Number System - Olympiad Level MCQ, Class 9 Mathematics - Class 9 MCQ

⇒ (73 x 75 x 78 x 57 x197 x 37) / 34

⇒ (5 x 7 x 10 x 23 x 27 x 3) / 34

[We have taken individual remainder, which means if 73 is divided by 34 individually, it will give remainder 5, 75 divided 34 gives remainder 7 and so on.]

⇒ (5 x 7 x 10 x 23 x 27 x 3)/34

⇒ (35 x 30 x 23 x 27) / 34 [Number Multiplied]
⇒ (35 x 30 x 23 x 27) / 34

⇒ (1 x -4 x -11 x -7) / 34

[We have taken here negative as well as positive remainder at the same time. When 30 divided by 34 it will give either positive remainder 30 or negative remainder -4. We can use any one of negative or positive remainder at any time.]

⇒ (1 x -4 x -11 x  -7) / 34

⇒ (28 x -11) / 34

⇒ (-6 x -11) / 34

⇒  66 / 34

⇒ R = 32.

Required remainder = 32.

(3x+ 1)^1/4 = 2

=> 3x+ 1 = (2²)²

=> 3x+ 1 = 16

=> 3x = 16-1

=> 3x = 15

=> x = 5

(256)^0.16 x (256)^0.09 = (256)^{(0.16 + 0.09)}

   = (256)^0.25

   = (256)^(25/100)

   = (256)^(1/4)

   = (4⁴)^(1/4)

   = 4^4(1/4)

   = 4¹

   = 4

(√2)² = 2 which is a rational number

(4x -7 )^1/3 -5 =0

(4x -7)^1/3 = 5

4x -7 = 5³

4x -7 =125

4x = 125 +7

4x = 132 

x = 132/4

 x = 33

In proper fraction, the  numerator is greater than it's denominator whereas in improper fraction denominator is greater than numerator.

1421 = 118*12 + 5 = 12k + 5, where k = 118.

1423 = 12k + 7

1425 = 12k + 9

N = (12k + 5)(12 k + 7)(12k + 9)

When N is divided by 12, the remainder is same as the remainder, when 5 * 7 * 9 = 315 is divided by 12.

5 * 7 * 9 = 35 * 9 = 315 

315 = 312 + 3 = 12 * 26 + 3 

The remainder is 3.

We have to find the Least number, therefore we find out the LCM of 8, 12, 16 and 20.
8 = 2*2*2;
12 = 2*2*3;
16 = 2*2*2*2;
20 = 2*2*5;
LCM = 2*2*2*2*3*5 = 240;
This is the least number which is exactly divisible by 8, 12, 16 and 20.
Thus,
required number which leaves remainder 5 is,
240+5 = 245.

1 * 2³ + 0 * 2² + 1 * 2¹ + 0 * 2^-1 +1 * 2^-2 + 1 * 2^-3 = 11.375
Hence, (1011.011)10 = 11.375

LHS = 1/[1 + p + q⁻¹] + 1/[1 + q + r⁻¹] + 1/[1 + r + p⁻¹] 

1/[1 + p + q⁻¹] = 1/[1 + p + 1/q] = q/ [q + pq + 1] -------(1)

1/[1 + q + r⁻¹] = 1/[1 + q + 1/r ]
= 1/[1 + q + pqr/r ] {from pqr = 1}
= 1/[1 + q + pq] = 1/[q + pq + 1] --------(2)

1/[1 + r + p⁻¹] = pqr/[pqr + r + pqr/p] 
= pqr/[pqr + r + qr] = pq/[pq + 1 + q] ------(3)

Add equations (1), (2) and (3) 
q[1 + pq + q] + 1/[q + pq + 1] + PQ/[pq + 1 + q] 
= [1 + pq + q]/[1 + pq + q] = 1 
Hence , LHS = RHS proved

The two digit number which gives a remainder of 3 when divided by 7 are:

10, 17, 24 94.

Now, these number are in AP series with 

1st Term, a = 10; 

Number of Terms, n = 13;

Last term, L = 94 and

Common Difference, d = 7.

Sum,

= n*(a+L)/2

= 13*52 = 676.

Let 2^x = 3y = 6 ^-z = k
⇒ 2^x = k

Similarly, 

From the following, we need to find the values of x,y, and z:

a x =b , b y =c , c z =a

From ax =  b, we get x  =b/a -------------------------- (i)
From by = c, we get y  =c/b ---------------------------(ii)
From cz = a, we get z = c/a ---------------------------(iii)

Multiplying the left hand sides and right hand sides of (i), (ii) and (iii) we get
xyz=(b/a)*(c/b)*(c/a)

When we simplify, we find that xyz=1. Hence proved.

Let the required numbers be x, 2x and 3x. Then, their H.C.F. = x. So, x = 12  ∴  The numbers are 12, 24 and 36  

Let's say the following polynomial be f(x) .

f(x) = | 17x - 8 | - 9 .

We know that Modulus value can be 0 at least and at most positive infinite. 

So, | 17x - 8 | = 0

17x - 8 = 0 
x = 8/17 .

Now, Finding 

f min ( x) = | 17(8/17)-8 | - 9 
= 0 - 9 
= -9.

Therefore, The minimum value of | 17x - 8 | - 9 is -9.

The correct option is Option A.

2a - 9 = b + a 

=> 2a - a = b + 9 

=> a = b + 9 

Now, |a-b| + |b-a|

= √a² + b² +√b² + a²

= √(b+9)² + b² +√b² + (b+9)²

= b + 9 + b + b + b + 9

= 4b +18

On multiplying the decimal number continuously by 2, the binary equivalent is obtained by the collection of the integer part. However, if it’s an integer, then it’s binary equivalent is determined by dividing the number by 2 and collecting the remainders.