# Free Algebra Miscellaneous - 01 Practice Test - CAT

### Question 1

Find the number of children in the class.

#### 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)

### Question 2

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

#### 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)

### Question 3

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 :

#### SOLUTION

Solution :C

Solve using assumption

Let a = 0, b = 1, c = 2. The expression becomes 3x

^{2}-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

x

^{2}- cx - bx + bc + x^{2}- cx - ax + ac + x^{2}- bx - ax + ab = 03x

^{2}- 2(a + b + c)x + (ab + bc + ca) = 0Discriminant = b

^{2}- 4ac4(a + b + c)

^{2}- 4 x 3 x (ab + bc + ca)4{a

^{2}+ b^{2}+ c^{2}+ 2(ab + bc + ca) - 3(ab + bc + ca)}4{a

^{2}+ b^{2}+ c^{2}- (ab + bc + ca)}The discriminant of this exp is positive. You will always get real roots.

### Question 4

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

#### 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.

x

^{2}– 3x + 2 = 0(x - 2) (x - 1) = 0

x = {1, 2}

If x is less than 0.

x

^{2}+ 3x + 2 = 0(x + 2) (x + 1) = 0

x = { - 1, - 2}

Option (a)

### Question 5

The largest interval for which x^{12} – x^{9} + x^{4} – x + 1 > 0 is:

#### SOLUTION

Solution :D

Use answer options to arrive at the answer at the earliest

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)

### Question 6

If a+b+c =0, then the quadratic equation 3ax^{2} + 2bx + c = 0 has:

#### 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)

### Question 7

The equation x−2x−1=1−2x−1 has:

#### SOLUTION

Solution :A

Option (a)

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

x2−x−2=x−1−2

x2−2x+1=0

(x−1)2=0

x = 1.

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

Hence, there are no roots.

### Question 8

If α and β are the roots of the equation x^{2} + px + q = 0 and α^{4} and β^{4} are the roots of the equation x^{2}– rx + s = 0, then the equation x^{2} – 4qx + 2q^{2} – r = 0 has always:

#### 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 x

^{2}– 3x + 2 = 0. So p = -3, q = 2, r = 17, s = 16. The new expression is x^{2}– 8x – 9 = 0. 1 root is positive and other is negative.

### Question 9

Determine k for which the roots of the equation 9x^{2} + 2kx + 4 = 0 are equal.

#### SOLUTION

Solution :C

Let the roots be P and P.

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

P = - k/9

and product of roots = P

^{2}= 4/9(- k/9)

^{2}= 4/9k

^{2}/81 = 4/9k

^{2}= 36k = ± 6

### Question 10

#### SOLUTION

Solution :B

### Question 11

For the equation 3x^{2 }+ px + 3 = 0, the roots are real and 1 root is the square of the other, Find p?

#### SOLUTION

Solution :C

Let roots of the given equation are a and a

^{2}.Sum of the roots (a + a

^{2}) = - P/3 .........................(1)Product of the roots (a)

^{3}= 1 => a = 1From equation (1),

1 + 1 = - P/3 => P = - 6

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

### Question 12

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

#### 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

### Question 13

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

#### 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

2

^{nd}method: - x^{2}– 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

### Question 14

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

^{-1}(x) f(x) = 1

^{-1}(x) f(x) = -1

^{-1}(x) = f(x)

^{-1}(x) - f(x) = 2

#### SOLUTION

Solution :B

option (b)

### Question 15

If x = 1 + a^{2}, y = 1 + b^{2}, z = 1 + c^{2} and (a + b + c)^{2} = 0, then ab + bc + ca =

#### 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 + a^{2}, =>a^{2 }= x - 1

y = 1 + b^{2},=>b^{2 }= y - 1

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

As we know,(a + b + c)^{2 }= a^{2}+ b^{2}+ c^{2}+ 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