Algebra - Aptitude Questions and Answers - in Hindi

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If x= √(a+b) - √(a-b) / √(a+b) + √(a-b) , then what is bx2 -2ax + b equals to?


यदि x= √(a+b) - √(a-b) / √(a+b) + √(a-b), तो bx2 -2ax + b किसके बराबर है?

A)

0


B)

1


C)

ab


D)

2ab



Correct Answer :

0


Explanation :

x = √(a+b) - √(a-b) / √(a+b) + √(a-b)
x = (√(a+b) - √(a-b))*(√(a+b) - √(a-b)) / (√(a+b) + √(a-b)) * (√(a+b) - √(a-b))
x = (√(a+b) - √(a-b))2 / (a+b) - (a-b)
x = (a+b) + (a-b) -2*√(a+b)*√(a-b) / 2b
x = 2*a -2*√(a2 - b2) / 2b
x = a - √(a2 - b2) / b
a - b*x= √(a2 - b2)
After squaring both the sides
(a - bx )2 = a2 - b2
a2 + b2x2 - 2abx = a2 - b2
b2x2 - 2abx + b2 = 0
bx2 - 2ax + b = 0

If x = t 1 / t-1 and y = t t / t-1 , t > 0, t ≠ 1 then what is the relation between x and y ?


यदि x = t 1 / t-1 और y = t t / t-1 , t > 0, t ≠ 1 है तो x और y के बीच क्या संबंध है?

A)

yx = x1/y


B)

x1/y = y1/x


C)

xy =yx


D)

xy = y1/x



Correct Answer :

xy =yx


Explanation :

Now, y = t t / t-1 = ( t 1 / t-1)t = xt …..(i)
y / x = t t / t-1 / t 1 / t-1
= t t / t-1 - 1 / t-1
= t t / t-1 - 1 / t-1
= t t - 1 / t - 1
y / x = t …..(ii)
From Eqs. (i) and (ii), we get y = x y/x
=> yx = xy

If x=2 + 21/3 + 22/3, then the value of x3- 6x2 + 6x is?


यदि x=2 + 21/3 + 22/3 , तो x3- 6x2 + 6x का मान है?

A)

3


B)

2


C)

1


D)

0



Correct Answer :

2


Explanation :

x = 2 + 21/3 + 22/3 …..(i)
x - 2 = 21/3 * (21/3 + 1)
After cubing both the sides
(x - 2)3 = 2* (21/3 + 1)3
x3 - 3*x2 *2 + 3*x*22 - 23 = 2*( (21/3)3 + 3*(21/3)2*1 + 3*21/3*12 + 13)
x3 - 6x2 + 12x - 8 = 2*(2 + 3*22/3 + 3*21/3 + 1)
x3 - 6x2 + 12x - 8 = 2*(3 + 3*22/3 + 3*21/3)
x3 - 6x2 + 12x - 8 = 6*(1+ 22/3 + 21/3)
From Eqs (i)
x3 - 6x2 + 12x - 8 = 6*(1+ x - 2)
x3 - 6x2 + 12x - 8 = 6(x - 1)
x3 - 6x2 + 12x - 8 = 6x - 6
x3 - 6x2 + 6x = 2

If √(x/y) = 24/5 + √(y/x) and x + y = 26, then what is the value of xy?


यदि √(x/y) = 24/5 + √(y/x) और x + y = 26 है, तो xy का मूल्य क्या है?

A)

5


B)

15


C)

25


D)

30



Correct Answer :

25


Explanation :

Let z=√(x/y) , then, z = 24/5 + 1/z
z = 24z + 5 / 5z
5z2 - 24z - 5 = 0
5z2 - 25z + z - 5 = 0
5z( z - 5) + 1( z - 5) = 0
( z - 5)(5z + 1) = 0
z = 5 or -1/5 that is √(x/y)= 5 or -1/5
Let consider
√(x/y)= 5
x/y= 25
x = 25y ....(i)

As is given x + y = 26
then 25y + y = 26 From Eqs (i)
26y = 26
y = 1
Hence x = 25
and xy = 25 * 1 = 25

If α and β are the roots of the equation x2 + px + q = 0, then what is α2 + β2 equal to?


यदि α और β समीकरण x2 + px + q = 0 के मुल हैं, तो α2 + β2 किसके बराबर है?

A)

p2 - 2q


B)

q2 - 2p


C)

p2 + 2q


D)

q2 - q



Correct Answer :

p2 - 2q


Explanation :

As α αnd β αre the roots of the equqtion x2 + px + q = 0
therefore α + β = - p αnd αβ = q
Now,
α2 + β2 = (α + β)2 - 2αβ
= (- p)2 - 2q
= p2 - 2q

If a3 = 335 + b3 and a = 5 + b, then what is the value of a + b (given that a > 0 and b > 0)?


यदि a3 = 335 + b3 और a = 5 + b है, तो a + b का मान क्या है (दिया गया है कि a> 0 और b> 0)?

A)

7


B)

9


C)

16


D)

49



Correct Answer :

9


Explanation :

As is given, a3 = 335 + b3 and a = 5 + b , thus
a3 - b3 = 335 …(i)
a - b = 5 …(ii)

As, (a - b)3 = a3 - b3 - 3ab(a - b)
From Eqs (i) and (ii)
53 = 335 - 3ab(5)
125 = 335 - 15ab
ab = 14
Also, (a + b)2 = (a - b)2 + 4ab
= 52 + 4*14
= 25 + 56
= 81

Hence a + b = 9

If 9x 3y = 2187 and 23x 22y - 4xy = 0 , then what can be the value of (x + y )?


यदि 9x 3y = 2187 और 23x 22y - 4xy = 0, तो (x + y) का मान क्या है?

A)

1


B)

3


C)

5


D)

7



Correct Answer :

5


Explanation :

9x * 3y = 2187
(32)x . 3y = 2187
32x + y = 37
2x + y = 7 …...(i)

Again,
23x 22y - 4xy = 0
23x * 22y = 4xy
23x +2y = (22)xy
3x + 2y = 2xy …...(ii)
From Eqs. (i) and (ii)
3x + 2(7 - 2x ) = 2x(7 - 2x )
3x + 14 - 4x = 14x -4x2
4x2 - 15x + 14 = 0
(x - 2)(4x - 7) = 0

Thus x = 2 or 7/4
y = 3 or 7/2

x + y = 5 or 21/4

The pair of linear equations kx + 3y +1 = 0 and 2x + y + 3 = 0 intersect each other, if?


युगल रैखिक समीकरण kx + 3y +1 = 0 और 2x + y + 3 = 0 एक दूसरे को प्रतिच्छेद करती है, यदि?

A)

k = 6


B)

K ≠ 6


C)

k = 0


D)

k ≠ 0



Correct Answer :

K ≠ 6


Explanation :

linear equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Has unique solution if a1/a2 ≠ b1/b2

for the euqation,
kx + 3y + 1 = 0
2x + y + 3 = 0
the unique solution,
k/2 ≠ 3/1
k 6

The values of x which satisfy the equation 51 + x +51 - x = 26 are?


समीकरण 51 + x + 51 - x = 26 को संतुष्ट करने वाले X के मान कौन से हैं?

A)

18 days


B)

20 days


C)

24 days


D)

25 days



Correct Answer :

20 days


Explanation :

51 + x + 51 - x = 26
(5*5x) + (5*5- x) = 26
(5*5x) + (5/5x) = 26
5(5x + 1/5x) = 26

Let 5x = y , then
5y2 - 26y +5 = 0
5y2 - 25y - y + 5 = 0
5y( y - 5) - 1( y - 5) = 0
( y - 5)(5y - 1) = 0
y = 5 , 1/5
that is , 5x = 5 or 5-1
therefore x = 1 , -1

If a + b = 5 and ab = 6, then what is the value of a3 + b3 ?


यदि a + b = 5 और ab = 6 है, तो a3 + b3 का मान क्या है?

A)

35


B)

40


C)

90


D)

125



Correct Answer :

35


Explanation :

a3 + b3 = (a + b)3 - 3ab(a + b)
= (5)3 - 3 * 6 * 5 = 125 - 90 = 35

If (7 - 12x) - (3x - 7) = 14, then the value of x is ?


यदि (7 - 12x) - (3x - 7) = 14, तो x का मान है?

A)

-4


B)

0


C)

5


D)

2



Correct Answer :

0


Explanation :

(7 - 12x) - (3x - 7) = 14
7 - 12x - 3x + 7 = 14
- 15x + 14 = 14
- 15x = 0
x = 0

Find the roots of the quadratic equation 6x2 - 11x - 35 = 0


द्विघात समीकरण के मूल को ज्ञात करें 6x2 - 11x - 35 = 0

A)

5/3, - 7/2


B)

- 5/3, 7/2


C)

- 3/5, 2/7


D)

3/5, - 2/7



Correct Answer :

- 5/3, 7/2


Explanation :

6x2 - 11x - 35 = 0
6x2 - 21x + 10x - 35 = 0
3x(2x - 7) + 5(2x - 7) = 0
(3x + 5)(2x - 7) = 0
x = -5/3, x = 7/2

The distance between the points (4,-8) and (k,0) is 10. Find k?


अंक (4, -8) और (k, 0) के बीच की दूरी 10 है. k का मान ज्ञात करे?

A)

k = 6 or - 2


B)

k = 10 or - 2


C)

k = 10 or - 4


D)

k = 6 or - 4



Correct Answer :

k = 10 or - 2


Explanation :

the distance between the points, c2 = (xA − xB)2 + (yA − yB)2

(K - 4)2 + (0 + 8)2 = (10)2
K2 + 16 - 8K + 64 = 100
K2 - 8K - 20 = 0
K2 - 10K + 2K - 20 = 0
K(K - 10) + 2(K - 10) = 0
(K + 2)(K - 10) = 0
K = - 2, K = 10

What is the equation of the line which passes through the points (2, 3) and (- 4, 1)?


बिन्दु (2, 3) और (- 4, 1) से होकर गुजरने रेखा का समीकरण क्या है ?

A)

x - 3y = - 7


B)

x + 3y = 7


C)

x - 3y = 7


D)

x + 3y = - 7



Correct Answer :

x - 3y = - 7


Explanation :

Let (2, 3) is (x1, y1) and (-4, 1) is (x2, y2)

The equation of a line passing through two points (x1, y1) and (x2, y2) is given by
y - y1 = m(x - x1) , m is the slope

there, m = (y2 - y1)/(x2 - x1) = (1-3)/(-4-2) = -2/-6 = 1/3

Then the equation is:
y - 3 = (1/3)*(x-2)
3y - 9 = x - 2
3y - x = 7
x - 3y = - 7

Aman and Alok attempted to solve a quadratic equation. Aman made a mistake in writing down the constant term and ended up in roots (4, 3). Alok made a mistake in writing down the coefficient of x to get roots (3, 2). The correct roots of the equation are?


अमन और आलोक ने एक द्विघात समीकरण को हल करने का प्रयास किया। अमन ने नियत शब्द लिखने में गलती की और मुल (4, 3) प्राप्त किए। आलोक ने x के गुणांक को लिखने में गलती की और मुल (3, 2) प्राप्त किए। समीकरण के सही मुल हैं?

A)

-4, -3


B)

6, 1


C)

4, 3


D)

-6, -1



Correct Answer :

6, 1


Explanation :

Let quadratic equation be
ax2 + bx + c = 0
If a and b are roots, then
α + β = -b/a
and αβ = c/a

Since Aman made a mistake in writing down the constant term.
α + β = 4 + 3 = 7
and Alok made a mistake in writing down the coefficient of x.
αβ = 3 * 2 = 6
So, equation will be
x2 - (α + β)x + αβ = 0
x2 - 7x + 6 = 0
(x - 6)(x - 1) = 0
x = 6, 1

The system of equations 2x + 4y = 6 and 4x + 8y = 8 is ?


समीकरणों की प्रणाली 2x + 4y = 6 और 4x + 8y = 8 है?

A)

consistent with a unique solution


B)

consistent with infinitely many solutions


C)

inconsistent


D)

None of the above



Correct Answer :

inconsistent


Explanation :

We have, 2x + 4y = 6 and 4x + 8y = 8

a1 = 2, b1 = 4, c1 = -6
and
a2 = 4, b2 = 8, c2 = -8
Now,
a1/a2 =2/4 =1/2
b1/b2 =4/8 =1/2
c1/c2 =-6/-8 =3/4

When there is no solution, the equations are called inconsistent. This happens, when the lines are parallel.

Here, a1/a2 = b1/b2 = 1/2 ≠ c1/c2

Hence, system of equation is inconsistent.

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