Applications of the Fundamental Theorem of Arithmetic are used to find the LCM and HCF of positive integers. So, this exercise deals with problems in finding the LCM and HCF by the prime factorisation method. Also, the relationship between LCM and HCF is explained in the RD Sharma Solutions Class 10 Exercise 1.4. For quick access to the complete solutions of this exercise, the RD Sharma Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.4 PDF is provided below.
RD Sharma Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.4
Access Answers to RD Sharma Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.4
1. Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = Product of the integers:
(i) 26 and 91
Solution:
Given integers are: 26 and 91
First, find the prime factors of 26 and 91.
26 = 2 × 13
91 = 7 × 13
∴ LCM (26, 91) = 2 × 7 × 13 = 182
And,
HCF (26, 91) = 13
Verification:
LCM × HCF = 182 x 13= 2366
And, product of the integers = 26 x 91 = 2366
∴ LCM × HCF = product of the integers
Hence verified.
(ii) 510 and 92
Solution:
Given integers are: 510 and 92
First, find the prime factors of 510 and 92.
510 = 2 × 3 × 5 × 17
92 = 2 × 2 × 23
∴ LCM (510, 92) = 2 × 2 × 3 × 5 × 23 × 17 = 23460
And,
HCF (510, 92) = 2
Verification:
LCM × HCF = 23460 x 2 = 46920
And, product of the integers = 510 x 92 = 46920
∴ LCM × HCF = product of the integers
Hence verified.
(iii) 336 and 54
Solution:
Given integers are: 336 and 54
First, find the prime factors of 336 and 54.
336 = 2 × 2 × 2 × 2 × 3 × 7
54 = 2 × 3 × 3 x 3
∴ LCM (336, 54) = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 7 = 3024
And,
HCF (336, 54) = 2 x 3 = 6
Verification:
LCM × HCF = 3024 x 6 = 18144
And, product of the integers = 336 x 54 = 18144
∴ LCM × HCF = product of the integers
Hence verified.
2. Find the LCM and HCF of the following integers by applying the prime factorisation method:
(i) 12, 15 and 21
Solution:
First, find the prime factors of the given integers: 12, 15 and 21
For, 12 = 2 × 2 × 3
15 = 3 × 5
21 = 3 × 7
Now, LCM of 12, 15 and 21 = 2 × 2 × 3 × 5 × 7
∴ LCM (12, 15, 21) = 420.
And, HCF (12, 15 and 21) = 3.
(ii) 17, 23 and 29
Solution:
First, find the prime factors of the given integers: 17, 23 and 29
For, 17 = 1 × 17
23 = 1 × 23
29 = 1 × 29
Now, LCM of 17, 23 and 29 = 1 × 17 × 23 × 29
∴ LCM (17, 23, 29) = 11339.
And, HCF (17, 23 and 29) = 1.
(iii) 8, 9 and 25
Solution:
First, find the prime factors of the given integers: 8, 9 and 25
For, 8 = 2 × 2 x 2
9 = 3 × 3
25 = 5 × 5
Now, LCM of 8, 9 and 25 = 23 × 32 × 52
∴ LCM (8, 9, 25) = 1800.
And, HCF (8, 9 and 25) = 1.
(iv) 40, 36 and 126
Solution:
First, find the prime factors of the given integers: 40, 36 and 126
For, 40 = 2 x 2 x 2 × 5
36 = 2 x 2 x 3 x 3
126 = 2 × 3 × 3 × 7
Now, LCM of 40, 36 and 126 = 23 × 32 × 5 × 7
∴ LCM (40, 36, 126) = 2520.
And, HCF (40, 36 and 126) = 2.
(v) 84, 90 and 120
Solution:
First, find the prime factors of the given integers: 84, 90 and 120
For, 84 = 2 × 2 × 3 × 7
90 = 2 × 3 × 3 × 5
120 = 2 × 2 × 2 × 3 × 5
Now, LCM of 84, 90 and 120 = 23 × 32 × 5 × 7
∴ LCM (84, 90, 120) = 2520.
And, HCF (84, 90 and 120) = 6.
(vi) 24, 15 and 36
Solution:
First, find the prime factors of the given integers: 24, 15 and 36
For, 24 = 2 × 2 x 2 x 3
15 = 3 × 5
36 = 2 × 2 × 3 × 3
Now, LCM of 24, 15 and 36 = 2 × 2 × 2 × 3 × 3 × 5 = 23 x 32 x 5
∴ LCM (24, 15, 36) = 360.
And, HCF (24, 15 and 36) = 3.
3. Given that HCF (306, 657) = 9, find LCM ( 306, 657 ).
Solution:
Given two integers are: 306 and 657
We know that,
LCM × HCF = Product of the two integers
⇒ LCM = Product of the two integers / HCF
= (306 x 657) / 9 = 22338.
4. Can two numbers have 16 as their HCF and 380 as their LCM? Give reason.
Solution:
On dividing 380 by 16, we get
23 as the quotient and 12 as the remainder.
Now, since the LCM is not exactly divisible by the HCF, it can be said that two numbers cannot have 16 as their HCF and 380 as their LCM.
5. The HCF of the two numbers is 145, and their LCM is 2175. If one number is 725, find the other.
Solution:
The LCM and HCF of the two numbers are 145 and 2175, respectively. (Given)
It is also given that one of the numbers is 725
We know that,
LCM × HCF = first number × second number
2175 × 145 = 725 × second number
⇒ Second number = (2175 × 145)/ 725 = 435
∴ The other number is 435.
6. The HCF of the two numbers is 16, and their product is 3072. Find their LCM.
Solution:
Given,
HCF of two numbers = 16
And, their product = 3072
We know that,
LCM × HCF = Product of the two numbers
LCM × 16 = 3072
⇒ LCM = 3072/ 16 = 192
∴ The LCM of the two numbers is 192.
7. The LCM and HCF of the two numbers are 180 and 6, respectively. If one of the numbers is 30, find the other number.
Solution:
Given,
The LCM and HCF of the two numbers are 180 and 6, respectively. (Given)
It is also given that one of the numbers is 30.
We know that,
LCM × HCF = first number × second number
180 × 6 = 30 × second number
⇒ Second number = (180 × 6)/ 30 = 36
∴ The other number is 36.
8. Find the smallest number that, when increased by 17, is exactly divisible by both 520 and 468.
Solution:
First, let’s find the smallest number, which is exactly divisible by both 520 and 468.
That is simply just the LCM of the two numbers.
By prime factorisation, we get
520 = 23 × 5 × 13
468 = 22 × 32 × 13
∴ LCM (520, 468) = 23 × 32 × 5 × 13 = 4680.
Hence, 4680 is the smallest number which is exactly divisible by both 520 and 468, i.e., we will get a remainder of 0 in each case. But, we need to find the smallest number, which, when increased by 17, is exactly divided by 520 and 468.
So that is found by,
4680 – 17 = 4663
∴ 4663 should be the smallest number which, when increased by 17, is exactly divisible by both 520 and 468.
9. Find the smallest number which leaves remainders 8 and 12 when divided by 28 and 32, respectively.
Solution:
First, let’s find the smallest number, which is exactly divisible by both 28 and 32.
Which is simply just the LCM of the two numbers.
By prime factorisation, we get
28 = 2 × 2 × 7
32 = 25
∴ LCM (28, 32) = 25 × 7 = 224
Hence, 224 is the smallest number which is exactly divisible by 28 and 32, i.e., we will get a remainder of 0 in each case. But, we need the smallest number, which leaves remainders 8 and 12 when divided by 28 and 32, respectively.
So that is found by,
224 – 8 – 12 = 204
∴ 204 should be the smallest number which leaves remainders 8 and 12 when divided by 28 and 32, respectively.
10. What is the smallest number that, when divided by 35, 56 and 91, leaves remainders of 7 in each case?
Solution:
First, let’s find the smallest number, which is exactly divisible by all 35, 56 and 91.
Which is simply just the LCM of the three numbers.
By prime factorisation, we get
35 = 5 × 7
56 = 23 × 7
91 = 13 × 7
∴ LCM (35, 56 and 91) = 23 × 7 × 5 × 13 = 3640
Hence, 3640 is the smallest number that, when divided by 35, 56 and 91, leaves the remainder of 7 in each case.
So that is found by,
3640 + 7 = 3647
∴ 3647 should be the smallest number that, when divided by 35, 56 and 91, leaves the remainder of 7 in each case.
11. A rectangular courtyard is 18m 72cm long and 13m 20 cm broad. It is to be paved with square tiles of the same size. Find the least possible number of such tiles.
Solution:
Given,
Length of courtyard = 18 m 72 cm = 1800 cm + 72 cm = 1872 cm (∵1 m = 100 cm)
Breadth of courtyard = 13 m 20 cm = 1300 cm + 20 cm = 1320 cm
The size of the square tile needed to be paved on the rectangular yard is equal to the HCF of the length and breadth of the rectangular courtyard.
Now, finding the prime factors of 1872 and 1320, we have
1872 = 24 × 32 × 13
1320 = 23 × 3 × 5 × 11
⇒ HCF (1872 and 1320) = 23 × 3 = 24
∴ The length of the side of the square tile should be 24 cm.
Thus, the number of tiles required = (area of the courtyard) / (area of a square tile)
We know that the area of the courtyard = Length × Breadth
= 1872 cm × 1320 cm
And, area of a square tile = (side)2 = (24cm)2
⇒ the number of tiles required = (1872 x 1320) / (24)2 = 4290
Thus, the least possible number of tiles required is 4290.
12. Find the greatest number of 6 digits exactly divisible by 24, 15 and 36.
Solution:
We know that the greatest 6-digit number is 999999.
Let’s assume that 999999 is divisible by 24, 15 and 36 exactly.
Then, the LCM (24, 15 and 36) should also divide 999999 exactly.
Finding the prime factors of 24, 15, and 36, we get
24 = 2 × 2 × 2 × 3
15 = 3 × 5
36 = 2 × 2 × 3 × 3
⇒ LCM of 24, 15 and 36 = 360
Since, (999999)/ 360 = 2777 × 360 + 279
Here, the remainder is 279.
So, the greatest number which is divisible by all three should be = 999999 – 279 = 999720
∴ 999720 is the greatest 6-digit number which is exactly divisible by 24, 15 and 36.
13. Determine the number nearest to 110000 but greater than 100000, which is exactly divisible by each of 8, 15 and 21.
Solution:
First, let’s find the LCM of 8, 15 and 21.
By prime factorisation, we have
8 = 2 × 2 × 2
15 = 3 × 5
21 = 3 × 7
⇒ LCM (8, 15 and 21) = 23 × 3 × 5 × 7 = 840
When 110000 is divided by 840, the remainder that is obtained is 800.
So, 110000 – 800 = 109200 should be divisible by each of 8, 15 and 21.
Also, we have 110000 + 40 = 110040 is also divisible by each of 8, 15 and 21.
⇒ 109200 and 110040 both are greater than 100000, but 110040 is greater than 110000.
Hence, 109200 is the number nearest to 110000 and greater than 100000, which is exactly divisible by each of 8, 15 and 21.
14. Find the least number that is divisible by all the numbers between 1 and 10 (both inclusive).
Solution:
From the question, it’s understood that
The LCM of 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10 will be the least number that is divisible by all the numbers between 1 and 10.
Hence, the prime factors of all these numbers are:
1 = 1
2 = 2
3 = 3
4 = 2 × 2
5 = 5
6 = 2 × 3
7 = 7
8 = 2 × 2 × 2
9 = 3 × 3
10 = 2 × 5
⇒ LCM will be = 23 × 32 × 5 × 7 = 2520
Hence, 2520 is the least number that is divisible by all the numbers between 1 and 10 (both inclusive).
15. A circular field has a circumference of 360 km. Three cyclists start together and can cycle 48,60 and 72 km a day around the field. When will they meet again?
Solution:
In order to calculate the time taken before they meet again, we must first find out the individual time taken by each cyclist to cover the total distance.
The number of days a cyclist takes to cover the circular field = (Total distance of the circular field) / (distance covered in 1 day by a cyclist).
So, for the 1st cyclist, number of days = 360 / 48 = 7.5 which is = 180 hours [∵1 day = 24 hours]
2nd cyclist, number of days = 360 / 60 = 6 which is = 144 hours
3rd cyclist, number of days = 360 / 72 = 5 which is 120 hours
Now, by finding the LCM (180, 144 and 120), we’ll get to know after how many hours the three cyclists meet again.
By prime factorisation, we get
180 = 22 x 32 x 5
144 = 24 x 32
120 = 23 x 3 x 5
⇒ LCM (180, 144 and 120) = 24 x 32 x 5 = 720
So, this means that after 720 hours, the three cyclists meet again.
⇒ 720 hours = 720 / 24 = 30 days [∵1 day = 24 hours]
Thus, all three cyclists will meet again after 30 days.
16. In a morning walk, three persons step off together, their steps measuring 80 cm, 85 cm and 90 cm, respectively. What is the minimum distance each should walk so that he can cover the distance in complete steps?
Solution:
From the question, it’s understood that the required distance each should walk would be the LCM of the measures of their steps, i.e., 80 cm, 85 cm, and 90 cm.
So, by finding LCM (80, 85 and 90) by prime factorisation, we get
80 = 24 × 5
85 = 17 × 5
90 = 2 × 3 × 3 × 5
⇒ LCM (80, 85 and 90) = 24 × 32 × 5 × 17 = 12240 cm = 122m 40 cm [∵ 1 m = 100 cm]
Hence, 122 m 40 cm is the minimum distance that each should walk so that all can cover the same distance in complete steps.
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