Quantitative Aptitude - Arithmetic Ability Questions

If you double the sides of a cube, the ratio of the surface areas of the old and new cubes will be 1: 4. The ratio of the volumes of the old and new cubes will be 1: 8.

Weight is proportional to volume. So, If the first weighs 6 pounds, the second weighs 6x8 pounds =48.

Sum =(B.D. x T.D.)/(B.D. - T.D.)

       = Rs. (72 x 60) / (72 - 60)

       = Rs. (72 x 60) /12

       = Rs. 360.

Let the original price of rice be Rs. x

Let rice a man can buy for Rs. 500 at rs. x/kg be = R kgs

From given data, for Rs. 500

x ---- R kgs

(x + 25x/100) ---- (R - 5) kgs

=> Rx = (R - 5)(x + 25x/100)

=> Rx = (R - 5)(125x/100)

=> 100Rx = 125Rx - 625x

=> R = 25

So at rate of Rs. x/kg, man get 25 kgs for Rs. 500

=> x = 500/25 = Rs. 20

Let ABC be the isosceles triangle and AD be the altitude 

Let AB = AC = x. Then, BC = (32 - 2x). 

Since, in an isosceles triangle, the altitude bisects the base,
so BD = DC = (16 - x).
In triangle ADC,AC2=AD2+DC2 

⇒x2=82+16-x2⇒x=10 

BC = (32- 2x) = (32 - 20) cm = 12 cm. 

Hence, required area =  12*BC*AD=12*12*8=48cm2

 Clearly, to find the average, we ought to know the number of boys, girls or students in the class, neither of which has been given.

So, the data provided is inadequate.

Let the number of buffaoes be x and number of ducks be y.

Then 4x + 2y = 2(x + y) + 24  

=> 2x = 24

=> x = 12.

Let x128=162x

Then, $$\begin{gathered} x2=128×162 =64×2×18×9 \\ =82×62×32 = 8×6×3 =144 \end{gathered}$$

∴x=12