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

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

(ii) 6, 8, 45

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

(iii) 24, 40, 84

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

(iv) 20, 36, 63, 67

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

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

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

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.

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

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.

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

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.

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

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.

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

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

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

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

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 :–

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