Word Problems with Systems: Additional Practice (Set 1)

Let $x$ be the larger and $y$ the smaller: $x + y = 96$ and $x - y = 24$. Add to eliminate $y$.

$2x = 120 \;\Rightarrow\; x = 60$

The larger number is $60$ (and the smaller is $36$).

Let $a$ and $c$ be adults and children: $a + c = 500$ and $9a + 5c = 3620$. Multiply the count equation by $9$ and subtract.

$(9a + 9c) - (9a + 5c) = 4500 - 3620 \;\Rightarrow\; 4c = 880 \;\Rightarrow\; c = 220$

So $220$ children visited (and $280$ adults).

Let $p$ and $a$ be pounds of peanuts and almonds: $p + a = 15$ and $3p + 8a = 5(15) = 75$. Substitute $p = 15 - a$.

$3(15 - a) + 8a = 75 \;\Rightarrow\; 5a = 30 \;\Rightarrow\; a = 6$

The mix uses $6$ pounds of almonds (and $9$ of peanuts).

Downstream $2(b + c) = 24$ gives $b + c = 12$; upstream $3(b - c) = 24$ gives $b - c = 8$. Add to eliminate $c$.

$2b = 20 \;\Rightarrow\; b = 10$

The still-water speed is $10$ mph (and the current is $2$ mph).

Let $x$ and $y$ be the amounts at $4\%$ and $7\%$: $x + y = 5000$ and $0.04x + 0.07y = 290$. Substitute $x = 5000 - y$.

$0.04(5000 - y) + 0.07y = 290 \;\Rightarrow\; 0.03y = 90 \;\Rightarrow\; y = 3000$

So $3000$ dollars is in the $7\%$ account (and $2000$ at $4\%$).

Let $L$ and $W$ be length and width: $L = W + 4$ and $2(L + W) = 52$, so $L + W = 26$. Substitute.

$(W + 4) + W = 26 \;\Rightarrow\; 2W = 22 \;\Rightarrow\; W = 11$

The width is $11$ cm (and the length is $15$ cm).

With the wind $480 / 3 = 160$, so $p + w = 160$; against it $480 / 4 = 120$, so $p - w = 120$. Subtract to eliminate $p$.

$(p + w) - (p - w) = 160 - 120 \;\Rightarrow\; 2w = 40 \;\Rightarrow\; w = 20$

The wind speed is $20$ mph (and the plane is $140$ mph in still air).

Set cost equal to revenue: $16x = 7x + 450$, so $9x = 450$ and $x = 50$. Then the revenue is $16x$.

$y = 16(50) = 800$

The revenue at break-even is $800$ dollars.

Let $s$ be the slower speed and $f$ the faster, both unknown: $f = s + 20$, and meeting after $2$ hours gives $2s + 2f = 240$, so $s + f = 120$. Substitute $s + 20$ for $f$.

$s + (s + 20) = 120 \;\Rightarrow\; 2s = 100 \;\Rightarrow\; s = 50$

Then the faster train goes $f = 50 + 20 = 70$ mph.

Let $y$ be the smaller and $x$ the larger: $x + y = 35$ and $x = 4y$. Substitute $4y$ for $x$.

$4y + y = 35 \;\Rightarrow\; 5y = 35 \;\Rightarrow\; y = 7$

The smaller number is $7$ (and the larger is $28$).

Let $b$ be the still-water speed and $c$ the current, both unknown. Downstream $3(b + c) = 60$ gives $b + c = 20$; upstream $5(b - c) = 60$ gives $b - c = 12$. Add to eliminate $c$.

$2b = 32 \;\Rightarrow\; b = 16$

The still-water speed is $16$ mph (and the current is $4$ mph).

A count of buses has to be a whole number, so a value of $7.5$ fails the check against the words even though it satisfies your equations.

$7.5 \text{ buses} \;\Rightarrow\; \text{fails the context check}$

Recheck both the translation into equations and the problem's own conditions and data, rather than rounding a wrong result.