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555-605 Level|   Algebra|   Word Problems|                     
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A certain fruit stand sold apples for $0.70 each and bananas for $0.50 17 Jan 2021, 07:23
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Video solution from Quant Reasoning starts at 16:58
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1
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A certain fruit stand sold apples for $0.70 each and bananas for $0.50 16 Feb 2021, 06:42
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IanStewart When we get 7a + 5b = 63 is there any way we could estimate to narrow down our possibilities?
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A certain fruit stand sold apples for $0.70 each and bananas for $0.50 16 Feb 2021, 07:21
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IanStewart When we get 7a + 5b = 63 is there any way we could estimate to narrow down our possibilities?
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For these types of questions, when the prices are very close together, estimation is usually the best strategy. Here by estimating, we only easily eliminate two answer choices -- if the person bought only apples, they would buy 9 apples, and if they bought only bananas, they would buy 12.6, so the answer must be between 9 and 12.6, and D and E must be wrong. If you combine that estimate with a bit of number theory, you can get the answer easily though:

7a + 5b = 63
2a + 5a + 5b = 63
2a + 5(a + b) = 63

Notice now that 2a is even, and 63 is odd, so 5(a + b) must be odd, so a+b is odd. And a + b is one of the answer choices, and the only odd answer we haven't eliminated yet (after estimating) is 11.

I still prefer the divisibility approach I explained on the previous page; with practice, that kind of approach can be used to solve this question in 15-30 seconds.
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A certain fruit stand sold apples for $0.70 each and bananas for $0.50 05 Aug 2021, 20:19
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Here's a slightly different solution from the one I posted a few years ago:

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A certain fruit stand sold apples for $0.70 each and bananas for $0.50 05 Mar 2022, 06:14
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Why can't we work this in the form of simultaneous equation? Equation #1 being 7x+5y=63 and equation #2 being x+y=1.2

I tried to do this way but x value was is decimal which means the number of apples were in decimal. which is not possible. But I don't understand where am I going wrong in framing the second equation?

Thanks,
Vaishali
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A certain fruit stand sold apples for $0.70 each and bananas for $0.50 13 Sep 2022, 11:19
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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
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I solved this question in the easiest way possible. As the question mentions that customer purchased both apples and bananas, therefore for each pair of banana and apple the customer would spend 1.2$. Lets start with maximum number of possible pairs (which will be 5 as 1.2*6=7.2$ which is greater than 6.3$ if you note)

Now, if we assume 5 such pairs then 1.2*5=6.0$, hence remaining 0.3$ cannot be put under apple or banana as their cost is 0.5$ and 0.7$ (which is greater than 0.3$). This means we overshot the possible number of pair.

Now, if we assume 4 such pairs then 1.2*4=4.8$, hence remaining 1.5$ can be accounted in only one way i.e. 3 apples (0.5*3=1.5$).

Thus the total number of fruits = 4*2 (4 pair of apple and banana each) + 3 (apples) = 11

Hope this helps!
Please hit like if you felt this easier than solving equations and testing out divisibility!!
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A certain fruit stand sold apples for $0.70 each and bananas for $0.50 13 Sep 2022, 11:42
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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
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A certain fruit stand sold apples for $0.70 each and bananas for $0.50 13 Oct 2025, 03:42
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