# QUESTIONS ON SIMPLIFICATION (PART-IV)

## QUESTIONS ON SIMPLIFICATION (PART-IV)

#### QUERY 31

(27 + √756)1/3 + (27 – √756)1/3 = ?

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

SHIV KISHOR
Let A = (27 + √756)1/3 + (27 – √756)1/3 = ?

Cubing
A³ = (27 + √756) + (27 – √756) + 3(27 + √756)1/3(27 -√756)1/3(A) ………….using identity (a+b)^3 = a³ + b³ + 3ab(a+b) i.e. a³ + b³ + 3ab(A)

=> A³ = 54 + 3[(27 + √756)(27 – √756)]1/3(A)

=> A³ = 54 + 3(729 – 756)1/3(A) …..using (a + b)(a – b) = a² – b²

=> A³ = 54 + 3(-27)1/3(A)

=> A³ = 54 + 3(-3)A

=> A³ + 9A – 54 = 0

=> A = 3 (option ‘C’)

#### QUERY 32

If xy + yz + zx = 0; then 1/(x² – yz) + 1/(y² – zx) + 1/(z² – xy) is equal to

A) 1
B) 0
C) 3
D) x+y+z

SHUCHI SHARMA
xy + yz + zx = 0 [given]
Dividing LHS and RHS both by xyz
1/x + 1/y + 1/z = 0 —-(i)
Taking LCM of the denominators and solving
(x+y+z)/xyz = 0 —-(ii)

The find expression can be re-written
1/x² + 1/y² + 1/z² – (xy+yz+zx)

= {(1/x + 1/y + 1/z)² – 2(1/xy + 1/yz +1/zx)} – 0 —-using identity (a+b+c)² = a² + b² + c² + 2(ab + bc + ca) and putting xy+yz+zx = 0

= [0 – 2{(x+y+z)/xyz}] —-from equation (i)

= -2*0 —-from equation (ii)

= 0 (option ‘B’)

See here also
If ab + bc + ca = 0, then 1/(a² – bc) +1/(b² – ca) +1/(c² – ab) = ?

ab + bc + ca = 0
=> ab = -(bc + ca)
and bc = -(ab + ca)
and ca = -(ab + bc)

#### QUERY 33

If 1x+1  +  2y+2  +  1009z+1009 = 1

the value of xx+1 + yy+2 + zz+1009 = ?

A) 0
B) 2
C) 3
D) 4

The given expression can be re-written as
1+x-xx+1 + 2+y-yy+2 + 1009+z-zz+1009 = 1

=> [x+1x+1 – xx+1] + [y+2y+2 – yy+2] + [z+1009z+1009 – z1009 = 1

=> 1 – xx+1 + 1 – yy+2 + 1 – z1009 = 1

=> – xx+1  –  yy+2  –  z1009 = -3 + 1

=> xx+1 + yy+2 + z1009 = 3 – 1 —-by changing signs of both sides of above

=2 (option ‘B’)

#### QUERY 34

If a – b = 3 and a³ – b³ = 117;
then |a+b|=?

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

SHIV KISHOR
a³ – b³ = 117
=> (a – b)(a² + b² + ab) = 117
=> a² + b² + ab = 117/3
=> a² + b² + ab = 39
=> (a -b)² + 3ab = 39
=> 3² + 3ab = 39
=> ab= 10

Now, (a + b)² = (a – b)² + 4ab
=> (a + b)² = 9 + 4*10 = 49
=> a + b = ±7
=> ।a + b। = 7 (option ‘A’)

TRICK
RAHUL SHARMA
This type of questions, many a time, become really a matter of seconds in objective pattern. If we see closely only and only a=5 and b=2 can satisfy the given conditions; therefore |a+b|=7 (option ‘A’)

#### QUERY 35

If a = b²/(b-a); then a³ + b³ is equal to

A) 0
B) 1
C) 2
D) 3

SHIV KISHOR
a = b²/(b-a)
=> ab – a² = b²
=> a² + b² – ab = 0

Now, a³ + b³ = [a+b][a² – ab + b²] = 0
= (a+b)*0 = 0 (option ‘A’)

#### QUERY 36

If a/x + y/b =1 and b/y + z/c = 1; then x/a + c/z = ?

A) 0
B) b/y
C) 1
D) y/b

MAHA GUPTA
From the first equation
x/a = b/(b –y)

And from the second equation
c/z = y/(y –b) = -y/(b –y)

Now x/a + c/z = b/(b –y) + [-y/(b – y)]
= (b – y)/(b – y) = 1 (option ‘C’)

#### QUERY 37

If 3x – 2y = 11 and (x+y)² = 48+(x-y)², then 27x³ – 8y³ will be equal to?

A) 3700
B) 3300
C) 3707
D) 3777

SHIV KISHOR
The find expression can be re-written as (3x)³ – (2y)³
We see that it’s in the form a³ – b³; and we know it’s equal to (a – b) (a² + y² + ab)

Therefore (3x)³ – (2y)³ = (3x – 2y)(9x² + 4y² + 6xy)

We see that out of above now we only need to know 9x² + 4y²; and xy

First take (x+y)² = 48 + (x-y)²
=> (x+y)² – (x-y)² = 48
=> 4xy = 48 ——using identity
=> xy = 12
Now squaing 3x – 2y = 11 we get 9x² + 4y² – 12xy = 121
means 9x² + 4y² = 12xy +121

Now; given expression 27x³ – 8y³
= (3x – 2y)*(9x² + 4y² + 6xy)
= (3x – 2y)*[(9x² + 4y²) + 6xy]
= 11*[(121 + 12X) + 6xy]
= 11*(121+ 18 xy)
= 11*(121+ 18*12)
= 11*337
= 3707 (option ‘C’)

#### QUERY 38 SHUBHANSHU MISHRA #### QUERY 39 A) 0
B) 1
C) 5
D) 2

MAHA GUPTA

If possible; try to do such sums by imagining smallest possible values of the variables (here a, b, c) so that the given equation is satisfied.

We can take, for example, a = 1; b = 0; c = 0. These values satisfy the given equation you’ll see

Now by putting these values in the find expression, it is 15/1= 1 (option ‘B’)

#### QUERY 40 MAHA GUPTA
4x – 4(x-1) = 24
=> 4(x–1)(4 -1) = 24
=> 4(x–1) = 8
=> 2(2x–2) = 8
=> (22x)/22 = 8
=> 22x = 32
=> 22x = 25
=> 2x = 5
=> x = 2.5

Therefore (2x)x = (2*2.5)2.5 = 52.5 = 52√5 = 25√5 (option ‘D’)

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