One of the very common and interesting topics taught to kids at an early age is Highest Common Factor or H.C.F. Those, who want to practice H.C.F Word Problems with Answers can get them all in one place. In this article, you will find Word Problems on H.C.F covering various questions. Know how questions are framed on the topic HCF by referring to the HCF Example Problems and get an idea on how to find HCF Quickly. The HCF Sums provided here include questions on finding the Highest Common Factor of 2 or more numbers.
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Word Problems Involving Highest Common Factor
Example 1:
Ram has 60 liters of oil and 90 liters of oil. He wants to sell oil by mixing oils in tins of equal volumes. What is the greatest of such tin?
Solution:
We have to fill the tins with equal volumes.
We have to find the bigger number which exactly divides 60, 90. i.e. nothing but the H.C.F of (60, 90).
H.C.F(60, 90)=30.
The first kind 60 liters is sold in two tins of volumes 30 liters in each tin.
The second kind of oil 90 liters is sold in three tins of volume 30 liters in each tin.
Therefore, the greatest volume of each tin is 30 liters.
Example 2:
A fruit juice owner has three types of juices. 320 liters of the first kind, 350 liters of the second kind, 400 liters of the third kind. Find the least possible number of bottles of equal size in which different types of juices can be filled without mixing?
Solution:
For the least possible number of bottles of equal size, the size of the bottle must be great volume.
To get the greatest volume of equal size, We have to find the greatest number that exactly divides 320 liters, 350 liters, 400 liters.
H.C.F(320,350,400)=10
Each bottle must be of the volume 10 liters.
Required no. of bottles= (320/10+350/10+400/10)
=32+35+40=107
Therefore, the least possible number of equal sizes of bottles required is 107.
Example 3:
Sailaja invited her friends. she has 20 samosas and 15 sweets. She wants to make all the plates similar without any food leftover, what is the greatest number of plates Sailaja can prepare?
Solution:
To make all the plates similar and find the greatest number of plates, we have to find the greatest number that can divide 20, 15.
H.C.F(20, 15)=5
20 samosas are filled in 5 plates with 4 samosas.
15 sweets are filled in 5 plates with 3 sweets.
On each plate, there will be 4 samosas, 3 sweets. All 5 plates will be similar.
Therefore, the greatest number of plates Sailaja can prepare is 5 plates.
Example 4:
Laxman has two pieces of wire, one is 8m long and another is 12m long. He wants to cut them of equal length with no pieces of wire left. Find the greatest length of wire he can make?
Solution:
When the two wires are cut into small pieces, each piece must of the same length, and also it has to be the possible greatest length.
8 feet wire can be cut into pieces of (2, 2, 2,2) or (4, 4)
12 feet wire can be cut into pieces of (2, 2, 2, 2, 2, 2 ) or (3, 3, 3, 3)
The length of each small piece must be the greatest length.
To find the possible greatest length, we have to find the greatest number which can divide both 8 and 12. That is H.C.F of (8,12).
H.C.F of (8,12)=4
Hence, the greatest length of each small piece will be 4 cm.
8 feet wire is cut into two pieces.
12 feet wire is cut into 3 pieces.
Example 5:
Raju has 25 white chairs and 15 red chairs. They want to arrange the chairs in such a way that each row contains an equal number of chairs and also each row should have only red chairs or green chairs. What is the greatest number of chairs that can be arranged in each row?
Solution:
In order to find the greatest number of chairs that can be arranged in equal rows, we have to find the HCF of two numbers.
Factors of 25 are 1, 5,25
Factors of 15 are 1,3,5,15.
The highest common factor is 5.
Therefore the greatest number of chairs can be arranged in a row is 5.
Example 6:
Laya has 10 chocolates and 15 biscuits. If she wants to give all her friends the same without any leftovers, Find how many maximum friends she can give?
Solution:
she wants to give all her friends the same, we have to find the greatest number which can divide 10,15.
H.C.F of (10, 15) is 5.
10 chocolates can be served to 5 people with 2 chocolates each.
15 biscuits can be served to 5 people 3 biscuits each.
Therefore, each member will be given 5 people and 3 biscuits.
Example 7:
A milkman has two kinds of milk 80 liters, 280 liters. He wants to sell milk by mixing two kinds of milk in tins of equal volumes. What is the greatest of such a tin?
Solution:
The volume of the milk filled in the tin is greatest.
We have to find the bigger number that exactly divides 80, 280. It is nothing but H.C.F of (80,280)
H.C.F(80,280)=40.
Water 80 liters is sold in two tins of volume 40 liters each.
Milk 280 liters are sold in seven tins of volume 40 liters each.
Therefore, the greatest volume of each tin is 40 liters.
Example 8:
The sum of two numbers is 100. HCF is 20. Find how many such pairs can be formed?
Solution:
As H.C.F is 20.
So the two numbers are 20x, 20y.
20x+20y=100
Divide each side by 20.
X+y=5
We have to find the values of x and y such that the sum is 5.
The possible values of x,y are (1,4),(2,3)
Here we have to check for Co-primes i.e. if they have no common positive factor other than 1.
In the above pairs (2,3) is Co-primes.
Hence the number of pairs is 1.