Two-step word problems in 3rd and 4th grade

The answer is No. A quick estimate gives about $30$ books, but Maya claims $300$. Since $300$ is much greater than $30$, the claim is not reasonable.

1. 1First, multiply to find how many books are on the shelves. Then subtract the books taken away.
2. 2Round to estimate. $12$ rounds to $10$ and $8$ rounds to $10$.
3. 3The estimate is about $30$ books. Maya claims $300$. Since $300$ is about $10$ times greater than $30$, the claim is far from the estimate.
4. 4The claim is not reasonable, so the answer is No.

Common mistake. A common mistake is adding an extra zero by accident, getting $300$ instead of about $30$. The correct estimate is $4 \times 10 - 10 = 30$, not $300$.

The idea underneath. When deciding if an answer is reasonable, round the numbers and make a quick estimate first. If the claim is far from your estimate, it is not reasonable — even if it looks like a number you'd get from a small mistake.

Both methods are correct. The supply closet has 369 pencils. Method A applies each change in order; Method B combines the gain and loss into a net change of 45 first. Both ways give the same final answer.

1. 1Read the problem. The supply closet starts with 324 pencils, gains 175 from a delivery, then loses 130 handed out to students. The final count is $324 + 175 - 130$.
2. 2Check Method A. Method A applies each change in story order: first add the delivery, then subtract the pencils handed out.
3. 3Check Method B. Method B combines the two changes first. Gaining 175 and losing 130 is the same as a net gain of 45. Then add 45 to the starting amount.
4. 4Compare the results. Method A gives 369 pencils and Method B gives 369 pencils. Both methods are correct.

Common mistake. A common mistake is thinking only the chronological method (Method A) is valid because it follows the order of the story. The order in which you apply a gain and a loss does not change the final total, so combining the two changes first works just as well.

The idea underneath. When a problem has one gain and one loss, you can apply the changes one at a time or combine them into a single net change first. Both approaches give the same final answer.

Answer. 37

There are $37$ students still in the classroom. Multiplying $7$ tables by $8$ students gives $56$ total students, then subtracting the $19$ who left gives $37$.

1. 1First, find the total number of students. There are $7$ tables with $8$ students at each table, so multiply.
2. 2Next, subtract the $19$ students who go home from the $56$ total.
3. 3There are $37$ students still in the classroom.

Common mistake. A common mistake is stopping after the first step and entering $56$ (the total students) instead of subtracting $19$ to find the final answer of $37$.

The idea underneath. Two-step word problems require both steps in the correct order. The first step gives an intermediate value; the final answer comes from applying the second operation to that intermediate value, not from the first step alone.

Liam's equation does not match the story. He multiplied the $6$ stickers by $4$ when he should have simply added $6$ after the first multiplication step.

1. 1Read the story and translate each part into math. Rosa has $9$ stickers. The cousin has '$4$ times as many,' so multiply: $4 \times 9$. Then the cousin finds $6$ more stickers, so just add $6$.
2. 2Build the equation that matches the story:
3. 3Now look at Liam's equation:
4. 4Liam wrote $4 \times 6$ for the stickers found, but the story says the cousin just finds $6$ more — that means add $6$, not multiply $6$ by $4$. The equation multiplies the $6$ stickers found by $4$ instead of just adding them.

Common mistake. A common mistake is applying the multiplier to both the starting count and the adjustment. 'Finds $6$ more' means add $6$ to the total — it does not mean multiply $6$ by $4$.

The idea underneath. When a story says to multiply first and then adjust by adding or subtracting, only the original starting count gets multiplied — the adjustment is added or subtracted on its own.