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Class 10 RD Sharma Solutions – Chapter 1 Real Numbers – Exercise 1.4
  • Difficulty Level : Hard
  • Last Updated : 21 Dec, 2020

Question 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:

As we know that LCM is “Lowest Common Multiple” and HCF is “Highest Common Factor”. Here we have to 
Verify that LCM x HCF = Product of the integers. 

Given integers in questions are: 26 and 91 
First, we will find the prime factors of 26 and 91. 
26 = 2 × 13 
91 = 7 × 13 
⇒ L.C.M (26, 91) = 2 × 7 × 13 = 182 

H.C.F (26, 91) = 13 

According to question we need to verify the condition: 
L.C.M × H.C.F = 182 × 13= 2366 
Product of the integers = 26 × 91 = 2366 
⇒ L.C.M × H.C.F = product of the integers 



(ii) 510 and 92

Solution:

As we know that LCM is “Lowest Common Multiple” and HCF is “Highest Common Factor”. Here we have to 
Verify that LCM x HCF = Product of the integers. 

Given integers in question are: 510 and 92 
We will find the prime factors of 510 and 92 to find LCM and HCF. 
510 = 2 × 3 × 5 × 17 
92 = 2 × 2 × 23 
⇒ L.C.M (510, 92) = 2 × 2 × 3 × 5 × 23 × 17 = 23460 

H.C.F (510, 92) = 2 

According to question we need to verify the condition: 

L.C.M × H.C.F = 23460 x 2 = 46920 
Product of the integers = 510 x 92 = 46920 
⇒ L.C.M × H.C.F = product of the integers.

(iii) 336 and 54

Solution: 

As we know that LCM is “Lowest Common Multiple” and HCF is “Highest Common Factor”. Here we have to 
Verify that LCM x HCF = Product of the integers. 



Given integers in question are: 336 and 54 
We will find the prime factors of 336 and 54 to find LCM and HCF. 
336 = 2 × 2 × 2 × 2 × 3 × 7 
54 = 2 × 3 × 3 × 3 
⇒L.C.M (336, 54) = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 7 = 3024 

H.C.F (336, 54) = 2 × 3 = 6 

According to question we need to verify the condition: 

L.C.M × H.C.F = 3024 × 6 = 18144 
Product of the integers = 336 × 54 = 18144 
⇒L.C.M × H.C.F = product of the integers 

Question 2. Find the LCM and HCF of the following integers by applying the prime factorization method:

(i) 12, 15 and 21

Solution: 

We will find the prime factors of the given integers: 12, 15 and 21 
Prime factor will help us to find LCM and HCF 
12 = 2 × 2 × 3 
15 = 3 × 5 
21 = 3 × 7 
Now for L.C.M we will find common multiples from each terms. 
L.C.M of 12, 15 and 21 = 2 × 2 × 3 × 5 × 7 
So L.C.M (12, 15, 21) = 420 
And, for H.C.F we will find the highest common factors from each term. 
H.C.F (12, 15 and 21) = 3

 (ii) 17, 23 and 29

Solution: 

We will find the prime factors of the given integers: 17, 23 and 29 
Prime factor will help us to find LCM and HCF 
17 = 1 × 17 
23 = 1 × 23 
29 = 1 × 29 
Now for L.C.M we will find common multiples from each terms. 
L.C.M of 17, 23 and 29 = 1 × 17 × 23 × 29 
So L.C.M (17, 23, 29) = 11339 

And, for H.C.F we will find highest common factors from each term. 
H.C.F (17, 23 and 29) = 1

(iii) 8, 9 and 25

Solution: 

We will the find the prime factors of the given integers: 8, 9 and 25 
Prime factor will help us to find LCM and HCF 
8 = 2 × 2 × 2 
9 = 3 × 3 
25 = 5 × 5 
Now for L.C.M we will find common multiples from each terms. 
L.C.M of 8, 9 and 25 = 23 × 32 × 52 
So L.C.M (8, 9, 25) = 1800 

And, for H.C.F we will find the highest common factors from each term. 
H.C.F (8, 9 and 25) = 1

(iv) 40, 36 and 126

Solution: 

We will find the prime factors of the given integers: 40, 36 and 126 
Prime factor will help us to find LCM and HCF 
40 = 2 × 2 × 2 × 5 
36 = 2 × 2 × 3 × 3 
126 = 2 × 3 × 3 × 7 
Now for L.C.M we will find common multiples from each terms. 
L.C.M of 40, 36 and 126 = 23 × 32 × 5 × 7 
So L.C.M (40, 36, 126) = 2520 
And, for H.C.F we will find highest common factors from each term. 
H.C.F (40, 36 and 126) = 2

(v) 84, 90 and 120

Solution: 

We will find the prime factors of the given integers: 84, 90 and 120 
Prime factor will help us to find LCM and HCF 
84 = 2 × 2 × 3 × 7 
90 = 2 × 3 × 3 × 5 
120 = 2 × 2 × 2 × 3 × 5 
Now for L.C.M we will find common multiples from each terms. 
L.C.M of 84, 90 and 120 = 23 × 32 × 5 × 7 
SO L.C.M (84, 90, 120) = 2520 
And, for H.C.F we will find highest common factors from each term. 
H.C.F (84, 90 and 120) = 6

(vi) 24, 15 and 36

Solution:

We will find the prime factors of the given integers: 24, 15 and 36 
Prime factor will help us to find LCM and HCF 
24 = 2 × 2 × 2 × 3 
15 = 3 × 5 
36 = 2 × 2 × 3 × 3 
Now for L.C.M we will find common multiples from each terms. 
LCM of 24, 15 and 36 = 2 × 2 × 2 × 3 × 3 × 5 = 23 x 32 x 5 
So LCM (24, 15, 36) = 360 
And, for H.C.F we will find highest common factors from each term. 
HCF (24, 15 and 36) = 3

Question 3. Given that HCF (306, 657) = 9, find LCM (306, 657)

Solution: 

Given integers in questions are: 306 and 657 
As we know that, 
LCM × HCF = Product of the two integers 
So here 
⇒ LCM = Product of the two integers / HCF (putting values in formula) 
= (306 × 657) / 9 = 22338

Question 4. Can two numbers have 16 as their HCF and 380 as their LCM? Give reason.

Solution:

If we divide 380 by 16 we will get, quotient as 23 and remainder as 12. 
As we know that, 
LCM × HCF = Product of the two integers 
Now, the L.C.M is not exactly divisible by the HCF it’s can be said that two numbers cannot have HCF and LCM as 16 and 380 respectively.

Question 5. The HCF of two numbers is 145 and their LCM is 2175. If one number is 725, find the other.

Solution:

Here LCM and HCF of two numbers are 145 and 2175 given respectively. 
One of the numbers is 725 
As we know that, 
LCM × HCF = First number × Second number 
2175 × 145 = 725 × Second number (putting values in formula) 
⇒ Second number = (2175 × 145)/ 725 = 435 
Then the Second number is 435. 

Question 6. The HCF of two numbers is 16 and their product is 3072. Find their LCM.

Solution:

Here we have, as given 
HCF of two numbers = 16 
Product of two numbwe = 3072 
As we know that, 
LCM × HCF = Product of the two numbers 
LCM × 16 = 3072 (putting values in formula) 
⇒ LCM = 3072/ 16 = 192 
So the LCM of the two numbers is 192. 

Question 7. The LCM and HCF of two numbers are 180 and 6 respectively. If one of the numbers is 30, find the other number.

Solution:

Here we have, 
L.C.M = 180 
H.C.F = 6 
One of the numbers is 30 
As we know that, 
LCM × HCF = first number × second number 

180 × 6 = 30 × second number 
(putting values in formula) 
Second number = (180 × 6)/ 30 = 36 
Then the Second number is 36.

Question 8. Find the smallest number which when increased by 17 is exactly divisible by both 520 and 468.

Solution:

Firstly we will find LCM of both numbers which will be smallest number which is exactly divisible by both 520 and 468. 
We will do prime factorization of given number, to find LCM: 
520 = 23 × 5 × 13 
468 = 22 × 32 × 13 
LCM (520, 468) = 23 × 32 × 5 × 13 = 4680 
Here 4680 is the smallest number which is exactly divisible by both 520 and 468. 
Now, we will get a remainder of Zero in each case. But, we need to find the smallest number which when increased by 17 is exactly divided by 520 and 468. 
So it can found by, subtracting, 
4680 – 17 = 4663 
4663 should be the smallest number which when increased by 17 is exactly divisible by both 
520 and 468.

Question 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 factorization, we get 
28 = 2 × 2 × 7 

32 = 25 
L.C.M (28, 32) = 25 
× 7 = 224 

We found that 224 is the smallest number which is divisible by 28 and 32 both So, we will get a remainder of 0 in each case. But, we need the smallest number which will leave remainder of 8 and 12 
When divided by 28 and 32 respectively. 
So it can found by, subtracting 8 and 12 from 224 
224 – 8 – 12 = 204 
Here 204 must be the smallest number which will leave remainder of 8 and 12 when divided by both 28 and 
32. 

Question 10. What is the smallest number that, when divided by 35, 56 and 91 leaves remainders of 7 in each case?

Solution:

Firstly we will find LCM of all three given numbers which will be smallest number which is exactly divisible by given three numbers 35, 56 and 91. 
Now we will do prime factorization, to get LCM 
35 = 5 × 7 
56 = 23 × 7 
91 = 13 × 7 
So L.C.M (35, 56 and 91) = 23 × 7 × 5 × 13 = 3640 
Here, we found that 3640 is the smallest number that can be divided by 35, 56 and 91 and gives remainder of 0 in each case. We require to find a number that leaves a remainder 7 in each case. 
So it can be done by adding 7 in 3640, 
3640 + 7 = 3647 
So 3647 should be the smallest number that when divided by 35, 56 and 91 and gives the remainder 
of 7 in each case. 

Question 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 data in the question are 
Length of courtyard = 18 m 72 cm = 1800 cm + 72 cm = 1872 cm (We know that 1 m = 100 cm) 
Breadth of courtyard = 13 m 20 cm = 1300 cm + 20 cm = 1320 cm 

Here we need to the size of the square tile paved on the rectangular yard which will equal to the HCF of the length and breadth of the rectangular courtyard. 
Now, we will find the prime factorization of 1872 and 1320 
1872 = 24 × 32 × 13 
1320 = 23 × 3 × 5 × 11 
So the HCF (1872 and 1320) = 23 × 3 = 24 
Here the length of side of the square tile is 24 cm. 
Here We need to find number of tiles required. So, the number of tiles required = (area of the courtyard) / (area of a square tile) 

As we know that, area of the courtyard = Length × Breadth 

= 1872 cm × 1320 cm 
(putting values in formula) 
Area of a square tile = (side of square)2 = (24cm)2 
The number of tiles required = (1872 x 1320) / (24)2 = 4290 
(putting values in formula) 
Here, we got the least possible number of tiles required is 4290. 

Question 12. Find the greatest number of 6 digits exactly divisible by 24, 15 and 36.

Solution:

The greatest 6 digit number is 999999. 
If we assume that, 24,15 and 36 divides 999999 without leaving any remainder. 
Then we need to find LCM (24, 15 and 36) because LCM will also divide 999999 exactly. 
Now the prime factorization of 24, 15, and 36, to get LCM 
24 = 2 × 2 × 2 × 3 
15 = 3 × 5 
36 = 2 × 2 × 3 × 3 
Then L.C.M of 24, 15 and 36 = 360 
If we divide (999999)/ 360 = 2777 × 360 + 279 
Then remainder is 279. 
Now to find greatest 6 digit number which is divisible by all three should be, we need to 
We need to subtract remainder from 6 digit greatest number. 
So, 999999 – 279 = 999720 
Here 999720 is the greatest 6 digit number which is exactly divisible by 24, 15 and 36.

Question 13. Determine the number nearest to 110000 but greater 100000 which is exactly divisible by each of 8, 15 and 21.

Solution:

Firstly we will find the L.C.M of 8, 15 and 21. 
By prime factorization, we will get LCM 
8 = 2 × 2 × 2 
15 = 3 × 5 
21 = 3 × 7 
Now L.C.M (8, 15 and 21) = 2 
3 × 3 × 5 × 7 = 840 
If we divide the 110000 by 840, then the remainder will be 800. 

Now, If we substract 800 from the 110000, it will be divisible by each of 8, 15 and 21. 
So, 110000 – 800 = 109200. 
If we add 40, it will be also divisible by each of 8, 15 and 21 
Now we have 110000 + 40 = 110040 
We got two result, 109200 and 110040 both are greater than 100000 but 110040 is greater than 110000. 
So, 109200 is the number nearest to 110000 and greater than 100000 which is exactly divisible by each of 8, 15 and 21.

Question 14. Find the least number that is divisible by all the numbers between 1 and 10 (both inclusive).

Solution:

According to the question we need to find the least number that is divisible by all the numbers between 1 and 10 
So, the L.C.M of 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10 will be the least number that will be divisible by all the numbers between 1 and 10. 
Hence, the prime factorization of all these numbers are, to get LCM: 
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 
L.C.M will be = 23 × 32 × 5 × 7 = 2520 
Here we got 2520 as the least number that is divisible by all the numbers between 1 and 10 (both inclusive)

Question 15. A circular field has a circumference of 360km. Three cyclists start together and can cycle 48, 60 and 72 km a day, round the field. When will they meet again?

Solution:

Here we need to calculate the time taken before they meet again , so we need to find the individual time taken by each cyclist in covering the total distance. 
Now, we will find Number of days a cyclist takes to cover the circular field by: 

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) 
1st cyclist, number of days = 360 / 48 = 7.5 which will = 180 hours [we know that,1 day = 24 hours] 
2nd cyclist, number of days = 360 / 60 = 6 which will = 144 hours 
3rd cyclist, number of days = 360 / 72 = 5 which will 120 hours 
Now, the LCM (180, 144 and 120) will help to know after how many hours the three cyclists meet again. 
By prime factorization, we get LCM 
180 = 22 x 32 x 5 
144 = 24 x 33 
120 = 23 x 3 x 5 
L.C.M (180, 144 and 120) = 24 x 32 x 5 = 720 
So, the after 720 hours the three cyclists meet again. 
720 hours = 720 / 24 = 30 days [we know that, 1 day = 24 hours] 
So, we can say that all the three cyclists will meet again after 30 days. 

Question 16. In a morning walk three persons step off together, their steps measure 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:

As per given in the question, that the required distance each should walk would be the 
LCM of the measures of their steps that is 80 cm, 85 cm, and 90 cm. 
So, L.C.M (80, 85 and 90) with the help of prime factorization, 
80 = 24 × 5 
85 = 17 × 5 
90 = 2 × 3 × 3 × 5 
L.C.M (80, 85 and 90) = 24 × 32 × 5 × 17 = 12240 cm = 122m 40cm [we know that, 1 m = 100cm] 

We got, 122m 40cm as the minimum distance that each should walk so that all can cover the same distance in complete steps.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

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