Word Problems with Linear Equations: Additional Practice (Set 2)

Let $d$ be the daughter's age now, so the father is $4d$. In $5$ years she is $d + 5$ and he is $4d + 5$, and he is then $3$ times her age.

$4d + 5 = 3(d + 5), \qquad 4d + 5 = 3d + 15, \qquad d = 10$

The daughter is $10$ and the father is $40$; in $5$ years they are $15$ and $45$, and $45 = 3 \times 15$.

Let $w$ be the width in centimeters, so the length is $4w + 2$. The perimeter is twice the sum of length and width.

$2(w + (4w + 2)) = 54, \qquad 10w + 4 = 54, \qquad w = 5$

The width is $5$ cm and the length is $22$ cm; check $2(5 + 22) = 54$ cm.

Let $q$ be the number of quarters, so the dimes number $24 - q$. Working in cents, the total value is $480$.

$25q + 10(24 - q) = 480, \qquad 15q + 240 = 480, \qquad q = 16$

There are $16$ quarters ($400$ cents) and $8$ dimes ($80$ cents); check $400 + 80 = 480$.

Three times the number decreased by $7$ is $3n - 7$, and the number increased by $5$ is $n + 5$.

$3n - 7 = n + 5$

Decreasing $3n$ by $7$ gives $3n - 7$, so the correct model is $3n - 7 = n + 5$.

Adding a $20\%$ tip makes the total $120\%$ of the meal's cost, so $1.20c = 42$.

$1.20c = 42, \qquad c = \frac{42}{1.20} = 35$

Check: $20\%$ of $35$ is $7$, and $35 + 7 = 42$, so the meal cost $35$ dollars.

Let $n$ be the number of nickels, so the dimes number $n + 6$. In cents, nickels are worth $5n$ and dimes $10(n + 6)$.

$5n + 10(n + 6) = 210, \qquad 15n + 60 = 210, \qquad n = 10$

There are $10$ nickels ($50$ cents) and $16$ dimes ($160$ cents); check $50 + 160 = 210$.

Let $w$ be the width in centimeters, so the length is $3w$. The perimeter is twice the sum of length and width.

$2(w + 3w) = 64, \qquad 8w = 64, \qquad w = 8$

The width is $8$ cm, so the length is $3 \times 8 = 24$ cm; check $2(8 + 24) = 64$ cm.

Let $t$ be the hours she jogs, so she walks for $t + 1$ hours. Jogging covers $8t$ miles and walking covers $2(t + 1)$ miles, for $22$ miles total.

$8t + 2(t + 1) = 22, \qquad 10t + 2 = 22, \qquad t = 2$

Check: jogging $2$ hours covers $16$ miles and walking $3$ hours covers $6$ miles, and $16 + 6 = 22$, so she jogs $2$ hours.

Let $m$ be her starting money in dollars. Spending $30\%$ leaves $70\%$, so $0.70m = 42$.

$0.70m = 42, \qquad m = \frac{42}{0.70} = 60$

Check: $30\%$ of $60$ is $18$, and $60 - 18 = 42$, so she started with $60$ dollars.

Let $t$ be the number of hours until they meet. One train covers $30t$ miles and the other $40t$ miles, and together they close the $210$-mile gap.

$30t + 40t = 210, \qquad 70t = 210, \qquad t = 3$

After $3$ hours one train has gone $90$ miles and the other $120$ miles, and $90 + 120 = 210$, so they meet in $3$ hours.

Let $s$ be the son's age now, so the mother is $s + 26$. In $3$ years the son is $s + 3$ and the mother is $s + 29$, and she is then $3$ times his age.

$s + 29 = 3(s + 3), \qquad s + 29 = 3s + 9, \qquad s = 10$

The son is $10$ and the mother is $36$; in $3$ years they are $13$ and $39$, and $39 = 3 \times 13$.

Let $s$ be each equal side in centimeters, so the base is $s - 4$. The perimeter adds the two equal sides and the base.

$2s + (s - 4) = 41, \qquad 3s - 4 = 41, \qquad s = 15$

Each equal side is $15$ cm, so the base is $15 - 4 = 11$ cm; check $15 + 15 + 11 = 41$ cm.