Quantitative Aptitude
QUADRATIC EQUATIONS MCQs
Quadratic Equations, 10th And 12th Grade Quadratic Equations
Total Questions : 102
| Page 3 of 11 pages
Answer: Option C. -> B alone can finish the work in 12 days
:
C
Suppose B alone can takesx days to finish the work.
Then A alone can finish it in (x−6) days.
B's 1 day's work = 1x.
A's 1 day's work = 1x−6.
(A+B)'s 1 day's work = 14
∴1x+1x−6=14
⇒x−6+xx(x−6)=14
⇒8x−24=x2−6x
⇒x2−14x+24=0
⇒x2−12x−2x+24=0
⇒(x−12)(x−2)=0
⇒x=12orx=2
But x cannot be less than 6.
⇒x=12
Hence, B alone can finish the work in 12 days
:
C
Suppose B alone can takesx days to finish the work.
Then A alone can finish it in (x−6) days.
B's 1 day's work = 1x.
A's 1 day's work = 1x−6.
(A+B)'s 1 day's work = 14
∴1x+1x−6=14
⇒x−6+xx(x−6)=14
⇒8x−24=x2−6x
⇒x2−14x+24=0
⇒x2−12x−2x+24=0
⇒(x−12)(x−2)=0
⇒x=12orx=2
But x cannot be less than 6.
⇒x=12
Hence, B alone can finish the work in 12 days
Answer: Option B. -> both of them are integers
:
B
Step 1:- For x2–2x–3=0, value of discriminant
D=(−2)2–4(−3)(1)=4+12=16.
Step 2:- Since, D is a perfect square, roots are rational and unequal.
Step 3:- Solving the equation,
x=−(−2)+√(16)2or−(−2)−√(16)2
x=3,−1
Thus, both of them are integers.
:
B
Step 1:- For x2–2x–3=0, value of discriminant
D=(−2)2–4(−3)(1)=4+12=16.
Step 2:- Since, D is a perfect square, roots are rational and unequal.
Step 3:- Solving the equation,
x=−(−2)+√(16)2or−(−2)−√(16)2
x=3,−1
Thus, both of them are integers.
Answer: Option D. -> 44.6 m & 14.20 m
:
D
Let, Length = x
Breadth =y
Given,
Area of Rectangle = 634m2
Length :x
Thrice the breadth: 3y
2 more that thrice : 3y+2
So, x=3y+2
Area oftherectangle=length×breadth
634=xy634=(2+3y)y634=2y+3y2∴3y2+2y−634=0
This equation resembles the general form of quadratic equation ax2+bx+c=0.
Lets find the values of y satisfying the equation.(Roots of the equation)
y=−b±√b2−4ac2a=−2±√22−4×3(−634)2×3y=−2±√4+76086=−2±√76126y=−2±87.2466y=−2+87.2466or−2−87.2466y=14.20or−14.87
Length is always positive.
∴y=14.20m
x=2+3y∴x=2+3(14.20)=44.6m
Length =44.6m
Breadth =14.2m
:
D
Let, Length = x
Breadth =y
Given,
Area of Rectangle = 634m2
Length :x
Thrice the breadth: 3y
2 more that thrice : 3y+2
So, x=3y+2
Area oftherectangle=length×breadth
634=xy634=(2+3y)y634=2y+3y2∴3y2+2y−634=0
This equation resembles the general form of quadratic equation ax2+bx+c=0.
Lets find the values of y satisfying the equation.(Roots of the equation)
y=−b±√b2−4ac2a=−2±√22−4×3(−634)2×3y=−2±√4+76086=−2±√76126y=−2±87.2466y=−2+87.2466or−2−87.2466y=14.20or−14.87
Length is always positive.
∴y=14.20m
x=2+3y∴x=2+3(14.20)=44.6m
Length =44.6m
Breadth =14.2m
Answer: Option C. -> 5 and -2
:
C
By using the method of factorisation,
x2−3x−10=0
⇒ x2−5x+2x−10=0
⇒x(x−5)+2(x−5)=0
⇒ (x−5)(x+2)=0
⇒x=5 and x=−2
:
C
By using the method of factorisation,
x2−3x−10=0
⇒ x2−5x+2x−10=0
⇒x(x−5)+2(x−5)=0
⇒ (x−5)(x+2)=0
⇒x=5 and x=−2
Answer: Option B. -> 3±√195
:
B
Multiply the equation by 5.
we get 25x2–30x–10=0
(5x)2–[2×(5x)×3]+32–32–10=0
(5x–3)2–9–10=0(5x–3)2–19=0(5x–3)2=195x−3=±√19x=3±√195
Roots of the equation are 3+√195&3−√195
:
B
Multiply the equation by 5.
we get 25x2–30x–10=0
(5x)2–[2×(5x)×3]+32–32–10=0
(5x–3)2–9–10=0(5x–3)2–19=0(5x–3)2=195x−3=±√19x=3±√195
Roots of the equation are 3+√195&3−√195
Answer: Option A. -> True
:
A
x+51=2x+10x−6
⇒(x + 5)(x - 6) = 2x + 10
⇒x2−6x+5x−30=2x+10
⇒x2−3x−40=0 is ofthe form of ax2+bx+c
Hence, is a quadratic equation.
:
A
x+51=2x+10x−6
⇒(x + 5)(x - 6) = 2x + 10
⇒x2−6x+5x−30=2x+10
⇒x2−3x−40=0 is ofthe form of ax2+bx+c
Hence, is a quadratic equation.
Answer: Option B. -> 13 , 14
:
B
Let the numbers be x&x+1
Square of consecutive number: x2&(x+1)2
Sum of square:x2+(x+1)2
∴x2+(x+1)2=365
Lets solve
x2+x2+2x+1=3652x2+2x−364=0x2+x−182=0
(Dividing by 2)
Factorise,
x2+14x−13x−182=0x(x+14)−13(x+14)=0(x+14)(x−13)=0∴x=13&−14
Since numbers are positive integers, x=13
So, required consecutive numbers are 13&14.
:
B
Let the numbers be x&x+1
Square of consecutive number: x2&(x+1)2
Sum of square:x2+(x+1)2
∴x2+(x+1)2=365
Lets solve
x2+x2+2x+1=3652x2+2x−364=0x2+x−182=0
(Dividing by 2)
Factorise,
x2+14x−13x−182=0x(x+14)−13(x+14)=0(x+14)(x−13)=0∴x=13&−14
Since numbers are positive integers, x=13
So, required consecutive numbers are 13&14.
Answer: Option C. -> 24
:
C
Given: A can do a piece of work in x days and B in x+16 days.
Work done by A in one day = 1x
Work done by Bin one day = 1x+16
Work done by A and B together in one day = 115
⇒ 1x + 1x+16 = 115
⇒ 2x+16x(x+16)=115
⇒ x2+16x=15(2x+16)
⇒ x2−14x−240=0
⇒ x2−24x+10x−240=0
⇒ x(x−24)+10(x−24)=0
⇒ (x−24)(x+10)=0
⇒ x=24orx=−10
⇒ x=24 as x cannot be negative.
:
C
Given: A can do a piece of work in x days and B in x+16 days.
Work done by A in one day = 1x
Work done by Bin one day = 1x+16
Work done by A and B together in one day = 115
⇒ 1x + 1x+16 = 115
⇒ 2x+16x(x+16)=115
⇒ x2+16x=15(2x+16)
⇒ x2−14x−240=0
⇒ x2−24x+10x−240=0
⇒ x(x−24)+10(x−24)=0
⇒ (x−24)(x+10)=0
⇒ x=24orx=−10
⇒ x=24 as x cannot be negative.
Answer: Option A. -> 8,9
:
A
Let the smaller numberbex. Then thelarger numberisx+1.
Their product is x(x+1).
⇒ x(x+1)=72
⇒ x2+x−72=0
⇒ x2+9x−8x−72=0
⇒ x(x+9)−8(x+9)=0
⇒ (x−8)(x+9)=0
⇒ x=8orx=−9
Since the given numbers are natural numbers, we get x=8
Hence the two consecutive natural numbers are 8 and 9.
:
A
Let the smaller numberbex. Then thelarger numberisx+1.
Their product is x(x+1).
⇒ x(x+1)=72
⇒ x2+x−72=0
⇒ x2+9x−8x−72=0
⇒ x(x+9)−8(x+9)=0
⇒ (x−8)(x+9)=0
⇒ x=8orx=−9
Since the given numbers are natural numbers, we get x=8
Hence the two consecutive natural numbers are 8 and 9.
Answer: Option A. -> x=−12,−198
:
A
Given:4(2x+3)2−(2x+3)−14=0
Substitute (2x+3)=y, Hence the given equation reduces to
4y2−y−14=0
⇒ 4y2−8y+7y−14=0
⇒ 4y(y−2)+7(y−2)=0
⇒ (4y+7)(y−2)=0
⇒ y=−74ory=2
When y=−74,
(2x+3)=−74
2x=−194
⇒x=−198
When y=2,
(2x+3)=2
2x=−1
⇒x=−12
:
A
Given:4(2x+3)2−(2x+3)−14=0
Substitute (2x+3)=y, Hence the given equation reduces to
4y2−y−14=0
⇒ 4y2−8y+7y−14=0
⇒ 4y(y−2)+7(y−2)=0
⇒ (4y+7)(y−2)=0
⇒ y=−74ory=2
When y=−74,
(2x+3)=−74
2x=−194
⇒x=−198
When y=2,
(2x+3)=2
2x=−1
⇒x=−12
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