Solved Problems on HCF and LCM
Aptitude questions with Explanations - HCF and LCM
Express $\frac{1168}{1095}$
in simple form?
a)$\frac{15}{16}$ b)$\frac{16}{15}$ c)$\frac{8}{9}$ d)$\frac{12}{17}$ e)None
Answer: B
Explanation: To express any fraction in simple (small form), we have to find
the HCF of numerator and denominator.
1095)1168(1
1095
73)1095(15
1095
———
0
HCF of 1095
and 1168 is 73
Therefore $\frac{\frac{1168}{73}}{\frac{1095}{73}}\,=\,\frac{16}{15}$
Quantitative Aptitude Questions- HCF and LCM
Find the HCF of the numbers a and b, where
Find the HCF of the numbers a and b, where
a = 25 x 32 x 76 x 114 and
n = 23x34x56x11x133?
a)396 b)408 c)792 d)1584 e)None of these
a)396 b)408 c)792 d)1584 e)None of these
Answer: C
Explanation: We use prime factorization method, because already
the numbers m and n written as product of prime factors
We take only common factors
with least powers
HCF = 23 x 32 x
11 = 792
Arithmetic Questions with Answers - HCF and LCM
Find the HCF of of 0.36, 0.48, 0.72 and 2.4
a)0.12 b)0.24 c)0.06 d)0.08 e)0.96
Answer: A
Explanation: Given numbers are decimal fractions.
Make the number of decimal
places in all the given numbers the same
i.e., 0.36, 0.48, 0.72 and
2.40 this they become
36, 48, 72, 240 (multiply all by 100)
HCF of 36,48,72,240,=1
Previously we multiplied the given numbers by 100. Now to get the answer divide the obtained HCF by 12
Previously we multiplied the given numbers by 100. Now to get the answer divide the obtained HCF by 12
HCF of 0.36,0.48,0.72 and 2.4= 0.12
Aptitude Questions with Explanations -HCF and LCM
What is the side of the largest possible square
brick which can be paved on the floor of a room 4m 96cm long
and 4m 3cm broad?
a)27 b)28 c)29 d)31 e)37
Answer: D
Explanation: This is the problem on applications of HCF.
Converting given measurements into centimeters
Converting given measurements into centimeters
4m 96 cm = 400+96 = 496 cm
4m 3cm =
400+3 =403 cm
To find the side of the largest possible brick , the side of brick must divide length as well as breadth exactly.
To find the side of the largest possible brick , the side of brick must divide length as well as breadth exactly.
Using division method
403)496(1
403
93)
403 (4
370
31) 93 (3
93
0
HCF = 31
Side of the largest possible square brick =31cm
Side of the largest possible square brick =31cm
Quantitative Aptitude Questions & Answers- HCF & LCM
In
a school, the number of students in Sections A , B and C of 10th class are
70,98 and 126 respectively. Due to overload of student in each class , the
administration wants to increase the number rooms . What is the minimum number
of rooms required, if in each room the same number of students are to seated
and all of them being in the same section?
a)18 b)20 c)21 d)23 e)25.
Answer : C
Explanation: This is the problem on real time applications of HCF
The number students in each room
= HCF of 70,98 and 126 =14
The number students in each room
= HCF of 70,98 and 126 =14
In each room , maximum 14 students can be
seated.
Total number
students = 70 + 98 +126 =294
Number of rooms required
= $\frac{294}{14}$ =21
Aptitude Questions with Explanations- HCF and LCM
Find
the greatest number that will divide 147,185 and 251 leaving the
remainders 3,5 and 11 respectively.
a) 12 b)15 c)18 d)21 e)23
Answer: A
Explanation: The greatest number when divide 147,185 and 251 leaving
the remainders 3,5 and 11.Means the greatest number is not exactly dividing the
given numbers. If we subtract the remainders from the respective numbers, then
they are exactly divisible by the greatest numbers.
The required greatest number = HCF of (147 -
3),(185-5) and ( 251-11)
=HCF of 144,180 and 240
=12
Quantitative Aptitude For GMAT & GRE - HCF & LCM
Find the
greatest number that will divide 172, 205 and 304 so as to leave the
same remainder in each case?
a)19 b)21 c)33 d)45 e)49
Answer : C
Explanation: The
greatest number that will divide 172,205 and 304 leaving the same
remainder in each case is HCF ((205-172), (304-205), (304-172))
= HCF of 33, 99, and 132
33=31x11
99=3 x 3 x 11 132=3 x 2 x 2 x11
Required number =3
x 11=33
Aptitude Questions for CRT - HCF and LCM
What is
the least number of that is exactly divisible by 4, 12, 15, and
18?
a) 120 b)108 c)180 d)240 e)360
Answer: C
Explanation: LCM of
(4, 12, 15, and 18)
2|4 12 15 18
3|2 6
15 9
2|2 2
5 3
1 1
5 3
Least number which is exactly divisible by 4,12,15 and 18 is =180
Least number which is exactly divisible by 4,12,15 and 18 is =180
Aptitude Questions with Explanations- HCF and LCM
LCM of
0.12, 0.15, 0.2 and 0.54?
a) 5.4 b)10.8 c)16.2 d)2.7 e)None of these
Answer: A
Explanation : Make the number of decimal
places in all the given numbers the same i.e., 0.12, 0.15, 0.2 and 0.54
LCM of 12, 15, 20 and 54
2|12 15
20 54
2|6
15 10 27
3|3
15 5 27
5|1
5 5 9
1 1
1 9
LCM=22×3×5×9=540
LCM=$ \frac{540}{10} $ =5.4
Aptitude Questions with Answers - HCF and LCM
A, B and
C start running around a circular stadium and complete one round in 27 s, 9 s
and 36 s, respectively. In how much time will they meet again at the
starting point?
a)1 hour b) 1 minute c) 1 minute 48 seconds
d) 2 minutes e)3 minutes
d) 2 minutes e)3 minutes
Answer: C
Explanation : This problem is on real time applications of LCM
LCM of 27, 9 and 36 = 108
LCM of 27, 9 and 36 = 108
So
they will meet again at the starting point after 108 s. i.e., 1 min 48 s.
Aptitude Questions and Answers - HCF and LCM
Three
friends Raju , Ramesh and Sunil start running around a circular stadium and
complete a single round in 24 s, 36 s and 40 s, respectively. After how many
minutes will they meet against at the starting point?
a)1 hour b) 30 minutes c)15 minutes d)6 minutes e)10 minutes
Answer: D
Explanation : This is the problem on appliations of LCM in real life.
24 = 3 × 2 × 2 × 2 = 3 × 2³
24 = 3 × 2 × 2 × 2 = 3 × 2³
36 = 3 × 3 × 2 × 2 = 3² × 2²
and 40 = 2 × 2 × 2 × 5 = 5¹ × 23
LCM of 24, 36 and 40 = 3² × 2³ × 5
= 9 × 8 × 5 = 360
Hence, they will meet again at
the starting point after 360 s, i.e., 6 min
Aptitude Questions for CAT - HCF and LCM
What is
the least number when divided by 4, 5, 6, 8 and 10 leaves 3 as
remainder?
a)63 b)117 c)123 d)243 e)363
Answer: C
Explanation: The least number which when divided by 4,5,6,8
and 10 is the LCM of these numbers. But each time to get 3 as remainder, we have to add the
remainder 3 to the obtained LCM.
LCM of
(4, 5, 6, 8, 10)+ 3
=120+3
=123
Quantitative Aptitude Questions - HCF and LCM
The least number which should be added to 2497 so that sum is divisible by 5, 6, 4, 3?
The least number which should be added to 2497 so that sum is divisible by 5, 6, 4, 3?
a)
43 b)23 c)45 d)37 e)46
Answer: B
Explanation: LCM of 5,6,4,3 is 60.
On dividing 2497 by 60 we get 37 as remainder.
Therefore number to be added is 60 - 37 =23.
On dividing 2497 by 60 we get 37 as remainder.
Therefore number to be added is 60 - 37 =23.
Answer is 23.
Aptitude Questions -HCF and LCM
The HCF
of 2 numbers is 11 and LCM is 693.If one of numbers is
77.find the other number ?
a)99 b)198 c)66 d)143 e)None of these
Answer: A
Explanation : Product
of two numbers= HCF x LCM
77 x a = 11 x 693
Other number a =
$\frac{11\times693}{77}$
=99.
Quantitative Aptitude
- HCF and LCM
Find largest number of four digits divisible by 12, 15, 18, 27 ?
Find largest number of four digits divisible by 12, 15, 18, 27 ?
a)
9820 b)9720 c)9960 d)9760 e)9620
Answer: B
Explanation : The largest 4 digit number is
9999.
We know
that , if LCM of given numbers divide a number N , then N is exactly divisible
by all the given numbers .
LCM of 12,15,18,27 is 540.
On dividing 9999 by 540 we get 279 as
remainder.
Therefore number =9999 - 279
=9720.
Aptitude Questions - HCF and LCM
Find the
smallest number of two digits which on being divided by 2, 3, 4 and 5 leaves 0,
1, 2 and 3 as remainder respectively.
a)
40 b)52 c)54 d)67 e)58
Answer : E
Explanation: LCM of 2, 3, 4 and 5 is 60.
The smallest number of two digits is 10.
But we also see that the smallest number divisible by 2, 3, 4 and 5 is
60. Therefore, 10 has no significance here.
The difference between divisor and remainder is 2 in each case. So, we
subtract 2 from 60.
60 - 2 = 58 Ans.
Aptitude Questions with Solutions- HCF and LCM
Aptitude Questions with Solutions- HCF and LCM
The least positive integer which is a perfect square
and also divisible by each of 6,12, 18 and 24?
a)36 b)100 c)144 d)196 e)108
Answer:
C
Explanation
: Perfect Square : A number when written as product of
prime factors , if each prime factor has even number as its power , then the
number is perfect square.
Writing the given number as
product of prime factors
6 = 2
x 3
12 = 23 x 3
18 = 2 x 32
24= 23 x 3
LCM = 23 x 32
= 72
But 72 is not the perfect
square. But multiples of 72 are also exactly divisible by given numbers.
72 x 2 =144 is the perfect
square. So answer is 144
Aptitude Questions with Answers - HCF and LCM
What is the greatest number that will divide 99, 147 and 219
leaving the same reminder in each case?
a) 396 b) 12 c)
369 d) 3 e) 24
Answer: E
Explanation : Given that, a = 99,
b = 147, c = 219
Required number = [H.C.F. of (a-b),
(b-c), (c-a)]
Now,
(a-b) = (99 – 147) = 48
(b-c) = (147 – 219) = 72
(c-a) = (219 – 99) = 120
Also
48 = 24 x 3¹
72 = 23 x 32
120 = 23 x 3¹ x 5¹
∴ HCF of 48,72 and
120 = 23 x 3¹ = 8 x 3 = 24
Aptitude Questions and Answers - HCF and LCM
What is the least number, which when divided by
5,6,8,10 and 12 leaves the remainder 2 in each case , but when divided by 13
leaves no remainder ?
a)262 b)362 c)960 d)962 e)None of these
a)262 b)362 c)960 d)962 e)None of these
Answer
: D
Explanation
: LCM of 5,6,8,10 and 12 = 120
But in each case , we have a remainder 2
.So the required number is 120n+2 which is exactly divisible by 13.
120 n + 2 = 13 x 9 n
+ 2
So clearly 3n+2 must be
divisible by 13.
For n=8, 3n +2 is exactly divisible by 13.
Therefore , the required
number = 120n + 2
= 120 x 8 + 2 = 962
Aptitude Questions with Explanations - HCF and LCM
Aptitude Questions with Explanations - HCF and LCM
HCF of two numbers is 27 and sum of the two numbers
is 216. How many such pairs of numbers are possible?
a)0 b)1 c)2 d)3 e)4
Answer
: C
Explanation
: HCF of two numbers =27
Let two numbers are in
the ratio a:b , then numbers are 27a and
27b.
Given that , sum of
two numbers is 216
27a + 27 b = 216
a+b = 216/27= 8
Writing the possibilities of a
and b to be sum is 8
( 1,7 ),
(2,6),(3,5) and (4,4).
Of these pairs, (1,7)
and (3,5) are relatively primes .
Therefore, 2 such pairs are possible.
And those required numbers are ( 1 x 27 , 7 x 27) and ( 3 x 27, 5 x 27)
= (27 , 189) and (81,135)
And those required numbers are ( 1 x 27 , 7 x 27) and ( 3 x 27, 5 x 27)
= (27 , 189) and (81,135)
Quantitative Aptitude for Bank Exams - HCF and LCM
Find the smallest number which when increased by 5 is divisible by 9, 21, 25 and 30?
Find the smallest number which when increased by 5 is divisible by 9, 21, 25 and 30?
a)3145 b)6305 c)3155 d)1605 e)6295
Answer:
Explanation
:
LCM of the given
numbers.
3
|9, 21, 25, 30
5
|3, 7, 25, 10
| 3, 7, 5, 2
LCM
= 3 × 5 × 3 × 7 × 5 × 2 = 3150
By
definition, LCM of a given set of numbers is their smallest common multiple. In
other words, it is the smallest numbers which is divisible by all the given
numbers.
So,
to get the answer , we have to deduct 5
from the LCM.
3150 – 5 = 3145 Ans.
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