There were some kiwis and pears in Box S and Box T. In Box S, the ratio of the kiwis to the number of pears was 7 : 1. In Box T, the ratio of the number of kiwis to the number of pears was 5 : 3. There were 3 times as many fruits in Box S as in Box T. After another 170 pears were put into Box T, the ratio of the number of kiwis to the number of pears in Box T became 1 : 4. How many fruits were there in Box T in the end?
Box S |
Box T |
3x8 = 24 u |
1x8 = 8 u |
Kiwis |
Pears |
Kiwis |
Pears |
7x3 |
1x3 |
5 |
3 |
21 u |
3 u |
5 u |
3 u |
The total number of fruits in Box S is repeated. Make the total number of fruits in Box S the same. LCM of 3 and 8 is 24.
The total number of fruits in Box T at first is repeated. Make the total number of fruits in Box T the same. LCM of 1 and 8 is 8.
|
Box S |
Box T |
|
Kiwis |
Pears |
Kiwis |
Pears |
Before |
21 u |
3 u |
5 u |
3 u |
Change |
|
|
|
+ 170 |
After
|
21 u
|
3 u
|
1x5 = 5 u |
4x5 = 20 u |
Number of kiwis in Box T remains unchanged. Make the number of kiwis in Box T the same. LCM of 5 and 1 is 5.
Number of pears put into Box T
= 20 u - 3 u
= 17 u
17 u = 170
1 u = 170 ÷ 17 = 10
Number of fruits in Box T in the end
= 5 u + 20 u
= 25 u
= 25 x 10
= 250
Answer(s): 250