Word Problems with Linear Equations: Additional Practice (Set 1)

Let $q$ be the number of quarters, so the nickels number $22 - q$. Working in cents, the total value is $370$.

$25q + 5(22 - q) = 370, \qquad 20q + 110 = 370, \qquad q = 13$

There are $13$ quarters ($325$ cents) and $9$ nickels ($45$ cents); check $325 + 45 = 370$.

Let $t$ be the time in hours. Going opposite ways, the distance apart is the sum $18t + 22t$.

$18t + 22t = 100, \qquad 40t = 100, \qquad t = 2.5$

Check: in $2.5$ hours they cover $45$ and $55$ miles, and $45 + 55 = 100$, so it takes $2.5$ hours.

Let $p$ be the price before markup in dollars. A $25\%$ markup makes the price $125\%$ of $p$, so $1.25p = 75$.

$1.25p = 75, \qquad p = \frac{75}{1.25} = 60$

Check: $25\%$ of $60$ is $15$, and $60 + 15 = 75$, so the price before was $60$ dollars.

Let $n$ be the number. Three fourths of it is $\frac{3}{4}n$ and one fourth is $\frac{1}{4}n$, with the three fourths being $12$ larger.

$\tfrac{3}{4}n = \tfrac{1}{4}n + 12, \qquad \tfrac{1}{2}n = 12, \qquad n = 24$

Check: three fourths of $24$ is $18$ and one fourth is $6$, and $18 = 6 + 12$, so the number is $24$.

Let $n$ be the smaller angle in degrees, so the larger is $n + 18$. Complementary angles sum to $90$.

$n + (n + 18) = 90, \qquad 2n + 18 = 90, \qquad n = 36$

The smaller is $36$ and the larger is $36 + 18 = 54$ degrees; check $36 + 54 = 90$.

Let $n$ be the smallest, so the integers are $n$, $n + 1$, $n + 2$, $n + 3$. Their sum is $58$.

$n + (n+1) + (n+2) + (n+3) = 58, \qquad 4n + 6 = 58, \qquad n = 13$

The integers are $13$, $14$, $15$, $16$, so the smallest is $13$; check $13 + 14 + 15 + 16 = 58$.

Let $g$ be the number of games played. Winning $60\%$ of them means $0.60g = 18$.

$0.60g = 18, \qquad g = \frac{18}{0.60} = 30$

Check: $60\%$ of $30$ is $18$, so the team played $30$ games.

Let $d$ be the daughter's age, so the mother is $d + 30$. Their ages sum to $54$.

$d + (d + 30) = 54, \qquad 2d + 30 = 54, \qquad d = 12$

The daughter is $12$ and the mother is $42$; check $12 + 42 = 54$ and $42 - 12 = 30$.

The width is $w$ and the length is $w + 7$. The perimeter adds two widths and two lengths.

$2w + 2(w + 7) = 50$

The second choice is only one width plus one length (half the perimeter), so the correct model is $2w + 2(w + 7) = 50$.

Let $n$ be the number. The sum of the number and $3$ is $n + 3$, and twice that is $2(n + 3)$, which equals $22$.

$2(n + 3) = 22, \qquad n + 3 = 11, \qquad n = 8$

Check: $2(8 + 3) = 22$, so the number is $8$.

Let $n$ be the smaller even integer, so the next is $n + 2$ and their sum is $2n + 2$. This sum is $6$ less than $3n$.

$2n + 2 = 3n - 6, \qquad 8 = n, \qquad n = 8$

The integers are $8$ and $10$ with sum $18$, and $3(8) - 6 = 18$, so the smaller is $8$.

Let $m$ be the number of marbles he started with. Giving away $40\%$ leaves $60\%$, so $0.60m = 54$.

$0.60m = 54, \qquad m = \frac{54}{0.60} = 90$

Check: $40\%$ of $90$ is $36$, and $90 - 36 = 54$, so he started with $90$ marbles.