# Class-6 Math Chapter-3: Excercise 3.7 NCERT Solutions

Chapter 3: Playing With Numbers

Exercise 3.7

Q1.

Ans.

Factors of 75 = 3 x 5 x 5

Factors of 69 = 3 x 69

H.C.F. = 3

Therefore the required weight is 3 kg.

Q2.

Ans.

L.C.M. of 63, 70 and 77 = 7 x 9 x 10 x 11 = 6930 cm.

Therefore, the minimum distance is 6930 cm.

Q3.

Ans.

7 | 63, 70, 77 |

9 | 9, 10, 11 |

10 | 1, 10, 11 |

11 | 1, 1, 11 |

| 1, 1, 1 |

Factors of 825 = 3 x 5 x 5 x 11

Factors of 675 = 3 x 5 x 5 x 3 x 3

Factors of 450 = 2 x 3 x 3 x 5 x 5

H.C.F. = 3 x 5 x 5 = 75 cm

Therefore, the longest tape is 75 cm.

Q4. Determine the smallest 3-digit number which is exactly divisible by 6, 8 and 12.

Ans.

2 | 6, 8, 12 |

2 | 3, 4, 6 |

2 | 3, 2, 3 |

3 | 3, 1, 3 |

| 1, 1, 1 |

The smallest 3-digit number = 100

To find the number, we have to divide 100 by 24

100 = 24 x 4 + 4

Therefore, the required number = 100 + (24 – 4) = 120.

Q5. Determine the largest 3-digit number which is exactly divisible by 8, 10 and 12.

Ans.

2 | 8, 10, 12 |

2 | 4, 5, 6 |

2 | 2, 5, 3 |

3 | 1, 5, 3 |

5 | 1, 5, 1 |

| 1, 1, 1 |

L.C.M. of 8, 10, 12 = 2 x 2 x 2 x 3 x 5 = 120

The largest three digit number = 999

Now,

Therefore, the required number = 999 – 39 = 960

Q6. 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?

Ans.

2 | 48, 72, 108 |

2 | 24, 36, 54 |

2 | 12, 18, 27 |

2 | 6, 9, 27 |

3 | 3, 9, 27 |

3 | 1, 3, 9 |

3 | 1, 1, 3 |

| 1, 1, 1 |

L.C.M. of 48, 72, 108 = 2 x 2 x 2 x 2 x 3 x 3 x 3 = 432 sec.

After 432 seconds, the lights change simultaneously.

432 second = 7 minutes 12 seconds

Therefore the time = 7 a.m. + 7 minutes 12 seconds

= 7:07:12 a.m.

Q7.

Ans.

Factors of 403 = 13 x 31

Factors of 434 = 2 x 7 x 31

Factors of 465 = 3 x 5 x 31

H.C.F. = 31

Therefore, 31 litres of container is required to measure the quantity.

Q8.

Ans.

2 | 6,15,18 |

3 | 3,15,9 |

3 | 1,5,3 |

5 | 1,5,1 |

| 1,1,1 |

L.C.M. of 6, 15 and 18 = 2 x 3 x 3 x 5 = 90

Therefore, the required number = 90 + 5 = 95

Q9.

Ans.

2 | 18, 24, 32 |

2 | 9, 12, 16 |

2 | 9, 6, 8 |

2 | 9, 3, 4 |

2 | 9, 3, 2 |

3 | 9, 3, 1 |

3 | 3, 1, 1 |

| 1,1,1 |

L.C.M. of 18, 24 and 32 = 2 x 2 x 2 x 2 x 2 x 3 x 3 = 288

The smallest four-digit number = 1000

Now,

Therefore, the required number is 1000 + (288 – 136) = 1152.

Q10.

(a) 9 and 4

(b) 12 and 5

(c) 6 and 5

(d) 15 and 4

Observe a common property in the obtained L.C.Ms. Is L.C.M. the product of two numbers in each case?

Ans.

(a)

2 | 9, 4 |

2 | 9, 2 |

3 | 9, 1 |

3 | 3, 1 |

| 1, 1 |

L.C.M. of 9 and 4

= 2 x 2 x 3 x 3

= 36

(b)

2 | 12, 5 |

2 | 6, 5 |

3 | 3, 5 |

5 | 1, 5 |

| 1, 1 |

L.C.M. of 12 and 5

= 2 x 2 x 3 x 5

= 60

(c)

2 | 6, 5 |

3 | 3, 5 |

5 | 1, 5 |

| 1, 1 |

L.C.M. of 6 and 5

= 2 x 3 x 5

= 30

(d)

3 | 9, 45 |

3 | 3, 15 |

5 | 1, 5 |

| 1, 1 |

L.C.M. of 15 and 4

= 2 x 2 x 3 x 5

= 60

Yes, the L.C.M. is equal to the product of two numbers in each case and L.C.M. is also the multiple of 3.

Q11.

(a) 5, 20

(b) 6, 18

(c) 12, 48

(d) 9, 45

What do you observe in the result obtained?

Ans.

(a)

2 | 5, 20 |

2 | 5, 10 |

5 | 5, 5 |

| 1, 1 |

L.C.M. of 5 and 20

= 2 x 2 x 5

= 20

(b)

2 | 6, 18 |

3 | 3, 9 |

3 | 1, 3 |

| 1, 1 |

L.C.M. of 6 and 18

= 2 x 3 x 3

= 18

(c)

2 | 12, 48 |

2 | 6, 24 |

2 | 3, 12 |

2 | 3, 6 |

3 | 3, 3 |

| 1, 1 |

L.C.M. of 12 and 48

= 2 x 2 x 2 x 2 x 3

= 48

(d)

3 | 9, 45 |

3 | 3, 15 |

5 | 1, 5 |

| 1, 1 |

L.C.M. of 9 and 45

= 3 x 3 x 5

= 45

From these all cases, we can conclude that if the smallest number if the factor of largest number, then the L.C.M. of these two numbers is equal to that of larger number.