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: B

The marbles used to tile the floor are square marbles.

Therefore, the length of the marble=width of the marble.the length of the marble=width of the marble.

As we have to use whole number of marbles, the side of the square should a factor of both 5 m 78 cm and 3m 74. And it should be the highest factor of 5 m 78 cm and 3m 74.

5 m 78 cm = 578 cm and 3 m 74 cm = 374 cm.
The HCF of 578 and 374 = 34.

Hence, the side of the square is 34.

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

Count each digit's appearances across the 900 three-digit numbers.

Zeros cannot occupy the hundreds place:

\[ \text{tens place} = 9 \times 10 = 90, \qquad \text{units place} = 9 \times 10 = 90 \]

\[ a = 180 \]

Fives may occupy any place:

\[ 100 \ (\text{hundreds}) + 90 + 90 = 280 \]

The same count applies to eights, so \( b = c = 280 \).

\[ b + c - a = 280 + 280 - 180 = 380 \]

Zero is the odd digit out because a three-digit number cannot begin with it.

Answer: C

All the students from each have to be accommodated in a certain number of rooms.

There should be no person left over (remainder) from any places who can be clubbed together with the persons left over from other places.

To have the minimum number of rooms, the capacity of each room is HCF of all the numbers.
HCF(42,60,210,90,84)=6HCF(42,60,210,90,84)=6

Thus Min number of rooms =(42+60+210+90+84) / 6=81

Answer: C

Treat every number from 0000 to 9999 as a four-digit string with leading zeros. That is 10,000 strings and 40,000 digit positions, each digit equally likely:

\[ \frac{40{,}000}{10} = 4000 \]

The padding adds only zeros, so the count of 9s is unaffected, and 10,000 itself contains no 9. So the digit 9 is written 4000 times.

Answer: A

Since the given number is divisible by 5, so 0 or 5 must come in place of #.

But, a number ending with 5 is never divisible by 8.

So, 0 will replace #.

Now, the number formed by the last three digits is 4@0, which becomes divisible by 8, if @ is replaced by 4.

Hence, digits in place of @ and # are 4 and 0 respectively.

Answer: B

While typing numbers from 1 to 1000, you have 9 single digit numbers from 1 to 9. Each of them requires one keystroke. That is 9 key strokes.

There are 90 two-digit numbers, from 10 to 99. Each of these numbers requires 2 keystrokes. Therefore, one requires 180 keystrokes to type the 2 digit numbers.

There are 900 three-digit numbers, from 100 to 999. Each of these numbers requires 3 keystrokes. Therefore, one requires 2700 keystrokes to type these 3 digit numbers.

Then 1000 is a four-digit number which requires 4 keystrokes.

Totally, therefore, one requires 9+180+2700+4=9+180+2700+4= 2893 keystrokes.

Answer: A

If the number is divisible by 5 and 11 it must be divisible by 55.

The numbers are less than 660.

Hence, dividing 659 by 55 gives the number of multiples of 55 = 11 (ignoring fraction part).

The 11 multiples of 55 which are less than 560, but of these 11 multiples some can be multiples of 3.

The numbers of such, multiples is the quotient of 11 by 3.
Quotient of 11/3=3.
Out of 11 multiples of 55, 3 are multiples of 3.

Hence, numbers less than 660 and divisible by 5 and 11 but not by 3=11−3=8

Answer: D

List all possible factors xx and yy whose product is 36, and calculate the corresponding sum x+yx+y:

From the table, the minimum sum is 12