Question

There were 100 banana and100 people which consist of children,young people, old age people. first ,childrens go and carry 4 banana then young people go and carry 2 banana and then there were few banana left so old people decide to divide one banana into four old person then how many children, young people, old age people are there to balance this

Answer 1

Let the number of children = x
the number of young people = y
the number of old people = z

x + y + z = 100
4x + 2y + 0.25z = 100
There is no third equation, which makes it difficult to solve but we have a condition that x, y and z are all integers as we can't have 1.7 children.

x + y +z = 100
⇒x = 100- y - z
Putting x in second equation
4(100-y-z) + 2y + 0.25z = 100
⇒400 - 4y - 4z + 2y + 0.25 z = 100
⇒2y + 3.75z = 300

⇒z = $\frac{300-2y}{3.75}$

Now I will solve by hit-and-trial method keeping certain points in mind.
1. Take only integral values of y.
2. z is an integer. So $\frac{300-2y}{3.75}$ is an integer or (300-2y) is a multiple of 3.75. 
3. (300-2y) will always be an even integer.
4. An even integer which is a multiple of 3.75 should be an multiple of 30.(This is an important conclusion to reduce the calculation).
5. After you get a answer, check it all conditions are satisfied.
6.Value of z is greater than x and y.
(If you don't understand any point, ask)
 take y = 15, so that 2y = 30 is a multiple of 30 and (300-30)=270 is also a multiple of 30.
then z = 270/3.75 = 72
x = 100-72-15 = 13
x+y+z = 100(satified)
4x+2y+0.25z= 100(satisfied)
Answer in the first attempt!!!

So children=13
young people = 15
old people = 72

Answer 2

Step-by-step explanation:

Let the number of children = x

the number of young people = y

the number of old people = z

x + y + z = 100

4x + 2y + 0.25z = 100

There is no third equation, which makes it difficult to solve but we have a condition that x, y and z are all integers as we can't have 1.7 children.

x + y +z = 100

⇒x = 100- y - z

Putting x in second equation

4(100-y-z) + 2y + 0.25z = 100

⇒400 - 4y - 4z + 2y + 0.25 z = 100

⇒2y + 3.75z = 300

⇒z = \frac{300-2y}{3.75}

3.75

300−2y

Now I will solve by hit-and-trial method keeping certain points in mind.

1. Take only integral values of y.

2. z is an integer. So \frac{300-2y}{3.75}

3.75

300−2y

is an integer or (300-2y) is a multiple of 3.75.

3. (300-2y) will always be an even integer.

4. An even integer which is a multiple of 3.75 should be an multiple of 30.(This is an important conclusion to reduce the calculation).

5. After you get a answer, check it all conditions are satisfied.

6.Value of z is greater than x and y.

(If you don't understand any point, ask)

take y = 15, so that 2y = 30 is a multiple of 30 and (300-30)=270 is also a multiple of 30.

then z = 270/3.75 = 72

x = 100-72-15 = 13

x+y+z = 100(satified)

4x+2y+0.25z= 100(satisfied)

Answer in the first attempt!!!

So children=13

young people = 15

old people = 72

thanks