The average of four numbers ‘a’, ‘b’, ‘c’ and ‘d’ is 60. The value of ‘c’ is 16.67% more than the value of ‘d’. The value of ‘a’ is 8 more than the value of ‘c’. The ratio between the values of ‘b’ and ‘d’ is 3:2 respectively. Which of the following statements is/are true? (It is assumed that each of the numbers is natural.)

(i) The value of ‘d’ is the multiple of 6.

(ii) The value of ‘a’ is 16 more than the value of ‘d’.

(iii) The value of ‘c’ is 54.

Solution

The average of four numbers ‘a’, ‘b’, ‘c’ and ‘d’ is 60. a+b+c+d = 60x4 = 240    Eq.(i) The value of ‘c’ is 16.67% more than the value of ‘d’.  Let’s assume ‘d’ = 6y. c = (100+16.67)% of 6y = 116.67% of 6y = (7/6) of 6y          [we know that 16.67% = (1/6).] = 7y The value of ‘a’ is 8 more than the value of ‘c’. a = (7y+8) The ratio between the values of ‘b’ and ‘d’ is 3:2 respectively.  b = (6y/2)x3 = 9y Put the values of ‘a’, ‘b’, ‘c’ and ‘d’ in terms of ‘y’ in Eq.(i). (7y+8)+9y+7y+6y = 240 29y+8 = 240 29y = 240-8 = 232 y = 8 So ‘a’ = (7x8+8) = (56+8) = 64 b = 9x8 = 72 c = 7x8 = 56 d = 6x8 = 48 (i) The value of ‘d’ is the multiple of 6. The above given statement is true. Because the value of ‘d’ is the multiple of 6. (ii) The value of ‘a’ is 16 more than the value of ‘d’. The above given statement is true. Because ‘a’ = d+16. 64 = 48+16 (iii) The value of ‘c’ is 54. The above given statement is not true. Because ‘c’ = 56.