**Question 1:**

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

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 2:**

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

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

y

x = 0 or -3 or -15 or -53

No. of valid values of x =

y

^{3}= 4(1-x) = 4 – 4xx = 0 or -3 or -15 or -53

No. of valid values of x =

**4. Option D****Question 3:**

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

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

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 =

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

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

^{2}– 3x + 2 = 0x = 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 4:**

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

x

u \(lo{g_2}x\) = 8

\(lo{g_2}x\) =\(\frac{8}{u}\)

u

(u -4)

u = 4

\(lo{g_2}x\)= \(\frac{8}{u}\) = 2

x = 2

We have

^{u}=256u \(lo{g_2}x\) = 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}= 0u = 4

\(lo{g_2}x\)= \(\frac{8}{u}\) = 2

x = 2

We have

**exactly one solution for x. Option B****Question 5:**

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

For these equations to have real roots

b^{2 }– 4ac ≥ 0

^{2 }≥ 4acc^{2 }– 4ab ≥ 0

^{2}≥ 4aba^{2} – 4ac ≥ 0

^{2}≥ 4acMultiplying 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**

- Algebra Practice Test 1
- Algebra Practice Test 2
- Algebra Practice Test 3
- Algebra Practice Test 4
- Algebra Practice Test 5
- Algebra Practice Test 6
- Algebra Practice Test 7
- Algebra Practice Test 8
- Algebra Practice Test 9
- Algebra Practice Test 10
- Algebra Practice Test 11
- Algebra Practice Test 12
- Functions Questions for CAT with Solutions
- [Difficult] Sequence and Series Questions
- TOUGH Quadratic Equations Questions
- CAT Logarithm Questions [Difficult]