Question

doug has 45 coins worth $9.00. The coins are all quarters or dimes. How many of each type of coin dose doug have?

Answer 1

Doug have 30 quarters and 15 dimes.

Step-by-step explanation:

Given,

Number of coins = 45

Worth of coins = $9.00 = 9.00*100 = 900 cents

Each quarter = 25 cents

Each dime = 10 cents

Let,

x be the number of quarters

y be the number of dimes

According to given statement;

x+y=45            Eqn 1

25x+10y=900   Eqn 2

Multiplying Eqn 1 by 10

$10(x+y=45)\\10x+10y=450\ \ \ Eqn\ 3$

Subtracting Eqn 3 from Eqn 2

$(25x+10y)-(10x+10y)=900-450\\25x+10y-10x-10y=450\\15x=450$

Dividing both sides by 15

$\frac{15x}{15}=\frac{450}{15}\\x=30$

Putting x=30 in Eqn 1

$30+y=45\\y=45-30\\y=15$

Doug have 30 quarters and 15 dimes.

Keywords: linear equation, elimination method

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