- Equations
- Progressions
- Functions
- Maxima Minima
- Logarithms

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 + 4tt =

**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 = 125Values 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 – 39.5H + 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) = 0.50.5(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 x

^{2}+ 4y

^{2}< 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:**

\(x = \frac{{4 - {y^3}}}{4} = 1 - \frac{{{y^3}}}{4}\)

y

^{3}= 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 x

^{3}+3x

^{2}+4x+5=0 and x

^{3}+2x

^{2}+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

x^{3}+3x^{2}+4x+5 = x^{3}+2x^{2}+7x+3

^{2}+4x+5 = 2x

^{2}+7x+3

x

^{2}– 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 x

^{u}=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:**

x

^{u}=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

^{3}= 64 – 48u + 12u

^{2}

u

^{3}– 12u

^{2}+ 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: ax

^{2}+ bx + c, bx

^{2}+ cx + a, and cx

^{2}+ ax + b.

[1] 4

[2] 5

[3] 6

[4] 0

**Solution:**

For these equations to have real roots

b^{2 }– 4ac ≥ 0

^{2 }≥ 4ac

c^{2 }– 4ab ≥ 0

^{2}≥ 4ab

a^{2} – 4ac ≥ 0

^{2}≥ 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) = x

^{4}+ax

^{2}+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 4^{th} root be r

Coefficient of x^{3} is - (Sum of the roots)

r = - 4

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

f(1) = (-1)×4×5×(-4) = 80a + 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 ax

^{2}+bx+1=0 having real roots is

[1] 10

[2] 7

[3] 6

[4] 12

**Solution:**

For the equation to have real roots

b^{2} – 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

x^{2}– 10x + 16 = 0

Manish’s equation

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

x^{2}+ 10x + 9 = 0

Correct equation is x^{2} – 10x + 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 0Both 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 px

^{2}+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+2

^{2/3}+2

^{1/3}, then the value of x

^{3}-6x

^{2}+6x is:

[1] 2

[2] -2

[3] 0

[4] 4

**Solution:**

x =2+2

^{2/3}+2

^{1/3}

x – 2 = 2

^{2/3}+2

^{1/3}

(x – 2)

^{3}= (2

^{2/3}+2

^{1/3})

^{3}

x

^{3 }– 6x

^{2 }+ 12x – 8 = 2

^{2}+ 3. 2

^{4/3}.2

^{1/3}+ 3. 2

^{2/3}.2

^{2/3}+ 2

x

^{3 }– 6x

^{2 }+ 12x – 8 = 4 + 3.2

^{5/3}+ 3.2

^{4/3}+ 2

x

^{3 }– 6x

^{2 }+ 12x – 8 = 6 + 6.2

^{2/3}+ 6.2

^{1/3}= (12 + 6.2

^{2/3}+ 6.2

^{1/3}) – 6

x

^{3 }– 6x

^{2 }+ 12x – 8 = 6x – 6

x

^{3 }– 6x

^{2 }+ 6x =

**2. Option A**

**Question 27:**

If the roots of the equation x

^{2}-2ax+a

^{2}+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) ≥ 0a ≤ 3

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

Put a = 0, we get the equation as x^{2} – 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 x

^{3}– 3x

^{2}+ 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 2z

^{3}+ 4z

^{2}-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 2z

^{3}+ 4z

^{2}-3z -1 =0, then

f(z) = 2z^{3} + 4z^{2} -3z -1 = (z - p) × (z - q) × (z - 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 4x

^{2}+ 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

b^{2} – 4ac > 0

^{2}– 4×4×(4 – 3p) > 0

p

^{2}– (4 – 3p) > 0

p

^{2}+ 3p – 4 > 0

(p + 4)(p – 1) > 0

**p < -4 or p > 1. Option A**

**Question 33:**

If x

^{2}+ 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:**

x

^{2}+ 4x + n > 13

x

^{2}+ 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) = a

_{0}+ a

_{1}x + a

_{2}x

^{2}… + a

_{100}x

^{100 }then the value of a

_{99}is equal to:

[1] 50

[2] 0

[3] -50

[4] -100

**Solution:**

a

_{100}= 1

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

a_{99} = - (– 1 + 2 – 3 + 4 – 5 + 6 … – 99 + 100) = **- 50. Option C**

**Question 35:**

If a, b, and c are the solutions of the equation x

^{3}– 3x

^{2}– 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 x

^{3}– 4x

^{2}+ 3x + 5 = 0, find (a + 1)(b + 1)(g + 1)

[1] -3

[2] 0

[3] 3

[4] 1

**Solution:**

f(x) = x

^{3}– 4x

^{2}+ 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] x

^{4}

[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 3x

^{5}+ 2x

^{4}– 3x

^{3}– x

^{2}+ 2x + 2 is divided by x

^{2}– 1.

[1] 3

[2] 2x – 2

[3] 2x + 3

[4] 2x – 1

**Solution:**

When 3x

^{5}+ 2x

^{4}– 3x

^{3}– x

^{2}+ 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 3x^{5} + 2x^{4} – 3x^{3} – x^{2} + 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

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

*x*

^{3}–

*ax*

^{2}+

*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}

^{3}= 8(m+1)

^{2}

Using options,

3 + 4^{2} + 5^{3} = 3 + 16 + 125 = 144 = 12^{2} = (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}= 8a

^{2}

a – 1 + a

^{3}+ 3a

^{2}+ 3a + 1 = 8a

^{2}

a

^{3}– 5a

^{2}+ 4a = 0

a(a

^{2}– 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

*n*th 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 + 0.15

*n*, on the

*n*th 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 = 110n = 140

On the 140^{th} 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 **20 ^{th} May. Option C**

**Question 47:**

The polynomial f(x)=x

^{2}-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 r

^{2}

Sum of the roots = r^{2} + r = 12

^{2}+ r – 12 = 0

r = -4, 3

Product of the roots = c = r^{3} = -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 < 30No. 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 \(x + \frac{1}{x} = y\)

6y

^{2}– 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 |x

^{2}-5x|<6 is:

[1] 10

[2] 15

[3] 20

[4] 0

**Solution:**

|x

^{2}-5x|<6

x

^{2 }- 5x – 6 < 0

(x – 6)(x + 1) < 0

- 1 < x < 6

x = {0, 1, 2, 3, 4, 5}

And

|x^{2}-5x|<6

^{2 }- 5x) – 6 < 0

x

^{2}– 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 a

^{3}+ b

^{3}+ c

^{3}=216

[1] 48

[2] 72

[3] 24

[4] 216

**Solution:**

a

^{3}+ b

^{3}+ c

^{3}- 3abc = (a + b + c)(a

^{2}+ b

^{2}+ c

^{2}- 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 = x^{2}

2x^{2} – 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] x

^{4}+ x

^{2}+ 1

[2] x

^{4}+ x

^{3}+ x + 1

[3] x

^{6}– 1

[4] x

^{6}+ 1

**Solution:**

Put x = 2, we get 9×7/3 = 21

We get 21 from **x ^{4} + x^{2} + 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)^{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**

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

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] 5x

^{2}-16x+7=0

[2] 7x

^{2}-16x+5=0

[3] 7x

^{2}-16x+8=0

[4] 3x

^{2}-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 x

^{2}+ ax + (b + 2) = 0 has real roots. What is the minimum value of a

^{2}+ b

^{2}?

[1] 0

[2] 1

[3] 2

[4] 4

**Solution:**

Condition for real roots is

a^{2} – 4(b+2) ≥ 0

^{2}+b

^{2}–(b

^{2}+4b+8) ≥ 0

a

^{2}+b

^{2}≥ (b

^{2}+4b+8)

a

^{2}+b

^{2}≥ (b+2)

^{2}+ 4

Minimum value of a

^{2}+b

^{2}will occur when b = -2 and it will be

**4. Option D**