###

For complete understanding of concept of this chapter,
go through **HCF and LCM Concept**
For more solved problems and examples, go
through HCF
and LCM Solved Problems

**Sample questions for Placement Papers - Latest Pattern Question 1**

Which is the largest integer that divides
all three numbers 23400,272304,205248 without leaving a remainder?

a. 48
b.
24
c.
96
d. 72
**Answer: b**
**Explanation:** To get the answer to this question, we have to find the
HCF of 23400, 272304 and 205248.
The HCF is 24

**Sample questions for
Placement Papers – Latest Pattern Question 2**
The H.C.F of two numbers is 12 and their
L.C.M is 360. If one of the numbers is 36. Find the other number?
a)100
b)120
c)160
d)180
**Answer: b**
**Explanation:
**
Given that, HCF =12 and LCM=360 and
one number a=36 another number be b
We have the product of two number s =
product of their HCF and LCM
a x b = HCF x LCM
36 x b = 12 x 360
b =120
Answer is 120.
**Sample questions for
Placement Papers – Latest Pattern Question 3**
A school has
120, 192 and 144 students enrolled for its science, arts and commerce courses.
All students have to be seated in rooms for an exam such that each room has
students of only the same course and also all rooms have an equal number of
students. What is the least number of rooms needed?
a)10
b)13
c)15
d)19
**Answer: b**
**Explanation:
**
To get least
number of students, we have to put the maximum number of students in the room.
So we have to
find the HCF of given students of each course, to know the number of students
in each room
Number of
students in each room = HCF OF 120 , 144 and 192
=24.

**Sample questions for Placement Papers – Latest Pattern Question 4**

Two cyclists begin training on an oval
racecourse at the same time. The professional cyclist completes each lap in 4
minutes; the novice takes 6 minutes to complete each lap. How many minutes
after the start will both cyclists pass at exactly the same spot where they
began to cycle?

a)10
b)8
c)14
d)12
**
****Answer: D**
**Explanation:
**
Professional
Cyclist completes each lap in 4 minutes
Novice Cyclist completes each lap in 6 minutes
Professional Cyclist completes laps => 4
min, 8 min, 12 min, and 16 min and so on
Novice cyclist completes laps => 6 min,
12 min, 18 min, 24 min and so on
So they both meet at the starting point at
the 12^{th} minute again.
In these kinds
of problems, we have to find the LCM of 4 and 6 = 12 min.

**Sample questions for Placement Papers 2015 - Latest Pattern Question 5**

Six bells commence tolling together and
toll at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. In 30 minutes,
how many times do they toll together?

a) 4
b) 10
c) 15
d) 16
**Answer: d**
**Explanation:
**
LCM OF 2,4,6,8,10 and 12 = 120 sec
Convert into minutes => 120 sec = 2
minutes
For every 2 minutes , all the six bells
toll together.
In 30 minutes ,
30
2
= 15 times + 1(
here 1 is added for the last toll)
=16 times

**Sample questions for Placement Papers 2014 - Latest
Pattern Question 6**
2 gears, one with 12 teeth and the other one with 14 teeth are
engaged with each other. One tooth in smaller gear and one tooth in a bigger gear
are marked and initially, those 2 marked teeth are in contact with each other.
After how many rotations of the smaller gear with the marked teeth in the other
gear will again come into contact for the first time?
a.
7
b. 12
c. Data
Insufficient
d. 84
**Answer: a**
**Explanation:
**
Let us assume the distance between them is 1 cm.
Then the circumference of first gear is 12
cm and the second is 14 cm.
To know, after how many rotations will they
come in contact, we must know after how much distance they will meet.
To know the distance, we have to find the
LCM of 12,14 =84cm
So to cover 84 cm, the first gear has to
rotate $ \frac{84}{12} $= 7 rounds
(the second
gear rotates $ \frac{84}{14} $ = 6 rounds as it is bigger)

**Sample questions for Placement Papers
Latter Pattern Question 7**
If the LCM of 2
natural numbers A and B is 1260, what is the minimum possible value of
A+B?
a) 65
b)71
c)83
d)94
**Answer:
b**
**Explanation: **
Solution:
Writing 1260 as product of prime factors = 2 x 2 x 3 x 3 x 7 x 5
The possible pairs whose product is 1260
are (630,2)(420,3)(315,4)(252,5)(210,6)
(180,7) (140,9)
(126,10)(105,12)(90,14)(84,15)(70,18)(63,20)(60,21)(45,28)(42,30)(36,35)’
Among these pairs
the minimum possible value of A+B= 35 +36 =71

Alternative
Solution
The given LCM
1260 can be expressed as

2 x 2 x 3 x 3 x 7 x 5= 28 x 46 or 42
x30 or 35 x 36 (These are in the form of A*B)

So, the minimum value of A+B is 35 +36 =71

**Sample questions for
Placement Papers Latest Pattern Question 8**
Arun wanted to
find the largest number of 4 digits such that when added to 7249 generated a number that gave a remainder 0 when divided by 54, 12, 14, 21, 33. Find the
number.
a)9624
b)9283 c)9927
d)9428
**Answer:
b**
**Explanation: **
To get a number
which leaves remainder 0, when divided by 12,14,21,33 and 54
= LCM of
12,14,21,33 and 54 =8316
Multiples of LCM
8316 also give the remainder zero.
Now taking 8316
,
We have to find
the least number of 4 digits, when added to 7249, gives remainder 0.
Taking
8316 =>
8316-7249=1047
Taking
8316 x 2, => 16632-7249=9383
Taking
8316 x 3 =>
24948-7249=17699( 17699 is a 5 digit number, so it cannot be answered)
So we have two 4
digit numbers 1047 and 9383. Of these largest is 9383.
So the answer is
9383

**Sample questions for
Placement Papers Latest Pattern Question 9**

What is the quotient when least common multiple of the
first, 86 natural numbers divided by the least common multiple of the first 83
natural numbers?
a)1
b)3
c)7
d)9
**Answer:
A**
**Explanation:** LCM of given numbers is always the
product of all the prime factors with their highest powers.

LCM of 1^{st} 96 natural numbers = 2^{6} x 3^{4} x 5^{2} x 7^{2} x 11
x 13 x 17
x 19
x 23 x 29 x 31
x 37 x 41 x 43
x 47
x 53
x 59
x 61 .
Therefore LCM of the first 83 natural numbers and LCM of the first 83 natural number is the same
(Because they
both have same prime factors)

so $ \frac{LCM\,of\,first\,86\,natural\,numbers}{LCM\,of\,first\,83\,natural\,numbers}\,
$ =1

Therefore the quotient=1

**Sample questions for
Placement Papers Latest Pattern Question 10**

Find the greatest possible length which can be
used to measure exactly the lengths 4 meters 95 cm, 9 meters and 16 meters 65 cm.

a) 45 cm b) 57 cm c) 63 cm d) 99 cm
**Answer: A**
**Explanation: **The greatest length which
can measure the lengths 4 m 95 cm, 9 m
and 16 m 65 cm is HCF of all the given lengths.
Converting
into centimeters, the lengths are 495
cm , 900 cm, 1665 cm .
Writing the given numbers as the product
of prime factors
495 = 3^{2} x 5^{1} x 11^{1}
900 = 2^{2} x 3^{2} x 5^{2}
1665 = 3^{2} x 5^{1} x 37^{1}
HCF= 3^{2} x 5^{1} = 45 CM

For complete understanding of concept of this chapter, go through

**HCF and LCM Concept**

**Sample questions for Placement Papers - Latest Pattern Question 1**

Which is the largest integer that divides
all three numbers 23400,272304,205248 without leaving a remainder?

a. 48
b.
24
c.
96
d. 72

**Answer: b**

**Explanation:**To get the answer to this question, we have to find the HCF of 23400, 272304 and 205248.

The HCF is 24

**Sample questions for Placement Papers – Latest Pattern Question 2**

The H.C.F of two numbers is 12 and their
L.C.M is 360. If one of the numbers is 36. Find the other number?

a)100
b)120
c)160
d)180

**Answer: b**

**Explanation:**

Given that, HCF =12 and LCM=360 and
one number a=36 another number be b

We have the product of two number s =
product of their HCF and LCM

a x b = HCF x LCM

36 x b = 12 x 360

b =120

Answer is 120.

**Sample questions for Placement Papers – Latest Pattern Question 3**

A school has
120, 192 and 144 students enrolled for its science, arts and commerce courses.
All students have to be seated in rooms for an exam such that each room has
students of only the same course and also all rooms have an equal number of
students. What is the least number of rooms needed?

a)10
b)13
c)15
d)19

**Answer: b**

**Explanation:**

To get least
number of students, we have to put the maximum number of students in the room.

So we have to
find the HCF of given students of each course, to know the number of students
in each room

Number of
students in each room = HCF OF 120 , 144 and 192

=24.

**Sample questions for Placement Papers – Latest Pattern Question 4**

Two cyclists begin training on an oval
racecourse at the same time. The professional cyclist completes each lap in 4
minutes; the novice takes 6 minutes to complete each lap. How many minutes
after the start will both cyclists pass at exactly the same spot where they
began to cycle?

a)10
b)8
c)14
d)12

**
**

**Answer: D**

**Explanation:**

Professional
Cyclist completes each lap in 4 minutes

Novice Cyclist completes each lap in 6 minutes

Professional Cyclist completes laps => 4
min, 8 min, 12 min, and 16 min and so on

Novice cyclist completes laps => 6 min,
12 min, 18 min, 24 min and so on

So they both meet at the starting point at
the 12

^{th}minute again.
In these kinds
of problems, we have to find the LCM of 4 and 6 = 12 min.

**Sample questions for Placement Papers 2015 - Latest Pattern Question 5**

Six bells commence tolling together and
toll at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. In 30 minutes,
how many times do they toll together?

a) 4
b) 10
c) 15
d) 16

**Answer: d**

**Explanation:**

LCM OF 2,4,6,8,10 and 12 = 120 sec

Convert into minutes => 120 sec = 2
minutes

For every 2 minutes , all the six bells
toll together.

In 30 minutes ,

30
2

= 15 times + 1(
here 1 is added for the last toll)

=16 times

**Sample questions for Placement Papers 2014 - Latest Pattern Question 6**

2 gears, one with 12 teeth and the other one with 14 teeth are
engaged with each other. One tooth in smaller gear and one tooth in a bigger gear
are marked and initially, those 2 marked teeth are in contact with each other.
After how many rotations of the smaller gear with the marked teeth in the other
gear will again come into contact for the first time?

a.
7
b. 12
c. Data
Insufficient
d. 84

**Answer: a**

**Explanation:**

Let us assume the distance between them is 1 cm.

Then the circumference of first gear is 12
cm and the second is 14 cm.

To know, after how many rotations will they
come in contact, we must know after how much distance they will meet.

To know the distance, we have to find the
LCM of 12,14 =84cm

So to cover 84 cm, the first gear has to
rotate $ \frac{84}{12} $= 7 rounds

(the second
gear rotates $ \frac{84}{14} $ = 6 rounds as it is bigger)

**Sample questions for Placement Papers Latter Pattern Question 7**

If the LCM of 2
natural numbers A and B is 1260, what is the minimum possible value of
A+B?

a) 65
b)71
c)83
d)94

**Answer: b**

**Explanation:**

Solution:
Writing 1260 as product of prime factors = 2 x 2 x 3 x 3 x 7 x 5

The possible pairs whose product is 1260
are (630,2)(420,3)(315,4)(252,5)(210,6)
(180,7) (140,9)
(126,10)(105,12)(90,14)(84,15)(70,18)(63,20)(60,21)(45,28)(42,30)(36,35)’

Among these pairs
the minimum possible value of A+B= 35 +36 =71

Alternative
Solution

The given LCM
1260 can be expressed as

2 x 2 x 3 x 3 x 7 x 5= 28 x 46 or 42
x30 or 35 x 36 (These are in the form of A*B)

So, the minimum value of A+B is 35 +36 =71

**Sample questions for Placement Papers Latest Pattern Question 8**

Arun wanted to
find the largest number of 4 digits such that when added to 7249 generated a number that gave a remainder 0 when divided by 54, 12, 14, 21, 33. Find the
number.

a)9624
b)9283 c)9927
d)9428

**Answer: b**

**Explanation:**

To get a number
which leaves remainder 0, when divided by 12,14,21,33 and 54

= LCM of
12,14,21,33 and 54 =8316

Multiples of LCM
8316 also give the remainder zero.

Now taking 8316
,

We have to find
the least number of 4 digits, when added to 7249, gives remainder 0.

Taking
8316 =>
8316-7249=1047

Taking
8316 x 2, => 16632-7249=9383

Taking
8316 x 3 =>
24948-7249=17699( 17699 is a 5 digit number, so it cannot be answered)

So we have two 4
digit numbers 1047 and 9383. Of these largest is 9383.

So the answer is
9383

**Sample questions for Placement Papers Latest Pattern Question 9**

What is the quotient when least common multiple of the
first, 86 natural numbers divided by the least common multiple of the first 83
natural numbers?

a)1
b)3
c)7
d)9

**Answer: A**

**Explanation:**LCM of given numbers is always the product of all the prime factors with their highest powers.

LCM of 1 3 5 7 11
x 13 x 17
x 19
x 23 x 29 x 31
x 37 x 41 x 43
x 47
x 53
x 59
x 61 .

^{st}96 natural numbers = 2^{6}x^{4}x^{2}x^{2}x
Therefore LCM of the first 83 natural numbers and LCM of the first 83 natural number is the same

(Because they
both have same prime factors)

so $ \frac{LCM\,of\,first\,86\,natural\,numbers}{LCM\,of\,first\,83\,natural\,numbers}\,
$ =1

Therefore the quotient=1

**Sample questions for Placement Papers Latest Pattern Question 10**

Find the greatest possible length which can be
used to measure exactly the lengths 4 meters 95 cm, 9 meters and 16 meters 65 cm.

a) 45 cm b) 57 cm c) 63 cm d) 99 cm

**Answer: A**

**Explanation:**The greatest length which can measure the lengths 4 m 95 cm, 9 m and 16 m 65 cm is HCF of all the given lengths.

Converting
into centimeters, the lengths are 495
cm , 900 cm, 1665 cm .

Writing the given numbers as the product
of prime factors

495 = 3

^{2}x 5^{1}x 11^{1}
900 = 2

^{2}x 3^{2}x 5^{2}
1665 = 3

^{2}x 5^{1}x 37^{1}
HCF= 3

^{2}x 5^{1}= 45 CM