BASIC ALGEBRA MCQs

Total Questions : 90 | Page 1 of 9 pages
Question 1.  How many different straight lines can be formed by joining 14 different points of which 6 are collinear and rest are non-collinear?
1.    76
2.    63
3.    64
4.    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=14C26C2+1=9115+1=77.
2nd method:- By selection, 8C2+8C1×6C1+1=77.
Question 2. A basket has 5 oranges and 4 apples. In how many ways can you make a selection if you have to take at least 1 orange and 1 apple?
1.    22
2.    345
3.    365
4.    465
:
D
At least 1 orange can be selected in251=31ways.
At least 1 apple can be selected in241=15ways.
Total no. of selections possible=31×15=465
Question 3.  If x3k2x2+(6k+5)x5k=0 has two roots 3 and 5, then which of the following is the third root?
1.    1
2.    2
3.    3
4.    4
:
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=1or89
Hence the 3rd root is 1.
Question 4. Suppose a and b are two roots of the equation x2(α4)x+α=0. Find out the maximum possible value of 5aba2b2.
1.    0
2.    1614
3.    39
4.    5
5.    Can’t be determined
:
B
a and b are the roots of the equation
a+b=(α4),ab=α
5aba2b2
5ab(a2+b2)
5ab[(a+b)22ab]
5α[(α4)22α]
5α[α210α+16]
α2+15α16=(α2+15α16)=(α2+16αα16)=[α(α+16)1(α+16)]=(α+16)(α1)
So,the maximum value=D4a=[1524(1)(16)]4×1=1614.
Question 5. For the equation x8+6x75x43x2+2x5=0, determine the maximum number of real roots possible.
1.    4
2.    3
3.    8
4.    7
5.    Can’t be determined
:
A
f(x)=x8+6x75x43x2+2x5
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)=x86x75x43x22x5. No. of sign changes = 1
Now check for zero as a root.f(0)=50
So, maximum number of real roots = 3 + 1 = 4. Hence option (a)
Question 6. How many distinct real roots are possible for the below equation?
x626x327=0
1.    6
2.    2
3.    3
4.    5
:
B
Let x3=y
y226y27=0
y227y+y27=0
y(y27)+1(y27)=0
(y+1)(y27)=0
The roots are y = -1, y = 27
x3=1,x=1
x3=27,x=3
Question 7. The A.M. of two positive numbers is 15 and their GM is 12. What is the bigger number?
1.    16
2.    22
3.    24
4.    18
:
C
Let the numbers be x and y.
x+y=30
xy = 144
(xy)2=(x+y)24xy
900(4×144)=324
hence (xy)=18
Solving:xy=18&x+y=30
x = 24, y = 6
So, the bigger number is 24.
Question 8. Find the roots of the following equations:
2x+3+2x=6
1.    -1, -2
2.    1, -2
3.    - 1, - 2
4.    1, 2
Answer: Option C. -> - 1, - 2
:
C
2x.23+2x=6
Let2x=z
8z+1z=6
8z2+1=6z
8z26z+1=0
8z24z2z+1=0
4z(2z1)1(2z1)=0
(4z1)(2z1)=0
z=14,
z=12
2x=14
x=2
2x=12
x=1
Roots are x=2,x=1
Question 9. Find the number of terms of the sequence 32, 24, 16, 8 ... for which the sum of the terms is zero.
1.    7
2.    8
3.    6
4.    9
:
D
Sn=(n2)[2a+(n1)d]
0=(n2)[64+(n1)(8)]
0=(n2)[64(n1)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.
Question 10.  In how many ways can a pack of 52 cards be divided equally among 4 players?
1.    (52!)[(4!)13]
2.    (52!)[(13!)4]4!
3.    (52!)[(13!)4]
4.    (52!)(4!)
5.    None of these