## RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1B

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 4 Triangles Ex 4B. You must go through **NCERT Solutions for Class 10 Maths** to get better score in CBSE Board exams along with RS Aggarwal Class 10 Solutions.

**Question 1.**

**Solution:**

(i) 36, 84

36 = 2 x 2 x 3 x 3 = 2² x 3²

84 = 2 x 2 x 3 x 7 = 2² x 3 x 7

HCF = 2² x 3 = 2 x 2 x 3 = 12

LCM = 2² x 3² x 7 = 2 x 2 x 3 x 3 x 7 = 252

Now HCF x LCM = 12 x 252 = 3024

and product of number = 36 x 84 = 3024

HCF x LCM = Product of given two numbers.

(ii) 23, 31

23 = 1 x 23

31 = 1 x 31

HCF= 1

and LCM = 23 x 31 = 713

Now HCF x LCM = 1 x 713 = 713

and product of numbers = 23 x 31 = 713

HCF x LCM = Product of given two numbers

(iii) 96, 404

96 = 2 x 2 x 2 x 2 x 2 x 3 = 2^{5} x 3

404 = 2 x 2 x 101 = 2² x 101

HCF = 2² = 2 x 2 = 4

LCM = 2^{5} x 3 x 101 = 32 x 3 x 101 = 9696

Now HCF x LCM = 4 x 9696 = 38784

and product of two numbers = 96 x 404 = 38784

HCF x LCM = Product of given two numbers

(iv) 144, 198

144 = 2 x 2 x 2 x 2 x 3 x 3 = 2^{4} x 32

198 = 2 x 3 x 3 x 11 = 2 x 3² x 11

HCF = 2 x 3^{2} = 2 x 3 x 3 = 18

LCM = 2^{4} x 3² x 11 = 16 x 9 x 11 = 1584

and product of given two numbers = 144 x 198 = 28512

and HCF x LCM = 18 x 1584 = 28512

HCF x LCM = Product of given two numbers

(v) 396, 1080

396 = 2 x 2 x 3 x 3 x 11 = 2² x 3² x 11

1080 = 2 x 2 x 2 x 3 x 3 x 3 x 5 = 2^{3} x 3^{3} x 5

HCF = 2² x 3² = 2 x 2 x 3 x 3 = 36

LCM = 2^{3} x 3^{3} x 11 x 5 = 2 x 2 x 2 x 3 x 3 x 3 x 5 x 11 = 11880

Now HCF x LCM = 36 x 11880 = 427680

Product of two numbers = 396 x 1080 = 427680

HCF x LCM = Product of two given numbers.

(vi) 1152, 1664

1152 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 = 2^{7} x 3²

1664 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 13 = 2^{7} x 13

HCF = 2^{7} = 2 x 2 x 2 x 2 x 2 x 2 x 2 = 128

LCM = 2^{7} x 3² x 13 = 128 x 9 x 13 = 128 x 117= 14976

Now HCF x LCM = 128 x 14976= 1916928

and product of given two numbers = 1152 x 1664 = 1916928

HCF x LCM = Product of given two numbers.

**Question 2.**

**Solution:**

**Question 3.**

**Solution:**

HCF of two numbers = 23

LCM =1449

One number = 161

Second number = 207

**Question 4.**

**Solution:**

HCF of two numbers = 145

LCM = 2175

One number = 725

Second number = 435

**Question 5.**

**Solution:**

HCF of two numbers = 18

and product of two numbers = 12960

LCM of two numbers = 720

**Question 6.**

**Solution:**

HCF= 18

LCM = 760

HCF always divides the LCM completely

760 – 18 = 42 and remainder 4

Hence, it is not possible.

**Question 7.**

**Solution:**

(a) \(\frac { 69 }{ 92 }\)

HCF of 69 and 92 = 23

**Question 8.**

**Solution:**

Numbers are 428 and 606 and remainder in each case = 6

Now subtracting 6 from each number, we get 438 – 6 = 432

and 606 – 6 = 600

Required number = HCF of 432 and 600 = 24

The largest required number is 24

**Question 9.**

**Solution:**

The numbers are 320 and 457

and remainders are 5 and 7 respectively

320 – 5 = 315 and 457 – 7 = 450

Now the required greatest number of 315 and 450 is their HCF

Now HCF of 315 and 450 = 45

**Question 10.**

**Solution:**

The numbers are given = 35, 56, 91 and the remainder = 7 in each case,

Now the least number = LCM of 35, 56, 91 = 3640

LCM = 7 x 5 x 8 x 13 = 3640

Required least number = 3640 + 7 = 3647

**Question 11.**

**Solution:**

Given numbers are 28 and 32

Remainders are 8 and 12 respectively

28 – 8 = 20

32 – 12 = 20

Now, LCM of 28 and 32 = 224

LCM = 2 x 2 x 7 x 8 = 224

Least required number = 224 – 20 = 204

**Question 12.**

**Solution:**

The given numbers are 468 and 520

Now LCM of 468 and 520 = 4680

LCM = 2 x 2 x 13 x 9 x 10 = 4680

When number 17 is increase then required number = 4680 – 17 = 4663

**Question 13.**

**Solution:**

LCM of 15, 24, 36 = 360

Required number = 9999 – 279 = 9720

**Question 14.**

**Solution:**

Greatest number of 4 digits is 9999

LCM of 4, 7 and 13 = 364

On dividing 9999 by 364, remainder is 171

Greatest number of 4 digits divisible by 4, 7 and 13 = (9999 – 171) = 9828

Hence, required number = (9828 + 3) = 9831

**Question 15.**

**Solution:**

LCM of 5, 6, 4 and 3 = 60

On dividing 2497 by 60, the remainder is 37

Number to be added = (60 – 37) = 23

**Question 16.**

**Solution:**

We can represent any integer number in the form of: pq + r, where ‘p is divisor, ‘q is quotient’, r is remainder*.

So, we can write given numbers from given in formation As :

43 = pq_{1} + r …(i)

91 = pq_{2} + r …(ii)

And 183 = pq_{3} + r …(iii)

Here, we want to find greatest value of ‘p’ were r is same.

So, we subtract eq. (i) from eq. (ii), we get

Pq_{2} – Pq_{1} = 48

Also, subtract eq. (ii) from eq. (iii), we get

pq_{3} – pq_{2} = 92

Also, subtract eq. (i) from eq. (iii), we get

Pq_{3} – Pq_{1} = 140

Now, to find greatest value of ‘p’ we find HCF of 48, 92 and 140 as,

48 = 2 x 2 x 2 x 2 x 3

92 = 2 x 2 x 2 x 2 x 2 x 3

and 140 = 2 x 2 x 5 x 7

So, HCF (48, 92 and 140) = 2 x 2 = 4

Greatest number that will divide 43, 91 and 183 as to leave the same remainder in each case = 4.

**Question 17.**

**Solution:**

Remainder in all the cases is 6, i.e.,

20 – 14 = 6

25 – 19 = 6

35 – 29 = 6

40 – 34 = 6

The difference between divisor and the corresponding remainder is 6.

Required number = (LCM of 20, 25, 35, 46) – 6 = 1400 – 6 = 1394

**Question 18.**

**Solution:**

Number of participants in Hindi = 60

Number of participants in English = 84

Number of participants in Mathematics =108

Minimum number of participants in one room = HCF of 60, 84 and 108 = 12

**Question 19.**

**Solution:**

Number of books in English = 336

Number of books in Mathematics = 240

Number of books in Science = 96

Minimum number of books of each topic in a stack = HCF of 336, 240 and 96 = 48

**Question 20.**

**Solution:**

Length of first piece of timber = 42 m

Length of second piece of timber = 49 m

and length of third piece of timber = 63 m

**Question 21.**

**Solution:**

Lengths are given as 7 m, 3 m 85 cm and 12 m 95 cm = 700 cm, 385 cm and 1295 cm

Greatest possible length that can be used to measure exactly = HCF of 700, 385, 1295 = 35 cm

**Question 22.**

**Solution:**

Number of pens =1001

and number of pencils = 910

Maximum number of pens and pencils equally distributed to the students = HCF of 1001 and 910 = 91

Number of students = 91

**Question 23.**

**Solution:**

Length of the room = 15 m 17 cm = 1517 cm

and breadth = 9 m 2 cm = 902 cm

Maximum side of square tile used = HCF of 1517 and 902 = 41 cm

**Question 24.**

**Solution:**

Measures of three rods = 64 cm, 80 cm and 96 cm

Least length of cloth that can be measured an exact number of times

= LCM of 64, 80, 96

= 960 cm

= 9 m 60 cm

= 9.6 m

LCM = 2 x 2 x 2 x 2 x 2 x 2 x 5 x 3 = 960

**Question 25.**

**Solution:**

Beep made by first devices after every = 60 seconds

Second device after = 62 seconds

Period after next beep together = LCM of 60, 62

LCM = 2 x 30 x 31 = 1860 = 1860 seconds = 31 minutes

Time started beep together, first time together = 10 a.m.

Time beep together next time = 10 a.m. + 31 minutes = 10 : 31 a.m.

**Question 26.**

**Solution:**

The traffic lights of three roads change after

48 sec., 72 sec. and 108 sec. simultaneously

They will change together after a period of = LCM of 48 sec., 72 sec. and 108 sec.

= 7 minutes, 12 seconds

First time they light together at 8 a.m. i.e., after 8 hr.

Next time they will light together = 8 a.m. + 7 min. 12 sec. = 8 : 07 : 12 hrs.

**Question 27.**

**Solution:**

Tolling of 6 bells = 2, 4, 6, 8, 10, 12 minutes

They take time tolling together = LCM of 2, 4, 6, 8, 10, 12 = 120 minutes = 2 hours

LCM of 2 x 2 x 2 x 3 x 5 = 120 min. (2 hr)

They will toll together after every 2 hours Total time given = 30 hours

Number of times, there will toll together in 30 hours = \(\frac { 30 }{ 2 }\) = 15 times

Total numbers of times = 15 + 1 (of starting time) = 16 times

Hope given RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers Ex 1B are helpful to complete your math homework.

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