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:
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:
4. Determine the smallest 3digit number which is exactly divisible by 6, 8 and 12.
Answer:
5. Determine the greatest 3digit number exactly divisible by 8, 10 and 12.
Answer:
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:
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:
8. Find the least number which when divided by 6, 15 and 18 leave remainder 5 in each
case.
Answer:
9. Find the smallest 4digit number which is divisible by 18, 24 and 32.
Answer:
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:
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:
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 :  




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 
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 :  




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 
The prime factorisations of 6,8 and 12 are  




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 3digit number exactly divisible by 6, 8 and 12 is 120 
The prime factorisations of 8 , 10 and 12 are  




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 3digit number exactly divisible by 8, 10 and 12 is 960 
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 :  




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 
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 :  




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 
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 :  


LCM of 6, 15 and 18= 2 ×3×3×5 = 90  
∴ required number is = LCM + given Remainder = 90 + 5= 95 
the smallest 4digit 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 :  


LCM of 18, 24 and 32= 2 ×2 ×2 ×2 ×2 ×3×3× = 288  
∴ required smallest 4digit number which is divisible by 18, 24 and 32number is = 4 × 288 = 1152 
(a) The prime factorisations of 9 and 4 are  



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  



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  



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  



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 
(a)The prime factorisations of 5 and 20 are  



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  



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  



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  



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. 
best man cool these are my ans that i want
ReplyDeleteDetermine the smallest 3digit number which is exactly divisible by 6, 8 and 12. Answer:
ReplyDeletehere u gave 72 as the ans, but the question is smallest 3 digit number then how can be the ans is 72? please explain.
ist take lcm 6,8,12
Deletethe lcm is 24
then find the multiple 24
24*5=120 is the lowest 2 digit no which is divisible by 6,8,12
G
DeleteWell that answer is wrong here. LCM is supposed to be 24 and final answer would be 120
ReplyDeleteSo much thank you because of this only now I am ready for my exam
ReplyDeleteawesome now i can solve these problems
ReplyDelete