# QUESTIONS ON SIMPLIFICATION (PART-I)

## QUESTIONS ON SIMPLIFICATION (PART-I)

#### QUERY 1

**If 2x + 2/x = 1, then find the value of x³ + 1/x³**

A) -8/11

B) -11/8

C) 3/2

D) 2/3

**JAYANTCHARAN CHARAN**

Whenever such type of question is asked …..first make it simple by converting the equation like: x + 1/x = 1/2

Hence 2x + 2/x = 1

=> x + 1/x = 1/2

Cubing it on both sides: (x + 1/x)**³** = (1/2)**³**

=> x**³** + (1/x)**³** + 3*x*(1/x)(x+1/x) = (1/2)**³**

=> x**³** + (1/x)**³** = 1/8 – 3/2

Hence the expression to be find = -11/8 (option ‘A’)

#### QUERY 2

** If a² = 2, then find the value of a+1**

A) a – 1

B) 2/(a -1)

C) (a + 1)/(3 – 2a)

D) (a – 1)/(3 – 2a)

**JAYANTCHARAN CHARAN
**Just put a+1 equal to all the options one by one till the result is a² = 2

a) a+1 = a - 1 impossible

b) a+1 = 2/(a – 1)

=> a² - 1 = 2

=> a² = 3 (no solution)

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

=> a = 3/2 (no solution)

We see first 3 solutions are not there; so last one (option ‘D’) is the desired option.

To me it seems the fastest method in such a question.

#### QUERY 3

**If 1.5x = 0.04y, then the value of (y² – x²)/(y² + 2xy + x²)**

A) 730/77

B) 73/77

C) 73/770

D) 74/77

**Mukesh Kumar Bansal
**1.5 X = 0.04 Y

Therefore, x/y = 2/75

The given expression = (y+x)(y-x)/(y+x)2

= (y – x)/(y + x)

= (1 – x/y)/(1 + x/y)

put the value of x/y=2/75

= (1 – 2/75)/(1 + 2/75)

= 73/75*75/77

= 73/77 (option ‘B’) Ans

#### QUERY 4

**If x² + y² – 2x + 6y + 10 = 0, then the value of x² + y² is?**

A) 5

B) 10

C) 15

D) 12

**PRAKASH G**

x² + y² – 2x + 6y + 10 = 0

=> (9x² – 2x + 1) + (y² + 6y + 9) = 0

=> (x – 1)² + (y + 3)² = 0

Here we see that both the term oh LHS are square of a number, hence both are positive. We also know that if sum of the two positive numbers is zero both are zero.

Therefore here both (x – 1)² = 0 and (y + 3)² = 0

=> x=1 , y=-3

Hence x² + y² = 10 (option ‘B’)

#### QUERY 5

**If x ^{4} + 1/x^{4} = 119; then x^{3} – 1/x^{3} = ?**

A) 11

B) 63

C) 36

D) 30

**RONNIE BANSAL
**x

^{4}+ 1/x

^{4}= 119

=> (x

^{2}+ 1/x

^{2})

^{2}– 2 = 119

=> (x

^{2}+ 1/x

^{2})² = 121

=> x

^{2}+ 1/x

^{2}= 11

=> (x – 1/x)

^{ 2}+ 2 = 11

=> (x – 1/x)

^{ 2}= 9

=> x – 1/x = 3

Now to be valued:

x^{3} – 1/x^{3} = ?

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

=> 3³ + 3*3

=> 27 + 9

36 (option ‘C’)

**CHANDAN DUTTA on the above solution**

How much time do you allot to one to solve and answer this question ?

If I were you I would allot maximum ONE Minute. Many steps one has to pass through mental calculation to get the answer looking to the constrain of allotted time.

**RONNIE BANSAL**

Of course such questions are to be solved within seconds. But here the steps are only to make the things clear; this doesn mean one has to go every step. Sometimes we have to solve a sum with the help of given answer options; sometimes other techniques have to be kept in mind. So many things are there actually. Also the net has it’s own limitations.

#### QUERY 6

**If X : Y=5 : 6, then (3x² – 2y²) : (y² – X²) is?**

A) 11:3

B) 3:11

C) 6:7

D) 1:2

**RONNIE BANSAL**

From the ratio of X and Y if X = 5 then Y = 6

Now putting these value of X and Y in in the other expression of ratios [3*5² – 2*6²] : (6² – 5²)

=> (75 – 72) : (36 – 25)

=> 3 : 11 (option ‘B)

#### QUERY 7

**If a : b : c = 2 : 3 : 4 and 2a – 3b + 4c = 33, then value of c is**

A) 6

B) 9

C) 12

D) 66/7

**JAYANTCHARAN CHARAN**

Put a, b, c as 2k, 3k, 4k in the given equation

Then 11k = 33

=> k = 3

=> But ‘c’ = 4k

=> ‘c’ = 4 x 3 = 12 (option ‘C’)

#### QUERY 8

**If x + y + z = 1, xy + yz + zx = -1, xyz = -1, then what is the value of x³ + y³ + z³ = ?**

A) 0

B) -1

C) -2

D) 1

**MADHAV JHA**

(x+y+z)² = x² + y² + z² + 2(xy+yz+zx)

=> x² + y² + z² = 3

we know,

x³ + y² + z² – 3xyz = (x+y+z) (x² + y² + z² – xy -yz -zx)

Now just put all the given & derived values, we get:

x² + y² + z² = 1 (option ‘D’)

TRICK

**Ronnie Bansal**

By looking at the answer option we can easily derive that values of x, y, & z must be in integers. Now xyz = -1 (given) ; means none of them can be ‘0’; and either ‘all three are -1’ or ‘only one of them is -1 and two are 1 each’. If we see x² + y² + z² it’s not possible that x, y, z all are -1. So two of them is 1 each and the third is -1. Therefore answer 1 is possible only. (option ‘D’)

#### QUERY 9

**If x ^{1/3} + y^{1/3} = z^{1/3}; then (x + y – z)³ + 27xyz is equal to?**

A) 0

B) 1

C) 2

D) 3

**SANDES….. HIGHLY INFLAMMABLE
**The given equation = x

^{1/3}+ y

^{1/3}= z

^{1/3}

Cubing both sides

(x^{1/3} + y^{1/3})^{3} = z

=> x + y + 3(x^{2/3})(y^{1/3}) + 3(x^{1/3})(y^{2/3}) = z

=> x + y + 3[(x^{2/3})(y^{1/3}) + (x^{1/3})( y^{2/3})] = z

=> x + y + 3(x^{1/3})(y^{1/3})(x^{1/3} + y^{1/3}) = z

putting x^{1/3} + y^{1/3} = z^{1/3}

=> x + y+ 3(x^{1/3})(y^{1/3})(z^{1/3}) = z

=> x + y – z = -3(x^{1/3})(y^{1/3})(z^{1/3})

=> (x + y – z )^{3} = -27 xyz (cubing both sides)

=> (x + y – z)^{3} + 27xyz = 0 (option ‘A’)

#### QUERY 10

**x² + y² + 1/x² + 1/y² = 4 then the value of x² + y² is?**

A) 4

B) 3

C) 1

D) 2

**Hemant Singh
**The given equation is = (x

**²**+ 1/x

**²**) + (y

**²**+ 1/y

**²**) = 4

=> [(x – 1/x)**²** + 2] + [(y – 1/y)**²** + 2] = 4

=> (x – 1/x)**²** + (y – 1/y)**²** + 4 = 4

=> (x – 1/x)**²** + (y – 1/y)**²** = 0

We see that both the left hand terms are positive and their sum is equal to zero

So (x – 1/x) = 0 and (y – 1/y) = 0

Solving both x**²** = 1 and y**²** = 1

Hence x**²** + y**²** = 2 (option ‘D’)