GCF AND LCM WORD PROBLEMS WORKSHEET

GCF 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 GCF and LCM.

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GCF and LCM Word Problems Worksheet - Problems

Problem 1 :

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 ?

Problem 2 :

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

Problem 3 :

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.

Problem 4 :

In two numbers, one number is a multiple of 6 and the other one is a multiple of 7. If their LCM is 84, then find the two numbers.

Problem 5 :

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 ?

Problem 6 :

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 ?

Problem 7 :

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 ?

GCF and LCM Word Problems Worksheet - Solutions

Problem 1 :

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 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.

Problem 2 :

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 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.

Problem 3 :

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 nothing but the G.C.F of (403, 434, 465)

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

Each cask must be of the volume 31 liters.

Req. No. of casks is

= (403/31) + (434/31) + (465/31)

= 13 + 14 + 15

= 42

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

Problem 4 :

In two numbers, one number is a multiple of 6 and the other one is a multiple of 7. If their LCM is 84, then find the two numbers.

Solution :

From the given information, the numbers can be assumed as 6x and 7x.

We can find LCM of 6x and 7x using synthetic division as given below.

Therefore, LCM of (6x, 7x) is

= x ⋅ 6 ⋅ 7

= 42x

Given : LCM of the two numbers is 84.

Then, we have

42x = 84

Divide each side by 42.

x = 2

Substitute 2 for x in 6x and 7x to find the two numbers.

6x = 6 ⋅ 2 = 12

7x = 7 ⋅ 2 = 14

So, the two numbers are 12 and 14.

Problem 5 :

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 nothing but G.C.F of 12 and 16.

G.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.

Problem 6 :

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 nothing but G.C.F of 8 and 12.

G.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

Problem 7 :

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 G.C.F of (6, 12).

G.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)