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mel-nik [20]
3 years ago
8

b. Sixty-five pounds of candy was divided into four different boxes. The second box contained twice the amount of the first box.

The third box contained two more pounds than the first box. The last box contained one-fourth the amount in the second box. How much candy was in each box? 

Mathematics
1 answer:
Ainat [17]3 years ago
8 0

First box = 14 pounds

Second box = 2x = 28 pounds

Third Box = x+2= 16 pounds

Fourth box = x/2 = 7 pounds

<u>Step-by-step explanation:</u>

Here we have , Sixty-five pounds of candy was divided into four different boxes. The second box contained twice the amount of the first box. The third box contained two more pounds than the first box. The last box contained one-fourth the amount in the second box. We need to find How much candy was in each box. Let's find out:

We have a total of 65 pounds of candy ! Let in first box we have x pounds so , second box contained twice the amount of the first box i.e.

⇒ 2x

The third box contained two more pounds than the first box i.e.

⇒ x+2

The last box contained one-fourth the amount in the second box i.e.

⇒ (\frac{1}{4})2x = \frac{x}{4}

Therefore , Sum of pounds of candy are :

⇒ \frac{x}{2} +x+2+2x+x=65

⇒ \frac{x}{2} +4x=63

⇒ \frac{9x}{2}=63

⇒ x=63(\frac{2}{9} )

⇒ x=14

Therefore , Candy in each box is :

First box = 14 pounds

Second box = 2x = 28 pounds

Third Box = x+2= 16 pounds

Fourth box = x/2 = 7 pounds

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