Quantitative Aptitude - SPLessons

Algebra Practice Quiz 3

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Algebra Practice Quiz 3

shape Introduction

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.

The article Algebra Practice Quiz 3 is useful for candidates preparing for different competitive examinations like RRB ALP/Technical Exams/Junior Engineer Recruitment Exams, SSC, IBPS PO Exams and etc.


shape Quiz

1.
Statement: M ≥ P T = M
Conclusions:
I. V ≥ P
II. T ≥ H


    A. If only conclusion I is true
    B. If only conclusion II is true
    C. If either conclusion I or II is true
    D. If neither conclusion I nor II is true
    E. If both conclusions I and II are true


Answer: Option A

Explanation:
M ≥ P T = M …… (ii)
Combining all statements, we get
V ≥ T = M ≥ P P is true.
Again, T ≥ H is not true.


2.
Statements: A ≥ B = C ≥ D, V ≥ G ≤ H = D
Conclusion:
I. C ≥ D
II. A ≥ H
III. B ≥ G
IV. C ≤ V


    A. Only I and II are true
    B. Only III and IV are true
    C. Only I, II and III are true
    D. All I, II and III are true
    E. None of these


Answer: Option C

Explanation:
A ≥ B = C ≥ D …..(i)
V ≥ G ≤ H = D ……(ii)
Combining both statement, we get
A ≥ B = C ≥ D = H ≥ G ≤ V
Thus, C ≥ G is true.
A ≥ H is true. B ≥ G is true. C ≤ V is not true.
Hence, only I, II and III are true.


3.
Statements: M ≤ N T ≥ Q
Conclusions:
I. R ≥ L
II. T ≤ N
III. L ≥ M
IV. R ≥ M


    A. Only III and IV are true
    B. Only III is true
    C. Only I and IV are true
    D. All I, II, III and IV are true
    E. None of these


Answer: Option B

Explanation:
M ≤ N T ≥ Q ……(ii)
Combining both the statements, we get
M ≤ N ≤ L ≥ Q ≤ T ≤ R
Thus, R ≥ L is not true.
T ≤ N is not true.
Again, M M is true.
R ≥ M is not true.


4.
Statement: M ≥ P T = M
Conclusions:
I. V ≥ P
II. T ≥ H


    A. If only conclusion I is true
    B. If only conclusion II is true
    C. If either conclusion I or II is true
    D. If neither conclusion I nor II is true
    E. If both conclusions I and II are true


Answer: Option A

Explanation:
M ≥ P T = M …… (ii)
Combining all statements, we get
V ≥ T = M ≥ P P is true.
Again, T ≥ H is not true.


5.
Statements: E = G ≥ H = N, C ≥ F ≥ M = N
Conclusions:
I. F ≥ E
II. E ≥ M
III. C ≥ G
IV. C ≥ H


    A. Only I and III are true
    B. All I, II, III and IV are true
    C. Only II and IV are true
    D. Only II is true
    E. None of these


Answer: Option C

Explanation:
E = G ≥ H = N ……(i)
C ≥ F ≥ M = N …….(ii)
Combining both the statements, we get
E = G ≥ H = N = M ≤ F H or H ≤ C is true.

In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.

1.
I. \(x^{5}\) – 41\(x^{3}\) + 400x = 0
II. \(y^{2}\) – 14y + 30 = −18


    A. if x ≥ y
    B. if x ≤ y
    C. if x ≥ y
    D. if x ≤ y
    E. if x = y or relationship between x and y can’t be established


Answer: Option B

Explanation:
I. \(x^{5}\) – 41\(x^{3}\) + 400x = 0
x(\(x^{4}\) – 41\(x^{2}\) + 400) = 0
x = 0 or \(x^{4}\) – 41\(x^{2}\) + 400 = 0
Let \(x^{2}\) = a
\(a^{2}\) – 41a + 400 = 0 ∴ \(a^{2}\) – 25a − 16a + 400 = 0
(a – 16)(a – 25) = 0
a = 16 or a = 25
\(x^{2}\) = 16 or \(x^{2}\) = 25
x = ±4 or x = ±5
x = −5 or −4 or 0 or 4 or 5
II. \(y^{2}\) – 14y + 30 = −18
\(y^{2}\) – 14y + 48 = 0
\(y^{2}\) – 8y − 6y + 48 = 0
(y – 6)(y – 8) = 0
y = 6 or y = 8
For both values of y


2.
I. \(x^{2}\) – 3 = 2x
II. \(y^{2}\) + 5y + 6 = 0


    A. if x ≥ y
    B. if x ≤ y
    C. if x ≥ y
    D. if x ≤ y
    E. if x = y or relationship between x and y can’t be established


Answer: Option B

Explanation:
I. \(x^{2}\) – 3 = 2x
\(x^{2}\) – 2x – 3 = 0
\(x^{2}\) – 3x + x – 3 = 0
(x + 1)(x – 3) = 0
x = 3 or x = −1
II. \(y^{2}\) + 5y + 6 = 0
\(y^{2}\) + 3y + 2y + 6 = 0
(y + 3)(y + 2) = 0
y = –3 or y = –2
For both values of x, x ≥ y


3.
I. 4\(x^{3}\) + 24\(x^{2}\) – 64x = 0
II. 3\(y^{2}\) + 39y + 126 = 0


    A. if x ≥ y
    B. if x ≤ y
    C. if x ≥ y
    D. if x ≤ y
    E. if x = y or relationship between x and y can’t be established


Answer: Option E

Explanation:
I. 4\(x^{3}\) + 24\(x^{2}\) – 64x = 0
4x(\(x^{2}\) + 6x – 16) = 0
4x = 0 or \(x^{2}\) + 6x – 16 = 0 (Here 4x = 0 is nothing but x = 0)
\(x^{2}\) + 6x – 16 = 0
\(x^{2}\) + 8x − 2x – 16 = 0
(x − 2)(x + 8) = 0
x = 2 or x = −8
x = −8, 0 or 2
II. 3\(y^{2}\) + 39y + 126 = 0
\(y^{2}\) + 13y + 42 = 0
\(y^{2}\) + 7y + 6y + 42 = 0
(y + 7)(y + 6) = 0
y = –7 or y = –6 When x = −8, x y Hence, no relation can be established between x and y.


4.
I. \(x^{2}\) – 25x + 114 = 0
II. \(y^{2}\) – 10y + 24 = 0


    A. if x ≥ y
    B. if x ≤ y
    C. if x ≥ y
    D. if x ≤ y
    E. if x = y or relationship between x and y can’t be established


Answer: Option C

Explanation:
I. \(x^{2}\) – 25x + 114 = 0
\(x^{2}\) – 19x – 6x + 114 = 0
(x – 6)(x – 19) = 0
x = 19 or x = 6
II. \(y^{2}\) – 10y + 24 = 0
\(y^{2}\) – 6y – 4y + 24 = 0
(y – 4)(y – 6) = 0
y = 6 or y = 4
When x = 19, x ≥ y
When x = 6, x ≥ y
Hence, x ≥ y


5.
I. 48\(x^{2}\) – 24x + 3 = 0
II. 55\(y^{2}\) + 53y + 12 = 0


    A. if x ≥ y
    B. if x ≤ y
    C. if x ≥ y
    D. if x ≤ y
    E. if x = y or relationship between x and y can’t be established


Answer: Option A

Explanation:
48\(x^{2}\) – 24x + 3 = 0 is of the form a\(x^{2}\) – bx + c = 0
Both roots of this equation are positive i.e. x ≥ 0
55\(y^{2}\) + 53y + 12 = 0 is of the form a\(y^{2}\) + by + c = 0
Both roots of this equation are negative i.e. y

1. If a two-digit number is added to a number obtained by reversing the digits of the given number, then the sum is always divisible by which one of the following numbers?

    A. 7
    B. 9
    C. 10
    D. 11


Answer: Option D

Explanation:
Let a two-digit number be (10x + y) and reversing number be (10y + x)
Required sum = 10x + y + 10y + x = 11x + 11y = 11(x + y)
Hence, it’s divisible by 11.


2. What is the sum of two numbers whose differences is 45 and the quotient of the greater number by the lesser number is 4?

    A. 100
    B. 90
    C. 80
    D. 75


Answer: Option D

Explanation:
Let the greater number be x and smaller number be y
x – y = 45 ….. (i)
x = 4y ….. (ii)
From eqs. (i) and (ii)
4y – y = 45
y = \(\frac{45}{3}\)= 15
On putting the value of y in Eq. (ii), we get
x = 4 × 15 = 60
Hence, required sum = x + y = 60 + 15 = 75.


3. If one-third of a two-digit number exceeds its one-fourth by 8, then what is the sum of the digits of the number?

    A. 6
    B. 13
    C. 15
    D. 17


Answer: Option C

Explanation:
Let the number be y,
\(\frac{y}{3}\)=\(\frac{y}{4}\) + 8
\(\frac{4y – 3y}{12}\)=8
y = 12 × 8 = 96
Sum of digits = 9 + 6 = 15.


4. A person bought 5 tickets from a station P to a station Q and 10 tickets from the station P to a station R. He paid Rs. 350. If the sum of a ticket from P to Q and a ticket from P to R is Rs. 42, then what is the fare from P to Q?

    A. Rs. 12
    B. Rs. 14
    C. Rs. 16
    D. Rs. 18


Answer: Option B

Explanation:
Let the fare from station P to station Q is ₹ x and the fare from station P to station R is ₹ y.
By given condition, x + y = 42 …..(i)
and 5x + 10y = 350 …. (ii)
On solving Eqs. (i) and (ii), we get
x = 14 and y = 28
Hence, fare from station P to station Q is Rs. 14.


5. If 6x – 5y = 13, 7x + 2y = 23 then 11x + 18y = ?

    A. – 15
    B. 51
    C. 33
    D. 15


Answer: Option B

Explanation:
6x – 5y = 13 …… (i)
7x + 2y = 23 …… (ii)
By equation (i) * 2 & (ii) * 5,
12x – 10y = 26 ….. (iii)
35x + 10y = 115 …… (iv)
by solving these equation we get
x = 3, y = 1
11x + 18y = 11 × 3 + 18 × 1 = 33 + 18 = 51.


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