  # Algebra Practice Quiz 5 Steps - 3 Clicks

# Algebra Practice Quiz

### Introduction

The article Algebra Practice Quiz provides information about Algebra, a important topic of Quantitative Aptitude section. Consists of different types Algebra questions with solutions useful for candidates preparing for different competitive examinations like RRB ALP/Technical Exams/Junior Engineer Recruitment Exams, SSC, IBPS PO Exams and etc. Algebra Practice Quiz is very useful to crack the questions from the algebra sections.

Algebra is a division of mathematics designed to help solve certain types of problems quicker and easier. Algebra is based on the concept of unknown values called variables, unlike arithmetic which is based entirely on known number values.

### Quiz

1. Which of the following equations has real roots?

A. $${3x}^{2} + 4x + 5 = 0$$
B. $${x}^{2} + x + 4 = 0$$
C. (x – 1) (2x – 5) = 0
D. $${2x}^{2} – 3x + 4 = 0$$

Explanation –

(x – 1) (2x – 5) = 0

x = 1, x = $$\frac{5}{2}$$

Both roots are real.

For real roots of quadratic equation

$${ax}^{2}$$ + bx + c = 0

we have $${b}^{2}$$ – 4ac ≥= 0

For $${3x}^{2}$$+ 4x + 5 = 0

$${b}^{2}$$ – 4ac = 16 – 4 * 3 * 5 = – 44

For $${x}^{2}$$ + x + 4 = 0

$${b}^{2}$$ – 4ac = 1 – 4 * 1 * 4 = – 15

For $${2x}^{2}$$ – 3x + 4 = 0

$${b}^{2}$$– 4ac = 9 – 4 * 2 * 4 = – 23

2. 3 chairs and 2 tables cost 700, while 5 chairs and 3 tables cost 1100. What is the cost of 2 chairs and 2 tables ?

A. 300
B. 350
C. 450
D. 600

Explanation –

Let cost of chair = x and cost of table = y

i.e, 3x + 2y = 700 …(i)

and 5x + 3y = 1100 …(ii)

Solving (i) and (ii) we get

x = 100 and y = 200

Now 2x + 2y = 200 + 400 = 600

3. If x = $$\frac{4ab}{a + b}$$, then value of $$\frac{x + 2a}{x – 2a}$$ + $$\frac{x + 2a}{x – 2a}$$ is equal to

A. 0
B. 1
C. 2
D. None of these

Explanation –

$$\frac{x + 2a}{x – 2a} + \frac{x + 2b}{x – 2b}$$ = $$\frac{\frac{6ab + {2a}^{2}}{a + b}}{\frac{2ab -{2a}^{2}}{a + b}} + \frac{\frac{2ab + {6b}^{2}}{a + b}}{\frac{2ab – {2b}^{2}}{a + b}}$$

$$\frac {2a(3b + a)}{2a(b – a)} + {2b(3b + a)}{2b(a – b)}$$

$$\frac {(3b + a)}{(b – a)} + {(3b + a)}{(a – b)}$$ = 0

4. If a = $$\frac{x}{x + y}$$ and b = $$\frac{y}{x – y}$$, then $$\frac{ab}{a – b}$$ is equal to

A. $$\frac{xy}{{x}^{2} + {y}^{2}}$$
B. $$\frac {{x}^{2} + {y}^{2}} {xy}$$
C. $$\frac{x}{x + y}$$
D. $$\frac{y}{x – y}$$

Explanation –

$$\frac{ab}{a + b}$$ = $$\frac{(\frac{x}{x + y}) * (\frac{x}{x – y})}{(\frac{x}{x + y}) * (\frac{y}{x – y})}$$

= $$\frac{xy}{{x}^{2} + {y}^{2}}$$

5. If $$\frac{y}{x – z}$$ = $$\frac{x + y}{z}$$ = $$\frac{x}{y}$$ determine the ratio of x : y : z

A. 1 : 2 : 3
B. 3 : 2 : 1
C. 4 : 2 : 3
D. 2 : 4 : 7

Explanation –

Substitute in the choices.

Using x = 4, y = 2, z = 3, we find the equation is satisfied

6. Sum of two numbers is 15 and sum of their reciprocals is $$\frac{15}{56}$$ The two numbers are

A. 4, 11
B. 5, 10
C. 6, 9
D. 7, 8

Explanation –

$$\frac{x + y + z}{4(x + y + z)}$$ = a = $$\frac{1}{4}$$

7. If a + b + c = 0, find the value of $$\frac{{a}^{2}}{{a}^{2} – {bc}} + \frac{{b}^{2}}{{b}^{2} – {ca}} + \frac{{c}^{2}}{{c}^{2} – {ab}}$$

A. 4
B. 3
C. 1
D. None of these

Explanation –

= $$\frac{{a}^{2}}{{a}^{2} – {bc}} + \frac{{b}^{2}}{{b}^{2} – {ca}} + \frac{{c}^{2}}{{c}^{2} – {ab}}$$

= $$\frac {a}{a}(\frac{{2a}^{2} – bc}{bc}) + \frac {b}{b} (\frac {{2b}^{2}-ac} – {ca}) + \frac{c}{c}(\frac{{2c}^{2} – ab} {ab})$$

[Applying componendo and dividendo on each terms]

= $$( \frac {{2a}^{3} – abc}{abc}) + ( \frac{{2b}^{3} – abc}{abc}) + ( \frac{{2c}^{3} – abc} {abc})$$

= $$2 ( \frac{{a}^{3} + {b}^{3} + {c}^{3}} {abc})$$

$$\frac{6abc – 3abc}{abc}$$ = 3

8. If p, q, r, s are in harmonic progression and p > s, then

A. $$\frac{1}{ps} < \frac{1}{qr}$$
B. -q + r = p + s
C. $$\frac{1}{p} + \frac{1}{q} = \frac{1}{r} + \frac{1}{s}$$
D.None of these

Explanation –

p, q, r, s are in H. P.

= $$\frac{1}{p}, \frac{1}{q}, \frac{1}{r}, \frac{1}{s}$$ are in A.P

= $$\frac{1}{p} – \frac{1}{q} = \frac{1}{r} – \frac{1}{s}$$

= $$\frac{1}{q} + \frac{1}{r} = \frac{1}{p} + \frac{1}{s}$$

9. If $${2x}^{2} – 7 xy + {3y}^{2}= 0$$, then the value of x : y is

A. 3 : 2
B. 2 : 3
C. 3 : 1 and 1 : 2
D. 5 : 6

Explanation –

$${2x}^{2} – 7 xy + {3y}^{2}= 0$$

= $$\frac{7 ± \sqrt{49 – 24} }{2 * 2}$$ = $$\frac{7 ± 5}{4} = 3 \frac{1}{2}$$

Thus x : y = 3 : 1 and 1 : 2

10. The difference between the logarithms of sum of the squares of two positive numbers A and B and the sum of logarithms of the individual numbers is a constant C. If A = B, then C is

A. 2
B. 1.3031
C. log 2
D. exp (2)

11. The arithmetic mean of the series 1, 2, 4, 8, 16, … $${2}^{n}$$ is

A. $$\frac{{2}^{n} – 1}{n}$$
B. $$\frac{{2}^{n + 1} – 1}{n + 1}$$
C. $$\frac{{2}^{n} + 1}{n + 1}$$
D. $$\frac{{2}^{n} – 1}{n + 1}$$

Explanation –

A.M. of the series 1, 2, 4, 8, 16, ….$${2}^{n}$$

= $$\frac{Sum of (n + 1) terms of the given series}{n + 1}$$

= $$\frac{\frac{{2}^{n + 1} – 1}{ 2 – 1}}{\frac{{2}^{n + 1} – 1}{n + 1}}$$

= $$\frac{{2}^{n + 1} – 1}{n + 1}$$

Since series is in G.P. with first term a = 1

and common ratio r = 2

Number of terms = n + 1

Sum of this series, $${S}_{n + 1}$$ = $$a \frac{{r}^{n} – 1}{r – 1}$$

r = 2 > 1

12. If the sum of the 6th and the 15th elements of an arithmetic progression is equal to the sum of the $${7}^{th}, {10}^{th}, {12}^{th}$$ elements of the same progression, then which elements of the series
should necessarily be equal to zero ?

A. $${10}^{th}$$
B. $${8}^{th}$$
C. $${1}^{th}$$
D. None of these

Explanation –

Let a be the first term and d be common ratio of an A.P.

i.e, (a + 5d) + (a + 14d)

= (a + 6d) + (a + 9d) + (a + 11d)

a + 7d = 0

Hence 8th term = 0

13. Let $${S}_{n}$$ denote the sum of the first ‘n’ terms of an A.P. $${S}_{2n} = {3S}_{n}$$. Then the ratio $${S}_{3n}{S}_{n}$$ is equal to

A. 4
B. 6
C. 8
D. 10

Explanation –

Let a be the first term and d be the common difference of an A. P.

$${S}_{n}$$ = $$\frac{n}{2}(2a + (n – 1)d)$$

$${S}_{2n}$$ = $$\frac{2n}{2}(2a + (2n – 1)d)$$

$${S}_{3n}$$ = $$\frac{3n}{2}(2a + (3n – 1)d)$$

$${S}_{2n}$$ = $${3S}_{n}$$

n{2a + (2n – 1) d} = 3 * $$\frac{n}{2}(2a + (2n – 1)d)$$

4a + (4n – 2) d = 6a + (3n – 3) d

(n + 1) d = 2a

d = $$\frac{2a}{n + 1}$$

i.e, $${S}_{n}$$ = $$\frac{n}{2}(2a + (n – 1) \frac{2a}{n + 1})$$ = $$\frac{{2an}^{2}}{n + 1}$$

and $${S}_{3n}$$ = $$\frac{3n}{2}(2a + (3n – 1)\frac{2a}{n + 1})$$ = $$\frac{{12an}^{2}}{n + 1}$$

$$\frac{{S}_{n}}{{S}_{3n}}$$ = 6

14. The term independent of x in the expansion of $${2x + \frac{1}{3x}}^{6}$$ is

A. $$\frac{160}{9}$$
B. $$\frac{80}{9}$$
C. $$\frac{160}{27}$$
D. $$\frac{80}{3}$$

15. Find the term independent of x in $${({3x}^{2} – \frac{2}{{x}^{2}})}^{20}$$

A. 126070 * $${3}^{18} * {2}^{9}$$
B. 184756 * $${3}^{10} * {2}^{10}$$
C. 120000 * $${3}^{20} * {2}^{19}$$
D. 320 * $${{3}^{4}}^{20} * {2}^{21}$$

1. Find the value of $${676}^{2} – {33}^{2}.$$

A. 3200
B. 3400
C. 3146
D. 3143

Explanation –

$${676}^{2} – {33}^{2}$$ = (67 + 33) (67 – 33)

= 100 × 34 = 3400

2. Given that log 2 = 0.3 approx., one billion would be approx

A. $${2}^{9}$$
B. $${2}^{10}$$
C. $${2}^{20}$$
D. $${2}^{30}$$

Explanation –

1 billion = $${10}^{9}$$

Let $${10}^{9}$$ = $${2}^{N}$$

Taking log both sides we get,

9 lot 10 = n log 2

9 = n × 0.3

n = 30

3. In how many different ways can 3 identical white balls and 2 identical red balls be arranged besides each other, in a straight line?

A. 6
B. 10
C. 12
D. 120

Explanation –

Required number of ways

$$\frac{5!}{3! * 2!}$$ = 10

4. What is the probability of getting 3 aces if three cards are drawn from a set of 52 playing cards?

A. $${52}^{3}$$
B. $$\frac{1}{{52}^{3}}$$
C. $$\frac{1}{52!}$$
D. $$\frac{4 * 3 * 2}{52 * 51 * 50}$$

Explanation –

Required probability

$$\frac{^{4}{C}_{3}}{^{52}{C}_{3}}$$ = $$\frac{4 * 3 * 2}{52 * 51 * 50}$$

5. In a class of 40 students, 25 are sports persons and 25 are mathematicians. What is the probability that the monitor of the class is both a sports person and a mathematician?

A. $$\frac{1}{40}$$
B. $$\frac{1}{25}$$
C. $$\frac{1}{4}$$
D. $$\frac{1}{50}$$

Explanation –

Number of boys who are both sports person and a mathematician = 254 + 25 – 10 = 40

Hence, required probability = $$\frac{10}{40} = \frac{1}{4}$$

6. Sum of two numbers is 15 and sum of their reciprocals is $$\frac{15}{56}$$. The two numbers are

A. 4, 11
B. 5, 10
C. 6, 9
D. 7, 8

Explanation –

a + b = 15 and $$\frac{1}{a} + \frac{1}{b} = \frac{15}{56}$$

$$\frac{a + b}{ab} = \frac{15}{56}$$

(a, b) = (7, 8)

7. If α, β are the roots of quadratic equation $${x}^{2}$$+ x + 1 = 0 , then $$\frac{1}{α}+ \frac{1}{β}$$ is

A. -1
B. 1
C. 0
D. None of these

Explanation –

$$\frac{1}{α} + \frac{1}{β} = \frac{α + β}{αβ}$$

given, α + β = –1 and αβ = 1

Hence, $$\frac{α + β}{αβ} = \frac{-1}{1} = -1$$

8. Value of $$\sqrt{6 + \sqrt{6 +\sqrt{6 + …..}}}$$ is

A. $$\frac{5}{2}$$
B. -2
C. 3
D.4

Explanation –

$$\sqrt{6 + \sqrt{6 +\sqrt{6 + …..}}}$$ = x

squaring both sides we get,

6 + x = $${x}^{2}$$

x = 3

9. If a, b, c, d, e and f are in arithmetic progression, then e – c is equal to

A. 2(b – a)
B. c – b
C. 2 (f – d)
D. 2(d – b)

e – c = (e – d) + (d – c)

= 2(b – a)

10. Find the median of the following numbers : 14, 23, 20, 12, 11, 15, 24, 17, 9, 21, 25

A. 15
B. 20
C. 17
D. 14

14, 23, 20, 12, 11, 15, 24, 17, 9, 21, 25

When written in ascending order becomes: 9, 11, 12, 14, 15, 17, 20, 21, 23, 24, 25

Hence, median or mid-term = 17

11. A student was asked to multiply a number by 12. By mistake he multiplied the number by 21 and got the answer 63 more than the correct answer. What is the correct answer?

A. 9
B. 8
C. 7
D. 84

Explanation –

set the number be x.

x × 21 = x + 12 + 63

x = 7

Hence, correct answer = 12 × 7 = 84

12. Find the value of $$\frac{{768}^{3} + {232}^{3}}{{768}^{2} – (768 * 232) + {232}^{2}}$$

A. 1000
B. 536
C. 500
D. 268

Explanation –

$$\frac{{768}^{3} + {232}^{3}}{{768}^{2} – (768 * 232) + {232}^{2}}$$

$$(768 + 232)(\frac{{768}^{2} + {232}^{2}}){{768}^{2} – (768 * 232) + {232}^{2}}$$

= 768 + 232

= 1000

13. Find the value of (1 + 2 + 3 + 4 +……….+ 45) :

A. 2140
B. 2070
C. 1035
D. 1280

Explanation –

Required sum = (1 + 4S) ×45

= 23 × 4S – 1035

14. In an examination, a student gets 4 marks for every correct answer and loses 1 mark for even’ wrong answer. If he attempts in all 60 questions and secures 130 marks, then find the number of questions he attempted correctly.

A. 42
B. 48
C. 36
D. 38

Explanation –

set the no. of correct questions be x

x × 4 – (60 – x) × 1 = 130

Sx – 60 = 130

x = 38

15. If $$\frac{x}{y} = \frac{6}{5}$$, then find the value of $$\frac{{x}^{2} + {y}^{2}}{{x}^{2} – {y}^{2}}$$:

A. 11
B. $$\frac{61}{11}$$
C. $$\frac{11}{5}$$
D. 6

Explanation –

$$\frac{x}{y} = \frac{6}{5}$$

$$\frac{{x}^{2} + {y}^{2}}{{x}^{2} – {y}^{2}}$$ = $$\frac{{6}^{2} + {5}^{2}}{{6}^{2} – {5}^{2}}$$ = $$\frac{61}{11}$$

1. $${log}_{4}^{5} * {log}_{5}^{6} * {log}_{6}^{7}$$ is equal to:

A. $$log \frac{7}{4}$$
B. $${log}_{4}^{7}$$
C. $$log \frac{4}{7}$$
D. $${log}_{7}^{4}$$

Explanation –

$${log}_{4}^{5} * {log}_{5}^{6} * {log}_{6}^{7}$$

$$\frac{{log}_{4}^{5}}{{log}_{10}^{4}} * \frac{{log}_{5}^{6}}{{log}_{10}^{4}} * \frac{{log}_{6}^{7}}{{log}_{10}^{4}}$$

$$\frac{{log}_{10}^{7}}{{log}_{10}^{4}} = {log}_{4}^{7}$$

2. The sum of first n odd natural numbers is:

A. $${n}^{2} – 1$$
B. $${n}^{2}$$
C. $${(n + 1)}^{2}$$
D. $${(n – 1)}^{2}$$

Explanation –

The sum of first n odd natural numbers. = n2

1 = $${1}^{2}$$

1 + 3 = $${2}^{2}$$

1 + 3 + 5 = $${3}^{2}$$

3. A person puts 1 grain of rice in the first square of a chess board. In the subsequent squares, he puts twice that of the previous square. How many grains would he need to put on all the squares of the chess board?

A. 64!
B. $${2}^{64} – 1$$
C. $${2}^{63} – 1$$
D. p(64, 2)

Explanation –

Number of grains = 1 + 2 + 4 + 8 + …

$${2}^{0} + {2}^{1} + {2}^{2} +….{2}^{63}$$

$$\frac{1 * ({2}^{64} – 1)}{(2 -1)} = {2}^{64} – 1$$

i.e, $${S}_{{n}_{(G.P)}}$$ = $$\frac{a * ({r}^{n} – 1)}{(r -1)}$$

4. Which of the following statement is correct?

A. n linear equations with n variables may have a unique solution
B. n linear equations with n variables may have no solution
C. Both (a) & (b) are correct
D. Both (a) & (b) are wrong

5. Find the next number in the series. 1,2,6,24, 120,___________.

A. 240
B. 480
C. 560
D. 720

Explanation –

Ans = 120 × 6

= 720

6. A coin is tossed two times. On both occasions, the result is heads. When the coin is tossed a
third time, what is the probability of getting a head?

A. 1
B. $$\frac{1}{2}$$
C. $$\frac{1}{4} * \frac{1}{2}$$
D. $$\frac{3}{4} * \frac{1}{2}$$

Explanation –
Probability of getting head in third occasion will be $$\frac{1}{2}$$, since it is independent of the first two tosses

7. If 3x + 7 = $${x}^{2}$$ + p = 7x + 5, then the value of ‘p’ will be :

A. $$\frac{1}{2}$$
B. 8$$\frac{1}{2}$$
C. 2 $$\sqrt{2}$$
D. 8 $$\frac{1}{4}$$

Explanation –

3x + 7 = $${x}^{2}$$ + p = 7x + 5

3x + 7 = 7x + 5

4x = 2

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

i.e, 2($$\frac{1}{2}$$) + 7 = ($${\frac{1}{2}}^{2}$$) + p

$$\frac{3}{2}$$ + 7 = $$\frac{1}{4}$$ + p

p = 8 $$\frac{1}{4}$$

8. If $${log}_{10}7$$ = x, then the value of $${log}_{10} (\frac{1}{70})$$ is equal to:

A. – (1 + x)
B. $${(1 + x)}^{-1}$$
C. $$\frac{x}{10}$$
D. $$\frac{1}{10x}$$

Explanation –

$${log}_{10}7$$ = x

$${log}_{10} (\frac{1}{70})$$ = $${log}_{10} (\frac{1}{7 * 10})$$

= –$${log}_{10}7 – {log}_{10}10$$

= – x – 1

9. Find the value of: (51 + 52 + 53+ 54 +…. +100)

A. 3775
B. 5050
C. 1275
D. 2525

Explanation –

S = (51 + 52 + … 100)

$$\frac{50}{2}(51 + 100)$$

$${sum}_{A.P}$$ = $$\frac{n}{2}(a + 1)$$

=3775

10. The sum of the squares of three consecutive odd numbers is 2531. Find these three odd numbers:

A. 21, 23, 25
B. 25, 27, 29
C. 27, 29, 31
D. 23, 25, 27

Explanation –

Let consecutive odd numbers be x, (x + 2), (x + 4)

According the question: $${x}^{2}$$ + $${(x + 2)}^{2}$$ + (x + 4) = 2531

27, 29, 31 satisfies

11. The present age of a father is 3 years more than three times the age of his son. After three years, father’s age will be 10 years more than twice the age of the son. Find the present age of the Father.

A. 30 years
B. 33 years
C. 36 years
D. 39 years

Explanation –

Let present age of father be x

Let age of son be y

According the question:

x = 3 + 3y

Also, (x + 3) = 10 + 2 (y + 3)  y = 10  x = 33

12. Find the value of $$^{10}{C}_{3}$$:

A. 720
B. 240
C. 120
D. 1000

Explanation –

$$^{n}{C}_{r}$$ = $$\frac{n!}{r!(n – r)!}$$

i.e, $$^{10}{C}_{3}$$ = $$\frac{10!}{3! * 7!}$$

= $$\frac{10 * 9 * 8 * 7!}{3 * 2 * 1* 7!}$$ = 120

13. 1250 articles were distributed among students of a class. Each student got twice as many articles as the number of students in that group. The number of students in the group was :

A. 25
B. 45
C. 50
D. 100

Explanation –

Let no. of students be x.

According the question:

x × (2x) = 1250

x = 25

14. The sum of two numbers is 2490. If 6.5% of one number is equal to 8.5% of the other, the greater number is :

A. 1079
B. 1380
C. 1411
D. 1250

Explanation –

x + y = 2490

$$x \frac{6.5}{100}$$ = $$y \frac{8.5}{100}$$

13x = 17y

x = 1411

15. Let α and β be the roots of $$\frac{x}{2}$$ + kx + 8 = 0, such that α – β = 2, then the value of k are

A. ± 3
B. ± 6
C. ± 4
D. ± 8

Explanation –$${x}^{2}$$ + kx + 8 = 0

αβ = $$\frac{8}{1}$$ = 8

i.e, Product of root = $$\frac{-c}{a}$$ for ($${ax}^{2}$$ + bx + c)

α – β = 2

α = 4 or α = -2

β = 2 or β = -4

k = $$\frac{-b}{2}$$ = sum of roots = ± 6