Find the Numbers
Out of three numbers, the first is twice the second and thrice the third. If their average is 88. Find the numbers?
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Let us assume the three numbers are x, y and z respectively.
given:
x = 2y therefore, y = x/2
also given
x = 3z therefore, z = x/3
Average is
(x + y + z) / 3 = 88
Substitute the value of y and z
x + x/2 + x/3 = 88 * 3
(6x + 3x + 2x) / 6 = 264
11x = 264 * 6
x = 264 * 6/11
x = 24 * 6
x = 144
Therefore, y = x/2 = 144/2 = 72
and z = x/3 = 144/3 = 48
Therefore, the three numbers are 144, 72 and 48
given:
x = 2y therefore, y = x/2
also given
x = 3z therefore, z = x/3
Average is
(x + y + z) / 3 = 88
Substitute the value of y and z
x + x/2 + x/3 = 88 * 3
(6x + 3x + 2x) / 6 = 264
11x = 264 * 6
x = 264 * 6/11
x = 24 * 6
x = 144
Therefore, y = x/2 = 144/2 = 72
and z = x/3 = 144/3 = 48
Therefore, the three numbers are 144, 72 and 48
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