ML Aggarwal Playing With Numbers Exe-4.5 Class 6 ICSE Maths Solutions

ML Aggarwal Playing With Numbers Exe-4.5 Class 6 ICSE Maths Solutions. We Provide Step by Step Answer of  Exe-4.5 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 Exe-4.5 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 Exe-4.5 Questions
Edition 2023-2024

Playing With Numbers Exe-4.5

ML Aggarwal Class 6 ICSE Maths Solutions

Page-86

Question 1. Find the L.C.M. of the given numbers by prime factorization method :

(i) 28, 98
(ii) 36, 40, 126
(iii) 108, 135, 162
(iv) 24, 28, 196.

Answer:

(i) Prime factorization of the given number are

28 = 2 x 2 x 7

98 = 2 x 7 x 7

Here 2 and 7 occurs as a prime factor maximum 2 times

so, LCM = 2x 2 x 7 x 7 = 196

(ii) Prime factorization of the given number are

36 = 2 x 2 x 3 x 3

40 = 2 x 2 x 2 x 5

126 = 2 x 3 x 3 x 7

2 occurs  as a prime factor maximum 3 times,

3 two times , 5 and 7 are one time

so, LCM = 2x 2 x 2 x 3 x 3 x 5 x 7 = 2520

(iii) Prime factorization of the given number are

108 = 2 x 2 x 3 x 3  x 3

135 = 3 x 3 x 3 x 5

162 = 2 x 3 x 3 x 3 x 3

2 occurs  as a prime factor maximum 2 times,

3 four times , 5 one time

so, LCM = 2x 2 x 3 x 3 x 3 x 3 x 5 = 1720

(iv) Prime factorization of the given number are

24 = 2 x 2 x 2 x 3

28 = 2 x 2 x 7

196 = 2 x 7 x 7 x 7

2 occurs  as a prime factor maximum 3 times,

3 one times , 7 two times

so, LCM = 2x 2 x 2 x 3 x 7 x 7 = 1176


Playing With Numbers Exe-4.5

ML Aggarwal Class 6 ICSE Maths Solutions

Page-87

Question 2. Find the L.C.M. of the given numbers by division method :

(i) 480, 672
(ii) 6, 8, 45
(iii) 24, 40, 84
(iv) 20, 36, 63, 67

Answer:

(i) 480, 672

Playing With Numbers ML Aggarwal Solutions for ICSE Class-6 Mathematics Exercise - 4.3 img 14

So, LCM = 2 x 2 x 2 x 2 x 2 x 3 x 5 x 7 = 3360

(ii) 6, 8, 45

Playing With Numbers ML Aggarwal Solutions for ICSE Class-6 Mathematics Exercise - 4.3 img 15

So, LCM = 2 x 2 x 2 x 3 x 3 x 5 = 360

(iii) 24, 40, 84

Playing With Numbers ML Aggarwal Solutions for ICSE Class-6 Mathematics Exercise - 4.3 img 16

So, LCM = 2 x 2 x 2 x 3 x 5 x 7 = 840

(iv) 20, 36, 63, 67

Playing With Numbers ML Aggarwal Solutions for ICSE Class-6 Mathematics Exercise - 4.3 img 17

So, LCM = 2 x 2 x 2 x 3 x 3 x 5 x 7 x 11 = 813860

 Question 3 . Find the least number which when increased by 15 is exactly divisible by 15, 35 and 48.

Answer:

First, we find the least number which is exactly divisible by the numbers  15, 35 and
48. For this, we find L.C.M. of 15, 35 and 48

Playing With Numbers ML Aggarwal Solutions for ICSE Class-6 Mathematics Exercise - 4.3 img 18

L.C.M. = 3 x 5 x 7 x 2 x 2 x 2 x 2 = 1680
According to given condition, the required number will be 15 less than 1680.

The required least number 1680 -15 = 1665

Question 4. Find the least number which when divided by 6, 15 and 18 leaves remainder 5 in each case.

Answer:

L.C.M. of 6, 15 and 18

= 2 x 3 x 3 x 5 = 90

Hence, the required number is  90 + 5  … (ex. 95)

Question 5. Find the least number which when divided by 24, 36, 45 and 54 leaves a remainder of 3 in each case.

Answer:

24, 36, 45 and 54

Playing With Numbers ML Aggarwal Solutions for ICSE Class-6 Mathematics Exercise - 4.3 img 19

So, LCM = 2 x 2 x 2 x 3 x 3 x 3 x 5 = 1080

According to given condition the number will be 3 more than 1080.

So, The required number = 1080 + 3 = 1083

Question-6. Find the greatest 3-digit number which is exactly divisible by 8, 20 and 24.

Answer:

We, find the LCM of 8, 20 and 24

Playing With Numbers ML Aggarwal Solutions for ICSE Class-6 Mathematics Exercise - 4.3 img 20

LCM of given numbers = 2 x 2 x 2 x 3 x 5 = 120
Greatest number of 3 digit is 999
We divide 999 by 120 and find the remainder.

Playing With Numbers ML Aggarwal Solutions for ICSE Class-6 Mathematics Exercise - 4.3 img 21

According to given condition, we need a greatest 3-digit number which is exactly
divisible by 120.

So, The required number = 999 – 39 = 960

Question 7. Find the smallest 4-digit number which is exactly divisible by 32, 36 and 48.

Answer:

We, find the LCM of 32, 36 and 48

Playing With Numbers ML Aggarwal Solutions for ICSE Class-6 Mathematics Exercise - 4.3 img 22

LCM of given numbers = 2 x 2 x 2 x 2 x 2 x 3 x 3 = 288
Smallest number of 4  digit is 1000
We divide 1000 by 288 and find the remainder.

Playing With Numbers ML Aggarwal Solutions for ICSE Class-6 Mathematics Exercise - 4.3 img 23

According to given condition, we need a least number which is exactly
divisible by 288.

So, The required number = 1000 + (288- 136) = 1152

Question 8. Find the greatest 4-digit number which is exactly divisible by each of 8, 12 and 20.

Answer:

We, find the LCM of 8, 12 and 20

Playing With Numbers ML Aggarwal Solutions for ICSE Class-6 Mathematics Exercise - 4.3 img 24

LCM of given numbers = 2 x 2 x 3 x 5 = 120
According to given condition, we need a greatest number which is exactly
divisible by 120

Greatest number of 4 digit = 9999

We divide 9999 by 120 and find the remainder.

Playing With Numbers ML Aggarwal Solutions for ICSE Class-6 Mathematics Exercise - 4.3 img 25

So, The required number = 9999 – 39  = 9960

Question 9. Find the least number of five digits which is exactly divisible by 32, 36 and 45.

Answer:

We, find the LCM of 32, 36 and 45

Playing With Numbers ML Aggarwal Solutions for ICSE Class-6 Mathematics Exercise - 4.3 img 26

LCM of given numbers = 2 x 2 x 2 x 2 x 3 x 3 x 5 = 1440

Smallest number of 5 digit = 10000

We divide 10000 by 1440 and find the remainder.

Playing With Numbers ML Aggarwal Solutions for ICSE Class-6 Mathematics Exercise - 4.3 img 27

According to given condition, we need a least number which is exactly
divisible by 1440.

So, The required number = 10000 + 1440 – 1360

= 10080

Question 10. Three boys step off together from the same spot. Their steps measure 63 cm, 70 cm and 77 cm respectively. What is the minimum distance each should cover so that all can cover the same distance in complete steps?

Answer:

LCM of 63, 70 and 77

Playing With Numbers ML Aggarwal Solutions for ICSE Class-6 Mathematics Exercise - 4.3 img 28

63 = 3 x 3 x 7

70 = 2 x 5 x 7

77 = 7 x 11

so, LCM = 3 x 3 x 2 x 5 x 7 x 11 = 6930

so The minimum distance each will cover is 6930 cm

ex. 69 m 30 cm

Question 11. Traffic lights at three different road crossing change after 48 seconds, 72 seconds and 108 seconds respectively. At what time will they change together again if they change simultaneously at 7 A.M.?

Answer:

LCM of 48, 72 and 108

Playing With Numbers ML Aggarwal Solutions for ICSE Class-6 Mathematics Exercise - 4.3 img 29

2 x 2 x 2 x 2 x 3 x 3 x 3  = 432

432 second = 7 minute 12 second past 7 AM

Question 12. If the product of two numbers is 4032 and their HCF is 12, find their LCM.

Answer

The product of two numbers is 4032

HCF = 12

LCM = 4032 / 12

336

Question 13. The HCF and LCM of two numbers are 9 and 270 respectively. If one of the numbers is 45, find the other number.

Answer:

HCF x LCM = one number x other number

9 x 270 = 45 x 2nd number

2430 =  45 x 2nd number

2430/45 = 162/3 54

So, 54 is the other number

Question 14. Find the HCF of 180 and 336. Hence, find their LCM.

Answer:

Division method : HCF of 180 and 336

Playing With Numbers ML Aggarwal Solutions for ICSE Class-6 Mathematics Exercise - 4.3 img 30

So, HCF of 180 and 336 = 12

Products of number LCM of 180 and 336 = Products of number/ their HCF

= (180 x 336)/ 12

=15 x 336

= 5040

Question 15. Can two numbers have 15 as their HCF and 110 as their LCM? Give reason to justify your answer.

Answer:

One dividing 110 by 15, we get

Playing With Numbers ML Aggarwal Solutions for ICSE Class-6 Mathematics Exercise - 4.3 img 31

7 is quotient and 5 is reminder

We find that the reminder ≠ 0

So, 110 is not exactly divisible by 15

So, HCF and LCM of two number can not be 15 and 115 respectively.

As LCM of two number is always exactly divisible by their HCF.

—  : End of ML Aggarwal Playing With Numbers Exe-4.5 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.