A fruit seller sells only pears and peaches. The ratio of pears to peaches is 2 : 3. A total of 70 fruits are added. The number of pears becomes 3 times as many as before and the number of peaches becomes 2 times as many as before. How many fruits are there at first?
| |
Comparing pears
at first and
in the end |
Pears |
Peaches |
Comparing peaches
at first and
in the end |
| Before |
1x2 =
2 u |
2 u |
3 u |
1x3 =
3 u |
| Change |
|
+ 70 |
|
| After |
3x2 =
6 u |
6 u |
6 u |
2x3 =
6 u |
The number of peaches at first is repeated. Make the number of peaches at first the same. LCM of 1 and 3 is 3.
The number of pears at first is repeated. Make the number of pears at first the same. LCM of 1 and 2 is 2.
Number of fruits at first
= 2 u + 3 u
= 5 u
Number of fruits in the end
= 6 u + 6 u
= 12 u
Number of fruits added
= 12 u - 5 u
= 7 u
7 u = 70
1 u = 70 ÷ 7 = 10
Number of fruits at first
= 5 u
= 5 x 10
= 50
Answer(s): 50
