Edburg bought some apples, mangoes, and bananas. He bought 42 fruits in all. The number of bananas he bought is less than half the number of apples he bought; the number of mangoes he bought is more than one-third the number of apples he bought but less than three-fourths the number of bananas he bought. How many bananas did he buy?
(A) 5
(B) 6
(C) 8
(D) 11
(E) 12
42 fruits in all
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Hi, here is my approach
A + M + B= 42----(1)
Also M < A/2 -----(2)
M> A/3-------------(3)
M< (3/4) B--------(4)
Based on the last equation , M < 3/4 B, I took B has to be a multiple of 4. Of the choices provided only C and E are possible.
Testing with B =8, M < 6. So it can be 1,2,3..5.
if M=1, A = 42 - (1+8)= 33. Not valid since eq 3 is not satisfied.
if M=5, A = 42 -( 5+8) =29. Eq 3 not satisfied.
So testing with B=12, M <9. M can be 1,2,...8
if M=1, B=12, A= 42 - (1+12) =29. Eq 3 not satisfied
if M=8, B=12, A= 42 - (8+12) = 22. All equations are satisfied.
Actually if we do not assume the whole number of bananas to start with 11 also works.
Both 11 and 12 works :
A + M + B= 42----(1)
Also M < A/2 -----(2)
M> A/3-------------(3)
M< (3/4) B--------(4)
Based on the last equation , M < 3/4 B, I took B has to be a multiple of 4. Of the choices provided only C and E are possible.
Testing with B =8, M < 6. So it can be 1,2,3..5.
if M=1, A = 42 - (1+8)= 33. Not valid since eq 3 is not satisfied.
if M=5, A = 42 -( 5+8) =29. Eq 3 not satisfied.
So testing with B=12, M <9. M can be 1,2,...8
if M=1, B=12, A= 42 - (1+12) =29. Eq 3 not satisfied
if M=8, B=12, A= 42 - (8+12) = 22. All equations are satisfied.
Actually if we do not assume the whole number of bananas to start with 11 also works.
Both 11 and 12 works :
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A + M + B = 42sanju09 wrote:Edburg bought some apples, mangoes, and bananas. He bought 42 fruits in all. The number of bananas he bought is less than half the number of apples he bought; the number of mangoes he bought is more than one-third the number of apples he bought but less than three-fourths the number of bananas he bought. How many bananas did he buy?
(A) 5
(B) 6
(C) 8
(D) 11
(E) 12
Because we are dealing with objects, all values must be positive integers.
Translating the second sentence into a series of inequalities, we have:
B < A/2, which, because we are dealing only with positive integers, can be safely rewritten as: A > 2B;
M > A/3, which can be rewritten as: 3M > A; and
M < 3B/4
Thus:
3M > A > 2B
we can see that 3M > 2B or M > 2B/3...and we also know that M < 3B/4
(ie, 2B/3 < M < 3B/4)
Now, we have to test the answer choices against the last two inequalities that I wrote (ie, the ones I've emboldened). When backsolving, in general, we should start from B or D. (although for this particular question, it wouldn't really matter).
Looking at B: if the number of Bananas is 6, then M > (2*6)/3 and M < (3*6)/4,
or 4 < M < 4.5, which makes the number of mangoes a non-integer, and therefore this choice is incorrect.
Looking at D: If B = 11, then M > (2*11)/3 and M < (3*11)/4 or 7.3333 < M < 8.25. Then, the number of mangoes is 8. If there are 8 mangoes, 11 bananas, and 42 fruit in total, then there are 23 apples. These numbers satisfy all of the inequalities. As there can only be one answer choice that satisfies all the conditions (ie, the correct answer), the correct answer is choice D.
Last edited by Testluv on Thu Jan 14, 2010 12:17 am, edited 1 time in total.
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Based on the last equation , M < 3/4 B, I took B has to be a multiple of 4. Of the choices provided only C and E are possible.
But it's not an equation. It's an inequality. And you can't apply this thinking when dealing with an inequality. That is, the number of mangoes can be less than three-fourths the number of bananas without the number of bananas being a multiple of 4. For example, if we had 10 bananas, and 2 mangoes, then, because three-fourths of 10 is 7.5, the inequality is satisfied even though the number of bananas (ie, 10) is not a multiple of 4.
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