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# How do I solve this?

This was a homework question and I need help with it:

A fruit salad consists of blueberries, raspberries, grapes, and cherries.The fruit salad has a total of 280 pieces of fruit. There are twice as many raspberries as blueberries, three times as many grapes as cherries, and four times as many cherries as raspberries. How many cherries are there in the fruit salad?

### 1 Answer by Expert Tutors

Sameer K. | Professional and Patient Tutor for Various TopicsProfessional and Patient Tutor for Vario...
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Hi Zara! So this can be made into a series of algebra equations and solved from there - it's probably the easiest way to do it. If you define B as the number of Blueberries, R as the number of Raspberries, G as the number of Grapes and C as the number of Cherries, you can make the following equations:

B + R + G + C = 280 ("The fruit salad has a total of 280 pieces of fruit.")
R = 2B ("There are twice as many raspberries as blueberries.")
G = 3C ("Three times as many grapes as cherries.")
C = 4R ("Four times as many cherries as raspberries.")

Now you can start substituting to get at the number of cherries there are, but first you need to do a little manipulation of these equations to get everything else in terms of the number of cherries. That third one, with grapes, is how you want it - solving for something else in terms of the number of cherries. So let's start with the last one, and divide both sides by 4.

C /4 = 4R /4
C /4 = 4R /4
C/4 = R    OR
R = C/4

Now let's work on the second one, with raspberries and blueberries. Since neither deal with cherries, you want to replace the number of either raspberries or blueberries with the number of cherries first. Since we just solved for R (above, R = C/4), we'll start with that replacement.

R = 2B
C/4 = 2B

Now we'll divide both sides by 2 to get B in terms of the number of cherries.

C/4 (/2) = 2B /2
C/8 = 2B/2
C/8 = B     OR
B = C/8

So now you've got every other letter in terms of C. We have R = C/4, B = C/8, and without changing that third equation, we have G = 3C. Now you go back to that first equation and put all three of these in, and you'll end up with just one variable (C).

B + R + G + C = 280
C/8 + C/4 + 3C + C = 280

Now you can simplify a bit by first combining all the Cs together.

C/8 + C/4 + 3C + C = 280
C/8 + (2)C/(2)4 + 3C + C = 280
C/8 + 2C/8 + 4C = 280
3C/8 + 4(8)C(/8) = 280 (Here it can be helpful to think of C as 8C/8 to help you combine the fraction into the whole number.)
3C/8 + 32C/8 = 280
35C/8 = 280

Now you just multiply both sides by 8 and then divide them by 35 to get rid of that fraction, and you get:

35C/8 = 280
35C/8 (*8) = 280 (*8)
35C/8 (*8) = 2240
35C = 2240
35C (/35) = 2240 (/35)
35C (/35) = 64
C = 64