HCF & LCM Test

Result

HCF & LCM Test - 2

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
2 / -0.83

The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is:

A
101

B
107

C
111

D
185

Solution

Let the numbers be 37a and 37b .
Then, 37a x 37b = 4107
ab = 3.
Now, co-primes with product 3 are (1, 3).
So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111).
Greater number = 111.
Question 2
2 / -0.83

Q. Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.

A
4

B
7

C
9

D
13

Solution

Given numbers are 43, 91, and 183.
Subtract smallest number from both the highest numbers.
We have three cases:
183 >43; 183 >91 and 91 >43
Therrefore,
183 –43 = 140
183 –91 = 92
91 –43 = 48
Now, we have three new numbers: 140, 48 and 92.
HCF (140, 48 and 92) = 4
The highest number that divides 183, 91, and 43 and leaves the same remainder is 4.
Question 3
2 / -0.83

The H.C.F. of two numbers is 23 and the other two factors of their L.C.M. are 13 and 14. The larger of the two numbers is:

A
276

B
299

C
322

D
345

Solution
Question 4
2 / -0.83

Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then sum of the digits in N is:

A
4

B
5

C
6

D
8

Solution

N = greatest number that will divide 105, 4665 and 6905 leaving the same remainder in each case
⇒N = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305)
⇒N = H.C.F. of 3360, 2240 and 5600 = 1120
Sum of digits in N = (1 + 1 + 2 + 0) = 4
Question 5
2 / -0.83

The greatest number of four digits which is divisible by 15, 25, 40 and 75 is:

A
9000

B
9400

C
9600

D
9800

Solution

Greatest number of 4 −digits is 9999.
Now, 15=3 ×5
25 = 5 ×5
40 = 2 ×2 ×2 ×5
and 75 = 3 ×5 ×5
L.C.M. of 15,25,40 and 75 is 2 ×2 ×2 ×3 ×5 ×5 = 600.
On dividing 9999 by 600, the remainder is 399.
Required number = (9999 −399) = 9600.
Question 6
2 / -0.83

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
4

B
10

C
15

D
16

Solution
Question 7
2 / -0.83

Three number are in the ratio of 3 : 4 : 5 and their L.C.M. is 2400. Their H.C.F. is:

A
40

B
80

C
120

D
200

Solution
Question 8
2 / -0.83

The Greatest Common Divisor of 1.08, 0.36 and 0.9 is:

A
0.03

B
0.9

C
0.18

D
0.108

Solution

Given numbers are 1.08 , 0.36 and 0.90
G.C.D. i.e. H.C.F of 108, 36 and 90 is 18
Therefore, H.C.F of given numbers = 0.18
Question 9
2 / -0.83

The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is:

A
1

B
2

C
3

D
4

Solution

Let the numbers 13a and 13b .
Then, 13a x 13b = 2028
⇒ ab = 12.
Two integers a and b are said to be coprime or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1
Now, the co-primes with product 12 are (1, 12) and (3, 4).
⇒ The required numbers are (13 x 1, 13 x 12) and (13 x 3, 13 x 4).
Question 10
2 / -0.83

The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:

A
74

B
94

C
184

D
364

Solution

L.C.M. of 6, 9, 15 and 18 is 90.
Let required number be 90k + 4, which is multiple of 7.
Least value of k for which (90k + 4) is divisible by 7 is k = 4.
∴Required number = (90 x 4) + 4 = 364.