# QUESTIONS ON SIMPLIFICATION (PART-II)

## QUESTIONS ON SIMPLIFICATION (PART-II)

#### QUERY 11

**If x² + y² + z² + 2= 2(y – x), then what is the value of x³ + y³ + z³ = ?**

A) 0

B) 1

C) 2

D) 3

**MADHAV JHA
**In such question one needs to be very vigil; by looking at the whole question we see that any formula involving x

**³**+ y

**³**+ z

**³**is not possible here. So we need to think elsewhere. If we see the given equation, it can be converted in 3 term easy quadratic equation like this:

(x

**²**+ 1+ 2x) + (y

**²**+ 1 – 2y) + z

**²**= 0

=> (x+1)**²** + (y -1)**²** + z**²** = 0

Here we see that the left hand side is the sum of squares of three positive terms, and also equal to zero.

This is only possible when each of these terms is equal to zero.

so, x = -1, y = 1 & z = 0.

Now putting these values in x**³**+y**³**+z**³** we get 0 (option ‘A’) which is the answer

#### QUERY 12

**If x + y = 2z, then the value of x/(x – z) + z/(y – z)**

A) 1/2

B) 2

C) 1

D) 3

**RONNIE BANSAL**

We can write x+y = 2z as x – z = z – y

Therefore the second term of the question z/(y – z) = -z/(z – y) = -z/(x – z) [z – y = x – z (shown above)]

Hence x/(x – z) + z/(y – z) = x/(x – z) – z/(x – z)

=> (x – z)/(x – z)

=> 1. So option ‘C’ is correct

#### QUERY 13

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

A) 15

B) 20

C) 10

D) 12

**KANW@LJEET
**x² + y² – 2x + 6y + 10 = 0

=> (x² – 2x + 1) + (y² + 6y +9|

=> (x – 1)² + (y + 3)² = 0

=> x – 1 =0 and y + 3 =0

=> x = 1 and y = -3

Therefore x² + y² = 1 + 9 = 10 (option ‘C’)

#### QUERY 14

**If x² = y+z, y² = z+x, z² = x+y; then the value of 1/(x+1) + 1/(y+1) + 1/(z+1) is ?**

A) 0

B) 1

C) 15

D) 2

**Hemant Singh
**x² = y+z

=> x = (y+z)/x

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

=> x+1 = (x+y+z)/x

=> 1/(x+1) = x/(x+y+z) ……..I

similarly

1/(y+1) = y/(x+y+z) and ……..II

1/(z+1) = z/(x+y+z) ……..III

on adding 1, 2 and 3

1/(x+1) + 1/(y+1) + 1/(z+1) = (x+y+z)/(x+y+z) = 1 (option ‘B’)

TRICK

This type of questions will surely appear in SSC. Though it is always good to know the basic method for solving these questions, but in the exam there isn’t that much time and therefore everything should be done in minimum time possible. So just put very small number as value of the variables such as 0, 1, 2 etc because they are framed in such a way that you will get the answer by putting these values surely almost always. If done like that you’ll see if 2 is taken as the value of each variable 1 is the answer of the question.

#### QUERY 15

**If x + 1/x = √3; then the value of x ^{18} + x^{12} + x^{6} + 1 is ?**

A) 0

B) 1

C) 2

D) 3

**Vaibhav Vats
**x + 1/x = √3

squaring both sides and taking LCM we get:

x

^{4}+ 1 = x

^{2}………………..(1)

multiply both sides by x² (We have to multiply it by x

^{2}so that the degree of the equation is same as the degree of the last term in x in the expression is, i.e. 6)

So the equation will be (x

^{4}*x

^{2}) + (1*x²) = (x²*x²)

=> x^{6} + x² = x^{4}

=> x^{6} = x^{4} – x²

Now the given expression can be re-written as:

(x^{6})^{3} + (x^{6})^{2} + x^{6} + 1

= (x^{4} – x^{2})^{3} + (x^{4} – x^{2})^{2} + (x^{4}^{ }– x^{2}) + 1 (by putting the value of x^{6})

from (1) we get x^{4} – x^{2} = -1

put it in the above and solve, answer = 0 (option ‘A’)

#### QUERY 16

**If a/b = b/c = c/d; then (b³ + d³ + c³)/(a³ + b³ + c³) = ?**

A) d/a

B) d/c

C) a/b

D) a/d

**Vijay Bharath Reddy
**Let a/b = b/c = c/d = k

=> c = dk, b = dk², a = dk**³**

Now the given expression

= (b**³** + d**³** + c**³**) /(a**³** + b**³** + c**³**)

= [(d^{3}k^{6}) + d^{3} + (d^{3}k^{3})]/[(d^{3}k^{9}) + (d^{3}k^{6}) + (d^{3}k^{3})]

= [d^{3}(k^{6} + 1+ k^{3})]/[d^{3}(k^{9} + k^{6} + k^{3})]

= (k^{6} + 1+ k^{3})/(k^{9} + k^{6} + k^{3})

= (k^{6} + 1+ k^{3})/[k^{3}(k^{6} + k^{3} + 1)

= 1/k^{3}

= (b/a)(c/b)(d/c)

= d/a (option ‘A’)

#### QUERY 17

**If a/b = b/c = c/d; then √[(a – b + c) (b – c + d)] /[√(ab) – √(bc) + √(cd)]**

A) -2

B) 0

C) 1

D) 2

**Vijay Bharath Reddy
**Let a/b = b/c = c/d = k

then,

c = dk

b = ck = dk²

a = bk = dk³

Now, by replacing a, b, c with these values in the given expression we will get the answer as 1 (option ‘C’)

#### QUERY 18

**a ^{4} + a^{2}b^{2} + b^{4} = 8,**

**and a**

^{2}+ ab + b^{2}= 4; then ab = ?A) 0

B) 1

C) 2

D) 3

**RONNIE BANSAL
**If we see all the three terms of the LHS of the first equation are square of all the terms of LHS of equation second. So better to apply this formula in the second equation: (x + y + z)

^{2}= x

^{2}+ y

^{2}+ z

^{2}+2xy + 2yz + 2za)

Therefore a^{2} + ab + b^{2} = 4

Squaring both sides

(a^{2} + ab + b^{2})^{2} = 4^{2}

=> (a^{4} + a^{2}b^{2} + b^{4)} + 2a³b + 2ab^{3} + 2a²b²=16

by putting a^{4} + a^{2}b^{2} + b^{4} = 8

8 +2ab(a^{2} + b² + ab) =16

by putting a^{2} + ab + b^{2} = 4

2ab*4 = 8 (by putting a^{2} + ab + b^{2} = 4)

=> ab = 1 (option ‘B’)

#### QUERY 19

**If a, b are rational numbers and (a – 1)√2 + 3 = b√2 + a; then the value of a+b is**

A) -3

B) 5

C) -5

D) 3

**JAYANTCHARAN CHARAN
**Bringing root terms of the given equation on one side we get.

√2(a – 1 – b) = a – 3

Since a and b both are rational; a – 1 – b must be equal to zero. And if it’s so a – 3 also is zero.

Therefore

a – 1 – b = 0 (i)

a – 3 = 0 (ii)

Solving both we get

a + b = 5 (option ‘B’)

#### QUERY 20

**1000 ^{9} ÷ 10^{24} = ?**

A) 900

B) 1000

C) 2000

D) 1500

**MAHA GUPTA
**1000

^{9}÷ 10

^{24}

= (10

^{3})

^{9}÷ 10

^{24}

= 10

^{27}÷ 10

^{24}

= 10

^{(27 – 24)}

= 10

^{3}

= 1000 (option ‘B’)