HCF and LCM of fractions

To find the HCF and LCM of several fractions:

Step 1: First express the fractions in their lowest terms.

Step 2: HCF = HCF of Numerator / LCM of denominator

Step 3: LCM = LCM of Numerator/ HCF of denominator



 Question:

The circumference of the fore wheels and hind wheels of a carriage are 2²/₅ and 3 ³/₇ feet respectively. A chalk mark is made at the point of contact of the wheels with the ground at any given point. How far will the carriage have traveled so that their chalk marks may be again on the ground at the same time?

 

tougher problems on LCM

A gardener had a number of shrubs to plant in rows. At first he tried to plant 5 in each row, then 6, then 8, then 9 and then 12, but had always 1 left. On trying 13 he had none left. What is the smallest number of shrubs that he could have had?    

Answer:

           

The LCM of 5,6,8,9 and 12 is 360.

So, the required number is 360y + 1

Given that 360y + 1 does not leave a remainder with 13.

360 = 13 x 27 + 9

So, 360y + 1 = (13 x 27 + 9)y + 1

= 13x27y + 9y+1

Since there is no remainder, 9y + 1 should be equal to 13 or its multiple.

For, y = 1, 9y +1 = 10 ..... (x)

Y = 2, 9y + 1 = 19 ..............(x)

Y = 3, 9y +1 = 28 ...............(x)

Y = 4, 9y +1 = 37 ...............(x)

Y = 5, 9y +1 = 46 ...............(x)

Y = 6, 9y +1 = 55 ...............(x)

Y = 7, 9y +1 = 64 ...............(x)

Y = 8, 9y +1 = 73 ...............(x)

Y = 9, 9y +1 = 82 ...............(x)

Y = 10, 9y +1 = 91 ...............(√) divisible by 13

So, the number is 360y + 1 = 360x10 + 1 = 3601

Tougher problems on HCF and LCM

In the HCF of two numbers which were prime to each other, the quotients are 2,2,7,2 on finding HCF by method of division. Find the two numbers.

 Answer:

Since the numbers are prime, the HCF is 1.

Now following the method of finding HCF by the method of division, we get

Final Dividend = 1x2 = 2

In the pen ultimate division, the divisor is 2, the quotient is 7 and the remainder is 1.

So, the dividend is = 2x7 + 1 = 15

In the prior division, the divisor is 15, the quotient is 2 and the remainder is 2.

So, the dividend is = 15x 2 + 2 = 32

Therefore, the smaller of the two numbers is 32.

The next number is the divisor = 32 x 2 + 15 = 79

So, the two numbers are 32 and 79.

                                                               

Check:

                               32) 79 (2

                                     64

                                       15) 32 (2

                                              30

                                                2)15 (7

                                                   14

  1) 2 (2

2

x

HCF by method of division

In finding the HCF of two numbers, the last divisor is 49 and the quotient 17, 3 and 2. Find the numbers?

Answer:

Since the last divisor is 49 and the quotient is 2, we have the dividend = 49 x 2 = 98. (Because the last divisor will not leave any remainder)

For the pen-ultimate, 98 will become the divisor and 3 will be the quotient and 49 will be the remainder. Therefore, the dividend will be 98 x 3 + 49 = 343

Now, 343 will become the divisor and 17 will be the quotient and 98 will be the remainder.

Dividend = 343 x 17 + 98 = 5929.

So, the two numbers are 5929 and 343.