Maths (Hcf Lcm)

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1. Two numbers are in the ratio 2 : 3. If their L.C.M. is 48. what is sum of thenumbers?

A. 28 B. 40

C. 64 D. 42

Here is the answer and explanation

Answer : Option B

Explanation :

Let the numbers be 2x and 3x

LCM of 2x and 3x = 6x (∵ LCM of 2 and 3 is 6. Hence LCM

of 2x and 3x is 6x)

Given that LCM of 2x and 3x is 48

=> 6x = 48

=> x = 48∕6 = 8

Sum of the numbers = 2x + 3x = 5x = 5 × 8 = 40

2. What is the greatest number of four digits which is divisible by 15, 25, 40 and 75 ?

A. 9800 B. 9600

C. 9400 D. 9200


Answer : Option B

Explanation :

Greatest number of four digits = 9999

LCM of 15, 25, 40 and 75 = 600

9999 ÷ 600 = 16, remainder = 399

Hence, greatest number of four digits which is divisible by 15, 25, 40 and 75

= 9999 399 = 9600

3. Three numbers are in the ratio of 2 : 3 : 4 and their L.C.M. is 240. Their H.C.F. is:

A. 40 B. 30

C. 20 D. 10


Answer : Option C

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Explanation :

Let the numbers be 2x, 3x and 4x

LCM of 2x, 3x and 4x = 12x

=> 12x = 240

=> x = 240∕12 = 20

H.C.F of 2x, 3x and 4x = x = 20

4. What is the lowest common multiple of 12, 36 and 20?

A. 160 B. 220

C. 120 D. 180


Answer : Option D

Explanation :

2 12, 36, 20

2 6, 18, 10

3 3, 9, 5

1, 3, 5

LCM = 2 × 2 × 3 × 1 × 3 × 5 = 180

5. What is the least number which when divided by 5, 6, 7 and 8 leaves a remainder3, but when divided by 9 leaves no remainder?

A. 1108 B. 1683

C. 2007 D. 3363


Answer : Option B

Explanation :

Solution 1LCM of 5, 6, 7 and 8 = 840

Hence the number can be written in the form (840k + 3) which is divisible by 9

If k = 1, number = (840 × 1) + 3 = 843 which is not divisible by 9

If k = 2, number = (840 × 2) + 3 = 1683 which is divisible by 9

Hence 1683 is the least number which when divided by 5, 6, 7 and 8 leaves aremainder 3,

but when divided by 9 leaves no remainder

Solution 2 Hit and Trial MethodJust see which of the given choices satisfy the given condtions

Take 3363. This is not even divisible by 9. Hence this is not the answer

Take 1108. This is not even divisible by 9. Hence this is not the answer

Take 2007. This is divisible by 9.

2007 ÷ 5 = 401, remainder = 2 . Hence this is not the answer



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Take 1683. This is divisible by 9.

1683 ÷ 5 = 336, remainder = 3

1683 ÷ 6 = 280, remainder = 3

1683 ÷ 7 = 240, remainder = 3

1683 ÷ 8 = 210, remainder = 3

Hence 1683 is the answer

6. The H.C.F. of two numbers is 5 and their L.C.M. is 150. If one of the numbers is 25,then the other is:

A. 30 B. 28

C. 24 D. 20


Answer : Option A

Explanation :

Product of two numbers = Product of their HCF and LCM.

7. 504 can be expressed as a product of primes as

A. 2 × 2 × 3 × 3 × 7 × 7 B. 2 × 3 × 3 × 3 × 7 × 7

C. 2 × 3 × 3 × 3 × 3 × 7 D. 2 × 2 × 2 × 3 × 3 × 7


Answer : Option D

Explanation :

It is clear that 504 = 2 × 2 × 2 × 3 × 3 × 7

8. Which of the following integers has the most number of divisors?

A. 101 B. 99

C. 182 D. 176


Answer : Option D

Explanation :

99 = 1 × 3 × 3 × 11

=> Divisors of 99 are 1, 3, 11, 9, 33 and 99

101 = 1 × 101

=> Divisors of 101 are 1 and 101

Let one number = x

⇒ 25 × x = 5 × 150

⇒ x = = 305 × 150

25




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182 = 1 × 2 × 7 × 13

=> Divisors of 182 are 1, 2, 7, 13, 14, 26, 91 and 182

176 = 1 × 2 × 2 × 2 × 2 × 11

=> Divisors of 176 are 1, 2, 11, 4, 22, 8, 44, 16, 88, 176

Hence 176 has most number of divisors

9. The least number which should be added to 28523 so that the sum is exactlydivisible by 3, 5, 7 and 8 is

A. 41 B. 42

C. 32 D. 37


Answer : Option D

Explanation :

LCM of 3, 5, 7 and 8 = 840

28523 ÷ 840 = 33 remainder = 803

Hence the least number which should be added = 840 803 = 37

10. What is the least number which when doubled will be exactly divisible by 12, 14,18 and 22 ?

A. 1286 B. 1436

C. 1216 D. 1386


Answer : Option D

Explanation :

LCM of 12, 14, 18 and 22 = 2772

Hence the least number which will be exactly divisible by 12, 14, 18 and 22 = 2772

2772 ÷ 2 = 1386

=> 1386 is the number which when doubled, we get 2772

Hence, 1386 is the least number which when doubled will be exactly divisible by 12,14, 18 and 22 ?

11. What is the greatest possible length which can be used to measure exactly thelengths 8 m, 4 m 20 cm and 12 m 20 cm?

A. 10 cm B. 30 cm

C. 25 cm D. 20 cm


Answer : Option D

Explanation :

Required length = HCF of 800 cm, 420 cm, 1220 cm = 20 cm

12. Which of the following fraction is the largest ?




8/5/2015 H.C.F. and L.C.M.




A. B.

C. D.


Answer : Option A

Explanation :

Solution 1

Solution 2

13. The product of two 2 digit numbers is 2028 and their HCF is 13. What are thenumbers ?

A. 26, 78 B. 39, 52

C. 13, 156 D. 36, 68


Answer : Option B

Explanation :

Let the numbers be 13x and 13y (∵ HCF of the numbers = 13)

13x × 13y = 2028

=> xy = 12

coprimes with product 12 are (1, 12) and (3, 4) (∵ we need to take only

1112

4150

2140

56

LCM of 6, 40, 50 and 12 = 600

=56

500600

=2140

315600

=4150

492600

=1112

550600

Clearly = is the largest among these fractions550600

1112

LCM of 6, 40, 50 and 12 = 600

= .8356

= .522140

= .824150

= .921112

Clearly .92 = is the largest among these1112



8/5/2015 H.C.F. and L.C.M.




coprimes with product 12. If we take two numbers with product 12, but not coprime,

the HCF will not remain as 13)

(Reference : Coprime Numbers)

Hence the numbers with HCF 13 and product 2028

= (13 × 1, 13 × 12) and (13 × 3, 13 × 4)

= (13, 156) and (39, 52)

Given that the numbers are 2 digit numbers

Hence numbers are 39 and 52

14. The product of two numbers is 2028 and their HCF is 13. What are the number ofsuch pairs?

A. 4 B. 3

C. 2 D. 1


Answer : Option C

Explanation :

Let the numbers be 13x and 13y (∵ HCF of the numbers = 13)

13x × 13y = 2028

=> xy = 12

coprimes with product 12 are (1, 12) and (3, 4) (∵ we need to take only

coprimes with product 12. If we take two numbers with product 12, but not coprime,

the HCF will not remain as 13)



= (13 × 1, 13 × 12) and (13 × 3, 13 × 4)

= (13, 156) and (39, 52)

So, there are 2 pairs of numbers with HCF 13 and product 2028

15. N is the greatest number which divides 1305, 4665 and 6905 and gives the sameremainder in each case. What is the sum of the digits in N?

A. 4 B. 3

C. 6 D. 5


Answer : Option A

Explanation :

If the remainder is same in each case and remainder is not given, HCF of thedifferences of the numbers is the required greatest number

6905 1305 = 5600

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6905 4665 = 2240

4665 1305 = 3360

Hence, the greatest number which divides 1305, 4665 and 6905 and gives the same remainder, N

= HCF of 5600, 2240, 3360

= 1120

Sum of digits in N

= Sum of digits in 1120

= 1 + 1 + 2 + 0

= 4

16. A boy divided the numbers 7654, 8506 and 9997 by a certain largest number andhe gets same remainder in each case. What is the common remainder?

A. 156 B. 199

C. 211 D. 231


Answer : Option B

Explanation :

If the remainder is same in each case and remainder is not given, HCF of thedifferences of the numbers is the required largest number

9997 7654 = 2343

9997 8506 = 1491

8506 7654 = 852

Hence, the greatest number which divides 7654, 8506 and 9997 and leaves same remainder

= HCF of 2343, 1491, 852

= 213

Now we need to find out the common remainder.

Take any of the given numbers from 7654, 8506 and 9997, say 7654

7654 ÷ 213 = 35, remainder = 199

17. Find the greatest common divisor of 24 and 16

A. 6 B. 2


8/5/2015 H.C.F. and L.C.M.






C. 4 D. 8


Answer : Option D

Explanation :

16) 24 (1 16

8) 16 (2 16

0

Hence, greatest common divisor of 24 and 16 = 8

18. A, B and C start at the same time in the same direction to run around a circularstadium. A completes a round in 252 seconds, B in 308 seconds and c in 198 seconds,all starting at the same point. After what time will they again at the starting point ?

A. 36 minutes 22 seconds B. 46 minutes 22 seconds

C. 36 minutes 12 seconds D. 46 minutes 12 seconds


Answer : Option D

Explanation :

LCM of 252, 308 and 198 = 2772

Hence they all will be again at the starting point after 2772 seconds

or 46 minutes 12 seconds

19. The ratio of two numbers is 4 : 5. If the HCF of these numbers is 6, what is theirLCM?

A. 30 B. 60

C. 90 D. 120


Answer : Option D

Explanation :

Let the numbers be 4k and 5k

HCF of 4 and 5 = 1

Hence HCF of 4k and 5k = k

Given that HCF of 4k and 5k = 6

=> k = 6

Hence the numbers are (4 × 6) and (5 × 6) = 24 and 30

LCM of 24 and 30 = 120

20. What is the HCF of 2.04, 0.24 and 0.8 ?

A. 1 B. 2

C. 0.02 D. 0.04





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Answer : Option D

Explanation :

Reference : How to calculate LCM and HCF for Decimals

Step 1 : Make the same number of decimal places in all the given numbers bysuffixing

zero(s) in required numbers as needed.

=> 2.04, 0.24 and 0.80

Step 2 : Now find the HCF of these numbers without decimal.

=>HCF of 204, 24 and 80 = 4

Step 3 : Put the decimal point in the result obtained in step 2 leaving as many digits

on its right as there are in each of the numbers.

i.e., here, we need to put decimal point in the result obtained in step 2 leaving

two digits on its right.

=> HCF of 2.04, 0.24 and 0.8 = 0.04

21. If HCF of two numbers is 11 and the product of these numbers is 363, what is thethe greater number?

A. 9 B. 22

C. 33 D. 11


Answer : Option C

Explanation :

Let the numbers be 11a and 11b

11a × 11b = 363

=> ab = 3

coprimes with product 3 are (1, 3)



= (11 × 1, 11 × 3)

= (11, 33)

Hence numbers are 11 and 33

The greater number = 33

22. What is the greatest number which on dividing 1223 and 2351 leaves remainders90 and 85 respectively?

A. 1133 B. 127

C. 42 D. 1100


Answer : Option A

Explanation :


http://www.careerbless.com/aptitude/qa/numbers_imp.php#imp11



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Required number

= HCF of (1223 90) and (2351 85)

= HCF of 1133 and 2266

= 1133

23. What is the least multiple of 7 which leaves a remainder of 4 when divided by 6,9, 15 and 18 ?

A. 364 B. 350

C. 343 D. 371


Answer : Option A

Explanation :

LCM of 6, 9, 15 and 18 = 90

Required Number = (90k + 4) which is a multiple of 7

Put k = 1. We get number as (90 × 1) + 4 = 94. But this is not a multiple of 7



Put k = 4. We get number as (90 × 4) + 4 = 364. This is a multiple of 7

Hence 364 is the answer.

24. Three numbers which are coprime to each other are such that the product of thefirst two is 119 and that of the last two is 391. What is the sum of the three numbers?

A. 47 B. 43

C. 53 D. 51


Answer : Option A

Explanation :

Since the numbers are coprime, their HCF = 1

Product of first two numbers = 119

Product of last two numbers = 391

The middle number is common in both of these products.

Hence if we take HCF of 119 and 391, we get the common middle number

HCF of 119 and 391 = 17

=> Middle Number = 17

First Number = 119∕17 = 7

Last Number = 391∕17 = 23

Sum of the three numbers = 7 + 17 + 23 = 47

25. Reduce to its lowest terms43294662

7 13



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A. B.

C. D.


Answer : Option C

Explanation :

We need to find out HCF of 4329 and 4662

4329) 4662 (1 4329

333) 4329 (13 4329

0

Hence, HCF of 4329 and 4662 = 333

4329 ÷ 333 = 13

4662 ÷ 333 = 14

26. What is the greatest number which divides 24, 28 and 34 and leaves the sameremainder in each case?

A. 1 B. 2

C. 3 D. 4


Answer : Option B

Explanation :

If the remainder is same in each case and remainder is not given, HCF of thedifferences ofthe numbers is the required greatest number

34 24 = 10

34 28 = 6

28 24 = 4

Hence, the greatest number which divides 24, 28 and 34 and gives the same remainder

= HCF of 10, 6, 4

= 2

27. Six bells start ringing together and ring at intervals of 4, 8, 10, 12, 15 and 20seconds respectively. how many times will they ring together in 60 minutes ?

A. 31 B. 15

C. 16 D. 30

713

1317

1314

712

Hence =43294662

1314



8/5/2015 H.C.F. and L.C.M.






Answer : Option A

Explanation :

LCM of 4, 8, 10, 12, 15 and 20 = 120

120 seconds = 2 minutes

Hence all the six bells will ring together in every 2 minutes

Hence, number of times they will ring together in 60 minutes

28. What is the least number which when divided by 8, 12, 15 and 20 leaves in eachcase a remainder of 5 ?

A. 125 B. 117

C. 132 D. 112


Answer : Option A

Explanation :

LCM of 8, 12, 15 and 20 = 120

Required Number = 120 + 5 = 125

29. The HCF of two numbers is 23 and the other two factors of their LCM are 13 and14. What is the largest number?

A. 312 B. 282

C. 299 D. 322


Answer : Option D

Explanation :

The HCF of a group of numbers will be always a factor of their LCM

HCF is the product of all common prime factors using the least power of eachcommon prime factor.

LCM is the product of highest powers of all prime factors

HCF of the two numbers = 23

=> Highest Common Factor in the numbers = 23

Since HCF will be always a factor of LCM, 23 is a factor of the LCM.

Other two factors in the LCM are 13 and 14.

Hence factors of the LCM are 23, 13, 14

So, numbers can be taken as (23 × 13) and (23 × 14)

= 1 + = 31602




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= 299 and 322

Hence, largest number = 322

[A more detailed explanation ...

we can take the numbers as (23 × 13) and (23 × 14) because of the

following reasons

HCF is given as 23.

The HCF of a group of numbers will be always a factor of their LCM.

Hence, 23 is a factor of the LCM

Given that other two factors of the LCM are 13 and 14.

Hence factors of the LCM are 23, 13, 14

Now assume that we take the numbers are (23 × 13) and (23 × 14).

If we write the numbers as the product of prime factors,

first number = (23 × 13)

second numbers = (23 × 14) = (23 × 2 × 7)

HCF = product of all common prime factors using the least power of each common prime factor

= 23

LCM is the product of highest powers of all prime factors

= 23 × 13 × 2 × 7 = 23 × 13 × 14

Clearly we get HCF as 23 and the factors in the LCM as 13, 14 and 23.

Hence every conditions are satisfied.

30. What is the smallest number which when diminished by 12, is divisible 8, 12, 22and 24?

A. 276 B. 264

C. 272 D. 268


Answer : Option A

Explanation :

Required Number = (LCM of 8, 12, 22 and 24) + 12 = 264 + 12 = 276

1 2 1


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31. What is the HCF of and ?

A. B.

C. D.


Answer : Option D

Explanation :

Read More ...

32. What is the LCM of and ?

A. B.

C. D.


Answer : Option D

Explanation :

Read More ...

Comments(58) Newest Sign in (optional)

showing 110 of 58 comments, sorted newest to the oldest

, 13

23

14

23

13

14

112

HCF for fractions = HCF of Numerators

LCM of Denominators

HCF of , and = =13

23

14

HCF (1, 2, 1)

LCM (3, 3, 4)112

, 23

56

49

310

320

103

203

LCM for fractions = LCM of Numerators

HCF of Denominators

LCM of , and = =23

56

49

LCM (2, 5, 4)

HCF (3, 6, 9)203

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Ashok Bhukhar 4 months ago

How to find out sum of numbers if their lcm given..?plz . tell me

surajsingh 4 months ago

The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is?

Jay 4 months ago

Take numbers as 13a and 13b (because 13 is the HCF)13a * 13b = 2028ab = 12





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12 can be written as a product of coprime numbers in the following waysa. 1*12b. 3*4(i.e., two ways)

So required number of ways = 2

The pairs are (13*1, 13*12) and (13*3, 13*4)

Priyanka 5 months ago

If LCM of 15,20,X=180 then what is the value of X???

Jay 5 months ago

Many values are possible.

15 = 3*520 = 2*2*5LCM of 15 and 20 = 3*2*2*5=60

(3*2*2*5) * 3 = 180ie, an additional 3 should be there to make the LCM 180

number can be 3*3=9, 3*5*3=45, 2*2*5*3*3 = 180, etc

priya 5 months ago

the HCF of 2 number is 98 and their LCM is 2352.the sum of the number may bea.1372b.1398c.1426d.1484

gaukaran 2 months ago

X*y= 98*2352X*y=98*98*12*2X*y= (98*2)*(98*12)As compair to x and y then X= 98*2X= 196Y = 98*12Y= 1176ThenX+y196+1176= 1372

Jay 5 months ago

In the question, 'none of these' is not given as an option and 1372 is the only numberdivisible by 98. So one can directly write the answer as 1372

siva 7 months ago

Hcf is equal to 103 LCM 19261Sum of numbers? Given numbers are four digits

Jay 6 months ago

let numbers be 103a and 103b

19261/103 = 187 = 11*17 (prime factorization)

So numbers can be 103*11 and 103*17.ie, 1133 and 1751

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