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Apples and Bananas

by mstone » Mon Feb 22, 2010 8:43 am
A certain fruit stand sold apples for .70 each and bananas for .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

the only equation i can set up for this is:

.70 a + .50 b = 6.30

how can i solve this without a second equation??? probably seems simple, but i'm confused.
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by KICKGMATASS123 » Mon Feb 22, 2010 9:08 am
mstone wrote:A certain fruit stand sold apples for .70 each and bananas for .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

the only equation i can set up for this is:

.70 a + .50 b = 6.30

how can i solve this without a second equation??? probably seems simple, but i'm confused.
We dont have the second equation. you're right about that.. however, you should subsitute the numbers and look for which of the 2nd equation would be possible. for example.. along with the 1st equation, you'd have A + B = 13 for D. This is the only possible answer. IMO. Solve it.
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by thephoenix » Mon Feb 22, 2010 10:09 am
mstone wrote:A certain fruit stand sold apples for .70 each and bananas for .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

the only equation i can set up for this is:

.70 a + .50 b = 6.30

how can i solve this without a second equation??? probably seems simple, but i'm confused.
here the q has given an eqn and has asked which out of the 5 can be a proper second eqn

here proper eqn is the one which will give an int value for a and b

best way is to eliminate

let us start with C a+b=12.........eqn1
0.7a+0.5b=6.30...............eqn2

solving b=2.1/0.2...not an int value not possible

for b)
a+b=11
solving b=1.4/0.2=7---->a=4
correct

for d)a+b=13

solving b=14 and a=-1....not possible

like wise imo B
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by harshavardhanc » Mon Feb 22, 2010 10:33 am
mstone wrote:A certain fruit stand sold apples for .70 each and bananas for .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

the only equation i can set up for this is:

.70 a + .50 b = 6.30

how can i solve this without a second equation??? probably seems simple, but i'm confused.
when you can only make one equation from the given information, remember, there will only be a single set of value which will satisfy that equation and the answer options.

Now, coming to our question, look at it in this way :

sum of a multiple of .7 and .5 whose units digit is .3 ?

multiples of .7 end with . 7, .4, .1, .8, .5, .2, .9, .6, .3, and 0

multiples of .5 will end with .5 or 0.

so you have to try left with X.5 + Y.8 or X.0 + Y.3

1)X.0 + Y.3
=0.0 + 6.3
cannot be the case as the customer purchased both fruits.

2)X.5 + Y.8
=4 * 7 + 7 * .5
=2.8+3.5
=6.3

:)

bingo!

there total fruits purchased = 7 + 4 = 11.
Regards,
Harsha
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by junis1 » Mon Feb 22, 2010 10:36 am
I tried to do it the following way, but the numbers aren't working out... Don't know where I'm going wrong, any help would be appreciated...

0.7A + 0.5B = 6.3

Also:
A = 1.4B ( Is this a valid deduction? )

So,

0.7(1.4B) + 0.5B = 6.3
0.98B + 0.5B = 6.3
1.48B = 6.3 => Which means that B has to be 4.

So,

0.7A + 0.5(4) = 6.3
0.7A = 4.3 => Which means A has to be 6.

But when substituting in A and B in the original equation I get 6.2 not 6.3:

0.7(6) + 0.5(4) = 6.2 (NOT 6.3)

Where did I go wrong???
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by shashank.ism » Mon Feb 22, 2010 11:12 am
mstone wrote:A certain fruit stand sold apples for .70 each and bananas for .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

the only equation i can set up for this is:

.70 a + .50 b = 6.30

how can i solve this without a second equation??? probably seems simple, but i'm confused.
Let a denotes the apple and b denotes the bananas\
so 0.7a +0.5 b = 6.3

also total no. = a+b
so this is not sufficient to solve the problem...is anything missing here....
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by harshavardhanc » Mon Feb 22, 2010 11:19 am
shashank.ism wrote: Let a denotes the apple and b denotes the bananas\
so 0.7a +0.5 b = 6.3

also total no. = a+b
so this is not sufficient to solve the problem...is anything missing here....
see my response...
Regards,
Harsha
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by shashank.ism » Mon Feb 22, 2010 11:21 am
harshavardhanc wrote:
when you can only make one equation from the given information, remember, there will only be a single set of value which will satisfy that equation and the answer options.

Now, coming to our question, look at it in this way :

sum of a multiple of .7 and .5 whose units digit is .3 ?

multiples of .7 end with . 7, .4, .1, .8, .5, .2, .9, .6, .3, and 0

multiples of .5 will end with .5 or 0.

so you have to try left with X.5 + Y.8 or X.0 + Y.3

1)X.0 + Y.3
=0.0 + 6.3
cannot be the case as the customer purchased both fruits.

2)X.5 + Y.8
=4 * 7 + 7 * .5
=2.8+3.5
=6.3

:)

bingo!

there total fruits purchased = 7 + 4 = 11.
there are other options possible
like .7 x 9 + .5 x0 = 6.3
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by harshavardhanc » Mon Feb 22, 2010 11:25 am
shashank.ism wrote:
harshavardhanc wrote:
when you can only make one equation from the given information, remember, there will only be a single set of value which will satisfy that equation and the answer options.

Now, coming to our question, look at it in this way :

sum of a multiple of .7 and .5 whose units digit is .3 ?

multiples of .7 end with . 7, .4, .1, .8, .5, .2, .9, .6, .3, and 0

multiples of .5 will end with .5 or 0.

so you have to try left with X.5 + Y.8 or X.0 + Y.3

1)X.0 + Y.3
=0.0 + 6.3
cannot be the case as the customer purchased both fruits.

2)X.5 + Y.8
=4 * 7 + 7 * .5
=2.8+3.5
=6.3

:)

bingo!

there total fruits purchased = 7 + 4 = 11.
there are other options possible
like .7 x 9 + .5 x0 = 6.3
nopes! not possible!
in other cases, you will have to select one variable as 0, meaning no quantity purchased, which will violate the condition.
Regards,
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by shashank.ism » Mon Feb 22, 2010 11:38 am
so in that case we can say 7 +4 = 11 is correct
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by harshavardhanc » Mon Feb 22, 2010 11:46 am
shashank.ism wrote:so in that case we can say 7 +4 = 11 is correct
precisely! :)
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by jeffedwards » Mon Feb 22, 2010 12:12 pm
Here's how I did it.

Saw that the total was $6.30 . To make it simple I made that number 63 and my two other number 7 and 5. Since I know that if I add 5, I'll either have a number ending in 5 or 0, so I need to figure out what combo will end in a 3. I know that we can add 7 until we get a number that ends in 3 or add 8 to 5 to get 3. I hope you're following

So then I just figured out the multiples of 7 (7, 14, 21, 28, 35, 52, 49, 56, and 63)

Great so I saw that 63 is 7*9; however 9 is not an option, because the person purchased both apples and bananas. So I looked for an 8. Found only one option with eight at 28.

Solving gave me the answer of 4 apples and 7 bananas, or in other words 13 pieces of fruit. Does that make sense?
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by thephoenix » Mon Feb 22, 2010 12:21 pm
jeffedwards wrote:Here's how I did it.

Saw that the total was $6.30 . To make it simple I made that number 63 and my two other number 7 and 5. Since I know that if I add 5, I'll either have a number ending in 5 or 0, so I need to figure out what combo will end in a 3. I know that we can add 7 until we get a number that ends in 3 or add 8 to 5 to get 3. I hope you're following

So then I just figured out the multiples of 7 (7, 14, 21, 28, 35, 52, 49, 56, and 63)

Great so I saw that 63 is 7*9; however 9 is not an option, because the person purchased both apples and bananas. So I looked for an 8. Found only one option with eight at 28.

Solving gave me the answer of 4 apples and 7 bananas, or in other words 13 pieces of fruit. Does that make sense?
u mean 7+4=11
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by jeffedwards » Mon Feb 22, 2010 12:38 pm
haha, thanks that's exactly what I meant :)

That's what i get for multi tasking at work :)

But the method made sense right?
thephoenix wrote:
jeffedwards wrote:Here's how I did it.

Saw that the total was $6.30 . To make it simple I made that number 63 and my two other number 7 and 5. Since I know that if I add 5, I'll either have a number ending in 5 or 0, so I need to figure out what combo will end in a 3. I know that we can add 7 until we get a number that ends in 3 or add 8 to 5 to get 3. I hope you're following

So then I just figured out the multiples of 7 (7, 14, 21, 28, 35, 52, 49, 56, and 63)

Great so I saw that 63 is 7*9; however 9 is not an option, because the person purchased both apples and bananas. So I looked for an 8. Found only one option with eight at 28.

Solving gave me the answer of 4 apples and 7 bananas, or in other words 13 pieces of fruit. Does that make sense?
u mean 7+4=11
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by harshavardhanc » Mon Feb 22, 2010 1:32 pm
jeffedwards wrote:haha, thanks that's exactly what I meant :)

That's what i get for multi tasking at work :)

But the method made sense right?
thephoenix wrote:
jeffedwards wrote:Here's how I did it.

Saw that the total was $6.30 . To make it simple I made that number 63 and my two other number 7 and 5. Since I know that if I add 5, I'll either have a number ending in 5 or 0, so I need to figure out what combo will end in a 3. I know that we can add 7 until we get a number that ends in 3 or add 8 to 5 to get 3. I hope you're following

So then I just figured out the multiples of 7 (7, 14, 21, 28, 35, 52, 49, 56, and 63)

Great so I saw that 63 is 7*9; however 9 is not an option, because the person purchased both apples and bananas. So I looked for an 8. Found only one option with eight at 28.

Solving gave me the answer of 4 apples and 7 bananas, or in other words 13 pieces of fruit. Does that make sense?
u mean 7+4=11
yes, makes sense!

'coz that's my method as well ! :) ;) :D
Regards,
Harsha
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