Each of the students in a class either has 3 pencils and 1 pen or has

Question

Each of the students in a class either has 3 pencils and 1 pen or has 1 pencil and 2 pens. The average (arithmetic mean) of pencils in total class is 1.6, what is the average (arithmetic mean) of pen?

A. 1.3
B. 1.4
C. 1.5
D. 1.6
E. 1.7

EgmatQuantExpert · Accepted Answer

Given

• Each of the students in a class either has 3 pencils and 1 pen or has 1 pencil and 2 pens.
• The average (arithmetic mean) of pencils in total class is 1.6.

To Find

• The average (arithmetic mean) of pen.

Approach and Working Out

• Let us assume there are ‘x’ number of students with the first set and ‘y’ number of students with the second set.
• Total pencil = 3x + y.
 o Average,  \frac{3x + y}{x + y}  = 1.6
o 3x + y = 1.6x + 1.6y
o  \frac{x}{y}  =  \frac{0.6}{1.4 } =  \frac{3}{7}  
• We can assume 3 and 7 as the numbers as ratio will get cancelled.
 o Average of pen =  \frac{3*1 + 7* 2}{3 + 7}  = 1.7 
 
  Correct Answer: Option E

VoltAmpere · Answer

Given: average pencils=1.6
2 options: 
a) 3 pencils AND 1 pen
b) 1 pencil AND 2 pens

since we are given the weighted average, it would be easier to use the teeter-totter method (by Manhattan Prep)
The average 1.6 is nearer to 1 than to 3, so the weight of 1 is bigger.
Weight of a: (1.6-1)/(3-1)=0.6/2=0.3
Weight of b: (3-1.6)/(3-1)=1.4/2=0.7

To check
is 1.6=0.3(3)+0.7(1)=1.6 CHECK

This weighting of the pencils is also the weighting of the corresponding pens.
Hence,
average pens=0.3(1)+0.7(2)=0.3+1.4=1.7

CORRECT ANSWER: E

stne · Answer

Hi  sabrinaZ ,

May be this soln. can help to some extent.

Let there be total  x  students out of which  n  have 3 pencils and 1 pen and  (x-n)  students have 1 pencil and 2 pens.

so total pencils are:

3n + (x-n)1 = 1.6x

.6x = 2n

x=\frac{10}{3}n   (i)

Similarly total pens are:

n+(x-n)2=xa  where a is the average of the pens

n+2x-2n=xa

n+2(\frac{10}{3}n)-2n=(\frac{10}{3}n)a  ... substituting the value of  x  from (i)

\frac{17}{3}n =\frac{10}{3}na

so  a = \frac{17}{3}*\frac{3}{10} = 1.7

Ans-E

Hope it's clear.

lucasd14 · Answer

Interesting approach! And since x:y = 3:7 are ratios, we can always consider those values since any other would eventually cancel each other the multiple. Thanks.

Madhvendrasinh · Answer

I did this question a little differently, 
the question has given that the avg. of pencils is 1.6 and every student has either 1 or 3 pencils.
Now we can say from this information that majority of the students in class had option 2 (1 pencils and 2 pens), because average is closer to 1 than 3.
So we know that majority of students had 1 pencil and 2 pens.

And since majority of students had 2 pens we know that the average will be more than that of pencils, but not significantly more because in the other option it is 3 pencils while in the case of pens it is 1 pen.
In the options there is only one option greater than 1.6, so I went with E.
Kindly correct me if I'm approacing this the wrong way.

Each of the students in a class either has 3 pencils and 1 pen or has

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