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  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

 

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thank you so much 

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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.

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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

 

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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

 

 

 

 

 

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pls kindly explain in algebra method. as i’m not familiar with model drawing method. thankyou.

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thanks for your detail explanation.

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