AskQ logo

STUCK ON HOMEWORK?

ASK FOR HELP FROM OUR KIASUPARENTS.COM COMMUNITY!

Question

There were 160 red and green pens in a box. When 2/5 of the red pens and 2/3 of the green pens were removed,the total number of red and green pens became 72. How many red pens were there in the box?
How to do?
Tks

Answer

Red -> 5u

Green -> 3p

5u + 3p = 160

3u + 1p = 72

9u + 3p = 216

4u = 216 – 160

= 56

1u -> 56 ÷ 4 = 14

5u -> 14 × 5 = 70 (Answer)

 

0 Replies 2 Likes ✔Accepted Answer

If there were 15u of red pens, there were (160 – 15u) green pens.

2/5 of the red pens were removed,   ( 1- 2/5)15u = 9u red pens remained.

2/3 of the green pens were removed,  (1 – 2/3)(160 – 15u) = (160/3 – 5u) green pens remained.

9u + (160/3 – 5u) = 72

160/3  + 4u = 72            multiply both sides by 3

160 + 12u = 216

12u = 56

u = 14/3

15u =70

9u = 42

There were 70 red pens in the box initially and finally there were 42 pens in the box.

0 Replies 2 Likes