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03 Apr 2020, 05:46
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C
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Difficulty:
85%
(hard)
Question Stats:
58% (03:13) correct
42%
(03:19)
wrong
based on 384
sessions
History
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A trader sells 10 litres of a mixture of paints A and B, where the amount of B in the mixture does not exceed that of A. The cost of paint A per litre is $8 more than that of paint B. If the trader sells the entire mixture for $264 and makes a profit of 10%, then what is the highest possible cost of paint B, in $ per litre?
A. 16
B. 18
C. 20
D. 22
E. 26
Are You Up For the Challenge: 700 Level Questions
A. 16
B. 18
C. 20
D. 22
E. 26
Are You Up For the Challenge: 700 Level Questions
03 Apr 2020, 09:34
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Solution:
Sold for $264 with 10 % profit.
Thus the original cost is \(264 * \frac{10}{11} = $ 240\).
The problem tells us "Where the amount of B in the mixture does not exceed that of A".
Thus 10 liters must be 5 from A and 5 from B.
Additionally, we know that \(Cost(a) = Cost(B)+8\) and in order to maximize the value of b we should minimize A.
So b<=a
Finally we have all the info, setup the equation with x:
\((x+8)*5 + x*5 = 240\)
\(5x +40 + 5x = 240\)
\(10x = 200\)
\(x= \frac{200}{10} =20 \)
Answer C.
Kudos if you appreciate my explanation.
Sold for $264 with 10 % profit.
Thus the original cost is \(264 * \frac{10}{11} = $ 240\).
The problem tells us "Where the amount of B in the mixture does not exceed that of A".
Thus 10 liters must be 5 from A and 5 from B.
Additionally, we know that \(Cost(a) = Cost(B)+8\) and in order to maximize the value of b we should minimize A.
So b<=a
Finally we have all the info, setup the equation with x:
\((x+8)*5 + x*5 = 240\)
\(5x +40 + 5x = 240\)
\(10x = 200\)
\(x= \frac{200}{10} =20 \)
Answer C.
Kudos if you appreciate my explanation.
General Discussion
02 May 2020, 08:33
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Bunuel
10=a+b, b=10-a, (b≤a);
10%=sale-cost/cost,
sale=264, cost=a(x+8)+(10-a)x;
1/10=264-c/c, 11c=2640, c=240;
a(x+8)+(10-a)x=240, (to max b we must min a)
a=5: 5(x+8)+5x=240, 10x=240-40, x=20
Ans (C)
