Practice Math word problem practice: 6 questions with answers Six math word problems, from easy to hard, with worked explanations. Pick an answer to check it instantly.
These questions come from our SHSAT, TACHS and SSAT practice apps. For parents
Word problems test whether you can turn a situation into math. Most of the work happens before any calculating.
A four-step method
Find the question. Underline what you’re asked to find, such as “how many tickets” or “how much change.”Name the unknown. Let t be the number of tickets.Write the math. “Each ticket costs $6 and the total was $42” becomes 6t = 42.Solve, then check it against the story. t = 7. Seven tickets at $6 each is $42, so the answer fits.
Watch for answers that are a step short, like the total when the question asks for what’s left. The last step, checking against the story, catches most of them.
The questions start easy and get harder. Try each one before you look at the explanation.
Practice questions Each question and explanation was reviewed by a person. Pick an answer to see if you're right.
Question 1 In a relay lineup, Ana finished ahead of Bo, and Bo finished ahead of Cai. Of these three runners, who finished last?
A Ana B Bo C Cai D Cannot be determined
"Ahead of" means finishing earlier. Ana is ahead of Bo, and Bo is ahead of Cai, giving the order Ana, then Bo, then Cai, so Cai is last of the three. "Cannot be determined" is what the two comparisons look like to a student who does not chain them: each one alone leaves a runner unplaced, and only putting them together fixes the order.
Question 2 Five runners, Ana, Bo, Cai, Dee and Eve, finished a race, and no two of them finished at the same time. Ana finished ahead of Bo. Bo finished ahead of Cai. Dee finished ahead of Cai. Eve finished behind Ana. Which statement must be true?
A Ana finished ahead of Cai. B Bo finished right behind Ana. C Dee finished ahead of Bo. D Ana finished in first place. E Eve finished in last place.
Ana finished ahead of Bo, and Bo finished ahead of Cai, so Ana finished ahead of Cai. That must be true in every order that fits the clues. None of the other statements is forced. The order Ana, Eve, Bo, Dee, Cai fits every clue, and in it Bo is not right behind Ana, Eve is not last, and Dee is not ahead of Bo. The order Dee, Ana, Bo, Cai, Eve also fits every clue, and in it Ana is not first.
Question 3 A river ferry can carry at most 28 cars on each trip. There are 375 cars waiting to cross. What is the least number of trips the ferry must make so that every car crosses the river?
A 11 11 11 B 12 12 12 C 13 13 13 D 14 14 14 E 15 15 15
Divide:
375 ÷ 28 = 13 375 \div 28 = 13 375 ÷ 28 = 13 remainder
11 11 11 , because
28 × 13 = 364 28 \times 13 = 364 28 × 13 = 364 and
375 − 364 = 11 375 - 364 = 11 375 − 364 = 11 . Thirteen full trips carry 364 cars, leaving 11 cars that still need one more trip, so the ferry needs
13 + 1 = 14 13 + 1 = 14 13 + 1 = 14 trips. Sanity check: 14 trips can carry up to
28 × 14 = 392 ≥ 375 28 \times 14 = 392 \ge 375 28 × 14 = 392 ≥ 375 cars, while 13 trips carry only
364 < 375 364 < 375 364 < 375 ; so the least number of trips is 14.
Question 4 A repair shop charges a
$ 45 \$45 $45 diagnostic fee plus
$ 32 \$32 $32 for each hour of labor. Mr. Lee's repair takes
3 3 3 hours, and he also buys a replacement part for
$ 58 \$58 $58 . A
10 % 10\% 10% tax is applied to the entire bill. What is the total amount Mr. Lee pays?
A $ 155.10 \$155.10 $155.10 B $ 169.40 \$169.40 $169.40 C $ 199.00 \$199.00 $199.00 D $ 209.00 \$209.00 $209.00 E $ 218.90 \$218.90 $218.90
Labor:
45 + 32 × 3 = 45 + 96 = 141 45+32\times 3 = 45+96 = 141 45 + 32 × 3 = 45 + 96 = 141 . Add the part:
141 + 58 = 199 141+58 = 199 141 + 58 = 199 subtotal. Apply the
10 % 10\% 10% tax:
199 × 1.10 = 218.90 199\times 1.10 = 218.90 199 × 1.10 = 218.90 . Check:
10 % 10\% 10% of
199 199 199 is
19.90 19.90 19.90 , and
199 + 19.90 = 218.90 199+19.90 = 218.90 199 + 19.90 = 218.90 ✓.
Question 5 Mr. Okafor has a box of pencils. When he puts the pencils into bundles of 3, he has 2 pencils left over. When he puts them into bundles of 5, he has 4 pencils left over. He has more than 20 but fewer than 40 pencils. How many pencils does he have?
A 14 14 14 B 21 21 21 C 22 22 22 D 23 23 23 E 29 29 29
List the numbers between 20 and 40 that leave 4 when divided by 5: 24, 29, 34 and 39. Check each for a leftover of 2 in bundles of 3:
24 = 8 × 3 24 = 8 \times 3 24 = 8 × 3 leaves 0,
29 = 9 × 3 + 2 29 = 9 \times 3 + 2 29 = 9 × 3 + 2 leaves 2,
34 = 11 × 3 + 1 34 = 11 \times 3 + 1 34 = 11 × 3 + 1 leaves 1, and
39 = 13 × 3 39 = 13 \times 3 39 = 13 × 3 leaves 0. Only 29 works. Check:
29 = 9 × 3 + 2 29 = 9 \times 3 + 2 29 = 9 × 3 + 2 and
29 = 5 × 5 + 4 29 = 5 \times 5 + 4 29 = 5 × 5 + 4 .
Question 6 A grocer makes a nut mix by combining almonds worth
$ 8 \$8 $8 per pound with peanuts worth
$ 3 \$3 $3 per pound. The grocer makes
20 20 20 pounds of mix that is worth
$ 5 \$5 $5 per pound. How many pounds of almonds are in the mix?
Let the almonds be
a a a pounds, so the peanuts are
20 − a 20-a 20 − a pounds. The total dollar value must match either way:
8 a + 3 ( 20 − a ) = 5 ( 20 ) = 100 8a+3(20-a)=5(20)=100 8 a + 3 ( 20 − a ) = 5 ( 20 ) = 100 . Then
8 a + 60 − 3 a = 100 8a+60-3a=100 8 a + 60 − 3 a = 100 , so
5 a = 40 5a=40 5 a = 40 and
a = 8 a=8 a = 8 pounds of almonds (with
12 12 12 pounds of peanuts). Check:
8 × 8 + 3 × 12 = 64 + 36 = 100 = 20 × 5 8\times 8 + 3\times 12 = 64+36 = 100 = 20\times 5 8 × 8 + 3 × 12 = 64 + 36 = 100 = 20 × 5 ✓. (Alligation confirms it: the ratio almonds:peanuts
= ( 5 − 3 ) : ( 8 − 5 ) = 2 : 3 =(5-3):(8-5)=2:3 = ( 5 − 3 ) : ( 8 − 5 ) = 2 : 3 , so almonds
= 2 5 × 20 = 8 =\tfrac{2}{5}\times 20 = 8 = 5 2 × 20 = 8 .)
For parents
These practice questions come from our SHSAT, TACHS and SSAT practice apps. Each app has more than 1,200 practice questions, every one with a worked explanation.
Checked against official sources. How we make this site Wednesday, September 30, 2026