Exams > Cat > Quantitaitve Aptitude
BASIC ALGEBRA MCQs
Total Questions : 90
| Page 4 of 9 pages
Answer: Option B. -> Real
:
B
D=(ab)2+4a2b2=5a2b2
a2b2has to be greater than or equal to 0 (i.e.,)D>0.So,roots are always real.
:
B
D=(ab)2+4a2b2=5a2b2
a2b2has to be greater than or equal to 0 (i.e.,)D>0.So,roots are always real.
Answer: Option C. -> 212
:
C
Sum of the roots:p+q=−5
Product of the roots:pq=2
pq+qp=(p2+q2)pq=[(p+q)2−2pq]pq=[(p+q)2]pq−2=252−2=212
:
C
Sum of the roots:p+q=−5
Product of the roots:pq=2
pq+qp=(p2+q2)pq=[(p+q)2−2pq]pq=[(p+q)2]pq−2=252−2=212
Answer: Option B. -> 2
:
B
[(x+1)(x+4)][(x+2)(x+3)]=24
(x2+5x+4)(x2+5x+6)=24
Letx2+5x=k
(k+4)(k+6)=24
k2+10k=0
k(k+10)=0
k=0,k=−10
x2+5x=0,x2+5x=−10
x(x+5)=0,x2+5x+10=0
x=0,x=−5;D=25−40=−15 No real roots.
So, roots are 0, -5.
:
B
[(x+1)(x+4)][(x+2)(x+3)]=24
(x2+5x+4)(x2+5x+6)=24
Letx2+5x=k
(k+4)(k+6)=24
k2+10k=0
k(k+10)=0
k=0,k=−10
x2+5x=0,x2+5x=−10
x(x+5)=0,x2+5x+10=0
x=0,x=−5;D=25−40=−15 No real roots.
So, roots are 0, -5.
Answer: Option A. -> 84
:
A
6,_,_,_,96
If the common ratio is r, 96=6(r)4
r4=16,r=2
So, the geometric means are: 12, 24 and 48.
a+b+c = 84
:
A
6,_,_,_,96
If the common ratio is r, 96=6(r)4
r4=16,r=2
So, the geometric means are: 12, 24 and 48.
a+b+c = 84
Answer: Option A. -> 1
:
A
The first two terms = a, ar
a+ar=53
a(1−r)=3
a=3(1−r)
a(1+r)=53
3(1−r)(1+r)=53
1−r2=59
r2=49
r=23
a=3(1−23)=1
First term is 1.
:
A
The first two terms = a, ar
a+ar=53
a(1−r)=3
a=3(1−r)
a(1+r)=53
3(1−r)(1+r)=53
1−r2=59
r2=49
r=23
a=3(1−23)=1
First term is 1.
Answer: Option B. -> 1023
:
B
Sn=[a(rn−1)r−1]
In the question a = 3, r = 4, n = 5
S5=3×(45−1)(4−1)=1023
:
B
Sn=[a(rn−1)r−1]
In the question a = 3, r = 4, n = 5
S5=3×(45−1)(4−1)=1023
Answer: Option C. -> −1384
:
C
The series is a GP with common ratior=(−1613)=−12
Eight term=ar7=13×(−12)7=−1384
:
C
The series is a GP with common ratior=(−1613)=−12
Eight term=ar7=13×(−12)7=−1384
Answer: Option A. -> −12,−13,2,3
:
A
Let x−1x=z
(x−1x)2=z2
(x2+1x2−2)=z2
x2+1x2=z2+2
Now substituting in the equation:
6(z2+2)−25z+12=0
6z2+12−25z+12=0
6z2−25z+24=0
6z2−16z−9z+24=0
2z(3z−8)−3(3z−8)=0
(2z−3)(3z−8)=0
z=32;z=83
x−1x=32
x−1x=83
2x2=3x−2=0
3x2−8x−3=0
2x2−4x+x−2=0
3x2−9x+x−3=0
2x(x−2)+1(x−2)=0
3x(x−3)+1(x−3)=0
(2x+1)(x−2)=0
(3x+1)(x−3)=0
x=−12,x=2
x=−13,x=3
So,the roots are:−12,2,−13,3
:
A
Let x−1x=z
(x−1x)2=z2
(x2+1x2−2)=z2
x2+1x2=z2+2
Now substituting in the equation:
6(z2+2)−25z+12=0
6z2+12−25z+12=0
6z2−25z+24=0
6z2−16z−9z+24=0
2z(3z−8)−3(3z−8)=0
(2z−3)(3z−8)=0
z=32;z=83
x−1x=32
x−1x=83
2x2=3x−2=0
3x2−8x−3=0
2x2−4x+x−2=0
3x2−9x+x−3=0
2x(x−2)+1(x−2)=0
3x(x−3)+1(x−3)=0
(2x+1)(x−2)=0
(3x+1)(x−3)=0
x=−12,x=2
x=−13,x=3
So,the roots are:−12,2,−13,3
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