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KEYE food drive

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by sanju09 » Thu Sep 09, 2010 4:58 am
KEYE food drive encourages grocery shoppers to donate to their local food bank by purchasing charity coupons worth $3 each or $7 each. Stacy, a shopper, purchased m charity coupons worth $3 each and n charity coupons worth $7 each towards the food drive and paid a total of $144 for the coupons. Which of the following cannot be m + n?
(A) 40
(B) 36
(C) 32
(D) 28
(E) 20

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by Gurpinder » Thu Sep 09, 2010 5:50 am
sanju09 wrote:KEYE food drive encourages grocery shoppers to donate to their local food bank by purchasing charity coupons worth $3 each or $7 each. Stacy, a shopper, purchased m charity coupons worth $3 each and n charity coupons worth $7 each towards the food drive and paid a total of $144 for the coupons. Which of the following cannot be m + n?
(A) 40
(B) 36
(C) 32
(D) 28
(E) 20

Source: [email protected]
(E) - 20
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by sanju09 » Thu Sep 09, 2010 5:54 am
E is not -20, your explanation Gurpinder?
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by Gurpinder » Thu Sep 09, 2010 6:02 am
sanju09 wrote:E is not -20, your explanation Gurpinder?
Oh, no no, I meant (E).

m --> $3 each
n --> $7 each


just to play around with the numbers, I tried to go for the lowest value of M+N. I can get the lowest value when I have fewer cupons. To have fewer coupons, I should have most of then the $7 ones and only the rest $3.

140/7=20

left with 4...... and 1 of the $3..... so I am still left with a $1 remaining.

This = 21 coupons with $1 remaining. If 21 doesent work out, there is not way you can get 20 (m+n).

That means that the total cannot be 20, so (E).

Does that make sense Sanju?
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by sanju09 » Thu Sep 09, 2010 6:16 am
Gurpinder wrote:
sanju09 wrote:E is not -20, your explanation Gurpinder?
Oh, no no, I meant (E).

m --> $3 each
n --> $7 each


just to play around with the numbers, I tried to go for the lowest value of M+N. I can get the lowest value when I have fewer cupons. To have fewer coupons, I should have most of then the $7 ones and only the rest $3.

140/7=20

left with 4...... and 1 of the $3..... so I am still left with a $1 remaining.

This = 21 coupons with $1 remaining. If 21 doesent work out, there is not way you can get 20 (m+n).

That means that the total cannot be 20, so (E).

Does that make sense Sanju?
Have you ever tried to take a coin out from the right cross-pocket of the pant you're wearing, using your left hand, or vise versa? You did exactly that here, and did it to perfection as well. I would still love to see more approaches, mate.
The mind is everything. What you think you become. -Lord Buddha



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by Gurpinder » Thu Sep 09, 2010 6:21 am
sanju09 wrote:
Gurpinder wrote:
sanju09 wrote:E is not -20, your explanation Gurpinder?
Oh, no no, I meant (E).

m --> $3 each
n --> $7 each


just to play around with the numbers, I tried to go for the lowest value of M+N. I can get the lowest value when I have fewer cupons. To have fewer coupons, I should have most of then the $7 ones and only the rest $3.

140/7=20

left with 4...... and 1 of the $3..... so I am still left with a $1 remaining.

This = 21 coupons with $1 remaining. If 21 doesent work out, there is not way you can get 20 (m+n).

That means that the total cannot be 20, so (E).

Does that make sense Sanju?
Have you ever tried to take a coin out from the right cross-pocket of the pant you're wearing, using your left hand, or vise versa? You did exactly that here, and did it to perfection as well. I would still love to see more approaches, mate.

yayy....... :D hahah
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by kmittal82 » Thu Sep 09, 2010 6:39 am
Very interesting question..

A closer look will reveal that the first set of numbers which will satisfy 144 is (3x41) + (7x3), next one is (3x34) + (7x6) and the one after is (3x27) + (7x9)... see the pattern? Multiple of 3 always decreases by 7, and multiple of 7 always increases by 3.

Using this information, 20 is not possible, hence (E)
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