Solve Problem
You are in the fair, You have to buy some animals from this fair. Price list as below....
*Rs 10/-- 1 Elephant
* Rs 1/-- 1 Horse
* Rs 1/- - 8 Camel
And you have Rs 100/-.Count of animals should be 100 only.
Considering above all animals is compulsory???
Solve it
Answers
Given : Rs 10/-- 1 Elephant , Rs 1/-- 1 Horse , Rs 1/- - 8 Camel
To find : you have Rs 100/-.Count of animals should be 100 only.
Solution:
Rs 10/-- 1 Elephant
* Rs 1/-- 1 Horse
* Rs 1/- - 8 Camel
Elephant = E
Horse = H
Camel = 100 - E - H
10E + H + (100 - E - H)/8 = 100
=> 80E + 8H + 100 - E - H = 800
=> 79E + 7H = 700
=> 79E = 700 - 7H
=> 79E = 7 ( 100 - H)
as 79 & 7 are co prime numbers hence E = 7 as elephants can not not be 0 or above 10 ( as 10 Elephant it self cost rs 100)
=> E = 7 & 100 - H = 79
=> H = 21
Camel = 100 - 7 - 21 = 72
Cost
Elephant = 7 70
Horse = 21 21
Camel = 72 9
Total 100 100
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The number of elephants, horse and camel are 1, 89 and 10 respectively.
Step-by-step explanation:
Let number of elephants be x
Let number of horse be y
Let number of camel be z
From question, the total number of animals is 100, thus,
x + y + z = 100 → (equation 1)
From question, the price of the total animals is also Rs. 100, thus,
10x + y + z/8 = 100
80x + 8y + z = 800 → (equation 2)
From the question, two equations with three unknowns are formed. Thus, the equations have infinite number of solutions.
On solving equation (1) and (2), we get,
79x + 7y = 700 → (equation 3)
Let's assume the value of x = 1, we get,
79 + 7y = 700
7y = 700 - 79
7y = 621
Therefore, x = 1 and y = 88.7 ≈ 89
On substituting value of y and z in equation (1), we get,
1 + 89 + z = 100
z = 100 - 90
∴ z = 10