Quantitative Aptitude - Arithmetic Ability Questions

The ratio of the ages of A and B is 3 : 5.
The ratio of the ages of B and C is 3 : 5.

B's age is the common link to both these ratio. Therefore, if we make the numerical value of the ratio of B's age in both the ratios same, then we can compare the ages of all 3 in a single ratio.

The can be done by getting the value of B in both ratios to be the LCM of 3 and 5 i.e., 15.

The first ratio between A and B will therefore be 9 : 15 and
the second ratio between B and C will be 15 : 25.

Now combining the two ratios, we get A : B : C = 9 : 15 : 25.

Let their ages be 9x, 15x and 25x.
Then, the sum of their ages will be 9x + 15x + 25x = 49x

The question states that the sum of their ages is 147.
i.e., 49x = 147 or x = 3.

Therefore, B's age = 15x = 15*3 = 45

Distance covered in one revolution = 88×10001000= 88m. 

2πR2=88 ⇒2×227×R=88 ⇒R=88×744

             A              :               B
(8000x4)+(4000x8) : (9000x6)+(6000x6)
         64000           :      90000
=>        32 : 45

Average =   (76 + 65 + 82 + 67 + 85 )/ 5 = 375/5   =  75.

On straight line means 180 degree angle.
180 = 11/2 min – 30 hrs
180 = 11/2 m – 30 × 3
180 = 11/2 m – 90
(180 + 90) 2 = 11 m
m = 540/11 = 49 1/11 minutes.