Quantitative Aptitude - Arithmetic Ability Questions

Let the distance and original speed be 'd' km and 'k' kmph respectively.

d/0.8k - d/k = 20/60 => 5d/4k - d/k = 1/3

=> (5d - 4d)/4k = 1/3 => d = 4/3 k

Time taken to cover the distance at original speed

= d/k = 4/3 hours = 1 hour 20 minutes.

Let the cost price be x and selling price be y.
Loss = x – y

When the cost price doubles, the loss gets tripled.
So it becomes like this, 2x – y = 3(x-y)

=> x = 2y

Loss % = (loss/ C.P) x 100 = [(2y-y)/2y] x 100= 50 %

Let A's age 10 years ago = x years.
Then, B's age 10 years ago = 2x years.
(x + 10) / (2x+ l0) = 3/4
=> x = 5.
So, the total of their present ages =(x + 10 + 2x + 10)
= (3x + 20) = 35 years.

Raj’s age today = 17 decades,
Hence, 5 decades back, he must be 12 years old.
Rahim must be 36 years old, Because (3×12).
5 years back Rahim must be 41 years old today. Because (36+5).

Volume of block=(6 x 9 x 12) cu.cm = 648 cu.cm 

Side of largest cube = H.C.F of 6,9,12 = 3cm 

Volume of the cube=(3 x 3 x 3)=27cu.cm 

Number of cubes=(648/27)=24

Let T.D. be Re. 1.

Then, B.D. = Rs.11/10= Rs. 1.10.

Sum = Rs. (1.10*1)/(1.10-1) = Rs. (110/10)= Rs. 11.

S.I. on Rs. 11 for 2 years is Rs. 1.10

Rate = [(100 x 1.10)/(11 x 2 )]% = 5%.

Suppose they meet x hrs after 8 a.m

then,

[Distance moved by first in x hrs] + [Distance moved by second in (x-1) hrs] = 330.

Therefore, 60x + 75(x-1) = 330.

 => x=3.

So,they meet at (8+3) i.e, 11a.m.