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.

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.

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.

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....

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.

so in that case we can say 7 +4 = 11 is correct

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?

haha, thanks that's exactly what I meant

That's what i get for multi tasking at work

But the method made sense right?