HCF & LCM

PrepBank · পাতা ৩ / ৩ · ২০১–২২৬ / ২২৭

২০১.

Which is the greatest three-digit number which when divided by 6, 9 and 12 leaves a remainder of 3 in each case?

ক
996
খ
975
গ
939
ঘ
972
ঙ
903

ব্যাখ্যা

Question: Which is the greatest three-digit number which when divided by 6, 9 and 12 leaves a remainder of 3 in each case?

Solution:
Greatest three digit number = 999
LCM of 6, 9 and 12 = 36

On dividing 999 by 36,
Remainder = 27.

∴ The greatest three digit number divisible by 6, 9, and 12 = (999 - 27) = 972

As per the question, the required number is (972 + 3) = 975.

২০২.

The H. C. F of (9/10), (12/20), (15/25), (27/50) is?

ক
7/5
খ
3/100
গ
18/100
ঘ
5/7

ব্যাখ্যা

Question: The H. C. F of (9/10), (12/20), (15/25), (27/50) is?

Solution:
Required H. C. F
= (H. C. F of 9, 12, 15, 27)/(L. C. M of 10, 20, 25, 50)
= 3/100

২০৩.

Two numbers have LCM = 180 and HCF = 6. If one number is 30, find the other.

ক
36
খ
54
গ
60
ঘ
72

ব্যাখ্যা

Question: Two numbers have LCM = 180 and HCF = 6. If one number is 30, find the other.

Solution:
Let the numbers be a=30 and b=?, with HCF = 6 and LCM = 180.
Use the formula connecting LCM and HCF:
a × b = HCF × LCM
30 × b = 6 × 180
b = 1080/30
b = 36

২০৪.

The maximum number of students among whom 1001 pens and 910 pencils can be distributed in such a way that each student gets the same number of pens and the same number of pencils is =?

ক
73
খ
97
গ
93
ঘ
91

ব্যাখ্যা

Question: The maximum number of students among whom 1001 pens and 910 pencils can be distributed in such a way that each student gets the same number of pens and the same number of pencils is =?

Solution:
Required number of students
= HCF of 1001 and 910
= 91
........................................................
........................................................
HCF of 1001 and 910:

We can find the HCF of 1001 and 910 using prime factorization.
First, let's find the prime factorization of both numbers:

1001 = 7 × 11 × 13
910 = 2 × 5 × 7 × 13

Now, to find the HCF, we take the product of the common prime factors with the lowest exponent:
The common prime factors are 7 and 13.

So, the product of the common prime factors is 7 × 13 = 91

Therefore, the HCF of 1001 and 910 is 91.

২০৫.

A factory manufactures products in batches of 16, 24, and 32 units. What is the minimum number of units the factory needs to produce so that each batch can be formed exactly?

ক
80
খ
64
গ
102
ঘ
96

ব্যাখ্যা

Question: A factory manufactures products in batches of 16, 24, and 32 units. What is the minimum number of units the factory needs to produce so that each batch can be formed exactly?

Solution:
To find the minimum number of units the factory needs to produce so that each batch size (16, 24, and 32) can be formed exactly, we need to find the least common multiple (LCM) of these batch sizes.

The prime factorization of each batch size is -
16 = 2 × 2 × 2 × 2 = 2⁴
24 = 2 × 2 × 2 × 3 = 2³ × 3
32 = 2 × 2 × 2 × 2 × 2 = 2⁵

Now,
So the highest power of 2 is 2⁵
The highest power of 3 is 3¹

So, the LCM of 16, 24, and 32 is = 2⁵ × 31 = 32 × 3 = 96.

So the minimum number of units the factory needs to produce is 96.

২০৬.

The traffic lights at three different road crossings change after every 40 sec, 72 sec and 108 sec respectively. If they all change simultaneously at 5 : 20 hours, then find the time at which they will change simultaneously.

ক
5 : 28 hrs
খ
5 : 30 hrs
গ
5 : 38 hrs
ঘ
5 : 40 hrs

ব্যাখ্যা

Question: The traffic lights at three different road crossings change after every 40 sec, 72 sec and 108 sec respectively. If they all change simultaneously at 5 : 20 hours, then find the time at which they will change simultaneously.

Solution:
Traffic lights at three different road crossings change after every 40 sec, 72 sec and 108 sec respectively.
Therefore, find the L.C.M. of 40, 72 and 108.
L.C.M. of 40, 72 and 108 = 1080
The traffic lights will change again after 1080 seconds = 18 min
The next simultaneous change takes place at 5 : 38 hrs.

২০৭.

What is the least number which, when tripled, becomes exactly divisible by 15, 20, 24, and 30?

ক
30
খ
36
গ
40
ঘ
45

ব্যাখ্যা

Question: What is the least number which, when tripled, becomes exactly divisible by 15, 20, 24, and 30?

Solution:
Let the required number be x.
Then, 3x must be divisible by 15, 20, 24, and 30.

Find the LCM of the numbers:
15 = 3 × 5
20 = 2² × 5
24 = 2³ × 3
30 = 2 × 3 × 5

∴ LCM = 2³ × 3 × 5 = 8 × 3 × 5 = 120

∴ 3x = 120
∴ x = 120/3 = 40

২০৮.

Three friends, Titu, Joyonto, and Rajib, start running around a circular track at the same time in the same direction. Titu completes a round in 120 seconds, Joyonto in 150 seconds, and Rajib in 180 seconds, all starting at the same point. After how much time will they meet again at the starting point?

ক
20 minutes
খ
25 minutes
গ
30 minutes
ঘ
40 minutes
ঙ
None

ব্যাখ্যা

Question: Three friends, Titu, Joyonto, and Rajib, start running around a circular track at the same time in the same direction. Titu completes a round in 120 seconds, Joyonto in 150 seconds, and Rajib in 180 seconds, all starting at the same point. After how much time will they meet again at the starting point?

Solution:
L.C.M. of 120, 150 and 180 = 1800
So, Titu, Joyonto, and Rajib will again meet at the starting point in 1800 sec
Now, 1800 sec = 1800/60 minutes
= 30 minutes

২০৯.

Four people are running around a circular ground from a point on the circumference at 8:00 am. For one round, these four persons take respectively 40, 50, 60 and 30 minutes. At what time will they meet together again ?

ক
6:20 PM
খ
6:00 PM
গ
8:00 PM
ঘ
7:30 PM

ব্যাখ্যা

Question: Four people are running around a circular ground from a point on the circumference at 8:00 am. For one round, these four persons take respectively 40, 50, 60 and 30 minutes. At what time will they meet together again ?

Solution:
L.C.M. of 40, 50, 60 and 30
= 600 minutes
= 10 hours
So, they meet again 10 hours after they start.
They meet together again = 8:00 am + 10 hours
= 6:00 pm

So they will meet together again at 6:00 PM.

২১০.

What is the H.C.F. of 4/9, 10/21 and 20/63?

ক
4/189
খ
6/63
গ
2/63
ঘ
20/21

ব্যাখ্যা

Question: What is the H.C.F. of 4/9, 10/21 and 20/63?

Solution:
H.C.F of 4/9, 10/21 and 20/63 = H.C.F of 4,10 and 20 / L.C.M of 9,21 and 63

H.C.F of 4, 10 and 20 = 2
& L.C.M. of 9, 21 and 63 = 63.

∴ Required H.C.F. = 2/63

২১১.

The LCM of two numbers is 2852 and their HCF is 4. If one of the number is 124, find the other number.

ক
92
খ
96
গ
104
ঘ
84

ব্যাখ্যা

Question: The LCM of two numbers is 2852 and their HCF is 4. If one of the number is 124, find the other number.

Solution:
1st number × 2nd number = L.C. M. × H.C.F

We have,
First number × second number = LCM × HCF
∴ Second number = (2852 × 4)/124
= 92

২১২.

The ratio of two numbers is 4 : 5 and their HCF is 4. Their LCM is -

ক
120
খ
40
গ
80
ঘ
240

ব্যাখ্যা

Question: The ratio of two numbers is 4 : 5 and their HCF is 4. Their LCM is -

Solution:
let,
the numbers are 4x and 5x
their HCF is = x
∴ x = 4

so, the numbers ar 16, 20.

the LCM is = 80

২১৩.

If x is an odd negative integer and y is an even integer, which of the following statements must be true?

ক
(3x - 2y) is odd
খ
xy² is an even negative integer
গ
(y² - x) is an odd negative integer
ঘ
All of the above
ঙ
None of the above

ব্যাখ্যা

Question: If x is an odd negative integer and y is an even integer, which of the following statements must be true?

Solution:
Let x = - 1, y = 2
Option A: (3x - 2y) = 3(- 1) - 2(2) = -7 is ODD
Option B: xy² = (-1)(2²) = - 4 is EVEN NEGATIVE
Option C. (y² - x) = 2² - (-1) = 5 is ODD POSITIVE

Since the question involves Even and Odd numbers, let us also consider y = 0.
Option A: (3x - 2y) = 3(-1) - 2(0) = -3 is ODD
Option B: xy² = (-1)(0²) = 0 is EVEN POSITIVE
Option C. This condition was proved false using the above values.

Hence only option A is satisfied.

২১৪.

The ratio of two numbers is 3 : 4 and their H.C.F is 6. Find their L.C.M.

ক
32
খ
72
গ
36
ঘ
42

ব্যাখ্যা

Question: The ratio of two numbers is 3 : 4 and their H.C.F is 6. Find their L.C.M.

Solution:
Let the two numbers be 3x and 4x.
∴ H.C.F = x = 6

∴ The two numbers are, 3 × 6 = 18 and 4 × 6 = 24

∴ Product of the two numbers = 18 × 24 = 432
And H.C.F = 6

We know,
L.C.M = (Product of two numbers)/H.C.F
= 432/6
= 72

∴ The L.C.M of the two numbers = 72.

২১৫.

Find the greatest number, which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively.

ক
127
খ
132
গ
114
ঘ
108

ব্যাখ্যা

Question: Find the greatest number, which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively.

Solution:
The number on dividing 1657 and 2037 leaves remainders 6 and 5 respectively.
Hence, make the dividend completely divisible by the divisor. This is possible, if we subtract remainder from the dividend.
Therefore,
1657 - 6 = 1651
2037 - 5 = 2032

H.C.F. of 1651 and 2032 is 127. 127 is the common factor.
127 × 13 = 1651
Thus by adding 6, we get 1651 + 6 = 1657
127 is the correct answer.

২১৬.

Two numbers are in the ratio 3 : 4. If their LCM is 240, the smaller of two number is

ক
40
খ
50
গ
60
ঘ
70
ঙ
None

ব্যাখ্যা

Question: Two numbers are in the ratio 3 : 4. If their LCM is 240, the smaller of two number is

Solution:
Let, these two numbers be 3x and 4x then their LCM = 12x
Now, according to question,
12x = 240
Or, x = 20

Thus, the numbers are (3x = 3 × 20) = 60 and (4x = 4 × 20) = 80
Then smaller in this two is 60

২১৭.

What is the least multiple of 11 which leaves a remainder of 5 when divided by 6, 8, and 10?

ক
605
খ
585
গ
550
ঘ
525

ব্যাখ্যা

Question: What is the least multiple of 11 which leaves a remainder of 5 when divided by 6, 8, and 10?

Solution:
L.C.M. of 6, 8, and 10 is 120.
Let required number be 120k + 5 which is multiple of 11.

Least value of k for which (120k + 5) is divisible by 11 is k = 5 .
Required number = (120 × 5) + 5 = 605

২১৮.

What is the least number which when tripled is exactly divisible by 10, 12, 15, and 18?

ক
80
খ
180
গ
40
ঘ
60

ব্যাখ্যা

Question: What is the least number which when tripled is exactly divisible by 10, 12, 15, and 18?

Solution:
Let the number be x.
tripled the number is 3x.

10 = 2 × 5
12 = 2 × 2 × 3
15 = 3 × 5
18 = 2 × 3 × 3

∴ LCM = 2 × 2 × 3 × 3 × 5
= 180

∴ x = 180/3 = 60

২১৯.

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?

ক
4
খ
10
গ
15
ঘ
16

ব্যাখ্যা

Question: 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?

Solution:
L.C.M. of 2, 4, 6, 8, 10, 12 is 120.
So, the bells will toll together after every 120 seconds = 2 minutes

∴ In 30 minutes, they will toll together 30/2 + 1 = 15 + 1 = 16 times.

২২০.

Find the least number which will leaves remainder 5 when divided by 8, 12, 16 and 20.

ক
245
খ
240
গ
235
ঘ
250

ব্যাখ্যা

Question: Find the least number which will leaves remainder 5 when divided by 8, 12, 16 and 20.

Solution:
We have to find the Least number, therefore we find out the LCM of 8, 12, 16 and 20.
8 = 2 × 2 × 2;
12 = 2 × 2 × 3;
16 = 2 × 2 × 2 × 2;
20 = 2 × 2 × 5

∴ LCM = 2 × 2 × 2 × 2 × 3 × 5 = 240

This is the least number which is exactly divisible by 8, 12, 16 and 20.
So, Required number which leaves remainder 5 is = 240 + 5 = 245.

২২১.

Four metal rods of lengths 78 cm, 104 cm, 117 cm and 169 cm are to be cut into parts of equal length. Each part must be as long as possible. What is the maximum number of pieces that can be cut?

ক
48
খ
36
গ
34
ঘ
42
ঙ
38

ব্যাখ্যা

Question: Four metal rods of lengths 78 cm, 104 cm, 117 cm and 169 cm are to be cut into parts of equal length. Each part must be as long as possible. What is the maximum number of pieces that can be cut?

Solution:
Four metal rods of lengths 78 cm, 104 cm, 117 cm and 169 cm
78 = 2 × 3 × 13
104 = 2 × 2 × 2 × 13
117 = 3 × 3 × 13
169 = 13 × 13

HCF of 78, 104, 117 and 169 = 13

Maximum length of each part = HCF of 78 cm, 104 cm, 117 cm, 169 cm = 13 cm

The maximum number of pieces,
78/13 = 6
104/13 = 8
117/13 = 9
169/13 = 13

The maximum number of pieces = 6 + 8 + 9 + 13 = 36

∴ The maximum number of pieces is 36.

২২২.

Find the largest number of 4-digits divisible by 12, 15 and 18.

ক
9900
খ
9750
গ
9450
ঘ
9000

ব্যাখ্যা

Question: Find the largest number of 4-digits divisible by 12, 15 and 18.

Solution:
Required largest number must be divisible by the L.C.M. of 12, 15 and 18
L.C.M. of 12, 15 and 18
12 = 2 × 2 × 3
15 =5 × 3
18 = 2 × 3 × 3
L.C.M. = 180
Now divide 9999 by 180, we get remainder as 99
The required largest number = (9999 - 99) = 9900
Number 9900 is exactly divisible by 180.

২২৩.

What is the smallest number when added by 5 the summation will be divided by 12, 16, 24 and 32?

ক
101
খ
96
গ
91
ঘ
89

ব্যাখ্যা

Question: What is the smallest number when added by 5 the summation will be divided by 12, 16, 24 and 32?

Solution:
The required number will be 5 less than L.C.M. of 12, 16, 24 and 32.

12 = 2 × 2 × 3
16 = 2 × 2 × 2 × 2
24 = 2 × 2 × 2 × 3
32 = 2 × 2 × 2 × 2 × 2

So the L.C.M of 12, 16, 24 and 32 = 96

Required number= 96 - 5 = 91

২২৪.

Find the H.C.F. of p(x) = 2x³ - 3x² - 2x + 3 and q(x) = 3x² + 8x + 5.

ক
(2x + 1)
খ
(x + 1)
গ
(2x - 3)
ঘ
(x - 1)

ব্যাখ্যা

Question: Find the H.C.F. of p(x) = 2x³ - 3x² - 2x + 3 and q(x) = 3x² + 8x + 5.

Solution:
Given that,
p(x) = 2x³ - 3x² - 2x + 3 and q(x) = 3x² + 8x + 5

Now,
The factors of p(x) = 2x³ - 3x² - 2x + 3
⇒ x²(2x - 3) - 1(2x - 3)
⇒ (x² - 1)(2x - 3)
⇒ (x + 1)(x - 1)(2x - 3)

And,
The factors of q(x) = 3x² + 8x + 5
⇒ 3x² + 5x + 3x + 5
⇒ x(3x + 5) + 1(3x + 5)
⇒ (3x + 5)(x + 1)

∴ The required H.C.F. is (x + 1).

২২৫.

The least number, which when divided by 12, 15, 20 and 54 leaves in each case a remainder of 8 is-

ক
536
খ
548
গ
480
ঘ
544

ব্যাখ্যা

Question: The least number, which when divided by 12, 15, 20 and 54 leaves in each case a remainder of 8 is-

Solution:
The number leaves a remainder 8 when divided by 12, 15, 20 and 54.
So the required number = LCM(12, 15, 20, 54) + 8

Now,
12 = 2 × 2 × 3
15 = 3 × 5
20 = 2 × 2 × 5
54 = 2 × 3 × 3 × 3

∴ LCM(12, 15, 20, 54) = 540

∴ Required Number = 540 + 8 = 548

২২৬.

H.C.F. of two numbers is 13. If these two numbers are in the ratio of 15: 11, then find the numbers.

ক
230, 140
খ
215, 130
গ
195, 143
ঘ
155, 115

ব্যাখ্যা

Question: H.C.F. of two numbers is 13. If these two numbers are in the ratio of 15: 11, then find the numbers.

Solution:
H.C.F. of two numbers = 13
The numbers are in the ratio of 15 : 11

Let the two numbers be 15y and 11y
H.C.F. is the product of common factors
Therefore, H.C.F. is y.
So y = 13

The two numbers are:
15y = 15 × 13 = 195
11y = 11 × 13 = 143

We can cross-check the answer using the trick. (Product of two numbers = Product of their H.C.F. and L.C.M.)
Product of H.C.F. and L.C.M. = 13 × 2145 = 27885
Product of two numbers = 195 × 143 = 27885
Hence, the calculated answer is correct.