Grade 4 Multiplication and Division

What this problem type covers

Nineteen problem types across five levels: totalling goods sold at two prices, converting a day and some hours into a distance, hours and minutes over several weeks, spacing objects around a pond or along a road, how many more are needed to share equally, the perimeter of one piece of a divided sheet, undoing a calculation done the wrong way, following a made-up operation, counting two-digit numbers with a given remainder, the smallest multiplier that makes a square, the largest quotient and the largest product from five number cards, the multiplier that lands closest to a target, every number meeting three conditions, hidden digits in a multiplication and in a division, the three-digit number whose quotient and remainder add to the most, and comparing two shops with different offers.

Example problems

One example from each difficulty band, with worked solutions.

These problems are generated automatically. The answers are derived as each problem is built and are checked by tests, but mistakes are still possible — please tell us if you find one.

1. 1 · ★1
   There were 7 dozen pencils. 5 dozen were sold at 360 each and the remaining 2 dozen at 290 each. If all were sold, how much was taken? (One dozen is 12 pencils.)
   • ①30,240
   • ②2,380
   • ③24,360
   • ④28,560
   • ⑤28,920
   Show solution

   ④28,560

   1. One dozen is 12, so 5 dozen is 60 pencils.
   2. 60 × 360 = 21,600.
   3. The other 2 dozen is 24 pencils: 24 × 290 = 6,960.
   4. Adding the two gives 28,560.
2. 2 · ★2
   A circular pond has a perimeter of 555 m. Bins are placed every 15 m around it. How many bins are needed? (Ignore the width of the bins.)
   • ①35
   • ②36
   • ③37
   • ④39
   • ⑤38
   Show solution

   ③37

   1. 555 ÷ 15 = 37.
   2. On a closed loop the first and last positions meet, so the count equals the number of gaps.
   3. On a straight line the two ends would add one more; on a circle they do not.
   4. So 37 are needed.
3. 3 · ★3
   A number was meant to be divided by 9, but it was multiplied by 33 instead, giving 1,617. What is the remainder of the correct calculation?
   • ①3
   • ②5
   • ③4
   • ④49
   • ⑤6
   Show solution

   ③4

   1. Find the original number first.
   2. Multiplying by 33 gave 1,617, so the number is 1,617 ÷ 33 = 49.
   3. Done correctly, 49 ÷ 9 = 5 remainder 4.
   4. The remainder is 4.
4. 4 · ★4
   Using each of the cards 9, 8, 0, 1, 5 once, a three-digit number is multiplied by a two-digit number. What is the largest possible product?
   • ①76,950
   • ②78,350
   • ③78,260
   • ④77,435
   • ⑤77,350
   Show solution

   ⑤77,350

   1. For the largest product, the biggest card goes in the tens place of the two-digit number.
   2. The second biggest goes in the hundreds place of the three-digit number.
   3. That gives 910 × 85.
   4. The product is 77,350.
5. 5 · ★5
   Each box holds a digit from 0 to 9. If □2□ ÷ 7 = □5 remainder 6, what do the three hidden digits add up to?
   • ①7
   • ②4
   • ③5
   • ④8
   • ⑤9
   Show solution

   ④8

   1. dividend = divisor × quotient + remainder.
   2. Try each digit for the tens of the quotient in 7 × quotient + 6.
   3. Only a quotient of 45 gives the right tens digit, making the dividend 321.
   4. The three hidden digits add to 8.

Where people usually go wrong

• Forgetting a dozen is twelve. The price is per pencil, not per dozen.
• Reading "one day and eight hours" as eight hours. It is 32.
• Reading leftover minutes as hours. 75 minutes is 1 hour 15 minutes, not 1.75 hours.
• Adding one on a closed loop. Around a pond the first and last positions meet, so the count equals the number of gaps.
• Forgetting to add one along a line. With both ends used there is always one more than the number of gaps.
• Answering with the remainder. If 5 are left over and there are 25 students, 20 more are needed, not 5.
• Answering the area when the perimeter was asked for.
• Dividing the wrong result straight away. Undo the wrong operation first, then do the right one.
• Trusting "biggest over smallest" for the largest quotient. Every way of dealing the cards has to be checked.
• Stopping at the last multiple below the target. The one just above it may be closer.