Four boys own a total of 150 baseball cards. If the first boy owns 28%

Question

Four boys own a total of 150 baseball cards. If the first boy owns 28% of the cards, the second owns 24% of the cards, and the third owns three times as many
cards as the fourth, what is the greatest number of cards owned by any one boy?

(A) 28
(B) 36
(C) 42
(D) 54
(E) 64

QuantMadeEasy · Answer

Let cards owned by third be x and cards owned by fourth = y 
x = 3y

First owned = 28% of the cards
Second owned = 24% of the cards
Total cards remaining = 48% of the cards
 Third owned = 75% of the remaining cards =75% of 48% of the cards = 36% of the cards, which is highest  
hence x = 0.75*0.48*150 = 54

IMO D

Aks111 · Answer

Let's calculate number of cards with 3rd and the 4th boy = 150 - (28%+24%) of 150 = 150 - (52% of 150) = 150 - 52*3/2 = 150-78 = 72.
Let 4th boy have ''x' cards. So, 3rd boy has '3x' cards. 
-> 3x+x = 72 -> 4th boy has 18 and 3rd boy has 54 cards.
1st and 2nd boy have 28% and 24% of 150. 
As, 30% of 150 is 45, the 1st and 2nd definitely have less number of cards than the 3rd boy has.
Answer: D - 54

Nipungupta9081 · Answer

IMO D = 54

a + b + c + d = 150
.28*150 + .24*150 + 3d + d = 150
78 + 4d = 150
4d = 72
d = 18

4*d = 4*18 = 54

CrackverbalGMAT · Answer

I think you meant to write 3d. 4d = 72

TarunKumar1234 · Answer

Given, 1st = 28%, 2nd = 24%, 3rd = 3p, 4th= p Now, 52+4p= 100, P=12%, so, 3rd = 36% (max) 150* 36% = 54. So, I think D.

MHIKER · Answer

The first boy  =150*\frac{28}{100}=42

The second boy  =150*\frac{24}{100}=36

Rest  =150-(42+36)=150-78=72

The Fourth boy  = x

The third boy  =3x

3x+x=72

x=18= Fourth boy

So, Third boy  =18*3=54

The highest  54 .

The answer is  D

Four boys own a total of 150 baseball cards. If the first boy owns 28%

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