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Quantitative Aptitude
HCF AND LCM
Important Points:
Factors : The numbers which exactly divide a given
number are called the factors of that number. For example,
factors of 15 are 1, 3, 5 and 15.
Common Factor : When a unique number divides exactly two or more given numbers then that number is called
the common factor of the given numbers. For example,
each of the numbers 2, 3 and 6 is a common factor of 12
and 18.
Multiple : When a number is exactly divisible by another number, then the former number is called the multiple of the latter number. Obviously, latter number is contained in the former. For example. 24 is a multiple of 1, 2,
3. 4, 6, 8, 12 and 24.
Prime Number: A number is said to be a prime number
if its factors are 1 and the number itself. For example, 2, 3,
5, 7. 11. 13, 17, 19, 23, 29, 31
are prime numbers.
Co-prime : Two numbers are said to be co-prime (prime
to each other) if there is no common factor other than 1
between them. It should be noted that the numbers which
are co-prime, are not necessarily prime numbers. Even
composite numbers may be mutually prime. For example,
5 and 7 are co-prime. Again, 16 and 25 are also co-prime
where neither of them is a prime number.
Highest Common Factor : The highest common factor (HCF) also known as greatest common divisor (CCD) or
greatest common measure (GCM) of two or more numbers
is the greatest number which divides each of them exactly.
The HCF of two co prime numbers is I. It can be taken as
a test to decide whether two given numbers are co-prime
or not. For example,
Lets consider two numbers 24 and 36.
All possible factors of 24 are 1, 2, 3. 4, 6, 8, 12 and 24
All possible factors of 36 are 1, 2, 3, 4. 6, 9, 12, 18
and 36.
The common factors of 24 and 36 are 1, 2.3,4,6 and 12
The greatest common factor is 12.
Hence, 12 is the HCF of 24 and 36.
METHODS TO FIND OUT HCF : There are two methods of determining the HCF of two or more numbers.
(i) Method of Prime Factors : First of all resolve the
given numbers into prime factors and then find the product of all the prime factors common to all the given numbers. The product is the required HCF.
Example : Find the HCF of 126, 396 and 5400
Sol.:
126 = 2 x 3 x 3 x 7
396 = 2 x 2 x 3 x 3 x 1 1
5400 = 2 x 2 x 2 x 3 x 3 x 3 x 5 x 5
Common factors are 2, 3 and 3.
. HCF= 2 x 3 x 3 = 18
(ii) Method of Division : Suppose we are given two
numbers whose HCF is to be determined. Divide the greater
number by the smaller one. Now, divide the divisor by the
remainder, divide the remainder by the next remainder and
so on, until no remainder (o-remainder) is left. The last
divisor 50 obtained is the required HCF of the two given
numbers.
If there are three or more numbers whose HCF is to
be determined, then first find the HCF of any two of the
three numbers and then find the HCF of the third number
and the HCF of these two numbers and so on. The last
HCF will be the required HCF.
Example : Find the HCF of 48, 168 and 324
Sol. : Firstly, we find the HCF of 48 and 168.
4 8 ) 168(3
144
24) 48 ( 2
48
x
. HCF of 48 and 168 = 2 4
Now, we find the HCF of 24 and 324.
24)324(13
24
84
72 /
12)24(2
24
x
.-. HCF of 48, 168 and 324 is 12.
Lowest Common Multiple : The lowest common multiple (LCM) of two or more given numbers is the least number which is exactly divisible by each of them.
For example.
Multiples of 12 are 12, 24, 36, 48, 60, 72
Multiples of 18 are 18, 36, 54. 72, ...
Common multiples are 36, 72.
.-. LCM = 36
METHOD TO FIND LCM : Put the numbers side by
side putting comma in a row. Divide by any of the prime
numbers 2, 3, 5, 7 which divides at least any two of the
given numbers exactly. Write down the quotients so obtained and the indivisible numbers side by side and proceed in the way until you get a row of numbers which are
prime to one another. The continued product of all the divisors and the numbers in the last row will be the required
LCM.
Example : Find the LCM of 12. 15. 20 and 54.
Sol.
L
C
M
HCF o f Denominators
Some Improtant Results:
(i) If the numbers are divided by their HCF, then relatively prime numbers are obtaiend as quotient. Hence, if
the numbers be x and y and their HCF be m then
x = ma and y - mb, where a and b are co-prime.
(ii) When two numbers divided by a third number leave
the same remainder, then the difference of two numbers is
exactly divisible by the third number. Accordingly the HCF
of the numbers will be same as that of their differences.
(iii) The product of two numbers is equal to the product of their HCF and LCM. i.e.,
First number x second number = HCF x LCM
(iv) The HCF of the sum of the numbers and their LCM
is equal to the HCF of the numbers.
SOLVED EXAMPLES
4.
Find Ihe least n u m b e r of soldiers in a regiment which
c a n allow it to be d r a w n up in a h o l l o w s q u a r e 8, 10,
12 or 15 d e e p a n d a l s o into a solid s q u a r e .
7.
W h e n a Ijkeap of p e b b l e s is a r r a n g e d into groups of
32 e a c h '10 p e b b l e s a r e left over. W h e n they are
a r r a n g e d in h e a p s of 40 e a c h , 18 pebbles are left
o v e r a n d w h e n in g r o u p s of 72 each, 50 are left
over. T h e least n u m b e r of p e b b l e s in the heap is
(1) 1450
(2) 1440
(3) 1418
(4) 1412
Sol. : N u m b e r o f s o l d i e r s r e q u i r e d t o b e d r a w n u p
into hollow s q u a r e of 8, 10, 12 or 15 d e e p = 4 x ( L C M
of 8, 10, 12, 15) b e c a u s e t h e y s t a n d o n l y on t h e
perimeter.
[ S S C T a x A s s i s t a n t E x a m , 13.12.20091
Sol. : H e r e ,
3 2 - 10 = 2 2
4 0 - 18 = 2 2
7 2 - 50 = 22
.*, R e q u i r e d n u m b e r of p e b b l e s
= ( L C M of 3 2 , 4 0 a n d 72) - 2 2 = 1440 - 2 2 = 1418
8.
.
T o t a l n u m b e r of soldiers r e q u i r e d for h o l l o w s q u a r e
= 4 x 4 x 30 = 480
B u t to be d r a w n up into solid s q u a r e , the total n u m b e r s h o u l d be a perfect s q u a r e as t h e s o l d i e r s o c c u p y the area included in t h e s q u a r e .
.-.
5.
R a t i o of t w o n u m b e r s is 5 : 7; if t h e L C M of those
t w o n u m b e r s be 140 t h e n . H C F of those two numb e r s is
(1) 4
(2) 14
(3) 15
(4) 20
[ S S C T a x A s s i s t a n t E x a m , 13.12.2009]
Sol. : Let t h e n u m b e r s b e 5 x a n d 7x.
.-. L C M = 5 x 7 x x = 3 5 x
T h e least n u m b e r of s o l d i e r s in t h e r e g i m e n t
= 4 x 4 x 3 0 x 3 0 = 14,400
T h e L C M o f t w o n u m b e r s i s 2 8 t i m e s their HCF. T h e
s u m o f their L C M a n d H C F i s 1740. I f o n e o f t h e
n u m b e r s i s 2 4 0 , find t h e o t h e r n u m b e r .
Sol. : Let t h e H C F of t w o n u m b e r s be H a n d t h e i r
L C M b e L.
EVALUATE YOURSELF
SUBJECTIVE TYPE QUESTIONS
1. F o u r d r u m s c o n t a i n i n g w a t e r h a v e capacities 135
litres, 2 0 5 litres, 165 litres a n d 2 4 0 litres. Find the
greatest capacity measure of the drums.
2. Find t h e least m u l t i p l e of 13 w h i c h w h e n divided by
6, 8 a n d 12 leaves 5, 7 a n d 11 as remainders respectively.
3. T h e r a t i o b e t w e e n t h e L C M a n d HCF" of t w o numb e r s is 28 : 1. If t h e n u m b e r s be 96 a n d 168, find
their L C M a n d HCF.
4. A labourer w a s engaged for a certain n u m b e r of days
for R s . 5 7 . 5 0 , b u t b e i n g a b s e n t on s o m e of those
d a y s he w a s paid only R s . 5 0 . P r o v e that his daily
w a g e s could not b e m o r e than R s . 2 . 5 0 .
6.
5. In finding the G C M of t w o n u m b e r s the last remaind e r is 35 a n d the q u o t i e n t s a r e 1,2, 1,3. Find the
numbers.
6. Find the least n u m b e r w h i c h is divisible by all the
n u m b e r s from 1 to 20 inclusive.
7. T h r e e m e n start t o g e t h e r to travel the s a m e way
a r o u n d a circular track of 11 k m s in circumference.
T h e i r s p e e d s are 4, 5Vi a n d 8 k m s per h o u r respectively. W h e n will t h e y m e e t at a s t a r t i n g point ?
8. T h e c i r c u m f e r e n c e of t h e w h e e l s of a carriage are
2m 4 d m 5 cm a n d 3m 5 d m , w h a t is the least dist a n c e in w h i c h b o t h the w h e e l s will m a k e an exact
n u m b e r of r e v o l u t i o n s ?
9. W h a t is t h e least n u m b e r t h a t m u s t be added and
w h a t is the greatest n u m b e r that m u s t be subtracted
from 9 0 9 0 0 that the results m a y be divisible by 777.
819 and 4329?
10.
A h e a p of pebbles can be m a d e up exactly into groups
of 2 5 , b u t w h e n m a d e up into g r o u p s of 18, 27 and
3 2 , there is in e a c h c a s e a r e m a i n d e r of 1 1 ; find the
least n u m b e r of p e b b l e s s u c h a h e a p c a n contain.
11. An electronic device makes a beep every 60 sec.
Another device makes a beep after every 62 sec.
They beeped together at 10 a.m. Find the time when
they will next make a beep together at the earliest.
12. The traffic lights at three different road crossings
change after every 48 s e c , 72 sec. and 108 sec.
respectively. If they all change simultaneously at 8 :
30 : 00 hours then they will again change simultaneously at what time?
13. The HCF and LCM of two numbers are respectively
12 and 2448 and their difference is 60. Find the
numbers.
14. Two cog wheels revolve together. One of the wheels
has 24 teeth and the other has 30 teeth. The smaller
one makes 200 revolutions in 1 min. Find the number of revolutions made by larger wheel in an hour.
15. A gardener was asked to plant flowers in a row containing equal number of plants. He tried to plant 6,
8, 10 and 12 in each row, but 5 plants left in each
case but when he planted 13 in a row, no plant was
left. Find the least no. of plants with him.
OBJECTIVE TYPE QUESTIONS
16. The H.C.F. and L.C.M. of two numbers are 44 and
264 respectively. If the first number is divided by
2. the quotient is 44. The other number is
(1) 33
(2) 66
(3) 132
f4) 264
17. The traffic lights at three different road crossings
change after every 48 s e c , 72 sec. and 108 sec.
respectively. If they all change simultaneously at
8 : 20 : 00 hrs: then they will again change simultaneously at
(1) 8 : 27 : 12 hrs.
(2) 8 r27 : 24 hrs.
(3) 8 : 27 : 36 hrs.
(4) 8 : 27 : 48 hrs.
18. What least number must be subtracted from 1294
so that the remainder when divided by 9, 11, 13
will leave in each case the same remainder 6 ?
(DO
(2) 1
(3) 2
(4) 3
19. The largest natural number which exactly divides
the product of any four consecutive natural numbers is
(1)6
(2) 12
(3)24
(4) 120
20. Six bells commence tolling together and toll at intervals of 2, 4, 6, 8, 10 and 12 seconds respectively.
IrL30 minutes, how many times do they toll together?
CI) 4
(2) 10
'(3) 15
(4) 16
23. A worker was engaged for a certain number of days
and was promised to be paid Rs. 1189. He remained
absent forsome days and was paid Rs. 1073 only.
What were his daily wages?
(1) Rs. 27
(2) Rs. 28
(3) Rs. 29
(4) Rs. 31
24. The H.C.F. and L.C.M. of a pair of numbers are 12
are 924 respectively. How many such distinct pairs
are possible?
(1) 1
(2) 2
(3) 3
(4) 4
25. If m = 2 . 3 . 5 and n = 2* 3 . 7 . then the
H.C.F. of m and n is given by
(1) 2 . 3
(2) 2 . 3 . 5
(3) 4 0 . 3 . 5 . 7
(4) 2 . 3 . 5 . 7
26. A man was employed on the promise that he will be
paid the highest wages per day. The contract money
to be paid was Rs. 1189. Finally he was paid only
Rs. 1073. For how many days did he actually work?
(1)39
(2) 40
5
5
7
1 0
8
7
5
7
7
1 0
1 2
1 2
7
5
8
1 0
1 2
(3) 37
(4) 35
27. An officer was appointed on maximum daily wages
on a contract money of Rs. 4956. But on being
absent for some days, he was paid only Rs. 3894.
For how many days was he absent ?
(1) 3
(2) 4
(3) 5
(4) 6
28. A merchant has three kinds of milk : 435 litres, 493
litres and 551 litres. Find the least number of casks
of equal size required to store all the milk without
mixing.
(1) 51
(2) 61
(3) 47
(4) 45
29. A wholesale tea dealer has 408 kg. 468 kg and 516
kg of three different qualities of tea. He wants it all
to be packed into boxes of equal size without mixing.
Find the capacity of the largest possible box.
(1) 50
(2) 36
(3)24
- (4) 112
30. What is the least number of cut pieces of equal length
that can be cut out of two lengths 10 metres 857
milllimetres and 15 metres 87 millimetres.
(1) 174
(2) 172
(3) 164
(4) 184
ANSWERS
1. 5 litres
3. 672 and 24
6.232792560
8. 24 m 50 cm
10. 875
12. 8 : 37 : 48 hours
14.9600
16.(3)
17.(1)
18.(2)
21. (3)
22. (4)
23. (3)
26.(3)
27.(1)
28.(1)
2. 143
5. 385 and 525
7. 22 hours
9. 9 and 60597
11. 10.31 am
13. 144 and 204
15. 845
19. (3)
24. (2)
20. (4)
25. (1)
2 0 . (41
3 0 . (4)