 34 ft.

Length of fencing = (l + 2b) = (20 + 68) ft = 88 ft.

Given ratio of initial mixture of milk and water in Q = 5 : 3

Let the initial quantity of mixture in vessel Q = 8x

Let quantity of Milk = 5x and

Let quantity of water = 3x

According to the question,

5x + 10 x 353x + 10 x 25 = 85

=> 25x + 30 = 24x + 32

=> x = 2

Required Initial quantity of milk = 5x = 5 x 2 = 10 lit.

Speed of train 1 = 48 kmph
Let the length of train 1 = 2x meter

Speed of train 2 = 42 kmph
Length of train 2 = x meter (because it is half of train 1's length)

Distance = 2x + x = 3x
Relative speed= 48+42 = 90 kmph = (90*5/18)  m/s = 25 m/s 

Time = 12 s

Distance/time = speed 

=>3x/12 = 25 (25*12)/3 

Length of the first train = 2x = 200 meter
Time taken to cross the platform= 45 s

Speed of train 1 = 48 kmph = 480/36 = 40/3 m/s

Distance = 200 + y     [where y is the length of the platform] 

x =100m => 200+y = 45* 40/3
y = 400m

Let Rajan's present age be 'x' years.
=> Then his daughter's present age is = x/3 years.
His mother's present age = 13x/9 years
Now, according to the question,
x + x/3 + 13x/9 = 125
=> x = 125x9/25 = 45
Therefore, required difference = 13x/9 - x/3 = 13x - 3x/9 = 10x/9
=> 10 x 45/9 = 50 years.

(24 + 31 + 18) days = 73 days = 1/5 year .

P = Rs 3000 and R = 18 % p.a.

 S.I=3000*18*15100

Let C.P.= Rs. 100.

Then, Profit = Rs.220, 

S.P. = Rs.320.

New C.P. = 125% of Rs. 100 = Rs. 125 

New S.P. = Rs.320.

Profit = Rs. (320 - 125) = Rs. 195

Required percentage = 195320×100== 60.9 =~ 61%