## RS Aggarwal Class 6 Math Second Chapter Factors And Multiples Exercise 2E Solution

## EXERCISE 2E

**Find the LCM of the numbers given below:**

**(1) 42, 63**

**Ans: **Multiples of 42 are: 42, 84, 126, 168, 210……

Multiples of 63 are: 63, 126, 189, 252, 315,…….

Common multiples of 42 and 63 are: 126,….

Lowest common multiple of 42 and 63 is 126.

Hence, LCM of 42 and 63= 126.

**(2) 60, 75**

**Ans: **Multiples of 60 are: 60, 120, 180, 240, 300……

Multiples of 75 are:75, 150, 225, 300, 375,…….

Common multiples of 60 and 75 are: 300,….

Lowest common multiple of 60 and 75 is 300.

Hence, LCM of 60 and 75= 300.

**(3) 12, 18, 20**

**Ans: **Multiples of 12 are: 12, 24, 36, 48, 180……

Multiples of 18 are: 18, 36, 54, 72, 90,180…….

Multiples of 20 are: 20, 40, 60, 80, 100,180….

Common multiples of 12, 18 and 20 are: 180,….

Lowest common multiple of 12, 18 and 20 is 180.

Hence, LCM of 12, 18 and 20= 180.

**(4) 36, 60, 72**

**Ans: **Multiples of 36 are 36 x10= 360, 36 x 11= 396,….

Multiples of 60 are 60 x 6= 360, 60 x 7= 420,…

Multiples of 72 are 72 x 5= 360, 72 x 6= 432,…

Common multiples of 36, 60 and 72 are 360,….

Lowest common multiple of 36, 60 and 72 is 360.

Hence, LCM of 36, 60 and 72= 360.

**(5) 36, 40, 126**

**Ans: **Multiples of 36 x 70= 2520,…

Multiples of 40 x 63= 2520,….

Multiples of 126 x 20= 2520,….

Common multiples of 36, 40 and 126 are 2520,….

Lowest common multiple of 36, 40 and 126 is 2520.

Hence, LCM of 36, 40 and 126= 2520.

**(6) 16, 28, 40, 77**

**Ans: **Multiples of 16 are 16 x 385= 6160,…

Multiples of 28 are 28 x 220= 6160,…

Multiples of 40 are 40 x 154= 6160,….

Multiple of 77= 77 x 80= 6160,….

Common multiples of 16, 28, 40 and 77= 6160.

Lowest common multiplie is 6160.

Hence, LCM of 16, 28, 40 and 77= 6160.

**(7) 28, 36, 45, 60**

**Ans: **Multiples of 28 are 28 x 45= 1260,…

Multiples of 36 x 35= 1260,…

Multiples of 45 x 28= 1260,…

Multiples of 60 are 60 x 21= 1260,…

Common multiples of 28, 36, 45 and 60 are 1260,…

Lowest common multiplies 1260.

Hence, LCM of 28, 36, 45 and 60= 1260.

**(8) 144, 180, 384**

**Ans: **Multiples of 144 are 144 x 40= 5760,…

Multiples of 180 are 180 x 32= 5760,…

Multiples of 384 are 384 x 15= 5760,…

Common multiples of 144, 180 and 384 are 5760,…

Lowest common multiple is 5760.

Hence, LCM of 144, 180 and 384= 5760.

**(9) 48, 64, 72, 96, 108**

**Ans:**

** **Multiples of 48 are 48 x 36= 1728,…

Multiples of 64 are 64 x 27= 1728,…

Multiples of 72 are 72 x 24= 1728,…

Multiples of 96 are 96 x 18= 1728,…

Common multiples of 48, 64, 72 and 96 are 1728,…

Lowest common multiplies 1728.

Hence, LCM of 48, 64, 72 and 96= 1728.

**Find the HCF and LCM of**

**(10) 117, 221**

**(11) 234, 572**

**Ans: **We first find the HCF of the given numbers.

Therefore, HCF= 26.

And LCM= 5148.

Therefore HCF= 26 and LCM= 5148.

**(12) 693, 1078**

**Ans: **We first find the HCF of the given numbers.

Therefore, HCF= 77.

And LCM= 9702.

Therefore HCF= 77 and LCM= 9702.

**(13) 145, 232**

**Ans: **We first find the HCF of the given numbers.

Therefore, HCF= 29.

And LCM= 1160.

Therefore HCF= 29 and LCM= 1160.

**(14) 861, 1353**

**Ans: **We first find the HCF of the given numbers.

Therefore, HCF= 123.

And LCM= 9471.

Therefore HCF= 123 and LCM= 9471.

**(15) 2923, 3239**

**Ans: **We first find the HCF of the given numbers.

Therefore, HCF= 79.

And LCM= 119843.

Therefore HCF= 79 and LCM= 119843.

**(16) For each pair of numbers, verify that their product= (HCF x LCM).**

**(i) 87, 145**

**Solution: **We have, 87= 3 x 29

and, 145= 5 x 29

So, the HCF of 87 and 145 is 29

And, the LCM of 87 and 145 is 3 x 5 x 29= 435.

Now, the product of the given numbers= 87 x 145= 12615.

Product of their HCF and LCM=(29 x 435)= 12615.

**(ii) 186, 403**

**(iii) 490, 1155**

**(17) The product of two numbers is 2160 and their HCF is 12. Find their LCM.**

**Solution: ** We know that

**(18) The product of two numbers is 2560 and their LCM is 320. Find their HCF.**

**Solution: **We know that

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

**Solution: **We know that,

(One number) x (other number)= (HCF X LCM)

Let the number is A.

725 x A= 145 x 2175

The other number is 435.

**(20) The HCF and LCM of two numbers are 131 and 8253 respectively. If one of the numbers is 917, find the other.**

**Solution: **We know that,

(One number) x (other number)= (HCF X LCM)

Let the number is A.

917 x A= 131 x 8253

The other number is 1179.

**(21) Find the least number divisible by 15, 20, 24, 32 and 36.**

**Ans: **15= 15 x 96= 1440.

20= 20 x 72= 1440.

24= 24 x 60= 1440.

32= 32 x 45= 1440.

36= 36 x 40= 1440.

Hence, the required number is 1440.

**(22) Find the least number which when divided by 25, 40 and 60 leaves 9 as the remainder in each case.**

**Solution: **25= 25 x 24= 600.

40= 40 x 15= 600.

60= 60 x 10= 600

Hence the required number is (600+9)= 609.

**(23) Find the least number of five digits exactly divisible by 16, 18, 24 and 30.**

**Solution: **16= 16 x 630= 10080.

18= 18 x 560= 10080.

24= 420= 10080.

30= 30 x 336= 10080.

Hence, the required number is 10080.

**(24) Find the greatest number of five digits exactly divisible by 9, 12, 15, 18 and 24.**

**Solution: **9= 9 x 11080= 99720.

12= 12 x 8310= 99720.

15= 15 x 6648= 99720.

18= 18 x 5540= 99720.

24= 24 x 4155= 99720.

Hence the required number is 99720.

**(25) Three bells toll at intervals of 9, 12, 15 minutes. If they start tolling together, after what time will they next toll together?**

**Solution: **

Therefore, LCM of 9, 12, 15= (3 x 3x 4 x5)= 180minutes

So, all the toll at intervals will start tolling together after 180 minutes. i.e. 3 hours.

**(26) Three boys step off together from the same place. If their steps measure 36 cm, 48 cm and 54 cm, at what distance from the starting point will they again step together?**

**Solution: **Therefore LCM of 36, 48, 54= 2 x 3 x 3 x 2 x 1 x 4 x 3= 432 cm =4m 32 cm

So their distance from the starting points is 4m 32cm.

** ****(27) The traffic lights at three different road crossing change after every 48 seconds, 72 seconds and 108 seconds. If they start changing simultaneously at 8 a.m., after how much time will they change again simultaneously?**

**Solution: **Therefore, LCM of 48, 72, 108= 2 x 2 x 3 x 3 x 2 x 2 x 3= 432 sec

So, they will change again simultaneously after 432 sec i.e. 7 minutes 12 sec.

**(28) Three measuring rods are 45cm, 50cm and 75cm in length. What is the least length (in meters) of a rope that can be measured by the full length of each of these three rods?**

**Solution: **Therefore, LCM of 45, 50, 75= 450cm

So the full length of each of these three rods is 450cm i.e., 4m 5cm.

**(29) An electrical device makes a beep after every 15 minutes. Another device makes a beep after every 20 minutes. They beeped together at 6 a.m. At what time will they next beep together?**

**Solution: ** Therefore, LCM of 15, 20= 60 minutes

Thus they beeped together of 6 a.m. that they will next beep together at (6a.m + 60 minutes)= 7 a.m.

**(30) The circumferences of four wheels are 50 cm, 60 cm, 75 cm and 100 cm. They start moving simultaneously. What least distance should they cover so that each wheel makes a complete number of revolutions?**

**Solution: **

Therefore, LCM of 50, 60, 75, 100= 5 x 5 x 2 x 3 x 2 x 1 x 1 x 1 x 1= 300 cm.

So, the least distance is 300 cm i.e. 3 m.