Test: Number System - 1 - Electronics and Communication Engineering (ECE) MCQ

Let the numerator of the fraction be x, so the denominator will be 11 - x

So, the fraction = x/11 - x

On adding 2 in both numerator and denominator, according to the question, the fraction will be -

(x + 2)/(11 - x + 2) = (x)/(11 - x) + 1/24

(x + 2)/(13 - x) - (x)/(11 - x) = 1/24

[11x + 22 - x² -2x - 13x + x²]/(13 - x) (11 - x) = 1/24

After solving the above equation, we will get

=> 528 - 96x = 143 - 24x + x²

x² + 72x - 385 = 0

(x + 77) (x - 5) = 0

So, x =5

So, numerator = 5

and, denominator = 11 - 5 = 6

Difference between both is = 6 - 5 = 1

The required number = Increase in result/(wrong multiplier - correct multiplier)

= 200/45 - 25

= 10

Let the numerator be 'a' so, denominator will be 'a + 3'.

According to the question,

(a + 7)/ ((a + 3) - 2) = 2/1

(a + 7)/(a + 1) = 2/1

2a + 2 = a + 7

=> a = 5

So, the fraction is a/a + 3 = 5/ 5 + 3

= 5/8

And the sum of the numerator and denominator of the fraction is 13.

 Let the numbers be a and b.

So, according to the question,

a * 3/5 = 70% of b

3a/5 = 70b/100

a/b = (70 * 5)/(100 * 3) = 7/6

So, the ration between a and b is 7 : 6.

Suppose a = (x - y), b = (y - z), and c = (z - x)

On adding a, b, and c we will get

a + b + c = x -y + y - z + z - x

=> a + b + c =0

So, a³ + b³ + c³ = 3abc [because if a + b + c = 0, then a³ + b³ + c³ = 3abc]

We can say that (x - y)³ + (y - z)³ + (z - x)³ = 3 (x - y) (y - z) (z - x)

Suppose the three consecutive positive numbers be x, x + 1, and x + 2

According to the question,

(x)² + (x + 1)² + (x + 2)² = 365

On expanding, we will get,

x² + x² + 1 + 2x + x² + 4 + 4x = 365

3x² + 6x = 360

Or, x² + 2x - 120 = 0

=> (x - 10) (x + 12) = 0

So, x = 10

First number x = 10

Second number x + 1 = 11

Third number x + 2 = 12

So, sum of the numbers is = 10 + 11 + 12 = 33

Given x = a(b - c), y = b(c - a), z = c(a - b)

x = a(b - c)

or, x/a = (b - c) …..(i)

y = b(c - a)

or, y/b = (c - a) …..(ii)

z = c(a - b)

or, z/c = (a - b) …..(iii)

Add equation (i), (ii), and (iii), we will get

x/a + y/b + z/c = b - c + c - a + a - b

x/a + y/b + z/c = 0

So, (x/a)³ + (y/b)³ + (z/c)³ = 3 (x/a) * (y/b) * (z/c) [because if a + b + c = 0, then a³ + b³ + c³ = 3abc]

= 3xyz/abc

Given 1397 x 1397

We can also write it as (1397)²

Or, (1400 - 3)²

On expanding we get,

= (1400)² + (3)² - (2 x 1400 x 3)

= 1960000 + 9 - 8400

= 1960009 - 8400

= 1951609

Here, we can apply the formula of a² - b² = (a + b) (a - b)

So, after applying the formula, we will get

(64 + 36) (64 - 36) = 20 * x

100 * 28 = 20 * x

x = 100 * 28/ 20 = 140

We can write x⁶ / x³ as,

x^{6 - 3} = x³