# 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 = 2^{4*.16 }× 2^{4*.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 a ^{12} + a^{6} + 1 is ?**

A -3

B) 1

C) 2

D) 3

**MAHA GUPTA**

a^{2} +1 = a

=> a^{2} – a + 1 = 0

We know a^{3} + b^{3} = (a + b)(a^{2} – ab + b^{2})

So a^{3} + 1 = (a+1)(a^{2} – a + 1)

But a^{2} – a + 1 = 0 (shown above)

Therefore a+1= 0

=> a = -1

Now putting this in a^{12} + a^{6} + 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

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

**MADHAV JHA**

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’)