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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
Here is the answer and explanation
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
Here is the answer and explanation
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
Here is the answer and explanation
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
Here is the answer and explanation
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
Here is the answer and explanation
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
Here is the answer and explanation
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
Here is the answer and explanation
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
Here is the answer and explanation
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
Here is the answer and explanation
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
Here is the answer and explanation
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 ?
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A. B.
C. D.
Here is the answer and explanation
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
Here is the answer and explanation
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
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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
Here is the answer and explanation
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)
(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)
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
Here is the answer and explanation
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
Here is the answer and explanation
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
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C. 4 D. 8
Here is the answer and explanation
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
Here is the answer and explanation
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
Here is the answer and explanation
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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Here is the answer and explanation
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
Here is the answer and explanation
Answer : Option C
Explanation :
Let the numbers be 11a and 11b
11a × 11b = 363
=> ab = 3
coprimes with product 3 are (1, 3)
(Reference : Coprime Numbers)
Hence the numbers with HCF 11 and product 363
= (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
Here is the answer and explanation
Answer : Option A
Explanation :
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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
Here is the answer and explanation
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 = 2. We get number as (90 × 2) + 4 = 184. But this is not a multiple of 7
Put k = 3. We get number as (90 × 3) + 4 = 274. 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
Here is the answer and explanation
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.
Here is the answer and explanation
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
Here is the answer and explanation
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
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Here is the answer and explanation
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
Here is the answer and explanation
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
Here is the answer and explanation
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
Here is the answer and explanation
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.
Here is the answer and explanation
Answer : Option D
Explanation :
Read More ...
32. What is the LCM of and ?
A. B.
C. D.
Here is the answer and explanation
Answer : Option D
Explanation :
Read More ...
Comments(58) Newest Sign in (optional)
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, 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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