Exams > Cat > Quantitaitve Aptitude
BASIC ALGEBRA MCQs
Total Questions : 90
| Page 3 of 9 pages
Answer: Option C. -> 3
:
C
Let y be the given expression, then
y=√(6+y)
y2=6+y
y2−y−6=0
(y – 3)(y + 2) = 0
y = 3, y = -2
So, the value of the given expression is 3.
:
C
Let y be the given expression, then
y=√(6+y)
y2=6+y
y2−y−6=0
(y – 3)(y + 2) = 0
y = 3, y = -2
So, the value of the given expression is 3.
Answer: Option C. -> 4
:
C
The equationsx2−kx−21=0andx2−3kx+35=0have a common root.
So, equating the two equations:
x2−kx−21=x2−3kx+35
2kx=56
k=28x
Putting in the equation:
x2−28−21=0
x2=49
x=+7
ork=+4,As,k>0,k=4.
:
C
The equationsx2−kx−21=0andx2−3kx+35=0have a common root.
So, equating the two equations:
x2−kx−21=x2−3kx+35
2kx=56
k=28x
Putting in the equation:
x2−28−21=0
x2=49
x=+7
ork=+4,As,k>0,k=4.
Answer: Option B. -> x2−7x+12=0
:
B
Substitute x by (x-2)
The equation becomes:(x−2)2−3(x−2)+2=0,x2−7x+12=0.
:
B
Substitute x by (x-2)
The equation becomes:(x−2)2−3(x−2)+2=0,x2−7x+12=0.
Answer: Option D. -> x2−6x+8=0
:
D
Substitute x by(x2)
The equation becomes:(x2)2−3(x2)+2=0,x2−6x+8=0.
:
D
Substitute x by(x2)
The equation becomes:(x2)2−3(x2)+2=0,x2−6x+8=0.
Answer: Option A. -> 7√32
:
A
For a quadratic equationax2+bx+c=0, if the roots are in ratiom:nthenmnb2=(m+n)2ac.
In the equation given:
4×3×b2=(4+3)2(3×3)
12b2=49×9
b=7×3(2√3)=7√32
:
A
For a quadratic equationax2+bx+c=0, if the roots are in ratiom:nthenmnb2=(m+n)2ac.
In the equation given:
4×3×b2=(4+3)2(3×3)
12b2=49×9
b=7×3(2√3)=7√32
Answer: Option A. -> 2x2−3x+1=0
:
A
Soln:
Interchange ‘a’ and ‘c’ in the equationax2+bx+c=0
The equation becomes:2x2−3x+1=0.Hence option (a)
:
A
Soln:
Interchange ‘a’ and ‘c’ in the equationax2+bx+c=0
The equation becomes:2x2−3x+1=0.Hence option (a)
Answer: Option B. -> 2
:
B
Soln:
a+1a=(√a−1√a)2+2
For minimum value ofa+1a,we want minimum value of(√a−1√a)2that will be zero.
So, minimum value ofa+1awill be 2. Hence option (b)
2nd method:-By, A.M≥G.Ma+1a2≥√(a×1athusa+1a>2(i.e.,)(a+1a)min will be 2 . option (b).
:
B
Soln:
a+1a=(√a−1√a)2+2
For minimum value ofa+1a,we want minimum value of(√a−1√a)2that will be zero.
So, minimum value ofa+1awill be 2. Hence option (b)
2nd method:-By, A.M≥G.Ma+1a2≥√(a×1athusa+1a>2(i.e.,)(a+1a)min will be 2 . option (b).
:
Let the two digit number be ab.
a×b=12
When 9 is added to the number
10a+b+9=10b+a
9a−9b+9=0
a−b+1=0
a−12a+1=0
a2+a−12=0
(a+4)(a−3)=0
a = -4 is not accepted. So, a = 3, b = 4
The number is 34 .
Answer: Option D. -> 17
:
D
Let the roots be a,b.
ab=k,a+b=8
a2+b2=30
(a+b)2=64,a2+b2+2ab=64
30+2k=64,k=17
:
D
Let the roots be a,b.
ab=k,a+b=8
a2+b2=30
(a+b)2=64,a2+b2+2ab=64
30+2k=64,k=17
Answer: Option D. -> x2−8x+15=0
:
D
Given equation:x2−3x+2=0
with roots p, q
When the roots become 2p and 2q,
the equation becomes:
(x2)2−3(x2)+2=0,x2−6x+8=0
When the roots become
(2p+1)&(2q+1)
the equation becomes:
(x−1)2−6(x−1)+8=0,x2−8x+15=0.
Hence option (D)
:
D
Given equation:x2−3x+2=0
with roots p, q
When the roots become 2p and 2q,
the equation becomes:
(x2)2−3(x2)+2=0,x2−6x+8=0
When the roots become
(2p+1)&(2q+1)
the equation becomes:
(x−1)2−6(x−1)+8=0,x2−8x+15=0.
Hence option (D)
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