ALGEBRA MCQ LIST

 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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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.

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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

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Question 13:
Given the system of equations $\{\begin{array}{c}2x+y+2z=4\\ x+2y+3z=-1\\ 3x+2y+z=9\end{array}$, find the value of x+y+z.
[1] -1
[2] 3.5
[3] 2
[4] 1

Solution:
The given equations are

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

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

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

Take the first and the second equation :

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

Now take equation (2) and (3)

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

$\begin{array}{c}\\ -3x-6y-9z=3\\ 3x+2y+z=9\end{array}$

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

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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

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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.

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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:
$\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\dots }}}}=x$
$\sqrt{2+x}=x$
$2+x=x^2$
$x^2-x-2=0$
$x=2,-1$
as the value of the expression will be positive, we can reject x=-1.
Hence, x=2.

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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:
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{ab+bc+ca}{abc}=\frac{\frac{Coefficientofx}{Coefficientof{x}^3}}{\frac{-Const}{Coefficientof{x}^3}}=-\frac{Coefficientofx}{Const}=-1$Option B

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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

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Question 31:
If $\alpha ,\beta$ and $\gamma$ are the roots of the equation $x^3-7x+3=0$ 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

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

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

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

$=49$

Then

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

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

Option D

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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

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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

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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

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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

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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

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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

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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

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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

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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:

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

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

$x^2+\frac{1}{x^2}=7$ Equation (2)

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

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

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

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

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

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

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

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

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

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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:
$\begin{array}{cc}& \sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+.....}}}}=10\\ ;& \sqrt{x+10}=10\end{array}$

x + 10 = 100

x = 90. Option B

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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,
${\alpha }^4+{\beta }^4=(\alpha +\beta {)}^4+2(\alpha \beta {)}^2-4\alpha \beta (\alpha +\beta {)}^2$
$=(1{)}^4+2(-6{)}^2-(4\times -6)(1{)}^2$
$=1+72+24$
$=97$

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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:
$\begin{array}{c}\sqrt{4-\sqrt{4+\sqrt{4-\sqrt{4+...}}}}=x\\ \sqrt{4-\sqrt{4+x}}=x\\ 4-\sqrt{4+x}=x^2\\ 4-x^2=\sqrt{4+x}\\ 16-8x^2+x^4=4+x\\ x^4-8x^2-x+12=0\end{array}$

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

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Question 44:
If the roots of the equation x3 – ax2 + 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

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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

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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

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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

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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

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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

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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:
$\begin{array}{c}\frac{5}{3+\frac{5}{3+\frac{5}{3+...}}}=x\\ \Rightarrow \frac{5}{3+x}=x\\ \Rightarrow 5=3x+x^2\\ \Rightarrow x^2+3x-5=0\\ \Rightarrow x=\frac{-3\pm \sqrt{3^2-4\times 1\times (-5)}}{2}\\ \Rightarrow x=\frac{-3\pm \sqrt{29}}{2}\\ \Rightarrow x=\frac{-3+\sqrt{29}}{2}\end{array}$

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Question 51:
If $a,b,c$ are the roots of $x^3-x^2-1=0,$ what's the value of $\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}$ ?
[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
$x^3-(a+b+c)x^2+(ab+ac+bc)x-abc=x^3-x^2-1$
Now you have to compare/equate the coefficients on both sides of $\left({}^{\ast }\right)$ and get
$a+b+c=1,ab+ac+bc=0,abc=1$
Using these and the identity $(a+b+c{)}^2=a^2+b^2+c^2+2(ab+ac+bc)$ for evaluation you get
$\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab}=\frac{a^2+b^2+c^2}{abc}$
$=\frac{(a+b+c{)}^2-2(ab+ac+bc)}{abc}$
$=1$

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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

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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

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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:
$\begin{array}{c}\sqrt{x+\sqrt{x+\sqrt{x}+....}}=\frac{3}{2}\\ \sqrt{x+\frac{3}{2}}=\frac{3}{2}\\ x+\frac{3}{2}=\frac{9}{4}\\ x=\frac{3}{4}\end{array}$

Option C

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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:
$\begin{array}{c}\sqrt{\frac{3}{2}+\sqrt{\frac{3}{2}+\sqrt{\frac{3}{2}}+....}}=x\\ \sqrt{\frac{3}{2}+x}=x\end{array}$

3/2 + x = x2

2x2 – 2x – 3 = 0

x = $\frac{1+\sqrt{7}}{2}$Option B

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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:
$\begin{array}{c}\sqrt{x^2+\sqrt{x^2+\sqrt{x^2+...}}}=9\\ \sqrt{x^2+9}=9\\ x^2+9=81\\ x^2=72\\ x=\pm 6\sqrt{2}\end{array}$     

Option D

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Question 57:
For x ≠ 1 and x ≠ -1, simplify the following expression: $\frac{(x^3+1\left)\right(x^3-1)}{(x^2-1)}$
[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

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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:
$\begin{array}{c}\sqrt{x}+\sqrt{y}=6\\ x+y+2\sqrt{xy}=36\\ x+y+4=36\\ x+y=32\end{array}$

Option C

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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:
$\begin{array}{c}\sqrt{1+\sqrt{21+12\sqrt{3}}}=\sqrt{1+\sqrt{9+12+2\times 3\times 2\sqrt{3}}}\\ =\sqrt{1+\sqrt{9+12+2\times 3\times 2\sqrt{3}}}=\sqrt{1+\sqrt{{(3+2\sqrt{3})}^2}}=\sqrt{1+3+2\sqrt{3}}\\ =\sqrt{{(1+\sqrt{3})}^2}=1+\sqrt{3}\end{array}$

Since a < b, b = 3 & a = 1. Option A

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Question 60:
One root of the following given equation $2x^5-14x^4+31x^3-64x^2+19x+130=0$ is
[1] 1
[2] 3
[3] 5
[4] 7

Solution:
f(x) = $2x^5-14x^4+31x^3-64x^2+19x+130=0$

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

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

f(5) = 2(5)5 -14(5)4 + 31(5)3  -64(5)2 + 19(5) + 130 =  0

f(7) = 2(7)5 -14(7)4 + 31(7)3  -64(7)2 + 19(7) + 130 = 7760

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

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

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

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

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Question 62:
If $x=\sqrt{7+4\sqrt{3}},$ then $x+\frac{1}{x}=$
[1] 4
[2] 6
[3] 3
[4] 2

Solution:
$x=\sqrt{7+4\sqrt{3}},$
= $\sqrt{2^2+{\left(3\sqrt{2}\right)}^2+2\times 2\sqrt{3}}$

=

substituting in the given equation :

$x+\frac{1}{x}=2+\sqrt{3}+\frac{1}{2+\sqrt{3}}=\frac{4+3+4\sqrt{3}+1}{2+\sqrt{3}}=\frac{8+4\sqrt{3}}{2+\sqrt{3}}=4$

Option A

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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

 $\begin{array}{c}\frac{\alpha +\beta }{2}=\frac{8}{5}\\ \to \alpha +\beta =\frac{16}{5}\end{array}$

Arithmetic Mean of the reciprocals is given by:

$\begin{array}{c}\frac{\frac{1}{\alpha }+\frac{1}{\beta }}{2}=\frac{8}{7}\\ \to \frac{\alpha +\beta }{2\alpha \beta }=\frac{8}{7}\\ \to \frac{16}{5\times 2\alpha \beta }=\frac{8}{7}\end{array}$

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 $5x^2-16x+7=0$

Answer is option A

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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