Telangana SCERT Class 5 Mathematics Chapter 14 Solution – Factors and multiples is available in this page.
Telangana State Board Class 5 Maths Chapter 14 Factors and multiples Solution:
1.) Which of these numbers are odd and which of these numbers are even?
23, 18, 65, 70, 47, 325, 610, 354, 289, 842, 169, 431, 400, 553, 724, 807, 999
Odd | Even |
23, 65, 47,325, 289, 169, 431, 553, 807, 999 | 18, 70, 610, 354, 842, 400, 724 |
2.) Encircle the numbers which are divisible by 5.
10, 25, 70, 52, 45, 68, 94, 85, 100, 71, 20, 58, 43, 235, 400, 353, 255, 91, 78, 420, 32, 99
Which of these numbers are also divisible by 10?
Answer – 10, 70, 100, 20, 400, 420.
3.) Write the first 10 multiples of 5 and 4
a) Multiples of 5 = 5, 10, 15, 20, 25,30,35, 40, 45, 50
b) Multiples of 4 = 4, 8, 12, 16, 20,24,28, 32,36, 40
c) Common multiples of 4 and 5 = 20 ,40
4.) Write the factors of the following numbers.
a) 14 – 1, 2, 7 and 14
b) 24–1, 2, 3, 4, 6, 8, 12, 24
c) 16 – 1, 2, 4, 8, 16
d) 42 – 1, 2, 3, 6, 7, 21, 42
5.) a) Write the first 6 multiples of 3.
Answer – 3, 6, 9, 12, 15, 18.
3 |
× | 1 | 3 |
3 | × | 2 |
6 |
3 |
× | 3 | 9 |
3 | × | 4 |
12 |
3 |
× | 5 | 15 |
3 | × | 6 |
18 |
b) Write the first 6 multiples of 9.
Answer – 9, 18, 27, 36, 45,54
9 |
× | 1 | 9 |
9 | × | 2 |
18 |
9 |
× | 3 | 27 |
9 | × | 4 |
36 |
9 |
× | 5 | 45 |
9 | × | 6 |
54 |
c) Are all the multiples of 3 also multiples of 9?
Answer –
Multipliers of 3 = 3, 6, 9, 12, 15, 18, 21, 28, 27, 30
Multipliers of 9 = 9, 18, 27, 36, 45, 54, 63, 72, 81, 90
No, all the multiples of 3 are not multiples of 9.
d) Are all the multiples of 9 also multiples of 3?
Answer – Yes, all the multiples of 9 are also multiples of 3.
5.)
A) Write the first 6 multiples of 12.
Answer –
First 6 multiples of 12 are = 12, 24, 36, 48, 60, 72.
B) Write the first 3 multiples of 4.
Answer –
First 3 multiples of 4 are = 4, 8, 12.
C) Are all the multiples of 4 also multiples of 12?
Answer –
Multiples of 4
4 |
× | 1 | 4 |
4 | × | 2 |
8 |
4 |
× | 3 | 12 |
4 | × | 4 |
16 |
4 |
× | 5 | 20 |
4 | × | 6 |
24 |
4 |
× | 7 | 28 |
4 | × | 8 |
32 |
4 |
× | 9 | 36 |
4 | × | 10 |
40 |
12 |
× | 1 | 12 |
12 | × | 2 |
24 |
12 |
× | 3 | 36 |
12 | × | 4 |
48 |
12 |
× | 5 | 60 |
12 | × | 6 |
72 |
12 |
× | 7 | 84 |
12 | × | 8 |
96 |
12 |
× | 9 | 108 |
12 | × | 10 |
120 |
So, as you can see all the multiples of 4 are not multiples of 12.
D) Are all the multiples of 12 also multiples of 4?
Answer – All the multiples of 12 are also multiples of 4.
7) Are all numbers which are divisible by 10 also divisible by 2 and 5?
Number |
Divisible by 2 | Divisible by 5 |
10 | ✅ |
✅ |
20 |
✅ | ✅ |
30 | ✅ |
✅ |
40 |
✅ | ✅ |
50 | ✅ |
✅ |
150 |
✅ | ✅ |
210 | ✅ |
✅ |
8.) The teacher gave Julie and Jasmine ribbon of equal length. Jasmine cut piecesof 5 inches each from her ribbon and Julie cut pieces of 7 inches each. Bothgirls had no ribbon left after they had cut their ribbons. What is the minimumlength of ribbon that the teacher could have given to the girls?
Answer –
Julie cut 7 inches pieces.
Julie cut 5 inches pieces.
So, if all the pieces are same kind of length then, minimum length the teacher could have given to the girls are,
(5×7) Inches = 35 inches long.
9) There are 10 boys and 15 girls in a class. The teacher wants to divide thechildren into groups such that each group contains an equal number of boysand girls. What is the greatest number of groups that she can make like this?
Answer –
First we have to know the factors of 10 and 15.
Factors of 10 = 1, 2, 5, 10
Factors of 15 ,= 1, 3, 5, 15
So, to have each team equal number of boys and girls,
Each group can have 2 boys then, (10÷2) = 5 groups.
To fit in 15 girls, each group should have (15÷3) = 5 groups.
So, maximum number of groups she can make is 5 groups, consisting 2 boys and 3 girls.
10) A truck can carry 12 sacks weighing 100 kg each at one time. Another truck can carry 15 sacks weighing 100 kg each at one time. If both trucks carry anequal number of bags in the day, then what is the minimum number of sacksthey could have carried?
Answer –
First truck can carry 12 sacks weighing 100 kg each.
Second truck can carry 15 sacks weighing 100 kg each.
So, to carry equal number of bags the minimum number of sacks they could carry
Factors of 12 = 1, 2, 3, 4, 6, 12
Factors of 15 = 1, 3, 5, 15
As you can see they have a common factor 3. So minimum number of sacks they could have carried is 3 sacks.
11) There are 3 clocks in a shop. One climes after every 5 minutes, the secondafter every 15 minutes and third after every 30 minutes. If all of them have climed at 10 o’clock for how many hours they all again clim (at what time)?
Answer –
1^{st} clock climes at every = 5 minutes.
2^{nd} clock climes at every = 15 minutes
3^{rd} clock climes at every = 30 minutes
All of the clocks climed at 10 o’clock.
So next they will clime at,
1^{st} clock = 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60
2^{nd} clock = 15, 30, 45, 60
3^{rd} clock = 30, 60
Next, all clocks will clime on 30 minutes and 60 minutes.
So, of they climed at 10 o’clock next time will be = 10.30 and 11 o’clock .
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