**HCF and LCM Word Problems Worksheet :**

Worksheet given in this section will be much useful to the students who would like to practice solving word problems on HCF and LCM.

Before look at the worksheet, if you would like to learn the stuff HCF and LCM,

**Problem 1 :**

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?(excluding the one at start)

**Problem 2 :**

A merchant has 120 ltrs of and 180 ltrs of two kinds of oil. He wants the sell oil by filling the two kinds of oil in tins of equal volumes. What is the greatest of such a tin.

**Problem 3 :**

Find the least number of square tiles by which the floor of a room of dimensions 16.58 m and 8.32 m can be covered completely.

**Problem 4 :**

The sum of two numbers is 588 and their HCF is 49. How many such pairs of numbers can be formed ?

**Problem 5 :**

The drama club meets in the school auditorium every 2 days, and the choir meets there every 5 days. If the groups are both meeting in the auditorium today, then how many days from now will they next have to share the auditorium ?

**Problem 6 :**

John is printing orange and green forms. He notices that 3 orange forms fit on a page, and 5 green forms fit on a page. If John wants to print the exact same number of orange and green forms, what is the minimum number of each form that he could print ?

**Problem 7 :**

Two number are in the ratio 3 : 5 and their HCF is 5. Find the numbers.

**Problem 8 :**

Two number are in the ratio 3 : 25 and their LCM is 150. How many such pairs of numbers can be formed ?

**Problem 1 : **

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 ? (excluding the one at start)

**Solution : **

**For example, let the two bells toll after every 3 secs and 4 secs respectively. **

**Then the first bell tolls after every 3, 6, 9, 12 seconds... Like this, the second bell tolls after every 4, 8, 12 seconds... So, if the two bell toll together now, again they will toll together after 12 seconds. This 12 seconds is nothing but the L.C.M of 3 seconds and 4 seconds The same thing happened in our problem. To find the time, when they will all toll together, we have to find the L.C.M of (2, 4, 8, 6, 10, 12). L.C.M of (2, 4, 8, 6, 10, 12) is 120 seconds = 2 minutes. **

**So, after every two minutes, all the bell will toll together. **

**For example, in 10 minutes, they toll together : **

**10/2 = 5 times **

**That is, after 2,4,6,8,10 minutes. It does not include the one at the start Similarly, in 30 minutes, they toll together :**

**30/2 = 15 times**

**(excluding one at the start).**

**Problem 2 :**

A merchant has 120 ltrs of and 180 ltrs of two kinds of oil. He wants the sell oil by filling the two kinds of oil in tins of equal volumes. What is the greatest of such a tin.

**Solution : **

**The given two quantities 120 and 180 can be divided by 10, 20,... exactly. That is, both the kinds of oils can be sold in tins of equal volume of 10, 20,... ltrs. **

**But, the target of the question is, the volume of oil filled in tins must be greatest. **

**So, we have to find the largest number which exactly divides 120 and 180.That is nothing but the H.C.F of (120, 180) **

**G.C.F of (120, 180) = 60 **

**The 1st kind 120 ltrs is sold in 2 tins of of volume 60 ltrs in each tin. **

**The 2nd kind 180 ltrs is sold in 3 tins of volume 60 ltrs in each tin. **

**Hence, the greatest volume of each tin is 60 ltrs.**

**Problem 3 :**

Find the least number of square tiles by which the floor of a room of dimensions 16.58 m and 8.32 m can be covered completely.

**Solution :**

**We require the least number of square tiles. So, each tile must be of maximum dimension. To get the maximum dimension of the tile, we have to find the largest number which exactly divides 16.58 and 8.32. That is nothing but the H.C.F of (16.58, 8.32).**

**To convert meters into centimeters, we have to multiply by 100. **

**16.58 **⋅ 1**00 = 1658 cm**

**8.32 **⋅ **100 = 832 cm **

**H.C.F of (1658, 832) = 2 **

**Hence the side of the square tile is 2 cm Required no. of tiles : **

** = (Area of the floor) / (Area of a square tile)**

**= (1658 **⋅ **832) / (2 **⋅ **2)**

**= 344,864 **

**Hence, the least number of square tiles required is 344,864.**

**Problem 4 :**

The sum of two numbers is 588 and their HCF is 49. How many such pairs of numbers can be formed ?

**Solution :**

**Because the H.C.F is 49, the two numbers can be assumed as 49x and 49y **

**Their sum is 588. So, we have **

**49x + 49y = 588 **

**Divide each side 49. **

**x + y = 12**

**We have to find the values of "x" and "y" such that their sum is 12.**

**The possibles pairs of values of (x, y) are **

**(1, 11), (2, 10), (3, 9), (4, 8), (5, 7), (6, 6) **

**Here, we have to check an important thing. That is, in the above pairs of values of (x, y), which are all co-primes ? **

**[Co-primes = Two integers are said to be co-primes or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1] **

**Therefore in the above pairs, (1, 11) and (5, 7) are the co-primes. **

**Hence, the number of pairs is 2.**

**Problem 5 :**

The drama club meets in the school auditorium every 2 days, and the choir meets there every 5 days. If the groups are both meeting in the auditorium today, then how many days from now will they next have to share the auditorium ?

**Solution : **

If the drama club meets today, again they will meet after 2, 4, 6, 8, 10, 12.... days.

Like this, if the choir meets today, again they will meet after 5, 10, 15, 20 .... days.

From the explanation above, If both drama club and choir meet in the auditorium today, again, they will meet after 10 days.

And also, 10 is the L.C.M of (2, 5).

Hence, both the groups will share the auditorium after ten days.

**Problem 6 :**

John is printing orange and green forms. He notices that 3 orange forms fit on a page, and 5 green forms fit on a page. If John wants to print the exact same number of orange and green forms, what is the minimum number of each form that he could print ?

**Solution :**

**The condition of the question is, the number of orange forms taken must be equal to the number of green forms taken. **

**Let us assume that he takes 10 orange and 10 green forms. **

**10 green forms can be fit exactly on 2 pages at 5 forms/page. But,10 orange forms can't be fit exactly on any number of pages. **

**Because, 3 orange forms can be fit exactly on a page. In 10 orange forms, 9 forms can be fit exactly on 3 pages and 1 form will be remaining. **

**To get the number of forms in orange and green which can be fit exactly on some number of pages, we have to find L.C.M of (3, 5). That is 15. **

**15 orange forms can be fit exactly on 5 pages at 3 forms/page. **

**15 green forms can be fit exactly on 3 pages at 5 forms/page. Hence,the smallest number of each form could be printed is 15.**

**Problem 7 :**

Two number are in the ratio 3 : 5 and their HCF is 5. How many such pairs of numbers can be formed ?

**Solution :**

From the ratio 3 : 5, the following two numbers can be assumed as 3a and 5a.

Let us decompose 3a and 5a into factors as given below.

When we look at the factors of 3a and 5a, we find "a" in common.

This common factor "a" is nothing but the H.C.F of 3a and 5a.

So, we have

a = 5

Then,

3a = 3 ⋅ 5 = 15

5a = 5 ⋅ 5 = 25

Hence, the two numbers are 15 and 25.

**Problem 8 :**

Two number are in the ratio 3 : 25 and their LCM is 150. How many such pairs of numbers can be formed ?

**Solution :**

From the ratio 3 : 25, the following two numbers can be assumed as 3a and 25a.

We can find LCM of 3a and 25a using synthetic division as given below.

Therefore, LCM of (3a, 25a) is

= a ⋅ 3 ⋅ 25

= 75a

**Given :** LCM is 150.

Then, we have

75a = 150

Divide each side by 75.

a = 2

Therefore,

3a = 3 ⋅ 2 = 6

25a = 25 ⋅ 2 = 50

Hence, the two numbers are 6 and 50.

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