Algebra

1)

If x6 + y6 = x3y3, then x9 + y9 = ?.

Show Solution

x9 + y9 = (x3)3 + (y3)3
=> (x3 + y3)(x6 + y6 – x3y3)
It is given that, x6 + y6 = x3y3
Therefore, (x3 + y3)(x3y3 – x3y3) = 0. Hence, (b).

2)

Solve the equation 4x2 – 13x + 9 = 0.

Show Solution

The given equation is a quadratic equation, by splitting up the middle term, we get 
4x2 – 13x + 9 = 0
4x2 – 4x – 9x + 9 = 0
4x(x – 1) – 9(x – 1) = 0
(4x – 9)(x – 1) = 0
4x – 9 = 0 or x – 1 = 0
x = (3/2) or 1. Hence, (a).

3)If x = (y2/y – x), then find the value of x3 + y3.

Show Solution
Consider x = (y2 /y – x)
=> x(y – x) = y2
=> xy – x2 = y2
=> x2 + y2 – xy = 0
We know that, x3 + y3 = (x + y)( x2 + y2 – xy)
x3 + y3 = (x + y)(0)
Therefore, x3 + y3 = 0. Hence, (d).

4)

If abc = 90, (1/a) + (1/b) + (1/c) = 9/10 and a2 + b2 + c2 = 162, then find the value of a + b + c.

Show Solution

(1/a) + (1/b) + (1/c) = 9/10
=> (bc + ac + ab)/abc = 9/10
=> bc + ac + ab = (9 x 90)/10 = 81
We know that, (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ac)
162 + 2(81)
162 + 162 = 324
(a + b + c)2 = 324
Therefore, a + b + c = 18. Hence, (b).

5)If a = bx; b = cy and c = az, then xyz = ?

Show Solution
c1 = az = (bx )z = (cy )xz
c1 = cxyz
Therefore, xyz = 1. Hence, (c).

6)If a + b = 8, then a3 + b3 + 24ab = ?

Show Solution
We know that, (a + b)3 = a3 + b3 + 3ab(a + b)
Since, a + b = 8, 
83 = a3 + b3 + 3ab(8)
512 = a3 + b3 + 24ab. Hence, (d).

7)If x3 + 3x2 + 3x = 26; then x2 + 2x = ?

Show Solution
(x + 1)3 = x3 + 3x2 + 3x + 1
 x3 + 3x2 + 3x = 26
Add and subtract 1 on LHS, so that the value does not change.
x3 + 3x2 + 3x + 1 – 1 = 26
(x + 1)3 = 27 = 33 .
x + 1 = 3.
(x + 1)2 = 9
(x + 1)2 = x2 + 1 + 2x = 9
x2 + 2x = 8. Hence, (b).

8)If \(20 \sqrt {p}= \sqrt {245}+ \sqrt {845}\), then what is the value of p?

Show Solution
\(20 \sqrt {p}= \sqrt {245}+ \sqrt {845}\)
=> \(20 \sqrt {p}=7 \sqrt {5}+13 \sqrt {5}\)
=>=> \(20 \sqrt {p}=20 \sqrt {5}\)
=>=> \( \sqrt {p}= \sqrt {5}\)
Squaring on both sides, we get,
p = 5. Hence, (c).

9)If x2 + y2 + z2 = 2(x – y – z) – 3, then 6x – 5y + 8z = ?

Show Solution
Consider x2 + y2 + z2 = 2(x – y – z) – 3
x2 – 2x + 1 + y2 + 2y + 1 + z2 + 2z + 1 = 0
(x – 1)2 + (y + 1)2 + (z + 1)2 = 0
As the sum of the squares is zero, each of the square must be zero.
This implies, x = 1, y = –1, z = –1.
Therefore, 6(1) – 5(–1) + 8(–1) = 3. Hence, (a).

10)

If a = 2017, b = 2018 and c = 2019, then a2 + b2 + c2 – ab – bc – ac = ?

Show Solution

(a – b)2 + (b – c)2 + (c – a)2 – 2(a2 + b2 + c2 – ab – bc – ac) = 0.
(a – b)2 + (b – c)2 + (c – a)2 = 2(a2 + b2 + c2 – ab – bc – ac) 
a2 + b2 + c2 – ab – bc – ac = ½ [(a – b)2 + (b – c)2 + (c – a)2 ]
Since, a, b and c are given,
½ (1 + 1 + 4)
½ (6) = 3. Hence, (c).

11)If (3/10)2y = 0.0081, then find the value of (1/2)2y.

Show Solution
Consider (3/10)2y = 0.0081.
(0.3)2y = (0.3)4
Since, the base is same, equate the powers
2y = 4 => y = 2.
(1/2)2y = (0.5)4 = 0.0625. Hence, (b).

12)If a + (1/a) = 4, then find the value of a5 + (1/a5).

Show Solution
[a + (1/a)]2 = 42
a2 + (1/a2 ) = 16 – 2 = 14
Carrying on further to get sum of inverse cubes,
a3 + (1/a3 ) = [a + (1/a)][a2 – 1 + (1/a2 )]
  • 4 x (14 – 1) = 52

  • [a2 + (1/a2 )][a3 + (1/a3 )] = [a5 + (1/a5 )][a + (1/a)]
    14 x 52 = [a5 + (1/a5 )] x 4
    a5 + (1/a5 ) = (14 x 52)/4 = 182. Hence, (d).

    13)

    If a2 + b2 – 4a + 8b + 20 = 0, then a4 + b4 = ?

    Show Solution

    a2 + b2 – 4a + 8b + 20 = 0
    a2 – 4a + 4 + b2 + 8b + 16 = 0
    (a – 2)2 + (b + 4)2 = 0
    a – 2 = 0 and b + 4 = 0
    a = 2 and b = – 4
    Therefore, a4 + b4 = (2)4 + (– 4)4 = 16 + 256 = 272. Hence, (b).

    14)If x + y + z = 2a, then \({a^{2}+(a-x)^{2}+(a-y)^{2}+(a-z)^{2} \over {x^{2}+y^{2}+z^{2}}}\) = ?

    Show Solution
    a 2   + (a  − x ) 2   +   ( a  − y ) 2   + (a  − z ) 2   = 4a 2   + x 2   + y 2   + z 2 – 2a(x  + y  + z )
    = 4a 2   + x 2   + y 2   + z 2 – 2a.2a   [substituting the value of x  + y  + z  = 2a ]
    = 4a 2   + x 2   + y 2   + z 2 – 4a 2
    =   x 2   + y 2   + z 2  
    \({a^{2}+(a-x)^{2}+(a-y)^{2}+(a-z)^{2} \over {x^{2}+y^{2}+z^{2}}}={x^{2}+y^{2}+z^{2} \over {x^{2}+y^{2}+z^{2}}}=1\) . Hence, (c).

    15)If x(4 – (3/x)) = (4/x), then find the value of x2 + (1/x2).

    Show Solution
    x(4 – (3/x)) = (4/x)
    4 – (3/x) = (4/x2 )
    Multiply by x on both sides, 
    4x – 3 = 4/x
    4x – 4/x – 3 = 0
    4(x – (1/x)) = 3
    x – (1/x) = 3/4
    Squaring on both sides, we get,
    x2 + (1/x2 ) – 2 = 9/16
    x2 + (1/x2 ) = (9/16) + 2 = 41/16. Hence, (a).