# Free Algebra Miscellaneous - 01 Practice Test - CAT

Find the number of children in the class.

A. 54
B. 42
C. 102
D. Cannot be determined

#### SOLUTION

Solution : B

S = I + 2II + 3III

X = I+ II +III

where I = Number of people who play one game

II = Number of people who play two games

III =Number of people who play three games

Given that II+III=24, and III=12, thus II= 12

S= 25+25+28= 78

S-X= II+2III

=>X= 78-36= 42. Option (b)

If it is known that number of students playing Hockey and Football only is 4. Find the number of  students playing only cricket.

A. 5
B. 9
C. 13
D. Cannot be determined

#### SOLUTION

Solution : A

Let x be the number of children playing only cricket

a+b= 12-4=8

Thus 25= 12+8+x => x= 5. Option (a)

Both the roots of the equation, (if given that a,b,c are real) (x-b)(x-c) + (x-a)(x-c) + (x-a)(x-b) = 0 are always :

A. Positive
B. Negative
C. Real
D. None of these

#### SOLUTION

Solution : C

Solve using assumption

Let a = 0, b = 1, c = 2. The expression becomes 3x2 -6x + 2 = 0. The discriminant of this exp is positive. You can try with some other values. You will always get real roots.

Alternatively: -

(x-b)(x-c) + (x-a)(x-c) + (x-a)(x-b) = 0

x2 - cx - bx + bc + x2 - cx - ax + ac + x2 - bx - ax + ab = 0

3x2 - 2(a + b + c)x + (ab + bc + ca) = 0

Discriminant = b2 - 4ac

4(a + b + c)2 - 4 x 3 x (ab + bc + ca)

4{a2 + b2 + c2 + 2(ab + bc + ca) - 3(ab + bc + ca)}

4{a2 + b2 + c2 - (ab + bc + ca)}

The discriminant of this exp is positive. You will always get real roots.

The number of real solutions of the equation |x|2 – 3|x| + 2 = 0 is:

A. 4
B. 1
C. 3
D. 2

#### SOLUTION

Solution : A

The expression can be rewritten as

(|x| - 1)(|x| - 2) = 0. x = {-1, +1, -2, +2}

Or, we can say, If x is greater than 0.

x2 – 3x + 2 = 0

(x - 2) (x - 1) = 0

x = {1, 2}

If x is less than 0.

x2 + 3x + 2 = 0

(x + 2) (x + 1) = 0

x = { - 1, - 2}

Option (a)

The largest interval for which x12 – x9 + x4 – x + 1 > 0 is:

A. -4 < x < 0
B. 0 < x < 1
C. -100 < x < 100
D.

#### SOLUTION

Solution : D

Since the highest power of x is even (12) , it will always be greater than its negative counterpart (9), hence, irrespective of the value of x, the net value will be positive. Answer is option (d)

If a+b+c =0, then the quadratic equation 3ax2 + 2bx + c = 0 has:

A. at least one root in (0,1)
B. one root in (2,3) and the other in (-2,-1)
C. Imaginary roots
D. None of these

#### SOLUTION

Solution : A

These questions are flexible. So, we can assume values of a, b, and c to make an other quardatic equation.

Use Assumption

Let a = 0, b = 1, c = -1. Equation becomes 2x -1 = 0, x = ½.

Now, we can eleminate options (b) and (c). It satisfies option (a)

The equation x2x1=12x1 has:

A. No root
B. One root
C. 2 equal roots
D. Infinite roots

#### SOLUTION

Solution : A

Option (a)

x(x - 1) - 2 = (x - 1) - 2

x2x2=x12

x22x+1=0

(x1)2=0

x = 1.

x = 1 is the only solution but it will make the denominator 0

Hence, there are no roots.

If α and β are the roots of the equation x2 + px + q = 0 and α4 and β4 are the roots of the equation  x2– rx + s = 0, then the equation x2 – 4qx + 2q2 – r = 0 has always:

A. 2 real roots
B. 2 positive roots
C. 2 negative roots
D. 1 positive and 1 negative root

#### SOLUTION

Solution : D

Given question is flexible. So, we can assume the roots and can make an other quardatic equation.

Let α = 1 and β = 2, then the expression becomes x2 – 3x + 2 = 0. So p = -3, q = 2, r = 17, s = 16. The new expression is x2 – 8x – 9 = 0. 1 root is positive and other is negative.

Determine k for which the roots of the equation 9x2 + 2kx + 4 = 0 are equal.

A. 6
B. - 6
C. Both (a) and (b)
D. None of these

#### SOLUTION

Solution : C

Let the roots be P and P.

Sum of the roots = 2P = - 2k/9

P = - k/9

and product of roots = P2 = 4/9

(- k/9)2 = 4/9

k2/81 = 4/9

k2 = 36

k = ± 6

A. No solution
B. One solution
C. Two solutions
D. More than two solutions

#### SOLUTION

Solution : B

For the equation 3x2 + px + 3 = 0, the roots are real and 1 root is the square of the other, Find p?

A. -3
B. -1
C. -6
D. 2/3

#### SOLUTION

Solution : C

Let roots of the given equation are a and a2.

Sum of the roots (a + a2) = - P/3 .........................(1)

Product of the roots (a)3 = 1 => a = 1

From equation (1),

1 + 1 =  - P/3 => P = - 6

Alternatively: - 1, 1 be the roots. Eqn. becomes x2 - 2x + 1 = 0 or 3 x2 - 6x + 3 = 0.

The number of values of k for which the system of equations have infinite solution- (k+1)x + 8y = 4k kx + (k+3)y = 3k – 1

A. 0
B. 1
C. 2
D. Infinite

#### SOLUTION

Solution : B

option (b)

Sol:- k+1/k = 8/k+3 = 4k/3k-1

Only k = 1 satisfies this.

2nd method: - For infinite solution

If a and b are the roots of the equation x2 – px + r = 0 and a/2 and 2b be the roots of the equation  x2 – qx + r = 0. Then the value of r is:

A.
B.
C.
D.

#### SOLUTION

Solution : D

1st method: -

Sum of the roots (a + b) = p and Product of the roots (ab) = r

for other equation, a/2 + 2b = q and (ab) = r

From above equations,

a = 2/3 (2p - q) and b = (2q - p)/3

The value of r is

2nd method: - x2 – 3x + 2 = 0 and α = 2 and β = 1 and α/2 = 1 and 2β = 2 forms the same expression. P = q = 3 and r = 2. Only d) gives r = 2

If f(x) = then which of the following is necessarily true?

A. f-1(x) f(x) = 1
B. f-1(x) f(x) = -1
C. f-1(x) = f(x)
D. f-1(x) - f(x) = 2

#### SOLUTION

Solution : B

option (b)

If x = 1 + a2, y = 1 + b2, z = 1 + c2 and (a + b + c)2 = 0, then ab + bc + ca =

A. [3 – (x + y + z)]/2
B. 1 – (x + y + z)/2
C. 1 + (x + y + z)/2
D. 1 – (x + y + z)

#### SOLUTION

Solution : A

Sol:- a = 0, b = 1, c = -1. x = 1, y = 2, c = 2. ab + bc + ca = -1. Only option a) satisfies

Alternatively: -

Given

x = 1 + a2, => a= x - 1

y = 1 + b2=> b= y - 1

z = 1 + c=> c= z - 1 and (a + b + c)2 = 0

As we know, (a + b + c)= a2 + b2 + c2 + 2(ab + bc + ca)

Put the given values.

0 = (x - 1) + (y - 1) + (z - 1) + 2(ab + bc + ca)

(ab + bc + ca) = [3 – (x + y + z)]/2