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.

Imension of the duster is assumed as 5x5x7 cm3

Volume of the duster below the water layer
= volume of water increased
= πxrxrxh = 3.14×52×2.14 ≈ 168 cm3

Total volume of the duster = 5×5×7 = 175 cm3

Percentage of the duster part below the layer = 168×100/175 = 96%

Percentage of the duster part above the layer = 4%

% Profit with respect to A = 40 + 50 + (40 x 50)/100 = 90 + 20 = 110 %.

Let present Anirudh's age be 'A' and Bhavana's age be 'B'

Given seven years, their ratio is 7:9

=> A-7 : B-7 = 7 : 9

=> 9A - 7B = 14 .......(1)

And given,

Chandhu is 12 years older than A nad 12 years younger than B

=> C = A + 12....(2)

=> B = C + 12

=> B = A + 12 + 12 .....(From (2))

=> B = A + 24 ....(3)

Put (3) in (1)

9A - 7(A+24) = 14

9A - 7A - 168 = 14

2A = 182

=> A = 91 years

=> C = A + 12 = 91 +12 = 103 years.

Using pythagarous theorem,
distance travelled by first train = 36x5/18x30 = 300m
distance travelled by second train = 48x5/18x30 = 400m
so distance between them =√( 90000 + 160000) = √250000 = 500mts.

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 Original capital amount be Rs. P

Given after 2 years the amount is Rs. 10000

Rate of interest = 30%

According to the question,

$$\begin{gathered} P x 130100 x 130100 = 10000 \\ => P = 1000000169 = 5917 \end{gathered}$$

Therefore, the capital amount P = Rs. 5917

Time taken by 8th tap = 2 x 2 x 2= 8 hours

Time taken by 12th tap = 2x (1/2) x (1/2)  = 1/2 hour

Ratio of time taken by 8th tap and 12th tap = 8 : 1/2 =16:1

Therefore, Ratio of efficiencies of 8th tap and 12th tap =1:16