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A certain fruit stand sold apples for $0.70 each and bananas for $0.50

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Re: A certain fruit stand sold apples for $0.70 each and bananas for $0.50  [#permalink]

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New post 30 Apr 2019, 21:32
Bunuel 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



I don't know whether this method is right or wrong but I somehow got the right answer.

I took the factor or 63, which is the total amount for apples & bananas. The factors are 7, 3 and 1 that sums to 11.

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Re: A certain fruit stand sold apples for $0.70 each and bananas for $0.50  [#permalink]

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New post 07 Jul 2019, 07:36
Let the no of apples sold = a
no of bananas sold = b
Question is a+b=?
Thus 70a + 50b = 630
7a + 5b = 63
Quick Tip- In order to find out the value of a & b its better to find the value of 'a' as "63- 7a" must leave a number which will end either with 0 or
with 5. (Think about it for a second)
Thus the only value which satisfies above equation is a=4 & b=7
a+b=11 (Other values of a & b will lie outside the answer choices)
Answer B
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A certain fruit stand sold apples for $0.70 each and bananas for $0.50  [#permalink]

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New post 07 Jul 2019, 09:45
could some one put a link about solving problem with Diophantine ???????

I think best approach to solve these problem without ANY risk !!
put here a link about fastest method of Diophantus .....
thanks
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A certain fruit stand sold apples for $0.70 each and bananas for $0.50  [#permalink]

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New post 07 Jul 2019, 17:32
Bunuel 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


we know the customer bought at least one apple-banana pair for $1.20
say she bought 5 pairs for $6.00, leaving her 30¢ no
say she bought 4 pairs for $4.80, leaving her $1.50 yes
she can buy 4 pairs for $4.80 plus 3 additional bananas for $1.50 for $6.30 total
11
B
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A certain fruit stand sold apples for $0.70 each and bananas for $0.50  [#permalink]

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New post 08 Jul 2019, 00:03
First, simplify the equation 0.7A + 0.5B = 6.30 by multiplying 10 on each side of the equation and you get
7A + 5B = 63

Next, find A and B by summing the last digit through trials.
The trick is to begin testing with the multiples that end with less variations.

multiples of 7A ends with digit - x7,x4,x1,x8,x5,x2,x9,x6,x0
multiples of 5B ends with digit - x0,x5

In this case, starts with multiple of 5B, which ends with 2 variations as compared to multiples of 7A, which ends with 9 variations.

Case1: last digit of 5B is 0,
5B + 7A = 63
_0 + _? = _3 ; (? = 3)
The multiples of 7 that ends with 3 is 9 ; 7X9 = 63.
This means A+B =9 since 5(0)+ 7(9) = 63;

Case2: last digit of 5B is 5,
5B + 7A = 63
_5 + _? = _3 ; (? = 8)
The multiples of 7 that ends with 8 is 4; 7X4 = 28.
This means A+B =11 since 5(7)+ 7(4) = 63

Base on the answer choices, A+B is more than 9, therefore, A+B = 11. B is the answer.
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Re: A certain fruit stand sold apples for $0.70 each and bananas for $0.50  [#permalink]

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New post 12 Jul 2019, 05:48
A is the number of apples
B is the number of bananas
---->What is being asked: A + B =?????
0.7*A + 0.5*B = 6.3
-->7*A + 5*B = 63 = 7*9
--->5*B = 7*9 - 7*A = 7*(9 - A)
From here, we realise that B should be an integer that has to be a multiple of 5 & 7, option can be only B = 7 and A = 4
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Re: A certain fruit stand sold apples for $0.70 each and bananas for $0.50   [#permalink] 12 Jul 2019, 05:48

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