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# PS - Certain Fruit stand - OG 12

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Math Forum Moderator
Joined: 20 Dec 2010
Posts: 2058
Followers: 122

Kudos [?]: 815 [0], given: 376

Re: PS - Certain Fruit stand - OG 12 [#permalink]  25 May 2011, 01:55
mrsmarthi wrote:
A certain fruit stand sold apples for \$0.70 each and bananas for \$0.50 each. If a customer purchased both apples and bananas from the stand for a total of \$6.30, what total number of apples and bananas did the customer purchase?

a) 10
b) 11
c) 12
d) 13
e) 14.

63:

Just recall table of 7 and stop where units digit is either 3 or 8.

7,14,21,28(We got it). Four 7's. Rest 5's.

7*9=63 but buyer purchased both. Ignore this.
_________________

~fluke

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Joined: 08 Jun 2011
Posts: 98
Followers: 1

Kudos [?]: 14 [0], given: 65

Re: PS - Certain Fruit stand - OG 12 [#permalink]  03 Dec 2011, 10:07
Here's a quick way to do this one.

7a + 5b=63

You basically subtract 7 from 63 which leads us to

7 + 56 = 63. Is 56 a multiple of 5? No. Then move on to
2 (7 ) + 49 = 63. Is 49 a multiple of 5? No. Then move on to
3 (7 ) + 42 = 63. Is 42 a multiple of 5? No. Then move on to
4 (7 ) + 35 = 63. Is 35 a multiple of 5? Yes. Save this answer
5 (7) + 28 = 63. Is 28 is a multiple of 5? No. Then move on to
6 (7) + 21 = 63. Is 21 a multiple of 5? No. Then move on to
7 (7) + 14 = 63. Is 14 a multiple of 5? No. Then move on to
8 (7) + 7 = 63. Is 7 a multiple of 5? No. Then move on to
9 (7) + 0. 63. Does not work because he/she bought both Apples and Bananas.

So the answer is 4 + 7 = 11.
The reason you start with 7 in your assessment is that it's the bigger number and 63 is a multiple of 7. This is my opinion, I hope it helps.
Re: PS - Certain Fruit stand - OG 12   [#permalink] 03 Dec 2011, 10:07
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