## 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.**

**Using prime factorization, find the HCF and LCM of**

**(i) 36, 84**

**(ii) 23, 31**

**(iii) 96, 404**

**(iv) 144, 198**

**(v) 396, 1080**

**(vi) 1152, 1664**

**In each case, verily that:**

**HCF x LCM = product of given numbers.**

**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.**

**Using prime factorization, find the HCF mid LCM of:**

**(i) 8, 9, 25**

**(ii) 12, 15, 21**

**(iii) 17, 23, 29**

**(iv) 24, 36, 40**

**(v) 30, 72, 432**

**(vi) 21, 28, 36, 45**

**Solution:**

**Question 3.**

**The HCF of two numbers is 23 and their LCM is 1449. If one of the number is 161, find die other.**

**Solution:**

HCF of two numbers = 23

LCM =1449

One number = 161

Second number = 207

**Question 4.**

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

**Solution:**

HCF of two numbers = 145

LCM = 2175

One number = 725

Second number = 435

**Question 5.**

**The HCF of two numbers is 18 and their product is 12960. Find their LCM.**

**Solution:**

HCF of two numbers = 18

and product of two numbers = 12960

LCM of two numbers = 720

**Question 6.**

**Is it possible to have two numbers whose HCF is 18 and LCM is 760. Give reason.**

**Solution:**

HCF= 18

LCM = 760

HCF always divides the LCM completely

760 – 18 = 42 and remainder 4

Hence, it is not possible.

**Question 7.**

**Find the simplest form of**

**(a) **

**(b) **

**(c) **

**(d) **

**Solution:**

(a)

HCF of 69 and 92 = 23

**Question 8.**

**Find the largest number which divides 438 and 606, leaving remainder 6 in each case.**

**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.**

**Find the largest number which divides 320 and 457, leaving remainders 5 and 7 respectively.**

**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.**

**Find the least number which when divided by 35, 56 and 91 leaves the same remainder 7 in each case.**

**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.**

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

**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.**

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

**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.**

**Find the greatest number of four digits which is exactly divisible by 15, 24 and 36.**

**Solution:**

LCM of 15, 24, 36 = 360

Required number = 9999 – 279 = 9720

**Question 14.**

**Find the largest four-digits number which when divided by 4, 7 and 13 leaves a remainder of 3 in each case.**

**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.**

**Find the least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3.**

**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.**

**Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.**

**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.**

**Find the least number which when divided by 20, 25, 35 and 40 leaves remainder 14, 19, 29 and 34 respectively.**

**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.**

**In a seminar, the number of participants in Hindi, English and Mathematics are 60, 84 and 108 respectively. Find the minimum number of rooms required, if in each room, the same number of participants are to be seated and all of them being in the same subject.**

**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.**

**Three sets of English, mathematics and science books containing 336, 240 and 96 books respectively have to be stacked in such a way that all the books are stored subject wise and the height of each stack is the same. How many stacks will be there?**

**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.**

Three pieces of timber 42 m, 49 m and 63 m long have to be divided into planks of the same length. What is the greatest possible length of each plank ? How many planks are formed?

**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.**

**Find the greatest possible length which can be used to measure exactly the lengths 7 m, 3 m 85 cm and 12 m 95 cm.**

**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.**

**Find the maximum number of students among whom 1001 pens and 910 pencils can be distributed in such a way that each student gets the same number of pens and the same number of pencils.**

**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.**

**Find the least number of square tiles required to pave the ceiling of a room 15 m 17 cm long and 9 m 2 cm broad.**

**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.**

**Three measuring rods are 64 cm, 80 cm and 96 cm in length. Find the least length of cloth that can be measured an exact number of times, using any of the rods.**

**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.**

**An electronic device makes a beep after every 60 seconds. Another device makes a beep after every 62 seconds. They beeped together at 10 a.m. At what time will they beep together at the earliest?**

**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.**

**The traffic lights at three different road crossings change after every 48 seconds, 72 seconds and 108 seconds respectively. If they all change simultaneously at 8 a.m., then at what time will they again change simultaneously?**

**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.**

**Six bells commence tolling together and toll at intervals of 2, 4, 6, 8, 10, 12 minutes respectively. In 30 hours, how many times do they toll together?**

**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 = = 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.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

Abhishek roushan says

April 18, 2019 at 8:04 pmSo nyc answer