LCM and HCF Question Types
In this article, we will try to cover all the types of aptitude questions that are framed on the concepts of LCM and HCF.
Type 1: Application of LCM
Type 1a
Q. There are two bells in a temple. One tolls 3 times per minute and the other tolls 4 times per minute, once the temple is opened. Calculate how many times will they toll together in first 30 minutes, after the temple is opened?
(a) 29 (b) 30 (c) 31 (d) 32
Explanation:
One bell tolls 3 times in a minute. Thus, it tolls every 60/3 = 20 seconds.
Second bell tolls 4 times in a minute. Thus, it tolls every 60/4 = 15 seconds.
The time interval in which they will toll together = LCM (15, 20) = 60 seconds = 1 minute.
Thus, in 30 minutes they will toll together 30 times.
Answer: (b)
Type 1b
Q. If LCM of two numbers is 198, then what is the least possible value of their sum?
(a) 30 (b) 31 (c) 32 (d) 29
Explanation:
We know that the two numbers are multiples of the factors of the given LCM. So, first we need to find out the factors of the given LCM.
Now, 198 = 2 × 3 × 3 × 11
Hence, 1, 2, 3, 6, 9, 11, 18, 22, 33, 66, 99 and 198 are the factors of 198.
In order to get the pair of numbers, such that their LCM is 198, and their sum is the least possible, they should be as close to each other as possible, i.e. 11 and 18.
Thus, the least possible sum = 11 + 18 = 29.
Answer: (d)
Type 1c
Q. A person X wants to distribute some pens among six children A, B, C, D, E and F. Suppose A gets twice the number of pens received by B, three times that of C, four times that of D, five times that of E and six times that of F. What is the minimum number of pens X should buy so that the number of pens each one gets is an even number?
(a) 147 (b) 150 (c) 294 (d) 300
Explanation:
Let the number of pens with A be the LCM of 2, 3, 4, 5, and 6 = 60
Then the number of pens with B = 60/2 = 30
The number of pens with C = 60/3 = 20
The number of pens with D = 60/4 = 15 (an odd number)
The number of pens with E = 60/5 = 12
The number of pens with F = 60/6 = 10
To ensure that all get an even number of pens, we need to double the number of pens bought by A, i.e. 60 × 2 = 120
So, total number of pens bought by X = 120 + 60 + 40 + 30 + 24 + 20 = 294
Answer: (c)
Type 2: Application of HCF
Q. Consider the information given below and answer the 2 (two) items that follow.
A milkman has three varieties of milk in quantities of 93 liter, 155 liter and 217 liter. He wants to fill them in bottles of equal size without mixing any of them.
Q1. What is the largest possible size of such bottles?
(a) 31 liters
(b) 33 liters
(c) 45 liters
(d) 29 liters
Q2. What is the minimum possible number of such bottles required?
(a) 20
(b) 18
(c) 16
(d) 15
Explanation:
Answer 1: (a)
Bottles have to be of equal size and the three varieties of milk should not be mixed. Thus the size of the bottle must be a factor of 93, 155 and 217.
Since, the maximum possible size of the bottle is to be found out, we will have to find the highest possible common factor or HCF of 93, 155 and 217.
HCF of (93, 155, 217) = 31
Thus, the largest possible size of such bottles = 31 liters
Answer 2: (d)
As minimum possible number of bottles is asked, the size of bottle has to be the maximum possible. We already know that the maximum possible size is 31 liters.
Thus, number of bottles needed to fill 93 liters of the first variety of milk = 93/31 = 3 bottles
Number of bottles needed to fill 155 liters of the second variety of milk = 155/31 = 5 bottles
Number of bottles needed to fill 217 liters of the third variety of milk = 217/31 = 7 bottles
Thus, minimum possible number of such bottles required = 3 + 5 + 7 = 15 bottles.
Type 3: Based on relationship between LCM and HCF
Q. Ratio of two numbers is 3 : 5 and their HCF is 10. What must be the half of their LCM?
(a) 150 (b) 300 (c) 75 (d) 225
Explanation:
The ratio of the two numbers is 3 : 5. So let’s assume the numbers to be 3x and 5x.
Now, HCF of 3x and 5x = x = 10 (given)
Thus, the two numbers are 30 and 50.
LCM of (30, 50) = 150
Thus, half of their LCM = 150/2 = 75.
Answer: (c)