Basic Algebra(Exams > Cat > Quantitaitve Aptitude ) Questions and answers for exam Preparation

Exams > Cat > Quantitaitve Aptitude

BASIC ALGEBRA MCQs

Total Questions : 90 | Page 1 of 9 pages

Question 1. How many different straight lines can be formed by joining 14 different points of which 6 are collinear and rest are non-collinear?

Discuss Question

Answer: Option D. -> 77
:
D
No. of lines formed by joining 14 points =

^{14} C_{2}

No. of lines formed by joining 6 points =

^{6} C_{2}

Out of the 14 points, when you join the 6 collinear points, you don’t get

^{6} C_{2}

lines, but you get only one line.
So, no. of lines

=^{14} C_{2} -^{6} C_{2} + 1 = 91 - 15 + 1 = 77

.
2nd method:- By selection,

^{8} C_{2} +^{8} C_{1} \times^{6} C_{1} + 1 = 77

Question 2. A basket has 5 oranges and 4 apples. In how many ways can you make a selection if you have to take at least 1 orange and 1 apple?

Discuss Question

Answer: Option D. -> 465
:
D
At least 1 orange can be selected in

2^{5} - 1 = 31

ways.
At least 1 apple can be selected in

2^{4} - 1 = 15

ways.
Total no. of selections possible

= 31 \times 15 = 465

Question 3. If

x^{3} - k^{2} x^{2} + (6 k + 5) x - 5 k = 0

has two roots 3 and 5, then which of the following is the third root?

Discuss Question

Answer: Option A. -> 1
:
A
Assume

α

be the 3rd root.
Sum of the roots, i.e.,

3 + 5 + α = k^{2}

Product of the roots, i.e.

3 \times 5 \times α = 5 k

So,

8 + α = 9 α^{2}

Then,

α = 1 o r - \frac{8}{9}

Hence the 3rd root is 1.

Question 4. Suppose a and b are two roots of the equation

x^{2} - (α - 4) x + α = 0

. Find out the maximum possible value of

5 a b - a^{2} - b^{2} .

0
1614
39
5
Can’t be determined

Discuss Question

Answer: Option B. -> 1614
:
B
a and b are the roots of the equation

a + b = (α - 4), a b = α

5 a b - a^{2} - b^{2}

5 a b - (a^{2} + b^{2})

5 a b - [(a + b)^{2} - 2 a b]

5 α - [(α - 4)^{2} - 2 α]

5 α - [α^{2} - 10 α + 16]

- α^{2} + 15 α - 16 = - (α^{2} + 15 α - 16) = - (α^{2} + 16 α - α - 16) = - [α (α + 16) - 1 (α + 16)] = - (α + 16) (α - 1)

So,the maximum value

= \frac{D}{4 a} = \frac{[15^{2} - 4 (- 1) (- 16)]}{4 \times 1} = \frac{161}{4} .

Question 5. For the equation

x^{8} + 6 x^{7} - 5 x^{4} - 3 x^{2} + 2 x - 5 = 0

, determine the maximum number of real roots possible.

4
3
8
7
Can’t be determined

Discuss Question

Answer: Option A. -> 4
:
A

f (x) = x^{8} + 6 x^{7} - 5 x^{4} - 3 x^{2} + 2 x - 5

Check the number of positive roots= no. of sign changes in f(x) = 3
Check the number of negative roots = no. of sign changes in f(-x)

f (- x) = x^{8} - 6 x^{7} - 5 x^{4} - 3 x^{2} - 2 x - 5

. No. of sign changes = 1
Now check for zero as a root.

f (0) = - 5 \neq 0

So, maximum number of real roots = 3 + 1 = 4. Hence option (a)

Question 6. How many distinct real roots are possible for the below equation?

x^{6} - 26 x^{3} - 27 = 0

Discuss Question

Answer: Option B. -> 2
:
B
Let

x^{3} = y

y^{2} - 26 y - 27 = 0

y^{2} - 27 y + y - 27 = 0

y (y - 27) + 1 (y - 27) = 0

(y + 1) (y - 27) = 0

The roots are y = -1, y = 27

x^{3} = - 1, x = - 1

x^{3} = 27, x = 3

Question 7. The A.M. of two positive numbers is 15 and their GM is 12. What is the bigger number?

Discuss Question

Answer: Option C. -> 24
:
C
Let the numbers be x and y.

x + y = 30

xy = 144

(x - y)^{2} = (x + y)^{2} - 4 x y

900 - (4 \times 144) = 324

hence

(x - y) = 18

Solving:

x - y = 18 & x + y = 30

x = 24, y = 6
So, the bigger number is 24.

Question 8. Find the roots of the following equations:

2^{x + 3} + 2^{- x} = 6

-1, -2
1, -2
- 1, - 2
1, 2

Discuss Question

Answer: Option C. -> - 1, - 2
:
C

2^{x} {.2}^{3} + 2^{- x} = 6

Let

2^{x} = z

8 z + \frac{1}{z} = 6

8 z^{2} + 1 = 6 z

8 z^{2} - 6 z + 1 = 0

8 z^{2} - 4 z - 2 z + 1 = 0

4 z (2 z - 1) - 1 (2 z - 1) = 0

(4 z - 1) (2 z - 1) = 0

z = \frac{1}{4},

z = \frac{1}{2}

2^{x} = \frac{1}{4}

x = - 2

2^{x} = \frac{1}{2}

x = - 1

Roots are

x = - 2, x = - 1

Question 9. Find the number of terms of the sequence 32, 24, 16, 8 ... for which the sum of the terms is zero.

Discuss Question

Answer: Option D. -> 9
:
D

S_{n} = (\frac{n}{2}) [2 a + (n - 1) d]

\Rightarrow 0 = (\frac{n}{2}) [64 + (n - 1) (- 8)]

\Rightarrow 0 = (\frac{n}{2}) [64 - (n - 1) 8]

\Rightarrow

n (36– 4n) = 0
n = 0; n=9
As, n can not be equal to 0, so, the number of terms required is 9.

Question 10. In how many ways can a pack of 52 cards be divided equally among 4 players?

(52!)[(4!)13]
(52!)[(13!)4]4!
(52!)[(13!)4]
(52!)(4!)
None of these

Discuss Question

Answer: Option C. -> (52!)[(13!)4]
:
C
Here as it has to be divided among 4 different players, the order is important.
No. of ways =

\frac{(52!)}{[(13!)^{4}]}

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