# SSC MTS Algebra Quiz 1

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# SSC MTS Algebra Quiz 1

### Introduction

Algebra is a division of mathematics designed to 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 SSC MTS Algebra Quiz 1 is useful for candidates preparing for SSC MTS and other competitive examinations like RRB ALP/Technical Exams/Junior Engineer Recruitment Exams, IBPS PO Exams and etc. SSC MTS Algebra Quiz 1 contains important questions matching the exact syllabus and pattern of upcoming exams. Aspirants can attempt the SSC MTS Algebra Quiz 1 for Upcoming exams to check the preparation level.

### Quiz

1. For what value of ‘y’, $${x}^{2}$$ + 112x +$${y}^{2}$$ is a perfect square?

A. $$\frac{1}{24}$$
B. $$\frac{1}{12}$$
C. $$\frac{1}{6}$$
D. $$\frac{1}{3}$$

Explanation:
$${x}^{2}$$ + 112x +$${y}^{2}$$ = $${x}^{2}$$ + 2(124)x + $${y}^{2}$$

By, $${(a + b)}^{2}$$ = $${a}^{2}$$ + 2ab + $${b}^{2}$$

Here, a = x and b = y = $$\frac{1}{24}$$

So, for $${x}^{2}$$ + 112x +$${y}^{2}$$ to be perfect square y = $$\frac{1}{24}$$

2. If x and y are natural numbers such that x + y = 2017, then what is the value of $${(-1)}^{x}$$ + $${(-1)}^{y}$$?

A. 2
B. -2
C. 0
D. 1

Explanation:
Sum of two numbers can be odd only if one of them is odd and other one is even. Let x be odd and y be even

⇒ $${(-1)}^{x}$$ + $${(-1)}^{y}$$ = -1 + 1 = 0

∴ The value is 0

3. A root of equation a$${x}^{2}$$ + bx + c = 0 (where a, b and c are rational numbers) is 5 + 3√3. What is the value of $$\frac{({a}^{2}+ {b}^{2} + {c}^{2})}{(a + b + c)}$$?

A. $$\frac{35}{3}$$
B. $$\frac{37}{3}$$
C. $$\frac{-105}{11}$$
D. $$\frac{-105}{13}$$

Explanation:
Since one root of the equation is irrational, so another root will be 5 – 3√3

According to the problem statement,

⇒ [x – (5 + 3√3)] [x – (5 – 3√3)] = 0

⇒$${x}^{2}$$ – 10x – 2 = 0

⇒ a = 1, b = -10 and c = -2, Put this in the required equations

∴ $$\frac{({a}^{2}+ {b}^{2} + {c}^{2})}{(a + b + c)}$$ = $$\frac{(1 + 100 + 4)}{ (1 – 10 – 2)}$$ = $$\frac{-105}{11}$$

4. If the arithmetic mean of two numbers is 7 and the geometric mean of the same two number is 2√10. Then find the numbers x and y respectively, such that x > y.

A. 4,10
B. 2,5
C. 5,2
D. 10,4

Explanation:
Given, Arithmetic Mean = 7

⇒ $$\frac{(x + y)}{2}$$ = 7

⇒ x + y = 14 —-(1)

And, Geometric Mean = 2√10

⇒ √xy = 2√10

⇒ xy = $${(2√10)}^{2}$$ = 40

Now,

⇒ $${(x – y)}^{2}$$ = $${(x + y)}^{2}$$ – 4xy = 142 – (4 × 40) = 196 – 160 = 36

⇒ x – y = 6 —-(2)

Now, adding equation (1) and (2) we get

⇒ 2x = 20

⇒ x = 10

Now, by equation (1)

⇒ y = 14 – x = 14 – 10 = 4

5. Cost of 8 pencils, 5 pens and 3 erasers is Rs. 111. Cost of 9 pencils, 6 pens and 5 erasers is Rs. 130. Cost of 16 pencils, 11 pens and 3 erasers is Rs. 221. What is the cost (in Rs) of 39 pencils, 26 pens and 13 erasers?

A. 316
B. 546
C. 624
D. 482

Explanation:
Let the price of single pencil, pen, and eraser be x, y, and z respectively

According to question,

8x + 5y + 3z = Rs. 111 —-(1)

9x + 6y + 5z = Rs. 130 —-(2)

16x + 11y + 3z = Rs. 221 —-(3)

Subtracting equation (1) from (3)

⇒ (16x + 11y + 3z) – (9x + 6y + 5z) = 221 – 111

⇒ 8x + 6y = 110

⇒ 4x + 3y = 55 —-(4)

Multiply the equation (2) by 3 and (3) by 5 and then subtracting equation (2) from (3)

⇒ (16x + 11y + 3z) × 5 – (9x + 6y + 5z) × 3 = 221 × 5 – 130 × 3

⇒ 80x + 55y + 15z – 27x – 18y – 15z = 1105 – 390

⇒ 53x + 37y = 715 —-(5)

Multiply the equation (4) by 53 and (5) by 4 and then subtracting equation (4) from (5)

⇒ 212x + 159y – 212x – 148y = 2915 – 2860

⇒ 11y = 55

⇒ y = 5

By putting the value of y = 5 in equation (4)

⇒ 4x + 3 × 5 = 55

⇒ x = 10

By putting the value of y = 5 and x = 10 in equation (1)

⇒ 8 × 10 + 5 × 5 + 3z = 111

⇒ 80 + 25 + 3z = 111

⇒ z = 2

∴ Cost of 39 pencils, 26 pens and 13 erasers is 39x + 26y + 13z =39 × 10 + 26 × 5 + 13 × 2 = Rs. 546

1. If a + b + c = 9, ab + bc + ca = 26, $${a}^{3}$$ + $${b}^{3}$$ = 91, $${b}^{3}$$ + $${c}^{3}$$ = 72 and $${c}^{3}$$ + $${a}^{3}$$ = 35, then what is the value of abc?

A. 48
B. 24
C. 36
D. 42

Explanation:
Using algebraic identities,

$${(a + b + c)}^{2}$$ = $${a}^{2}$$ + $${b}^{2}$$ + $${c}^{2}$$ + 2ab + 2bc + 2ca

By putting the respective values given in question,

⇒ $${(9)}^{2}$$ = $${a}^{2}$$ + $${b}^{2}$$ + $${c}^{2}$$ + 2(ab + bc + ca) [∵ ab + bc + ca = 26]

⇒ $${(9)}^{2}$$ = $${a}^{2}$$ + $${b}^{2}$$ + $${c}^{2}$$ + 2(26)

⇒ $${a}^{2}$$ + $${b}^{2}$$ + $${c}^{2}$$ = 81 – 52 = 29

Given equations,

$${a}^{3}$$ + $${b}^{3}$$ = 91 —-(1)

$${b}^{3}$$ + $${c}^{3}$$ = 72 —-(2)

$${c}^{3}$$ + $${a}^{3}$$ = 35 —-(3)

On adding (1), (2) and (3)

$${a}^{3}$$ + $${b}^{3}$$ + $${b}^{3}$$ + $${c}^{2}$$ + $${c}^{2}$$ + $${a}^{3}$$ = 91 + 72 + 35

⇒ 2($${a}^{3}$$ + $${b}^{3}$$ + $${c}^{3}$$) = 198

⇒ $${a}^{3}$$ + $${b}^{3}$$ + $${c}^{3}$$ = 99

Using algebraic identities,

$${a}^{3}$$ + $${b}^{3}$$ + $${c}^{3}$$ – 3abc = (a + b + c) ($${a}^{2}$$ + $${b}^{2}$$ + $${c}^{2}$$ – ab – bc – ca)

By putting the respective values,

⇒ 99 – 3abc = 9 (29 – 26) [∵ ab + bc + ca = 26 and a + b + c = 9]

⇒ 3abc = 99 – 27

⇒ abc = $$\frac{72}{3}$$

∴ abc = 24

2. If $$\frac{x}{a}$$ + $$\frac{y}{b}$$ = 3 and $$\frac{x}{b}$$ – $$\frac{y}{a}$$ = 9, then what is the value of $$\frac{x}{y}$$?

A. $$\frac{(b + 3a)}{(a – 3b)}$$
B.$$\frac{(a + 3b)}{(b – 3a)}$$
C. $$\frac{(1 + 3a)}{ (a + 3b)}$$
D.$$\frac{(a + 3{{b}^{2}})}{(b – 3{{a}^{2}})}$$

Explanation:
Here$$\frac{x}{a}$$ + $$\frac{y}{b}$$ = 3

⇒ bx + ay – 3ab = 0 —- (1)

Here$$\frac{x}{b}$$ – $$\frac{y}{a}$$ = 9

⇒ ax – by – 9ab = 0 —- (2)

Multiplying equation 1 by ‘a’ and equation 2 by ‘b’

⇒ abx + $${a}^{2}$$y – 3$${a}^{2}$$b = 0 —- (3)

⇒ abx – $${b}^{2}$$y – 9a$${b}^{2}$$ = 0 —- (4)

Subtracting equation 4 from 3,

⇒ y =$$\frac{3ab(a – 3b)}{({a}^{2} + {b}^{2})}$$,

substituting value of y in equation 3,

⇒ x =$$\frac{ 3ab(b + 3a)}{({a}^{2}+ {b}^{2})}$$
⇒ $$\frac{x}{y}$$ = $$\frac{(b + 3a)}{(a – 3b)}$$

∴ The value of $$\frac{x}{y}$$ = $$\frac{(b + 3a)}{(a – 3b)}$$

3. The value of x for which the expressions 19 – 5x and 19x + 5 become equal is ______.

A. $$\frac{7}{12}$$
B. $$\frac{-7}{12}$$
C. $$\frac{12}{7}$$
D. $$\frac{-12}{7}$$

Explanation:
According to the given information,

19 – 5x = 19x + 5

⇒ 19 – 5 = 19x + 5x

⇒ 14 = 24x

∴ x = $$\frac{14}{24}$$ = $$\frac{7}{12}$$

4. What is the sum of all the natural numbers from 1 to 35?

A. 620
B. 630
C. 650
D. 680

Explanation:
We know that Formula for Sum of natural no’s beginning from 1

= $${n}^{2}$$ × (1+n)

where n = last no. i.e. 35

⇒ Sum = 352 × (1 + 35)

⇒ Sum = 35 × 18 = 630

5. The sum of three consecutive odd natural numbers is 93. The smallest of three numbers is:

A. 29
B. 31
C. 23
D. 27

Explanation:
Let the three consecutive odd numbers be a, a + 2, a + 4.

Given, the sum of three consecutive odd natural numbers is 93.

∴ a + a + 2 + a + 4 = 93

⇒ 3a = 87

⇒ a = 29

1. The price of 10 chairs is equal to that of 4 tables. The price of 15 chairs and 3 tables together is Rs. 4500. The total price of 12 chairs and 5 tables is:

A. Rs. 4750
B. Rs. 4840
C. Rs. 4500
D. Rs. 4900

Explanation:
Let the price of chair be ‘a’.

Given, the price of 10 chairs is equal to that of 4 tables.

∴ 10 × a = 4 × price of the table

⇒ Price of table = 2.5a

Given, the price of 15 chairs and 3 tables together is Rs. 4500

∴ 15a + 7.5a = 4500

⇒ 22.5a = 4500

⇒ a = Rs. 200

Price of table = 2.5a = Rs. 500

Total price of 12 chairs and 5 tables = 12 × 200 + 5 × 500 = Rs. 4900

2. A number is greater than twice its reciprocal by $$\frac{31}{4}$$. Find the number.

A. 7
B. 8
C. 9
D. 6

Explanation:

Let the number be ‘x’.

A number is greater than twice its reciprocal by $$\frac{31}{4}$$

⇒ x = 2($$\frac{1}{x}$$) + $$\frac{31}{4}$$

⇒ x – $$\frac{2}{x}$$= $$\frac{31}{4}$$

⇒ 4$${(x -2)}^{2}$$ = 31x

⇒ 4$${x}^{2}$$ – 31x – 8 = 0

⇒ 4$${x}^{2}$$ – 32x + x – 8 = 0

⇒ 4x(x – 8) + 1(x – 8) = 0

⇒ (x – 8)(4x + 1) = 0

x = 8 or x = – $$\frac{1}{4}$$

∴ x = 8

3. If x = 2 then the value of $${x}^{3}$$ + 27$${x}^{2}$$ + 243x + 631 is:

A. 1211
B. 1231
C. 1233
D. 1321

Explanation:
Given equation,

f(x) = $${x}^{3}$$ + 27$${x}^{2}$$ + 243x + 631

⇒ x($${x}^{2}$$ + 27x+ 243) + 631

Now, put the value of x = 2

⇒ 2(22 + 27 × 2 + 243) + 631

⇒ 2 (4 + 54 + 243) + 631

⇒ 2(301) + 631 = 602 + 631 = 1233.

4. If a = 2.234, b = 3.121 and c = –5.355, then the value of $${a}^{3}$$ + $${b}^{3}$$ + $${c}^{3}$$ – 3abc is

A. -1
B. 0
C. 1
D. 2

Explanation:
a + b + c = 2.234 + 3.121 – 5.355 = 0

If a + b + c = 0, then$${a}^{3}$$ + $${b}^{3}$$ + $${c}^{3}$$ – 3abc = 0, which can be proved as under

a + b = – c

Cubing both sides, we get

⇒ $${(a + b)}^{3}$$= $${(-c)}^{3}$$

⇒ $${a}^{3}$$ + $${b}^{3}$$ + 3ab(a + b) = – $${c}^{3}$$

⇒ $${a}^{3}$$ + $${b}^{3}$$ + 3ab(– c) = – $${c}^{3}$$

⇒ $${a}^{3}$$ + $${b}^{3}$$ – 3abc = – $${c}^{3}$$

⇒ $${a}^{3}$$ + $${b}^{3}$$ + $${c}^{3}$$ – 3abc = 0

5. If $${x}^{2}$$ + $${y}^{2}$$ + 1 = 2x, then the value of $${x}^{3}$$ + $${y}^{5}$$ is

A. 2
B. 0
C. -1
D. 1

Explanation:
$${x}^{2}$$ + $${y}^{2}$$ + 1 = 2x

⇒ $${x}^{2}$$ + $${y}^{2}$$ + 1 – 2x = 0

⇒ $${x}^{2}$$ – 2x + 1 + $${y}^{2}$$ = 0

⇒ $${(x – 1)}^{2}$$ + $${y}^{2}$$ = 0

In the above eq. the L.H.S. can only become zero when the base of terms; (x – 1) and y becomes zero because for any other value the sum of their squares will always be a positive integer.

Taking (x – 1) = 0
and y = 0

Therefore, x = 1 and y = 0

∴ $${x}^{3}$$ + $${y}^{5}$$ = 1 + 0 = 1.

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