Exams > Cat > Quantitaitve Aptitude
BASIC ALGEBRA MCQs
Total Questions : 90
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Answer: Option D. -> 77
:
D
No. of lines formed by joining 14 points =14C2
No. of lines formed by joining 6 points =6C2
Out of the 14 points, when you join the 6 collinear points, you don’t get6C2lines, but you get only one line.
So, no. of lines=14C2−6C2+1=91−15+1=77.
2nd method:- By selection, 8C2+8C1×6C1+1=77.
:
D
No. of lines formed by joining 14 points =14C2
No. of lines formed by joining 6 points =6C2
Out of the 14 points, when you join the 6 collinear points, you don’t get6C2lines, but you get only one line.
So, no. of lines=14C2−6C2+1=91−15+1=77.
2nd method:- By selection, 8C2+8C1×6C1+1=77.
Answer: Option D. -> 465
:
D
At least 1 orange can be selected in25−1=31ways.
At least 1 apple can be selected in24−1=15ways.
Total no. of selections possible=31×15=465
:
D
At least 1 orange can be selected in25−1=31ways.
At least 1 apple can be selected in24−1=15ways.
Total no. of selections possible=31×15=465
Answer: Option A. -> 1
:
A
Assumeαbe the 3rd root.
Sum of the roots, i.e.,3+5+α=k2
Product of the roots, i.e.3×5×α=5k
So,8+α=9α2
Then,α=1or−89
Hence the 3rd root is 1.
:
A
Assumeαbe the 3rd root.
Sum of the roots, i.e.,3+5+α=k2
Product of the roots, i.e.3×5×α=5k
So,8+α=9α2
Then,α=1or−89
Hence the 3rd root is 1.
Answer: Option B. -> 1614
:
B
a and b are the roots of the equation
a+b=(α−4),ab=α
5ab−a2−b2
5ab−(a2+b2)
5ab−[(a+b)2−2ab]
5α−[(α−4)2−2α]
5α−[α2−10α+16]
−α2+15α−16=−(α2+15α−16)=−(α2+16α−α−16)=−[α(α+16)−1(α+16)]=−(α+16)(α−1)
So,the maximum value=D4a=[152−4(−1)(−16)]4×1=1614.
:
B
a and b are the roots of the equation
a+b=(α−4),ab=α
5ab−a2−b2
5ab−(a2+b2)
5ab−[(a+b)2−2ab]
5α−[(α−4)2−2α]
5α−[α2−10α+16]
−α2+15α−16=−(α2+15α−16)=−(α2+16α−α−16)=−[α(α+16)−1(α+16)]=−(α+16)(α−1)
So,the maximum value=D4a=[152−4(−1)(−16)]4×1=1614.
Answer: Option A. -> 4
:
A
f(x)=x8+6x7−5x4−3x2+2x−5
Check the number of positive roots= no. of sign changes in f(x) = 3
Check the number of negative roots = no. of sign changes in f(-x)
f(−x)=x8−6x7−5x4−3x2−2x−5. No. of sign changes = 1
Now check for zero as a root.f(0)=−5≠0
So, maximum number of real roots = 3 + 1 = 4. Hence option (a)
:
A
f(x)=x8+6x7−5x4−3x2+2x−5
Check the number of positive roots= no. of sign changes in f(x) = 3
Check the number of negative roots = no. of sign changes in f(-x)
f(−x)=x8−6x7−5x4−3x2−2x−5. No. of sign changes = 1
Now check for zero as a root.f(0)=−5≠0
So, maximum number of real roots = 3 + 1 = 4. Hence option (a)
Answer: Option B. -> 2
:
B
Let x3=y
y2−26y−27=0
y2−27y+y−27=0
y(y−27)+1(y−27)=0
(y+1)(y−27)=0
The roots are y = -1, y = 27
x3=−1,x=−1
x3=27,x=3
:
B
Let x3=y
y2−26y−27=0
y2−27y+y−27=0
y(y−27)+1(y−27)=0
(y+1)(y−27)=0
The roots are y = -1, y = 27
x3=−1,x=−1
x3=27,x=3
Answer: Option C. -> 24
:
C
Let the numbers be x and y.
x+y=30
xy = 144
(x−y)2=(x+y)2−4xy
900−(4×144)=324
hence (x–y)=18
Solving:x−y=18&x+y=30
x = 24, y = 6
So, the bigger number is 24.
:
C
Let the numbers be x and y.
x+y=30
xy = 144
(x−y)2=(x+y)2−4xy
900−(4×144)=324
hence (x–y)=18
Solving:x−y=18&x+y=30
x = 24, y = 6
So, the bigger number is 24.
Answer: Option C. -> - 1, - 2
:
C
2x.23+2−x=6
Let2x=z
8z+1z=6
8z2+1=6z
8z2−6z+1=0
8z2−4z−2z+1=0
4z(2z−1)−1(2z−1)=0
(4z−1)(2z−1)=0
z=14,
z=12
2x=14
x=−2
2x=12
x=−1
Roots are x=−2,x=−1
:
C
2x.23+2−x=6
Let2x=z
8z+1z=6
8z2+1=6z
8z2−6z+1=0
8z2−4z−2z+1=0
4z(2z−1)−1(2z−1)=0
(4z−1)(2z−1)=0
z=14,
z=12
2x=14
x=−2
2x=12
x=−1
Roots are x=−2,x=−1
Answer: Option D. -> 9
:
D
Sn=(n2)[2a+(n−1)d]
⇒0=(n2)[64+(n−1)(−8)]
⇒0=(n2)[64−(n−1)8]
⇒n (36– 4n) = 0
n = 0; n=9
As, n can not be equal to 0, so, the number of terms required is 9.
:
D
Sn=(n2)[2a+(n−1)d]
⇒0=(n2)[64+(n−1)(−8)]
⇒0=(n2)[64−(n−1)8]
⇒n (36– 4n) = 0
n = 0; n=9
As, n can not be equal to 0, so, the number of terms required is 9.
Answer: Option C. -> (52!)[(13!)4]
:
C
Here as it has to be divided among 4 different players, the order is important.
No. of ways =(52!)[(13!)4].
:
C
Here as it has to be divided among 4 different players, the order is important.
No. of ways =(52!)[(13!)4].
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