1. Primary 4
Multiplication
and
Division Word Problems

2. Multiplication & Division Problem Sums
There were 4 times as many children as adults in a
party. After 16 adults left, the number of children in
the party was 20 times the number of the remaining
adults. How many people attended the party?

3. Multiplication & Division Problem Sums
There were 4 times as many children as adults in a
party. After 16 adults left, the number of children in
the party was 20 times the number of the remaining
adults. How many people attended the party?
Extract Info:
C -> 4u Draw the models
A -> 1u according to your
extracted info.
Notice the number of children
remains unchanged.
Therefore, we multiply 4u by
5 to get 20u for the children.
Logically, we also multiply 1u
by 5 for the adults.
C -> 20u ?
A -> 5u - 16

4. Multiplication & Division Problem Sums
There were 4 times as many children as adults in a
party. After 16 adults left, the number of children in
the party was 20 times the number of the remaining
adults. How many people attended the party?
Extracted Info
C -> 20u
A -> 5u - 16
C
?
A
16

5. Multiplication & Division Problem Sums
There were 4 times as many children as adults in a
party. After 16 adults left, the number of children in
the party was 20 times the number of the remaining
adults. How many people attended the party?
Extracted Info
C -> 20u
A -> 5u - 16
C
?
A
16
4u -> 16
1u -> 16 ÷ 4 = 4

6. Multiplication & Division Problem Sums
There were 4 times as many children as adults in a
party. After 16 adults left, the number of children in
the party was 20 times the number of the remaining
adults. How many people attended the party?
Extracted Info
C -> 20u
A -> 5u - 16
C
?
A
16
4u -> 16
1u -> 16 ÷ 4 = 4
25u -> 25 x 4 = 100
100 people attended the party.

7. Multiplication & Division Problem Sums
Mark went shopping with twice as much money as
Paul. After Paul spent $60, Mark had 6 times as
much money as Paul. How much more money did Mark
bring along for shopping than Paul?

8. Multiplication & Division Problem Sums
Mark went shopping with twice as much money as
Paul. After Paul spent $60, Mark had 6 times as
much money as Paul. How much more money did Mark
bring along for shopping than Paul?
Extract Info:
Mark -> 2u Draw the models
Paul -> 1u according to your
extracted info.
Notice Mark's money remains
unchanged. Therefore, we
multiply 2u by 3 to get 6u for
Mark. Logically, we also multiply
1u by 3 for Paul.
Mark -> (2u x 3) = 6u
Paul -> (1u x 3) = 3u - 60

9. Multiplication & Division Problem Sums
Mark went shopping with twice as much money as
Paul. After Paul spent $60, Mark had 6 times as
much money as Paul. How much more money did Mark
bring along for shopping than Paul?
Extract Info:
Mark -> (2u x 3) = 6u
Paul -> (1u x 3) = 3u - 60
Mark
?
Paul
60

10. Multiplication & Division Problem Sums
Mark went shopping with twice as much money as
Paul. After Paul spent $60, Mark had 6 times as
much money as Paul. How much more money did Mark
bring along for shopping than Paul?
Extract Info:
Mark -> (2u x 3) = 6u
Paul -> (1u x 3) = 3u - 60
Mark 2u -> 60
?
Paul
60

11. Multiplication & Division Problem Sums
Mark went shopping with twice as much money as
Paul. After Paul spent $60, Mark had 6 times as
much money as Paul. How much more money did Mark
bring along for shopping than Paul?
Extract Info:
Mark -> (2u x 3) = 6u
Paul -> (1u x 3) = 3u - 60
Mark 2u -> 60
1u -> 60 ÷ 2 = 30
?
Paul
60

12. Multiplication & Division Problem Sums
Mark went shopping with twice as much money as
Paul. After Paul spent $60, Mark had 6 times as
much money as Paul. How much more money did Mark
bring along for shopping than Paul?
Extract Info:
Mark -> (2u x 3) = 6u
Paul -> (1u x 3) = 3u - 60
Mark 2u -> 60
1u -> 60 ÷ 2 = 30
? 3u -> 30 x 3 = 90
Paul
60 Mark brought $90 more
than Paul.

13. Multiplication & Division Problem Sums
The total cost of 3 erasers and 5 pens is $52. Each
pen costs $8. Find the cost of each eraser.
Extract Info:
3E + 5P = 52 (Total cost)
P = 8
We know the cost of 1 pen. So let's find the cost of
5 pens.

14. Multiplication & Division Problem Sums
The total cost of 3 erasers and 5 pens is $52. Each
pen costs $8. Find the cost of each eraser.
Extract Info:
3E + 5P = 52 (Total cost)
P = 8
P -> $8
5P -> 5 x $8 = $40

15. Multiplication & Division Problem Sums
The total cost of 3 erasers and 5 pens is $52. Each
pen costs $8. Find the cost of each eraser.
Extract Info:
3E + 5P = 52 (Total cost)
P = 8
P -> $8
5P -> 5 x $8 = $40
Cost of 3 erasers -> Total cost – Cost of 5 pens
-> $52 – $40
-> $12

16. Multiplication & Division Problem Sums
The total cost of 3 erasers and 5 pens is $52. Each
pen costs $8. Find the cost of each eraser.
Extract Info:
3E + 5P = 52 (Total cost)
P = 8
P -> $8
5P -> 5 x $8 = $40
Cost of 3 erasers -> Total cost – Cost of 5 pens
-> $52 – $40
-> $12
Cost of 1 eraser -> $12 ÷ 3 = $4

17. Multiplication & Division Problem Sums
The total length of 4 ropes and 6 strings is 334cm.
The length of each rope is 46cm. Find the length of
each string.
Extract Info:
4R + 6S = 334 (Total length)
R = 46
We know the length of 1 rope. So let's find the
length of 4 ropes.

18. Multiplication & Division Problem Sums
The total length of 4 ropes and 6 strings is 334cm.
The length of each rope is 46cm. Find the length of
each string.
Extract Info:
4R + 6S = 334 (Total length)
R = 46
We know the length of 1 rope. So let's find the
length of 4 ropes.
R -> 46 cm
4R -> 46 x 4 = 184 cm

19. Multiplication & Division Problem Sums
The total length of 4 ropes and 6 strings is 334cm.
The length of each rope is 46cm. Find the length of
each string.
Extract Info:
4R + 6S = 334 (Total length)
R = 46
We know the length of 1 rope. So let's find the
length of 4 ropes.
R -> 46 cm
4R -> 46 x 4 = 184 cm
Length of 6 strings -> Total length – 4R
-> 334 – 184
-> 150 cm

20. Multiplication & Division Problem Sums
The total length of 4 ropes and 6 strings is 334cm.
The length of each rope is 46cm. Find the length of
each string.
Extract Info:
4R + 6S = 334 (Total length)
R = 46
We know the length of 1 rope. So let's find the
length of 4 ropes.
R -> 46 cm
4R -> 46 x 4 = 184 cm
Length of 6 strings -> Total length – 4R
-> 334 – 184
-> 150 cm
Length of 1 string -> 150 ÷ 6 = 25 cm

21. Multiplication & Division Problem Sums
The total mass of 3 watermelons and 8 mangoes is
76kg. Each watermelon weighs 12kg. Find the mass of
each mango.
Extract Info:
3W + 8M = 76 (Total mass)
W = 12
We know the mass of 1 watermelon. So let's find the
mass of 3 watermelons.

22. Multiplication & Division Problem Sums
The total mass of 3 watermelons and 8 mangoes is
76kg. Each watermelon weighs 12kg. Find the mass of
each mango.
Extract Info:
3W + 8M = 76 (Total mass)
W = 12
We know the mass of 1 watermelon. So let's find the
mass of 3 watermelons.
W -> 12 kg
3W -> x =
Mass of 8 mangoes -> Total mass – 3W
Mass of 1 mango -> Mass of 8 mangoes ÷ 8 =