Chapter 8, Problem 1RVS

To determine

To calculate: The number of each type of coins if a collection of nickels and quarters is worth $\$9.35$ and there are 59 coins in total and also determine the correct match for the equation from provided option.

Answer to Problem 1RVS

Solution:

The number of nickels are 27 and the number of quarters are 32 and the correct option is

$\begin{array}{l}J.\text{ 0}\text{.0}5x+0.25y=9.35\\ x+y=59\end{array}$.

Explanation of Solution

Given Information:

The provided options are,

$\begin{array}{l}A.5x+25y=9.35\\ x+y=59\\ B.4^2+x^2=8^2\\ C.x\left(x+26\right)=568\\ D.8=x\cdot 24\\ E.\frac{75}{x}=\frac{105}{x+5}\\ F.\frac{75}{x}=\frac{55}{x+5}\\ G.\text{ 2}x\text{+2}\left(x+26\right)=568\\ H.x+3x\text{+}\left(x+3x-17\right)=180\end{array}$

And,

$\begin{array}{l}I.x\text{+3}x+\left(3x-17\right)=180\\ J.\text{ 0}\text{.0}5x+0.25y=9.35\\ x+y=59\\ K.8^2+8^2=x^2\\ L.x^2+{\left(x+26\right)}^2=568\\ M.\frac{1}{4}+\frac{1}{x}=\frac{1}{55}\\ N.\frac{1}{55}+\frac{1}{75}=\frac{1}{x}\\ O.x+5\%\cdot x=8568\end{array}$

Formula used:

According to the elimination method, a set of equations is simplified by using the addition principle.

(1) The coefficient of one variable, of one equation, is converted to the opposite sign of the coefficient of the other equation.

(2) The equations are added and one variable is removed from the set of equations.

(3) The value of the variable is used to find the value of the other variable.

(4) Check by substituting the values of the variables in the equations.

According to the distributive property, for numbers a, b and c,

$a\cdot b+a\cdot c=a\cdot \left(b+c\right)$

Calculation:

Consider the number of nickels to be x and the number of quarters to be y.

The value of a nickel is $\$0.05$ and the value of a quarter is $\$0.25$.The collection of nickels and quarters is worth $\$9.35$, so,

$0.05x+0.25y=9.35$

Multiply both sides by $\frac{100}{5}$ using the distributive property,

$\begin{array}{c}\frac{100}{5}\left(0.05x+0.25y\right)=9.35\cdot \frac{100}{5}\\ \frac{100}{5}\cdot \left(0.05x\right)+\frac{100}{5}\left(0.25y\right)=\frac{935}{5}\\ \frac{5x}{5}+\frac{25y}{5}=187\\ x+5y=187\end{array}$ …… (1)

There are 59 coins in total, so,

$x+y=59$ …… (2)

Hence, the correct option is J.

Multiply the equation (2) by $-1$ using the distributive property,

$\begin{array}{c}\left(-1\right)\left(x+y\right)=-1\cdot 59\\ \left(-1\right)\left(x\right)+\left(-1\right)\left(y\right)=-1\cdot 59\\ -x-y=-59\end{array}$ …… (3)

The coefficients of the variable x are same and of opposite signs, thus, apply the elimination method by adding (1) and (3),

$\frac{\begin{array}{c}x+5y=187\\ -x-y=-59\end{array}}{4y=128}$

Divide both sides by 4,

$\begin{array}{c}\frac{4}{4}y=\frac{128}{4}\\ y=32\end{array}$

Substitute 32 for y in (2),

$x+32=59$

Subtract 32 from both sides,

$\begin{array}{c}x+32-32=59-32\\ x=27\end{array}$

Thus, there are 32 quarters are 27 nickels. Check by substituting 27 for x and 32 for y in the equations (1) and (2).

For equation (1),

$\begin{array}{r}27+5\left(32\right)\stackrel{?}{=}187\\ 27+160\stackrel{?}{=}187\\ 187\stackrel{?}{=}187\end{array}$

This is true. Now, for equation (2),

$\begin{array}{r}27+32\stackrel{?}{=}59\\ 59\stackrel{?}{=}59\end{array}$

This is true.

Hence, there are 32 quarters are 27 nickels and correct option is J.