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Problem

Jamil always throws loose change into a pencil holder on his desk and

takes it out every two weeks. This time it is all nickels and dimes. There

are 2 times as many dimes as nickels, and the value of the dimes is $1.65

more than the value of the nickels. How many nickels and dimes does

Jamil have?

My work

Let the number of dimes Jamil throws into the pencil holder be Y

Let the number of nickel Jamil throws into the pencil holder be X

Since there are 2 times as many dimes as nickel’s then we would have;

Y=2X

A dimes value is $0.05

A nickel’s value is $ 0.10

Since we have twice as many dimes as nickles then this would imply;

The sum of availables dimes and nickle is equal to $1.65

Converting Y=2X to a simple equation we would have;

Y- 2X=0 ……………………………..i

0.10X-0.05Y=1.65 …………….ii multiplying equation ii by 100 to get whole numbers then it will be;

Y- 2X=0 ………………………………i

10X+5Y = 165 ………………………iii

The least common multiple of 2 and 5 is 10. Multiply the first equation by 5 and the second equation by 2, by doing so that will have the y variable eliminated when the second equation is subtracted from the first equation.

(5)Y- (5)2X=(5)0 which solves to 5Y-10X=0

(2)10X+(2)5Y = (2)165 which solves to 20X+10Y=230

Subtract the second equation from the first and solve for x.

5Y-10X=0

20Y-10X=330

-15x= -330divide by sides by 12 to solve for x

-15 -15

Y = 22

From the equation we can be able to conclude that jamil has 22 dimes. We can go ahead and replace the number of dimes into 1 of the equations so that we can solve for the number of nickles. Taking the second equation.

20X+10Y=230 since we have computed for y=22

20(22)-10X=330then subtract 29.92 from both sides of the equation

440 – 10X=330

10X= 440-330

10X=110 dividing both sides by 10 we get;

10 10

X= 11 therefore jamil had 11 nickles

This would imply that jamil throwings in the pencil holder was 11 nickles and 22 dimes

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