RRB JE LCM AND HCF Questions with Detailed Solutions

Given:
Numbers: 84, 126, 210
Formula Used:
HCF (Highest Common Factor) is found using the prime factorization method.
The HCF is the product of the smallest powers of common prime factors.
Solution:
Prime factorization:
​$$84 = 2^2 \times 3 \times 7\\126 = 2 \times 3^2 \times 7\\210 = 2 \times 3 \times 5 \times 7\\$$​​
HCF = $$2 \times 3 \times 7 = 42$$​​
Thus, HCF = 42.

Given:

The time intervals at which the bells toll: 4 seconds, 8 seconds, 12 seconds, 16 seconds, and 20 seconds.

They toll together initially at 12:00 p.m. and continue tolling at their respective intervals.

Formula Used:

Number of times they toll together = $$\frac{\text{Time period}}{\text{LCM of intervals}}$$​ + 1

Solution:

Prime factorizations:
4 = 2²
8 = 2³
12 = 2²×3¹
16 = 2⁴

20 = 2²×5

LCM = 2⁴ × 3 × 5 = 16 × 3 × 5 = 240 seconds

Time period = 3 hours = 10,800 seconds.

Number of times they toll together: $$\frac{10,800}{240}$$​ = 45

Including the initial toll at 12:00 p.m., they toll together 46 times.

Concept used :

To find the maximum length of cloth that can be cut equally from both pieces without any remainder, we need to determine the Greatest Common Divisor (GCD) of the lengths of the two pieces of cloth (96 m and 144 m).

Solution :

Prime factorization of 96:

96 = $$2^5 \times 3$$​​

Prime factorization of 144:

144 = $$2^4 \times 3^2$$​​

Identify the common factors:

​$$\text{GCD} = 2^4 \times 3 = 16 \times 3 = 48$$​​

The maximum length of cloth that can be cut equally from both pieces is 48 metres.

Given:

Sum of LCM and HCF = 372

LCM = 92 × HCF

One of the numbers = 368

Formula Used:

Sum of LCM and HCF = 372

Product of numbers = LCM × HCF

Solution:

Let the other number be y.

Then, LCM = 92 × x.

Let HCF = x

LCM = 92 × x

Sum of LCM and HCF = 372

x + 92 × x = 372

93 × x = 372

x = 372 / 93

x = 4

HCF = 4

LCM = 92 × 4 = 368

Product of numbers = LCM × HCF

368 × y = 368 × 4

y = (368 × 4) / 368

y = 4

The other number is 4.

Given:
The numbers given are 2.05, 1.05, and 2.
Solution:
Convert to fractions: 205/100, 105/100, and 200/100.
Find LCM of numerators (205, 105, 200) and divide by 100.

Solution:
We know, 935 = 5 × 11 × 17
If (p > q)
Let p be 935, then q = 1
=> Sum of q = 1
Again,
Suppose, p = 935, then q = 5
=> Sum of digit q = 5
Again,
Suppose, p = 935, then q = 11 = 11
=> Sum of digit q = 1 + 1 = 2
Again,
Let p = 935 and q = 17
=> Sum of digit of q = 1 + 7 = 8
Again,
Let p = 935 and q = 55
=> Sum of digit q = 5 + 5 = 10
Again,
Let p = 935 and q = 85
=> Sum of digit q = 8 + 5 = 13
Again,
Let p = 935 and q = 187
=> Sum of digit of q = 1 + 8 + 7 = 16
Thus, the maximum sum of digits of q is 16, when p = 935 & q = 187.

Given:

LCM is 2400, HCF is 16.

one number is 480

Formula Used:

First number × Second number = LCM × HCF

Calculation:

Using the formula,

=> 480 × Second number = 2400 × 16

=> Second number =$$\frac {2400 × 16}{480}$$​

Therefore, the second number is 80.

Solution:
​For all positive numbers, the least value of HCF will always be 1. For all negative numbers, the least value of HCF will always be -1. ∴0cannot be the HCF of any set of numbers.​

Given:

We are tasked with finding the least perfect square that is divisible by the numbers 21, 36, and 66.

Concept Used:

To find the least perfect square divisible by multiple numbers, we first calculate the LCM (Least Common Multiple) of the given numbers. After that, we adjust the LCM by adding any necessary factors to make it a perfect square.

Solution:

Prime factorization of the given numbers.

​$$21 = 3 \times 7 \\36 = 2^2 \times 3^2 \\66 = 2 \times 3 \times 11$$

the LCM of 21, 36, and 66 by taking the highest powers of all prime factors involved.

​$$LCM = 2^2 \times 3^2 \times 7 \times 11 = 2772$$​​

​The LCM to make it a perfect square

The prime factorization of 2772

For a number to be a perfect square, all the powers in its prime factorization must be even.
To make the powers of 7 and 11 even, we multiply by 7 and 11.

Thus, the least perfect square = 2772 × 7 × 11 = 213444

The least perfect square divisible by 21, 36, and 66 is 213444.

Solution:

To find the least number divisible by all the given numbers, we need to calculate their Least Common Multiple (LCM).

Prime factorization of the given numbers:

​$$12 = 2^2 \times 3 \\18 = 2 \times 3^2 \\21 = 3 \times 7 \\30 = 2 \times 3 \times 5$$

The LCM is found by taking the highest powers of all prime factors present in the factorizations:

​$$LCM = 2^2 \times 3^2 \times 5 \times 7 \\LCM = 4 \times 9 \times 5 \times 7 \\LCM = 1260$$​​

​Therefore, the least number that is divisible by 12, 18, 21, and 30 is 1260​

Given
First number = $$2^3 × 6^2 × 11$$​​
Second number = $$2^2 × 11^2 × 17$$​​
Third number = $$6^3 × 11^2 × 17^2$$​​
Formula Used
LCM = Product of the highest powers of all prime factors involved.
Solution
First, express all the base numbers as prime factors:
First number = $$2^3 × (2 × 3)^2 × 11 = 2^3 × 2^2 × 3^2 × 11^1 = 2^5 × 3^2 × 11^1$$​​
Second number = $$2^2 × 11^2 × 17^1$$​​
Third number = $$(2 × 3)^3 × 11^2 × 17^2 = 2^3 × 3^3 × 11^2 × 17^2$$​​
Now, find the highest power of each prime factor (2, 3, 11, 17) among all the numbers:
Highest power of 2 = $$2^5$$​​
Highest power of 3 = $$3^3$$​​
Highest power of 11 = $$11^2$$​​
Highest power of 17 = $$17^2$$​​
Multiply these highest powers to get the LCM:
LCM = $$2^5 × 3^3 × 11^2 × 17^2$$​​
Final Answer
So the correct answer is (b)

Given
Number of mango trees = 95
Number of banyan trees = 171
Number of banana trees = 152
Formula Used
Maximum trees per row = Highest Common Factor (HCF) of the number of trees
Total number of rows =  $$\frac{\text{Total number of each tree}}{\text{HCF}}$$​​
Solution
Prime factorization:
95 = 5 × 19
171 = 9 × 19
152 = 8 × 19
The Highest Common Factor (HCF) is 19.
So, maximum trees per row = 19.
Number of rows for mango trees $$= \frac{95}{19} = 5$$​​

Number of rows for banyan trees =$$\frac{171}{19} = 9$$​

Number of rows for banana trees = $$\frac{152}{19} = 8$$​​
Total minimum rows required = 5 + 9 + 8 = 22
Final Answer
So the correct answer is (d)