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11 Dec 2020, 03:56
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
79% (01:31) correct
21%
(02:07)
wrong
based on 57
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The average of three equally spaced numbers is 12. If the largest number equals twice the smallest, what is the value of the smallest number?
(A) 4
(B) 6
(C) 8
(D) 10
(E) 12
(A) 4
(B) 6
(C) 8
(D) 10
(E) 12
11 Dec 2020, 04:07
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SOLUTION:
Let the three numbers be (x-d),x,( x +d)
"d" is the space.
Average = (x-d)+x+(x+ d)/3 = 12
=> x=36/3=12
According to the question:
(x+d) = 2(x-d)
=> x=3d
Using x=12 from (i) we have d=4
Substituting the values-
(12+4) , 12 , (12-4)
16 , 12 , 8
Hence the smallest number is 8. OPTION (C)
Hope this helps
Devmitra Sen (Math)
Let the three numbers be (x-d),x,( x +d)
"d" is the space.
Average = (x-d)+x+(x+ d)/3 = 12
=> x=36/3=12
According to the question:
(x+d) = 2(x-d)
=> x=3d
Using x=12 from (i) we have d=4
Substituting the values-
(12+4) , 12 , (12-4)
16 , 12 , 8
Hence the smallest number is 8. OPTION (C)
Hope this helps
Devmitra Sen (Math)
13 Dec 2020, 22:08
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Since the 3 numbers are evenly spaced,
I used: x, (3/2)x, 2x
The middle number is (3/2)x because it must be the mean of the smallest and largest numbers.
Add them all together and equal them to 36.
Answer is 8
I used: x, (3/2)x, 2x
The middle number is (3/2)x because it must be the mean of the smallest and largest numbers.
Add them all together and equal them to 36.
Answer is 8
