RD Sharma Solutions Class 10 Real Numbers Exercise 1.4

RD Sharma Class 10 Solutions Chapter 1 Ex 1.4 PDF Free Download

Exercise 1.4

 

Q.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

(ii) 510 and 92

(iii) 336 and 54

 

Sol:

Note: From the questions,

To Find: LCM and HCF of given pairs of integers

For Verification: L.C.M × H.C.F = product of the numbers

(i) 26 and 91

First, find the factors of 26 and 91

So,

  • factors of 26 = 2 × 13
  • factors of 91 = 7 × 13

∴ L.C.M of 26 and 91 = 2 × 7 × 13

Or, L.C.M (26, 91) = 182

And,

H.C.F of 26 and 91 = 182

Verification: L.C.M × H.C.F = First number × Second number

182 × 13 = 26 × 91 = 2366 = 2366 (hence verified).

 

(ii) 510 and 92

Find the factors of 510 and 92

  • Factors of 510 = 2 × 3 × 5 × 17
  • Factors of 92 = 2 × 2 × 23

Now, L.C.M of 510 and 92 = 2 × 2 × 3 × 5 × 23 × 17

∴ L.C.M (510, 92) = 23460

Also,

H.C.F of 510 and 92 = 2

Vverification: L.C.M × H.C.F = First Number × Second Number

23460 × 2 = 510 × 92

46920 = 46920 (hence verified).

 

(iii)  336 and 54

  • Factors of 336 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 7
  • Factors of 54 = 2 × 3 × 3 x3

Now, L.C.M of 336 and 54 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 7

Or, L.C.M (336, 54) = 3024

And,

H.C.F of 336 and 54 = 6

Verification: L.C.M × H.C.F = First Number × Second Number

3024 × 6 = 336 × 54

18144 = 18144 (hence verified).

 

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

(i) 12, 15 and 21

(ii) 17, 23 and 29

(iii) 8, 9 and 25

(iv) 40, 36 and 126

(v) 84, 90 and 120

(vi) 24, 15 and 36

 

Sol:

Method: For prime factorization, the prime factors of the given pairs have to be taken.

(i) 15, 12 and 21

Find the prime factors of 15, 12 and 21

  • Prime Factors of 12 = 2 × 2 × 3
  • Prime Factors of 15 = 3 × 5
  • Prime Factors of 21 = 3 × 7

Now, L.C.M of 12, 15 and 21 = 2 × 2 × 3 × 5 × 7

Or, L.C.M (12, 15, 21) = 420

And, H.C.F of 12, 15 and 21 = 3

 

(ii) 17, 23 and 29

Find the factors of 17, 23 and 29

  • Prime Factors of 17 = 1 × 17
  • Prime Factors of 23 = 1 × 23
  • Prime Factors of 29 = 1 × 29

So, L.C.M of 17, 23 and 29 = 1 × 17 × 23 × 29

Or, L.C.M (17, 23, 29) = 11339

And, H.C.F of 17, 23 and 29 = 1

 

(iii) 8, 9 and 25

Find the factors of 8, 9 and 25

  • Prime Factors of 8 = 2 × 2 x2
  • Prime Factors of 9 = 3 × 3
  • Prime Factors of 25 = 5 × 5

So, L.C.M of 8, 9 and 25 =23 × 32 × 52

Or, L.C.M (8, 9, 25) = 1800

And, H.C.F of 8, 9 and 25 = 1

 

(iv) 40, 36 and 126

Find the factors of 40, 36 and 126

  • Prime Factors of 40 = 23 × 5
  • Prime Factors of 36 = 23 × 32
  • Prime Factors of 126 = 2 × 3 × 3 × 7

So, L.C.M of 40, 36 and 126 = 23 × 32 × 5 × 7

Or, L.C.M (40, 36, 126) = 2520

And, H.C.F of 40, 36 and 126 = 2

 

(v) 84, 90 and 120

Find the factors of 84, 90 and 120

  • Prime Factors of 84 = 2 × 2 × 3 × 7
  • Prime Factors of 90 = 2 × 3 × 3 × 5
  • Prime Factors of 120 = 2 × 2 × 2 × 3 × 5

So, L.C.M of 84, 90 and 120 = 23 × 32 × 5 × 7

Or, L.C.M (84, 90, 120) = 2520

And, H.C.F of 84, 90 and 120 = 6

 

(vi) 24, 15 and 36

Find the factors of 24, 15 and 36

  • Prime Factors of 24 = 23 × 3
  • Prime Factors of 15 = 3 × 5
  • Prime Factors of 36 = 2 × 2 × 3 × 3

So, LCM of 24, 15 and 36 = 2 × 2 × 2 × 3 × 3 × 5

Or, LCM (24, 15, 36) = 360

And, HCF of 24, 15 and 36 = 3

 

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

 

Sol:

In the question, it is given that,

The HCF of two numbers 306 and 657 is 9

To find LCM, the following formula can be used:

LCM × HCF = first number × second number

LCM × 9 = 306 × 657

LCM = (306 × 657)/ 9 = 22338

 

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

 

Sol:

No. On dividing 380 by 16 we get 23 as the quotient and 12 as the remainder. Now, since LCM is not exactly divisible by the HCF, two numbers cannot have 16 as their HCF and 380 as their LCM.

 

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

 Sol:

Given:

LCM and HCF of two numbers 145 and 2175 respectively.

It is given that, one number is 725

Now, LCM × HCF = first number × second number

2175 × 145  = 725 × second number

LCM = (2175 × 145)/ 725 = 435

 

Q.6: The HCF of two numbers is 16 and their LCM is 3072. Find the LCM.

 

Sol:

It is given that HCF of two numbers is 16.

And, the product of the numbers is 3072.

Now, to find LCM of number,

LCM × HCF = first number × second number

LCM × 16 = 3072

LCM = 3072/ 16 = 192

  

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

 

Sol:

Given:

LCM (first number, second number) = 180

HCF (first number, second number) = 6

Among them, one number is 30.

Now to find the other number, the following formula can be used:

LCM × HCF = first number × second number

180 × 6 = 30 × second number

∴ Second number = (180 × 6)/ 30 = 36 

 

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

 

Sol:

We know

520 = 23 × 5 × 13

468 = 2 × 2 × 3 × 3 × 13

So, LCM (520, 468) = 23 × 32 × 5 × 13 = 4680

Hence 4680 is the least number which exactly divides 520 and 468  i.e. we will get a remainder of 0 in this case. But we need the smallest number which when increased by 17 is exactly divided by 520 and 468.

And, 4680 —17 = 46631

∴ 4663 is the smallest number which when increased by 17 is exactly divisible by both 520 and 468.

 

Q.9: Find the smallest number which leaves remainders 8 and 12 when divided by 28 and 32 respectively.

 

Sol:

We know,

To find L.C.M of 28 and 32, find their prime factors

  • Prime factors of 28 = 2 × 2 × 7
  • Prime factors of 32 = 25

L.C.M (28, 32) = 25 × 7 = 224

So, 224 is the least number which exactly divides 28 and 32 i.e. we will get a remainder of 0 in this case. But we need the smallest number which leaves remainders 8 and 12 when divided by 28 and 32 respectively.

So, 224 – 8 – 12 = 204

∴ 2041 is the smallest number which leaves remainders 8 and 12 when divided by 28 and 32 respectively

 

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

 

Sol:

Find the prime factors of 35, 56, and 91

  • Prime factors of 35 = 5 × 7
  • Prime factors of 56 = 23 × 7
  • Prime factors of 91 = 13 × 7

L.C.M of 35, 56 and 91 = 23 × 7  × 5 × 13 = 3640

Hence 84 is the least number which exactly divides 28, 42 and 84 i.e. we will get a remainder of 0 in this case. But we need the smallest number that, when divided by 35, 56 and 91 leaves the remainder of 7 in each case.

So, 3640+ 7 = 3647

∴36471 is the smallest number that, when divided by 35, 56 and 91 leaves the remainder of 7 in each case.

 

Q.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.

 

Sol:

Given,

Length of the yard = 18 m 72 cm = 1800 cm + 72 cm = 1872 cm (therefore, 1 m = 100 cm)

Breadth of the yard = 13 in 20 cm = 1300 cm + 20 cm = 1320 cm

The size of the square tile of same size needed to the pave the rectangular yard is equals the HCF of the length and breadth of the rectangular yard.

Now find the prime factors of 1872 and 1320

  • Prime factorization of 1872 = 24 × 32 × 13
  • Prime factorization of 1320 = 23 × 3 × 5 × 11

⇒ HCF of 1872 and 1320 = 23 × 3 = 24

∴ Length of side of the square tile = 24 cm

Number of tiles required = Area of the courtyard,

We know, Area of each tile = Length × Breadth

Side2 = 1872 cm × 1320 cm 24 cm = 4290.

Thus, the least possible number of tiles required is 4290.

 

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

 

Sol:

we know the greatest 6 digit number is 999999

Now, let 999999 divide 24, 15 and 36 exactly

Now, find the prime factors of 24, 15, and 36

  • 24 = 2 × 2 × 2 × 3
  • 15 = 3 × 5
  • 36 = 2 × 2 × 3 × 3

∴ L.C.M of 24, 15 and 36 = 360

Since, (999999)/ 360 = 2777 × 360 + 279

Here, the remainder is 279.

So, the number which is divisible by all three = 999999 – 279 = 9997201

∴ 9997201 is the greatest 6 digit number which is exactly divisible by 24, 15 and 36.

 

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

 

Sol:

L.C.M of 8, 15 and 21.

  • Prime factors of 8 = 2 × 2 × 2
  • Prime factors of 15 = 3 × 5
  • Prime factors of 21 = 3 × 7

L.C.M of 8, 15 and 21 =23 × 3 × 5 × 71= 840

When 110000 is divided by 840, the remainder is obtained as 800.

Now, 110000 — 800 = 109200 is divisible by each of 8, 15 and 21.

Also, 110000 + 40 = 110040 is divisible by each of 8, 15 and 21.

109200 and 110040 are greater than 100000.

Hence, 110040 is the number nearest to 110000 but greater than 100000 which is exactly divisible by each of 8, 15 and 21.

  

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

 

Sol:

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 2520

Hence 2520 is the least number that is divisible by all the numbers between 1 and 10 (both inclusive)

  

Q.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?

 

Sol:

In order to calculate the time when they meet, we first find out the time taken by each cyclist in covering the distance.

Number of days 1st cyclist took to cover 360 km = Total distance covered in 1 day = 36048 = 7.5 = 7510 = 152 days

Similarly, number of days taken by 2nd cyclist to cover same distance = 36060 = 6 days

Also, number of days taken by 3rd cyclist to cover this distance = 36072 = 5 days

Now, LCM of 152, 6 and 5 = LCM of numerators

HCF of denominators = 301 = 30 days

Thus, all of them will take 30 days to meet again.

 

 Q.16: In a morning walk three persons step off together, their steps measure 80cm, 85 cm and 90 cm respectively. What is the minimum distance each should walk so that he can cover the distance in complete steps? 

 

Sol:

The distance covered by each of them is required to be same as well as a minimum. The required distance each should walk would be the L.C.M of the measures of their steps i.e. 80 cm, 85 cm, and 90 cm,

So we have to find the L.C.M of 80 cm, 85 cm, and 90 cm.

80 = 24 × 5

85 = 17 × 5

90 = 2 × 3 × 3 × 5

L.C.M of 80, 85 and 90 = 24 × 3 × 3 × 5 × 17 = 12240 cm

Hence minimum 12240 cm distance each should walk so that all can cover the same distance in complete steps.

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