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A coin box contains $26.50 in dimes and quarters. There are 157 coins altogether. How many of each

type of coin are in the box?

2 Answers

3 votes

Answer:

so 85 dimes and 72 quarter coins are in the box

Explanation:

Given; Total number of coin =1557 Total amount money =26.50 To find Number of dimes and quarters in the box. In order to find the solution, make equation of number of coin and amount money. solve for unknown.Let, Number of dimes = x Number of quarters = y As, it is given that total number of coin is 157. x + y = 157 ......(a) Also, the total money is $ 26.50. .. 0.10x+0.25y = 26.50 ..(b)To, solve equation (a) & (b), put value of y from (a) to (b); From (b); From (a); y= 157 0.10x+0.25(157 - x) = 26.50 0.10x+39.25 - 0.25x = 26.50 0.15x+39.25 = 26.50 39.25 26.50 = 0.15x 12.75 = 0.15x 12.75 0.15 = x 85 = x Now, put value of x in equation (a); x + y 85+ y = y= 72

User Stefan Koenen
by
8.1k points
4 votes

Answers: 85 dimes and 72 quarters

======================================================

Work Shown:

x = number of dimes

y = number of quarters

x+y = 157 coins total

-------------------------------

$26.50 = 2650 cents

1 dime = 10 cents

x dimes = 10x cents

1 quarter = 25 cents

y quarters = 25y cents

total value = 10x+25y = 2650 cents

-------------------------------

We have this system of equations

x+y = 157

10x+25y = 2650

Let's solve for y in the first equation

x+y = 157

y = 157-x

Then plug this into the second equation and solve for x.

10x+25y = 2650

10x+25(157-x) = 2650

10x+25(157)+25(-x) = 2650

10x+3925-25x = 2650

-15x+3925 = 2650

-15x = 2650-3925

-15x = -1275

x = -1275/(-15)

x = 85 is the number of dimes

Then use this to find the value of y

y = 157-x

y = 157-85

y = 72 is the number of quarters

-------------------------------

Check:

85 dimes + 72 quarters = 157 coins total

85 dimes = 85*10 = 850 cents

72 quarters = 72*25 = 1800 cents

850 cents + 1800 cents = 2650 cents = $26.50

This confirms the answers.

User Adalle
by
7.5k points

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