Number System Questions and Answers

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,
\(\left ( 64 - 12 \right )^{2} + 4 × 64 × 12\)
\(=\left ( 64 + 12 \right )^{2}\)
\(=76^{2}\)
\(=5776\)

Answer: B

Write the numerator as a power of ten.

\[ (1000)^{9} = (10^{3})^{9} = 10^{27} \]

\[ \frac{10^{27}}{10^{24}} = 10^{3} = 1000 \]

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

A number is divisible by 11, if the difference between the sum of digits at even places and odd places is either 0 or divisible by 11.

Numbers 5,3,6,4,7 is a multiple of 11

∵(5+6+7)−(3+4)=11, which is a multiple of 11

The number at odd places 5,6,7 can be arranged in 3! ways and 3,4 can be arranged in 2! ways.

Therefore, the total number of ways of such numbers =3!×2!=12

Answer: A

Since xx is the product of four consecutive integers, it is always divisible by 4, i,.e., it is always even. So, 1+x is always odd.

n=1+x

x=(y−1)(y)(y+1)(y+2)=y(y2−1)(y+2)=(y3−y)(y+2)=y4+2y3−y2−2y

⇒1+x=y4+2y3−y2−2y+1=y4+y2+1+2y3−2y2−2y=(y2+y−1)2

So, 1+x is a perfect square as we can see. Hence, option A is the correct choice.

Answer: D

A perfect square never ends in 2, 3, 7 or 8.

\[ 143642 \text{ ends in } 2 \]

so it cannot be a square. The last digit of a square is always 0, 1, 4, 5, 6 or 9 - a single glance settles this without any arithmetic.

Answer: D

Count the 8s by the place they occupy.

Units digit 8 - the numbers 268, 278, …, 488:

\[ \frac{488 - 268}{10} + 1 = 23 \]

Tens digit 8 - the blocks 280–289, 380–389 and 480–489:

\[ 3 \times 10 = 30 \]

No number in the range has 8 in the hundreds place, since the hundreds digits are only 2, 3 and 4.

\[ 23 + 30 = 53 \]

288 and 388 and 488 are each counted twice - once for each 8 - which is correct, since the question asks how many 8s are written.

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: B

We are given that x=2×3×7×a=42ax=2×3×7×a=42a and y=2×2×8×b=32by=2×2×8×b=32b

We are given that the values of both xx and yy lie between 120 and 130 (not including the two).

The only multiple of 42 in this range is 42×3=12642×3=126.

Hence, x=126x=126 and a=3.a=3.
The only multiple of 32 in this range is 32×4=12832×4=128.

Hence, y=128y=128 and b=4b=4.

Hence, a−b=3−4=a−b=3−4= -1

Answer: A

From \( 7! \) onward every factorial contains the factor 7, so all of them leave remainder 0.

Only the first six terms matter:

\[ 1 + 2 + 6 + 24 + 120 + 720 = 873 \]

\[ 873 = 7 \times 124 + 5 \]

So the remainder is 5.