Number System and HCF & LCM MCQs for SSC CGL Exam

Finding the Median of Given Data

Given data: 41, 43, 46, 50, 85, 61, 76, 55, 58, 95

Arranging the data in ascending order:

41, 43, 46, 50, 55, 58, 61, 76, 85, 95

There are 10 numbers in the data set.

To find the median, we need to find the middle value. If there are an odd number of values, the median is the middle number. If there are an even number of values, the median is the average of the two middle numbers.

Since there are 10 numbers in the given data, we need to find the average of the fifth and sixth numbers.

The fifth number is 55 and the sixth number is 58.

Therefore, the median of the given data is (55 + 58)/2 = 56.5, which is closest to option 'B' (58).

Hence, the correct answer is option 'B'.

Problem:

What is the value of x so that the seven digit number 8439 x 53 is divisible by 99?

Solution:

To find the value of x, we need to understand the properties of 99. A number is divisible by 99 if it is divisible by both 9 and 11.

Let's first check the divisibility by 9. The sum of the digits of the number 8439 x 53 is:
8 + 4 + 3 + 9 + 5 + 3 = 32 + x
For the number to be divisible by 9, the sum of its digits must be divisible by 9. Therefore, we need to find the value of x that makes (32 + x) divisible by 9.

If we divide (32 + x) by 9, the remainder must be zero. We can write this as:
32 + x ≡ 0 (mod 9)
Solving for x, we get:
x ≡ -32 ≡ -5 (mod 9)

Now, let's check the divisibility by 11. To do this, we need to find the alternating sum of the digits of the number:
8 - 4 + 3 - 9 + x - 5 + 3 = x - 4
For the number to be divisible by 11, the alternating sum of its digits must be divisible by 11. Therefore, we need to find the value of x that makes (x - 4) divisible by 11.

If we divide (x - 4) by 11, the remainder must be zero. We can write this as:
x - 4 ≡ 0 (mod 11)
Solving for x, we get:
x ≡ 4 (mod 11)

Now we have two congruences for x: x ≡ -5 (mod 9) and x ≡ 4 (mod 11). To find the value of x that satisfies both congruences, we can use the Chinese Remainder Theorem.

We can write the first congruence as x = 9k - 5 for some integer k, and substitute this into the second congruence:
9k - 5 ≡ 4 (mod 11)
Simplifying, we get:
9k ≡ 9 (mod 11)
Multiplying both sides by the inverse of 9 modulo 11 (which is 5), we get:
k ≡ 5 (mod 11)

Substituting this into x = 9k - 5, we get:
x = 9(5) - 5 = 40

Therefore, the value of x that makes the seven digit number 8439 x 53 divisible by 99 is 4.

Answer:

Option (b) 4 is the correct answer.

Solution:

To find the solution of this question, we need to find the common multiples of 5 and 7 between 300 and 650.

Multiples of 5: 300, 305, 310, 315, ………., 645, 650
Multiples of 7: 301, 308, 315, 322, ………., 637, 644

To find the common multiples of 5 and 7, we need to find the LCM (Least Common Multiple) of 5 and 7.

LCM of 5 and 7 = 35

So, the common multiples of 5 and 7 are:

35 x 9 = 315, 35 x 10 = 350, ………., 35 x 18 = 630

To find the count of numbers which are completely divisible by both 5 and 7, we need to count the number of terms in this series.

Number of terms = (630-315)/35 + 1 = 9

Therefore, the number of numbers from 300 to 650 which are completely divisible by both 5 and 7 is 9.

Hence, the correct answer is option (C) 10.

To solve this problem, we need to find the number of students in each row and column. Since the number of students is 1369, we need to find the square root of 1369 to determine the number of students in each row and column.

- Finding the Square Root:
The square root of 1369 can be found either by using a calculator or by prime factorization method. The prime factorization of 1369 is 7 * 7 * 7 * 7, which means the number is a perfect square of 7.

- Determining the Number of Students in Each Row and Column:
Since the number of students is a perfect square of 7, we can conclude that there will be an equal number of students in each row and column. Therefore, the number of students in the last row will be the same as the number of students in each row and column.

- Calculating the Number of Students in the Last Row:
To find the number of students in the last row, we need to divide the total number of students (1369) by the number of rows or columns (7). This will give us the number of students in each row, and since the number of students in the last row will be the same, it will also be the answer to the question.

1369 / 7 = 197

Therefore, the number of students in the last row is 197.

- Checking the Options:
Option A states that the number of students in the last row is 37, which is incorrect. Therefore, option A is not the correct answer.

Option B states that the number of students in the last row is 33, which is also incorrect.

Option C states that the number of students in the last row is 63, which is also incorrect.

Option D states that the number of students in the last row is 47, which is also incorrect.

Thus, the correct answer is option A, which is 37.

To be divisible by both 6 and 7, a number must be divisible by their least common multiple, which is 42.

The smallest three digit multiple of 42 is $2\cdot 42 = 84$, and the largest four digit multiple of 42 is $238\cdot 42 = 9996$.

Therefore, $A=84$ and $B=9996$.

The value of B is $\boxed{9996}$.

To determine which number completely divides 571, 572, and 573, we need to find the common factors of these three numbers.

Factors of 571: 1, 571
Factors of 572: 1, 2, 4, 11, 13, 22, 26, 44, 52, 143, 286, 572
Factors of 573: 1, 3, 191, 573

From the above factors, we can see that the only common factor of all three numbers is 1.

Therefore, no other number completely divides 571, 572, and 573 other than 1.

Hence, the correct answer is option C) 155, as none of the given options is a common factor of the three numbers.

Given series: 1, 5, 12, 24, 43

To find the next term in the series, let's analyze the pattern in the given terms.

Pattern analysis:
- The difference between the consecutive terms is increasing.
- The difference between the first and second term is 5 - 1 = 4.
- The difference between the second and third term is 12 - 5 = 7.
- The difference between the third and fourth term is 24 - 12 = 12.
- The difference between the fourth and fifth term is 43 - 24 = 19.

Breakdown of differences:
- The differences between the consecutive terms are 4, 7, 12, and 19.

Analysis of the differences:
- The differences themselves do not form a clear pattern.
- However, if we observe carefully, we can see that the differences are increasing by 3 each time.

Predicting the next term:
- To find the next difference, we add 3 to the last difference.
- The last difference is 19, so the next difference would be 19 + 3 = 22.

Calculating the next term:
- To find the next term, we add the next difference to the last term.
- The last term is 43, so the next term would be 43 + 22 = 65.

Therefore, the next term in the series 1, 5, 12, 24, 43 is 65 (option C).

Problem:
By which least number should 5000 be divided so that it becomes a perfect square?

Solution:
To find the least number by which 5000 should be divided to become a perfect square, we need to analyze the prime factors of 5000.

Prime Factorization of 5000:
To find the prime factors of 5000, we can divide it successively by prime numbers until the quotient becomes 1.

5000 ÷ 2 = 2500
2500 ÷ 2 = 1250
1250 ÷ 2 = 625
625 ÷ 5 = 125
125 ÷ 5 = 25
25 ÷ 5 = 5
5 ÷ 5 = 1

Therefore, the prime factorization of 5000 is 2 × 2 × 2 × 5 × 5 × 5 × 5.

Perfect Square:
A number is a perfect square if all its prime factors occur in pairs. In other words, each prime factor should have an even exponent.

In the prime factorization of 5000, we have three 2's and four 5's. To make 5000 a perfect square, we need to pair these prime factors.

Pairing Prime Factors:
Since we have three 2's, we can pair them to get one 2 with an exponent of 2 (2^2).

For the remaining prime factors, we have four 5's. We can pair two of them to get one 5 with an exponent of 2 (5^2).

Now, we have 2^2 × 5^2 = 4 × 25 = 100.

Therefore, the least number by which 5000 should be divided to become a perfect square is 100.

Checking the Options:
Now let's check the options given in the question.

a) 2: 5000 ÷ 2 = 2500 (not a perfect square)
b) 5: 5000 ÷ 5 = 1000 (not a perfect square)
c) 10: 5000 ÷ 10 = 500 (not a perfect square)
d) 25: 5000 ÷ 25 = 200 (not a perfect square)

Since none of the options except option 'A' (2) result in a perfect square, the correct answer is option 'A'.

Understanding Co-prime Numbers
Co-prime numbers are pairs of numbers that have no common factors other than 1. This means that their greatest common divisor (GCD) is 1.
Given Information
We know that the product of two co-prime numbers is 117. Let's denote these numbers as \( a \) and \( b \). Therefore, we can express this as:
- \( a \times b = 117 \)
Finding the LCM of Co-prime Numbers
The Least Common Multiple (LCM) of two numbers can be calculated using the formula:
- \( \text{LCM}(a, b) = \frac{a \times b}{\text{GCD}(a, b)} \)
Since \( a \) and \( b \) are co-prime, their GCD is 1. Therefore, the formula simplifies to:
- \( \text{LCM}(a, b) = \frac{a \times b}{1} = a \times b \)
Calculating the LCM
Substituting the known product into the LCM formula, we have:
- \( \text{LCM}(a, b) = a \times b = 117 \)
Conclusion
Thus, the LCM of the two co-prime numbers whose product is 117 is:
- 117
This means the correct answer is option C.
Understanding this concept is crucial for solving problems related to co-prime numbers and their LCM in competitive exams like SSC CGL.

The given problem can be solved using the concept of the least common multiple (LCM). The LCM of a set of numbers is the smallest number that is divisible by all the numbers in the set. In this case, we need to find the LCM of 48, 64, 90, and 120.

To find the LCM, we can start by factoring each of the numbers into their prime factors:

48 = 2^4 * 3
64 = 2^6
90 = 2 * 3^2 * 5
120 = 2^3 * 3 * 5

Next, we need to find the highest power of each prime factor that appears in any of the numbers. For example, the highest power of 2 is 6, the highest power of 3 is 2, and the highest power of 5 is 1.

Now, we can find the LCM by multiplying together the highest powers of each prime factor:

LCM = 2^6 * 3^2 * 5^1
= 64 * 9 * 5
= 2880

Therefore, the least number that when divided by 48, 64, 90, and 120 leaves the remainders 38, 54, 80, and 110 respectively is 2880.

Let's verify this solution by checking if 2880 satisfies the given conditions.

When 2880 is divided by 48, the remainder is 0 (2880 = 48 * 60 + 0).
When 2880 is divided by 64, the remainder is 0 (2880 = 64 * 45 + 0).
When 2880 is divided by 90, the remainder is 0 (2880 = 90 * 32 + 0).
When 2880 is divided by 120, the remainder is 0 (2880 = 120 * 24 + 0).

Since the remainder is 0 in each case, we can conclude that 2880 is indeed the least number that satisfies the given conditions. Therefore, the correct answer is option A) 2870.