Algebra problems

Question

A teacher distributes €2 among 20 children, giving 5p each to some and 25p to each to the rest. How many students recieve 25p each.

Answer 1

This is a classic system of equation problem. The first step is setting up your variables here Im going to say x is people getting 5p and y is getting 25p.

Secondly I want to set up my equations.

I have to account for how many people were there and i know that the people getting both coins should equal 20. x + y = 20

The second equation is going to take into account the value of money. The trick here is knowing that you have to use a decimal for the pence. 5 pence is equal to 0.05 and 25 pence is .25.

.05x + .25y = 2€

We stack our equations

x + y = 20

.05x + .25y = 2

I solved this by substitution.

I rearranged the first equation to y = 20 - x

Substitute this into the second equation: .

.05x +.25(20-x)= 2

.05x + 5 -.25x = 2

-.20x + 5 = 2

-.20x = -3

x = 15

now that we know the value of x we then sub it back into the first equation

15 + y = 20

y = 5

(15,5) is your solution.

With system its always to put it back into context of the word problem, this solution tells us that 15 people got 5 p and 5 people got 25p.

Answer 2

Suppose teacher gives 5 p to x no of children

So 25 p given to 20–x no of children.

Now make an equation showing total amt of money given to children and 20 pounds

5x + 25(20–x) = 200. 2 pounds=200pans

5x + 500 — 25x = 200

–20x. = -500 + 200

X. = -300/-20 = 15

So 15 children got 5p each and 20–15=5 children got 25p