Five friends Alastair, Bell, Cook, Darren and Eoin appeared in

Question

Five friends Alastair, Bell, Cook, Darren and Eoin appeared in an aptitude test. Alastair scored exactly \(\frac{1}{2}\)of Darren’s score, whose score was \(\frac{1}{5}\)th more than Cook’s score. Eoin scored \(\frac{2}{5}\)th more than Cook and Darren’s score was \(\frac{3}{2}\) times that of Bell’s score. If the average score (arithmetic mean) of the group was 50, what was the range of the scores of the group?

A. 25
B. 30
C. 35
D. 40
E. 50

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Official Answer

D

A. 25
B. 30
C. 35
D. 40
E. 50

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gracie · Accepted Answer

Five friends Alastair, Bell, Cook, Darren and Eoin appeared in an aptitude test. Alastair scored exactly \(\frac{1}{2}\)of Darren’s score, whose score was \(\frac{1}{5}\)th more than Cook’s score. Eoin scored \(\frac{2}{5}\)th more than Cook and Darren’s score was \(\frac{3}{2}\) times that of Bell’s score. If the average score (arithmetic mean) of the group was 50, what was the range of the scores of the group?

A. 25
B. 30
C. 35
D. 40
E. 50

Show Spoiler

Thanks,
Saquib
Quant Expert
e-GMAT

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Show Answer

Official Answer

D

A. 25
B. 30
C. 35
D. 40
E. 50

the 5 scores form a progression:
C-2/5, C-1/5, C, C+1/5, C+2/5
if C=50,
then progression is 30, 40, 50, 60, 70
range=70-30=40
D

EgmatQuantExpert · Answer

Hey,

PFB the official solution.

Given

o So, total score of the group \(= 50*5 = 250\)

To Find:

o Range of the score = Highest score – Lowest Score

Working Out

So, the range of the scores of the group is \(40\).

Answer: D

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GMATGuruNY · Answer

To determine the ratio of the scores, plug in numbers and reduce the resulting ratio as much as possible.
Let A = Alastair, B = Bell, C = Cook, D = Darren, and E = Eoin.
Let B = the product of the denominators = 2*5*5*2 = 100.

Darren’s score was 3/2 times that of Bell’s score .
D = 3/2(B) = (3/2)(100) = 150.

Alastair scored exactly 1/2 of Darren’s score. 
A = (1/2)D = (1/2)(150) = 75.

Darren’s score was 1/5th more than Cook’s score. 
In other words, Darren's score of 150 is 6/5 Cook's score:
150 = (6/5)C
C = (5/6)150 = 125.

Eoin scored 2/5th more than Cook. 
In other words, Eoin's score is 7/5 Cook's score of 125:
E = (7/5)125 = 175.

A : B : C : D : E = 75:100:125:150:175 = 3:4:5:6:7.
When the values in the ratio are fully reduced, biggest - smallest = 7-3 = 4.
Implication:
The range of the values must be a MULTIPLE OF 4.

D

EgmatQuantExpert · Answer

Reserving this space to post the official solution.

adityapareshshah · Answer

IMO, the Answer is B..The explanation is as follows. Consider the scores of Alastair to be A, Bell to be B, Cook to be C,Darren to be D and Eoin to be E.
Given:-

A=1/2D
C=D-1/5
B=2/3D
E=D+1/5

Since average is 50. Summing A,B,C,D and E and taking average gives us value of D to be sixty.

Hence the lowest value in the set =1/2D=30
Biggest value in the set=D+1/5=60.2
Hence range =60.2- 30=30.2..which is approximated to 30. 
 Hence answer is Option B

EgmatQuantExpert · Answer

Hey, The calculation is not that long..why not give it a try...?

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EgmatQuantExpert · Answer

Hey, There are two things, which I want to point out - 1. The final value of A, B, C, D and E are all integers. So we don't need to approximate anything. 2. Also, the relation between C and D and a few others are not written correctly. Let me take an example - "Darren’s score, whose score was \frac{1}{5} th more than Cook’s score" This line means - D = C + \frac{1}{5} *C D = \frac{6C}{5} Similarly, check the other ratios once. I am sure you will figure out the minor errors.

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subrataroy0210 · Answer

A = (1/2) D; D = (6/5) C; E = (2/5) C; D = (3/2) B; 
Also, (6/5) C = (3/2) B
=> B = (4/5) C

A + B + C + D + E = 250
(3/5) C + (4/5) C + C + (6/5) C + (2/5) C = 250
C = 125/2

Highest -> D=(6/5) C => D = 75
Lowest -> E=(2/5) C => E = 25
Range -> 75-25 = 50

EgmatQuantExpert · Answer

Hey Subrata, You made a slight mistake while writing the relation between the variables - Eoin scored \frac{2}{5} th more than Cook. Therefore, the relation between E and C would be E = 7/5C and not 2/5C. Because of this mistake, the answer that you have got is not correct.

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kannu44 · Answer

( A+B+C+D+E )/5 = 50
 
putting a ,b,c and e in term of d

( D/2 + 2D/3 +5D/6 + D +7/6D  )/5 = 50

solving for D =60

lowest A=30
highest E=70
Range = 40

mbah191 · Answer

Not a hard question but lot's of computation involved. I don't see how this can be done carefully in 2 minutes. Just reading, understanding, and planning how to attack this question takes a solid minute. Then there is a good 2-3 minutes of math involved (5 equations, then combining them, then solving for the variable, then determining the range).

Would it be fair to say that the GMAT likely wouldn't have a question this math heavy? Or is there a shortcut? Would questions like this typically have a shortcut? How long did it take people to solve this?

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generis · Answer

This problem looks a lot worse than it is. Solved for C, used decimals instead of fractions, and ignored A and B until the end (b/c the way the question is written suggests C as jumping off point).

Stage I

A =  \frac{1}{2} D

B = ??

C = ??

D =  \frac{1}{5}  more than C = 1.2C
 
E =  \frac{2}{5}  more than C = 1.4C
-----
Stage II

2 of 5 variables are in terms of C. Add C itself, and that's 3 of 5.

A and B in terms of C?
-----
A =  \frac{1}{2} or .5D.

D = 1.2C.

1.2C x .5 = .6C
-----
B is hardest.

D = 1.5B. D also = 1.2C

1.5B = 1.2C

B =  \frac{1.2}{1.5} C
 =  \frac{12}{15} C
 =  \frac{4}{5} C 
 = .8C
---
Stage III --Now the list is

A = .6C
B = .8C
C = 1.0C
D = 1.2C
E = 1.4C

And now there's an evenly spaced set/progression where median = mean, so C = 50.

Largest - smallest = range. E is 1.5 x 50 = 70. A is .6 x 50 = 30.

Range is 70 - 30 = 40. Answer D

papagorgio · Answer

This question took me about 2 minutes to figure out..

Intuition: Since more than one fraction was referring to Cook's score, I used Cook as the base case and set him/her equal to  100 ..

Step 1: 
Given an average of 50 and 5 participants, all the fractions should add to 250.

Step 2: 
Alastair + Bell + Cook + Darren + Eoin = 250

Therefore, based on the instructions above:  \frac{1}{2}\frac{6}{5}100+(\frac{6}{5}100)/\frac{3}{2})+100+\frac{6}{5}100+\frac{7}{5}100 =250

Therefore: 60+80+100+120+140=250  //Since it looks like 100 as a starting point for Cook was too high, I'll half all the figures equally

Therefore the figures become: 30+40+50+60+70=250 //This works!

Therefore the range is 70-30=40 (largest-smallest)

devarshi9283 · Answer

From the question;
A=1/2D
D=6/5C 
E=7/5C
B=2/3D

Looking at the average (=50) and answer choices (25,30,35,40,40)
it is clear that all the scores are multiples of 5..
Now D is also a multiple of 6...
So, D can be 6*5=30 or 6*5*2=60...as average is 50 and D is better than A,B and C...60 seems better 
Assume D=60...We get E=70, B=40, A=30, C=50....and voila...the avg is 50...
So, range is 70-30=40

We can also take E..which is a multiple of 7 and 5...so probably 35 or 70..as E is highest..it is greater than 50...So take 70
again we get D=60, B=40, A=30, C=50..

Testing the values comes with practice...and by scanning the information and answer choices..but it does wonders

elainetianfong · Answer

First, I just wrote down all of what I could extract from the question.
A = (1/2) D 
 D  = (1 + 1/5)C =  (6/5)C 
 E  = (1 + 2/5)C =  (7/5)C  =>  C  =  (5/7)E 
 D =  (3/2)B

=> A = (1/2) D  = (1/2)x (6/5)C = (1/2)x (3/2)B  = (1/2)x (6/5) x (5/7)E

=> A = (1/2) D  = (3/5) C  = (3/4) B  = (3/7) E

=> A : D : C : B : E = 1 : 2 : 5/3 : 4/3 : 7/3

OR we could say: A : D : C : B : E = 3 : 6 : 5 : 4 :7

From the above ratio, we can conclude that the range of the score would be the difference between A & E.
The range would be equal to (50x5:18)x(7-3) = 40.
Answer D.

Krishaa12 · Answer

Payal mam how should I represent "if question was asked...... Eoin scored 2/5 th of Cook" ..?please help

EgmatQuantExpert · Answer

Hey Krishaa12 , Let me put the representations of both the statements: the one given in the question and the new statement that you mentioned. Given that, " Eoin scored \frac{2}{5} th more than Cook. " Hence, we can write Eoin's score = Cook’s score + \frac{2}{5} * Cook’s score However, if the statement says " Eoin scored \frac{2}{5} th of what Cook scored " Then we could have written it as Eoin's score = \frac{2}{5} * Cook’s score Hope this answers your query.

Akshaynandurkar · Answer

Lets say Darren's score is x then A = x/2 , 
Darren's score is 1/5th MORE than Cook's score (not 1/5th of the Cook's score)
x = C+1/5C 
C = 5x/6 
E's score is 2/5th more than C score (not 2/5th of the C score)
E = 2/5(5x/6) + 5x/6
E = 7x/6
and B = 2x/3 
Summation of these five scores = 50*5 = 250
summation = 25x/6 = 250 gives x = 60
range = 7x/6 - x/2 = 2x/3 = 2*60/3 = 40
answer is 40

saynchalk · Answer

After rearranging information in the stem:

A = 1/2 D
B = 2/3 D
C = 5/6 D
E = 7/6 D

A+B+C+D+E = 250, solving for which gives D = 60. Range = 70-30 = 40.

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