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Algebra Practice Questions for CAT with Solutions

Approximately 10-12 questions on Algebra are featuring in CAT in recent years. Most of these questions are from the following areas:To give CAT asirants a hands on experience on the variety of Algebra questions which frequently appear in CAT, we have listed around 60 questions practice on important topics from Algebra.
All these algebra questions are with detailed explanations.

Question 1:
For the given pair (x, y) of positive integers, such that 4x-17y=1 and x<1000 how many integer values of y satisfy the given conditions?
[1] 56
[2] 57
[3] 58
[4] 59
Solution:
We first need to find out a solution for x & y. Once we get a solution, values of x would be in an AP with a common difference of 17 whereas values of y would be in an AP with a common difference of 4.

Valid Solutions:

x = 13, y = 3

x = 30, y = 7

x = 47, y = 11

.

.

x = 999, y = 235

No. of terms =\(\frac{{999 - 13}}{{17}} + 1 = \) = 58 + 1 = 59. Option D


Question 2:
One year payment to the servant is Rs. 90 plus one turban. The servant leaves after 9 months and receives Rs. 65 and turban. Then find the price of the turban
[1] Rs.10
[2] Rs.15
[3] Rs.7.5
[4] Cannot be determined
Solution:
Payment for 12 months = 90 + t {Assuming t as the value of a turban}

Payment for 9 months should be ¾(90 + t)

Payment for 9 months is given to us as 65 + t

Equating the two values we get

¾(90 + t) = 65 + t

270 + 3t = 260 + 4t
t = 10 Rs. Option A

Question 3:
In CAT 2007 there were 75 questions. Each correct answer was rewarded by 4 marks and each wrong answer was penalized by 1 mark. In how many different combination of correct and wrong answer is a score of 50 possible?
[1] 14
[2] 15
[3] 16
[4] None of these
Solution:
Correct (c) + Wrong (w) + Not attempted (n) = 75

4c – w + 0n= 50

Adding the two equations we get

5c + n = 125
Values of both c & n will be whole numbers in the range [0, 50]
c (max) = 25; when n = 0
c (min) = 13; when n = 60 {Smallest value of ‘c’ which will take the marks from correct questions greater than or equal to 50}
No. of valid combinations will be for all value of ‘c’ from 13 to 25 = 13. Option D

Question 4:
How many integer solutions exist for the equation 8x – 5y = 221 such that \(x \times y < 0\)
[1] 4
[2] 5
[3] 6
[4] 8
Solution:
We first need to find out a solution for x & y. Once we get a solution, values of x would be in an AP with a common difference of 5 whereas values of y would be in an AP with a common difference of 8.

Valid Solutions:

x = 32; y = 7

x = 37; y = 15

x = 42; y = 23

But we need the solutions where one variable is negative whereas the other one is positive. so, we will move in the other direction.

x = 27; y = -1

x = 22; y = -9

x = 17; y = -17

x = 12; y = -25

x = 7; y = -33

x = 2; y = -41

So, number of integer solutions where \(x \times y < 0\) is 6. Option C


Question 5:
How many integer solutions exists for the equation 11x + 15y = -1 such that both x and y are less than 100?
[1] 15
[2] 16
[3] 17
[4] 18
Solution:
Valid Solutions:

x = 4; y = -3

x = 19; y = -14

.

.

x = 94; y = -69

So, there are 7 solutions of positive values of ‘x’.

x = -11; y = 8

x = -26; y = 19

.

.

x = __; y = 96

So, there are 9 solutions for positive values of ‘y’.

Total number of integer solutions = 7 + 9 = 16. Option B


Question 6:
The number of ordered pairs of natural numbers (a, b) satisfying the equation 2a + 3b = 100 is:
[1] 13
[2] 14
[3] 15
[4] 16
Solution:
Valid solutions:

a = 2; b = 32

a = 5; b = 30

.

.

a = 47; b = 2

No. of solutions = 16. Option D


Question 7:
For how many positive integral values of N, less than 40 does the equation 3a – Nb = 5, have no integer solution
[1] 13
[2] 14
[3] 15
[4] 12
Solution:
If N is a multiple of 3, then the LHS would be divisible by 3 and RHS won’t be. Number of positive integral values less than 40 which are multiple of 3 = 13. Option A

Question 8:
What are the number of integral solutions of the equation 7x + 3y = 123 for x,y > 0
[1] 3
[2] 5
[3] 12
[4] Infinite
Solution:
Valid Solution:

x = 3; y = 34

x = 6; y = 27

.

.

x = 15; y = 6

Number of integral solutions such that x, y > 0 are 5. Option B


Question 9:
The cost of 3 hamburgers, 5 milk shakes, and 1 order of fries at a certain fast food restaurant is $23.50. At the same restaurant, the cost of 5 hamburgers, 9 milk shakes, and 1 order of fries is \(\$ 39.50\). What is the cost of 2 hamburgers, 2 milk shakes, and 2 orders of fries at this restaurant?
[1] 10
[2] 15
[3] 7.5
[4] Cannot be determined
Solution:
3H + 5M + 1F = 23.50

5H + 9M + 1F = 39.50

2H + 2M + 2F = ?

Calculate 2(Equation 1) – (Equation 2)

H + M + F = 2×23.5 – 39.5
H + M + F = 7.5
2H + 2M + 2F = 15. Option B

Question 10:
How many integer solutions are there for the equation: |x| + |y| =7?
[1] 24
[2] 26
[3] 14
[4] None of these
Solution:
x can take any integer value from [-7,7].

So, there are 15 valid values of x.

For each of these values, there are 2 corresponding values of y. eg: For x = 3; y can be 4 or -4.

Except when x = 7 or -7; where the only possible value of y is 0.

Total valid values of x = 13×2 + 1 + 1 = 28. Option D


Question 11:
A shop stores x kg of rice. The first customer buys half this amount plus half a kg of rice. The second customer buys half the remaining amount plus half a kg of rice. Then the third customer also buys half the remaining amount plus half a kg of rice. Thereafter, no rice is left in the shop. Which of the following best describes the value of x?
[1] 2 ≤ x ≤ 6
[2] 5 ≤ x ≤ 8
[3] 9 ≤ x ≤ 12
[4] 11 ≤ x ≤ 14
Solution:
After first customer, amount of rice left is 0.5x – 0.5

After second customer, amount of rice left is 0.5(0.5x -0.5) – 0.5

After third customer, amount of rice left is 0.5(0.5(0.5x -0.5) – 0.5) – 0.5 = 0

0.5(0.5(0.5x -0.5) – 0.5) = 0.5
0.5(0.5x -0.5) – 0.5 = 1
0.5x -0.5 = 3
x = 7. Option B

Verification for better understanding:

Originally there were 7 kgs of rice.

First customer purchased 3.5kgs + 0.5kgs = 4 kgs.

After first customers, amount of rice left is 3 kgs.

Second customer purchased 1.5kgs + 0.5 kgs = 2 kgs.

After second customer, amount of rice left is 1 kg.

Third customer purchased 0.5kgs + 0.5kgs = 1 kg.

No rice is left after the third customer.


Question 12:
If p and Q are integers such that \(\frac{7}{10}<\frac{p}{q}<\frac{11}{15} \) , find the smallest possible value of q.
[1] 13
[2] 60
[3] 30
[4] 7
Solution:
The fraction lies in the range (0.7,0.733333)

We know that \(\frac{8}{{11}}\) = 0.727272.. is valid value.

The smallest value of q has to be less than or equal to 11. Only 7 fits in the range.

With a little hit and trial we get a valid value of \(\frac{p}{q}\) as \(\frac{5}{7}\)

Smallest value of q = 7. Option D


Question 13:
Given the system of equations \(\left\{ {\begin{array}{*{20}{c}}{2x + y + 2z = 4}\\{x + 2y + 3z = - 1}\\{3x + 2y + z = 9}\end{array}} \right. \), find the value of x+y+z.
[1] -1
[2] 3.5
[3] 2
[4] 1
Solution:
The given equations are

\(2x + y + 2z = 4 \ldots \left( 1 \right)\)

\(x + 2y + 3z =  - 1 \ldots \left( 2 \right)\)

\(3x + 2y + z = 9 \ldots \left( 3 \right)\)

Take the first and the second equation :

\(\begin{array}{*{20}{c}}{2x + y + 2z = 4}\\{x + 2y + 3z =  - 1}\\{}\end{array}\)  multiply equation 2 by -2 , thus the 2 equations we get after multiplying are \(\begin{array}{*{20}{c}}{2x + y + 2z = 4}\\{ - 2x - 4y - 6z = 2}\\{}\end{array}\), on solving this we get \( - 3y - 4z = 6\) … (4)

Now take equation (2) and (3)

\(\begin{array}{*{20}{c}}{}\\{x + 2y + 3z =  - 1}\\{3x + 2y + z = 9}\end{array}\)    multiply equation (2) by -3, thus the equations will be

\(\begin{array}{*{20}{c}}{}\\{ - 3x - 6y - 9z = 3}\\{3x + 2y + z = 9}\end{array}\)

On solving the above 2 equations we get  -4y-8z=12  ie.  \( - y - 2z = 3\) (5)

Again on multiplying equation (5) by -3 we get   \( - 3y + 6z =  - 9\).

Adding equations (4)(5) :

We get z=-1.5 and y=0 , substituting these values in any of the 3 main equations , we get x = ½ or 0.5

Adding x + y + z = 0.5+0-1.5 = -1, Option A


Question 14:
If x and y are positive integers and x+y+xy=54, find x+y
[1] 12
[2] 14
[3] 15
[4] 16
Solution:
With x + y = 12, maximum value possible is 6 + 6 + 6×6 = 48

With x + y = 14, maximum value possible is 7 + 7 + 7×7 = 63

6 + 8 + 6×8 = 62

5 + 9 + 5×9 = 59

4 + 10 + 4×10 = 54

So, x + y = 14. Option B

Alternatively,

x + y + xy = 54

1 + x + y + xy = 55
(1 + x)(1 + y) = 55
55 can be split as 5 and 11
So x and y can be 4 and 10
x + y = 14. Option B

Question 15:
How many pairs of integers (x, y) exist such that x2 + 4y2 < 100?
[1] 95
[2] 90
[3] 147
[4] 180
Solution:
y will lie in the range [-4, 4]

When y = 4 or – 4, x will lie in the range [-5, 5] = 11 values. Total pairs = 22

When y = 3 or – 3, x will lie in the range [-7, 7] = 15 values. Total pairs = 30

When y = 2 or – 2, x will lie in the range [-9, 9] = 19 values. Total pairs = 38

When y = 1 or – 1, x will lie in the range [-9, 9] = 19 values. Total pairs = 38

When y = 0, x will lie in the range [-9, 9] = Total 19 pairs.

Total pairs = 22 + 30 + 38 + 38 + 19 = 147.


Question 16:
A test has 20 questions, with 4 marks for a correct answer, –1 mark for a wrong answer, and no marks for an unattempted question. A group of friends took the test. If all of them scored exactly 15 marks, but each of them attempted a different number of questions, what is the maximum number of people who could be in the group?
[1] 3
[2] 4
[3] 5
[4] more than 5
Solution:
c + w + n = 20

4c – w = 15

Adding the two equations, we get 5c + n = 35

c(max) = 7, when n = 0 & w = 13

c(min) = 4, when n = 15 & w = 1

Maximum number of people who could be in the group = Number of possible values of ‘c’ = 4. Option B


Question 17:
How many integers x with |x|< 100 can be expressed as \(x = \frac{{4 - {y^3}}}{4} \) for some positive integer y?
[1] 0
[2] 3
[3] 6
[4] 4
Solution:
\(x = \frac{{4 - {y^3}}}{4} = 1 - \frac{{{y^3}}}{4}\)
y3 = 4(1-x) = 4 – 4x
x = 0 or -3 or -15 or -53
No. of valid values of x = 4. Option D

Question 18:
The number of roots common between the two equations x3+3x2+4x+5=0 and x3+2x2+7x+3=0 is:
[1] 0
[2] 1
[3] 2
[4] 3
Solution:
For the roots to be common to the two equation, both the equations must be equal to 0 and hence equal to each other at those values of x

x3+3x2+4x+5 = x3+2x2+7x+3

3x2+4x+5 = 2x2+7x+3
x2 – 3x + 2 = 0
x = 1 or 2
At x = 1 and at x = 2 both the equations become equal to each other
But at x = 1 or at x = 2 none of the original equations become 0.
Number of common roots = 0. Option A

Question 19:
Let u= \({({\log _2}x)^2} - 6{\log _2}x + 12 \) where x is a real number. Then the equation xu=256, has:
[1] no solution for x
[2] exactly one solution for x
[3] exactly two distinct solutions for x
[4] exactly three distinct solutions for x
Solution:
xu=256
u \(lo{g_2}x\) = 8
\(lo{g_2}x\) =\(\frac{8}{u}\)

Putting this in the first equation

u = (8/u)2 – 6×\(\frac{8}{u}\) + 12

u3 = 64 – 48u + 12u2
u3 – 12u2 + 48u – 64 =0
(u -4)3 = 0
u = 4
\(lo{g_2}x\)= \(\frac{8}{u}\) = 2
x = 2
We have exactly one solution for x. Option B

Question 20:
Let a, b, and c be positive real numbers. Determine the largest total number of real roots that the following three polynomials may have among them: ax2 + bx + c, bx2 + cx + a, and cx2 + ax + b.
[1] 4
[2] 5
[3] 6
[4] 0
Solution:
For these equations to have real roots

b2 – 4ac ≥ 0

b2 ≥ 4ac

c2 – 4ab ≥ 0

c2 ≥ 4ab

a2 – 4ac ≥ 0

a2 ≥ 4ac

Multiplying the three we get

(abc)2 ≥ 64(abc)2

This is not possible for positive values of a, b & c.

So, there are no real roots for the three given polynomials. Option D


Question 21:
Given that three roots of f(x) = x4+ax2+bx+c are 2, -3, and 5, what is the value of a+b+c?
[1] -79
[2] 79
[3] -80
[4] 80
Solution:
We have to find out a + b + c

f(1) is 1 + a + b + c

So, we need to find out f(1) – 1

Let the 4th root be r

Coefficient of x3 is - (Sum of the roots)

0 = - (r + 2 -3 + 5)
r = - 4

So, f(x) = (x – 2) (x + 3) (x + 4)(x – 5)

f(1) = (-1)×4×5×(-4) = 80
a + b + c = f(1) – 1 = 79. Option B

Question 22:
If both a and b belong to the set (1, 2, 3, 4), then the number of equations of the form ax2+bx+1=0 having real roots is
[1] 10
[2] 7
[3] 6
[4] 12
Solution:
For the equation to have real roots

b2 – 4a ≥ 0

b = 1, No equation exists

b = 2, a = 1. 1 equation exists

b = 3, a = 1 or 2. 2 equations exist

b = 4, a = 1 or 2 or 3 or 4. 4 equations exist

Total equations = 0 + 1 + 2 + 4 = 7. Option B


Question 23:
Rakesh and Manish solve an equation. In solving Rakesh commits a mistake in constant term and finds the root 8 and 2. Manish commits a mistake in the coefficient of x and finds the roots -9 and -1. Find the correct roots.
[1] 9,1
[2] -9,1
[3] -8,-2
[4] None of these
Solution:
Rakesh’s equation

(x – 8)(x – 2) = 0

x2 – 10x + 16 = 0

Manish’s equation

(x + 9)(x + 1) = 0

x2 + 10x + 9 = 0

Correct equation is x2 – 10x + 9 = 0

(x – 1)(x – 9) = 0
Roots are 9, 1. Option A

Question 24:
The number of quadratic equations which are unchanged by squaring their roots is
[1] 2
[2] 4
[3] 6
[4] None of these.
Solution:
This would happen if and only if the roots and their squares are the same value.

The roots can be 0 or 1 or a combination of these.

So, valid equations will be formed when

Both roots are 0
Both roots are 1
One root is 0 and the other root is 1

Number of equations = 3. Option D


Question 25:
If the roots of px2+qx+2=0 are reciprocals of each other, then
[1] p = 0
[2] p = -2
[3] p= +2
[4] p = √2
Solution:
If the roots are reciprocals of each other, product of the roots is 1
2/p = 1
p = 2. Option C

Question 26:
If x =2+22/3+21/3, then the value of x3-6x2+6x is:
[1] 2
[2] -2
[3] 0
[4] 4
Solution:
x =2+22/3+21/3
x – 2 = 22/3+21/3
(x – 2)3 = (22/3+21/3)3
x3 – 6x2 + 12x – 8 = 22 + 3. 24/3.21/3 + 3. 22/3.22/3 + 2
x3 – 6x2 + 12x – 8 = 4 + 3.25/3 + 3.24/3 + 2
x3 – 6x2 + 12x – 8 = 6 + 6.22/3 + 6.21/3 = (12 + 6.22/3 + 6.21/3) – 6
x3 – 6x2 + 12x – 8 = 6x – 6
x3 – 6x2 + 6x = 2. Option A

Question 27:
If the roots of the equation x2-2ax+a2+a-3=0 are real and less than 3, then
[1] a < 2
[2] 2 < a < 3
[3] 3 < a < 4
[4] a > 4
Solution:
For the roots to be real

\(4a^{2}-4(a^{2} + a-3) \ge 0\)

=> – (a – 3) ≥ 0
a ≤ 3

Answer could be Option (a) or Option (b)

Put a = 0, we get the equation as x2 – 3 = 0. This equation has real roots and both of them are less than 3. So, a = 0 is valid solution.

a = 0, is not a part of the solution 2 < a < 3 but it is a part of a < 2. Option A


Question 28:
Find the value of \(\sqrt {2 + \sqrt {2 + \sqrt {2 + \sqrt {2 + .....} } } } \)
[1] -1
[2] 1
[3] 2
[4] \(\frac{{\sqrt 2 + 1}}{2} \)
Solution:
$\sqrt { 2 + \sqrt { 2 + \sqrt { 2 + \sqrt { 2 + \ldots } } } } = x$
$\sqrt { 2 + x } = x$
$2 + x = x ^ { 2 }$
$x ^ { 2 } - x - 2 = 0$
$x = 2 , - 1$
as the value of the expression will be positive, we can reject x=-1.
Hence, x=2.

Question 29:
If a, b and c are the roots of the equation x3 – 3x2 + x + 1 = 0 find the value of \(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \)
[1] 1
[2] -1
[3] 1/3
[4] -1/3
Solution:
\(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{{ab + bc + ca}}{{abc}} = \frac{{\frac{{Coefficient\;of\;x}}{{Coefficient\;of\;{x^3}}}}}{{\frac{{ - \;Const}}{{Coefficient\;of\;{x^3}}}}} =  - \frac{{Coefficient\;of\;x}}{{Const}} =  - 1\)Option B

Question 30:
If p, q and r are the roots of the equation 2z3 + 4z2 -3z -1 =0, find the value of (1 - p) × (1 - q) × (1 - r)
[1] -2
[2] 0
[3] 2
[4] None of these
Solution:
If p, q and r are the roots of the equation 2z3 + 4z2 -3z -1 =0, then

f(z) = 2z3 + 4z2 -3z -1 = (z - p) × (z - q) × (z - r)

f(1) = 2 + 4 – 3 – 1 = (1 - p) × (1 - q) × (1 - r)
(1 - p) × (1 - q) × (1 - r) = 2. Option C

Question 31:
If $\alpha, \beta$ and $\gamma$ are the roots of the equation $x^{3}-7 x+3=0$ what is the value of $\alpha^{4}+\beta^{4}+\gamma^{4}$ ?
[1] 0
[2] 199
[3] 49
[4] 98
Solution:

Writing the equation as

$(x-\alpha)(x-\beta)(x-\gamma)=0,$ expanding and equating coefficients we get :

$\alpha \beta \gamma=-3$

$\alpha \beta+\alpha \gamma+\beta \gamma=-7$

$\alpha+\beta+\gamma=0$

From

$\alpha^{2}+\beta^{2}+\gamma^{2}=(\alpha+\beta+\gamma)^{2}-2(\alpha \beta+\alpha \gamma+\beta \gamma)=14$

$\alpha^{2} \beta^{2}+\alpha^{2} \gamma^{2}+\alpha^{2} \beta^{2}=(\alpha \beta+\alpha \gamma+\beta \gamma)^{2}-2\left(\alpha^{2} \beta \gamma+\alpha \beta^{2} \gamma+\alpha \beta \gamma^{2}\right)$

$=(\alpha \beta+\alpha \gamma+\beta \gamma)^{2}-2 \alpha \beta \gamma(\alpha+\beta+\gamma)$

$=49$

Then

$\alpha^{4}+\beta^{4}+\gamma^{4}=\left(\alpha^{2}+\beta^{2}+\gamma^{2}\right)^{2}-2\left(\alpha^{2} \beta^{2}+\alpha^{2} \gamma^{2}+\alpha^{2} \beta^{2}\right)$

$=14^{2}-2.49=98$

Option D

Question 32:
For what values of p does the equation 4x2 + 4px + 4 –3p = 0 have two distinct real roots?
[1] p < -4 or p > 1
[2] -1 < p < 4
[3] p < -1 or p > 4
[4] –4 < p < 1
Solution:
For the roots to be distinct and real

b2 – 4ac > 0

(4p)2 – 4×4×(4 – 3p) > 0
p2 – (4 – 3p) > 0
p2 + 3p – 4 > 0
(p + 4)(p – 1) > 0
p < -4 or p > 1. Option A

Question 33:
If x2 + 4x + n > 13 for all real number x, then which of the following conditions is necessarily true?
[1] n > 17
[2] n = 20
[3] n > -17
[4] n < 11
Solution:
x2 + 4x + n > 13
x2 + 4x + 4 + n > 13 + 4 {Adding 4 to both sides}
(x + 2)2 + n > 17
Minimum value (x + 2)2 can take is 0 when x = – 2
For this to be true for all real values of x, n > 17. Option A

Question 34:
If (x + 1)×(x – 2)×(x + 3)×(x – 4)×(x + 5)…(x – 100) = a0 + a1x + a2x2… + a100x100 then the value of a99 is equal to:
[1] 50
[2] 0
[3] -50
[4] -100
Solution:
a100 = 1

Sum of the roots = \(-\frac{{{a}_{99}}}{{{a}_{100}}}=-\ {{a}_{99}}\)

a99 = - (– 1 + 2 – 3 + 4 – 5 + 6 … – 99 + 100) = - 50. Option C


Question 35:
If a, b, and c are the solutions of the equation x3 – 3x2 – 4x + 5 = 0, find the value of \(\frac{1}{{ab}} + \frac{1}{{bc}} + \frac{1}{{ca}} \)
[1] 3/5
[2] -3/5
[3] -4/5
[4] 4/5
Solution:
\(\frac{1}{{ab}} + \frac{1}{{bc}} + \frac{1}{{ca}} = \frac{{a + \;b + c}}{{abc}} = \frac{{ - ( - 3)}}{{ - 5}} =  - \;\frac{3}{5}\)

Option B


Question 36:
If a, b, and g are the roots of the equation x3 – 4x2 + 3x + 5 = 0, find (a + 1)(b + 1)(g + 1)
[1] -3
[2] 0
[3] 3
[4] 1
Solution:
f(x) = x3 – 4x2 + 3x + 5 = (x – a)(x – b)(x – g)
f(-1) = – 1 – 4 – 3 + 5 = (– 1 – a)(– 1 – b)(– 1 – g) =  – (a + 1)(b + 1)(g + 1)
(a + 1)(b + 1)(g + 1) = 3. Option C

Question 37:
Let A = (x – 1)4 + 3(x – 1)3 + 6(x – 1)2 + 5(x – 1) + 1. Then the value of A is:
[1] (x – 2)4
[2] x4
[3] (x + 1)4
[4] None of these
Solution:
A = f(x) = (x – 1)4 + 3(x – 1)3 + 6(x – 1)2 + 5(x – 1) + 1
f(1) = 1
f(2) = 1 + 3 + 6 + 5 + 1 = 16
f(3) = 16 + 24 + 24 + 10 + 1 = 75

Now check the options which satisfy these values. Put x = 3, we get

Option A is 1. Option B is 81. Option C is 256. All of them are invalid. None of these. Option D


Question 38:
Find the remainder when 3x5 + 2x4 – 3x3 – x2 + 2x + 2 is divided by x2 – 1.
[1] 3
[2] 2x – 2
[3] 2x + 3
[4] 2x – 1
Solution:
When 3x5 + 2x4 – 3x3 – x2 + 2x + 2 is divided by (x – 1), the remainder can be obtained by putting the value of x as 1 = 3 + 2 – 3 – 1 + 2 + 2 = 5

When 3x5 + 2x4 – 3x3 – x2 + 2x + 2 is divided by (x + 1), the remainder can be obtained by putting the value of x as – 1 = – 3 + 2 + 3 – 1 – 2 + 2 = 1

Check with the options by putting in the value of x as 1 & –1

Option A is 3. Invalid

Option B is 0 and – 4. Invalid

Option C is 5 & –1. So, the valid value of the remainder is 2x + 3. Option C


Question 39:
A quadratic function f(x) attains a maximum of 3 at x = 1. The value of the function at x = 0 is 1. What is the value of f (x) at x = 10?
[1] -105
[2] -119
[3] -159
[4] -110
Solution:
If the function attains the maximum of 3 at x = 1

f(x) = p(x – 1)2 + 3

f(0) = p + 3 = 1 {It is given as 1}
p = –2
f(x) = –2(x – 1)2 + 3
f(10) = –2(10 – 1)2 + 3 = –162 + 3 = –159. Option C

Question 40:
\(x + \frac{1}{x} = 3\) then, what is the value of \({x^5} + \frac{1}{{x{}^5}}. \)
[1] 123
[2] 144
[3] 159
[4] 186
Solution:

$x+\frac{1}{x}=3$

$\left.x^{2}+\frac{1}{x^{2}}+2=9 \text { \{Squaring both sides }\right\}$

$x^{2}+\frac{1}{x^{2}}=7 \quad$ Equation (2)

$\left(x^{2}+\frac{1}{x^{2}}\right)\left(x+\frac{1}{x}\right)=7 \times 3$

$x^{3}+x+\frac{1}{x}+\frac{1}{x^{3}}=21$

$x^{3}+\frac{1}{x^{3}}=21-\left(x+\frac{1}{x}\right)=21-3$

$x^{3}+\frac{1}{x^{3}}=18 \quad \ldots$ Equation (3)

$x^{4}+\frac{1}{x^{4}}+2=49 \quad\{\text { Squaring both sides of Equation }(2)\}$

$x^{4}+\frac{1}{x^{4}}=47 \quad \ldots$ Equation (4)

$\left(x^{4}+\frac{1}{x^{4}}\right)\left(x+\frac{1}{x}\right)=47 \times 3$

$x^{5}+x^{3}+\frac{1}{x^{3}}+\frac{1}{x^{5}}=141$

$x^{5}+\frac{1}{x^{5}}=141-\left(x^{3}+\frac{1}{x^{3}}\right)=141-18=123$


Question 41:
If \(\sqrt {x + \sqrt {x + \sqrt {x + ....} } } = 10. \)What is the value of x?
[1] 80
[2] 90
[3] 100
[4] 110
Solution:
\(\begin{align}& \sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+.....}}}}=10 \\;& \sqrt{x+10}=10 \\ \end{align}\)

x + 10 = 100

x = 90. Option B


Question 42:
If $\alpha$ and $\beta$ are the roots of the quadratic equation $x^{2}-x-6,$ then find the value of $\alpha^{4}+\beta^{4} ?$
[1] 1
[2] 55
[3] 97
[4] none of these
Solution:
$\alpha+\beta=1,$ and $\alpha \beta=-6$
Then, using the identity,
$\alpha^{4}+\beta^{4}=(\alpha+\beta)^{4}+2(\alpha \beta)^{2}-4 \alpha \beta(\alpha+\beta)^{2}$
$=(1)^{4}+2(-6)^{2}-(4 \times-6)(1)^{2}$
$=1+72+24$
$=97$

Question 43:
Find the value of \(\sqrt {4 - \sqrt {4 + \sqrt {4 - \sqrt {4 + ...} } } } \)
[1] \(\frac{{\sqrt {13} - 1}}{2} \)
[2] \(\frac{{\sqrt {13} + 1}}{2} \)
[3] \(\frac{{\sqrt {11} + 1}}{2} \)
[4] \(\frac{{\sqrt {15} - 1}}{2} \)
Solution:
\(\begin{array}{l}\sqrt {4 - \sqrt {4 + \sqrt {4 - \sqrt {4 + ...} } } }  = x\\\sqrt {4 - \sqrt {4 + x} }  = x\\4 - \sqrt {4 + x}  = {x^2}\\4 - {x^2} = \sqrt {4 + x} \\16 - 8{x^2} + {x^4} = 4 + x\\{x^4} - 8{x^2} - x + 12 = 0\end{array}\)

We can say that the answer is a bit bigger than 2 which eliminates option A and option D.

Using the options we can say the answer is \(\frac{{\sqrt {13}  + 1}}{2}\). Option B


Question 44:
If the roots of the equation x3ax2 + bx c =0 are three consecutive integers, then what is the smallest possible value of b?
[1] -1/√3
[2] -1
[3] 0
[4] 1/√3
Solution:
b is sum of product of the roots taken 2 at a time which will be minimum when the roots are -1, 0 & 1

b = -1×0 + 0×1 + (-1)×1 = -1. Option B


Question 45:
Three consecutive positive integers are raised to the first, second and third powers respectively and then added. The sum so obtained is a perfect square whose square root equals the total of the three original integers. Which of the following best describes the minimum, say m, of these three integers?
[1] 1 ≤ m ≤ 3
[2] 4 ≤ m ≤ 6
[3] 7 ≤ m ≤ 9
[4] 10 ≤ m ≤ 12
Solution:
We are given

m + (m+1)2 + (m+2)3 = (3m+3)2 = 9(m+1)2

m + (m+2)3 = 8(m+1)2

Using options,

3 + 42 + 53 = 3 + 16 + 125 = 144 = 122 = (3 + 4 + 5)2

So, m = 3. Option A

Alternatively

Let us take the three numbers as a – 1, a and a+1

(a-1) + (a+1)3 = 8a2
a – 1 + a3 + 3a2 + 3a + 1 = 8a2
a3 – 5a2 + 4a = 0
a(a2 – 5a + 4) = 0
a = 0, 1, 4
0 and 1 are invalid values because a – 1 should be a positive integer
a = 4
m = a – 1 = 3. Option A

Question 46:
The price of Darjeeling tea (in rupees per kilogram) is 100 + 0.10 n, on the nth day of 2007 (n = 1, 2, ..., 100), and then remains constant. On the other hand, the price of Ooty tea (in rupees per kilogram) is 89 + 0.15n, on the nth day of 2007 (n = 1, 2, ..., 365). On which date in 2007 will the prices of these two varieties of tea be equal?
[1] May 21
[2] April 11
[3] May 20
[4] April 10
Solution:
100 + 0.1n = 89 + 0.15n
0.05n = 11
n = 220

But for Darjeeling tea n cannot be more than 100.

Maximum price of Darjeeling tea = 100 + 0.1×100 = 110

Price of Ooty tea should also be 110

89 + 0.15n = 110
n = 140

On the 140th day of 2007, the prices of the Darjeeling tea and Ooty tea will be equal

140 = 31 (Jan) + 28 (Feb) + 31 (March) + 30 (April) + 20 (May)

So, price will be equal on 20th May. Option C


Question 47:
The polynomial f(x)=x2-12x+c has two real roots, one of which is the square of the other. Find the sum of all possible value of c.
[1] -37
[2] -12
[3] 25
[4] 91
Solution:
Let the roots be r and r2

Sum of the roots = r2 + r = 12

r2 + r – 12 = 0
r = -4, 3

Product of the roots = c = r3 = -64 or 27

Sum of values of c = -64 + 27 = -37. Option A


Question 48:
Two sides of a triangle have lengths 10 and 20. How many integers can take the value of the third side length:
[1] 18
[2] 19
[3] 20
[4] 21
Solution:
Let the third side be x

 20 – 10 < x < 10 + 20

10 < x < 30

No. of integral values of x = 19. Option B


Question 49:
Which of the following is a solution to: \(6{\left( {x + \frac{1}{x}} \right)^2} - 35\left( {x + \frac{1}{x}} \right) + 50 = 0 \)
[1] 1
[2] 1/3
[3] 4
[4] 6
Solution:
Let \(x + \frac{1}{x} = y\)
6y2 – 35y + 50 = 0
(3y – 10)(2y – 5) = 0
y = 10/3, 5/2

From the options only valid value of x = 1/3. Option B


Question 50:
Find x if \(\frac{5}{{3 + \frac{5}{{3 + \frac{5}{{3 + ...}}}}}} = x. \) \( \)
[1] \(\frac{{ - 3 + \sqrt {29} }}{2} \)
[2] \(\frac{{3 + \sqrt {29} }}{2} \)
[3] \(\frac{{ - 1 + \sqrt 5 }}{2} \)
[4] \(\frac{{1 + \sqrt 5 }}{2} \)
Solution:
\(\begin{array}{l}\frac{5}{{3 + \frac{5}{{3 + \frac{5}{{3 + ...}}}}}} = x\\ \Rightarrow \frac{5}{{3 + x}} = x\\ \Rightarrow 5 = 3x + {x^2}\\ \Rightarrow {x^2} + 3x - 5 = 0\\ \Rightarrow x = \frac{{ - 3\; \pm \;\sqrt {{3^2} - 4 \times 1 \times ( - 5)} }}{2}\\ \Rightarrow x = \frac{{ - 3\; \pm \;\sqrt {29} }}{2}\\ \Rightarrow x = \frac{{ - 3\; + \;\sqrt {29} }}{2}\end{array}\)

Question 51:
If $a, b, c$ are the roots of $x^{3}-x^{2}-1=0,$ what's the value of $\frac{a}{b c}+\frac{b}{c a}+\frac{c}{a b}$ ?
[1] -1
[2] 1
[3] 2
[4] -2
Solution:

Under the precondition you can write
$(x-a)(x-b)(x-c)=0=x^{3}-x^{2}-1$
Expanding the product on the left gives
$x^{3}-(a+b+c) x^{2}+(a b+a c+b c) x-a b c=x^{3}-x^{2}-1$
Now you have to compare/equate the coefficients on both sides of $\left(^{*}\right)$ and get
$a+b+c=1, a b+a c+b c=0, a b c=1$
Using these and the identity $(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(a b+a c+b c)$ for evaluation you get
$\frac{a}{b c}+\frac{b}{a c}+\frac{c}{a b}=\frac{a^{2}+b^{2}+c^{2}}{a b c}$
$=\frac{(a+b+c)^{2}-2(a b+a c+b c)}{a b c}$
$=1$


Question 52:
The sum of the integers in the solution set of |x2-5x|<6 is:
[1] 10
[2] 15
[3] 20
[4] 0
Solution:
|x2-5x|<6
x2 - 5x – 6 < 0
(x – 6)(x + 1) < 0
- 1 < x < 6
x = {0, 1, 2, 3, 4, 5}

And

|x2-5x|<6

-(x2 - 5x) – 6 < 0
x2 – 5x + 6 > 0
(x – 2)(x - 3) > 0
x < 2 or x > 3

Values of x common to both {0, 1, 4, 5}

Sum of values of x = 0 + 1 + 4 + 5 = 10. Option A


Question 53:
Find abc if a+b+c = 0 and a3+ b3+ c3=216
[1] 48
[2] 72
[3] 24
[4] 216
Solution:
a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca)
216 – 3abc = 0
abc = 72. Option B

Question 54:
Solve for x: \(\sqrt {x + \sqrt {x + \sqrt x + ....} } = \frac{3}{2} \)
[1] Empty Set
[2] 3/2
[3] 3/4
[4] 3/16
Solution:
\(\begin{array}{l}\sqrt {x + \sqrt {x + \sqrt x  + ....} }  = \frac{3}{2}\\\sqrt {x + \frac{3}{2}}  = \frac{3}{2}\\x + \frac{3}{2} = \frac{9}{4}\\x = \frac{3}{4}\end{array}\)

Option C


Question 55:
Solve for x \(\sqrt {\frac{3}{2} + \sqrt {\frac{3}{2} + \sqrt {\frac{3}{2}} + ....} } = x \)
[1] \(\frac{{1 \pm \sqrt 7 }}{2} \)
[2] \(\frac{{1 + \sqrt 7 }}{2} \)
[3] \(\frac{{\sqrt 7 }}{2} \)
[4] \(\frac{3}{2} \)
Solution:
\(\begin{array}{l}\sqrt {\frac{3}{2} + \sqrt {\frac{3}{2} + \sqrt {\frac{3}{2}}  + ....} }  = x\\\sqrt {\frac{3}{2} + x}  = x\end{array}\)

3/2 + x = x2

2x2 – 2x – 3 = 0

x = \(\frac{{1 + \sqrt 7 }}{2}\)Option B


Question 56:
What is/are the value(s) of x if \(\sqrt {{x^2} + \sqrt {{x^2} + \sqrt {{x^2} + ...} } } = 9 \)
[1] 6√2
[2] 3√10
[3] ±3√10
[4] ±6√2
Solution:
\(\begin{array}{l}\sqrt {{x^2} + \sqrt {{x^2} + \sqrt {{x^2} + ...} } }  = 9\\\sqrt {{x^2} + 9}  = 9\\{x^2} + 9 = 81\\{x^2} = 72\\x =  \pm 6\sqrt 2 \end{array}\)     

Option D


Question 57:
For x ≠ 1 and x ≠ -1, simplify the following expression: \(\frac{{{\rm{(}}{{\rm{x}}^{\rm{3}}} + 1)({{\rm{x}}^3} - 1)}}{{({{\rm{x}}^2} - 1)}} \)
[1] x4 + x2 + 1
[2] x4 + x3 + x + 1
[3] x6 – 1
[4] x6 + 1
Solution:
Put x = 2, we get 9×7/3 = 21

We get 21 from x4 + x2 + 1. Option A


Question 58:
If √x + √y = 6 and xy = 4 then for: x>0, y>0 give the value of x+y
[1] 2
[2] 28
[3] 32
[4] 34
Solution:
\(\begin{array}{l}\sqrt x  + \sqrt y  = 6\\x + y + 2\sqrt {xy}  = 36\\x + y + 4 = 36\\x + y = 32\end{array}\)

Option C


Question 59:
Find a for which a<b and \(\sqrt {1 + \sqrt {21 + 12\sqrt 3 } } = \sqrt a + \sqrt b \)
[1] 1
[2] 3
[3] 4
[4] None of these
Solution:
\(\begin{array}{l}\sqrt {1 + \sqrt {21 + 12\sqrt 3 } }  = \sqrt {1 + \sqrt {9 + 12 + 2 \times 3 \times 2\sqrt 3 } } \\ = \sqrt {1 + \sqrt {9 + 12 + 2 \times 3 \times 2\sqrt 3 } }  = \sqrt {1 + \sqrt {{{(3 + 2\sqrt 3 )}^2}} }  = \sqrt {1 + 3 + 2\sqrt 3 } \\ = \sqrt {{{(1 + \sqrt 3 )}^2}}  = 1 + \sqrt 3 \end{array}\)

Since a < b, b = 3 & a = 1. Option A


Question 60:
One root of the following given equation \(2{x^5} - 14{x^4} + 31{x^3} - 64{x^2} + 19x + 130 = 0 \) is
[1] 1
[2] 3
[3] 5
[4] 7
Solution:
f(x) = \(2{x^5} - 14{x^4} + 31{x^3} - 64{x^2} + 19x + 130 = 0\)

f(1) = 2-1+31-64+19+130 = 117

f(3)= 2(3)5 – 14(3)4 + 31(3)-64(3)2 + 19(3) + 130 = -200

f(5) = 2(5)5 -14(5)4 + 31(5)-64(5)2 + 19(5) + 130 =  0

f(7) = 2(7)5 -14(7)4 + 31(7)-64(7)2 + 19(7) + 130 = 7760

Therefore the root in the above equation is 5. Option C


Question 61:
The equation \(x + \frac{2}{{1 - x}} = 1 + \frac{2}{{1 - x}}, \) has
[1] No real root
[2] One real root
[3] Two equal roots
[4] Infinite roots
Solution:
\(x + \frac{2}{{1 - x}} = 1 + \frac{2}{{1 - x}}\)=>\(\frac{{x - {x^2} + 2}}{{(1 - x)}}{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt}  = {\kern 1pt} \frac{{1 - x + 2}}{{(1 - x)}}\)

=>\(\frac{{{x^2} - x - 2 + 1 - x + 2}}{{(1 - x)}} = 0\)

On solving the above equation : \(\frac{{{x^2} - 2x + 1}}{{1 - x}} = 0\)

The only valid solution could have been x = 1 but that is ruled out because then the denominator would become 0.
No real root exists. Option  A

Question 62:
If \(x = \sqrt {7 + 4\sqrt 3 } , \) then \(x + \frac{1}{x} = \)
[1] 4
[2] 6
[3] 3
[4] 2
Solution:
\(x=\sqrt{7+4\sqrt{3}},\)
= \(\sqrt {{2^2} + {{(3\sqrt 2 )}^2} + 2 \times 2\sqrt 3 } \)

=

substituting in the given equation :

\(x + \frac{1}{x} = 2 + \sqrt 3  + \frac{1}{{2 + \sqrt 3 }} = \frac{{4 + 3 + 4\sqrt 3  + 1}}{{{\rm{2}} + \sqrt {\rm{3}} }} = \frac{{8 + 4\sqrt 3 }}{{2 + \sqrt 3 }} = 4\)

Option A


Question 63:
If A.M. of the roots of a quadratic equation is 8/5 and A.M. of their reciprocals is 8/7, then the equation is
[1] 5x2-16x+7=0
[2] 7x2-16x+5=0
[3] 7x2-16x+8=0
[4] 3x2-12x+7=0
Solution:
Arithmetic Mean of the roots is

 \(\begin{array}{l}\frac{{\alpha  + \beta }}{2} = \frac{8}{5}\\ \to \alpha  + \beta  = \frac{{16}}{5}\end{array}\)

Arithmetic Mean of the reciprocals is given by:

\(\begin{array}{l}\frac{{\frac{1}{\alpha } + \frac{1}{\beta }}}{2} = \frac{8}{7}\\ \to \;\frac{{\alpha  + \beta }}{{2\alpha \beta }} = \frac{8}{7}\\ \to \;\frac{{16}}{{5 \times 2\alpha \beta }} = \;\frac{8}{7}\end{array}\)

On solving this equation , we get \(\alpha \beta  = \frac{7}{5}\)

The equation thus formed will be :

\({x^2} - \frac{{16x}}{5} + \frac{7}{5} = 0\) ie \(5{x^2} - 16x + 7 = 0\)

Answer is option A


Question 64:
The equation x2 + ax + (b + 2) = 0 has real roots. What is the minimum value of a2 + b2?
[1] 0
[2] 1
[3] 2
[4] 4
Solution:
Condition for real roots is

a2 – 4(b+2) ≥ 0

a2 +b2 –(b2 +4b+8) ≥ 0
a2 +b2 ≥ (b2 +4b+8)
a2 +b2 ≥ (b+2)2 + 4
Minimum value of a2 +b2 will occur when b = -2 and it will be 4. Option D

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