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

Playing With Numbers ML Aggarwal Solutions for ICSE Class-6 Mathematics Exercise - 4.3 img 36

(ii) 84, 120, 156

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

Playing With Numbers ML Aggarwal Solutions for ICSE Class-6 Mathematics Exercise - 4.3 img 37

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

Playing With Numbers ML Aggarwal Solutions for ICSE Class-6 Mathematics Exercise - 4.3 img 38

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

Playing With Numbers ML Aggarwal Solutions for ICSE Class-6 Mathematics Exercise - 4.3 img 39

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

Playing With Numbers ML Aggarwal Solutions for ICSE Class-6 Mathematics Exercise - 4.3 img 40

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

Playing With Numbers ML Aggarwal Solutions for ICSE Class-6 Mathematics Exercise - 4.3 img 41

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

Playing With Numbers ML Aggarwal Solutions for ICSE Class-6 Mathematics Exercise - 4.3 img 42

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

Playing With Numbers ML Aggarwal Solutions for ICSE Class-6 Mathematics Exercise - 4.3 img 43

So, the other number is 76

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

Return to   ML Aggarwal Maths Solutions for ICSE Class -6

Thanks

 Share with your friends

Leave a Comment

This site uses Akismet to reduce spam. Learn how your comment data is processed.