## Algebra Questions

This page consists some of the Algebra Questions and Answers that will help you to level up your confidence in this topic. Algebra starts with a methodical learning of the uses along with the rules of arithmetic. The operations of multiplication, subtraction, addition as well as division functions as the foundation for all mathematical calculations.

### Rules of Algebra Questions and Answers

For achieving simplification, characters of the alphabet are taken in algebra in order to signify values or digits. The alphabets like x, y, a, or b are used for a specific value (known or unknown), or else it could represent for any value at all. An alphabet or variable that signifies a random number is known as a variable. The difference, quotient, sum, and product of two digits, x and y, can be represented in the form of x + y, x – y, xy and x÷y.

Linear equations are of the form of ax + b = c, ax + by + c = 0, ax + by + cz + d = 0. Elementary algebra based on the degree of the variables, branches out into quadratic equations and polynomials. Representation of a quadratic equation in a general form is ax2 + bx + c = 0, and for a polynomial equation, it is axn + bxn-1+ cxn-2+ …..k = 0

Now let us take a look on the Algebra Questions and Answers to make the concept more clear and understandable.

### Algebra Identities:

• $(a+b)^{2}=a^{2}+b^{2}+2ab$
• $(a-b)^{2}=a^{2}+b^{2}-2ab$
• $a^{2}-b^{2}=(a+b)(a-b)$
• $a^{2}+b^{2}=(a+b)^{2}-2ab = (a-b)^{2}+2ab$
• $a^{3}+b^{3}=(a+b)(a^{2}-ab+b^{2})$
• $a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})$
• $(a+b)^{3}= a^{3}+3ab(a+b)+b^{3}$
• $(a-b)^{3}= a^{3}-3ab(a+b)-b^{3}$

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1. If l + m + n = 0 then l3+m3+n3 would be equal to:

3lmn

3lmn

75.08%

(lm+ln+mn)/lmn

(lm+ln+mn)/lmn

7.6%

lmn(ln+lm+mn)

lmn(ln+lm+mn)

9.42%

(l^2+m^2+n^2)/lmn

(l^2+m^2+n^2)/lmn

7.9%

l +m= -n

= (l + m)3 = (-n)3

= l3+m3+3lm (l+ m) + n3= 0

= l3+m3+3lm(-n) + n3= 0

Hence l3+ m3 + n3= 3lmn

2. If a-(1/a) =2 then calculate the value of a4+1/a4?

14

14

22.56%

85

85

9.02%

34

34

56.02%

zero

zero

12.41%

a –  1/a =2

Squaring both the sides,

= (a – 1/a)2  = 4

= a2 + 1/a2 - 2 = 4

Or, (a2+1/a2) = 6

Squaring both the sides again,

(a2+1/a2)2 = (6)2

(a4+1/a4+2) = 36

Therefore,

a4+1/a4= 34

3. If b –1/b = 3 then calculate the value of b6+1/b6?

1298

1298

62.98%

1422

1422

19.71%

1413

1413

11.06%

1523

1523

6.25%

First, let's square the given equation:

$\left ( \frac{b-1}{b} \right )^{2} = 3^{2}$

Expanding the left side of the equation:

$b^{2} - 2(b)\frac{1}{b} + \left ( \frac{1}{b} \right )^{2} = 9$

Simplifying further:

$b^{2} - 2 + \left ( \frac{1}{b} \right )^{2} = 9$

Now, let's rearrange the terms:

$b^{2} + \left ( \frac{1}{b} \right )^{2} = 9 + 2$

$b^{2} + \left ( \frac{1}{b} \right )^{2} = 11$

Now, let's square this equation again:

$\left (b^{2} + \left ( \frac{1}{b} \right )^{2} \right )^{2} = 11^{2}$

Expanding the left side:

$b^{4} + 2(b^{2}) \left (\frac{1}{b^{2}} \right ) + \left ( \frac{1}{b} \right )^{4} = 121$

Simplifying further:

$b^{4} + 2 + \left ( \frac{1}{b} \right )^{4} = 121$

Now, let's rearrange the terms:

$b^{4} + \left ( \frac{1}{b} \right )^{4} = 121- 2$

$b^{4} + \left ( \frac{1}{b} \right )^{4} = 119$

Now, we can multiply the equation b^2 + 1/b^2 = 11 by the equation b^4 + 1/b^4 = 119:

(b^2 + 1/b^2)(b^4 + 1/b^4) = 11 * 119

Expanding both sides:

b^6 + (b^2)(1/b^4) + (1/b^2)(b^4) + 1/b^6 = 1309

Simplifying further:

b^6 + b^2/b^4 + b^4/b^2 + 1/b^6 = 1309

b^6 + 1/b^2 + b^2 + 1/b^6 = 1309

Now, we can substitute the value of b^2 + 1/b^2 from the previous equation (b^2 + 1/b^2 = 11):

b^6 + 11 + 1/b^6 = 1309

Simplifying further:

b^6 + 1/b^6 = 1309 - 11

b^6 + 1/b^6 = 1298

Therefore, the value of b^6 + 1/b^6 is 1298.

4. Calculate the value of p, if f(p)= P2 – 28p+196.

74

74

7.98%

14

14

80.67%

52

52

6.3%

12

12

5.04%

Given that f(p)= P2 – 28p+196

Or,

P2 – 28p + 196= 0

= (p-14)(p-14)= 0

Therefore p= 14

5. The unique solution for equation pm+qn= 1 and pm-qn = 1 will be:

p=1 and q= 1

p=1 and q= 1

45.04%

P= 0 and q- 2

P= 0 and q- 2

11.16%

P= 0 and q= -1

P= 0 and q= -1

14.88%

None

None

28.93%

P1m+ q1n= 1

P2m- q2n= 1

In order to get a unique solution we know  p1/p2≠ q1/-q2

Now by looking at the options, it is evident that since the value of p and q cannot be 0, so option b and c gets eliminated.

6. Calculate the value of r from the equation ab - 3a + 5b + r?

-15

-15

63.58%

7

7

17.88%

-9

-9

13.25%

31

31

5.3%

On factorizing the equation ab -3a + 5b + r we get

= a(b-3)+5(b-3)= ab – 3a + 5b - 15

Therefore r= -15

7. Adam is twice good in working as Smith. If by working together on the same task, they can finish the same in 14 days. If Adam is not well than Smith does the work alone. How much time would he require to finish the task?

41

41

7.93%

43

43

14.63%

42

42

69.51%

40

40

7.93%

Since Adam is a twice as good worker as Smith hence

His work is done per day 2x

And Smith work is done per day= x

Hence, Adam+Smith’s 1 day work= x+2x= 1/14

3x= 1/14

Therefore,

x= 1/14*3= 1/42 days

Hence Smith’s work done per day= 1/42

Adam’s efficiency is twice that of Smith, so time taken by him will be half = 42/2 =21

Hence Adam takes 21 days to finish the task, and Smith will take 2*21= 42 days to finish the same task if Adam is ill.

8. A mother said to her daughter, "When you were born, I was as old as you are now.” If the mother's age is 36 years at present, find the daughter’s age four years back.

12 years

12 years

10.93%

18 years

18 years

15.85%

14 years

14 years

66.12%

21 years

21 years

7.1%

If we mark a daughter’s current age as x.

Then, her mother’s age at the time of her daughter’s birth will be,

(36-x) = x

2x = 36

X= 18.

Therefore, if the daughter is 18 years old at present then her age four years back will be,

18 - 4 = 14 years

9. 30 students can write a thesis in 18 days when they study for 7 hours. How many days will 21 students who study for 8 hours take to finish that thesis assignment?

22.5

22.5

67.32%

22.1

22.1

12.42%

20

20

9.8%

22.6

22.6

10.46%

Let the days be X

30*7*18 = 21*8*X

3780= 168X

Therefore,

X= 3780/168= 22.5 days

10. Sam purchased some tennis balls for Rs 450. He could have managed to buy 5 more balls by spending the same amount of money if each ball was marked Rs 15 low. Calculate the total number of balls bought by Sam?

15

15

23.91%

10

10

52.9%

45

45

15.22%

50

50

7.97%

Let the number of balls be b

Then,

450/b – 450/b+5= 15

= 1/b – 1/(b+5) = 15/450

= (b+5-b)/b(b+5)= 1/30

= b2 + 5b -150 = 0

= (b+15) (b-10) = 0

Therefore,

b= 10

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