In a certain group of people, the average (arithmetic mean) weight of

Question

In a certain group of people, the average (arithmetic mean) weight of the males is 180 pounds and of the females, 120 pounds. What is the average weight of the people in the group?

(1) The group contains twice as many females as males.
(2) The group contains 10 more females than males.

DS47602.01

shridhar786 · Accepted Answer

the average weight of the males = 180
the average weight of the females = 120

STATEMENT (1)--The group contains twice as many females as males.
let the total number of male =x
the total number of female = 2x
the average weight of the males =  \frac{sum of weight of males}{total number of male} 
180 =  \frac{sum of weight of males}{total number of male} 
sum of weight of males = 180x

the average weight of the females =  \frac{sum of weight of females}{total number of female} 
120 =  \frac{sum of weight of females}{total number of female} 
sum of weight of females = 240x

What is the average weight of the people in the group?
=  \frac{sum of total}{number of total}  =  \frac{420x}{3x}  = 140
SUFFICIENT

STATEMENT (2)--The group contains 10 more females than males.
let males = x
females = x+10
sum of weights of male = 180x
sum of weights of female = 120(x+10)

average of total =  \frac{180x+120(x+10)}{2x+10} 
we dont know value of x so we cant find average
INSUFFICIENT

A is the answer

BrentGMATPrepNow · Answer

We can solve this question using  weighted averages 
  Weighted average of groups combined = (group A proportion)(group A average) + (group B proportion)(group B average) + (group C proportion)(group C average) + ...

Given : In a certain group of people, the average (arithmetic mean) weight of the males is 180 pounds and of the females, 120 pounds.

Target question :    What is the average weight of the people in the group?

Statement 1 : The group contains twice as many females as males.  
This statement is SUFFICIENT because it provides us with information about the proportion of people in each group. 
The above information tells us that, out of every three people, two are females and one is male. 
In other words 2/3 of the group are female and 1/3 are male. 
Adding this to our weighted average formula we get:  average weight of the people in the group = (2/3)(120) + (1/3)(180)  
Since we COULD evaluate this expression, we COULD answer the answer the  target question  with certainty
Statement 1 is SUFFICIENT

Statement 2 : The group contains 10 more females than males. 
This information does not tell us the proportion of people in each group (males and females)
For example, it could be the case that there are 14 people all together, with a breakdown of 12 females and 2 males. In this case, 12/14 of the people are females and 2/14 are males. 
Or it could be the case that there are 16 people all together, with a breakdown of 13 females and 3 males. In this case, 13/16 of the people are females and 3/16 are males. 
As you might imagine, the two scenarios, when plugged into are weighted averages formula, would yield different average weights for the group. 
Since we cannot answer the  target question  with certainty, statement 2 is NOT SUFFICIENT

Answer: A

Cheers,
Brent

ZoltanBP · Answer

This is a weighted average problem. We know the different partial averages, so we could determine the global average if we knew the weights. In the weighted arithmetic average form the weights would be the number of males and the number of females. However, in any weighted average it's the ratio of the weights that really matters, not their actual values.

The rephrased question: M:F=?

1)  We know that M:F=1:2. Thus, the answer to the rephrased question is a unique value.  \implies   Sufficient

2)  We know that F=M+10. If M=2, then M:F=1:6. However, if M=10, then M:F=1:2. Thus, we can't get a unique value to answer the rephrased question.  \implies   Insufficient

Answer: A

MathRevolution · Answer

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution. Visit https://www.mathrevolution.com/gmat/lesson for details. Assume m and f are the numbers of males and females, respectively. The question ask what the value of \frac{180m+120f}{m+f} . When a question asks for a ratio, if one condition includes a ratio and the other condition just gives a number, the condition including the ratio is most likely to be sufficient. This tells us that A is most likely to be the answer to this question. Condition 1) Since f = 2m , we have \frac{180m + 120f}{m+f} = \frac{180m+240m}{m+2m} = \frac{420m}{3m} = 140 . Since condition 1) yields a unique solution, it is sufficient. Condition 2) Since f = m + 10 , we have \frac{180m + 120f}{m+f} = \frac{180m + 120m + 1200}{m + m + 10} = \frac{300m + 1200 }{ 2m+10} = \frac{150m + 600 }{ m+5} . If m = 5 , then we have the average \frac{150*5+600)}{10} = \frac{1350}{10} = 135 . If m = 15 , then we have the average \frac{150*15+600}{20} = \frac{2850}{20} = 142.5 . Since condition 2) does not yield a unique solution, it is not sufficient Therefore, A is the answer as expected.

babloorudra · Answer

though it doesn't matter, I would like to know if it is F=2M or M=2F. Is there any resource for deciphering such statements. I find it a bit difficult to convert the english into Math.

hemantbafna · Answer

Please find attached the answer to the question.

DeeptiManyaExpert · Answer

When questions talks about average weight, height, age, etc. of 2 groups, think in terms of mixtures and alligations. Check the image given below for solution.

Beko2021 · Answer

Males average weight 180p
Females average weight 120p
1) 2M=F (numbers)
2) F=M+10 (numbers); So, 2M=M+10; M=10 and F=20
(180p+120p)/2=150p
The average weight in the group is 150 pounds ( ~68 kilograms).

Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.

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In a certain group of people, the average (arithmetic mean) weight of

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