Arun Sharma Test: Number System- 1 - CAT MCQ

(a) −√4 = −2 is a Real Number as well as a Rational Number
(b) −√3 is a Real number and an irrational number
(c) √−4 is an imaginary number

► All the digits written once = 2116
► All the digits without Bar written once = 21
► Number of digits with bar after decimal = 2
► Number of digits without bar after decimal = 0
► Rational form = 

► √16 = 4, which is a natural number, rational number as well as a real number.

(16a + 6) will always be even, whether the a is odd or even.

• For a = 1 (odd)
  (16 * 1) + 6 = 16 + 6 = 22 (even)
• For a = 2 (even)
  (16* 2) + 6 = 32 + 6 = 38 (even)

As it is given, 7a - 3 is odd

⇒ 7a = odd +3 i.e. even (say, x)
∵ a is an integer
⇒ For 7a = x, x should be completely divisble by 7
⇒ a = x/7 i.e. even/odd = even

We do not need to do the multiplication in this question.
► 1837 is Odd
► 1712 is Even
► 1839 is Odd

Odd x Even x Odd = Even

Only option 1 is Even rest all are Odd. 

The first two-digit number divisible by 3 is 12
The second two-digit number divisible by 3 is 15
The third is 18

We can see that an AP is formed here, whose first term is 12 and the common difference is 3

By hit and trial we can find the last two-digit number divisible by 3 will be 99, which means the nth term of an AP is 99

Here, a = 12, d = 3 and aₙ = 99

By putting the values in the formula,
99 = 12 + (n - 1)*3
n - 1 = 29
n = 29 + 1 = 30

The two-digit numbers divisible by 3 are 30

Similarly, by taking 10 as the first and 95 as the last digit we could find the number of two-digit numbers divisible by 5,
Here, a = 10, d = 5 and aₙ = 95

By putting the values in the formula,
95 = 10 + (n - 1)*5
n - 1 = 17
n = 17 + 1 = 18

The two-digit numbers divisible by 5 is 18

It is asked in the question that the two-digit number should either be divisible by 3 or 5 but not by both. The number of two-digit numbers divisible by 3 can be found by forming an AP (Arithmetic Progression), as done above. Here, the first term and the common difference will be L.C.M.(3, 5) = 15.

By hit and trial, the last term will be 90. Using the nth-term formula of an A.P.,
90 = 15 + (n - 1) × 15
75 = (n - 1) × 15
5 = n - 1
n = 6

The two-digit numbers divisible by both 3 and 5 are 6.

So, the total number of two-digit numbers divisible by 3 or 5 will be:
(Number divisible by 3) + (Number divisible by 5) - (Number divisible by both 3 and 5)
= 30 + 18 - 6
= 42

► 999 is the Largest composite number, it is divisible by 1, 3, 9, 111, 333, 999. 

Given:
The sum of seven consecutive natural numbers = 1617

Calculation:
Let the numbers be n, n+1, n+2, n+3, n+4, n+5, n+6 respectively
=> 7n + 21 = 1617
=> 7n = 1596
=> n = 228

The numbers are 228, 229, 230, 231, 232, 233, 234
Out of these, 229 and 233 are prime numbers

Therefore, the required prime numbers count is 2.

► 517^th Whole number = 516
► 516^th Natural Number = 516

• 516 - 516 = 0

GIVEN:
Four bells ring simultaneously at starting and an interval of 6 sec, 12 sec, 15 sec and 20 sec respectively.

CONCEPT:
LCM: It is a number which is a multiple of two or more numbers.

CALCULATION:
LCM of (6, 12, 15, 20) = 60
All 4 bells ring together again after every 60 seconds

Now,
In 2 Hours, they ring together = [(2 × 60 × 60) / 60] times + 1 (at the starting) = 121 times

Hence, in 2 hours they ring together for 121 times.

We know that,
product of two numbers = L.C.M × H.C.F of those numbers

Let the second number be x.

24 × x = 168 × 6
x = 6 × 7
x = 42

HCF = 57
LCM = 114

One number is 9/2 times the other.

Formula used:

Product of two numbers = HCF × LCM

Let the two numbers be x and (9/2)x.

Calculations:

Step 1: Use the formula for the product of two numbers

x × (9/2)x = HCF × LCM

(9/2)x² = 57 × 114

x² = (57 × 114 × 2) / 9

x² = 14436 / 9

x² = 1604

x = √1604

x = 38

Step 2: Calculate the other number

The other number = (9/2)x = (9/2) × 38 = 171

Given:
The 6-digit number is 910300.

Formula Used:
A number is completely divisible by 11 if the difference between the sum of its digits at odd positions and the sum of its digits at even positions is either 0 or a multiple of 11.

Calculation:
Sum of digits at odd positions (9, 0, 0): 9 + 0 + 0 = 9
Sum of digits at even positions (1, 3, 0): 1 + 3 + 0 = 4
Difference = 9 - 4 = 5

We need to add a 1-digit number to the original number so that this difference becomes a multiple of 11.

Let the 1-digit number be x. Then,
5 + x = 11 (since 11 is the smallest multiple of 11 greater than 5)
=> x = 11 - 5
=> x = 6

Therefore, the smallest 1-digit number to be added to 910300 so that it is completely divisible by 11 is 6.

The smallest natural number which is divisible by all given numbers is the Least Common Multiple (LCM).

Calculation:

Prime factorization:
16 = 2⁴
88 = 2³ × 11
12 = 2² × 3
22 = 2 × 11

LCM is found by taking the highest power of each prime:
LCM = 2⁴ × 3 × 11
⇒ LCM = 16 × 3 × 11
⇒ LCM = 48 × 11
⇒ LCM = 528

The smallest natural number which is divisible by 16, 88, 12, and 22 is 528.