A store owner has 88 more chikoos than pears. After he sold 75% of the chikoos and 20% of the pears, he is left with 22 more pears than chikoos. What fraction of the fruits left were pears?
| |
Chikoos |
Pears |
Comparing chikoos
and pears at first |
88 more |
|
| Before |
4x5 =
20 u |
+ 88 |
5x4 =
20 u |
| Change |
- 3x5 =
- 15 u |
- 66 |
- 1x4 =
- 4 u |
| After |
1x5 =
5 u |
+ 22 |
4x4 =
16 u |
Comparing chikoos
and pears in the end |
|
22 more |
75% =
75100 =
34 75% x 88 =
34 x 88 = 66
88 - 66 = 22
20% =
20100 =
15 The number of pears is 22 more than the number of chikoos in the end. So, if another 22 chikoos are added, the number of chikoos and pears in the end will be the same.
5 u + 22 + 22 = 16 u
5 u + 44 = 16 u
16 u - 5 u = 44
11 u = 44
1 u = 44 ÷ 11 = 4
Number of pears left
= 16 u
= 16 x 4
= 64
Total number of fruits left
= 5 u + 22 + 16 u
= 21 u + 22
= 21 x 4 + 22
= 84 + 22
= 106
Pears leftTotal fruits left =
64106 =
3253 Answer(s):
3253 
