# ML Aggarwal Playing With Numbers Check Your Progress Class 6 ICSE Maths Solutions

ML Aggarwal Playing With Numbers Check Your Progress Class 6 ICSE Maths Solutions. We Provide Step by Step Answer of Check Your Progress Questions for Playing With Numbers as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-6.

## ML Aggarwal Playing With Numbers Check Your Progress Class 6 ICSE Maths Solutions

Board | ICSE |

Publications | Avichal Publishig Company (APC) |

Subject | Maths |

Class | 6th |

Chapter-4 | Playing With Numbers |

Writer | ML Aggarwal |

Book Name | Understanding |

Topics | Solution of Check Your Progress Questions |

Edition | 2023-2024 |

**Playing With Numbers Check Your Progress**

**ML Aggarwal Class 6 ICSE Maths Solutions**

Page-90

**Question 1. Write all factors of:**

(i) 88

(ii) 105

(iii) 96

**Answer:**

(i) 88 = {1,2,4,8,11,22,44,88}

(ii) 105 = {1,3,5,7,15,21,35,105}

(iii) 96 = {1,2,3,4,6,8,12,16,24,32,48,96}

**Question 2. Find the common multiples of 8 and 12.**

**Answer:**

The multiple of 8 are 8, 16,24,32,40,48,56,64,72,80,88,96,104

The multiple of 12 are 12,24,36,48,60,72,84,96,108

The common multiple of 8 and 12 are 24,48,72,96

**Question 3. Which of the following pairs of numbers are co-prime?**

(i) 25 and 105

(ii) 59 and 97

(iii) 161 and 192

**Answer:**

**(i) 25 and 105**

The factor of 25 are 1,5,25

The factor of 105 are 1,3,5,7,15,21,35,105

The common factors of 25 and 105 are 1,5

So, They are not co-prime

**(ii) 59 and 97**

The factor of 59 are 1,59

The factor of 97 are 1,97

The common factors of 59 and 97 are 1

So, They are co-prime

**(iii) 161 and 192**

The factor of 161 are 1,161

The factor of 192 are 1,2,3,4,6,8,12,16,24,32,48,64,96,192

The common factors of 161 and 192 are 1

So, They are co-prime

**Playing With Numbers Check Your Progress**

**ML Aggarwal Class 6 ICSE Maths Solutions**

Page-91

**Question 4. Using divisibility tests, determine which of the following numbers are divisible by 4, 6, 8, 9 or 11:**

(i) 197244

(ii) 613440

(iii) 4100448

**Answer:**

(i) 197244 divisible by 4, 6, 9.

It is divisible by 4 as last two digits is divisible by 4.

It is divisible by 6 as last digit of given number is divisible 2 and their sum is also divisible

by 3

Sum of digits 1 +9+7+2 +4 +4 = 27

Which is divisible by 3.

It is not divisible by 8 as the sum of last

three digit 2 +4 + 4= 10, is not divisible by 8.

It is divisible by 9 as the sum of its digits 27 is divisible by 9.

It is not divisible by 11 as the difference of the sum of alternate number 1+7 +4 = 12 and 9+2+4 15, (15 12) = 3 is not divisible by 11.

197244 is divisible by 4, 6 and 9.

(ii) 613440: divisible by 4, 6, 8, 9.

It is divisible by 4 as last two digit is divisible by 4

It is divisible by 6 as the sum of all digits (6 + 1 + 3 + 4 + 4 +0) = 18 is divisible by

and by 2 also as last digit is 0.

it is divisible by 8 as the sum of last three digits (4 + 4+0) = 8 is divisible 8.

It is also divisible by 9 as the sum ot its digits 18 is divisible by 9.

It is not divisible by 11 as the difference of the sum of alternate number 6+3+4 = 13 and 1+4+0=5, (13-5) = 8 is not divisible by 11

613440 is divisible by 4, 6, 8 and 9.

(iii) 4100448: divisible by 4, 6, 8,11

It is divisible by 4 as last two digit is divisible by 4.

It is divisible by 6 as the sum of all digits 4+ 1+0+0+4+4+8 = 21 is divisible by 3

and also last digit is divisible by 2.

It is divisible by 8 as the sum of last three digits 4 + 4 +8= 16 is divisible by 8.

It is not divisible by 9 as the sum of its digit 21 is not divisible by 9.

It is divisible by 11 as the difference of the sum of alternate number 4 +0 +4 +8 = 16

and 1 +0+4 = 5,

(16-6) = 11 which is divisible by 11.

So, 4100448 is divisible by 4, 6, 8, 11.

**Question 5. In 92 * 389, replace * by a digit so that the number formed is divisible by 11.**

**Answer:**

The given number is 92 * 389

Here, * occur at odd place

Sum of digits at odd place 9+8 =17

Sum of digits at even place 2+3+9 = 14

Their difference 17- 14 = 3

If * is replaced by 8, then sum of digits at odd place = 9+8+8 =25

Their difference (Sum of digits at odd places – Sum of digits at even places)

= 25 – 14 = 11

Which is divisible by 11

So, * is to be replaced by the digit 8.

**Question 6. Find the prime factorization of the following numbers:**

(i) 168

(ii) 2304

**Answer:**

(i) 168

Factor of 168 = 2 x 2 x 2 x 3 x 7

(ii) 2304

Factor of 2304 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3

**Question 7. Find the G.C.D. of the given numbers by prime factorization method :**

(i) 24,45

(ii) 180, 252, 324

**Answer:**

(i) 24,45

Factor of 24 = 2 x 2 x 2 x 3

Factor of 45 = 3 x 3 x 5

The common factor is 3

GDC = 3

(ii) 180, 252, 324

Factor of 180 = 2 x 2 x 3 x 3 x 5

Factor of 252 = 2 x 2 x 3 x 3 x 7

Factor of 324 = 2 x 2 x 3 x 3 x 3 x 3

GDC = 2 x 2 x 3 x 3 = 36

**Question 8. Find the H.C.F of the given numbers by division method.**

(i) 54, 82

(ii) 84, 120, 156

**Answer:**

**(i) 54, 82**

H.C.F of the given numbers is = 2

**(ii) 84, 120, 156**

H.C.F of the given numbers is = 12

**Question 9. Find the L.C.M of the given numbers by prime factorization method.**

(i) 27, 90

(ii) 36, 48, 210

**Answer:**

(i) 27, 90

L.C.M of the given numbers is = 2 x 3 x 3 x 3 x 5 = 270

(ii) 36, 48, 210

L.C.M of the given numbers is = 2 x 2 x 3 x 3 x 2 x 2 x 5 x 7 = 5040

**Question 10. Find the L.C.M of the given numbers by division method:**

(i) 48, 60

(ii) 112, 168, 266

**Answer:**

(i) 48, 60

L.C.M of the given numbers is = 2 x 2 x 3 x 4 x 5 = 240

(ii) 112, 168, 266

L.C.M of the given numbers is = 2 x 2 x 2 x 7 x 2 x 3 x 19 = 6384

**Question 11. Find the greatest number which divides 2706, 7041 and 8250 leaving remainder 6, 21 and 42 respectively.**

**Answer:**

When 2706 is divided by the required number, 6 is left as a remainder.

So, 2706 – 6 = 2700

ex. 2700 is exactly divisible by that number.

Similarly, 7041 – 21 = 7020 is exactly divisible by that number.

Similarly, also, 8250 – 42 = 8208 is exactly divisible by that number

**Therefore,**

2700, 7020 and 8208 are divisible by that number.

Thus, the required number is the H.C.F. of 2700, 7020 and 8208

First, we find H.C.F. of 270Land 7020

So, the HCF of 2700, 7020 and 8208 is 108.

Hence, the required number is 108

**Question 12. Find the least number which on decreasing by 20 is exactly divisible by 18, 21, 28 and 30.**

**Answer:**

We, Find the least number which is exactly divisible by the numbers 18, 21, 28, and 30. For this, we find the LCM of 198, 21 28 and 30

So, L.C.M of the given numbers is = 2 x 2 x 3 x 3 x 5 x 7 = 1260

The required number = 1260 + 20 = 1280

**Question 13. There are three heaps of rice weighing 120 kg, 144 kg and 204 kg. Find the maximum capacity of a bag so that the rice of each heap can be packed in exact number of bags.**

**Answer:**

Weight of three heaps = 120 kg 144 kg and 204 kg

So, maximum capacity of a bag, which is exactly divides the heaps in exact number HCF of 120, 144, 204 = 12

So, Required capacity of bags = 12 kg

**Question 14. Three bells are ringing continuously at intervals of 30, 36 and 45 minutes respectively. At what time will they ring together again if they ring simultaneously at 8 a.m.**

**Answer:**

L.C.M of the given numbers is = 2 x 3 x 3 x 2 x 5 = 180

After 180 minute at 11: 00 am

**Question 15. Three brands A, B, C of biscuits are available in packets of 12, 15, and 21 biscuits respectively. If a shopkeeper wants to buy equal number of biscuits of each brand, what is the minimum number of packets of each brand he should buy ?**

**Answer:**

For a minimum number of biscuits, first, we find the LCM of 12, 15 and 21.

The prime factors of the following numbers are:

12=2×2×3

15=3×5

21=3×7

The maximum number of times the prime factor 2 occurs is 2.

The maximum number of times the prime factor 3 occurs is 1.

The maximum number of times the prime factor 5 occurs is 1.

The maximum number of times the prime factor 7 occurs is 1.

∴ LCM= 3 × 4 × 5 × 7 = 420

∴ Minimum number of packets of brand A =420/12=35

Minimum number of packets of brand B =420/15=28

And minimum number of packets of brand C =420/21=20

**Question 16. Two numbers are co-prime and their L.C.M. is 4940. If one of the numbers is 65, find the other number.**

**Answer:**

One number = 65

and other number = x

Two numbers are co-prime if their HCF is 1

Now, HCF x LCM of two numbers = Product of given two numbers

1 x 4940 = 65 x x

4940 = 65 x x

65 x x = 4940

x = 4940 / 65 = 76

So, the other number is 76

— : End of ML Aggarwal Playing With Numbers Check Your Progress Class 6 ICSE Maths Solutions :–

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