Quantitative Aptitude - Arithmetic Ability Questions

P = Rs. 900, R = 3 ½ % = 7/2 %, T = 2 years. 

Therefore, 

S.I. = PTR/100

S.I. = Rs. (900 x 7/2 x 2/100) = Rs. 63

Now, P = Rs. 200, S.I. = Rs. 63, R = 5 %

Time = ((100 x 63) / (200 x 5) ) years = 6.3 years.

55 min. spaces are covered in 60 min.

60 min. spaces are covered in min. = min.

Loss in 64 min. = min.

Loss in 24 hrs = min. = min.

Let the cost price be Rs. 100
Then, market price is Rs. 140
Now, the first discount is of 28.57% Rs.140 =~ Rs. 40
Hence, its selling price = 140 - 40 = Rs. 100
Now since Rana is selling at cost price, any further discount will be equal to loss percentage.
The next dicount of 12.5% will be the loss percentage to Rana.

let the side of the square be x meters
length of two sides = 2x meters
diagonal = 2x= 1.414x m

saving on 2x meters = .59x m

saving % = 0.59x2x*100%

= 30% (approx)

Vol of silver = πR2h  = 66 C.C  

Given D= 1mm => R = D/2 = 1/20 cm  

=> 227×1202×h = 66

= 66 x 400 x (7/22) = 8400 cm = 84 mtrs.

In the word 'MATHEMATICS', we treat the vowels AEAI as one letter.

Thus, we have MTHMTCS (AEAI).

Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different.

Number of ways of arranging these letters = 8!/(2! x 2!)= 10080.

Now, AEAI has 4 letters in which A occurs 2 times and the rest are different.

Number of ways of arranging these letters =4!/2!= 12.

Required number of words = (10080 x 12) = 120960

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.

The distance to be covered = Sum of their lengths = 200 + 300 = 500 m.

Relative speed = 72 -36 = 36 kmph = 36 x 5/18 = 10 mps.

Time required = d/s = 500/10 = 50 sec.