pls kindly explain in algebra method. thank you so much.
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Question
Answer
SAHMom
| Tank X | Tank Y (-trfr) | End in X | Rework units in Tank X | |
| G | 3u = 15q | 2p | 5r = 15q |
Know that no. of Goldfish in X remains the same. Let it be 15q (LCM of 3 and 5) |
| A |
1u +1p =5q + 28 |
5p -1p | 4r =12q | 12q= 5q + 28 |
| Total |
168 = 6p p=28 |
27q | ||
|
See above and deduce 7q=28 q=4 |
||||
| Total in Tank X |
27q = 108 |
ATM82517
thank you so much
alfretztay
In Tank X,
goldfish : angelfish
3u : 1u
In Tank Y,
goldfish : angelfish
2p : 5p
Mack transferred 20% of the angelfish from Tank Y to tank X.
100% – 20% = 80%
2p + 80% x 5p = 6p ——- 168
1p —— 168/6 = 28
3u/(1u + 28) = 5/4
(1u + 28) x 5 = 5u + 140 ——- 3u x 4 = 12u
1u ——- 140/(12 – 5) = 20
4u + 28 ——- 4 x 20 + 28 = 108
Ans : 108 fish.
Khong Pek Mao
This is Before After question.
In Tank X,
LCM of 3 and 5 is 15, goldfish no change
3 : 1 = 15 : 3
5 : 4 = 15 : 12
| Tank X | Goldfish | Angelfish | total |
| Before | 15u | 5u | |
| After | 15u | 12u | 27u |
20% = 12u – 5u = 7u
100% = 7u × 5 = 35u
| Tank Y | Goldfish | Angelfish | |
| Before | 14u | 35u | |
| After | 14u | 28u |
14u + 28u = 168
42u = 168
1u = 4
27u = 27 × 4 = 108
Khong Pek Mao
This is Before After question.
In Tank X,
LCM of 3 and 5 is 15, goldfish no change
3 : 1 = 15 : 3
5 : 4 = 15 : 12
| Tank X | Goldfish | Angelfish | total |
| Before | 15u | 5u | |
| After | 15u | 12u | 27u |
20% = 12u – 5u = 7u
100% = 7u × 5 = 35u
| Tank Y | Goldfish | Angelfish | |
| Before | 14u | 35u | |
| After | 14u | 28u |
14u + 28u = 168
42u = 168
1u = 4
27u = 27 × 4 = 108
