RD Sharma Solutions: Exercise 2.10- Playing With Numbers Class 6 Notes | EduRev

Mathematics (Maths) Class 6

Class 6 : RD Sharma Solutions: Exercise 2.10- Playing With Numbers Class 6 Notes | EduRev

The document RD Sharma Solutions: Exercise 2.10- Playing With Numbers Class 6 Notes | EduRev is a part of the Class 6 Course Mathematics (Maths) Class 6.
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Q.1. What is the smallest number which when divided by 24, 36 and 54 gives a remainder of 5 each time?

Ans: We have to find prime factorisation of 24, 36, and 54.

Prime factorisation of 24 = 2 × 2 × 2 × 3

Prime factorisation of 36 = 2 × 2 × 3 × 3

Prime factorisation of 54 = 2 × 3 × 3 × 3

∴ Required LCM = 2 × 2 × 2 × 3 × 3 × 3 = 216

Thus, 216 is the smallest number exactly divisible by 24, 36, and 54.

To get the remainder as 5:

Smallest number = 216 + 5 = 221

Thus, the required number is 221.


Q.2. What is the smallest number that both 33 and 39 divide leaving remainders of 5?

Ans: We have to find prime factorisation of 33 and 39.

Prime factorisation of 33 = 3 × 11

Prime factorisation of 39 = 3 × 13

∴ Required LCM = 3 × 11 × 13 = 429

Thus, 429 is the smallest number exactly divisible by 33 and 39.

To get the remainder as 5:

Smallest number = 429 + 5 = 434

Thus, the required number is 434.


Q.3. Find the least number that is divisible by all the numbers between 1 and 10 (both inclusive).

Ans: To find the required least number, we have to find the LCM of  the numbers from 1 to 10.

We know that 2, 3, 5,  and 7 are prime number.

Prime factorisation of 4 = 2 × 2

Prime factorisation of 6 = 2 × 3

Prime factorisation of 8 = 2 × 2 × 2

Prime factorisation of 9 = 3 × 3

Prime factorisation of 10 = 2 × 5

∴ Required least number = 2 × 2 × 2 × 3 × 3 × 5 × 7 = 2,520


Q.4. What is the smallest number that, when divide by 35, 56 and 91 leaves remainders of 7 in each case?

Ans: We have to find the prime factorisation of 35, 56, and 91.

Prime factorisation of 35 = 5 × 7

Prime factorisation of 56 = 2 × 2 × 2 × 7

Prime factorisation of 91 = 7 × 13

∴ Required LCM = 2 × 2 × 2 × 5 × 7 × 13 = 3,640

Thus, 3,640 is the smallest number exactly divisible by 35, 56, and 91.

To get the remainder as 7:

Smallest number = 3,640 + 7 = 3,647

Thus, the required number is 3,647.


Q.5. In a school there are two sections - section A and section B of Class VI. There are 32 students in section A and 36 in section B. Determine the minimum number of books required for their class library so that they can be distributed equally among students of section A or section B.

Ans: We have to find the LCM of 32 and 36.

Prime factorisation of 32 = 2 × 2 × 2 × 2 × 2

Prime factorisation of 36 = 2 × 2 × 3 × 3

Required LCM = 2 × 2 × 2 × 2 × 2 × 3 × 3 = 288

∴ Minimum number of books required = LCM of 32 and 36 = 288 books


Q.6. In a morning walk three persons step off together. Their steps measure 80 cm, 85 cm, and 90 cm respectively. What is the minimum distance each should walk so that he can cover the distance in complete steps?

Ans: We have to find the LCM of 80 cm, 85 cm, and 90 cm.

Prime factorisation of 80 = 2 × 2 × 2 × 2 × 5

Prime factorisation of 85 = 5 × 17

Prime factorisation of 90 = 2 × 3 × 3 × 5

∴ Required LCM = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 17 = 12,240

∴ Required minimum distance = LCM of 80 cm, 85 cm, and 90 cm

                                               = 12,240 cm

                                               = 122 m 40 cm     (since 1 m =100 cm)


Q.7. Determine the number nearest to 100000 but greater than 100000 which is exactly divisible by each of 8, 15 and 21.

Ans: First, we have to find the LCM of 8, 15, and 21.

Prime factorisation of 8 = 2 × 2 × 2

Prime factorisation of 15 = 3 × 5

Prime factorisation of 21 = 3 × 7

Therefore, required LCM =  2 × 2 × 2 × 3 × 5 × 7 = 840.

The number nearest to 1,00,000 and exactly divisible by each of 8, 15, and 21 should also be exactly divisible by their LCM (i.e. 840).

We have to divide 1,00,000 by 840.

RD Sharma Solutions: Exercise 2.10- Playing With Numbers Class 6 Notes | EduRev

Remainder = 40

∴ Number just greater than 1,00,000 and exactly divisible by 840 

= 1,00,000 + (840 − 40)

= 1,00,000 + 800 = 1,00,800

∴ Required number = 1,00,800


Q.8. A school bus picking up children in a colony of flats stops at every sixth block of flats. Another school bus starting from the same place stops at every eighth blocks of flats. Which is the first bus stop at which both of them will stop?

Ans: First bus stop at which both the buses will stop together =  LCM of 6th block and 8th block

Prime factorisation of 6 = 2 × 3

Prime factorisation of 8 = 2 × 2 × 2

∴ Required LCM = 2 × 2 × 2 × 3 = 24

Hence, the first bus stop at which both the buses will stop together will be at the 24th block.


Q.9. Telegraph poles occur at equal distances of 220 m along a road and heaps of stones are put at equal distances of 300 m along the same road. The first heap is at the foot of the first pole. How far from it along the road is the next heap which lies at the foot of a pole?

Ans: We have to find the LCM of 220 m and 300 m.

Prime factorisation of 220 = 2 × 2 × 5 × 11

Prime factorisation of 300 = 2 × 2 × 3 × 5 × 5

∴ Required LCM = 2 × 2 × 3 × 5 × 5 × 11 = 3,300

Hence, 3,300 m far is the next heap that lies at the foot of a pole.


Q.10. Find the smallest number which leaves remainders 8 and 12 when divided by 28 and 32 respectively.

Ans: First, we have to find the LCM of 28 and 32.

Prime factorisation of 28 = 2 × 2 × 7

Prime factorisation of 32 = 2 × 2 × 2 × 2 × 2

∴ Required LCM = 2 × 2 × 2 × 2 × 2 × 7 = 224

It is given that when we divide the number by 28, the remainder is 8 and when we divide the number by 32, the remainder is 12.

We observe:

28 − 8 = 20

32 − 12 = 20

∴ Required number = 224 − 20 = 204

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