Problems on HCF and LCM - Aptitude Questions and Answers

Problems on HCF and LCM Questions and Answers

Learn and practice the chapter "Problems on HCF and LCM" with these solved Aptitude Questions and Answers. Each question in the topic is accompanied by a clear and easy explanation, diagrams, formulae, shortcuts and tricks that help in understanding the concept.

Use of Problems on HCF and LCM

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1. Find the LCM of following three fractions:
36,48,72
22515065

a.72
225
b.36
65
c.144
5
d.288
5

Answer: c.144
5

Explanation:

Tip:
LCM of fraction =LCM of numerators
HCF of denominators


Numerators = 36, 48 and 72.
72 is largest number among them. 72 is not divisible by 36 or 48
Start with table of 72.
72 x 2 = 144 = divisible by 72, 36 and 48
∴ LCM of numerators = 144

Denominators = 225, 150 and 65
We can see that they can be divided by 5.
On dividing by 5 we get 45, 30 and 13
We cannot divide further.
So, HCF = GCD = 5
LCM of fraction =144
5


2. Find the HCF of following three fractions:
36,48,72
75150135

a.12
1350
b.72
225
c.150
36
d.1350
36

Answer: a.12
1350

Explanation:

Tip:
LCM of fraction =LCM of numerators
HCF of denominators


Numerators = 36, 48 and 72.
We can see that they can be divided by 12.
On dividing by 12 we get 3, 4 and 6.
We cannot divide further.
∴ HCF = GCD of numerators = 12

150 is largest number among them. 75 can divide 150, so neglect 75
Let's find LCM of 150 and 135
----------------------------------
5           150           135
----------------------------------
3           30             27
3           10             9
             10             3
----------------------------------
∴ LCM of denominators = 5 x 3 x 3 x 3 x 10 = 1350
HCF of fraction =12
1350


3. Given : Three numbers 17, 42 and 93
Find the largest number to divide all the three numbers leaving the remainders 4, 3, and 15 respectively at the end?


a. 13
b. 17
c. 78
d. 89

Answer: a. 13

Explanation:

Here greatest number that can divide means the HCF
Remainders are different so simply subtract remainders from numbers
17 - 4 = 13; 42 - 3 = 39; 93 - 15 = 78

Now let's find HCF of 13, 39 and 78
By direct observation we can see that all numbers are divisible by 13.
∴ HCF = 13 = required greatest number


4. Find the smallest number which leaves the remainders 13, 41 and 29 at the end when divided by 20, 48 and 36 respectively.

a. 187
b. 713
c. 720
d. 727

Answer: b. 713

Explanation:

Here least number is needed that means we need the Least Common Multiple i.e. LCM
We must now first find LCM of 20, 36 and 48
-----------------------------------------
4           20           36           48
-----------------------------------------
3           5             9             12
             5             3             4
-----------------------------------------
∴ LCM = 4 x 3 x 5 x 3 x 4 = 720

But this is not the answer because there are remainders as per the given condition.
If we observe closely, the difference between the given numbers and remainders is same
20 - 13 = 7; 48 - 1 = 7; 36 - 29 = 7
Difference is same = 7
So simply subtract this difference from LCM.
Number = 720 - 7 = 713


5. A number when divided by 36, 24 and 16, leaves the remainder 11 in each case. Find the smallest value of this number.

a. 36
b. 133
c. 144
d. 155

Answer: d. 155

Explanation:

Here smallest (least) number is needed that means we need the LCM

We must first find LCM of 36, 24 and 16
--------------------------------------
4           16           24           36
--------------------------------------
3           4             6              9
2           4             2              3
             2             1              3
--------------------------------------
∴ LCM = 4 x 3 x 2 x 2 x 1 x 3 = 144

Since remainder is same just add it to this LCM
Number = 144 + 11 = 155