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 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.
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.
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)
Question 5
The largest interval for which x12 – x9 + x4 – 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 3ax2 + 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 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:
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.
Question 9
Determine k for which the roots of the equation 9x2 + 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 = P2 = 4/9
(- k/9)2 = 4/9
k2/81 = 4/9
k2 = 36
k = ± 6
Question 10
SOLUTION
Solution : B
Question 11
For the equation 3x2 + 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 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.
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 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:




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
Question 14
If f(x) = then which of the following is necessarily true?
SOLUTION
Solution : B
option (b)
Question 15
If x = 1 + a2, y = 1 + b2, z = 1 + c2 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 + 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