When ‘x + 5’ numbers having an average of 50 is added with ‘x + 15’ numbers having an average of ‘x + 10’, then average of all the numbers together becomes 50. Which of the following statements can be true such that
x > 5 x  >  5 ?
I. The area of a square is ‘16x’ cm² and its perimeter is ‘1.6x’
cm.
II. When ‘x’ is subtracted from a positive number, then the number becomes 150 times of its reciprocal and if the number is divided by 6, then it becomes a perfect square.
III. When 1000 is divided by ‘x + 5’, then 10 is obtained as remainder.

Solution

According to question; {50(x + 5) + (x + 15)(x + 10)}/(x + 5 + x + 15) = 50 Or, {50x + 250 + x² + 25x + 150}/(2x + 20) = 50 Or, x² + 75x + 400 = 100x + 1000 Or, x² – 25x – 600 = 0 Or, (x – 40)(x + 15) = 0 Or, x = 40 For I: Area = 16 × 40 = 640, so side = 8√10 cm. Perimeter = 4 × 8√10 = 32√10 cm. Value of 1.6x = 1.6 × 40 = 64 cm. So, ‘I’ cannot be true. For II: Let the number be ‘a’. a – 40 = 150/a Or, a² – 40a – 150 = 0

This does not give a suitable positive integer value of ‘a’.

So, ‘II’ cannot be true. For III: Since, x + 5 = 40 + 5 = 45 And, 1000 = 45 × 22 + 10 Therefore, remainder = 10 So, ‘III’ can be true. Hence, option c.