The average of four numbers is 50. If 8 is added to the first number, the second number is multiplied by 4, 8 is subtracted from the third number, and the fourth number is divided by 4, then all the resulting values become equal. Find the difference between the second-highest number and the smallest number.

ATQ, Let the common resulting value be y. When 8 is added to the first number, the result is y. So, first number = y − 8 When the second number is multiplied by 4, the result is y. So, second number = y/4 When 8 is subtracted from the third number, the result is y. So, third number = y + 8 When the fourth number is divided by 4, the result is y. So, fourth number = 4y Average of four numbers = 50 So, total of four numbers = 50 × 4 = 200 Now add the four numbers: (y − 8) + y/4 + (y + 8) + 4y = 200 (2y) + 4y + y/4 = 200 6y + y/4 = 200 (24y + y)/4 = 200 25y/4 = 200 25y = 800 y = 800/25 y = 32 First number = 32 − 8 = 24 Second number = 32/4 = 8 Third number = 32 + 8 = 40 Fourth number = 4 × 32 = 128 Numbers are 24, 8, 40 and 128 Smallest number = 8 Second-highest number = 40 Required difference = 40 − 8 = 32