HCF and LCM Test

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HCF and LCM Test - 1

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Question 1
1 / -0

A B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 126 seconds, B in 154 seconds and c in 99 seconds, all starting at the same point. After what time will they again at the starting point?

A
26 minutes and 18 seconds

B
23 minutes and 6 seconds

C
45 minutes

D
46 minutes and 12 seconds

Solution

We have to find the L.C.M. of 126, 154, 99
2 126,154,99
3 63,77,99
11 21,77,33
3 21,7,3
7 7,7,1
1,1,1

LCM = 2 x 3 x 11 x 3 x 7 x 1 x 1 x 1 = 1386

So, A, B, and C will again meet at the starting point in 1386 sec. i.e., 23 min. 6 sec.
Question 2
1 / -0

LCM of 5, 12 and 24 is:

A
120

B
140

C
160

D
80

Solution

So 2 x 2 x 2 x 3 x 5 = 120
Question 3
1 / -0

The H.C.F. and the L.C.M. of two numbers are 80 and 320 respectively. If one of the numbers is 640, find the other one.

A
40

B
60

C
120

D
140

Solution

Given
H.C.F. = 80
L.C.M. = 320
One number = 640
We know that,
Product of H.C.F. and L.C.M. of two numbers is equal to product of those two numbers
For other number,
80 × 320 = 640 × other number
25600 / 640 = other number
40 = other number
Hence, the other number is 40
Question 4
1 / -0

The LCM of 154 and 98 is

A
1124

B
1078

C
1486

D
1248

Solution

2 154,98
7 77,49
7 11,7
11 11,1
1,1
Question 5
1 / -0

H.C.F of (200, 300) is:

A
150

B
250

C
100

D
50

Solution

• Find the prime factorization of 200
200 = 2 × 2 × 2 × 5 × 5
• Find the prime factorization of 300
300 = 2 × 2 × 3 × 5 × 5
• To find the H.C.F, multiply all the prime factors common to both numbers:
Therefore, HCF = 2 × 2 × 5 × 5
• HCF = 100
Question 6
1 / -0

In a morning walk, three persons step off together. Their steps measure 160 cm, 170 cm, and 180 cm respectively. What is the minimum distance each should walk so that all can cover the same distance in complete steps?

A
73140 cm

B
7234 cm

C
73440 cm

D
6187 cm

Solution

We have to find the LCM of 80 cm, 85 cm, and 90 cm
2 160,170,180
2 80,85,90
5 40,85,45
3 8,17,9
LCM = 2 x 2 x 5 x 3 x 8 x 17 x 9= 73440 cm.
The required minimum distance is 73440 cm.
Question 7
1 / -0

HCF and LCM of two numbers is 25 and 30 respectively. One of the numbers is 50, find the other number.

A
35

B
15

C
20

D
30

Solution

H.C.F of two numbers = 25
L.C.M of two numbers = 30
One of the number = 50
Other number = H.C.F x L.C.M / One of the number
= 25 x 30 / 50
= 750/50 = 15
Question 8
1 / -0

The product of two numbers is 16,400 and their H.C.F. is 25. Find their L.C.M.

A
650

B
656

C
482

D
584

Solution

Given
Product of two numbers = 16400 and H.C.F. = 25
We know that,
Product of H.C.F. and L.C.M. of two numbers is equal to product of those two numbers
Now, to find L.C.M.
25 × L.C.M. = 16400
L.C.M. = 16400 / 25
L.C.M. = 656
Hence, L.C.M. = 656
Question 9
1 / -0

Find the least number which when divided by 8, 14, 16 and 20 leaves a remainder 3 in each case.

A
280

B
454

C
563

D
720

Solution

Here, we will find LCM of 8, 14, 16 and 20.
2 8,14,16,20
2 4,7,8,10
2 2,7,4,5
2 1,7,2,5
5 1,7,1,5
7
1,7,1,1
1,1,1,1
Thus, LCM = 2 x 2 x 2 x 2 x 5 x 7 = 560
560 is the least number which when divided by the given numbers will leave remainder 0 in each case.
Therefore, the required number is 3 more than 560. The required least number = 560 + 3 = 563
Question 10
1 / -0

What is the L.C.M of 11 and 7?

A
7

B
11

C
77

D
111

Solution

7 7,11
11 1,11
1,1

So 7 x 11 = 77
Question 11
1 / -0

In a morning walk, three persons step off together from the same spot. Their steps measure 72cm, 64 cm and 80 cm respectively. What is the minimum distance each should cover so that all can cover the distance in complete steps?

A
3260

B
2880

C
3620

D
4200

Solution

The distance covered by each one of them is required to be the same as well as minimum. The required minimum distance each should walk would be the lowest common multiple of the measures of their steps. Thus, we find the LCM of 72, 64 and 80.
LCM = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 5 = 2880 cm.
The required minimum distance is 2880 cm.
Question 12
1 / -0

Determine the greatest 4-digit number exactly divisible by 12, 15 and 21.

A
9660

B
8440

C
7560

D
6540

Solution

Here, we will find LCM of 12, 15 and 21.
2 12,15,21
2 6,15,21
3 3,15,21
5 1,5,7
7 1,1,7
1,1,1

Thus, LCM = 2 x 2 x 3 x 5 x 7 = 420
We need to find greatest 4 digit number multiple of 420
420 x 1 = 420
420 x 2 = 840
Continue this process till we find the greatest 4 digit no..
420 x 23 = 9660
420 x 24 = 10080
Therefore, the greatest 4 digit divisible by 12, 15 and 12 is 9660.
Question 13
1 / -0

HCF of two numbers is 6 and product of two numbers is 4182. Find the LCM of two numbers.

A
482

B
546

C
697

D
528

Solution

H.C.F of two numbers = 6
Product of the two numbers = 4182
L.C.M = Product of two numbers/ H.C.F
= 4182/6 = 697
Question 14
1 / -0

The product of two numbers is 336 and their L.C.M. is 56. Find their H.C.F.

A
14

B
28

C
8

D
6

Solution

Given
Product of two numbers = 336 and L.C.M.= 56
We know that,
Product of H.C.F. and L.C.M. of two numbers is equal to product of those two numbers.
Now, to find H.C.F
H.C.F. × 56 = 336
H.C.F.= 336 / 56
H.C.F. = 6
Hence, H.C.F. = 6
Question 15
1 / -0

In a school, there are two Sections A and B of class X. There are 48 students in Section A and 60 students in Section B. Determine the least number of books required for the library of the school so that the books can be distributed equally among all students of each Section. (2017 OD)

A
140

B
230

C
240

D
180

Solution

Since the books are to be distributed equally among the students of Section A and Section B. therefore, the number of books must be a multiple of 48 as well as 60.
Hence, the required number of books is the LCM of 48 and 60.

48 = 2⁴ × 3
60 = 2² × 3 × 5
LCM = 2⁴× 3 × 5 = 16 × 15 = 240
Hence, the required number of books is 240.
Question 16
1 / -0

If P_nmeans prime factors of n, find P₅₀.

A
4 and 5

B
3 and 6

C
2 and 5

D
5 and 10

Solution

2 50
5 25
5 5
1

So , Prime factors of 50 are 2 and 5.
Question 17
1 / -0

Find the smallest 4 digit number which is divisible by 8, 12 and 14.

A
1126

B
1008

C
1002

D
1140

Solution

Here, we will find LCM of 8, 12 and 14.
2 8,12,14
2 4,6,7
2 2,3,7
3 1,3,7
7 1,1,7
1,1,1

Thus, LCM = 2 x 2 x 2 x 3 x 7 = 168
We need to find smallest 4 digit number multiple of 168
168 x 1 = 168
168 x 2 = 336
168 x 3 = 504
168 x 4 = 672
168 x 5 = 840
168 x 6 = 1008
Therefore, the smallest 4 digit divisible by 8, 12 and 14 is 1008
Question 18
1 / -0

The length, breadth and height of a cardboard box are 55 cm, 65 cm and 75 cm respectively. Find the longest tape which can measure the three dimensions of the room exactly.

A
7

B
6

C
5

D
8

Solution

Longest tape = HCF of 75 cm, 85 cm and 95 cm.

5 55,65,75
11,13,15

Therefore, longest tape = 5 cm
Question 19
1 / -0

Two Petrol tankers contain 750 litres and 900 litres of petrol respectively. Find the maximum capacity of a container which can contain the maximum amount of petrol to fill both the tankers.

A
150

B
250

C
100

D
50

Solution

To find maximum capacity of a container, we will find the HCF of 750 and 900.
2 750,900
3 375,450
5 125,150
5 25,30
5,6
HCF = 2 x 3 x 5 x 5 = 150
Therefore, maximum capacity of the required container is 150 litres. It will fill the first container in 750 ÷ 150 = 5 and the second in 900 ÷ 150 = 6 refills.
Question 20
1 / -0

Six bells commence tolling together and toll at intervals of 2, 4, 6, 8 10, and 12 seconds respectively. In 30 minutes, how many times do they toll together?

A
14 times

B
31 times

C
11 times

D
18 times

Solution

We have to find the L.C.M. of 2, 4, 6, 8, 10, 12
2 2,4,6,8,10,13
2 1,2,3,5,6
3 1,1,3,2,5,3
3 1,1,1,2,5,1
LCM = 2 x 2 x 3 x 3 x 1 x 1 x 1 x 2 x 5 x 1 = 360
So, the bells will toll together after every 360 seconds(6 minutes).
In 60 minutes, they will toll together $\frac{60}{2}$ + 1 = 31 times.

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Re-Attempt Weekly Quiz Competition

2	8,14,16,20
2	4,7,8,10
2	2,7,4,5
2	1,7,2,5
5	1,7,1,5
7	1,7,1,1 1,1,1,1

2	126,154,99
3	63,77,99
11	21,77,33
3	21,7,3
7	7,7,1
	1,1,1

2	154,98
7	77,49
7	11,7
11	11,1
	1,1

2	160,170,180
2	80,85,90
5	40,85,45
3	8,17,9

7	7,11
11	1,11
	1,1

HCF and LCM Test - 1

Result

HCF and LCM Test - 1

Score

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Rank

-

SHARING IS CARING

Question 1

26 minutes and 18 seconds

23 minutes and 6 seconds

45 minutes

46 minutes and 12 seconds

Solution

Question 2

120

140

160

80

Solution

Question 3

40

60

120

140

Solution

Question 4

1124

1078

1486

1248

Solution

Question 5

150

250

100

50

Solution

Question 6

73140 cm

7234 cm

73440 cm

6187 cm

Solution

Question 7

35

15

20

30

Solution

Question 8

650

656

482

584

Solution

Question 9

280

454

563

720

Solution

Question 10

7

11

77

111

Solution

Question 11

3260

2880

3620

4200

Solution

Question 12

9660

8440

7560

6540

Solution