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by pzazz12 » Wed Sep 29, 2010 11:03 pm
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


Help me to solve it...............
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by Rahul@gurome » Wed Sep 29, 2010 11:27 pm
Let the customer purchase 'a' apples and 'b' bananas.
So 0.70a + 0.50b = 6.3.
Or 7a + 5b = 63.
It has to be noted that 'a' and 'b' are positive integers.
7a= 63-5b = k (say).
Since 5b is positive 63 - 5b < 63.
Or k < 63.
So k is a positive multiple of 7 < 63.
So k can be 56, 49, 42, 35, 28, 21, 14, 7.
Also since 63 - k = 5b.
So 63-k is a multiple of 5.
Only k = 28 satisfies this criteria since 63 - 28 = 35 is a multiple of 5.
So 7a = 28 or a = 4.
So b is (63 - 28)/5 = 7.
Or a + b = 11.

The correct answer is B.
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by ankur.agrawal » Thu Sep 30, 2010 2:37 am
This statement converts into:

.70*x+ .50*y= 6.30--------(1)

Now we have a single equation & two variables.But we need to know x+y.

7x+5y=63 now either we can try hit & trial method. or better we follow the method given by rahul.

Rahul in your explaination below it might take sumtime to figure out that Only k = 28 satisfies this criteria since 63 - 28 = 35 is a multiple of 5. Is it you picked up nos vaguely or you had sum strategy in your mind.
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by GMATGuruNY » Thu Sep 30, 2010 2:49 am
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


Help me to solve it...............
How can we spend 630 cents if we can use only multiples of 70 and multiples of 50?

9*70 = 630, but all the answers are bigger than 9.

80+50 = 130, so let's try a multiple of 70 whose last 2 digits are 80 and a multiple of 50 whose last 2 digits are 50.

4*70 = 280.
630-280 = 350.
350/50 = 7.
So we could buy 4 apples for 70 cents each and 7 bananas for 50 cents each.
4+7 = 11.

The correct answer is B.
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by ankur.agrawal » Thu Sep 30, 2010 3:01 am
GMATGuruNY 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


Help me to solve it...............
How can we spend 630 cents if we can use only multiples of 70 and multiples of 50?

9*70 = 630, but all the answers are bigger than 9.

80+50 = 130, so let's try a multiple of 70 whose last 2 digits are 80 and a multiple of 50 whose last 2 digits are 50.

4*70 = 280.
630-280 = 350.
350/50 = 7.
So we could buy 4 apples for 70 cents each and 7 bananas for 50 cents each.
4+7 = 11.

The correct answer is B.
I think this hit & trial method comes more naturally to us. Thanx again guru.
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by sanabk » Thu Sep 30, 2010 10:17 am
70a+50b=630
-> 7a+5b=63
-> a=(63-5b)/7
-> a=9-(5/7)b

Here a and b should be integers.
if b=7, then a= 4 and a+b=11
if b=14, then a=-1 (Impossible scenario)

So, a+b=11 is the correct answer.
B
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by Ian Stewart » Thu Sep 30, 2010 11:33 am
I discussed this question before, but we have:

0.7a + 0.5b = 6.3

7a + 5b = 63

Now, we're adding a multiple of 7 and 5b, and we're getting a multiple of 7. That means 5b must also be a multiple of 7 (if this isn't clear, you can see this by factoring: 5b = 63 - 7a = 7(9 - a), so 5b is equal to a multiple of 7).

Now, if 5b is divisible by 7, then b must be divisible by 7. b cannot be 14, since then the bananas alone cost more than $6.30. So b = 7, and we can then find that a = 4.
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by pzazz12 » Fri Oct 01, 2010 4:24 am
sanabk wrote:70a+50b=630
-> 7a+5b=63
-> a=(63-5b)/7
-> a=9-(5/7)b

Here a and b should be integers.
if b=7, then a= 4 and a+b=11
if b=14, then a=-1 (Impossible scenario)

So, a+b=11 is the correct answer.
B
thank u....but from this step..
if b=7, then a= 4 and a+b=11
can you explain me clearly........
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by sanabk » Fri Oct 01, 2010 8:23 am
Here the catch is "a" and "b" are integers.

a=9-(5/7)b
integer=9-(integer)

"a" should be an integer and now think for what integer values of "b" the term "(5/7)b" will be an integer. "b" has to be multiples of 7.

Substitute values for b to check:
b=1, (5/7)1=5/7 (not an integer)
b=2, (5/7)2=10/7 (not an integer)
b=3, (5/7)3=15/7 (not an integer)
.
.
b=7, (5/7)7=5 (an integer)

if b=7, then a= 9-(5/7)7=9-5=4 and a+b=7+4=11
if b=14, then a=-1 (Impossible scenario)

Hope this work helps!!!
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