A fruit seller sells only pears and persimmons. The ratio of pears to persimmons is 2 : 3. A total of 200 fruits are added. The number of pears becomes 3 times as many as before and the number of persimmons becomes 3 times as many as before. How many fruits are there in the end?
| |
Comparing pears
at first and
in the end |
Pears |
Persimmons |
Comparing persimmons
at first and
in the end |
| Before |
1x2 =
2 u |
2 u |
3 u |
1x3 =
3 u |
| Change |
|
+ 200 |
|
| After |
3x2 =
6 u |
6 u |
9 u |
3x3 =
9 u |
The number of persimmons at first is repeated. Make the number of persimmons 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 + 9 u
= 15 u
Number of fruits added
= 15 u - 5 u
= 10 u
10 u = 200
1 u = 200 ÷ 10 = 20
Number of fruits in the end
= 15 u
= 15 x 20
= 300
Answer(s): 300
