Practicing with Maths Mela Class 5 Solutions Chapter 9 Coconut Farm Question Answer NCERT Solutions improves a student’s confidence in the subject.
Class 5 Maths Chapter 9 Coconut Farm Question Answer Solutions
Coconut Farm Class 5 Maths Solutions
Class 5 Maths Chapter 9 Solutions
(NCERT Pg 119)
Question 1.
Think and answer
35 ÷ 1 =…………
Answer:
We have, 35 ÷ 1 = 35
Question 2.
Write the appropriate multiplication fact for the array shown below. Write two division facts that follow from the multiplication fact.
Answer:
We have,
Then,
8 × 4 = 32
32 ÷ 4 = 8
32 ÷ 8 = 4
Let Us Play (NCERT Pg 120)
Question 3
Identify the numbers that can fill the circles such that the numbers in the squares are the products or the quotients of the numbers in the circles.
Answer:
Let Us Do (NCERT Pg 120-121)
Question 4.
Solve the following multiplication problems. Write two division statements in each case.
Answer:
(i) We have, 30 × 30 = 900
900 ÷ 30 = 30
900 ÷ 30 = 30
(ii) We have, 15 × 60 = 900
900 ÷ 15 = 60
900 ÷ 60 = 15
(iii) We have, 400 × 8 = 3200
3200 ÷ 8 = 400
3200 ÷ 400 = 8
(iv) We have, 200 × 16 = 3200
3200 ÷ 200 = 16
3200 ÷ 16 = 200
In each case, we observe N = D × Q.
Question 5.
Solve the following division problems. Notice the patterns and discuss in class.
Answer:
(i) As there are fifty 3s in 150,
i.e. 50 × 3 = 150
So, 150 + 3 = 50.
(ii) As there are twenty 4s in 80,
i.e. 20 × 4 = 80
So, 80 * 4 = 20
(iii) As there are one hundred 5s in 500,
i.e. 5 × 100 = 500
So, 500 + 5 = 100
(iv) Do same as part (i).
Answer: 10
(v) Do same as part (i).
Answer: 3
(vi) Do same as part (i).
Answer: 10
(vii) Do same as part (i).
Answer: 10
(viii) Do same as part (i).
Answer: 10
(ix) Do same as part (i).
Answer: 10
Patterns in Division and Place Value (NCERT Pg 121)
Question 6.
Answer:
(i) Here, 10 × 100 = 1000
1000 + 10 = 100
(ii) Here, 100 × 10 = 1000
1000 ÷ 100 = 10
(iii) Here, 4 × 400 = 1600
1600 + 4 = 400^n-
(iv) ‘Here, 2 × 1000 = 2000
2000 ÷ 2 = 1000
(v) Here, 20 × 100 = 2000
2000 ÷ 20 = 100
(vi) Here, 37 × 100 = 3700
3700 + 37 = 100
(vii) Here, 3 × 1100 = 3300
3300 ÷ 3 = 1100
(viii) Here, 300 × 11 = 3300
3300 + 300 = 11
(ix) Here, 40 × 100 = 4000
4000 + 40 = 100
Question 7.
(i) Now fill the place value chart.
(a) What is happening to the quotients in each case? Discuss.
(b) What patterns do you notice here?
Answer:
We have,
(a) When a number is divided by 10,100 ,…, then the digits in the quotient shift to the right and the number of zeroes removed from the dividend is same as the number of zeroes in the divisor.
When a number with zeroes at the end is divided by multiples of 10, then the quotient is obtained simply by removing same number of zeroes from the dividend as in the divisor.
(b) The pattern, we observe, is that when a number is divided by 10, 100, 1000,……. and its multiples, the digits in the quotient shift based on the number of zeroes in the divisor and dividend.
(ii)
Answer:
We have,
Let Us Do (NCERT Pg 122)
Question 8.
Sabina cycles 160 km in 20 days and the same distance each day. How many kilometres does she cycle each day?
Answer:
Given, the distance cycled by Sabina in 20 days = 160 km
Then, the distance cycled by Sabina in 1 day = (160 + 20) km
= 8 km
So, Sabina cycles 8 km each day.
Question 9.
How many notes of ₹ 100 does Seema need to carry if she wants to buy coconuts worth ₹ 4200 ?
Answer:
Given, the total cost of the coconuts
= ₹ 4200.
Seema wants to carry only ₹ 100 notes. Then, the required number of ₹ 100 notes
= 4200 ÷ 100
= 42
So, Seema needs to carry 42 notes of ₹ 100.
Question 10.
(i) The owner of an electric store has decided to distribute ₹ 5500 equally amongst 5 of his employees as a Diwali gift. What amount will each employee get?
(ii) What will happen if he distributes the same amount of money among 10 employees? Will each employee get more or less? How much money would he have to distribute if everyone must get the same amount as earlier?
Answer:
(i) Given, the total amount to be distributed
= ₹ 5500
and the number of employees = 5
Then, the amount received by each
employee = ₹(5500 ÷ 5)
= ₹ 1100
(ii) Now, if number of employees = 10 Then, the amount received by each
employee = ₹(5500 ÷ 10)
= ₹ 550
Since, ₹ 550 < ₹ 1100.
So, each employee will get less money. Since, earlier, each employee got ₹ 1100.
Then, total amount to be distributed among 10 employees (₹ 1100 each)
= 10 × 1100
= ₹ 11000
So, the owner would have to distribute ₹ 11000 if everyone must get the same amount of ₹ 1100.
Question 11.
Place the numbers 1 to 8 in the following boxes so that all the four operations, division, multiplication, addition and subtraction are correct. No number must be repeated.
How did you think about solving this?
Is there more than one answer?
Answer:
We have,
Yes, there can be more than one answer.
Question 12.
Fill in the blanks
(i) ________ ÷ 18 = 100.
(ii) ________ ÷ 10 = 610.
(iii) ________ + 100 = 72.
(iv) ________ + 100 = 10.
(v) 870 + ________ = 87.
(vi) ________ + 100 = 70.
(vii) 200 ÷ ________ = 2.
(viii) 130 ÷ ________ = 13.
Answer:
We have,
(i) 1800 + 18 = 100.
(ii) 6100 + 10 = 610.
(iii) 7200 + 100 = 72.
(iv) 1000 ÷ 100 = 10.
(v) 870 ÷ 10 = 87.
(vi) 7000 + 100 = 70.
(vii) 200 ÷ 100 = 2.
(viii) 130 + 10 = 13.
Try It! (NCERT Pg 123)
Question 13.
Answer:
Let Us Solve (NCERT Pg 124)
Question 14.
Solve the following problems using strategies used in the previous question.
(i) 256 ÷ 4
(ii) 545 ÷ 5
(iii) 147 ÷ 7
(iv) 1212 ÷ 6
(v) 648 ÷ 12
(vi) 9648 ÷ 48
(vii) 775 ÷ 25
(viii) 796 ÷ 4
Answer:
We have,
Susie’s Farm in Kerala (NCERT Pg 124)
Question 15.
Susie and Sunitha have a large coconut farm and they have harvested 1,117 coconuts in April. They sold 582 coconuts equally to 6 regular customers. How many coconuts did each customer get? They sold 582 + 6 coconuts to each customer.
Estimate the answer first. Do you realise that each customer will likely get less than 100 coconuts?
Each customer gets 97 coconuts.
Do you think Sunitha’s method is better? Discuss which one you would prefer and why.
Answer:
Yes, Sunitha’s method is better than Susie’s.
We would prefer Sunitha’s method because it is more efficient and direct; involving fewer steps to get the solution in the long division process.
Let Us Learn to Divide (NCERT Pg 125)
Question 16.
(i)
(ii) Sometimes, the divisor (D) does not completely divide the dividend (N) and leaves a remainder (R). What is the relationship between the dividend (N) and divisor (D), quotient (Q) and remainder (R)? Try to find out!
(iii) Is 726 = 4 × 181 ? Yes / No.
So, 726 = 4 × 181 + ________
(iv) Is 902 = 16 × 56 ? Yes / No.
So, 902 = 16 × 56 + ________
Answer:
(i) (a) We have,
Since, 4 × 200 = 800, which is greater than 726.
∴ No, we could not have written 200, in place of 100.
(b) We have,
(ii) The relation between dividend (M), divisor (D), quotient (Q) and remainder (R) is
N = D × Q + R
(iii) No.
Because, 726 = 4 × 181 + 2
(iv) No.
Because, 902 = 16 × 56 + 6
Let Us Solve (NCERT Pg 126)
Question 17.
Rani is planning to host a party. She estimates that 250 guests will attend. She plans to serve one samosa to each guest. Samosas are available in packs of 6 or 8. Which pack should Rani buy? Explain your answer.
Answer:
Given, the number of guests who will attend the party = 250, the number of samosas to be served to each guest = 1 and samosas are available in packs of 6 or 8. Now, the number of packs containing 6 samosas needed is 250 ÷ 6.
So, Rani needs 41 packs and 1 more pack for remaining 4 samosas i.e. 42 packs. Also, the number of packs containing 8 samosas needed is 250 + 8.
So, Rani needs 31 packs and 1 more pack for remaining 2 samosas i.e. 32 packs.
Clearly, in pack of 6 , she will get two extra samosas and in pack of 8 , she will get six extra samosas.
So, Rani should buy 42 packs of 6.
Question 18.
342 students from a school are going on a trip to the Science Park. Each bus can carry a maximum of 41 students.
How many buses does the school need to arrange?
Answer:
Given, total number of students = 342 and the maximum number of students that a bus can carry = 41.
Then, the number of buses needs to be arranged is 342 ÷ 41.
So, the, school needs 8 buses and 1 more bus for remaining 14 students i.e. 9 buses.
Question 19.
Sofia has only ₹ 50 and ₹ 20 notes. She needs to pay ₹ 520 using these notes. How many ₹ 50 and ₹ 20 notes does she need to make ₹ 520 ? Find out the different possible combinations.
Answer:
Given, the total amount to be paid = ₹ 520 and available notes are of ₹ 50 and ₹ 20. Then, the possible combinations of ₹ 50 and ₹ 20 notes are as follows
Question 20.
Three friends decide to split the money spent on their picnic equally. They buy snacks and sweets for ₹ 157 , juice and fruits for ₹ 124 and pulav and paratha for ₹ 136. How much should each person pay to share the cost equally?
Answer:
Given, the price of snacks and sweets
= ₹ 157,
the price of juice and fruits
= ₹ 124
and the price of pulav and paratha = ₹ 136
Then, the total expenses
= ₹ 157 + ₹ 124 + ₹ 136
= ₹ 417
Since, the number of friends = 3
So, the amount to be paid by all of three equally is 417 ÷ 3.
So, each person should pay ₹ 139.
Question 21.
Identify the remainder, if any. Check if
N = D × Q + R
(i) 887 ÷ 3
(ii) 283 ÷ 8
(iii) 745 ÷ 5
(iv) 767 + 26
(v) 530 ÷ 41
(vi) 888 ÷ 67
Answer:
We know that
Dividend (N) = Divisor (D) × Quotient (Q) + Remainder (R)
(i) We have, 887 + 3
So, R = 2
Also,
D × Q + R = 3 × 295 + 2 = 885 + 2
= 887 = N
(ii) We have, 283 + 8
So, R = 3
Also,
D × Q + R = 8 × 35 + 3
= 280 + 3 = 283 = N
(iii) We have, 745 + 5
So, R = 0
Also,
D × Q + R = 5 × 149 + 0
= 745 + 0 = 745 = N
(iv) We have, 767 + 26
So, R = 13
Also, D × Q + R = 26 × 29 + 13 = 754 + 13
= 767 = N
(v) We have, 530 + 41
So, R = 38
Also, D × Q + R = 41 × 12 + 38
= 482 + 38 = 530 = N
(vi) We have, 888 ÷ 67
So, R = 17
Also, D × Q + R = 67 × 13 + 17
= 871 + 17 = 888 = N
(NCERT Pg 127-128)
Question 22.
How much will they earn if they sell the oil at ₹ 175 for 1 L ?
They will earn ₹ 547 × 175. Find out.
Answer:
We have,
So, they will earn ₹ 95,725.
Question 23.
In the hot summer months, tender coconuts are sold for ₹ 35. Ibrahim earns ₹ 8,890 in a week. How many tender coconuts did he sell?
(i) The number of tender coconuts sold by Ibrahim is 8,890 ÷ 35.
.
Ibrahim sold ________ tender coconuts.
Answer:
We have,
So, Ibrahim sold 254 tender coconuts.
(ii) Ibrahim had-bought the tender coconuts for ₹ 20 each. How much extra money did he earn by selling the coconuts at ₹ 35 ?
The cost of ________ coconuts at ₹ 20 each = ________ × ₹ 20 = ₹ ________ He earned ₹ 8,890 from the sale.
The extra amount he earned is ₹ 8890 – ₹…. = ₹ ________
Answer:
Given, the cost price of each tender coconut = ₹ 20
and the selling price of each tender coconut = ₹ 35
Since, he sold 254 tender coconuts. Then, the cost of 254 coconuts at ₹ 20 each
= 254 × ₹ 20 = ₹ 5080.
Also, given, he earned ₹ 8890 from the sale. Then, the extra amount he earned is ₹ 8890 – ₹ 5080 = ₹ 3810.
Division Using Place Value (NCERT Pg 128-129)
Question 24.
Can you tell just by looking at the divisor and dividend, how many digits the quotient would have? Look at the problems above and find this out. Explain your thoughts.
Answer:
Yes, it is possible to estimate the number of digits in the quotient by looking at the divisor and dividend.
First, look at the dividend and try dividing the first digit or first two digits by the divisor. If the division is possible for the first digit then the number of digits from that first digit till end gives the number of digits in quotient.
If not, then try combining first two terms and repeat the process.
e.g. (i) 324 + 3
Here, 3 divides 3 (Hundreds place), then number of digits in quotients will be 3.
(ii) 136 ÷ 6
Here, 6 doesn’t divide (1 Hundreds place), then try 13 which can be divisible by 6.
So, the number of digits in the quotients will be 2.
Let Us Divide (NCERT Pg 130-131)
Question 25.
(i) 7,032 + 6
(ii) 3,005 ÷ 5
(iii) 2,874 ÷ 14
(iv) 9,805 + 32
Compare both solutions. Also, remember to put 0 in the right places.
Answer:
(i) We have,
(ii)
(iii)
(iv)
Let Us Do (NCERT Pg 131)
Question 26.
Find the missing numbers such that there is no remainder. Remember, there could be more than one solution.
Answer:
Question 27
I am a 3-digit number.
- If you divide me by 5, you get 42.
- If you multiply me by 2, you get 420.
What number am I?
Answer:
Given, number + 5 = 42
and number × 2 = 420
Then, number = 42 × 5 = 210
Also, 210 × 2 = 420
So, the number is 210.
Let Us Solve (NCERT Pg 132)
Question 28.
A theatre company can accommodate 45 people during one show.
(i) A total of 475 people bought tickets for a puppet show. How many shows are needed to seat all the people who bought tickets?
(ii) There are 2 shows in a day. How many days will be needed to accommodate all the people?
Answer:
(i) Given, the number of persons accommodated during one show = 45.
Here, the number of persons who bought tickets = 475.
Then, the number of shows needed to seat all the 475 persons is 475 + 45.
So, the company needs to arrange 10 shows and 1 more show for remaining 25 people i.e. 11 shows.
(ii) Given, the number of shows per day = 2 and total number of shows = 11.
Then, the number of days to accommodate all the 475 people in 11 shows is 11 + 2.
So, the number of days will be 5 and 1 more for remaining 1 show i.e. 6 days to accommodate all the people.
Question 29.
Naina bought 5 kg of ice cream as a birthday treat for her 23 friends. 400 g ice-cream was left after everyone had an equal share. How much ice-cream did each of her friends eat?
Answer:
Given, total weight of the ice-cream
= 5 kg
= 5 × 1000 g
= 5000 g
and the weight of the left over ice-cream
= 400 g.
Then, the total weight of the ice-cream which was equally shared
= 5000 g – 400 g = 4600 g
Since, the number of her friends = 23
So, the weight of the ice-cream, which each of her friends ate equally is 4600 + 23.
Therefore, each friend ate 200 g the ice-cream.
Question 30.
Megha packs 15 packets of ragi-oats biscuits for a 4 day group trip. Each packet contains 8 biscuits. There are 6 people in the group. If distributed evenly, how many biscuits can one person have each day.
Answer:
Given, total number of packets = 15 and the number of biscuits in a packet = 8
So, the total number of biscuits = 15 × 8 = 120
Also given, the number of persons in group = 6
and the number of days for which trip is planned = 4
So, the total number of servings = 6 × 4 = 24
Then, the number of biscuits that each person can have on each day is 120 ÷ 24.
So, each person can have 5 biscuits each day.
Question 31.
Solve the following and identify the remainder, if any. Check whether N = D × Q + R in each case.
(i) 9,045 + 5
(ii) 1,034 ÷ 4
(iii) 2,504 ÷ 7
(iv) 8,900 ÷ 15
(v) 9,876 + 32
(vi) 7,506 + 24
Answer:
(i) We have, 9,045 ÷ 5
So, R = 0
Also, D × Q + R = 5 × 1809 + 0 = 9045 = N
(ii) We have, 1,034 + 4
So, R = 2
Also,
D × Q + R = 4 × 258 + 2 = 1032 + 2
= 1034 = N
(iii) We have, 2,504 ÷ 7
So, R = 5
Also,
D × Q + R = 7 × 357 + 5 = 2499 + 5
= 2504 = N
(iv) We have, 8,900 * 15
So, R = 5
Also,
D × Q + R = 15 × 593 + 5 = 8895 + 5
= 8900 = N
(v) We have, 9,876 ÷ 32
So, R = 20
Also, D × Q + R
= 32 × 308 + 20
= 9856 + 20
= 9876 = N
(vi) We have, 7,506 ÷ 24
So, R = 18
Also, D × Q + R = 24 × 312 + 18
= 7488 + 18
= 7506 = N
Question 32.
Find the solutions for part A. Observe the relations between the quotient, divisor and dividend and use it to answer parts B and C.
Answer:
A. We have,
(a) 340 + 34 = 10
(b) 340 ÷ 17 = 20
(c) 680 ÷ 17 = 40
(d) 680 ÷ 34 = 20
(e) 170 + 17 = 10
(f) 680 + 68 = 10
We observe that when a divisor is halved, then the quotient becomes double, keeping the dividend same and when a divisor is double, then the quotient becomes half, keeping the dividend same.
Also, note that when dividend is doubled, then the quotient also gets doubled, keeping the divisor same and when the dividend is doubled, then the quotient becomes four times, keeping the half of divisor.
B. We have,
(a) 192 + 4 = 48
(b) 192 ÷ 8 = 24
(c) 384 ÷ 8 = 48
(d) 384 + 4 = 96
(e) 384 ÷ 8 = 48
(f) 86 + 2 = 43
C. We have,
(a) 352 ÷ 11 = 32
(b) 704 + 22 = 32
(c) 704 + 11 = 64
(d) 352 + 22 = 16
(e) 1,408 + 44 = 32
Question 33.
A company in Mumbai organises cycle rallies from Mumbai to Ratnagiri, Goa every year. They aim to cover 576 km in 12 days.
(i) How much distance should they cycle every day, to cover the distance evenly?
(ii) After reaching Ratnagiri, they rest for 1 day. How much distance should they cycle each day to reach Goa in 4 days? Assume that they cover the distance evenly.
Answer:
Given, the total distance = 576 km and the total number of days to cover 576 km} = 12
(i) Then, the distance that they should cycle each day evenly is 576 ÷ 12.
So, they need to cycle 48 km each day to cover 576 km in 12 days evenly.
(ii) Given, the distance between Mumbai and Ratnagiri = 344 km and the distance between Ratnagiri and Goa = 232 km.
So, in order to reach Goa in 4 days from Ratnagiri, the distance needs to be covered each day is 232 ÷ 4.
So, they need to cycle 58 km each day to reach Goa in 4 days after resting for one day in Ratnagiri.
Question 34.
Given below are a few problems. You may need some additional Information to solve these. Identify the missing information. Write the missing information and find the answer.
(i) A fruit vendor sells 6 baskets of mangoes. Each basket contains 12 mangoes. How much did the vendor earn in total?
(ii) A school has 8 classrooms and each classroom has an equal number of desks. How many desks are there in each classroom?
(iii) Rahul buys 5 cricket bats for his team. The total bill is ₹ 3,500. How much does one bat cost?
(iv) A restaurant serves 125 plates of idlis in a day. The total earnings from selling all the idli plates is ₹ 6250. How many idlis are there in each plate?
Answer:
(i) Given, the number of baskets of mangoes = 6 and the number of mangoes in a basket = 12 Then, the total number of mangoes
= 12 × 6 = 72
The missing information is the price of a mango.
Let the price of a mango be ₹ 10.
Then, the total amount earned = ₹ 10 × 72
= ₹ 720
(ii) Given, the number of classrooms = 8
The missing information is the total number of desk in the school.
Let the total number of desks in the school be 160.
Since, each classroom has same number of desks.
Then, the number of desks in each
classroom = 160 + 8 = 20
(iii) Given, the number of bats bought by Rahul = 5
and the total amount paid = ₹ 3,500.
The missing information is that each bat has the same price.
Then, the cost of a bat = ₹ 3,500 + 5 = ₹ 700
(iv) Given, the number of plates of idlis served in a day = 125
and the total earnings from selling all the idlis = ₹ 6,250
The missing information is the price of an idli.
Let the price of an idlie be ₹ 10.
Then, the total number of idlis = 6,250 + 10
= 625
and number of idlis in each plate
= 625 + 125 = 5
Question 35.
To make one bookshelf, a carpenter needs the following things-
4 long wooden panels
8 short wooden panels
16 small clips
4 large clips
32 screws
The carpenter has a stock of 264 long wooden panels, 306 short wooden panels, 2,400 small clips, 120 large clips and 2,800 screws. How many bookshelves can the carpenter make? Discuss your thoughts.
Answer:
We first write the given data in tabular form
So, it allows to make 87 bookshelves.
So, the number of bookshelves that the carpenter can make
= minimum of 66,38,150,30 and 87
= 30.
Vegetable Market (NCERT.Pg 134)
Question 36.
Munshi Lal has a big farm in Bihar. Every Saturday, he sells the vegetables from his farm at Sundar Sabzi Mandi. Munshi ji maintains a detailed record of the quantity of vegetables he sends to the Mandi and the cost of each vegetable. The following table shows his record book on one Saturday.
His naughty grandson has erased some numbers from his record book. Help Munshi Lal complete the table.
What information is recorded in this table?
Answer:
Given, the cost of 1 kg radish = ₹ 26
Then, the cost of 78 kg radish = ₹ 26 × 78 = ₹ 2,028
Given, the cost of 1 kg potato = ₹ 20
Then, the quantity supplied in
₹ 2,240 = 2,240 ÷ 20 = 112 kg
Given, the cost of 1 kg cabbage = ₹ 32
Then, the cost of 56 kg cabbage = ₹ 32 × 56 = ₹ 1,792
Given, the quantity of green peas supplied = 125 kg
Then, the cost of 1 kg green peas = 3,125 + 125 = ₹ 25.
Now, the total money earned through the sale
= ₹ 2,028 + ₹ 2,240 + ₹ 1,792 + ₹ 3,125
= ₹ 9,185.
Therefore, the table with filled blanks is given below
This table gives the total money earned from the sale of all vegetables.
Let Us Solve (NCERT Pg 134)
Question 37.
Divide the following. Try dividing using place values, whenever you can. Identify the remainder, if any and check whether N = D × Q + R.
(i) 506 ÷ 5
(ii) 918 + 8
(iii) 8,126 ÷ 7
(iv) 9,324 ÷ 4
(v) 876 ÷ 6
(vi) 7,008 + 3
(vii) 934 ÷ 12
(viii) 829 ÷ 23
(ix) 705 ÷ 18
(x) 8,704 ÷ 32
(xi) 6,790 ÷ 45
(xii) 5,074 ÷ 21
Answer:
(i) We have,
Here, N = 506, D = 5, Q = 101 and R = 1.
Now, D × Q + R = 5 × 101 + 1 = 505 + 1
= 506 = N
(ii) Do same as part (i).
Answer: R = 6
(iii) Do same as part (i).
Answer: R = 6
(iv) Do same as part (i).
Answer: R = 0
(v) Do same as part (i).
Answer: R = 0
(vi) Do same as part (i).
Answer: R = 0
(vii) Do same as part (i).
Answer: R = 10
(viii) Do same as part (i).
Answer: R = 1
(ix) Do same as part (i).
Answer: R = 3
(x) Do same as part (i).
Answer: R = 0
(xi) Do same as part (i).
Answer: R = 40
(xii) Do same as part (i).
Answer: R = 13
Mathematical Statements (NCERT Pg 135)
Question 38.
Find out whether the following statements are True (T) or False (F).
A true sentence is one where both sides of the ‘ = ‘ sign have the same value.
(i) 8 × 9 = 70 + 2
(ii) 20-6 = 7 × 3
(iii) 48 ÷ 3 = 4 × 4
(iv) 89-9 = 90 + 0
(v) 25 + 10 = 45 – 10
Answer:
(i) We have, LHS = 8 × 9 = 72
RHS = 70 + 2 = 72
So, statement is T.
(ii) We have, LHS = 20-6 = 14
RHS = 7 × 3 = 21
So, statement is F.
(iii) We have, LHS = 48 + 3 = 16
RHS = 4 × 4 = 16
So, statement is T.
(iv) We have, LHS = 89-9 = 80
RHS = 90 + 0 = 90
So, statement is F.
(v) We have, LHS = 25 + 10 = 35
RHS = 45-10 = 35
So, statement is T.
Question 39
Complete the following statements such that they are true.
(i) 7 × 6 = ________ + 17
(ii) 87 + 6 = ________ × 31
(iii) 63 + ________ = 74 – 4
(iv) ________ + 9 = 16 ÷ 2
Answer:
We have,
(i) 7 × 6 = 42 = 25 + 17
(ii) 87 + 6 = 93 = 3 × 31
(iii) 63 + 7 = 70 = 74 – 4
(iv) 72 + 9 = 8 = 16 + 2
Question 40.
Think about the following statements and find examples as suggested below.
(i) “When two odd numbers are added, the sum is even.” Find 5 examples for the above statement. Can you find an example to show that the statement can be false?
Always true
(ii) “Multiplying a number by 2 can give an odd number.” Give some example for this statement. Can you find any?
Never true
(iii) “Halving a number always leads to an even number.” Give 3 examples for the statement. Can you find 3 examples when this is not true? Sometimes true
Answer:
(i) We can choose the following 5 such pair of numbers
and 3 + 5 = 8 (even)
9 + 11 = 20 (even)
7 + 13 = 20 (even)
103 + 21 = 124 (even)
29 + 3 = 32 (even)
No, we can’t find any such example to show that the sum of two odd numbers is odd.
(ii) No, it is not possible to find any number, whose multiplication with 2 gives an odd number.
e.g. 4 (even) × 2 = 8 (even)
3 (odd) × 2 = 6 (even)
(iii) We can have the following 3 numbers, whose halve are even numbers.
Yes, we cañ find such examples as well where the given statement is not true.
Question 41.
Tick in the appropriate cell for the following statements.
Answer:
(i) The given statement is sometimes true. e.g. 24 + 10 = 34, which is not a multiple of ten, but 20 + 10 = 30, which is a multiple of ten.
(ii) The given statement is never true. e.g. 5 – 3 = 2, but 3 – 5 = -2.
(iii) The given statement is always true.
e.g. 4 × 6 = 24 and (4 × 2) × (6 ÷ 2) = 8 × 3 = 24.
(iv) The given statement is sometimes true. e.g. 3 × 2 = 6 (even), but 3 × 5 = 15 (odd).
(v) The given statement is sometimes true. e.g. 5 × 10 = 50, but 5 × 5 = 25.