Problems On H.C.F And L.C.M

Answer: Option A
Solution:
To find the greatest number that will divide 43, 91, and 183 while leaving the same remainder in each case, follow these steps:

Step 1: Find the differences between the numbers
Since the remainder is the same in each case, subtracting one number from another will cancel out the remainder. Thus, we compute:

\[
91 – 43 = 48
\]

\[
183 – 91 = 92
\]

\[
183 – 43 = 140
\]

Step 2: Find the greatest common divisor (GCD) of the differences
We need to find the GCD of 48, 92, and 140.

1. Prime factorization:
– \( 48 = 2^4 \times 3 \)
– \( 92 = 2^2 \times 23 \)
– \( 140 = 2^2 \times 5 \times 7 \)

2. Identify common factors:
– The only common factor in all three numbers is \( 2^2 = 4 \).

Step 3: Conclusion
The greatest number that will divide 43, 91, and 183 while leaving the same remainder in each case is 4.

Answer: Option C
Solution:
We are given:

– The HCF (GCD) of two numbers = 23
– The other two factors of their LCM = 13 and 14

Step 1: Express the Relationship between HCF and LCM
We use the fundamental relation:

\[
\text{Product of two numbers} = \text{HCF} \times \text{LCM}
\]

Since the LCM contains the HCF as a factor, we express the LCM as:

\[
\text{LCM} = 23 \times 13 \times 14
\]

Step 2: Find the Two Numbers
Let the two numbers be 23a and 23b, where a and b are co-prime (i.e., they have no common factors other than 1).

From the given data:

\[
\text{LCM} = 23 \times 13 \times 14
\]

Since LCM of 23a and 23b must be 23 × 13 × 14, the values of a and b must be 13 and 14.

Thus, the two numbers are:

\[
23 \times 13 = 299
\] \[
23 \times 14 = 322
\]

Step 3: Identify the Larger Number
The larger of the two numbers is 322.

Final Answer: 322.

Answer: Option D
Solution:
To determine how many times the six bells toll together in 30 minutes, follow these steps:

Step 1: Find the Least Common Multiple (LCM) of the Given Intervals
The bells toll at intervals of 2, 4, 6, 8, 10, and 12 seconds. The first time they toll together is at time = 0 seconds. After that, they toll together every LCM of (2, 4, 6, 8, 10, 12) seconds.

1. Prime factorization:
– \(2 = 2^1\)
– \(4 = 2^2\)
– \(6 = 2^1 \times 3\)
– \(8 = 2^3\)
– \(10 = 2^1 \times 5\)
– \(12 = 2^2 \times 3\)

2. Taking the highest powers of all prime factors:
– Highest power of 2: \(2^3 = 8\)
– Highest power of 3: \(3^1 = 3\)
– Highest power of 5: \(5^1 = 5\)

\[
\text{LCM} = 8 \times 3 \times 5 = 120 \text{ seconds} = 2 \text{ minutes}
\]

Step 2: Find How Many Times They Toll Together in 30 Minutes
– The first tolling happens at t = 0.
– After that, they toll together every 120 seconds (2 minutes).
– In 30 minutes, the number of times they toll together:

\[
\frac{30}{2} + 1 = 15 + 1 = 16
\]

Final Answer:
16 times.

Answer: Option A
Solution:
N = H.C.F. of (4665 – 1305), (6905 – 4665) and (6905 – 1305)
= H.C.F. of 3360, 2240 and 5600 = 1120.
Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4

Answer: Option C
Solution:
To find the greatest four-digit number that is divisible by 15, 25, 40, and 75, follow these steps:

Step 1: Find the LCM of 15, 25, 40, and 75
We first determine the least common multiple (LCM) of the given numbers.

Prime factorizations:
– \( 15 = 3 \times 5 \)
– \( 25 = 5^2 \)
– \( 40 = 2^3 \times 5 \)
– \( 75 = 3 \times 5^2 \)

Taking the highest powers of all prime factors:
– Highest power of 2: \( 2^3 = 8 \)
– Highest power of 3: \( 3^1 = 3 \)
– Highest power of 5: \( 5^2 = 25 \)

\[
\text{LCM} = 8 \times 3 \times 25 = 600
\]

Step 2: Find the Greatest Four-Digit Number Divisible by 600
The greatest four-digit number is 9999. We need to find the largest multiple of 600 that is ≤ 9999.

\[
\frac{9999}{600} \approx 16.665
\]

Taking the integer part:
\[
16 \times 600 = 9600
\]

Final Answer:
The greatest four-digit number divisible by 15, 25, 40, and 75 is 9600.
\[
\boxed{9600}
\]

Answer: Option C
Solution:
Let the numbers be 37a and 37b
Then, 37a x 37b = 4107
⇒ ab = 3
Now, co-primes with product 3 are (1, 3)
So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111)
∴ Greater number = 111

Answer: Option A
Solution:
We are given:

– Three numbers in the ratio 3:4:5.
– Their LCM = 2400.
– We need to find their HCF.

Step 1: Express the Numbers in Terms of a Common Factor
Let the numbers be:

\[
3x, 4x, 5x
\]

where x is the common factor.

Step 2: Find the LCM in Terms of \( x \)
The LCM of \( 3x, 4x, 5x \) is:

\[
\text{LCM} = \text{LCM}(3x, 4x, 5x) = \text{LCM}(3,4,5) \times x
\]

Since 3, 4, and 5 are co-prime, their LCM is:

\[
\text{LCM}(3,4,5) = 3 \times 4 \times 5 = 60
\]

Thus,

\[
\text{LCM} = 60x
\]

We are given:

\[
60x = 2400
\]

Solving for \( x \):

\[
x = \frac{2400}{60} = 40
\]

Step 3: Find the HCF
Since the numbers are 3x, 4x, 5x, their HCF is simply \( x \), which is:

\[
\mathbf{40}
\] Final Answer:
\[
\boxed{40}
\]

Answer: Option C
Solution:
To find the GCD (Greatest Common Divisor) of 1.08, 0.36, and 0.9, follow these steps:

Step 1: Convert to Whole Numbers
Since all numbers have decimals, multiply each by 100 to convert them to whole numbers:

\[
1.08 \times 100 = 108
\]

\[
0.36 \times 100 = 36
\]

\[
0.90 \times 100 = 90
\]

Now, we find the **GCD of 108, 36, and 90.

Step 2: Find the GCD of 108, 36, and 90
Prime factorizations:
– \( 108 = 2^2 \times 3^3 \)
– \( 36 = 2^2 \times 3^2 \)
– \( 90 = 2^1 \times 3^2 \times 5 \)

Identify the common factors:
– Smallest power of 2: \( 2^1 = 2 \)
– Smallest power of 3: \( 3^2 = 9 \)

Thus, the GCD is:

\[
2 \times 9 = 18
\]

Step 3: Convert Back to Decimal Form
Since we originally multiplied by **100**, divide the result by 100:

\[
\frac{18}{100} = 0.18
\] Final Answer:
\[
\boxed{0.18}
\]

Answer: Option B
Solution:
Let the numbers 13a and 13b
Then, 13a x 13b = 2028
⇒ ab = 12
Now, the co-primes with product 12 are (1, 12) and (3, 4)
[Note: Two integers a and b are said to be co-prime or relatively prime if they have no common positive factor other than 1 or, equivalently,
if their greatest common divisor is 1 ] So, the required numbers are (13 x 1, 13 x 12) and (13 x 3, 13 x 4)
Clearly, there are 2 such pairs.

Answer: Option D
Solution:
To find the least multiple of 7 that leaves a remainder of 4 when divided by 6, 9, 15, and 18, follow these steps:

Step 1: Find the LCM of 6, 9, 15, and 18
First, determine the Least Common Multiple (LCM) of the given divisors.

Prime factorizations:
– \( 6 = 2 \times 3 \)
– \( 9 = 3^2 \)
– \( 15 = 3 \times 5 \)
– \( 18 = 2 \times 3^2 \)

Take the highest powers of each prime factor:
– Highest power of 2: \( 2^1 = 2 \)
– Highest power of 3: \( 3^2 = 9 \)
– Highest power of 5: \( 5^1 = 5 \)

\[
\text{LCM} = 2 \times 9 \times 5 = 90
\]

Step 2: Find the Smallest Number of the Form \( 90k + 4 \) That is a Multiple of 7
We need to find the smallest integer \( k \) such that:

\[
90k + 4 \equiv 0 \pmod{7}
\]

or equivalently,

\[
90k \equiv -4 \pmod{7}
\]

Since \( 90 \equiv 6 \pmod{7} \), we substitute:

\[
6k \equiv -4 \pmod{7}
\]

Step 3: Solve for \( k \)
Rewriting \(-4\) in modulo 7:

\[
-4 \equiv 3 \pmod{7}
\]

Thus, we need to solve:

\[
6k \equiv 3 \pmod{7}
\]

Finding the modular inverse of 6 modulo 7:

\[
6^{-1} \equiv 6 \pmod{7} \quad \text{(since \(6 \times 6 = 36 \equiv 1 \pmod{7}\))}
\]

Multiplying both sides by 6:

\[
k \equiv 3 \times 6 \pmod{7}
\]

\[
k \equiv 18 \equiv 4 \pmod{7}
\]

Thus, the smallest positive \( k = 4 \).

Step 4: Compute the Number
\[
90k + 4 = 90(4) + 4 = 360 + 4 = 364
\] Final Answer:
\[
\boxed{364}
\]

Answer: Option C
Solution:
To find the lowest common multiple (LCM) of 24, 36, and 40, we first find the prime factorization of each number:

– \( 24 = 2^3 \times 3 \)
– \( 36 = 2^2 \times 3^2 \)
– \( 40 = 2^3 \times 5 \)

To find the LCM, take the highest power of each prime that appears:

– The highest power of 2 is \( 2^3 \)
– The highest power of 3 is \( 3^2 \)
– The highest power of 5 is \( 5 \)

Now, multiply these together:

\[
LCM = 2^3 \times 3^2 \times 5 = 8 \times 9 \times 5 = 360
\]

So, the lowest common multiple of 24, 36, and 40 is 360.

Answer: Option C
Solution:
To find the least number that should be added to 2497 so that the sum is divisible by 5, 6, 4, and 3, we need to determine the least common multiple (LCM) of 5, 6, 4, and 3 first.

– Prime factorizations:
– \( 5 = 5 \)
– \( 6 = 2 \times 3 \)
– \( 4 = 2^2 \)
– \( 3 = 3 \)

The LCM is the highest power of each prime factor:

– The highest power of 2 is \( 2^2 \)
– The highest power of 3 is \( 3 \)
– The highest power of 5 is \( 5 \)

So, the LCM of 5, 6, 4, and 3 is:

\[
LCM = 2^2 \times 3 \times 5 = 4 \times 3 \times 5 = 60
\]

Now, we need to find the remainder when 2497 is divided by 60:

\[
2497 \div 60 = 41 \text{ remainder } 37
\]

To make the sum divisible by 60, we need to add the difference between 60 and the remainder:

\[
60 – 37 = 23
\]

Therefore, the least number that should be added to 2497 is 23.

Answer: Option C
Solution
To reduce the fraction \( \frac{128352}{238368} \) to its lowest terms, we need to find the greatest common divisor (GCD) of the numerator (128352) and the denominator (238368), and then divide both by the GCD.

Let’s find the GCD of 128352 and 238368.

The greatest common divisor (GCD) of 128352 and 238368 is 18336.

Now, when we divide both the numerator and denominator by 18336, the fraction \( \frac{128352}{238368} \) reduces to:

\[
\frac{7}{13}
\] So, \( \frac{128352}{238368} \) in its lowest terms is \( \frac{7}{13} \).

Answer: Option B
Solution:
To solve this problem, we need to find the smallest number \( x \) that satisfies the following conditions:

1. When \( x \) is divided by 5, 6, 7, and 8, it leaves a remainder of 3.
2. When \( x \) is divided by 9, it leaves no remainder.

Step 1: Translate the conditions into mathematical equations.

– The first condition can be written as:
\[
x \equiv 3 \pmod{5}, \quad x \equiv 3 \pmod{6}, \quad x \equiv 3 \pmod{7}, \quad x \equiv 3 \pmod{8}
\] This means that \( x – 3 \) is divisible by 5, 6, 7, and 8. So, \( x – 3 \) is a multiple of the least common multiple (LCM) of 5, 6, 7, and 8.

Step 2: Find the LCM of 5, 6, 7, and 8.

The prime factorizations are:
– \( 5 = 5 \)
– \( 6 = 2 \times 3 \)
– \( 7 = 7 \)
– \( 8 = 2^3 \)

The LCM is the highest power of each prime:
\[
LCM = 2^3 \times 3 \times 5 \times 7 = 8 \times 3 \times 5 \times 7 = 840
\] So, \( x – 3 \) must be a multiple of 840. Therefore, we can express \( x \) as:
\[
x = 840k + 3
\] where \( k \) is some integer.

Step 3: Use the second condition.

The second condition states that \( x \) must be divisible by 9:
\[
x \equiv 0 \pmod{9}
\] Substitute \( x = 840k + 3 \) into this condition:
\[
840k + 3 \equiv 0 \pmod{9}
\] This simplifies to:
\[
840k \equiv -3 \pmod{9}
\] Now, reduce 840 modulo 9:
\[
840 \div 9 = 93 \text{ remainder } 3
\] So, \( 840 \equiv 3 \pmod{9} \). Therefore:
\[
3k \equiv -3 \pmod{9}
\] This simplifies to:
\[
k \equiv -1 \pmod{3}
\] which is equivalent to:
\[
k \equiv 2 \pmod{3}
\]

Step 4: Find the smallest \( k \).

The smallest value of \( k \) that satisfies \( k \equiv 2 \pmod{3} \) is \( k = 2 \).

Step 5: Find the smallest \( x \).

Now, substitute \( k = 2 \) into \( x = 840k + 3 \):
\[
x = 840(2) + 3 = 1680 + 3 = 1683
\]

Thus, the least number that satisfies all the conditions is 1683.

Answer: Option D
Solution:
L.C.M. of 252, 308 and 198 = 2772
So, A, B and C will again meet at the starting point in 2772 sec. i.e., 46 min. 12 sec.

Answer: Option C
Solution:
We can use the following relationship between the HCF (Highest Common Factor), LCM (Least Common Multiple), and the two numbers:

\[
\text{HCF} \times \text{LCM} = \text{Number 1} \times \text{Number 2}
\]

Given:
– HCF = 11
– LCM = 7700
– One number = 275

Let the other number be \( x \). Using the formula:

\[
11 \times 7700 = 275 \times x
\]

Now, solve for \( x \):

\[
x = \frac{11 \times 7700}{275}
\] Let’s calculate that.
The other number is 308.

Answer: Option B
Solution:
To find the least number \( x \) such that when doubled, it is divisible by 12, 18, 21, and 30, we need to follow these steps:

Step 1: Find the least common multiple (LCM) of 12, 18, 21, and 30.

We first find the prime factorizations of the numbers:

– \( 12 = 2^2 \times 3 \)
– \( 18 = 2 \times 3^2 \)
– \( 21 = 3 \times 7 \)
– \( 30 = 2 \times 3 \times 5 \)

The LCM is the highest power of each prime factor:

– The highest power of 2 is \( 2^2 \)
– The highest power of 3 is \( 3^2 \)
– The highest power of 5 is \( 5 \)
– The highest power of 7 is \( 7 \)

So, the LCM is:

\[
LCM = 2^2 \times 3^2 \times 5 \times 7 = 4 \times 9 \times 5 \times 7 = 1260
\]

Step 2: Find the least number \( x \).

We are looking for the least number \( x \) such that \( 2x \) is divisible by the LCM. This means \( 2x \) must be divisible by 1260. Therefore:

\[
2x = 1260 \quad \Rightarrow \quad x = \frac{1260}{2} = 630
\] Conclusion:
The least number \( x \) is 630.

Answer: Option D
Solution:
We are given the ratio of two numbers as 3:4 and their HCF as 4. We can use the following relationship to find their LCM:

\[
\text{HCF} \times \text{LCM} = \text{Number 1} \times \text{Number 2}
\]

Let the two numbers be \( 3x \) and \( 4x \), where \( x \) is a common factor.

Step 1: Apply the given values.

We know that:
– HCF = 4
– Number 1 = \( 3x \)
– Number 2 = \( 4x \)

Using the formula:

\[
\text{HCF} \times \text{LCM} = (3x) \times (4x)
\]

Substitute the values:

\[
4 \times \text{LCM} = 3x \times 4x
\] \[
4 \times \text{LCM} = 12x^2
\]

Step 2: Find the LCM.

Since the HCF of \( 3x \) and \( 4x \) is \( 4 \), the value of \( x \) must be \( \frac{4}{1} = 4 \).

Now, substitute \( x = 4 \) into the equation:

\[
4 \times \text{LCM} = 12 \times 4^2 = 12 \times 16 = 192
\]

Thus,

\[
\text{LCM} = \frac{192}{4} = 48
\] Conclusion:
The LCM of the two numbers is 48.

Answer: Option B
Solution:
To find the smallest number that, when diminished by 7, is divisible by 12, 16, 18, 21, and 28, we can follow these steps:

Step 1: Find the least common multiple (LCM) of 12, 16, 18, 21, and 28.

First, we find the prime factorizations of the numbers:

– \( 12 = 2^2 \times 3 \)
– \( 16 = 2^4 \)
– \( 18 = 2 \times 3^2 \)
– \( 21 = 3 \times 7 \)
– \( 28 = 2^2 \times 7 \)

To find the LCM, take the highest powers of all prime factors:

– The highest power of 2 is \( 2^4 \)
– The highest power of 3 is \( 3^2 \)
– The highest power of 7 is \( 7 \)

Thus, the LCM is:

\[
LCM = 2^4 \times 3^2 \times 7 = 16 \times 9 \times 7 = 1008
\]

Step 2: Find the smallest number.

We are looking for the smallest number \( x \) such that \( x – 7 \) is divisible by 1008. This means:

\[
x – 7 = 1008k
\]

for some integer \( k \). To find the smallest \( x \), set \( k = 1 \):

\[
x – 7 = 1008 \quad \Rightarrow \quad x = 1008 + 7 = 1015
\] Conclusion:
The smallest number is 1015.

Answer: Option A
Solution:
To express 252 as a product of primes, we need to find its prime factorization.

Step 1: Divide by the smallest prime numbers.

1. Divide 252 by 2 (the smallest prime number):
\[
252 \div 2 = 126
\]

2. Divide 126 by 2:
\[
126 \div 2 = 63
\]

3. Divide 63 by 3 (next smallest prime):
\[
63 \div 3 = 21
\]

4. Divide 21 by 3:
\[
21 \div 3 = 7
\]

5. Finally, 7 is a prime number.

Step 2: Write the prime factorization.

We divided 252 by 2 twice, by 3 twice, and ended with 7. So, the prime factorization of 252 is:
\[
252 = 2^2 \times 3^2 \times 7
\] Thus, 252 can be expressed as the product of primes \( 2^2 \times 3^2 \times 7 \).

Answer: Option C
Solution:
To find the greatest possible length that can be used to measure exactly 7 meters, 3 meters 85 cm, and 12 meters 95 cm, we need to calculate the greatest common divisor (GCD) of the three lengths. First, we need to express all the lengths in the same unit, say in centimeters.

Step 1: Convert the lengths to centimeters.

– \( 7 \, \text{m} = 700 \, \text{cm} \)
– \( 3 \, \text{m} \, 85 \, \text{cm} = 300 \, \text{cm} + 85 \, \text{cm} = 385 \, \text{cm} \)
– \( 12 \, \text{m} \, 95 \, \text{cm} = 1200 \, \text{cm} + 95 \, \text{cm} = 1295 \, \text{cm} \)

Now, we need to find the GCD of 700, 385, and 1295.

Step 2: Find the GCD.

We will use the Euclidean algorithm to find the GCD.

Let’s first find the GCD of 700 and 385:
\[
700 \div 385 = 1 \text{ (quotient)}, \text{ remainder } 700 – 385 = 315
\]

Now, find the GCD of 385 and 315:
\[
385 \div 315 = 1 \text{ (quotient)}, \text{ remainder } 385 – 315 = 70
\]

Next, find the GCD of 315 and 70:
\[
315 \div 70 = 4 \text{ (quotient)}, \text{ remainder } 315 – 4 \times 70 = 315 – 280 = 35
\]

Now, find the GCD of 70 and 35:
\[
70 \div 35 = 2 \text{ (quotient)}, \text{ remainder } 70 – 2 \times 35 = 70 – 70 = 0
\]

So, the GCD of 700 and 385 is 35.

Step 3: Find the GCD of 35 and 1295.

Now, find the GCD of 35 and 1295:
\[
1295 \div 35 = 37 \text{ (quotient)}, \text{ remainder } 1295 – 37 \times 35 = 1295 – 1295 = 0
\]

So, the GCD of 35 and 1295 is 35.
Conclusion:
The greatest possible length that can be used to measure exactly the lengths 7 m, 3 m 85 cm, and 12 m 95 cm is 35 cm.

Answer: Option C
Solution
Let the three numbers be \( a \), \( b \), and \( c \), and we are given the following information:

– The product of the first two numbers: \( a \times b = 551 \)
– The product of the last two numbers: \( b \times c = 1073 \)

Also, the three numbers are co-prime to each other, meaning that their pairwise greatest common divisor (GCD) is 1.

Step 1: Find the common factor \( b \)

We know that \( a \times b = 551 \) and \( b \times c = 1073 \), so \( b \) must be a common factor of 551 and 1073.

We will find the GCD of 551 and 1073 using the Euclidean algorithm:

– \( 1073 \div 551 = 1 \text{ (quotient)}, \text{ remainder } 1073 – 551 = 522 \)
– \( 551 \div 522 = 1 \text{ (quotient)}, \text{ remainder } 551 – 522 = 29 \)
– \( 522 \div 29 = 18 \text{ (quotient)}, \text{ remainder } 522 – 18 \times 29 = 522 – 522 = 0 \)

So, the GCD of 551 and 1073 is 29. Therefore, \( b = 29 \).

Step 2: Find \( a \) and \( c \)

Now that we know \( b = 29 \), we can find \( a \) and \( c \) using the given products:

– \( a \times 29 = 551 \), so \( a = \frac{551}{29} = 19 \)
– \( 29 \times c = 1073 \), so \( c = \frac{1073}{29} = 37 \)

Step 3: Find the sum of the three numbers

The three numbers are \( a = 19 \), \( b = 29 \), and \( c = 37 \). The sum of these numbers is:

\[
19 + 29 + 37 = 85
\] Conclusion:
The sum of the three numbers is 85.

Answer: Option C
Solution:
To find the highest common factor (HCF) of 36 and 84, we need to follow these steps:

Step 1: Find the prime factorizations of 36 and 84.

– \( 36 = 2^2 \times 3^2 \)
– \( 84 = 2^2 \times 3 \times 7 \)

Step 2: Identify the common factors.

The common prime factors between 36 and 84 are:
– The highest power of 2 is \( 2^2 \).
– The highest power of 3 is \( 3 \).

Step 3: Multiply the common factors.

The HCF is the product of the common factors:
\[
HCF = 2^2 \times 3 = 4 \times 3 = 12
\] Conclusion:
The highest common factor (HCF) of 36 and 84 is 12.

Answer: Option A
Solution:
To determine which of the fractions \( \frac{7}{8} \), \( \frac{13}{16} \), \( \frac{31}{40} \), and \( \frac{63}{80} \) is the largest, we can compare them by converting each fraction to a decimal or by finding a common denominator. Let’s convert them to decimal values:

1. \( \frac{7}{8} = 0.875 \)
2. \( \frac{13}{16} = 0.8125 \)
3. \( \frac{31}{40} = 0.775 \)
4. \( \frac{63}{80} = 0.7875 \)

Step 2: Compare the decimal values.

– \( \frac{7}{8} = 0.875 \)
– \( \frac{13}{16} = 0.8125 \)
– \( \frac{31}{40} = 0.775 \)
– \( \frac{63}{80} = 0.7875 \)

Conclusion:
The largest fraction is \( \frac{7}{8} \).

Answer: Option D
Solution:
We are asked to find the least number which, when divided by 12, 15, 20, and 54, leaves a remainder of 8 in each case.

Step 1: Express the problem mathematically.

Let the required number be \( x \). According to the problem, we can write:

\[
x \equiv 8 \pmod{12}, \quad x \equiv 8 \pmod{15}, \quad x \equiv 8 \pmod{20}, \quad x \equiv 8 \pmod{54}
\]

This implies that \( x – 8 \) is divisible by 12, 15, 20, and 54. Therefore, \( x – 8 \) is a common multiple of these numbers.

Step 2: Find the Least Common Multiple (LCM) of 12, 15, 20, and 54.

First, find the prime factorizations of the numbers:

– \( 12 = 2^2 \times 3 \)
– \( 15 = 3 \times 5 \)
– \( 20 = 2^2 \times 5 \)
– \( 54 = 2 \times 3^3 \)

Now, find the LCM by taking the highest power of each prime factor:

– The highest power of 2 is \( 2^2 \)
– The highest power of 3 is \( 3^3 \)
– The highest power of 5 is \( 5 \)

Thus, the LCM is:

\[
LCM = 2^2 \times 3^3 \times 5 = 4 \times 27 \times 5 = 540
\]

Step 3: Find the smallest \( x \).

Since \( x – 8 \) is divisible by 540, we can express \( x \) as:

\[
x = 540k + 8
\]

where \( k \) is some integer. To find the smallest \( x \), set \( k = 1 \):

\[
x = 540 \times 1 + 8 = 540 + 8 = 548
\] Conclusion:
The least number which, when divided by 12, 15, 20, and 54, leaves a remainder of 8 is 548.

Answer: Option B
Solution:
We are asked to find the greatest number that, when dividing 1657 and 2037, leaves remainders 6 and 5, respectively.

Step 1: Express the problem mathematically.

Let the required number be \( d \). According to the problem:

\[
1657 \equiv 6 \pmod{d}, \quad 2037 \equiv 5 \pmod{d}
\]

This implies that:

\[
1657 – 6 = 1651 \quad \text{is divisible by } d
\] \[
2037 – 5 = 2032 \quad \text{is divisible by } d
\]

Therefore, \( d \) must be a common divisor of 1651 and 2032.

Step 2: Find the greatest common divisor (GCD) of 1651 and 2032.

We will use the Euclidean algorithm to find the GCD.

1. \( 2032 \div 1651 = 1 \text{ (quotient)}, \text{ remainder } 2032 – 1651 = 381 \)
2. \( 1651 \div 381 = 4 \text{ (quotient)}, \text{ remainder } 1651 – 4 \times 381 = 1651 – 1524 = 127 \)
3. \( 381 \div 127 = 3 \text{ (quotient)}, \text{ remainder } 381 – 3 \times 127 = 381 – 381 = 0 \)

Since the remainder is 0, the GCD of 1651 and 2032 is 127.
Conclusion:
The greatest number that, when dividing 1657 and 2037, leaves remainders 6 and 5, respectively, is 127.

Answer: Option C
Solution:
99 = 1 × 3 × 3 × 11
101 = 1 × 101
176 = 1 × 2 × 2 × 2 × 2 × 11
182 = 1 × 2 × 7 × 13
So, divisors of 99 are 1, 3, 9, 11, 33, 99
Divisors of 101 are 1 and 101
Divisors of 176 are 1, 2, 4, 8, 11, 16, 22, 44, 88 and 176
Divisors of 182 are 1, 2, 7, 13, 14, 26, 91 and 182
Hence, 176 has the most number of divisors.

Answer: Option C
Solution:
We are given that the LCM of two numbers is 48 and that the numbers are in the ratio 2:3. Let’s solve for the numbers and their sum.

Step 1: Represent the numbers in terms of a common variable.

Let the two numbers be \( 2x \) and \( 3x \), where \( x \) is a common factor.

Step 2: Use the LCM formula.

The LCM of two numbers is given by:

\[
\text{LCM} = \frac{\text{Product of the numbers}}{\text{GCD of the numbers}}
\]

The GCD of \( 2x \) and \( 3x \) is \( x \) because the only common factor between 2 and 3 is 1, and both numbers share \( x \) as a common factor.

Therefore, the formula for the LCM becomes:

\[
\text{LCM} = \frac{(2x) \times (3x)}{x} = 6x^2 / x = 6x
\]

Step 3: Use the given LCM to solve for \( x \).

We are told that the LCM is 48:

\[
6x = 48 \quad \Rightarrow \quad x = \frac{48}{6} = 8
\]

Step 4: Find the two numbers.

Now that we know \( x = 8 \), the two numbers are:

– \( 2x = 2 \times 8 = 16 \)
– \( 3x = 3 \times 8 = 24 \)

Step 5: Find the sum of the numbers.

The sum of the two numbers is:

\[
16 + 24 = 40
\] Conclusion:
The sum of the two numbers is 40.

Answer: Option C
Solution:
To find the Highest Common Factor (HCF) of the fractions \( \frac{9}{10}, \frac{12}{25}, \frac{18}{35}, \) and \( \frac{21}{40} \), we will find the HCF of the numerators and the Least Common Multiple (LCM) of the denominators.

Step 1: Find the HCF of the numerators.
The numerators are 9, 12, 18, and 21. Let’s find the HCF of these numbers:

– \( 9 = 3^2 \)
– \( 12 = 2^2 \times 3 \)
– \( 18 = 2 \times 3^2 \)
– \( 21 = 3 \times 7 \)

The common factor is \( 3 \), so the HCF of the numerators is:

\[
\text{HCF of the numerators} = 3
\]

Step 2: Find the LCM of the denominators.
The denominators are 10, 25, 35, and 40. Let’s find the LCM of these numbers:

– \( 10 = 2 \times 5 \)
– \( 25 = 5^2 \)
– \( 35 = 5 \times 7 \)
– \( 40 = 2^3 \times 5 \)

To find the LCM, take the highest power of each prime factor:

– The highest power of 2 is \( 2^3 \)
– The highest power of 5 is \( 5^2 \)
– The highest power of 7 is \( 7 \)

Thus, the LCM of the denominators is:

\[
\text{LCM of the denominators} = 2^3 \times 5^2 \times 7 = 8 \times 25 \times 7 = 1400
\]

Step 3: Write the HCF of the fractions.

The HCF of the fractions is:

\[
\text{HCF} = \frac{\text{HCF of numerators}}{\text{LCM of denominators}} = \frac{3}{1400}
\] Conclusion:
The HCF of \( \frac{9}{10}, \frac{12}{25}, \frac{18}{35}, \) and \( \frac{21}{40} \) is \( \frac{3}{1400} \).

Answer: Option C
Solution:
We are given the following:

– The sum of two numbers \( a \) and \( b \) is 55: \( a + b = 55 \).
– The HCF of \( a \) and \( b \) is 5.
– The LCM of \( a \) and \( b \) is 120.

We are asked to find the sum of the reciprocals of the numbers, i.e.,

\[
\frac{1}{a} + \frac{1}{b}
\]

Step 1: Use the relationship between HCF, LCM, and the product of two numbers.

We know the formula for the product of two numbers in terms of their HCF and LCM:

\[
\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b
\]

Substituting the given values:

\[
5 \times 120 = a \times b \quad \Rightarrow \quad a \times b = 600
\]

Step 2: Use the sum and product of the numbers.

We now have the following system of equations:

1. \( a + b = 55 \)
2. \( a \times b = 600 \)

We need to find the sum of the reciprocals:

\[
\frac{1}{a} + \frac{1}{b} = \frac{a + b}{a \times b}
\]

Substitute the known values:

\[
\frac{1}{a} + \frac{1}{b} = \frac{55}{600} = \frac{11}{120}
\] Conclusion:
The sum of the reciprocals of the two numbers is \( \frac{11}{120} \).

Answer: Option D
Solution:
We can use the formula that relates the HCF, LCM, and the two numbers:

\[
\text{HCF} \times \text{LCM} = \text{Number 1} \times \text{Number 2}
\]

Given:
– HCF = 8
– LCM = 48
– One number = 24

Let the other number be \( x \). Using the formula:

\[
8 \times 48 = 24 \times x
\]

\[
384 = 24x
\]

\[
x = \frac{384}{24} = 16
\]

So, the other number is 16.

Answer: Option C
Solution:
To find the least number that leaves a remainder of 2 when divided by 4, 6, 8, 12, and 16, follow these steps:

Step 1: Find the LCM of the divisors
The given divisors are 4, 6, 8, 12, and 16. First, find their Least Common Multiple (LCM).

– Prime factorizations:
– \( 4 = 2^2 \)
– \( 6 = 2 \times 3 \)
– \( 8 = 2^3 \)
– \( 12 = 2^2 \times 3 \)
– \( 16 = 2^4 \)

– The LCM takes the highest powers of all prime factors:
– Highest power of 2: \( 2^4 = 16 \)
– Highest power of 3: \( 3^1 = 3 \)

\[
\text{LCM} = 16 \times 3 = 48
\]

Step 2: Add the remainder
The required number is **2 more than the LCM**, so:

\[
48 + 2 = 50
\]

Thus, the least number that leaves a remainder of 2 when divided by 4, 6, 8, 12, and 16 is 50.

Answer: Option A
Solution:
We use the relationship between the HCF, LCM, and the product of two numbers:

\[
\text{HCF} \times \text{LCM} = \text{Number 1} \times \text{Number 2}
\]

Given:
– LCM = 495
– HCF = 5
– Sum of numbers = 100

Let the two numbers be x and y. Using the formula:

\[
5 \times 495 = x \times y
\]

\[
2475 = x \times y
\]

We also know:

\[
x + y = 100
\]

Step 1: Solve for \( x \) and \( y \)
We solve the quadratic equation:

\[
t^2 – 100t + 2475 = 0
\]

Using the quadratic formula:

\[
t = \frac{-(-100) \pm \sqrt{(-100)^2 – 4(2475)}}{2}
\]

\[
t = \frac{100 \pm \sqrt{10000 – 9900}}{2}
\]

\[
t = \frac{100 \pm \sqrt{100}}{2}
\]

\[
t = \frac{100 \pm 10}{2}
\]

\[
t = \frac{110}{2} = 55 \quad \text{or} \quad t = \frac{90}{2} = 45
\]

So, the two numbers are 55 and 45.

Step 2: Find the Difference
\[
55 – 45 = 10
\] Thus, the difference between the two numbers is 10.

Answer: Option C
Solution:
99 = 1 × 3 × 3 × 11
101 = 1 × 101
176 = 1 × 2 × 2 × 2 × 2 × 11
182 = 1 × 2 × 7 × 13
So, divisor of 99 are 1, 3, 9, 11, 33 and 99
Divisor of 101 are 1 and 101
Divisor of 176 are 1, 2, 4, 8, 11, 16, 22, 44, 88 and 176
Divisor of 182 are 1, 2, 7, 13, 14, 26, 91 and 182
Hence, 176 has the most number of divisors.

Answer: Option D
Solution:
HCF = 8
Now LCM should have a factors 8
So check also the option we have only 60 which does not have a factor 8. So it will never be the LCM.

Answer: Option A
Solution:
To find the ratio of the LCM and HCF of 28 and 42, we first determine their HCF and LCM.

Step 1: Find HCF of 28 and 42
The prime factorizations are:

\[
28 = 2^2 \times 7
\] \[
42 = 2 \times 3 \times 7
\]

The common factors are \(2^1\) and \(7^1\), so:

\[
\text{HCF} = 2 \times 7 = 14
\]

Step 2: Find LCM of 28 and 42
The LCM takes the highest power of each prime factor:

\[
\text{LCM} = 2^2 \times 3^1 \times 7^1 = 84
\]

Step 3: Find the Ratio
\[
\frac{\text{LCM}}{\text{HCF}} = \frac{84}{14} = 6:1
\] Final Answer:
The ratio of LCM to HCF is 6:1.

Answer: Option A
Solution:
To simplify the fraction \( \frac{116690151}{427863887} \), we need to find the greatest common divisor (GCD) of the numerator and denominator and divide both by this value.
Let’s compute the GCD of 116690151 and 427863887.
The simplest form of \( \frac{116690151}{427863887} \) is \( \frac{3}{11} \).

Answer: Option C
Solution:
We use the relationship between HCF, LCM, and the two numbers:

\[
\text{HCF} \times \text{LCM} = \text{Number 1} \times \text{Number 2}
\]

Given:
– HCF = 26
– LCM = 1820
– One number= 130
– Let the other number be \( x \)

Using the formula:

\[
26 \times 1820 = 130 \times x
\]

\[
47320 = 130x
\]

\[
x = \frac{47320}{130} = 364
\] Final Answer:
The other number is 364.

Answer: Option A
Solution:
We use the fundamental relation between HCF, LCM, and the product of two numbers:

\[
\text{HCF} \times \text{LCM} = \text{Number 1} \times \text{Number 2}
\] Given:
– HCF= 478
– Numbers= 12906 and 14818
– LCM= ?
Step 1: Use the Formula
\[
\text{LCM} = \frac{\text{Product of the numbers}}{\text{HCF}}
\]

\[
\text{LCM} = \frac{12906 \times 14818}{478}
\] Let’s calculate this value.
The LCM of 12906 and 14818 is 400086.

Answer: Option C
Solution:
We need to find the greatest five-digit number that leaves a remainder of 2 when divided by 3, 5, 8, and 12.

Step 1: Find LCM of the Divisors
The given divisors are 3, 5, 8, and 12. First, we compute their LCM.

– Prime factorization:
– \( 3 = 3 \)
– \( 5 = 5 \)
– \( 8 = 2^3 \)
– \( 12 = 2^2 \times 3 \)

– The LCM takes the highest powers of all prime factors:
– Highest power of 2: \( 2^3 = 8 \)
– Highest power of 3: \( 3^1 = 3 \)
– Highest power of 5: \( 5^1 = 5 \)

\[
\text{LCM} = 8 \times 3 \times 5 = 120
\]

Step 2: Find the Largest Five-Digit Number
The largest five-digit number is 99999. We now find the largest multiple of 120 less than or equal to 99999.

\[
\frac{99999}{120} \approx 833.325
\]

So, the largest multiple of 120 is:

\[
120 \times 833 = 99960
\]

Step 3: Add the Remainder
Since the required number leaves a remainder of 2, we add 2:
\[
99960 + 2 = 99962
\] Final Answer:
The greatest five-digit number that satisfies the given condition is 99962.

Answer: Option A
Solution:
To find the HCF (Highest Common Factor) of the three numbers, we need to take the lowest powers of the common prime factors present in each number.

Given Numbers:

1. \( 2^4 \times 3^2 \times 5^3 \times 7 \)
2. \( 2^3 \times 3^3 \times 5^2 \times 7^2 \)
3. \( 3^1 \times 5^1 \times 7^1 \times 11^1 \)

Step 1: Find the Common Prime Factors
The prime factors in these numbers are 2, 3, 5, 7, and 11.

We now look for the minimum power of each prime factor common to all three numbers:

– For 2: The first number has \( 2^4 \), the second has \( 2^3 \), and the third has no factor of \( 2 \). So, the minimum power of 2 is \( 2^0 \) (no factor).
– For 3: The first number has \( 3^2 \), the second has \( 3^3 \), and the third has \( 3^1 \). So, the minimum power of 3 is \( 3^1 \).
– For 5: The first number has \( 5^3 \), the second has \( 5^2 \), and the third has \( 5^1 \). So, the minimum power of 5 is \( 5^1 \).
– For 7: The first number has \( 7^1 \), the second has \( 7^2 \), and the third has \( 7^1 \). So, the minimum power of 7 is \( 7^1 \).
– For 11: The first and second numbers have no factor of \( 11 \), but the third has \( 11^1 \). So, the minimum power of 11 is \( 11^0 \) (no factor).

Step 2: Compute the HCF
The HCF is the product of the common prime factors with the lowest powers:

\[
\text{HCF} = 3^1 \times 5^1 \times 7^1 = 3 \times 5 \times 7 = 105
\] Final Answer:
The HCF of the three numbers is 105.

Answer: Option D
Solution:
42. The H.C.F of 4 × 27 × 3125, 8 × 9 × 25 × 7 and 16 × We are given three numbers:

1. \( 4 \times 27 \times 3125 \)
2. \( 8 \times 9 \times 25 \times 7 \)
3. \( 16 \times 81 \times 5 \times 11 \times 49 \)

To find the HCF (Highest Common Factor) of these three numbers, we need to express each number in terms of its prime factorization.

Step 1: Prime Factorization of Each Number

1. For \( 4 \times 27 \times 3125 \):
– \( 4 = 2^2 \)
– \( 27 = 3^3 \)
– \( 3125 = 5^5 \)

So, \( 4 \times 27 \times 3125 = 2^2 \times 3^3 \times 5^5 \)

2. For \( 8 \times 9 \times 25 \times 7 \):
– \( 8 = 2^3 \)
– \( 9 = 3^2 \)
– \( 25 = 5^2 \)
– \( 7 = 7^1 \)

So, \( 8 \times 9 \times 25 \times 7 = 2^3 \times 3^2 \times 5^2 \times 7^1 \)

3. For \( 16 \times 81 \times 5 \times 11 \times 49 \):
– \( 16 = 2^4 \)
– \( 81 = 3^4 \)
– \( 5 = 5^1 \)
– \( 11 = 11^1 \)
– \( 49 = 7^2 \)

So, \( 16 \times 81 \times 5 \times 11 \times 49 = 2^4 \times 3^4 \times 5^1 \times 11^1 \times 7^2 \)

Step 2: Find the Common Prime Factors
Now, we compare the prime factorizations of all three numbers:

– For 2: The exponents are \( 2^2, 2^3, 2^4 \). The minimum exponent is \( 2^2 \).
– For 3: The exponents are \( 3^3, 3^2, 3^4 \). The minimum exponent is \( 3^2 \).
– For 5: The exponents are \( 5^5, 5^2, 5^1 \). The minimum exponent is \( 5^1 \).
– For 7: The exponents are \( 7^1, 7^1, 7^2 \). The minimum exponent is \( 7^1 \).
– For 11: The exponents are \( 11^0, 11^1, 11^1 \). The minimum exponent is \( 11^0 \) (no factor in the first number).

Step 3: Compute the HCF
The HCF is the product of the common prime factors with their lowest exponents:

\[
\text{HCF} = 2^2 \times 3^2 \times 5^1 \times 7^1
\]

Now, calculate this:

\[
2^2 = 4, \quad 3^2 = 9, \quad 5^1 = 5, \quad 7^1 = 7
\]

\[
\text{HCF} = 4 \times 9 \times 5 \times 7 = 36 \times 5 \times 7 = 180 \times 7 = 1260
\] Final Answer:
The HCF of the three numbers is 1260.81 × 5 × 11 × 49 is = ?

Answer: Option A
Solution:
Since all the numbers formed are even, 2 is a common factor.
Also, H.C.F. of two of the numbers i.e., 48 and 490, is 2.
So, the H.C.F. of all the numbers formed is 2.

Answer: Option B
Solution:
LCM of (15, 18, 21, 24)
⇒ 5 × 3 × 6 × 7 × 4 = 2520
⇒ In such type of questions,we take the difference between given number and remainder of that number.
Number Remainder
⇒ (15 – 11) = 4
⇒ (18 – 14) = 4
⇒ (21 – 17) = 4
⇒ (24 – 20) = 4
It will be same always
Now, Largest 4 digit number is 9999
⇒ On dividing 9999 by LCM (2520) we get remainder
⇒ 2439
Subtract remainder from 9999 we get largest 4 digit number, which is divisible by given number
= 9999 – 2439 = 7560 (9999 but required number gives difference on dividing)
So, our required number = 7560 – 4(difference) = 7556

Answer: Option C
Solution:
To solve this problem, we need to find the greatest number that divides 989 and 1327 leaving remainders 5 and 7, respectively.

Steps:
1. Subtract the remainder from both numbers:
– For 989, subtract 5: \( 989 – 5 = 984 \)
– For 1327, subtract 7: \( 1327 – 7 = 1320 \)

2. Now, we need to find the greatest common divisor (GCD) of 984 and 1320, because the number we’re looking for must divide both of these numbers.
We can calculate the GCD of 984 and 1320 using the Euclidean algorithm.
Let me calculate that.
The greatest number that divides both 989 and 1327, leaving remainders 5 and 7 respectively, is 24.

Answer: Option C
Solution:
To find the maximum capacity of the container that can measure the exact amount of milk from either of the containers, we need to find the Greatest Common Divisor (GCD) of 75 liters and 45 liters. The GCD represents the largest number that divides both 75 and 45 exactly.
Let me calculate the GCD for you.
The maximum capacity of the container that can measure the exact amount of milk from either of the containers is 15 liters.

Answer: Option B
Solution:
H.C.F. of 18 and 25 is 1
So, they are co-primes

Answer: Option C
Solution:
To find the Least Common Multiple (LCM) of 22, 54, 108, 135, and 198, we need to find the smallest number that is divisible by all of them.

The LCM can be found by:
1. Factorizing each number into prime factors.
2. Taking the highest power of each prime factor that appears.
Let me calculate the LCM for you.
The Least Common Multiple (LCM) of 22, 54, 108, 135, and 198 is 5940.

Answer: Option A
Solution:
When two numbers are in a given ratio and their HCF is known, the relationship between the HCF, LCM, and the numbers can be used to find the LCM. Here’s how:

Given:
– The ratio of the two numbers is \( 3 : 4 \).
– Their HCF is \( 4 \).

Let the numbers be \( 3x \) and \( 4x \), where \( x \) is a common factor. Since their HCF is \( 4 \), we know that \( x = 4 \).

So, the two numbers are:
– \( 3x = 3 \times 4 = 12 \)
– \( 4x = 4 \times 4 = 16 \)

Now, the formula for the product of two numbers is:
\[
\text{HCF} \times \text{LCM} = \text{Product of the two numbers}
\] Thus:
\[
4 \times \text{LCM} = 12 \times 16
\] Let me calculate the LCM using this formula.
The Least Common Multiple (LCM) of the two numbers is 48.

Answer: Option A
Solution:
We can use the relationship between the LCM, HCF, and the two numbers to find the other number.

The formula is:
\[
\text{LCM} \times \text{HCF} = \text{Product of the two numbers}
\]

Given:
– LCM = 2079
– HCF = 27
– One number = 189

Let the other number be \( x \). Then:
\[
2079 \times 27 = 189 \times x
\] We can solve for \( x \).
Let me calculate that for you.
The other number is 297.