1. Let $c$ be the number of correct answers and $i$ be the number of incorrect answers. We have the system:
$c + i = 100$
$2c - i = 80$
Adding the equations: $3c = 180$, so $c = 60$. Substituting back, $60 + i = 100$, so $i = 40$. The student answered 40 questions incorrectly.
2. Let $w$ be the number of white chocolate bars and $d$ be the number of dark chocolate bars. We have the system:
$w + d = 15$
$2w + 2.5d = 34$
Multiply the first equation by 2: $2w + 2d = 30$. Subtract from the second equation: $(2w + 2.5d) - (2w + 2d) = 34 - 30$, so $0.5d = 4$, thus $d = 8$. You bought 8 bars of dark chocolate.
3. Profit is zero when Revenue equals Cost: $R = C$.
$50n = 10n + 300$
$40n = 300$
$n = 7.5$
The company will have no profit when 7.5 phones are sold. (Since selling half a phone isn't practical, this is the break-even point mathematically).
4. Let $x$ be the larger number and $y$ be the smaller number. We have the system:
$x + y = 17$
$x - y = 7$
Adding the equations: $2x = 24$, so $x = 12$. Substituting back, $12 + y = 17$, so $y = 5$. The value of the smaller number is 5.
5. Let $q$ be the number of quarters and $n$ be the number of nickels. We have the system:
$q + n = 25$
$0.25q + 0.05n = 5.05$
Multiply the second equation by 20 to eliminate decimals: $5q + n = 101$. Subtract the first equation: $(5q + n) - (q + n) = 101 - 25$, so $4q = 76$, thus $q = 19$. John has 19 quarters.
6. Let $r$ be the number of students with red shirts and $y$ be the number with yellow shirts. We have the system:
$r + y = 32$
$r = y + 12$
Substitute the second equation into the first: $(y + 12) + y = 32$, so $2y + 12 = 32$, thus $2y = 20$, so $y = 10$. Then $r = 10 + 12 = 22$. 22 students have red shirts.
Parent Tip: Review the logic above to help your child master the concept of equation word problems worksheet.
