ARITHMETICMATHSSIMPLIFICATION

QUESTIONS ON SIMPLIFICATION (PART-III)

QUESTIONS ON SIMPLIFICATION (PART-III)

QUERY 21

If x = √3/2; then the value of (1+x)/(1+√(1+x) + (1-x)/(1 – √(1-x) is equal to

A) √3/2
B) √3
C) 0
D) 1

RONNIE BANSAL
x = √3/2
=> 1 + x = 1 + √3/2
=> 1 + x = (2 + √3)/2
Solving further
√(1 + x) = (√3 + 1)/2

Similarly we can find
1 – x = (2 – √3)/2
and √(1 – x) = (√3 -1)/2

Now substituting all these four values in the given expression and solving
the value of the expression = 1 (option ‘D’)

QUERY 22

If x = 1+√2 + √3; then find 2x4 – 8x3 – 5x2 + 26x – 28

A) 6 √6
B) 3 √6
C) 2 √6
D) 0

JAYANTCHARAN CHARAN
x = 1 + √2 + √3
=> x – 1 = √2 + √3
squaring both sides we get
x² – 2x + 1= 5 + 2√6

again squaring we get
x4 – 4x3 + 6x2 – 4x + 1 = 49 + 20√6

now as the first two terms of the question can be obtained by multiplying the above by 2……..we double it and adjust the remaining terms so that we may find the expression in question on the LHS of the above equation as follows:

2(x4 – 4x3 + 6x2 – 4x + 1) – 17x2 + 34x – 30 = 2(49 + 20√6) – 17x2 + 34x – 30

=> 2x4 – 8x3 + 12x2 – 8x + 2 – 17(x2 – 2x + 1) – 13 = 98 + 40√6 – 17(x2 – 2x + 1) – 13

So LHS = 98 + 40√6 – 17(5 + 2√6) -13……………………..(by putting value of x2 – 2x +1)

=98 + 40√6 – 85 – 34√6 – 13

= 6√6 (option ‘A’)

QUERY 23

If (x – 2a)(x – 5a)(x – 8a)(x – 11a) + ka4 is a perfect square; then the value of k is?

A) 81
B) 49
C) 64
D) 72

RONNIE BANSAL
If you think over it carefully; for the expression to be a perfect square x must be zero
Therefore after replacing x by zero the expression is
2a*5a*8a*11a + ka4

= 880a4 + ka4

= (880 + k)a4

Therefore 880 + k is a perfect square
Though k can have several values for 800 + k to be a perfect square
But numbers in the answer options only 81 is such value of k
Therefore k = 81 (option ‘A’)

QUERY 24

If (x + 1/x)2 = 3; then value of x206 + x200 + x90 + x84 + x12 + x6 +1 is?

A) 84
B) 206
C) 0
D) 1

JAYANTCHARAN CHARAN
Notice that there is a difference of 6 between the exponents of every set of two terms of the given expression
So the given expression can be rewritten: x203 (x3 + 1/x3) + x87(x3 + 1/x3) + x9 (x3 + 1/x3) +1

Now (x + 1/x)² = 3 (given)
=> x + 1/x = √3 (squaring root of both the above sides)

Now cubing both side we get
(x³ + 1/x³) + 3 (x + 1/x) = 3√3
=> (x³ + 1/x³) + 3 √3 = 3√3 (putting x + 1/x = root 3 as shown above)
=> (x³ + 1/x³) = 0

By putting it in the asked expression, you’ll see the answer is 1 (option ‘D’)

QUERY 25

If (x + 1/x)2 = 3; then the value of x206 + x200 + x90 + x84 + x18 + x12 + x6 +1 is

A) 84
B) 206
C) 0
D) 1

RONNIE BANSAL
If you see you’ll find an extra term i.e. x^18 in the query than in the query made just above
So according to the above method the solution is:

Notice that there is a difference of 6 between the exponents of every set of two terms of the given expression
So the given expression can be rewritten:
x203 (x3 + 1/x3) + x87(x3 + 1/x3) + x15(x3 + 1/x3) + x3(x3 + 1/x3)

Now (x + 1/x)2 = 3 (given)
=> x + 1/x = √3               (squaring root of both the above sides)

Now cubing both side we get
(x3 + 1/x3) + 3 (x + 1/x) = 3√3
=> (x3 + 1/x3) + 3√3 = 3√3                    (putting x + 1/x = √3 as shown above)
=> (x3 + 1/x3) = 0

By putting it in the asked expression, you’ll see the answer is 0 (option ‘C’)

QUERY 26

If x = √[(√5 + 1)/(√5 – 1)]; then 5x² – 5x – 1 = ?

A) 1
B) 2
C) 4
D) 10

KANW@LJEET
Just rationalize the value of x; you’ll get x = (√5 + 1)/2

By putting this in the other expression solve it further
On solving 5x² – 5x – 1 = 4 (option ‘C’)

QUERY 27

If x²+2 = 2x, then the value of x4 – x3 + x2 + 2

A) 0
B) 1
C) -1
D) √2

RONNIE BANSAL
Given x² + 2 = 2x
=> x² = 2x – 2

Now the find expression
= x4 – x3 + x² + 2
= (x²)² – x²*x + x² +2

(by substituting x² = 2x – 2 in the above)
= (2x – 2)² – (2x – 2)*x + 2x – 2 + 2
= (2x – 2)² – (2x – 2)*x + 2x
= 2x² – 4x + 4 (on solving)
= 2(x² – 2x + 2)

Again by substituting x² = 2x – 2
= 2(2x – 2 – 2x + 2)
= 2*0
= 0 (option ‘A’)

MEGHA DIXIT
x4 – x3 + x² + 2

= x²(x² + 2) -x3 + 2 – x²

Putting x² + 2 = 2x
x²(2x) -x3 + 2 – x²

= x– x² + 2

=> x(x² + 2) – x² + 2 – 2x

Again putting x² + 2 = 2x
x(2x) – x² + 2 – 2x

=> x² + 2 – 2x = 0 (option ‘A’)                               [x² + 2 – 2x = 0 is equal to the given equation x² + 2 = 2x]

QUERY 28

If A+ B+ C = 1 & AB + BC + CA = 1/3; then A : B : C is?

A) 1 : 2 : 2
B) 2 : 1 : 2
C) 1 : 1 : 1
D) 1 : 2 : 1

RAMESH JASWAL
To satisfy both A+ B+ C = 1 & AB + BC + CA = 1/3, it’s a must that A, B & C each equal to 1/3
So the ratio is 1 : 1 : 1 (option ‘C’)

QUERY 29

If x² + y² + z² = 2(x – y -z) – 3; then value of 2x – 3y + 4z is (assume x, y, z all are the real numbers)

A) 0
B) 1
C) 7
D) 3

SHUCHI SHARMA
The given equation can be written as
( x² – 2x + 1) + (y² + 2y + 1) + (z² + 2z + 1) = 0

=> (x- 1)² + (y + 1)² + (z+1)² = 0

The LHS is the sum of all square expressions; and we know that sum of squares of expressions can be equal to zero only when every expression is zero.

So x = 1, y = -1 and z = -1

substituting these values of x, y and z in 2x – 3y + 4z
we get 2(1) – 3(-1) + 4(-1) = 1 (option ‘B’)

QUERY 30

If x = 2 + 21/3 + 22/3, then find x3 – 6x2 + 6x

A) 0
B) 1
C) 2
D) 8

SHIV KISHOR
Given that x  = 2 + 21/3 + 22/3

=> x – 2 = 21/3 + 22/3

=> (x – 2)³ = (21/3 + 22/3 

=> x³ – 8 – 6x(x – 2)  = (21/3)³ + (22/3)³ + 3(21/3 x 22/3)(21/3 + 22/3)

=> x³ – 8 – 6x(x – 2)  = 2 + 4 + 6(21/3 + 22/3)

=> x³ – 8 – 6x² + 12x = 6 + 6(21/3 + 22/3)

=> x³ – 6x² + 6x = 8 + 6 + 6(21/3 + 22/3) – 6x

Putting x = 2 + 21/3 + 22/3 on the RHS

=> x³ – 6x² + 6x  = 14 + 6(21/3 + 22/3) – 6(2 + 21/3 + 22/3)

=> x³ – 6x² + 6x  = 14 + 6(21/3 + 22/3) – 12 – 6(21/3 + 22/3)

=> x³ – 6x² + 6x  = 14 – 12

=> x³ – 6x² + 6x  = 2 (option ‘C’)

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Maha Gupta

Maha Gupta

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