Number System Questions and Answers

Answer: D

The decimal expansion sorts the real numbers into two kinds:

• Rational - the expansion either stops or falls into a repeating block, as in \( \tfrac{1}{4} = 0.25 \) or \( \tfrac{1}{3} = 0.333\ldots \)
• Irrational - it runs on for ever without ever repeating, as in \( \sqrt{2} = 1.41421356\ldots \) or \( \pi \)

A non-terminating, non-repeating expansion is therefore irrational. Natural and whole numbers are particular rationals with expansions that stop at once, so none of the other options can apply.

Answer: A

Divide the largest four-digit number by 88 and drop the remainder.

\[ 9999 = 88 \times 113 + 55 \]

\[ 9999 - 55 = 9944 = 88 \times 113 \]

9988 on the list is even and looks plausible, but \( 9988 \div 88 = 113.5 \).

Answer: C

Rather than computing the power in full, reduce the base first. Since \( 3^{2} = 9 \equiv 1 \pmod 8 \), every even power of 3 is 1 modulo 8:

\[ 3^{7} = \left(3^{2}\right)^{3} \cdot 3 \equiv 1^{3} \cdot 3 \equiv 3 \pmod 8 \]

The direct route agrees: \( 3^{7} = 2187 \) and \( 2187 = 8 \times 273 + 3 \).

Spotting that the base squares to 1 is what makes this quick, and the same trick handles any power of 3 modulo 8: odd powers leave 3, even powers leave 1.

Answer: B

Ten's digit = 7 ⇒ units digit =8 ⇒ Hundred's digit =1,2,3,4,5,6.
⇒ Number of ways =1×6

Ten's digit = 6 ⇒ units digit =7,8 ⇒ Hundred's digit =1,2,3,4,5.
⇒ Number of ways =2×5

Ten's digit = 5 ⇒ units digit =6,7,8 ⇒ Hundred's digit =1,2,3,4
⇒ Number of ways =3×4

Ten's digit = 4 ⇒ units digit =5,6,7,8 ⇒ Hundred's digit =1,2,3.
⇒ Number of ways =4×3

Ten's digit = 3 ⇒ units digit =4,5,6,78 ⇒ Hundred's digit =1,2
⇒ Number of ways =5×2

Ten's digit = 2 ⇒ units digit =3,4,5,6,7,8 ⇒ Hundred's digit =1
⇒ Number of ways =6×1

Total number of ways =6+10+12+12+10+6= 56.

Answer: C

We have given that the number 2ab52ab5 is divisible by 25.

Any number divisible by 25 ends with the last two digits 00, 25, 50, or 75.
So, b5b5 should equal 25 or 75.

Hence, b=2b=2 or 77.
Since aa is now free to take any digit from 0 through 9, abab can have multiple values.

We also have that abab is divisible by 13.
The multiples of 13 are 13,26,39,52,65,7813,26,39,52,65,78 and 491491.

Among these, the only number ending with 2 or 7 is 52.

Hence, ab=52

Answer: A

Every piece must weigh the same and be as heavy as possible, which is exactly the highest common factor of the three coins:

\[ 780 = 2^{2}\cdot3\cdot5\cdot13, \quad 840 = 2^{3}\cdot3\cdot5\cdot7, \quad 960 = 2^{6}\cdot3\cdot5 \]

\[ \operatorname{HCF} = 2^{2}\cdot3\cdot5 = 60 \text{ g} \]

Counting the pieces each coin yields:

\[ \frac{780}{60} + \frac{840}{60} + \frac{960}{60} = 13 + 14 + 16 = 43 \]

Each piece is shared by two people, so

\[ 43 \times 2 = 86 \text{ persons} \]

43 is the trap for stopping at the piece count and forgetting the sharing.

Answer: A

Any prime number greater than 3 will be in the form of 6x+1 or 6x−1.

Thus, both prime number are twins:

Let first be 6x−1
and 2nd be 6x+1
Sum=12x

Thus it is always divisible by 12.

Answer: C

Let the divisor be dd. 

When 242 is divided by the divisor, let the quotient be 'xx' and we know that the remainder is 8.
Therefore, 242=xd+8242=xd+8

Similarly, let yy be the quotient when 698 is divided by dd.
Then, 698=yd+9698=yd+9.

242+698=940=xd+yd+8+9242+698=940=xd+yd+8+9
940=xd+yd+17940=xd+yd+17

As xd and ydxd and yd are divisible by dd, the remainder when 940 is divided by dd should have been 17.

However, as the question states that the remainder is 4, it would be possible only when 17/d leaves a remainder of 4.

If the remainder obtained is 4 when 17 is divided by dd, then dd has to be 13

Answer: C

From 100 to 200 there are 101 numbers.
There are 100, 1's in the hundred place.
10, 1's in tens place and10, 1's in unit place.

Thus, the answer is 100+10+10=100+10+10= 120.

Answer: D

Given statement is like \(\left ( a - b \right )^{2} + 4ab\) where \(a=64\) and \(b=12\)
\(\left ( a - b \right )^{2} + 4ab\)
\(=\left (a^{2} - 2ab+b^{2}\right )+4ab\)
\(=a^{2}+2ab+b^{2}\)
\(=\left ( a + b \right )^{2}\)
Hence,
\(\left ( 64 - 12 \right )^{2} + 4 × 64 × 12\)
\(=\left ( 64 + 12 \right )^{2}\)
\(=76^{2}\)
\(=5776\)