Number System Questions and Answers

Answer: C

Let the number be \( 10t + u \), with \( t \) the tens digit and \( u \) the units digit. Adding 75 per cent of it multiplies it by 1.75, and the result is the reversed number:

\[ 1.75\left(10t + u\right) = 10u + t \]

\[ 17.5t + 1.75u = 10u + t \]

\[ 16.5t = 8.25u \quad\Longrightarrow\quad \frac{u}{t} = \frac{16.5}{8.25} = 2 \]

So the units digit is twice the tens digit, a ratio of 2 : 1. The smallest number fitting is 12, and indeed \( 12 + 0.75(12) = 21 \).

Read the order carefully: the question asks unit to ten, so 1 : 2 is the same relationship stated backwards.

Answer: C

We need to find sum of S of the digits of PP.
Till 999 each of the digits from 1 to 9 occurs equal number of times.

Total digit of all the numbers from 1 to 999 will be multiple of ∑9=45.

Hence,

S=45k+(1+0+0+0)
⇒ Required remainder = 1.

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

Only the multiples of 3 in the set contribute, and each contributes as many threes as it contains.

Odd numbers from 1 to 55. The odd multiples of 3 are 3, 9, 15, 21, 27, 33, 39, 45, 51, carrying exponents 1, 2, 1, 1, 3, 1, 1, 2, 1:

\[ 1+2+1+1+3+1+1+2+1 = 13 \]

Even numbers from 56 to 150. These are the multiples of 6 in the range, 60 through 150, sixteen of them, with exponents 1, 1, 2, 1, 1, 2, 1, 1, 3, 1, 1, 2, 1, 1, 2, 1:

\[ 1+1+2+1+1+2+1+1+3+1+1+2+1+1+2+1 = 22 \]

\[ \therefore\; \text{index} = 13 + 22 = 35 \]

Note that 27 and 108 each supply three, and 9, 45, 72, 90, 126 and 144 each supply two; counting one per multiple would give 25 and miss the higher powers.

Answer: D

Given 0 x and y are either both positive or both negative.

Also given (x,y) is not equal to (y,x).
If x=1, y can take values from 1 to 99
⇒ we have 99×2=198 pairs but (1,1) is repeated
Thus can take 198−1=197 pairs.

If x=2 , yy can take values from 2 to 49
[(2,1) and (1,2) are also covered in 197 pairs above].
⇒ 48×2−1=95 pairs [(2,2) is repeated]

Similarly,
If x=3 or y=3 we have 61 pairs
If x=4 or y=4 we have 41 pairs
If x=5 or y=5 we have 29 pairs
If x=6 or y=6 we have 21 pairs
If x=7 or y=7 we have 15 pairs
If x=8 or y=8 we have 9 pairs
If x=9 or y=9 we have 5 pairs

We have total 473 pairs when x and y are positive.
We will have 473 pairs when a and b are negative.
⇒ We have a total of 946 ordered pairs.

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: 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.