A certain fruit stand sold apples for $0.70 each and bananas for $0.50 : Problem Solving (PS)

pzazz12

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

01 Oct 2010, 05:06

Bunuel wrote:

pzazz12 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. 15

Given: $0.7b+0.5a=6.3$ Question: $a+b=?$

$0.7a+0.5b=6.3$ --> $7a+5b=63$. After some trial and error you'll get that only two integer pairs of (a,b) satisfy this equation: (9,0) and (4,7) as we are told that "a customer purchased both apples and bananas" then the first pair is out and we'll have: $a=4$ and $b=7$ --> $a+b=11$.

Answer: B.

thank you, but can you explain me how this (9,0) and (4,7) to be solve...

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

03 Nov 2010, 13:08

Hi all

I can't understand MGMAT's OG guide answer for this one...

1st question: why can't you exchange apples with bananas 1 for 1 until you get the right answer?

2nd question: They're banging on about 5 apples having the same value as 7 bananas, then taking 5 apples away from the 'partial solution ~(i.e. when A = 9 and B = 0) and adding 7 bananas... Why would you do that? It makes no sense to me... If you set the vlaue of the apples = bananas you get them both equalling $3.50 in value which equals $7 between both of them... which is clearly over $63 ...

this is really doing my head in!

thanks

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

10 Dec 2010, 20:14

ezhilkumarank wrote:

pzazz12 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. 15

Without calculating anything in paper you could approach this problem.

Know -- Some multiple of 7 + Some multiple of 5 should yield 63. To get to a some multiple of 5, we should ensure that a 3 or 8 (5+3) should be a multiple of 7.

63 is a direct multiple of 7, however in this case there won't be any bananas. Hence the next option is to look for a multiple of 7 that has 8 as the unit digit. 28 satisfies this hence no. of apples is 4 and no of bananas is 7 -- Answer 11 (B). -- 35 seconds straight.

i get some multipule of 5 and 7 make 63... but why the multiple of 7 with a 3 or 8?

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

22 Aug 2014, 10:39

VeritasPrepKarishma wrote:

n2739178 wrote:

this is really doing my head in!

I have put up this theory on this link:
https://gmatquant.blogspot.com/2010/11/integral-solutions-of-ax-by-c.html

See if it makes sense now.
If there are doubts, get back to me on my blog itself or here...

Hi Karishma,

Interesting post -- makes complete sense. A question though: In your hypothetical question about "- And, a trickier thing to think about - how many integral solutions would 3x - 5y = 42 have?" -- both have to go up, right? So x would have to go up from 14, to 19, to 24 etc. Conversely, y would also go up from 0 to 3, to 6 etc. Neither of those values can be negative since we have the positive integer constraint. Am I correct?

Can you recommend other questions similar to this? Thanks!

L

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

24 Aug 2014, 22:50

Expert Reply

russ9 wrote:

VeritasPrepKarishma wrote:

n2739178 wrote:

this is really doing my head in!

I have put up this theory on this link:
https://gmatquant.blogspot.com/2010/11/integral-solutions-of-ax-by-c.html

See if it makes sense now.
If there are doubts, get back to me on my blog itself or here...

Hi Karishma,

Interesting post -- makes complete sense. A question though: In your hypothetical question about "- And, a trickier thing to think about - how many integral solutions would 3x - 5y = 42 have?" -- both have to go up, right? So x would have to go up from 14, to 19, to 24 etc. Conversely, y would also go up from 0 to 3, to 6 etc. Neither of those values can be negative since we have the positive integer constraint. Am I correct?

Can you recommend other questions similar to this? Thanks!

Yes, the first easy solution would be 14, 0. Both x and y will move in same direction. Since neither can be negative, they must move up only.

L

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

07 Oct 2017, 06:43

Expert Reply

Top Contributor

pzazz12 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. 15

Here's an approach where we test the POSSIBLE CASES.

FACT #1: (total cost of apples) + (total cost of bananas) = 630 CENTS
FACT #2: total cost of bananas is DIVISIBLE by 50, since each banana costs 50 cents.

Now let's start testing POSSIBLE scenarios.

Customer buys 1 apple.
1 apple costs 70 cents, which means the remaining 560 cents was spent on bananas.
Since 560 is NOT divisible by 50, this scenario is IMPOSSIBLE

Customer buys 2 apples.
2 apple costs 140 cents, which means the remaining 490 cents was spent on bananas.
Since 490 is NOT divisible by 50, this scenario is IMPOSSIBLE

Customer buys 3 apples.
3 apple costs 210 cents, which means the remaining 520 cents was spent on bananas.
Since 520 is NOT divisible by 50, this scenario is IMPOSSIBLE

Customer buys 4 apples.
4 apple costs 280 cents, which means the remaining 350 cents was spent on bananas.
Since 350 IS divisible by 50, this scenario is POSSIBLE
350 cents buys 7 bananas.
So, the customer buys 4 apples and 7 bananas for a total of [spoiler]11[/spoiler] pieces of fruit

Answer:

Show Spoiler

B

Cheers,
Brent

D

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

15 Jan 2019, 04:39

Expert Reply

Consider the following equation:
2x + 3y = 30.

If x and y are nonnegative integers, the following solutions are possible:
x=15, y=0
x=12, y=2
x=9, y=4
x=6, y=6
x=3, y=8
x=0, y=10

Notice the following:
The value of x changes in increments of 3 (the coefficient for y).
The value of y changes in increments of 2 (the coefficient for x).
This pattern will be exhibited by any fully reduced equation that has two variables constrained to nonnegative integers.

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

70x + 50y = 630
7x + 5y = 63

In accordance with the pattern illustrated above, we get the following nonnegative solutions for x and y:
x=9, y=0
x=4, y=7

Here -- since apples and bananas are both purchased -- x and y must both be positive.
Thus, only the option in green is viable, with the result that x+y = 4+7 = 11.

Show Spoiler

B

G

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

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

S

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

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

P

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

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

S

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

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.

B

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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

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

G

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

21 Jul 2019, 05:22

pzazz12 wrote:

Bunuel wrote:

pzazz12 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. 15

Given: $0.7b+0.5a=6.3$ Question: $a+b=?$

$0.7a+0.5b=6.3$ --> $7a+5b=63$. After some trial and error you'll get that only two integer pairs of (a,b) satisfy this equation: (9,0) and (4,7) as we are told that "a customer purchased both apples and bananas" then the first pair is out and we'll have: $a=4$ and $b=7$ --> $a+b=11$.

.

thank you, but can you explain me how this (9,0) and (4,7) to be solve...

Let me try a different way to solve this >
If an Apple is 70 cents, a Banana is 50 cents, and the total price is 6.3 dollars,
so 7A plus 5B equals 63;

63-7A=5B
=> 5B= 7(9-A)
If (9 - A) is a 5, that means that A = 4,

and the total number of apples and bananas is 11.

The correct answer chose is B.

Answer: B

L

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

21 Jul 2019, 08:26

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 have done so many DS questions in last few days that after reading the question I was like okay let's see if only (1) or only (2) is sufficient until I saw options and I am like meh I have to solve this one and there is surely an unique solution. Haha! Bunuel has provided the best solution.

B

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

20 Nov 2019, 10:36

VeritasKarishma wrote:

n2739178 wrote:

Hi all

I can't understand MGMAT's OG guide answer for this one...

1st question: why can't you exchange apples with bananas 1 for 1 until you get the right answer?

2nd question: They're banging on about 5 apples having the same value as 7 bananas, then taking 5 apples away from the 'partial solution ~(i.e. when A = 9 and B = 0) and adding 7 bananas... Why would you do that? It makes no sense to me... If you set the vlaue of the apples = bananas you get them both equalling $3.50 in value which equals $7 between both of them... which is clearly over $63 ...

this is really doing my head in!

thanks

I do not know what exactly your book says but I am guessing this is how they have solved it:

7a + 5b = 63
Such equations have infinite solutions. We can get a single solution under particular constraints. (Will explain this later)
One thing we notice right away is that one solution to this problem is a = 9 and b = 0 because 63 is divisible by 7.
7a + 5b = 63
a = 9, b = 0
a = 4, b = 7 (To get this solution, subtract 5, co-efficient of b, from a above and add 7, co-efficient of a, to b above)
a = -1, b = 14 (Again, do the same to the solution above)
a = 13, b = -7 (You will also get solutions when you add 5 to a of any other solution and subtract 7 from b of the same solution)
Hence there are infinite solutions.
Here the constraints are that a and b should not be negative. Also, they should not be 0 since he buys at least 1 apple and at least 1 banana. Only 1 solution satisfies these constraints so answer is a = 4 and b =7.
Why this works is because when you reduce a by 5, the reduction in 7a is offset by the increase in 5b when you increase b by 7. Let this suffice for now. This is the theory of Integral solutions to equations in two variables. I will explain you the complete theory soon.

Hi VeritasKarishma,
Is my approach is correct ?
given 0.7A+0.5B=6.30
If all apples purchased, than max 9 apples can be purchsed
on other hand if all banana purchased than max 12 banana purchased
as given both apples and banana purchased, hence number of bananas and apples purchsed will be 9< (A+B)<12 so only 11 satifies it.

B

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

13 May 2020, 19:39

We can also eliminate answer choices D & E since if B>13, and the price of bananas = 0.50, then this will exceed $6.30

L

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

23 Jun 2020, 03:39

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

A > 0 and B > 0 both are positive integers. Hence (A,B) = (9,0) is ruled out.
7A + 5B = 63
So, the unit digits of 7A and 5B should yield 3. As 5B would only result in 0 and 5 as unit place, 7A should result in either 3 or 8 as units digit.
7A results in units digit as 3 only when A = 9 which is ruled out. Hence 7A must result in 8 as units place.
This can be done as only when A = 4(under given conditions) i.e. 7*4 = 28.
Thus,
5B = 35 => B = 7
A + B = 7 + 4 = 11

Answer B.

gmatclubot

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

23 Jun 2020, 03:39

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