### Quantitative Aptitude

Total Questions : 102 | Page 1 of 11 pages
Question 1. One root of the following given equation 2x514x4+31x364x2+19x+130=0 is
1.    1
2.    3
3.    5
4.    7
:
C
We can solve this problem using options given -
On putting x = 5, we get
2(5)514(5)4+31(5)364(5)2+19(5)+130=0
Hence x = 5 satisfies the given equation.
Thus 5 is a root of the equation.
Other options will not satisfy the given equation so the answer is x = 5.
9x29(a+b)x+(2a2+5ab+2b2)=0
1.    The roots are 2a+b3 and a+2b3
2.    The roots are 2a+b3 and a−2b3
3.    The roots are 5a+b3 and a+2b3
4.    The roots are 2a+b3 and a−2b4
Answer: Option A. -> The roots are 2a+b3 and a+2b3
:
A
We have,
9x29(a+b)x+(2a2+5ab+2b2)=0
Comparing this equation with Ax2+Bx+C=0, we have
A=9,B=9(a+b)andC=2a2+5ab+2b2
D=B24AC
D=81(a+b)236(2a2+5ab+2b2)
D=81(a2+b2+2ab)(72a2+180ab+72b2)
D=9a2+9b218ab
D=9(a2+b22ab)
D=9(ab)20
D0
So, the roots of the given equation are real and are given by
α=B+D2A=9(a+b)+3(ab)18=12a+6b18=2a+b3
and, β=BD2A=9(a+b)3(ab)18=6a+12b18=a+2b3
Question 3. One leg of a right angled triangle exceeds the other leg by 4 inches. The hypotenuse is 20 inches. Find the length of the shorter leg of the triangle.
1.    12 inches
2.    7 inches
3.    20 inches
4.    14 inches
Answer: Option A. -> 12 inches
:
A
Let thelength of the legs of the triangle be x and x+4inches.
Applying Pythagoras' theorem,
x2+(x+4)2=(20)2x2+x2+8x+16=4002x2+8x384=0x2+4x192=0
Length can't be negative, hence
x=12inches
The shortest side(leg) is 12 .
Question 4. If x=23 is a solution of the quadratic equation 7x2+mx3=0, find the value of m.
1.    m=16
2.    m=−13
3.    m=−16
4.    m=13
:
C
Given: x=23 is a solution of the quadratic equation 7x2+mx3=0.
Substitutingx=23 in the given quadratic equation we get
7(23)2+m(23)3=0
7(49)+m(23)3=0
289+2m33=0
28+6m27=0
m=16
Question 5. If the equation 9x2+6kx+4=0 has equal roots, then find the value of k =
1.    0
2.    2 , 0
3.    -2, 0
4.    ± 2
Answer: Option D. -> ± 2
:
D
Given, the roots of quadratic equation 9x2+6kx+4=0are real and equal.
Discriminant,Δ=b24acΔ=36k24(4)(9)=36k24(36)
The roots of quadratic equation are real and equal thenΔ=036k24(36)=0k24=0k2=4k=4k=±2
Question 6. For what value of m does the equation, x2+2x+m=0 have two distinct real roots?
1.    m = 1
2.    m
3.    m > 1
4.    m = 2
:
B
Step 1 :For, x2+2x+m=0,
a=1,b=2,c=m
We know that
D=b24ac
D=(2)24m
=44m
Step 2 :The roots of a quadratic equation are real and distinct only when D>0
Step 3 : 44m>0
4>4m
m<1
Question 7. If the speed of a train is reduced by 40 km/hr, it takes 20 minutes more to cover 1200 km.  Find the speed of the train.
1.    400 km/hr
2.    360 km/hr
3.    420 km/hr
4.    380 km/hr
Answer: Option A. -> 400 km/hr
:
A
Let the speed of the train be x km/hr.
Time=DistanceSpeed
Time takento cover 1200 km = 1200x
Reduced speed = (x40) km/hr
Time takento cover 1200 km in reduced speed = 1200x40
Relation between the time taken to cover 1200 km with speed x and with speed (x - 40) km/hr is as given below
1200x401200x=13 (20minutes=13hours)
48000x(x40)=13
144000=x240x
x240x144000=0
x2400x+360x144000=0
x(x400)+360(x400)=0
(x+360)(x400)=0
x=360orx=400
Since speed cannot be negative, we get x = 400 km/hr.
The speed of the train is 400 km/hr.
Question 8. The roots of 2x26x+8=0 are ___________.
1.    real, unequal and rational
2.    real, unequal and irrational
3.    real and equal
4.    imaginary
:
D
Step 1:For 2x26x+8=0, value of discriminant
D=(6)24(2)(8)=3664=28
Step 2:Since D<0, roots are imaginary.
Question 9. If α and β are the roots of the equation 4x2+3x+7=0, then 1α+1β=
1.    −37
2.    37
3.    −35
4.    35
:
A
Given equation 4x2+3x+7=0,
α+β=34andαβ=74
Now 1α+1β=α+βαβ=3474=34×47=37
Question 10. The coefficient of x in the equation x2 + px + q = 0 was taken as 17 in place of 13, its roots were found to be -2 and -15, the roots of the original equation are

1.    3, 10
2.    - 3, - 10
3.    - 5, - 18
4.    18,5
Answer: Option B. -> - 3, - 10
:
B
Let the equation (inwritten form) bex2 + 17x + q = 0.
Roots are -2, -15.
So q = 30,
And correct equation isx2 + 13x + 30= 0.
Hence roots are -3, -10.