5 marks

Question 15 Marks

Find the length of the longest rope that can be used to measure exactly the ropes of length 1 m 20 cm, 3 m 60 cm and 4 m

Answer

1 m 20 cm = 120 cm
3 m 60 cm = 360 cm
4 m = 400 cm
This is a HCF related problem.
So, we need to find the HCF of 120, 360 and 400

2	120
2	60
2	30
3	15
	5

2	360
2	180
2	90
3	45
3	15
	5

2	400
2	200
2	100
2	50
5	25
	5

120 = 2 × 2 × 2 × 3 × 5
360 = 2 × 2 × 2 × 3 × 3 × 5
400 = 2 × 2 × 2 × 2 × 5 × 5
HCF = 2 × 2 × 2 × 5 = 40
The length of the longest rope = 40 cm

View full question & answer→

Question 25 Marks

The product of $2$ two digit numbers is $300$ and their HCF is $5.$ What are the numbers?

Answer

Given that HCF of 2 numbers is 5The numbers may like $5 x$ and $5 y$Also given their product $=300$
$ 5 x \times 5 y=300$
$\Rightarrow 25 x y=300$
$\Rightarrow x y=\frac{300}{25}$
$\Rightarrow x y=12 $
The possible values of $x$ and $y$ be $(1,12)(2,6)(3,4)$The numbers will be $(5 x, 5 y)$
$ =(5 \times 1,5 \times 12)=(5,60)$
$=(5 \times 2,5 \times 6)=(10,30)$
$=(5 \times 3,5 \times 4)=(15,20) $
$(5,60)$ is impossible because the given the numbers are two digit numbers.The remaining numbers are $(10,30)$ and $(15,20)$But given that HCF is 5
$(10,30)$ is impossible, because its HCF $=10$The numbers are $15,20$

View full question & answer→

Question 35 Marks

The LCM of two numbers is 210 and their HCF is 14. How many such pairs are possible?

Answer

Product of two numbers = LCM × HCF
x × y = 210 × 14
x × y = 2940
(12, 245), (20, 147)
Two pairs possible.

2	2940
2	1470
5	735
7	147
7	21
	3

View full question & answer→

Question 45 Marks

The traffic lights at three different road junctions change after every 40 seconds, 60 seconds and 72 seconds respectively. If they changed simultaneously together at 8 a.m at the junctions, at what time will they simultaneously change together again?

Answer

This is a HCF related problem.
So, we need to find the HCF of 80, 100, 120

2	80
2	40
2	20
2	10
	5

2	100
2	50
5	25
	5

2	120
2	60
2	30
3	15
	5

80 = 2 × 2 × 2 × 2 × 5
100 = 2 × 2 × 5 × 5
120 = 2 × 2 × 2 × 3 × 5
HCF of 80, 100 and 120 = 2 × 2 × 5 = 20
Volume of the vessel = 20 litres

View full question & answer→

Question 55 Marks

What is the greatest possible volume of a vessel that can be used to measure exactly the volume of milk in cans (in full capacity) of 80 litres, 100 litres and 120 litres?

Answer

This is a HCF related problem.
So, we need to find the HCF of 80, 100, 120

2	80
2	40
2	20
2	10
	5

2	100
2	50
5	25
	5

2	120
2	60
2	30
3	15
	5

80 = 2 × 2 × 2 × 2 × 5
100 = 2 × 2 × 5 × 5
120 = 2 × 2 × 2 × 3 × 5
HCF of 80, 100 and 120 = 2 × 2 × 5 = 20
Volume of the vessel = 20 litres

View full question & answer→

Question 65 Marks

Find the HCF and the LCM of the numbers 154, 198 and 286

Answer

Prime factorisation of 154 = 2 × 7 × 11
Prime Factorisation of 198 = 2 × 3 × 3 × 11
Prime Factorisation of 286 = 2 × 11 × 13

2	154
7	77
	11

To find HCF
Product of common factors of 154, 198 and 286
= 2 × 11
= 22
HCF (154, 198, 286) = 22

2	198
3	99
3	33
	11

2	286
11	143
	13

To find LCM
Product of common factors of atleast two numbers = 2 × 11 = 22
Product of other factors = 7 × 3 × 3 × 13 = 819
LCM (154, 198, 286) = Product of common factors × Product of other factors
= 22 × 819
= 18,018
LCM (154, 198, 286) = 18,018
819
22
1638
1638
18,018

View full question & answer→

Question 75 Marks

Find the prime factorisation number by factor tree method and division method.
60

Answer

∴ 60 = 2 × 2 × 3 × 5
Also

2	60
2	30
3	15
5	5
	1

60 = 2 × 30
= 2 × 2 × 15
∴ 60 = 2 × 2 × 3 × 5

View full question & answer→

Maths 5 marks

Test yourself on this topic