Word Problems with Systems: Additional Practice (Set 2)

Cost $y = 4x + 900$ and revenue $y = 13x$. Set them equal.

$13x = 4x + 900 \;\Rightarrow\; 9x = 900 \;\Rightarrow\; x = 100$

Break-even is at $100$ copies.

Cost $y = 2x + 180$ and revenue $y = 11x$. Set them equal.

$11x = 2x + 180 \;\Rightarrow\; 9x = 180 \;\Rightarrow\; x = 20$

Selling $20$ cakes breaks even.

Let $x$ be the larger and $y$ the smaller: $x + y = 84$ and $x = 5y$. Substitute $5y$ for $x$.

$5y + y = 84 \;\Rightarrow\; 6y = 84 \;\Rightarrow\; y = 14$

Then the larger is $x = 5(14) = 70$.

Let $t$ and $u$ be the tens and units digits: $t + u = 12$ and $u = t + 2$. Substitute $t + 2$ for $u$.

$t + (t + 2) = 12 \;\Rightarrow\; 2t = 10 \;\Rightarrow\; t = 5$

Then $u = 7$, so the number is $57$.

The two amounts add to $2000$ dollars, and the $8\%$ amount was $1300$.

$2000 - 1300 = 700$

So $700$ dollars is invested at $5\%$.

Let $A$ and $B$ be their present ages: $A = B + 5$ and $(A + 3) + (B + 3) = 33$, which gives $A + B = 27$. Substitute.

$(B + 5) + B = 27 \;\Rightarrow\; 2B = 22 \;\Rightarrow\; B = 11$

Ben is $11$ now (and Ana is $16$).

Let $x$ be the larger and $y$ the smaller: $x + y = 30$ and $\tfrac{1}{3}x = \tfrac{1}{2}y$. Clearing fractions gives $2x = 3y$, so $y = \tfrac{2}{3}x$; substitute.

$x + \tfrac{2}{3}x = 30 \;\Rightarrow\; \tfrac{5}{3}x = 30 \;\Rightarrow\; x = 18$

The larger number is $18$ (and the smaller is $12$).

Let $f$ and $t$ be the counts of $5$- and $20$-dollar bills: $f + t = 32$ and $5f + 20t = 400$. Multiply the count equation by $5$ and subtract.

$(5f + 20t) - (5f + 5t) = 400 - 160 \;\Rightarrow\; 15t = 240 \;\Rightarrow\; t = 16$

There are $16$ twenty-dollar bills (and $16$ five-dollar bills).

Set cost equal to revenue: $10x = 6x + 240$, so $4x = 240$ and $x = 60$.

$10x = 6x + 240 \;\Rightarrow\; x = 60$

Beyond $60$ units revenue outgrows cost, so selling more is a profit.

Let $x$ be the larger and $y$ the smaller: $x = 3y - 1$ and $x + y = 43$. Substitute $3y - 1$ for $x$.

$(3y - 1) + y = 43 \;\Rightarrow\; 4y = 44 \;\Rightarrow\; y = 11$

Then the larger is $x = 3(11) - 1 = 32$.

A multiple of the first equation is the same line, so it adds no new information.

$\text{same line} \;\Rightarrow\; \text{infinitely many solutions}$

The two conditions are not independent, so the problem is under-determined and does not have a unique answer.

The two amounts add to $5000$ dollars, and the $7\%$ amount was $3000$.

$5000 - 3000 = 2000$

So $2000$ dollars is in the $4\%$ account.