Free Algebra Miscellaneous - 01 Practice Test - CAT 

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)

Let x be the number of children playing only cricket

a+b= 12-4=8

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

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 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)

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)

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)

Option (a)

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

$x^2-x-2=x-1-2$

$x^2-2x+1=0$

$(x-1{)}^2=0$

x = 1. 

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

Hence, there are no roots.

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.

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

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.

option (b)

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

Only k = 1 satisfies this.

2nd method: - For infinite solution

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: - 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

option (b)

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, => a² = x - 1

y = 1 + b², => b² = y - 1

z = 1 + c² => c² = z - 1 and (a + b + c)² = 0

As we know, (a + b + c)² = 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