Word Problems on Multiplication and Division of Whole Numbers

These word problems are from multiplication and division and would help you assess the areas where we will have to multiply or divide by reading the story of each sums. When we are usually finding the value of lesser units then we actually need to find out the value of 1 unit first and for that division is needed. Multiplication is another form of repeated addition so when the values of more units are required we adhere to multiplication.

1. There are 250 toffees in each packet. Find the number of toffees in 25 such packets.

Solution:

                                TH      H       T        O

1 packet contains =              2       5        0

25 packet contains =       ×            2        5                                 

                               1         2        5         0

                               5         0        0         0  

Total no. of toffees    6         2        5         0

EXPLANATION:

Here in this sum it is said that there are 250 toffees in each packet therefore the number of toffees in 25 such packets will be more and more means we have to do either addition or multiplication. If we opt for addition then we will have to add 250 toffees 25 times for 25 packets which is repeated addition. Instead of repeated addition we can do multiplication. In the first line with 5 the multiplier 250 is multiplied and in the second line 250 is multiplied with 2. Then both the products are added to get the final answer.

2. 135000 m of rope is to be cut into 15 pieces. Find the length of each piece.

Solution:

Length of 15 pieces = 135000 m

Length of 1 piece     = 135000 ÷ 15

Therefore, the length of each piece is 900 m

EXPLANATION:

Here the total length of 15 piece of rope is given. Now this rope has to be cut into 15 pieces. Therefore the length of each piece will be of smaller length or we can say that the whole length of rope has to be divided into 15 pieces. Hence the length of each piece will be of smaller value than the total length. So we will have to do division. Here 13500 is the dividend and 15 is the divisor. 9 times 15 is 135 and then there are two zeros which will come in the quotient. Hence the answer is 900 m.

3. The cost of 5 pens is $ 100. Find the cost of 200 such pens.

Solution:

Cost of 5 pens =             $ 100

Cost of 1 pen will be = $ 100 ÷  5

Cost of 1 pen =    $ 20

Cost of 200 pens = $ 200 × 20

                  TH      H        T       O

                                    2          0          0

                                 ×          2          0

                                     0          0          0

                         4          0          0          0

                         4          0          0          0

Therefore the cost of 200 pens is $ 4000

EXPLANATION:

Here the cost of 5 pens is given as $ 200 and it is asked to find the price of 200 such pens. So firstly we have to find the price of each pen by dividing 200 by 5. As the cost of each pen will be less so division.  Now after finding the cost of each pen we will have to find the cost of 200 such pens. We will have to multiply as the cost of 200 such pens will be more.

From Word Problems on Multiplication and Division of Whole Numbers to HOME PAGE

ShareFacebookTwitterPinterestWhatsAppMessenger