# Mathematics (RD Sharma Solution) - Class 6 / Grade 6

- Playing With Numbers (RD Sharma –Chapter 2-Solution of Exercise 2.8)
### Playing With Numbers (RD Sharma –Chapter 2-Solution of Exercise 2.8)

Tags: RD Sharma Class 6 Math Solution Chapter 2 Playing with Numbers Exercise 2.8, Solution of RD Sharma for class 6, RD Sharma solutions Grade VI, Solution of RD Sharma exercises for sixth standard, R.D. Sharma Mathematics solutions and practice pages, Playing With Numbers Exercise 2.8 (Some Applications of H.C.F)Playing With Numbers

Exercise 2.8 (Some Application of H.C.F)

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

Solution:

Clearly, the required number is the H.C.F of the numbers.

615-6 = 609, 963-6 = 957

Now, first we find the H.C.F of 609 and 957.

The H.C.F of 609 and 957 is 87. Therefore, required number is 87.

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

Solution:

Clearly, the required number is the H.C.F of the numbers.

285-9 = 276, 1249-7 = 1242

Now, first we find the H.C.F of 285 and 1242.

The H.C.F of 276 and 1242 is 138. Therefore, required number is 138.

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

Solution:

Clearly, the required number is the H.C.F of the numbers.

626-1 = 625, 3127-2 = 3125, 15628-3 = 15625

Now, first we find the H.C.F of 625, 3125 and 15628.

Let’s take 625 and 3125 for the H.C.F

The H.C.F of 625, 3125 and 15628 is 625. Therefore, required number is 625.

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

Solution:

Length = 8m 25cm = 825 cm

Breadth = 6m 75cm = 675 cm

Height = 4m 50 cm = 450 cm

Now we find the H.C.F of 825, 675 and 450

5

825, 675, 450

5

165, 135, 90

3

33, 27, 18

11, 9, 6

Thus, H.C.F. of 825, 675 and 450 = 5x5x3 = 75

Therefore, longest rod which can measure the three dimensions of the room exactly= 75 cm

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

Solution:

Length = 20m 16cm = 2016cm

Breadth = 15m 60cm = 1560cm

Now we find the H.C.F of 2016 and 1560

2

2016, 1560

2

1008, 780

2

504, 390

3

252, 195

84, 65

H.C.F 2x2x2x3 = 24 cm

Therefore, least number of stones = Area of courtyard/Area of square stone

= 2016 x 1560 / 24 x 24

= 5460

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

Solution:

Length = 7m = 700 cm

Breadth = 3m 85cm = 385 cm

Height = 12m 95 cm = 1295 cm

Now we find the H.C.F of 700, 385 and 1295

5

700, 385, 1295

7

140, 77, 259

20, 11, 37

H.C.F = 5X7 = 35cm

Longest tape which can be used to measure exactly the lengths 7m, 3m 85cm and 12m 95cm is 35cm.

Question 7 (Refer Book - Mathematics Class VI R.D. Sharma)

Solution:

No. of goats = 105

No. of donkeys = 140

No. of cows = 175

Now we find the H.C.F of 105, 140 and 175.

5

105, 140, 175

7

21, 28, 35

3, 4, 5

H.C.F = 5X7 = 35

Therefore, 35 animals went in each trip.

Question 8 (Refer Book - Mathematics Class VI R.D. Sharma)

Solution:

No. of chocolate of brand 1=24

No. of chocolate of brand 2 = 15

The required number is the L.C.M of the numbers.

3

24, 15

5

8, 5

8

8, 1

1, 1

L.C.M = 3X5X8 = 120

No. of boxes of kind I = 120/24 = 5

No. of boxes of kind II = 120/15 = 8

Question 9 (Refer Book - Mathematics Class VI R.D. Sharma)

Solution:

No. of pencils in each packet = 24

No. of crayons in each packet = 32

The required number is the L.C.M of the numbers.

2

24, 32

2

12, 16

2

6, 8

2

3, 4

2

3, 2

3

3, 1

1, 1

L.C.M = 2x2x2x2x2x3 = 96

No. of pencils packets = 96/24 = 4

No. of crayons packets = 96/32 = 3

Question 10 (Refer Book - Mathematics Class VI R.D. Sharma)

Solution:

i. 161/207

23x7/23x9 = 7/9

ii. 296/481

37x8/37x9 = 8/9

Question 11 (Refer Book - Mathematics Class VI R.D. Sharma)

Solution:

I kind of oil =120 l

II kind of oil = 180 l

III kind of oil = 240 l

The required number is the H.C.F of the numbers.

2

120, 180, 240

2

60, 90, 120

3

30, 45, 60

5

10, 15, 20

2, 3, 4

H.C.F = 2X2X3X5 = 60 l

60 l is the capacity of each tin.