CBSE Class 6th ( VI) Mathematics Chapter 3. Playing with Numbers: Exercise 3.7 Solved

1. Renu purchases two bags of fertiliser of weights 75 kg and 69 kg. Find the maximum value of weight which can measure the weight of the fertiliser exact number of times.

Answer:

The Highest Common Factor (HCF) of 75kg and 69kg by weight will be the maximum value of weight which can measure the weight of the fertiliser exact number of times. To find HCF first we have to find prime factorisation of given numbers by weight i.e. 75 and 69
5 75 5 15 3 3 1	3 69 23 23 1
Prime Factor of 75 = 5 ×5 ×3
Prime Factor of 69 = 3 ×23
Highest Common Factors of 75 and 69 is 3
∴ Hence 3kg is the maximum value of weight which can measure the weight of the fertiliser in both the bags weighing 75kg and 69kg respectively, exact number of times

2. 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 distance in complete steps? Answer:

The minimum distance each should cover so that all can cover the distance in complete steps will be simply the Lowest Common Multiple(LCM) of theirs each steps.To find LCM we have to first find the prime factorisations of 63,70 and 77 respectively as given below :
3 63 3 21 7 7 1	2 70 5 35 7 7 1	7 77 11 11 1
Prime factors of 63 = 3 ×3× 7
Prime factors of 70 = 2 ×5 ×7
Prime factors of 77 = 7 ×11
In these prime factorisations, the maximum number of times the prime factors occurring for all the numbers 63,70 and 77 are =2,3,3,5,7,11
∴ Lowest Common Multiple of 63, 70 and 77 is = 2 × 3 × 3 × 5 × 7 × 11× = 6930 Hence,the minimum distance each should cover so that all can cover the distance in complete steps is 6930 cm

3. The length, breadth and height of a room are 825 cm, 675 cm and 450 cm respectively. Find the longest tape which can measure the three dimensions of the room exactly. Answer:

The longest tape which can measure the three dimensions of the room exactly will be simply the HCF of the given length, breadth and height of a room i.e. 825 cm, 675 cm and 450 cm.To find HCF, we have to first find the prime factorisations of 825,675 and 450 as given below :
3 825 5 275 5 55 11 11 1	3 675 3 225 3 75 5 25 5 5 1	2 450 3 225 3 75 5 25 5 5 1
Prime factorisation of 825 = 3 ×5 ×5× 11
Prime factorisation of 675 =3×3 ×3×5×5
Prime factorisation of 450 = 2 ×3×3×5×5
In these prime factorisations, the common factors occurring for both all the numbers 825,675 and 450 are = 3, 5, 5
∴ HCF of the numbers 825,675 and 450 is = 3 × 5 × 5 = 75 ∴The longest tape which can measure given length, breadth and height of a room is 75 cm

4. Determine the smallest 3-digit number which is exactly divisible by 6, 8 and 12. Answer:

The prime factorisations of 6,8 and 12 are
2 6 3 3 1	2 8 2 4 2 2 1	2 12 2 6 3 3 1
6 = 2 × 3
8 = 2×2 ×2
12 = 2 ×2× 3
In these prime factorisations, the maximum number of times the prime factors occurring for  the numbers 6,8 and 12 are = 2,2,2,3
∴ Lowest Common Multiple of 6 , 8 and 12 is = 2 × 2 × 2 × 3  =24 Further multiple of LCM 24  i.e. 48, 72, 96, 120, 144  will be exactly divisible by 6, 8 and 12 . Clearly, above the smallest 3-digit number exactly divisible by 6, 8 and  12 is 120

5. Determine the greatest 3-digit number exactly divisible by 8, 10 and 12. Answer:

The prime factorisations of 8 , 10 and 12 are
2 8 2 4 2 2 1	2 10 5 5 1	2 12 2 6 3 3 1
Factor of 8 = 2 ×2 ×2
Factor of 10 = 2 ×5
Factor of 12 = 2 × 2 × 3
∴ Lowest Common Multiple of  8, 10 and 12 are =2 × 2 × 2× 5 × 3=120
LCM 120 is  exactly divisible by 8, 10 and 12 Further multiple of LCM 120  i.e. 240, 360, 480, 600, 720, 840, 960, 1080 will be exactly divisible by 8, 10 and 12 . Clearly, above the greatest 3-digit number exactly divisible by 8, 10 and  12 is 960

6. The traffic lights at three different road crossings change after every 48 seconds, 72 seconds and 108 seconds respectively. If they change simultaneously at 7 a.m., at what time will they change simultaneously again? Answer:

All three traffic lights with changing times of 48 seconds, 72 seconds and 108 seconds respectively, after starting at 7 a.m., will change simultaneously again after a period of time which will be the Lowest Common Multiple(LCM) of theirs chaging times respectively.To find LCM we have to first find the prime factorisations of 48,72 and 108 respectively as given below :
2 48 2 24 2 12 2 6 3 3 1	2 72 2 36 2 18 3 9 3 3 1	2 108 2 54 3 27 3 9 3 3 1
Prime factors of 48 = 2 ×2×2 ×2×3
Prime factors of 72 = 2 ×2 ×2 ×3 ×3 ×7
Prime factors of 108 = 2 ×2 ×3 ×3 ×3
The prime factor 2 appears maximum number of four times in the prime factorisation of 48, the prime factor 3 occurs maximum number of three times in the prime factorisation of 108. Taking the maximum number of times the prime factors occurring for all the numbers 48, 72 and 108 Lowest Common Multiple(LCM) =2 ×2 ×2 ×2 × 3 ×3 ×3 ×=432
∴ The three lights will change simultaneously again after a period of 432 seconds or 432 ÷ 60 = 7 Min 12 Second

7. Three tankers contain 403 litres, 434 litres and 465 litres of diesel respectively. Find the maximum capacity of a container that can measure the diesel of the three containers exact number of times. Answer:

The maximum capacity of a container that can measure the diesel of the three containers exact number of times. will be simply the HCF of the given capcities of Three tankers i.e. 403 litres, 434 litres and 465 litres of diesel respectively. To find HCF, we have to first find the prime factorisations of 403,434 and 465 as given below :
13 403 31 31 1	2 434 7 217 31 31 1	3 465 5 155 31 31 1
Prime factorisation of 403 = 13 ×31
Prime factorisation of 434 = 2×7 ×31
Prime factorisation of 465= 3×5×31
In these prime factorisations, the only common factor occurring for all the numbers 403, 434 and 465 is = 31 which is HCF also
∴ HCF of the numbers 403, 434 and 465 = 31 Hence, The maximum capacity of a container that can measure the diesel of the three containers exact number of times is 31 Litres

8. Find the least number which when divided by 6, 15 and 18 leave remainder 5 in each case. Answer:

The least number which when divided by 6, 15 and 18 leave remainder 5 in each case, will be simply the sum of LCM of the given of Three 6, 15 , 18 and remainder 5. LCM of 6, 15 and 18 is as given below :

2 6 15 18 3 3 15 9 3 1 5 3 5 1 1 1

LCM of 6, 15 and 18= 2 ×3×3×5 = 90

∴ required number is = LCM + given Remainder = 90 + 5= 95

9. Find the smallest 4-digit number which is divisible by 18, 24 and 32. Answer:

the smallest 4-digit number which is divisible by 18, 24 and 32., will be simply the 4 digit number necessarily a multiple of LCM of the given numbers i.e. 18, 24 and 32.below :

2 18 24 32 2 9 12 16 2 9 6 8 2 9 3 4 2 9 3 2 3 9 3 1 3 3 1 1 1 1 1

LCM of 18, 24 and 32= 2 ×2 ×2 ×2 ×2 ×3×3× = 288

∴ required smallest 4-digit number which is divisible by 18, 24 and 32number is = 4 × 288 = 1152

10. Find the LCM of the following numbers : (a) 9 and 4 (b) 12 and 5 (c) 6 and 5 (d) 15 and 4 Observe a common property in the obtained LCMs. Is LCM the product of two numbers in each case? Answer:

(a) The prime factorisations of 9 and 4 are
3 9 3 3 1	2 4 2 2 1
9 = 3 ×3
4 = 2 ×2
In these prime factorisations, the maximum number of times the prime factors occurring for both the numbers 9 and 4 are = 3,3,2,2
∴ Lowest Common Multiple of 9 and 4 is = 3 × 3 × 2 × 2= 36

(b) The prime factorisations of 12 and 5 are
2 12 2 6 3 3 1	5 5 1
12 = 2 ×2 ×3
5 = 1 ×5
In these prime factorisations, the maximum number of times the prime factors occurring for both the numbers 12 and 5 are =2,2,3,5
∴ lowest Common multiple of 12 and 5 is = 2 × 2 × 3 × 5= 60

(c) The prime factorisations of 6 and 5 are
2 6 3 3 1	5 5 1
6 = 2 × 3
5 = 1 ×5
In these prime factorisations, the maximum number of times the prime factors occurring for both the numbers 6 and 5 are =2,3,5
∴ lowest Common multiple of 6 and 5 is = 2 × 3 × 5= 30

(d) The prime factorisations of 15 and 4 are
3 15 5 5 1	2 4 2 2 1
15 = 3 × 5
4 = 2 ×2
In these prime factorisations, the maximum number of times the prime factors occurring for both the numbers 15 and 4 are =3,5,2,2
∴Lowest Common multiple of 9 and 4 is = 3 × 5 × 2 × 2= 60
Common property in the obtained LCMs. we have observed : Here, in each case LCM i.e. 36, 60, 30, 60 is a multiple of 3 Yes, in each case LCM = the product of two numbers i.e. 36=9×4, 60=12×5, 30=6×5, 60=15×4

11. Find the LCM of the following numbers in which one number is the factor of the other. (a) 5, 20 (b) 6, 18 (c) 12, 48 (d) 9, 45 What do you observe in the results obtained? Answer:

(a)The prime factorisations of 5 and 20 are
5 5 1	2 20 2 10 5 5 1
5 = 5
20 = 2 ×2 ×5
In these prime factorisations, the maximum number of times the prime factors occurring for both the numbers 5 and 20 are =2,2,5
∴ Lowest Common Multiple of 5 and 20 is = 2 × 2 × 5= 20

(b) The prime factorisations of 6 and 18 are
2 6 3 3 1	2 18 3 9 3 3 1
6 = 2 ×3
18 = 2 ×3×3
In these prime factorisations, the maximum number of times the prime factors occurring for both the numbers 6 and 18 are =2,3,3
∴ lowest Common multiple of 6 and 18 is = 2 × 3 × 3 ×= 18

(c) The prime factorisations of 12 and 48 are
2 12 2 6 3 3 1	2 48 4 24 6 6 1
12 = 2 × 3 × 3
48 = 2 ×4 ×6
In these prime factorisations, the maximum number of times the prime factors occurring for both the numbers 12 and 48 are =2,4,6
∴ lowest Common multiple of 12 and 48 is = 2×4×6=48

(d) The prime factorisations of 9 and 45 are
3 9 3 3 1	5 45 9 9 1
9 = 3 × 5
45 = 2 ×2
In these prime factorisations, the maximum number of times the prime factors occurring for both the numbers 9 and 45 are =5,9
∴Lowest Common multiple of 9 and 45 is = 5 × 9 ×= 45
We have observed in the results obtained that the LCM of the given numbers in each case is the larger of the two numbers.