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Please can anyone help me solver this maths conundrum?

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Reddevil1976 | 16:17 Sat 21st Apr 2012 | Quizzes & Puzzles
6 Answers
My little cousin has asked me to help him solve this maths question and it has got us both stumped. Even the wife, who is an accountant, has got stumped and I look at her as a maths expert :)). So here is the question and equation to solve:

A young farmer has to stock his new farm and has allocated £100 to purchase livestock.The eagerly awaited day arrives when he is able to attend market. He discovers that Horses cost £10 each, Ducks are 8 for £1 and Sheep are £1 each. He sat down with a pen and paper to calculate what combination of each animal he could buy to meet his target of exactly 100 animals while spending all of his budget.


Do the same calculation to find out how many of each animal he bought, where
A= Horses, B= Ducks and C= Sheep.

The equations to solve are:

25.(A x 9) + (B x 7) + (C x 8) + 116

21.(A x 4) + (B x 3) + (C x 5) + 47

We need both equations solving and, if possible, workings shown. Many thanks in advance for any and all help.

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total cost: 10A + (D/8) + C = 100
so multiply by 8: 80A + C + 8C = 800

total number of animals: A + B + C = 100

subtract that from the previous equation to get: 79A + 7C = 700

re-arrange to get C = (700 - 79A)/7

because there can only be a positive whole number of each type of animal simply try different values of A (starting with 8 and working...
16:28 Sat 21st Apr 2012
Should the last + be an = ??
total cost: 10A + (D/8) + C = 100
so multiply by 8: 80A + C + 8C = 800

total number of animals: A + B + C = 100

subtract that from the previous equation to get: 79A + 7C = 700

re-arrange to get C = (700 - 79A)/7

because there can only be a positive whole number of each type of animal simply try different values of A (starting with 8 and working downwards)

you will quickly find that the only solution is A = 7 and so C = 21, which leaves B = 72

now substitute those values in the 2 equations
sorry for the typos in the initial lines:-

10A + (B/8) + C = 100
thus: 80A + B + 8C = 800
Surely another solution is A=0, B=0, C=100
i assumed that there had be some of each i.e. positive values only
Same answer as bibblebub, but a slightly different calculation:

Cost: 10A + B/8 + C = 100

Numbers: A + B + C = 100

Subtracting the second equation from the first:

9A - 7B/8 = 0

9A = 7B/8

72A = 7B

7 is a prime number, therefore the only possibilities are:

A=7, B=72

To make up 100, C=21

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