Quantitative Aptitude - Arithmetic Ability Questions

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

720 + 1024 x ? = 1232

? = 1232 - 720/1024

? = 512/1024 = 0.5

? = 0.5 

a^3 + b^3 = (a+ b) * (a^2 + b^2 - ab)

[ (10.3)^3 + (1)^3 ] / [(10.3)^2  + (1)^2 - (1 * 10.3)]

 => a + b = 10.3 + 1 = 11.3 

Let the lent at 5% be 'A'

(A x 5 x 1)/100 + [(1500 - A)x 6 x 1]/100 = 85

5A/100 + 90 – 6A/100 = 85

A/100 = 5

=> A = 500

1/2 as a percentage means

1/2 = 0.5 in decimal

Now, in percentage means 0.5 x 100% = 50%.

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

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

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

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]