# QUESTIONS ON SIMPLIFICATION (PART-VIII)

## QUESTIONS ON SIMPLIFICATION (PART-VIII)

#### QUERY 71

16.16 × 16.004  ×2.2 = ?

A) 2.856
B) 2.586
C) 3.856
D) 4

MAHA GUPTA
If you see every base of the question can be converted into base 2
So, the given expression = 24*.16 × 24*.004 × 2.2
= 2.64 × 2.016 ×2.2
= 2(.64+.016+.2)
= 2.856 (option ‘A’)

#### QUERY 72

If a² + 1 = a; then the value of a12 + a6 + 1 is ?

A -3
B) 1
C) 2
D) 3

MAHA GUPTA
a2 +1 = a
=> a2 – a + 1 = 0
We know a3 + b3 = (a + b)(a2 – ab + b2)
So a3 + 1 = (a+1)(a2 – a + 1)

But a2 – a + 1 = 0 (shown above)
Therefore a+1= 0
=> a = -1

Now putting this in a12 + a6 + 1
1 + 1 + 1 = 3 (option ‘D’)

NOTE: as (a = -1) does not hold true if it’s put in the original equation the question seems to be faulty in a way.

#### QUERY 73

The minimum value of (x-2)(x-9) is?

A) -11/4
B) 49/4
C) 0
D) -49/4

MAHA GUPTA
(x – 2)(x – 9) = x² – 11x + 18
We see that this expression is a quadratic one.

Note that in a quadratic expression when the coefficient of x² is greater than 0 the minimum value of it is (-D/4a) and maximum is infinitive & when the coefficient of x² is smaller than 0 the minimum value of it is infinitive and maximum is ( -D/4a)

Here, as the coefficient of x² is greater than 0, the minimum value of it
= ( -D/4a)
= -(b² – 4ac)/4a
= -(121 – 4*1*18)/4*1
= -49/4 (option ‘D’)

#### QUERY 74

If (x + 7954 × 7956) be a square number, then the value of ‘x’ is

A) 1
B) 16
C) 9
D) 4

MAHA GUPTA
Square of any number = (previous number × next number) +1
So 1 (option ‘A’) is the answer

#### QUERY 75

If a, b, c are non-zero, a+1/b = 1 and b+1/c = 1, then the value of abc is ?

A) -3
B) 1
C) -1
D) 3

MAHA GUPTA
From the given equations
ab +1 = b ————————i)
bc + 1 = c————————ii)

Now by multiplying both
ab²c + bc + ab + 1 = bc
=> ab²c = -(ab + 1)

Dividing both sides by ‘b’
abc = -(a + 1/b)
=> abc = -1       (Putting by a + 1/b = 1) [option ‘B’]

TRICK
Put b = 2 in the first equation ………… obviously ‘a’ will be equal to 1/2. ‘c’ will get its value by putting the value taken of ‘b’ in the second equation i.e. it will be -1. In fact we can take any value of ‘b’ except ‘0’ and 1 to get relative values of ‘a’ and ‘c’. Put them in abc to get its value which will be -1 in each case.

#### QUERY 76 Then find x² + x = ?

A) o
B) 1
C) 10
D) 15

KUMAR SAURABH
The given equation can be re-written

x(x+1)(x-1)(x+2)+1=0
=> (x² + x)(x² + x – 2) + 1 = 0
Now, let x² + x = m
By putting this in the equation given above
m(m – 2) + 1 = 0
=> m² – 2m + 1 = 0
From above m = 1
But m = x² + x also (assumed above)
Therefore x² + x = 1 (option ‘B’)

#### QUERY 77

If x = 4ab/(a+b), then value of (x+2a)/(x-2a) + (x+2b/x-2b)

A) a
B) b
C) 2ab
D) 2

RONNIE BANSAL
(x+2a)/(x-2a) + (x+2b/x-2b)
= x/2a + x/2b (by componendo & dividendo)

Now substituting x = 4ab/(a+b) in the above
[4ab/(a+b)]/2a + [4ab/(a+b)]/2b

= 2b/(a+b) + 2a/(a+b)

= 2(a+b)/(a+b)

=2 (option ‘D’)

#### QUERY 78

If x² + y² + z² = xy + yz + zx, (x is not equal to zero), then 4x + 2y – 9z / 2x = ?

A) 0
B) 3/2
C) 1
D) 1/2

By looking at x² + y² + z² = xy + yz + zx it can easily be derived that only and only values of x, y & z can be 1 each. So by putting this in the question 3/2 is correct. (option ‘B’)

#### QUERY 79

If A and B are the HCF and LCM of two algebraic expressions x and y, and A+B = x+y, then the value of A³ + B³ is

A) y³
B) x³ + y³
C) x³ – y³
D) x³

abhishek bhardwaj
We know the multiplication of two numbers is equal to the multiplication of their HCF and LCM
Therefore A*B = x*y
In other words AB = xy

Now given A+B = x+y
As we have to find the value of sum of the cubes of the numbers we must find the whole cube of the sum of those numbers to get it.

Therefore (A+B)³ = (x+y)³
=> A³ + B³ + 3AB(A+B) = x³ + y³ + 3xy(x + y)
=> A³ + B³ + 3xy(x + y) = x³ + y³ + 3xy(x + y)
=> A³ + B³ = x³ + y³ (option ‘B’)

#### QUERY 80

x + 1/x = 3, then x³ + 1/x³ = ?

A) 9
B) 27
C) 18
D) 6

Lokesh Shravan
Given x + 1/x = 3
cube both side
x³ + 1/x³ + 3(x + 1/x) = 27                                     [using (a+b)³ = a³ + b³ + 3ab(a+b)]
=> x³ + 1/x + 3(3) = 27
=> x³ + 1/x³ = 18 (option ‘C’)

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