Primary Mathematics Chapter 2 The Four Operations Of Whole Numbers
Chapter 2 The Four Operations Of Whole Numbers Exercise 2.8 Solutions Page 55 Exercise 2.8 Problem 1
We are given to estimate and multiply for 1893 × 4.
So, the nearest round figure for 1893 is 2000.
So, the estimation will be 2000 × 4 = 8000.
Therefore, the estimated product is 2000 × 4 = 8000.
Page 55 Exercise 2.8 Problem 2
We are given to estimate and multiply for 4036×7.
So, the nearest round figure for 4036 is 4000
So, the estimation will be 4000×7=28000
Therefore, the estimated product is 4000×7=28000.
Page 55 Exercise 2.8 Problem 3
We are given to estimate and multiply for 5987×8.
So, the nearest round figure for 6000 is 5987.
So, the estimation will be 4000 × 7 = 28000.
Therefore, the estimated product is 6000 × 8 = 48000.
Primary Mathematics 4A Chapter 2 Step-By-Step Solutions For Exercise 2.8 Page 55 Exercise 2.8 Problem 4
We are given to estimate and multiply for 8195 × 9.
So, the nearest round figure for 8000 is 8195.
So, the estimation will be 8000 × 9 = 72000.
Therefore, the estimated product is 8000 × 9 = 72000.
Page 56 Exercise 2.9 Problem 1
We are given that the hotel received 1320 stems of flowers each month and it is also given that the hotel received for 6 months.
We have to find the total number of stems of flowers did manager received.
Which is the product of a number of stems received per month and a number of months.
= 1320 × 6
= 7920
Therefore, the total number of flowers that the hotel received altogether is 7920.
The Four Operations exercise 2.8 Primary Mathematics Workbook answers Page 56 Exercise 2.9 Problem 2
We are given that a bottle contains red beads and white beads.
The number of red beads is three times the number of white beads.
If there are 1875 red beads then we have to find white beads.
Which is three times of red beads.
= 1875 × 3
= 5625
Therefore, the total number of white beads are 5625.
Page 57 Exercise 2.9 Problem 3
We are given that Mrs Wiley earned $2350 in January.
She earned $500 in February.
In March, she earned twice as much as in February.
We have to find the amount earned by Wiley in March.
Money earned in February is = $2350 + $500 = $2850.
Hence money earned in March is = 2 × $2850 = $5700.
Therefore, the money earned by Wiley in March is $5700.
Common Core Primary Mathematics 4A Chapter 2 Solved Examples For 2.8 Page 58 Exercise 2.9 Problem 4
It is given David bought 2 computers at $3569 each.
He had $2907 left with the money.
We are asked how much money he have at first.
The total money that he have at first is the sum of cost of the computers and the remaining amount.
Cost of two computers is = 2 × 3569
= $7138
The total money that he have at first is = 7138 + 2907
= $10045
Therefore, the total money that he have at first is $10045.
Solutions For The Four Operations Exercise 2.8 In Primary Mathematics 4A Page 58 Exercise 2.9 Problem 5
It is given that 5 people shared a sum of money.
2 of them received Others received
We are asked to find the sum of the money.
Sum of the money is the sum of the two people’s money and the other’s people money.
Two people’s money is = 2 × 4356 = 8712.
Other people’s money is = 3 × 3807 = 11421.
Therefore, total money is = 8712 + 11421 = 20133.
Therefore, the sum of money is $20133.
Page 59 Exercise 2.10 Problem 1
We are given to estimate and then divide for the given numbers.
Given: 292 ÷ 4
We have to round of the large number to the nearest second number’s multiple for estimating.
By rounding off, 292 will become 280.
So
= 280 ÷ 4 = 70
Therefore, by estimating and dividing the given number, we get 70.
Detailed Solutions For Exercise 2.8 The Four Operations In 4A Workbook Page 59 Exercise 2.10 Problem 2
We are given to estimate and then divide for the given numbers.
Given: 378 ÷ 6
We have to round of the large number to nearest second number’s multiple for estimating.
By rounding off, 378 will become 360.
So
= 360 ÷ 6 = 60
Therefore, by estimating and dividing the given number, we get 70.
Page 59 Exercise 2.10 Problem 3
We are given to estimate and then divide for the given numbers.
Given: 651 ÷ 7
We have to round of the large number to nearest second number’s multiple for estimating.
By rounding off, 651 will become 630.
So
= 630 ÷ 7 = 90
Therefore, by estimating and dividing the given number, we get 90.
Step-By-Step Guide For The Four Operations Exercise 2.8 In 4A Workbook Page 59 Exercise 2.10 Problem 4
We are given to estimate and then divide for the given numbers.
Given: 801 ÷ 9
We have to round of the large number to nearest second number’s multiple for estimating.
By rounding off, 801 will become 810.
So
= 810 ÷ 9 = 90
Therefore, by estimating and dividing the given number, we get 90.
Page 60 Exercise 2.10 Problem 5
We are asked to fill the empty cells in the given table using the answers.
Let us find the quotients and remainders.
After that let us fill the table with the correct divisors.
Let us solve the divisions
For 68 ÷ 4, the remainder is 0 and the quotient is 17.
For 96 ÷ 8 remainder is 0 and the quotient is 12.
For 98 ÷ 7 remainder is 0 and the quotient is 14.
For 635 ÷ 5 remainder is 0 and the quotient is 127.
For 963 ÷ 9 remainder is 0 and the quotient is 107.
Therefore the filled table is
Primary Mathematics Workbook 4A Exercise 2.8 The Four Operations Page 61 Exercise 2.11 Problem 1
We are given to estimate and then divide for the given numbers.
Given: 2475÷5
We have to round of the large number to nearest second number’s multiple for estimating.
By rounding off, 2475 will become 2500.
So
= 2500 ÷ 5 = 500
Therefore, by estimating and dividing the given number, we get 500
Page 61 Exercise 2.11 Problem 2
We are given to estimate and then divide for the given numbers.
Given: 3594÷6
We have to round of the large number to nearest second number’s multiple for estimating.
By rounding off, 3594 will become 3600.
So
= 3600 ÷ 6 = 600
Therefore, by estimating and dividing the given number, we get 600.
Chapter 2 The Four Operations Exercise 2.8 Breakdown With Solutions Page 61 Exercise 2.11 Problem 3
We are given to estimate and then divide for the given numbers.
Given: 4214 ÷ 7
We have to round of the large number to nearest second number’s multiple for estimating.
By rounding off, 4214 will become 4200.
So
= 4200 ÷ 7 = 600
Therefore, by estimating and dividing the given number, we get 600.
Page 61 Exercise 2.11 Problem 4
We are given to estimate and then divide for the given numbers.
Given: 6480 ÷ 9
We have to round of the large number to nearest second number’s multiple for estimating.
By rounding off, 6490 will become 6300.
So
= 6300 ÷ 9 = 700
Therefore, by estimating and dividing the given number, we get 700.
Common Core 4A Chapter 2 exercise 2.8 Solutions Page 62 Exercise 2.11 Problem 5
We are given to multiple or divide the given numbers.
Multiplication:
⇒ 4023 × 3 = 12069
⇒ 2370 × 5 = 11850
⇒ 3208 × 9 = 28872
⇒ 7248 × 6 = 43488
Division:
⇒ 5208 ÷ 4 = 1302
⇒ 9207 ÷ 8 = 3069
⇒ 1936 ÷ 8 = 242
⇒ 2520 ÷ 10 = 252
Therefore, by doing multiplication and division, we get
Multiplication:
⇒ 4023 × 3 = 12069
⇒ 2370 × 5 = 11850
⇒ 3208 × 9 = 28872
⇒ 7248 × 6 = 43488
Division:
⇒ 5208 ÷ 4 = 1302
⇒ 9207 ÷ 8 = 3069
⇒ 1936 ÷ 8 = 242
⇒ 2520 ÷ 10 = 252
Common Core 4A Chapter 2 exercise 2.8 solutions Page 63 Exercise 2.11 Problem 6
We are given to replace letter n with a number to make equation true.
Given: n × 6 = 6474
To find – The value of n .
Divide both sides by 6, and we get
⇒ \(\frac{n \times 6}{6}\) = \(\frac{6474}{6}\)
⇒ n = 1079
Therefore, value of n is 1079.
Page 63 Exercise 2.11 Problem 7
We are given to replace letter n with a number to make equation true.
Given: n × 7 = 8904.
To find – The value of n .
Divide both sides by 7, and we get
⇒ \(\frac{n \times 7}{7}\)= \(\frac{8904}{7}\)
⇒ n = 1272
Therefore, value of n is 1272.
Page 63 Exercise 2.11 Problem 8
We are given to replace letter n with a number to make equation true.
Given: n × 5 = 3290.
To find – The value of n.
Divide both sides by 5, and we get
⇒ \(\frac{n \times 5}{5}\)= \(\frac{3290}{5}\)
⇒ n = 658
Therefore, the value of n is 658.
Page 63 Exercise 2.11 Problem 9
We are given to replace letter n with a number to make equation true.
Given: n ÷ 2 = 3845.
To find – The value of n.
Multiply both sides by 2, and we get
\(\frac{n \times 2}{2}\)= 3845 × 2
⇒ n= 7690
Therefore, the value of n is 7690.
Page 63 Exercise 2.11 Problem 10
We are given to replace letter n with a number to make equation true.
Given: n ÷ 3 = 4095.
To find – The value of n.
Multiply both sides by 3, and we get
\(\frac{n \times 3}{3}\) = 4095 × 3
⇒ n = 12285
Therefore, the value of n is 12285.
Page 64 Exercise 2.12 Problem 1
It is given that a baker made meat pies 4
Times the number of vegetable pies.
It is given that there are 4864 meat pies.
We are asked to find how many meat pies are made more than vegetable pies.
For that we have to do number of meat pies – number of vegetable pies.
Let us consider x for vegetable pies.
Hence, from the question
4x = 4864
Or x = \(\frac{4864}{4}\)= 1216
Number of meat pies made more than vegetable pies are
⇒ 4864 − 1216
⇒ 3648
Therefore, 3648 meat pies are made more than vegetable pies.
Page 64 Exercise 2.12 Problem 2
It is given that Margo sold three times as many pears as apples in a week.
She sold 3456 apples.
We are asked to find the number of pears did she sell.
Let us consider number of pears as x.
From the question
3x = 3456
Or x= \(\frac{3456}{3}\)
x = 1152
Therefore, 1152 pears are sold.
Page 65 Exercise 2.12 Problem 3
It is given sue, Pat and Jerry shared $387 equally among themselves.
Sue gave $42 and Pat gave $28.
We are asked to find the money did Pat have in the end.
It is given that $387 is shared equally which means
= \(\frac{387}{3}\)
= 129
Therefore, everyone have $129 with them at start.
Now, Sue gave Pat $42 means
= $129+$42
= 172
And now pat sent $28 back.
= $172 − $28 = 144
Therefore, P at have $144 at the end.
Page 65 Exercise 2.12 Problem 4
It is given that a sum of 1040 was divided into 8 equal parts.
It is given that Alice got 4 parts and bob got 1 part and remaining are shared equally to the other 5 members,
By dividing 8 parts, we get
= \(\frac{1040}{8}\)
= 130
From the question we can say that Bob gets 130.
The remaining are 3 × 130 = 390.
Now Gloria have \(\frac{390}{5}\)
= 78
The amount that shared between them is 78 per each.
The number that bob exceeded than Gloria is
= 130 − 78
= 52
Therefore, Bob has 52 more than Gloria.
Page 67 Exercise 2.13 Problem 1
We are given to find the multiplications for the given numbers.
For up:
B − 21 × 13 = 273.
D − 17 × 39 = 663
F − 37 × 24 = 888.
G − 83 × 79 = 6557.
For down:
A − 28 × 31 = 868.
B − 53 × 45 = 2385.
C − 59 × 63 = 3717.
E − 49 × 14 = 686.
Therefore the products of the given numbers will be
For up:
B − 21 × 13 = 273.
D − 17 × 39 = 663
F − 37 × 24 = 888.
G − 83 × 79 = 6557.
For down:
A − 28 × 31 = 868.
B − 53 × 45 = 2385.
C − 59 × 63 = 3717.
E − 49 × 14 = 686.
Page 69 Exercise 2.14 Problem 1
We are given to estimate and multiply for 52 × 39.
So, the nearest round figure for 52 and 39 is 50 and 40 respectively.
So, the estimation will be 50 × 40=2000.
Therefore, the estimated product is 50 × 40 = 2000.
Page 69 Exercise 2.14 Problem 2
We are given to estimate and multiply for 78 × 33.
So, the nearest round figure for 78 and 33 is 80 and 30 respectively.
So, the estimation will be 80 × 30 = 2400.
Therefore, the estimated product is 80 × 30 = 2400.
Page 69 Exercise 2.14 Problem 3
We are given to estimate and multiply for 29 × 87.
So, the nearest round figure for 29 and 87 is 30 and 90 respectively.
So, the estimation will be 30 × 90 = 2700.
Therefore, the estimated product is 30 × 90 = 2700
Page 69 Exercise 2.14 Problem 4
We are given to estimate and multiply for 92 × 71.
So, the nearest round figure for 92 and 71 is 90 and 70 respectively.
So, the estimation will be 90 × 70 = 6300.
Therefore, the estimated product is 90 × 70 = 6300.
Page 70 Exercise 2.14 Problem 5
We are given to estimate and multiply for 218 × 37.
So, the nearest round figure for 218 and 37 is 200 and 40 respectively.
So, the estimation will be 200 × 40 = 8000.
Therefore, the estimated product is 200 × 40 = 8000.
Page 70 Exercise 2.14 Problem 6
We are given to estimate and multiply for 483×59.
So, the nearest round figure for 483 and 59 is 500 and 60 respectively.
So, the estimation will be 500 × 60 = 30000.
Therefore, the estimated product is 500 × 60 = 30000.
Page 70 Exercise 2.14 Problem 7
We are given to estimate and multiply for 372 × 64.
So, the nearest round figure for 372 and 64 is 400 and 60 respectively.
So, the estimation will be 400 × 60 = 24000.
Therefore, the estimated product is 400 × 60 = 24000.
Page 70 Exercise 2.14 Problem 8
We are given to estimate and multiply for 648 × 78.
So, the nearest round figure for 648 and 78 is 600 and 80 respectively.
So, the estimation will be 600 × 80 = 48000.
Therefore, the estimated product is 600 × 80 = 48000.
Page 71 Exercise 2.15 Problem 1
We are asked to find the value of 36 × 9.
Which means 9 times of 36.
So, now let us find 10 times of 36 and then we can subtract 36 from it.
= 36 × 10 − 36
= 360 − 36
= 324
Therefore, the product of 36 × 9 is 324.
Page 71 Exercise 2.15 Problem 2
We are asked to find the value of 58 × 99.
Which means 99 times of 58.
So, now let us find 100 times of 58 and then we can subtract 58 from it.
= 58 × 100 − 58
= 5800 − 58
= 5742
Therefore, the product of 58 × 99 is 5742.
Page 71 Exercise 2.15 Problem 3
We are asked to find the value of 69×99
Which means 99 times of 69.
So, now let us find 100 times of 69 and then we can subtract 69 from it.
= 69 × 100 − 69
= 6900 − 69
= 6831
Therefore, the product of 69 × 99 is 6831.
Page 71 Exercise 2.15 Problem 4
We are asked to find the value of 87×99.
Which means 99 times of 87.
So, now let us find 100 times of 87 and then we can subtract 87 from it.
= 87 × 100 − 87
= 8700 − 87
= 8613
Therefore, the product of 87 × 99 is 8631.
Page 71 Exercise 2.15 Problem 5
We are asked to find the value of 68 × 50.
Which means 68 times of 50.
Now let us make 50 as 100 by rewriting 68 as 34×2.
We get
= 34 × 2 × 50
= 34 × 100
= 3400
Therefore, the product of 68 × 50 is 3400.
Page 71 Exercise 2.15 Problem 6
We are asked to find the value of 32 × 25.
Which means 32 times of 25.
Now let us make 25 as 100 by rewriting 32 as 4 × 8.
We get
= 8 × 4 × 25
= 8 × 100
= 800
Therefore, the product of 32 × 25 is 800.
Page 71 Exercise 2.15 Problem 7
We are asked to find the value of 84 × 25.
Which means 84 times of 25.
Now let us make 25 as 100 by rewriting 84 as 4 × 21.
We get
= 21 × 4 × 25
= 21 × 100
= 2100
Therefore, the product of 84 × 25 is 2100.
Page 71 Exercise 2.15 Problem 8
We are asked to find the value of 25 × 56 .
Which means 56 times of 25.
Now let us make 25 as 100 by rewriting 84 as 4 × 14.
We get
= 14 × 4 × 25
= 14 × 100
= 1400
Therefore, the product of 25 × 56 is 1400.
Page 72 Exercise 2. 15 Problem 9
We are asked to answer for the given multiplications.
Across:
⇒ 118 × 23 = 2714
⇒ 249 × 31 = 7719
⇒ 329 × 18 = 5922
⇒ 167 × 17 = 2839
⇒ 138 × 11 = 1518
⇒ 249 × 25 = 6225
Down:
⇒ 895 × 31 = 27745
⇒ 676 × 62 = 7719
⇒ 346 × 28 = 9688
⇒ 406 × 53 = 21518
⇒ 119 × 29 = 3451
⇒ 135 × 65 = 8775
Therefore, the product of the given numbers is:
Across:
⇒ 118 × 23 = 2714
⇒ 249 × 31 = 7719
⇒ 329 × 18 = 5922
⇒ 167 × 17 = 2839
⇒ 138 × 11 = 1518
⇒ 249 × 25 = 6225
Down:
⇒ 895 × 31 = 27745
⇒ 676 × 62 = 7719
⇒ 346 × 28 = 9688
⇒ 406 × 53 = 21518
⇒ 119 × 29 = 3451
⇒ 135 × 65 = 8775
Page 73 Exercise 2.16 Problem 1
It is given that Natalie made 14 jars of butter biscuits and 16 jars of jam biscuits.
And it is also given that they are 48 biscuits in each jar.
We have to find the total number of biscuits she made which is equal to product of number of jars and number of biscuits.
The total number of jars is
14 + 16 = 30
Therefore, the total no of biscuits are 30 × 48 = 1440.
Therefore, Natalie made 1440 biscuits altogether.
Page 73 Exercise 2.16 Problem 2
It is given that William bought 12 packets of yoghurt.
Each pack contains 465ml of yoghurt. He used 2500ml of yoghurt.
We have to find the remaining yoghurt.
The remaining yoghurt is equal to subtraction of total yoghurt and used yoghurt.
Total yoghurt is
= 12 × 465
= 5580
Therefore, the remaining is
= 5580 − 2000
= 3580ml
Therefore, 3580ml of yoghurt is left after usage.