Create New Account
Create New Account

Sign in to Innings2

or

New Account     Reset Password     

    Exercise 6.3.7

Powered by Innings 2

Reset Progress

This will delete your progress and chat data for all chapters in this course, and cannot be undone!

Glossary

Select one of the keywords on the left…

6th class > > Exercise 6.3.7

Exercise 6.3.7

Find the maximum value of weight.

  • 1. Renu purchases two bags of fertiliser of weights 75 kg and 69 kg.
  • we need to determine the Highest Common Factor (HCF) of the two weights.
  • prime factors of 75 are: x x
  • prime factors of 69 are: x
  • The common prime factor between 75 and 69 is
  • The highest power of the common prime factor is.
  • Therefore, the maximum value of weight which can measure the weight of the fertiliser exactly for both bags is.
  1. Three boys step off together from the same spot. Their steps measure 63 cm, 70 cm and 77 cm respectively. What is the minimum distance each should cover so that all can cover the distance in complete steps?

To find the distance we need to know Lcm of each body.

Step measure of 1st boy = 63cm

Step measure of 2nd boy = 70cm

Step measure of 3rd boy = 77cm

LCM = x x x x = 6930. Therefore, the minimum distance each boy should cover so that all can cover the distance in complete steps is 6930 cm.

Find the longest tape.

  • 3.The length, breadth and height of a room are 825 cm, 675 cm and 450 cm respectively.
  • We have to factorize all three given dimensions.
  • Prime factors of 825: 3 x x x 11
  • Prime factors of 675: 3 x 3 x x x 5
  • Prime factors of 450: 2 x x 3 x x
  • Common factors are : x x 5.
  • Lowest power for 3 is 3^1.
  • lowest power for 5 is 5^2.
  • Calulate common factors 3 x 25 = .
  1. Determine the smallest 3-digit number which is exactly divisible by 6, 8 and 12

First, we have to find the LCM of 6,8 and 12.

Common multipes of 6 is : 6,,18,,30...

Common multipes of 8 is : 8,,24,32..

Common multipes of 12 is : 12,24,,48...

It is observed that is the least common multipes.

Now, smallest 3 digit number is 100, Divide 100 to 24 the remainder should be subtracted from the dividend and 24 will be added to it to make it perfectly divisible.

=100/24=4

=(100-4) +

=+ =120

Hence, the smallest 3-digit number which is exactly divisible by 6,8 and 12 is 120.
5. Determine the greatest 3-digit number exactly divisible by 8, 10 and 12.
prime factors of 8: x x 2
prime factors of 10: 2 x
prime factors of 12: 2 x 2 x
common factors are: , ,
LCM = .
The largest 3-digit number is 999. We need to find the largest multiple of 120 that is less than or equal to 999.
= 990/ = 8.3
= 8 x 120 = 960.
Therefore, the greatest 3-digit number that is exactly divisible by 8, 10, and 12 is 960.
6. The traffic lights at three different road crossings change after every 48 seconds, 72 seconds and 108 seconds respectively. If they change simultaneously at 7 a.m., at what time will they change simultaneously again?
prime factors of 48: 2 x x x x
prime factors of 72: 2 x x x x 3.
prime factors of 108: 2 x x x x 3.
common factors are: , .
LCM = 432.
The LCM of 48 seconds, 72 seconds, and 108 seconds is 432 seconds.
= 432/ = 7.2
So, 432 seconds is equivalent to 7 minutes and 12 seconds.
7:00 am + 7 min and 12 sec = 7:07 am and 12 sec.

Find the longest tape.

  • 7. Three tankers contain 403 litres, 434 litres and 465 litres of diesel respectively. Find the maximum capacity of a container that can measure the diesel of the three containers exact number of times.
  • Prime factorization of 403 : It is divisible by 13 . so 403 / = 31.
  • Prime factorization of 434 : It is divisible by 2 . so 434 / = 217.
  • 217 will only divisble by 7. so 217/7 = 31
  • 434 = x x
  • prime factorization of 465 : It is divisble by 3 . so 465 / = 155.
  • 155 will only divisible by 5. so 155/ = 31.
  • Common factor is : 31
  • The maximum capacity of a container that can measure the diesel in the three tankers exactly is 31 litres.
8. Find the least number which when divided by 6, 15 and 18 leave remainder 5 in each case.
prime factors of 6: 2 x
prime factors of 15: x
prime factors of 18: 2 x x .
LCM = 2 x 3 x 3 x 5 = .
So, the required number is 90 + 5 = .
Therefore, when 95 is divided by the given numbers, it will leave a remainder of 5.
  1. Find the smallest 4-digit number which is divisible by 18, 24 and 32.

First, we have to find the LCM of 18, 24 and 32.

Common multipes of 18 is : 2 x x .

Common multipes of 24 is : 2 x x 2 x 3

Common multipes of 32 is : 2 x 2 x x x

The LCM of 18, 24, and 32 is 288.

= 1000/=3.472

= x 288 = 1152.

Therefore, the smallest 4-digit number that is exactly divisible by 18, 24, and 32 is 1152.
  1. Find the LCM of the following numbers : (a) 9 and 4 (b) 12 and 5 observe a common property in the obtained LCMs. Is LCM the product of two numbers in each case?

Prime factors of 9 and 4 : 9 - x , 4 - x .

LCM = 2^ x 3^ = .

Prime factors of 12 and 5: 12 - x x , 5 - .

LCM = 2^ x 3 x 5 = .

In each case, the LCM is the product of the two numbers. This happens because the pairs of numbers are co-prime, meaning they have no common prime factors.
  1. Find the LCM of the following numbers in which one number is the factor of the other. (a) 5, 20 (b) 6, 18 (c) 12, 48 What do you observe in the results obtained

a. LCM of 5 and 20: prime factors of 5 - , 20 - x x 5. Since 5 is a factor of 20, the LCM is the larger number, 20.

b. LCM of 6 and 18: prime factors of 6 - x , 18 - x x . Since 6 is a factor of 18, the LCM is the larger number, 18.

c. LCM of 12 and 48: prime factors of 12 - x 2 x , 48 - x 2 x x x . Since 12 is a factor of 48, the LCM is the larger number, 48.