All questions of LCM and HCF for Civil Engineering (CE) Exam

(USE THIS METHOD ONLY IF ANY ONE OF THE OPTIONS REMAINDERS AS 0 and don't do this method if all the options got the remainder as 0).
So the question was by dividing 11 there will be no remainder. I divided the numbers in the option with 11 and I got remainders as
a) remainder is 6
b)remainder is 5
c)remainder is 8
d) remainder is 0
e) remainder is different

so the option d leaves a remainder is 0 the answer is 363.
I solved this problem by using this method.

Number= LCM of5,6,7&8+3=1683
5++6+8+3=18
which is divisible by 9 hence no reminder

Let the Nos be X and y.
then it is quite clear that,
11x*11y=11*385. (X*Y=HCF*LCM)

•. XY=35 and X and y should be. co-prime
therefore 5*7=x*y would be the required
pair

hence first no is 55 and second no is 77
therefore 77 is the absolute. answer for this question

The given question is related to finding the greater number out of two numbers whose product is given and the highest common factor (HCF) is also given. Let's solve it step by step.

Given:
Product of two numbers = 4107
HCF of the two numbers = 37

Step 1: Prime Factorization of 4107
To find the prime factorization of 4107, we divide it by prime numbers starting from 2 until we cannot divide any further. The prime factors of 4107 are:

4107 ÷ 3 = 1369
1369 ÷ 37 = 37

So, the prime factorization of 4107 is 3 × 37 × 37.

Step 2: Determining the Numbers
Now that we have the prime factorization, we can determine the two numbers whose product is 4107 and whose HCF is 37. Since the HCF is 37, one of the numbers should be a multiple of 37. Let's assign the factors as follows:

Number 1 = 37 × a
Number 2 = 37 × b

Multiplying the two numbers, we get:
Number 1 × Number 2 = (37 × a) × (37 × b) = 4107

Expanding the equation:
(37 × 37) × (a × b) = 4107

Simplifying:
1369 × (a × b) = 4107

Dividing by 1369 on both sides:
a × b = 3

Step 3: Finding the Greater Number
We need to find the greater number out of the two. To do this, we need to consider the values of a and b. Since a and b are positive integers and their product is 3, the possible values for a and b are:

a = 1, b = 3
a = 3, b = 1

Substituting the values back into the equation, we get the two numbers:

Number 1 = 37 × a = 37 × 1 = 37
Number 2 = 37 × b = 37 × 3 = 111

Comparing the two numbers, we can see that the greater number is 111.

Therefore, the correct answer is option C) 111.

Problem: Two containers contain 50 and 125 litres of water respectively. Find the maximum capacity of a container which can measure the water in each container an exact number of times(in litres)

Solution:
We can find the maximum capacity of a container by finding the HCF of the given volumes of water.

Step 1: Prime Factorization
50 = 2 x 5^2
125 = 5^3

Step 2: Identify the common factor(s)
The only common factor is 5.

Step 3: Find the HCF
HCF = 5

Therefore, the maximum capacity of a container is 5 litres, as it can measure 50 litres (10 times) and 125 litres (25 times) an exact number of times.

Answer: (a) 25

Problem: To find the largest size of a square tile that can be used to completely pave a rectangular courtyard.

Given: Length of courtyard = 140cm, width of courtyard = 525cm.

Solution:

To find the largest size of the tile, we need to find the common factor of both the length and width of the courtyard.

Step 1: Find the HCF (Highest Common Factor) of 140 and 525.

We can use the prime factorization method to find the HCF.

140 = 2 × 2 × 5 × 7
525 = 3 × 5 × 5 × 7

Multiplying the common factors, we get:

HCF = 5 × 7 = 35

Step 2: The largest size of the tile that can be used is equal to the HCF.

Therefore, the largest size of the tile that can be used is 35cm.

Hence, option (B) is the correct answer.

Note: It is important to find the HCF of the length and width of the courtyard to ensure that the entire area is covered without leaving any gaps. Using a tile size that is not a factor of both the length and width will result in incomplete paving.

Explanation:
- To find the greatest possible length of each piece, we need to find the greatest common divisor (GCD) of the lengths of the boxes.
- The lengths of the boxes are 60m, 30m, and 45m.
- We can find the GCD of these lengths by finding the GCD of the pairwise combinations of the lengths.
- GCD(60,30) = 30
- GCD(30,45) = 15
- GCD(60,45) = 15
- Therefore, the greatest common divisor of 60, 30, and 45 is 15m.
- So, the greatest possible length of each piece that can be cut from the boxes is 15m.
Therefore, the correct answer is option A) 15m.

LCM (6, 8, 15, 30) – 4 = 120 – 4 = 116.

Solution:

To find the length of the longest possible pieces that can be cut from the three bars, we need to find the greatest common divisor (GCD) of the three lengths.

Step 1: Find the factors of the lengths of the bars

The factors of 12 are 1, 2, 3, 4, 6, and 12.

The factors of 16 are 1, 2, 4, 8, and 16.

The factors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24.

Step 2: Find the common factors

The common factors of the three lengths are 1, 2, 4, and 6.

Step 3: Find the largest common factor

The largest common factor is 6. Therefore, the longest possible pieces that can be cut from each of the three bars will be 6 meters long.

Answer: Option D (6m)

Understanding the Problem
To solve for the two numbers given their HCF, LCM, and sum, we can use the relationship between HCF, LCM, and the product of the two numbers.
Key Relationships
- HCF (Highest Common Factor) = 5
- LCM (Least Common Multiple) = 275
- Sum of the two numbers = 80
Finding the Product of the Numbers
Using the relationship:
- HCF × LCM = Product of the two numbers
Therefore, we calculate:
- 5 × 275 = 1375
Let the two numbers be x and y. Hence, we have:
- x + y = 80
- xy = 1375
Forming the Quadratic Equation
From the equations, we can form:
- t^2 - (x+y)t + xy = 0
- t^2 - 80t + 1375 = 0
Calculating the Roots
Using the quadratic formula:
- t = [80 ± sqrt((80)^2 - 4 × 1 × 1375)] / 2
Calculating the discriminant:
- (80)^2 - 4 × 1375 = 6400 - 5500 = 900
Thus, we have:
- t = [80 ± 30] / 2
This gives us:
- t = 55 and t = 25
So, the two numbers are 55 and 25.
Finding the Sum of the Reciprocals
The sum of the reciprocals of the numbers is given by:
- (1/x) + (1/y) = (y + x) / (xy)
Substituting the values:
- (55 + 25) / (55 × 25) = 80 / 1375 = 16 / 275
Conclusion
The sum of the reciprocals of the two numbers is:
- 16/275
Thus, the correct answer is option 'B'.

Approach:

- The least number of students in the school will be the LCM of 9, 7, and 12.
- We can find the LCM by prime factorizing the numbers and taking the highest power of each prime factor.
- Once we have the LCM, we can check if it satisfies the given conditions.

Solution:

- Prime factorizing 9: 3^2
- Prime factorizing 7: 7
- Prime factorizing 12: 2^2 * 3
- Taking the highest power of each prime factor: 2^2 * 3^2 * 7 = 252
- So the least number of students in the school is 252.
- We can check if 252 satisfies the given conditions:
- 252/9 = 28 (rows of 9 students)
- 252/7 = 36 (rows of 7 students)
- 252/12 = 21 (rows of 12 students)
- So 252 satisfies the given conditions.
- Therefore, the correct answer is option D) 252.