## Topic outline

• ### Playing With Numbers (RD Sharma –Chapter 2-Solution of Exercise 2.11)

• Playing With Numbers

Exercise 2.11 (Some Properties of H.C.F and L.C.M of Given Numbers)

Question 1 (Refer Book - Mathematics Class VI R.D. Sharma)

Solution:

i. 25, 65

We compute the H.C.F and L.C.M of 25 and 65.

 5 25, 65 5,  13

H.C.F = 5  and  L.C.M = 5x5x13 = 325

Now, H.C.F x L.C.M = 5x325 = 1625

And, Product of the numbers = 25x65 = 1625

Therefore, Product of the numbers = Product of their H.C.F and L.C.M

ii. 117, 221

 13 117, 221 9,    17

H.C.F = 13  and  L.C.M = 9x13x17 = 1989

Now, H.C.F x L.C.M = 13x1989 = 25857

And, Product of the numbers = 117x221 = 25857

Therefore, Product of the numbers = Product of their H.C.F and L.C.M

iii. 35, 40

 5 35, 40 7,  8

H.C.F = 5  and  L.C.M = 5x7x8 = 280

Now, H.C.F x L.C.M = 5x280 = 1400

And, Product of the numbers = 35x40 = 1400

Therefore, Product of the numbers = Product of their H.C.F and L.C.M

iv. 87, 145

 29 87, 145 3,   5

H.C.F = 29  and  L.C.M = 3x5x29 = 435

Now, H.C.F x L.C.M = 29x435 = 12615

And, Product of the numbers = 87x145 = 12615

Therefore, Product of the numbers = Product of their H.C.F and L.C.M

v. 490, 1155

 5 490, 1155 98,   231

H.C.F = 5  and  L.C.M = 5x98x231 = 113190

Now, H.C.F x L.C.M = 5x113190 = 565950

And, Product of the numbers = 490x1155 = 565950

Therefore, Product of the numbers = Product of their H.C.F and L.C.M

Question 2 (Refer Book - Mathematics Class VI R.D. Sharma)

Solution:

i. 117, 221

 13 117, 221 9,    17

H.C.F = 13  and  L.C.M = 9x13x17 = 1989

ii. 234, 572

 26 234, 572 9,    22

H.C.F = 26  and  L.C.M = 9x22x26 = 5148

iii. 145, 232

 29 145, 232 5,    8

H.C.F = 29  and  L.C.M = 5x8x29 = 1160

iv. 861, 1353

 123 861, 1353 7,      11

H.C.F = 123  and  L.C.M = 7x11x123 = 9471

Question 3 (Refer Book - Mathematics Class VI R.D. Sharma)

Solution:

H.C.F of two numbers = 6

L.C.M of two numbers = 180

One of the number = 30

Other number = H.C.F x L.C.M / One of the number

= 6x180 / 30

= 1080/30

= 36

Question 4 (Refer Book - Mathematics Class VI R.D. Sharma)

Solution:

H.C.F of two numbers = 16

Product of the two numbers = 3072

L.C.M = Product of two numbers/ H.C.F

= 3072/16

= 192

Question 5 (Refer Book - Mathematics Class VI R.D. Sharma)

Solution:

H.C.F of two numbers = 145

L.C.M of two numbers = 2175

One of the number = 725

Other number = H.C.F x L.C.M / One of the number

= 145x2175 / 725

= 435

Question 6 (Refer Book - Mathematics Class VI R.D. Sharma)

Solution:

We know that the H.C.F of given numbers must divide their L.C.M completely. But, 16 do not divide 380 completely.

So, there can be no two numbers with 16 as their H.C.F and 380 as their L.C.M