The proof consists merely in reversing the positions of the multiplicand and multiplier, and performing the multiplication in the usual way. If the product is the same in both cases the work is correct.

TALK:

PROBLEMS IN MULTIPLICATION.

The child's early work in numbers has taught him how to find the cost of a number of things when he has the cost of one.

So in his work with measures, he has learned to find the number of small units in any number of larger units by multiplying, though perhaps he did it without calling it multiplication.

Many facts similar to these have been made familiar to him in his work with small numbers. When he comes to deal with problems involving large numbers he may be confused. At least some children will be. All do not grasp ideas equally well at the same stage of the work. Then, too, pupils are not equally bright every day.

Simple problems should be given the pupil throughout his work on any of the operations. Large numbers may be used as he gets familiar with their use in examples.

For pupils who have difficulty with problems in multiplication, the following pages are included. The simple, oral ones at first, with solutions, may be valuable for the slower pupils. All may be benefited by the solutions to the written problems given here.

Impress that where the cost of one of anything is given, to find the cost of many of the same, multiply. Explain why, frequently or until all understand.

EXAMPLE embraces any work =, etc. When a concrete ques

NOTE. In this book the term merely indicated, as,

42 × 25

tion for computation is asked it is termed a PROBLEM.

PROBLEM:

ORAL PROBLEMS.

Yesterday I bought a pencil for 34. To-day I want 5 more. What will they cost me?

SOLUTIONS:

1. Three pencils will cost me 3 times 5 = 15%. 2. They will cost me 3×5 = 15o.

NOTE. Either one is good.

PROBLEM:

I buy a book for 30. What will three more of the same kind cost me ?

SOLUTION:

They will cost me 3×30 = 90%.

WORK:

30%

× 3

90%

PROBLEM:

In one week there are 7 days. How many days in 4 weeks?

SOLUTION:

In 4 weeks there are 4×7 days = 28 days.

PROBLEM:

How many days in 9 weeks?

SOLUTION:

There are 9×7 days = 63 days.

PROBLEM:

I have 7 marbles. John has 5 times as many. How many has John?

PROBLEM:

SOLUTION:

He has 5×7 marbles = 35 marbles.

What is the cost of 20 sheep at $6 each?

Find the cost of 85 chickens @ 40%.

What is the cost of 45 A. of land @ $35?

SOLUTION:

45 acres will cost 45 × $35 = $1575.

WORK:

$35

X 45

175

140

$1575

PROBLEM:

What will Jones pay for 640 A. of land @ $34?

SOLUTION:

He will pay 640 × $34 = $21,760.

WORK:

$34

× 640

1360

204

$21,760

PROBLEM:

What is the yield from 187 A., if the average yield is 17 bushels?

NOTE. Such problems as this often give difficulty because the child does not understand the meaning. Be careful to explain all such.

SOLUTION:

The yield will be 187 × 17 bu.=3179 bu.

WORK:
187

X 17 bu.

1309

3179 bu.

PROBLEM:

The area of Belgium is 11,373 square miles. It has 551 people to the square mile. How many people in Belgium?

SOLUTION:

There are 11,373 × 551 people = 6,266,523 people.

WORK:

11,373
× 551

11373

56865

56865

6,266,523

NOTE. If there is trouble here lead up to an understanding of it as before. How many people to 2 square miles? 3 square miles ? It may this way:

be illustrated to the pupils in

Let each square represent a square mile.
Find how many people on 4

square miles, etc.

Devices like this at this time should not be necessary. If they are, use them.

Explain how the result in such a problem is only approximately the population of Belgium.

PROBLEM:

From Albany to Buffalo is 297 miles. It takes 158 tons of steel rails to lay one mile of track. How many tons of rails will it take to lay a track from one city to the other!

SOLUTION:

It will take 297 times 158 tons = 46,926 tons.