A store owner has 28 less pears than chikoos. After he sold 75% of the pears and 40% of the chikoos, he is left with 63 less pears than chikoos. What fraction of the fruits left were pears?
| |
Pears |
Chikoos |
Comparing pears
and chikoos at first |
|
28 less |
| Before |
4x5 =
20 u |
+ 28 |
5x4 =
20 u |
| Change |
- 3x5 =
- 15 u |
- 21 |
- 2x4 =
- 8 u |
| After |
1x5 =
5 u |
+ 7 |
3x4 =
12 u |
Comparing pears
and chikoos in the end |
63 less |
|
75% =
75100 =
34 75% x 28 =
34 x 28 = 21
28 - 21 = 7
40% =
40100 =
25 The number of pears is 63 less than the number of chikoos in the end. So, if another 63 pears are added, the number of pears and chikoos in the end will be the same.
5 u + 7 + 63 = 12 u
5 u + 70 = 12 u
12 u - 5 u = 70
7 u = 70
1 u = 70 ÷ 7 = 10
Number of pears left
= 5 u + 7
= 5 x 10 + 7
= 50 + 7
= 57
Total number of fruits left
= 5 u + 7 + 12 u
= 17 u + 7
= 17 x 10 + 7
= 170 + 7
= 177
Pears leftTotal fruits left =
57177 =
1959 Answer(s):
1959 
