HCF and LCM word problems play a major role in quantitative aptitude test. There is no competitive exam without the questions from this topic. We have already learned this topic in our lower classes.Even though we have been already taught this topic in our lower classes, we need to learn some more short cuts which are being used to solve the problems in the above topic.

The only thing we have to do is, we need to apply the appropriate short cut and solve the problems in a limited time. This limited time will be one minute or less than one minute in most of the competitive exams

Students who are preparing to improve their aptitude skills and those who are preparing for competitive exams must prepare this topic in order to have better score. Because, today there is no competitive exam without questions from the HCF and LCM word problems. Whether a person is going to write placement exam to get placed or a student is going to write a competitive exam in order to get admission in university, they must be prepared to solve HCF and LCM word problems. This is the reason for why people must study this topic.

As we mentioned in the above paragraph, a person who wants to get placed in a company must write placement test and a student who wants to get admission in university for higher studies must write entrance exam. To meet the above requirements, it is very important to score more marks in the above mentioned competitive exams. To score more marks, they have to prepare this topic. Preparing this topic would definitely improve their marks in the above exams. Preparing this topic is not difficult task. We are just going to remember the stuff that we have already learned in our lower classes

Students have to learn few basic operations in this topic and some additional tricks. Already we are much clear with the four basic operations which we often use in math. They are addition, subtraction, multiplication and division. Even though we are much clear with these four basic operations, we have to be knowing some more stuff to do the problems which are being asked from this topic in competitive exams. The stuff which I have mentioned above is nothing but the tricks and shortcuts which need to solve the problems in a very short time.

Short cut is nothing but the easiest way to solve word problems related to HCF amd LCM. In competitive exams, we will have very limited time to solve each problem. Then only we will be able to attend all the questions. If we do problems in competitive exams in perfect manner with all the steps, it will definitely take much time and we may not able to attend the other questions. So we need some other way in which the problems can be solved in a very short time. The way we need to solve the problem quickly is called as shortcut.

**Here, we are going to have some word problems on HCF and LCM . You can check your answer online and see step by step solution.**

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)

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)

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)

2. The traffic lights at three different road crossings change after every 48 sec, 72 sec and 108 sec respectively. If they all change simultaneously at 8:20:00 hrs, when will they again change simultaneously?

For example, let the two signals change after every 3 secs and 4 secs respectively.

Then the first signal changes after 3, 6, 9, 12 seconds...

Like this, the second signal changes after 4, 8, 12 seconds...

So, if the two signals change simultaneously now, again they will change simultaneously 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 change simultaneously, we have to find the L.C.M of (48,72,108).

L.C.M of (48,72,108) is 432 seconds = 7 min 12 sec

So, after every 7 min 12 sec, all the signals will change simultaneously.

At 8:20:00 hrs, if all the three signals change simultaneously, again they will change simultaneously after 7 min 12 sec. That is at 8:27:12 hrs.

Hence, three signals will change simultaneously at 8:27:12 seconds.

Then the first signal changes after 3, 6, 9, 12 seconds...

Like this, the second signal changes after 4, 8, 12 seconds...

So, if the two signals change simultaneously now, again they will change simultaneously 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 change simultaneously, we have to find the L.C.M of (48,72,108).

L.C.M of (48,72,108) is 432 seconds = 7 min 12 sec

So, after every 7 min 12 sec, all the signals will change simultaneously.

At 8:20:00 hrs, if all the three signals change simultaneously, again they will change simultaneously after 7 min 12 sec. That is at 8:27:12 hrs.

Hence, three signals will change simultaneously at 8:27:12 seconds.

3. 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.

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)

H.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 of volume 60 ltrs in each tin.

Hence, the greatest volume of each tin is 60 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)

H.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 of volume 60 ltrs in each tin.

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

4. Find the least number of soldiers in a regiment such that they stand in rows of 15, 20, 25 and form a perfect square.

To answer this question, we have to find the least number which is exactly divisible by the given numbers 15,20 and 25.That is nothing but the L.C.M of (15,20,25)

L.C.M of (15,20,25) = 300

So, we need 300 soldiers such that they stand in rows of 15, 20 , 25.

But, it has to form a perfect square (as per the question)

To form a perfect square, we have to multiply 300 by some number such that it has to be a perfect square.

To make 300 as perfect square, we have to multiply 300 by 3.

Then, it is 900 which is a perfect square.

Hence, the least number of soldiers required is 900.

L.C.M of (15,20,25) = 300

So, we need 300 soldiers such that they stand in rows of 15, 20 , 25.

But, it has to form a perfect square (as per the question)

To form a perfect square, we have to multiply 300 by some number such that it has to be a perfect square.

To make 300 as perfect square, we have to multiply 300 by 3.

Then, it is 900 which is a perfect square.

Hence, the least number of soldiers required is 900.

5. 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.

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)

16.58X100 = 1658 cm and 8.32X100 = 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)

= (1658X832)/2X2 = 344864

Hence, the least number of square tiles required = 344864

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)

16.58X100 = 1658 cm and 8.32X100 = 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)

= (1658X832)/2X2 = 344864

Hence, the least number of square tiles required = 344864

6. A wine seller had three types of wine. 403 liters of 1st kind, 434 liters of 2nd kind and 465 liters of 3rd kind. Find the least possible number of casks of equal size in which different types of wine can be filled without mixing.

For the least possible number of casks of equal size, the size of each cask must be of the greatest volume.

To get the greatest volume of each cask, we have to find the largest number which exactly divides 403, 434 and 465.That is nothing but the H.C.F of (403,434,465)

The H.C.F of (403,434,465) = 31 liters

Each cask must be of the volume 31 liters.

Req. No. of casks = (403/31)+(434/31)+(465/31)= 13+14+15 = 42

Hence, the least possible number of casks of equal size required is 42

To get the greatest volume of each cask, we have to find the largest number which exactly divides 403, 434 and 465.That is nothing but the H.C.F of (403,434,465)

The H.C.F of (403,434,465) = 31 liters

Each cask must be of the volume 31 liters.

Req. No. of casks = (403/31)+(434/31)+(465/31)= 13+14+15 = 42

Hence, the least possible number of casks of equal size required is 42

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

Since the H.C.F is 49, the two numbers could be 49x and 49y

Their sum is 588. So, 49x+49y = 588 ===> x+y = 12

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

The possibles values of (x,y) = (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

Their sum is 588. So, 49x+49y = 588 ===> x+y = 12

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

The possibles values of (x,y) = (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

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

Since the H.C.F is 13, the two numbers could be 13x and 13y

Their product is 2028. So, (13x)X(13y) = 2028 ===> xy = 12

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

The possibles values of (x,y) = (1,12),(2,6),(3,4).

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, (1,12) and (3,4) are the co-primes.

Hence, the number of pairs is 2

Their product is 2028. So, (13x)X(13y) = 2028 ===> xy = 12

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

The possibles values of (x,y) = (1,12),(2,6),(3,4).

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, (1,12) and (3,4) are the co-primes.

Hence, the number of pairs is 2

9. Lenin
is preparing dinner plates. He has 12 pieces of chicken and 16 rolls.
If he wants to make all the plates identical without any food left over,
what is the greatest number of plates Lenin can prepare?

To make all the plates identical and find the greatest number of plates, we have to find the greatest number which can divide 12 and 16 exactly.

That is nothing but H.C.F of 12 and 16.

H.C.F of (12, 16) = 4

That is, 12 pieces of chicken would be served in 4 plates at the rate of 3 pieces per plate.

And 16 rolls would be served in 4 plates at the rate of 4 rolls per plate.

In this way, each of the 4 plates would have 3 pieces of chicken and 4 rolls. And all the 4 plates would be identical.

Hence, the greatest number of plates Lenin can prepare is 4

That is nothing but H.C.F of 12 and 16.

H.C.F of (12, 16) = 4

That is, 12 pieces of chicken would be served in 4 plates at the rate of 3 pieces per plate.

And 16 rolls would be served in 4 plates at the rate of 4 rolls per plate.

In this way, each of the 4 plates would have 3 pieces of chicken and 4 rolls. And all the 4 plates would be identical.

Hence, the greatest number of plates Lenin can prepare is 4

10. 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?

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.

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.

11. 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?

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.

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.

12. Lily
has collected 8 U.S. stamps and 12 international stamps. She wants to
display them in identical groups of U.S. and international stamps, with
no stamps left over. What is the greatest number of groups Lily can
display them in?

To make all the groups identical and find the greatest number of groups, we have to find the greatest number which can divide 8 and 12 exactly.

That is nothing but H.C.F of 8 and 12.

H.C.F of (8, 12) = 4

That is, 8 U.S stamps can be displayed in 4 groups at 2 stamps/group.

And 12 international stamps can be displayed in 4 groups at 3 stamps/group.

In this way, each of the 4 groups would have 2 U.S stamps and 3 international stamps. And all the 4 groups would be identical.

Hence, the greatest number of groups can be made is 4

That is nothing but H.C.F of 8 and 12.

H.C.F of (8, 12) = 4

That is, 8 U.S stamps can be displayed in 4 groups at 2 stamps/group.

And 12 international stamps can be displayed in 4 groups at 3 stamps/group.

In this way, each of the 4 groups would have 2 U.S stamps and 3 international stamps. And all the 4 groups would be identical.

Hence, the greatest number of groups can be made is 4

13. Abraham has two pieces of wire, one 6 feet long and the other 12 feet long. If he wants to cut them up to produce many pieces of wire that are all of the same length, with no wire left over, what is the greatest length, in feet, that he can make them?

When the two wires are cut in to small pieces, each piece must of same length and also it has to be the possible greatest length.

6 feet wire can be cut in to pieces of (2, 2, 2) or (3, 3)

12 feet wire can be cut in to pieces of (2, 2, 2, 2, 2, 2 ) or (3, 3, 3, 3)

The length of each small piece must be of possible greatest length.

To find the possible greatest length, we have to find the greatest number which can divide both 6 and 12. That is H.C.F of (6, 12).

H.C.F of (6, 12) = 6.

Hence, the greatest length of each small piece will be 6 ft.

(That is, 6 feet wire is not cut in to small pieces and it is kept as it is. Only the 12 feet wire is cut in to 2 pieces at the length of 6 feet/piece)

6 feet wire can be cut in to pieces of (2, 2, 2) or (3, 3)

12 feet wire can be cut in to pieces of (2, 2, 2, 2, 2, 2 ) or (3, 3, 3, 3)

The length of each small piece must be of possible greatest length.

To find the possible greatest length, we have to find the greatest number which can divide both 6 and 12. That is H.C.F of (6, 12).

H.C.F of (6, 12) = 6.

Hence, the greatest length of each small piece will be 6 ft.

(That is, 6 feet wire is not cut in to small pieces and it is kept as it is. Only the 12 feet wire is cut in to 2 pieces at the length of 6 feet/piece)

**Click here to know more about H.C.F and L.C.M**

**Analytical Geometry Calculators**

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