Question

A bakery gave out coupons to celebrate its grand opening. each coupon was worth either $1$, $3$, or $5$. Twice as many $1$ coupons were given out as $3$ coupons, and $3$ times as many$3$ coupons were given out as $5$ coupons. The total value of all the coupons given out was $360$. How many $3$ coupons were given out?

• A. $40$
• B. $45$
• C. $48$
• D. $54$

Answer 1

The correct option is D $54$

Let$x=$ number of $1$ coupons

$y=$ number of $3$ coupons

$z=$ number of $5$ coupons

Twice as many $1$ coupons were given out as $3$ coupons,

$\Rightarrow x=2y$

$3$ times as many $3$ coupons were given as $5$ coupons.

$\Rightarrow y=3z$

The total value of all the coupons given out was $360$

$\Rightarrow x\times 1+y\times 3+z\times 5=360$

The system results

$x-2y+0=0$

$\Rightarrow x=2y$ .....$\left(1\right)$

$0+y-3z=0$

$\Rightarrow y=3z$ or $z=\frac{y}{3}$ ......$\left(2\right)$

$x+3y+5z=360$ ......$\left(3\right)$

Substituting equation $\left(2\right)$ and $\left(1\right)$ into $\left(3\right)$

$\Rightarrow 2y+3y+\frac{5y}{3}=360$

$\Rightarrow 5y+\frac{5y}{3}=360$

$\Rightarrow \frac{20y}{3}=360$

$\Rightarrow y=18\times 3=54$