R.D. Sharma Solutions Class 10th: Ch 1 Real Numbers Exercise 1.4

Chapter 1 Real Numbers R.D. Sharma Solutions for Class 10th Math Exercise 1.4

Exercise 1.4

Level 1

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 

Answer

(i) Factors of 26 and 91
26 = 2 ×13
91 = 7 ×13

L.C.M of 26 and 91 = 2 × 7 ×13 = 182
H.C.F of 26 and 91 = 13

A/q,
L.C.M×H.C.F = First number × Second number
⇒ 182×13 = 26×91
⇒ 2366 = 2366
Hence verified.

(ii) Factors of 510 and 92
510 = 2×3×5×17
92 = 22×23

L.C.M of 510 and 92 = 22×3×5×17×23 = 23460
H.C.F of 510 and 92 = 2

A/q,
L.C.M×H.C.F = First number × Second number
⇒ 23460×2 = 510×92
⇒ 46920 = 46920
Hence verified.

(iii) Factors of 336 and 54
336 = 24×3×7
92 = 2×33

L.C.M of 336 and 54 = 24×33×7 = 3024
H.C.F of 336 and 54 = 6

A/q,
L.C.M×H.C.F = First number × Second number
⇒ 3024×6 = 336×54
⇒ 18144 = 18144
Hence verified.

2. Find the LCM and HCF of the following integers by applying the prime factorisation 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


Answer

(i) Factors of 12,15 and 21
12 = 22×3
15 =3×5
21 = 3×7
L.C.M of 12, 15 and 21 = 22×3×5×7 = 420
H.C.F of 12, 15 and 21 = 3

(ii) Factors of 17, 23 and 29
17 = 1×17
23= 1×23
29= 1×29
L.C.M of 17,23 and 29 = 1×17×23×29 = 420
H.C.F of 17,23 and 29 = 1

(iii) Factors of 8,9 and 25
8 = 23
9 = 32
25 = 52
L.C.M of 8,9 and 25 = 23×32×52 = 1800
H.C.F of 8,9 and 25 = 1

(iv) Factors of 40,36 and 126
40 = 23×5
36= 22×32
126 = 2×32×7
L.C.M of 40, 36 and 126= 23×32×5×7 =2520
H.C.F of 40, 36 and 126 = 2


(v) Factors of 84,90 and 120
84 = 22×3×7
90 = 2×32×5
120 = 23×3×5
L.C.M of 84,90 and 120 = 23×32×5×7 = 2520
H.C.F of 84,90 and 120 = 6

(vi) Factors of 24,15 and 36.
24 = 23×3
15 = 3×5
36 = 22×32
L.C.M of 24,15 and 36 = 23×32×5 = 360
H.C.F of 24,15 and 36 = 3

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

L.C.M. × H.C.F = First Number × Second Number
L.C.M.×9 = 306×657
L.C.M. = (306 × 657)/9
L.C.M. = 22338

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

Answer

LCM = 380
HCF = 16
When we divide 380 by 16, we get 23 as the quotient and 12 as the remainder,
LCM is not exactly divisible by the H.C.F, two numbers cannot have 16 as their H.C.F and 380 as their L.C.M

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

Answer

L.C.M × H.C.F = First Number x Second Number
2175 × 145 = 725 × Second Number
∴ Second Number = (2175×145)/725 = 4351

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

Answer

L.C.M x H.C.F = First Number x Second Number
L.C.M ×16 =3072
∴ L.C.M = 3072/16 = 1921


7. The LCM and HCF of two numbers are 180 and 6 respectively. Hone of the numbers is In find the other number.

Answer

Second Number = 361
L.C.M × H.C.F = First Number × Second Number
180×6 = 30 × Second Number
Second Number = (180×6)/30 = 36

Level 2

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

Answer

520 = 23×5×13
468 = 22×32×13
LCM of 520 and 468 = 23×32×5×13 = 4680

Hence, 4680 is the least number which exactly divides 520 and 468 which will leave 0 as remainder. A/q,
We need the smallest number which when increased by 17 is exactly divided by 520 and 468. So, we will subtract 17 from 4680.
∴ 4680 -17 = 46631
4663 is 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.

Answer

L.C.M of 28 and 32,
28 = 22×7
32 = 25
L.C.M of 28 and 32 = 25×7 = 224

Hence, 224 is the least number which exactly divides 28 and 32 which will leave 0 as remainder.
A/q,
We need the smallest number which leaves remainders 8 and 12 when divided by 28 and 32 respectively. So, we will subtract 8 and 12 from 224.
∴ 224 - 8 - 12 = 204
204 is 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?

Answer

L.C.M OF 35,56 and 91
35 = 5×7
56 = 23×7
91 = 13×7
L.C.M of 35, 56 and 91 =2×5×7×13 = 3640
Hence, 84 is the least number which exactly divides 28, 42 and 84 which will leave 0 as remainder.
A/q,
We need the smallest number which when divided by 35, 56 and 91 leaves remainder of 7 in each case. So, we will add 7 to 3640.
∴ 3640 + 7 = 3647
3647 is smallest number that, when divided by 35, 56 and 91 leaves remainder of 7 in each case.

11. A rectangular courtyard is 18 m 72 cm long and 13 m 20 cm broad. It is to be paved with square tiles of the same size. Find the least possible number of such tiles.

Answer

Length of the yard =18 m 72 cm = 1800 cm + 72 cm = 1872 cm
Breadth of the yard =13 m 20 cm = 1300 cm + 20 cm = 1320 cm
Area of the yard = 1872×1320 = 2471040

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

Prime factorisation of length and breadth.
∴1872 = 24×32×13
∴1320 = 23×3×5×11
HCF of 1872 and 1320 = 23×3 = 24

Therefore, length of side of the square tile = 24 cm
Area of the tile = 24×24 = 576 cm2

Number of tiles required = Area of courtyard/Area of each tile = 2471040/576 = 4290

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

Answer

The greatest 6 digit number be 999999
Factors of 24, 15 and 36
24 = 23×3
15 = 3×5
36 = 22×32
LCM of 24, 15 and 36 = 23×32×5 = 360
A/q,
999999/360 = 2777×360 + 279
Remainder is 279. So, we need to subtract 279 from 999999 to get the required number.
∴ 999999-279 = 999720
Thus, 999720 is the greatest number which can be 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.

Answer

Factors of 8, 15 and 21.
8 = 23
15 = 3×5
21 = 3×7
LCM of 8, 15 and 21 = 23×3×5×7=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.

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

Answer

We will find the L.C.M of all the numbers between 1 and 10 (both inclusive)
1 = 1
2 = 2
3 = 3
4 = 22
5 = 5
6 = 2×3
7 = 7
8 = 23
9 = 32
10 = 2×5
LCM of all the numbers between 1 and 10 = 2520
2520 is he least number that is divisible by all the numbers between 1 and 10.

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

Answer

We need to find LCM of no. of days taken by cyclist to cover 360 days.
We will find the time taken by each cyclist in covering the distance in order to calculate the time when they meet, .

Number of days 1st cyclist took to cover 360 km = Total distance/Distance covered in 1 day
= 360/48 = 15/2

Number of days 2nd cyclist took to cover 360 km = 360/60 = 6

Number of days 3rd cyclist took to cover 360 km = 360/72 = 5

LCM of (15/2, 6 and 5) = LCM of numerators/HCF of denominators
= 30/1 = 30

Thus, all of them will meet after 30 days.

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? 

Answer

Minimum distance each should walk so that all can cover the same distance in complete steps.

The distance covered by each of them is required to be same as well as 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,
Factors of 80, 85 and 90.
80 = 24×5
85 = 17×5
90 = 2×32×5
LCM of 80, 85 and 90 = 24×32×5×17 = 12240
12240 is the minimum distance each should walk so that he can cover the distance in complete steps.

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