Number System Questions and Answers

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: 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: 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: A

Factorise the expression.

\[ x^{2}(x^{2}-1) = (x-1)\,x\,x\,(x+1) \]

\( (x-1)x(x+1) \) is a product of three consecutive integers, so it is divisible by both 2 and 3, hence by 6; the extra factor of \( x \) supplies a second factor of 2 whenever \( x \) is even, and when \( x \) is odd both neighbours are even.

Either way the product is divisible by 12. It is not always divisible by 24: \( x = 2 \) gives 12.

Answer: D

Work outward from the second statement.

\[ b = 3 \times 2 + 1 = 7 \]

\[ a = 3b + 2 = 3 \times 7 + 2 = 23 \]

Checking: \( 23 \div 3 = 7 \) remainder 2, and \( 7 \div 3 = 2 \) remainder 1.

Answer: D

\(x+(x+36)+y=100 \\ \Rightarrow 2x+y=64\)
Therefore \(y\) must be even prime, which is 2.
Therefore,
\(2x+2=64 \\ \Rightarrow x=31\)
Third prime number
\(= (x+36)\\= (31+36)\\= 67\)

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: A

The last possible card number is 996, and the useful observation is that this is just short of the modulus:

\[ 996 \equiv -1 \pmod{997} \]

The rule Pinaki wrote down says the product leaves remainder 123, so with \( P \) for the PIN

\[ 996 \times P \equiv 123 \pmod{997} \quad\Longrightarrow\quad -P \equiv 123 \pmod{997} \]

\[ P \equiv -123 \equiv 997 - 123 = 874 \pmod{997} \]

Check: \( 996 \times 874 = 870{,}504 \) and \( 870{,}504 = 997 \times 873 + 123 \). Replacing 996 by \( -1 \) turns a six-figure multiplication into a subtraction.

Answer: D

Ten cigarettes give 10 stub.

From 10 stubs 3 more cigarettes can be made (2 stubs would be obtained from 2 cigarettes formed by joining 8 stubs.)

So he could smoke 13 cigarettes in all.

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