Quantitative Aptitude - Arithmetic Ability Questions

Let Shilpa's present age is p years

According to question,

p = 4(x+4) – 4(x–4)

p = 4x+16 – 4x+16

p = 32

Shilpa's present age = 32 years

Rajeev is four years older than Shilpa

=> Shilpa's present age = Rajeev's age 4 years before.

=> Rajeev's age 4 years before = 32 years.

     D _  C  _ A          --------(1)

     D > B > C           --------(2)

from (1) and (2)

     D > B > C > A   ---------(3)

Again      E _ B _ A

But B > A, from (3)

So          E > D > B > C > A      [ Since B is the average of E and A so it is eqidistant from both E and A]

Let the sum be Rs. P
P{ 1+81002- 1 } = 2828.80

It is in the form of  a2-b2 = a+ba-b

P(8/100)(2 + 8/100) = 2828.80

P = 2828.80 / (0.08)(2.08)

= 1360/0.08 = 17000

Principal + Interest = Rs. 19828.80

Exact interest, I= prt = $8000 × 0.085 × 90/365 = 167.67

Ordinary Interest, I= Prt = $8000 x 0.085 x 90/360 = 170

let original radius = r and  new radius = (50/100) r = r/2

original area = πr2 and new area = πr/22

decrease in area =  3πr2/4*1πr2 *100 = 75%

Let marks obtained by N = x

M's marks = 0.80x

L's marks = 1.25 (0.80x)

K's marks = (0.90) (1.25) (0.80x)

= 0.9x

But 0.9x = 360 ⇒ x = 400

Percentage of Marks obtained by N

= 400 x 100/500 = 80%

80x900/100 + 256 x 4 x ? = 1331 - 99

720 + 1024 x ? = 1232

? = 1232 - 720/1024

? = 512/1024 = 0.5

? = 0.5 

Time taken by one tap to fill half of the tank = 3 hrs.

Part filled by the taps in 1 hour = 4 x 1/6 = 2/3

Remaining part = 1 - 1/2 = 1/2

2/3 : 1/2 :: 1 : p

p = 1/2 x 1 x 3/2 = 3/4 hrs. i.e., 45 min

So, total time taken = 3 hrs 45 min.