HCF AND LCM WORD PROBLEMS

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 and 4 seconds 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 is the least common multiple (LCM) of 3 and 4.

The same thing happened in our problem. To find the time, when they will all toll together, we have to find the LCM of (2, 4, 8, 6, 10, 12).

LCM (2, 4, 8, 6, 10, 12) is 120

That is, 120 seconds or 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 :

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 hours, when will they again change simultaneously ?

Solution :

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 is the least common multiple (LCM) of 3 and 4.

The same thing happened in our problem. To find the time, when they will all change simultaneously, we have to find the LCM of (48, 72, 108).

LCM (48, 72, 108)  =   432 seconds  or  7 min 12 sec

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

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

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

Problem 3 :

A merchant has 120 liters and 180 liters of two kinds of oil. He wants to sell the oil by filling the two kinds in tins of equal volumes. Find the greatest volume 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,... liters.

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 the highest common factor (HCF) of (120, 180).

HCF (120, 180)  =  60 liters

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

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

Hence, the greatest volume of the tin is 60 liters.

Problem 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.

Solution :

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 the least common multiple of (15, 20, 25).

LCM (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.

Problem 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.

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 the highest common factor (HCF) of (16.58, 8.32).

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

16.58 ⋅ 100  =  1658 cm

8.32 ⋅ 100  =  832 cm

HCF (1658, 832)  =  2 cm

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 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.

Solution :

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 the highest common factor (HCF) of (403, 434, 465).

HCF (403, 434, 465)  =  31 liters

Each cask must be of the volume 31 liters.

Required number casks :

=  (403/31) + (434/31) + (465/31)

=  13 + 14 + 15

=  42

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

Problem 7 :

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

Solution :

Because the highest common factor is 49, the two numbers can be assumed as 49x and 49y.

Their sum is 588. Then,

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 possible pairs of values of (x, y) are

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

In the above pairs of values, only co-primes will meet the condition given in the question.

[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].

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

Hence, the number of pairs is 2.

Problem 8 :

The product of two numbers is 2028 and their highest common factor (HCF) is 13. Find the number of such pairs.

Solution :

Since the highest common factor is 13, the two numbers could be 13x and 13y.

Their product is 2028. Then

(13x) ⋅ (13y)  =  2028

169xy  =  2028

Divide each side by 169.

xy  =  12

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

The possible pairs of values of (x, y) are

(1, 12), (2, 6), (3, 4)

In the above pairs of values, only co-primes will meet the condition given in the question.

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

Hence, the number of pairs is 2.

Problem 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 ?

Solution : 

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 the highest common factor of 12 and 16.

HCF (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

Problem 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 ?

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 least common multiple of (2, 5).

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

Problem 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 ?

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 the least common multiple of (3, 5).

LCM (3, 5)  =  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 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 ?

Solution :

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 the highest common factor of (8, 12).  

HCF (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.

Problem 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 ?

Solution : 

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 the highest common factor of (6, 12).

HCF (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)

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