# QUESTIONS ON SIMPLIFICATION (PART-VI)

## QUESTIONS ON SIMPLIFICATION (PART-VI)

#### QUERY 51

**If x + y + z = 7 and xy + yz + zx = 10; find the maximum value of x?**

A) 2

B) 7

C) √29

D) 5

**MAHA GUPTA**

x + y + z = 7

Squaring both sides

(x + y + z)² = (7)²

=> x² + y² + z² + 2xy + 2yz +2zx = 49

=> x² + y² + z² +2(xy + yz + zx) = 49

=> x² + y² + z² +2(10) = 49 —-Replacing given value of xy + yz + zx

=> x² + y² + z² = 29

To know the maximum value of x², we must know the minimum values of y² and z². As both of them are in squares, they must not be less than ‘0’ each. So minimum value of y² and z² is ‘0’ each.

So, for maximum value, x²= 29

=> x = √29 or x < √29

But √29 is not satisfying the given equations, therefore x must be smaller than this.

We see only 5 among the given options satisfies this, so the answer (option ‘D’)

SHORT

Try options you’ll get 5 only

QUERY 52

**If x – 5 < 1; which values of x will satisfied this equation**

A) less than 5

B) greater than 5

C) only 5

D) none

**MAHA GUPTA**

x – 5 < 1

=> x < 1 + 5

=> x < 6

Therefore NONE (option ‘D’) is correct. All other options are incorrect.

QUERY 53

**MAHA GUPTA
** The given expression is (1/4)*(2/6)*(3/8)*(4/10)*…*(31/64)

We see that numerator of every fraction is exactly dividing its previous denominator by 2 here, and the total terms are 31

Thus, the expression reduces to

(1/2)*(1/2)*(1/2)*(1/2)*…*(1/64), where 1/2 is 30 times

So the above expression = (1/2^{30})*(1/2^{6})

= 1/2^{36}

So comparing this to 1/2^{x}; x = 36 (option ‘C’)

QUERY 54

**If x + 1/x = √3; then x ^{18} + x^{24} = ?**

A) 0

B) 1

C) -2

D) 2

**MAHA GUPTA
**x + 1/x = √3 (given)

Cubing both sides

(x^{3} + 1/x^{3}) + 3(x + 1/x) = 3√3

=> (x^{3} + 1/x^{3}) + 3√3 = 3√3 (putting x + 1/x = √3 as given)

=> x^{3} + 1/x^{3} = 0

Now x^{18} + x^{24} = x^{(21 – 3)} + x^{(21 + 3)}

= (x^{21})/x^{3} + (x^{21})*x^{3}

= (x^{21})*(x^{3} + 1/x^{3})

= (x^{21})*0 —-putting x^{3} + 1/x^{3} = 0

= 0 (option ‘A’)

QUERY 55

**SUBHASH CHANDRA
** The given equation

Now let a – b + c = k, where k ≠ 0

Thus, b – c = a – k

a + c = b + k

a – b = k – c

Now putting these values in the given equation

=> (1 – k/a) + (1 + k/b) + (-1 + k/c) = 1

=> -k/a + k/b) + k/c = 0

=> k(-1/a + 1/b + 1/c) = 0

As k is not zero

therefore, -1/a + 1/b + 1/c = 0

=> 1/a = 1/b + 1/c (option ‘D’)

SHORT

**Natalia Romanova**

As a – b + c is not equal to 0, you can assume a =1, b = 2, c = 2 and check the options which one fits according to these values

Of course it is 1/a = 1/b = 1/c (option ‘D’)

QUERY 56

(Option ‘C’)

#### QUERY 57

**MAHA GUPTA
** We know that if a³ + b³ + c³ = 3abc, then a + b + c = 0

=> a³ + b³ + c³ – 3abc 0, then a + b + c = 0

Now by conversion according to the given equation

a³ – b³ – c³ – 3abc = 0, then a + (-b) + (-c) = 0

From the second equation above

a = b + c (option ‘D’)

#### QUERY 58

**MAHA GUPTA
** Such questions are generally easily done imagining a numerical value, Better to take that value the smallest one in whole numbers possibly. Here you can’t take 0 as it will be spoiling everything, not 1 also as it will be giving 3 correct options. So, let’s take it 2 for each a, b and c

Now, putting a = b = c = 2 in the dividend given above we get 72

And the divisor that we get = 12

So the quotient = 72/12 = 6, which is possible in only option ‘D’

QUERY 59

**MAHA GUPTA
**The given expression (x

^{5}+ 1/x

^{5}) is in form of (x

^{n}+ y

^{n}). We see here n is odd, when such is the case the expression is always divisible by (x +y).

So, it’s divisible by (x + 1/x), means (x^{5} + 1/x^{5}) is a multiple of (x + 1/x). But (x + 1/x) = 3, so it’s a multiple of 3. Here 123 (option ‘D’) is the only number which is multiple of 3, so it’s our answer.

#### QUERY 60

**If a + b + c = 13. Find the maximum value of (a – 3)(b – 2)(c – 1)**

A) 27/343

B) 433/27

C) 344/27

D) 343/27

**KUMAR SAURABH**

First we convert the terms of the given equation as are in the find expression

Now, a + b + c = 13

=> (a – 3) + (b – 2) + (c- 1) = 13 – 6

=> (a – 3) + (b – 2) + (c – 1) = 7

Now terms are same, we know that the arithmetic mean of non-negative real numbers is greater than or equal to the geometric mean if terms/items of two things are same, and geometric mean of three numbers = the cubic root of their product

So by above, {(a – 3) + (b – 2) + (c – 1)}/3 > or = {(a – 3)(b – 2)(c – 1)}^{1/3}

=> 7/3 > or = {(a – 3)(b – 2)(c – 1)}^{1/3 }—putting (a – 3) + (b – 2) + (c – 1) = 7

=> {(a – 3)(b – 2)(c – 1)} < or = (7/3)³

Hence the maximum value of the find expression = (7/3)³ = 343/27 (option ‘D’)