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Ratio Proportion

by guerrero » Sun Nov 11, 2012 12:51 pm
Two numbers are in the ratio 1:2. The ratio of addition of these two numbers to the third number is 4:3.The ration of addition of these three numbers to the fourth number is 3:2 .If the average of the four numbers is 35,find the average of the three numbers excluding the smallest number .

1)40 1/3
2)61 1/3
3)35
4)29
5)41 1/3

I could not follow the question, appreciate a step by step approach on this . Also, can i expect a question of this level of difficulty in the Exam ?

thanks in advance !
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by GMATGuruNY » Sun Nov 11, 2012 4:17 pm
guerrero wrote:Two numbers are in the ratio 1:2. The ratio of addition of these two numbers to the third number is 4:3.The ration of addition of these three numbers to the fourth number is 3:2 .If the average of the four numbers is 35,find the average of the three numbers excluding the smallest number .

1)40 1/3
2)61 1/3
3)35
4)29
5)41 1/3

I could not follow the question, appreciate a step by step approach on this . Also, can i expect a question of this level of difficulty in the Exam ?

thanks in advance !
Let the 4 numbers be a, b, c and d.
Work from the end of the problem to the beginning.

The average of the four numbers is 35:
Determine the sum.
a+b+c+d = number*average = 4*35 = 140.

The ratio of the sum of these three numbers to the fourth number is 3:2:
(a+b+c) : d = 3:2.
Since the sum of the parts of this ratio = 3+2 = 5, and a+b+c+d = 140, the multiplier for this ratio = 140/5 = 28.
Thus, (a+b+c) : d = 28(3:2) = 84:56.
Thus, a+b+c = 84 and d=56.

The ratio of the sum of these two numbers to the third number is 4:3:
(a+b) : c = 4:3.
Since the sum of the parts of this ratio = 4+3 = 7, and a+b+c = 84, the multiplier for this ratio = 84/7 = 12.
Thus, (a+b) : c = 12(4:3) = 48:36.
Thus, a+b = 48 and c=36.

Two numbers are in the ratio 1:2:
a:b = 1:2.
Since the sum of the parts of this ratio = 1+2 = 3, and a+b = 48, the multiplier for this ratio = 48/3 = 16.
Thus, a : b = 16(1:2) = 16:32.
Thus, a=16 and b=32.

Thus, the average of the 3 greatest numbers = (b+c+d)/3 = (32+36+56)/3 = 124/3 = 41 1/3.

The correct answer is E.
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by jkaustubh » Mon Nov 12, 2012 1:54 am
GMATGuruNY wrote:
guerrero wrote:Two numbers are in the ratio 1:2. The ratio of addition of these two numbers to the third number is 4:3.The ration of addition of these three numbers to the fourth number is 3:2 .If the average of the four numbers is 35,find the average of the three numbers excluding the smallest number .

1)40 1/3
2)61 1/3
3)35
4)29
5)41 1/3

I could not follow the question, appreciate a step by step approach on this . Also, can i expect a question of this level of difficulty in the Exam ?

thanks in advance !
Let the 4 numbers be a, b, c and d.
Work from the end of the problem to the beginning.

The average of the four numbers is 35:
Determine the sum.
a+b+c+d = number*average = 4*35 = 140.

The ratio of the sum of these three numbers to the fourth number is 3:2:
(a+b+c) : d = 3:2.
Since the sum of the parts of this ratio = 3+2 = 5, and a+b+c+d = 140, the multiplier for this ratio = 140/5 = 28.
Thus, (a+b+c) : d = 28(3:2) = 84:56.
Thus, a+b+c = 84 and d=56.

The ratio of the sum of these two numbers to the third number is 4:3:
(a+b) : c = 4:3.
Since the sum of the parts of this ratio = 4+3 = 7, and a+b+c = 84, the multiplier for this ratio = 84/7 = 12.
Thus, (a+b) : c = 12(4:3) = 48:36.
Thus, a+b = 48 and c=36.

Two numbers are in the ratio 1:2:
a:b = 1:2.
Since the sum of the parts of this ratio = 1+2 = 3, and a+b = 48, the multiplier for this ratio = 48/3 = 16.
Thus, a : b = 16(1:2) = 16:32.
Thus, a=16 and b=32.

Thus, the average of the 3 greatest numbers = (b+c+d)/3 = (32+36+56)/3 = 124/3 = 41 1/3.

The correct answer is E.

Is there any other quick approach for this question?
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by GMATGuruNY » Mon Nov 12, 2012 3:01 pm
jkaustubh wrote:
guerrero wrote:Two numbers are in the ratio 1:2. The ratio of addition of these two numbers to the third number is 4:3.The ration of addition of these three numbers to the fourth number is 3:2 .If the average of the four numbers is 35,find the average of the three numbers excluding the smallest number .

1)40 1/3
2)61 1/3
3)35
4)29
5)41 1/3
Is there any other quick approach for this question?
Another approach:

Sum of all 4 numbers = number*average = 4*35 = 140.
Let the 4 numbers be a, b, c and d.

To combine ratios with a COMMON ELEMENT, the common element must be represented by the SAME VALUE in each ratio.

It is given that a:b = 1:2 and (a+b):c = 4:3.
The COMMON ELEMENT in the two ratios is a+b.
First ratio:
a : (a+b) = 1:(1+2) = 1:3 = 4:12.
Second ratio:
(a+b) : c = 4:3 = 12:9.
Combining the ratios:
a : a+b : c = 4:12:9.

In the resulting ratio, a+b+c = 12+9 = 21.
Since (a+b+c) : d = 3:2, we get:
21/d = 3/2
d=14.

Combining the ratios above, we get:
a : a+b : c : d = 4:12:9:14.

Since a=4 and a+b+c+d = 12+9+14 = 35, a = 4/35 of the sum of all 4 numbers.
Thus, b+c+d is 31/35 of the sum:
b+c+d = (31/35)(140) = 124.

Average of b+c+d = 124/3 = 41 1/3.
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