InfyTQ Algebra Quiz 2

Correct answer: Option III

Solution:

Applying the formula = 2ab = (a² + b²) – (a-b)²

=> 58- 36 = 2ab

ab = \frac{22}{2} = 11

Solution:

Applying the remainder theorem,

A polynomial x is divided by (x-a), where ‘a’ is the real number, then the remainder will be calculated by the value of the polynomial at x= a

Hence, remainder = (2)⁵ - 3(2)³ + 10(2) - 20

= 32- 24 + 20 – 20 = 8

Correct answer:

Solution:

10a – 30 = 15 + a

9a = 15 + 30

a = \frac{45}{9} = 5

Correct answer: Option II

Solution:

20a + 20b + 18 = -20a + 20b -22

40a = -22 – 18

40 a = - 40

a = -1

Correct answer: Option III

Solution:

As stated in the question that,

4m + 6b = 1500 first equation

6m + 4b = 1400 second equation

Subtracting the equations we get,

-2m+ 2 b = 100,

2b – 2m = 100

b-m = 50

Correct answer: Option II

Solution:

Let the total number of stardust notes of each denomination be y

Then y + 10y + 20y = 589

= 31y = 589

= y = 19

Therefore, the total number of stardust notes = 3y = 57

Solution:

First of all we reciprocate the equation one to have a number value at the other side, which will be a² – 2a = -2

Putting this way we get,

= a² (a² – 2a) + 2a³ – a³ + 2a = a³- 2a² + 2a

By simplifying the equation using a²+ 2 = 2a

So we get,

= a(a²- 2a) + 2a = -2a + 2a = 0

So, the value will be 0

Correct answer: Option I

Solution: let the age of Karan and Ritu be a and b years for now

Then, a-b = 45 years ------------ first equation

And a- 10 = 10(b-10)

= a-10 = 10b- 100

= a – 10b = -100 + 10

= a – 10 b = -90------------- second equation

Solving, both the equation

Apply (1) – (2) that is

a-b= 45

a – 10b = -90

9b = 135

B = 15

Substitute b value in equation 1

• 15 = 45

Solution:

Let the number be a, then student did

\frac{(a + 24)}{12} = 224

a = 12 x 224 – 24

a = 2664

Then, putting in the equation,

=\frac{a}{12} + 24 =\frac{2664}{12} + 24

= 222+24 = 246

Solution:

Let the capacity of one container be a

Then, the capacity of reservoir = 50a

New capacity container = \frac{4}{10} of a

Therefore, required number of Containers = \frac{50a}{\frac{4a}{10}}

= (50a \times  \frac{10}{4a})

\frac{500}{4}

= 125