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Algebra Level 2 Test 3
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Question 1 |
If the roots of the equation (c2 - ab) x2 - 2 (a2 - bc) x + (b2 - ac) = 0 for a is not equal to 0 are real and equal, then the value of a3 + b3 + c3 is
abc | |
3abc | |
zero | |
None of these |
Question 1 Explanation:
We are given by that the roots are real and equal, that means we have
= b2- 4ac = 0
Therefore the equation becomes
[-2(a2 - cb) ]2 - 4 (c2 - ba) (b2 - ac) = 0
On solving we have
4a (a3 + b3 + c3 – 3abc) = 0
Therefore the value of a3 + b3 + c3 = 3abc
= b2- 4ac = 0
Therefore the equation becomes
[-2(a2 - cb) ]2 - 4 (c2 - ba) (b2 - ac) = 0
On solving we have
4a (a3 + b3 + c3 – 3abc) = 0
Therefore the value of a3 + b3 + c3 = 3abc
Question 2 |
If 2x2 -7xy + 3y2 = 0, then the value of x: y is
3 : 2 | |
2 : 3 | |
3 : 1 and 1 : 2 | |
5 : 6 |
Question 2 Explanation:
2x2 -7xy + 3y2 = 0
2 (x/y)2 – 7(x/y) +3 = 0
x/y = {-b + ^(b2 – 4ac) } / 2a
= {7 + ^(49 – 24)} / 2 x 2 = (7+ 5) / 4 = 3, ½
Therefore the value of x/y = 3/1 and x/y = ½
2 (x/y)2 – 7(x/y) +3 = 0
x/y = {-b + ^(b2 – 4ac) } / 2a
= {7 + ^(49 – 24)} / 2 x 2 = (7+ 5) / 4 = 3, ½
Therefore the value of x/y = 3/1 and x/y = ½
Question 3 |
If (a + b + 2c + 3d) (a - b – 2c + 3d) = (a - b + 2c - 3d) x (a + b – 2c - 3d), then 2bc is equal to
3ad | |
3/2 | |
a2d2 | |
3a/2d |
Question 3 Explanation:
(a + b + 2c + 3d) (a - b – 2c + 3d) = (a - b + 2c - 3d) x (a + b – 2c - 3d)
Therefore [(a+b) + (2c -3d) x (a + b – 2c - 3d)
=   [(a-b) + (2c -3d)] x [(a + b – (2c - 3d)]
=(a+b)(a-b) – (a+b)(2c – 3d) + (a – b)(2c – 3d) – (2c +3d)(2c -3d)
On further solving
= 2ac +3ad -2bc -3bd = 2ac - 3ad + 2bc - 3bd
= 2bc = 3ad
Therefore [(a+b) + (2c -3d) x (a + b – 2c - 3d)
=   [(a-b) + (2c -3d)] x [(a + b – (2c - 3d)]
=(a+b)(a-b) – (a+b)(2c – 3d) + (a – b)(2c – 3d) – (2c +3d)(2c -3d)
On further solving
= 2ac +3ad -2bc -3bd = 2ac - 3ad + 2bc - 3bd
= 2bc = 3ad
Question 4 |
If X=4ab/(a+b), a+-b, then the value of (x +2a)/(x – 2a) + (x +2b)/(x – 2b) is
0 | |
1 | |
2 | |
None of these |
Question 4 Explanation:
(x +2a)/(x – 2a) + (x +2b)/(x – 2b
Put values of x in the expression we will get
[{(6ab + 2a2)/(a+b)} / {(2ab – 2a2)/(a+b)}] + [{(6b2 + 2ab)/(a+b)} / {(2ab – 2b2)/(a+b)}]
= {2a(3b + a)/2a(b – a)} + {2b (3b+a) / 2b (a – b)}
(3b + a)/ (b – a) – (3b + a)/ (b – a) = 0
Put values of x in the expression we will get
[{(6ab + 2a2)/(a+b)} / {(2ab – 2a2)/(a+b)}] + [{(6b2 + 2ab)/(a+b)} / {(2ab – 2b2)/(a+b)}]
= {2a(3b + a)/2a(b – a)} + {2b (3b+a) / 2b (a – b)}
(3b + a)/ (b – a) – (3b + a)/ (b – a) = 0
Question 5 |
Richa's age is1/6th of Natasha 'sage. Natasha 's age will be twice of Sona 's age after 10 yr. If Sona' s eight birthday was celebrated two yr ago, then the present age of Richa  must be
5 yr | |
10 yr | |
15 yr | |
20 yr |
Question 5 Explanation:
Given that Age of Richa = 1/6 age of Natasha
Present age of Sona = 10 years
That means
Natasha + 10 = 2( Sona + 10)
Therefore Natasha + 10 = 40
Natasha = 30
Therefore the age of Richa = 5 years
Present age of Sona = 10 years
That means
Natasha + 10 = 2( Sona + 10)
Therefore Natasha + 10 = 40
Natasha = 30
Therefore the age of Richa = 5 years
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