Quizzes & Puzzles6 mins ago
Please can anyone help me solver this maths conundrum?
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.
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.
Answers
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...
so multiply by 8: 80A + C + 8C = 800
total number of animals: A + B + C = 100
15:28 Sat 21st Apr 2012
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
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
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