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 Help with 2 Mathematics question ! PLEASE help !!
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By: verachen (offline)  Tuesday, September 17 2013 @ 09:53 AM CDT (Read 894 times)  

1 ) At 10 am, a car was travelling from Town X to Town Y. At the same time, a lorry was travelling from Town Y to Town X along the same route. The car travelling at a speed of 25km/h faster, passed the lorry 15km from the midpoint between the 2 towns. a) At what time did they pass each other? b) The car took another 24 min to reach Town Y. What time would the lorry reach Town X?

a) 11.12am
b) 2.48pm
[ I am not very sure so I need someone to help me see if my answer is correct ]

2 ) A farmer had some chickens and ducks in his farm. If 50 chickens were sold, the ratio of number of chickens left to the number of duck would be 8:7. If 7 ducks were sold, the ratio of the number of chickens to the number of ducks left would be 10:7. a) How many chickens were there in the farm? b) When 422 ducks in the farm died, what percentage of the ducks were left?

[ I have no idea how to do Q2 ]

Your help is greatly appreciated !!

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By: jo sarah (offline)  Wednesday, September 18 2013 @ 08:54 PM CDT  
jo sarah


If there be 1u chickens and 1p ducks
Then, the first scenario results in:
(1u - 50)/(1p) = 8/7,
Which gives you: 7(1u - 50) = 8p,
I.e., 7u - 350 = 8p,
Or, 7u = 8p + 350..................[1]

Second scenario gives:
(1u)/(1p - 7) = 10/7
I.e., 7u = 10p - 70..................[2]

[1] and [2] gives:
8p + 350 = 10p -70
So that, 2p = 350 + 70 = 420
And, 1p = 210
Therefore, 1u = (1/7) (10 x 210 - 70) = 290

This, there are 290 chickens.

I am not sure if something is amiss in (b), since there are only 210 ducks.

Check for (a): (290-50/210=8/7. And, 290/(210-7)=10/7.

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