01 August 2021 - Answers

 1. Three people started walks along a 360 km circular track from a given point. The first person walks 48 kms per day.The second person walks 60 kms per day.The third person walks 72 kms per day. In how many days will they meet at the starting point?

Answer: 30 days

Solution: To complete one round, 

A takes 360/48 = 7.5 days = 180 hours.

B takes 360/60 = 6 days = 144 hours.

C takes 360/72 = 5 days = 120 hours.

L.C.M of 180,144 and 120 = 720 hours. Hence they will meet after 720/24 = 30 days.

2. There is some liquid in a 60 litre can and a 45 litre can. The maximum capacity of another container which can measure the liquid in both the cans exact number of times is

Answer: 15 litres

Solution: HCF of 60 and 45 = 15.

3. The smallest number which when divided by 24,30 and 40 leaves remainders as 5 is

Answer: 125

Solution: The required number = LCM of 24,30 and 40 + 5

= 120 + 5 = 125

4. A scooter and an auto can run around a circular ground in 24 mins and 16 mins. They start from a given point at 6.30 pm. When will they meet again at the same starting point?

Answer: 7.18 pm

Solution: LCM of 24 and 16 is 48

So they will meet after 6.30 pm + 48 minutes = 7.18 pm

5. The least perfect square, which is divisible by each of 21, 36 and 66 is:

Answer: 213444 

Solution: L.C.M. of 21, 36, 66 = 2772.

Now, 2772 = 2 x 2 x 3 x 3 x 7 x 11 To make it a perfect square, it must be multiplied by 7 x 11. So, required number = 2^2 x 3^2 x 7^2 x 11^2 = 213444

6. What is the greatest number that will leave the same remainder when dividing the numbers 25, 52 and 43?

Answer: None of these

Solution: None of the above (As answer is 9)

Required number = HCF of (52-25, 52-43, 43-25)

= HCF of (27,9,18) = 9

7. 272 blue caps, 374 pink caps and 410 red caps are distributed equally among some kids with 2 red caps left over. What is the greatest possible number of kids?

Answer: 34

Solution: Biggest possible number of kids = HCF of 272, 374 and 408(410-2)  {As 2 red caps were left).

HCF of 272, 374 and 408 is 34.

8. The smallest number which when divided by 24,32 and 36 leaves remainders as 19, 27 and 31 respectively is

Answer: 283

Solution: The common difference is 5 (24-19 = 32-27 = 36-31 = 5).

Hence the required number = LCM of 24,32,36  - 5

= 288 - 5 = 283

9. Two numbers are in the ratio  3:5 and their LCM is 225. The larger number is

Answer: 75

Solution: Let the numbers be 3x ad 5x.

3*5*x = 225, x=15

Larger number = 5x = 75

10. The greatest number less than 1500 which is divisible both by 16 and 18 is

Answer: 1440

Solution: LCM of 16 and 18 is 144

Remainder when 1500 is divided by 144 is 60.

So the reqd greatest number = 1500 - 60 = 1440

11. The least number which is a perfect square and is divisible by each of the numbers 16, 20 and 24, is

Answer: 3600

Solution: LCM of 16,20,24 = 240

Expressing 240 in terms of prime factors it is 2*2*2*2*3*5

To make this a perfect square, we need to multiply it with 3*5

Hence required number = 240 * 3*5 = 3600

12. The largest number which when  dividing 29, 60 and 103 leaves remainders as 5,12 and 7 respectively is

Answer: 24

Solution: Required number = HCF of (29-5),(60-12),(103-7)

= HCF of 24,48,96

= 24

13. At 9.00 pm three colour lights in a festival change after every 16, 64 and 80 seconds respectively. After how many seconds will they change again simultaneously?

Answer: 320

Solution: nterval of change =  LCM of 16, 64 and 80  = 320

After 320 seconds they change again simultaneously.

14. The LCM of two numbers is 864 and their HCF is 12.  If the sum of the two numbers is 204, their difference is

Answer: 12

Solution: Let the numbers be x,y.

xy = LCM * HCF = 864*12

(x-y)2 = (x+y)2 - 4xy = 204*204 - 4*864*12

(x-y)2 = 144,

x-y = 12

15. What is the HCF of 6/7,  5/14 and 10/21?

Answer: 1/42

Solution: HCF of fractions = HCF of numerators/LCM of denominators

= HCF of (6,5,10)/ LCM of (7,14,21)

= 1/42

16. Three bells rang simultaneously at 9am (They ring at regular intervals of 20 mins, 30 mins and 40 mins respectively). The next time the three bells will ring together is

Answer: 11:00 am

Solution: LCM of 20,30,40 = 120 mins = 2 hours

Hence next time they will ring together is = 9am + 2 hrs = 11am

17. The LCM of two numbers is 1470 and their HCF is 7. If one of the numbers is 49, the other number is

Answer: 210

Solution: Let the other number be x.

Product of two numbers = LCM * HCF

x * 49 = 1470 * 7,

x = 210

18. How may numbers less than 5000 are there which are divisible by 21,35 and 63?

Answer: 15

Solution: LCM of 21,35 and 63 is 315

Integral quotient whe 5000 is divided by 315 is 15.

Hence 15 such numbers are there.

19. The product of two numbers is 2028 and their HCF is 13. The count of such combination of numbers is

Answer: 2

Solution: Let the numbers be 13x and 13y where x and y are co-primes

13x*13y = 2028

xy = 12

12 can be written as a combination of (1,12) (2,6) (3,4) (4,3) (6,2) (12,1)

But as x and y are to be co-primes only (1,12) (3,4). Hence the number of combinations is 2 (13,156 and 39,52)

Note: The number of combination is 2 but there are 4 possible pairs when we interchange the values

20. A shop keeper has three cakes of weight 12 kgs, 16kgs and 20 kgs. If he wants to cut these cakes into smaller pieces having equal weight without wastage, what is the maximum possible weight of each piece?

Answer: 4 kgs

Solution: Max possible weight of the pieces = HCF of 12,16 and 20 = 4kgs.

21. The LCM of three numbers is 120. Which of the following number must not be their HCF?

Answer: 16

Solution: Among the given options, 16 is not HCF as it cannot be formed from the factors 2*2*2*3*5.

22. A small bus interchange has 2 feeder services that start simultaneously at 9am. Bus number 801 leaves the interchange at 15-min intervals, while bus number 802 leaves at 20-min intervals. On a particular day, how many times did both services leave together from 9 am to 12 noon inclusive?

Answer: 4

Solution: L.C.M of leaving interval of bus 801 and 802 is = L.C.M of 15 and 20 = 60 mins.

 

Hence both services leave together at 9am, 10 am, 11 am and 12 noon (as it is inclusive). Hence answer is 4 times.

23. Four ropes of different lengths, 256 cms, 896 cms, 384 cms and 576 cms are to be cut into equal lengths. What is the greatest possible length of each piece?

Answer: 64 cms

Solution: The greatest possible length is the HCF of the given lengths. HCF of 256,896,384,576 is 64 cms.

24. The LCM of three different numbers is 120. Which of the following cannot be their HCF?

Answer: 35

Solution: 120 can be written in the form of prime factors as  2*2*2*3*5

Going through the optios, 35 cannot be the HCF as it cannot be formed as a product of the prime factors given above.

25. In a hostel, food ration is distributed to each person. If a day’s overall ration is 264 packets of biscuits, 396 packets of instant noodles and 1056 bottles of water, how many are there in the hostel?

Answer: 132

Solution: Number of people in hostel = H.C.F of 264, 396 and 1056 which is 132.