Let the original speed of train is x km/h
Time taken to cover 63 km with speed x km/h,
Time=distancespeed=63x hours
After 63 km, speed of train becomes(x + 6) km/h
Time taken to cover 72km with speed (x + 6) km/h
Time=distancespeed=72x+6 hours
Now as per the question
63x+72x+6=3
21x+24x+6=1
(21x+126+24x)x(x+6=1
45x+126=x2+6x
x2−39x−126=0
(x + 3)(x - 42) = 0
∴x = 42 and -3 , but x ≠ -3 as speed cannot be negative
Hence, original speed of train = 42 km/h
Let n and n-2 be the two consecutive negative even integers
then
n(n−2)=24
n2−2n−24=0
or
(n−6)(n+4)=0
n=6 or -4
Since we want the negative integers only, n=-4
other number =n-2= -4-2 = -6
So numbers are (-6,-4)
Let x= the length of the rectangle and y= its width
As per the question
x=3y−2
Given ,The area of this rectangle is 16
Now
Area=xy=16
or
(3y−2)y=16
3y2−2y−16=0
Factoring
(3y−8)(y+2)=0
y = 8/3 and -2
Since we can't have a negative width, y= 8/3
Now x=3y-2 = 6
Perimeter = 2x + 2y = 12 + 16/3 = 64/3
Let x be the smaller number
x+2 could be the larger number
x(x+2)=224
x2+2x−224=0
Factoring the quadratics
(x+16)(x−14)=0
x= -16 or 14
For positive Numbers
x= 14
Second number= x+2= 16
So the numbers are 14 and 16
For Negative Numbers
x=-16
Second number= x+2= -14
So the numbers are -16 and -14
Let p and q are days required for X and Y to complete the project alone
Now p =q-5
Also
1p+1q=16
or
1q−5+1q=16
q+q−5q(q−5)=16
6(2q−5)=q2−5q
q2−17q+30=0
factoring this quadratics
q=15 or 2
It can not be 2 as then p will be negative
So q=15 days and p=10 days
Let the speed of the stream be x km/hr.
Distance upstream = Distance downstream = 24 km
Speed of the boat going upstream=18 - x
Speed of the boat going downstream=18+x
Time taken going upstream = 24/18-x
Time taken going downstream = 24/18+x
Now as per question
2418−x=1+2418+x
2418−x=42+x18+x
24(18+x)=(42+x)(18−x)
x2+48x−182=0
x=-54 or 6
Rejecting negative value, the speed of stream is 6 km/hr
let x be the base ,the altitude will be x-7
Now as per pythagorus theorem
x2+(x−7)2=132
x2−7x−69=0
x=12 or -5
Since x cannot be negative
x=12
kx(x−2√5)+10=0
kx(x−2√5)+10=0
or
kx2−2√5kx+10=0
For equal roots, Discriminant should be zero
b2−4ac=0
here b=−2√5k, a=k ,c=10
(−2√5k)2−4×k×10=0
20k2−40k=0
or k=0 or 2
k cannot be zero, So correct value of k=2
Let denominator be x ,the numerator x-3.
Fraction will be = x−3x
Now when 2 is added
New fraction= x−3+2x+2=x−1x+2
According to the question
x−3x+x−1x+2=2920
(x−3)(x+2)+x(x−1)x(x+2)=2920
2x2−2x−6x(x+2)=2920
20(2x2−2x−6)=29x(x+2)
40x2−40x−120=29x2+58x
11x2−98x−120=0
Factoring the quadratics
x=10 or -12/11
So fraction isx−3x=710
Let x and y hour be the time taken by larger pipe and smaller pipe to fill the swimming pool
Now it is given y−x=10 or y=x+10
If smaller pipe takes x hours to fill the pool, it will fill 1/x part in 1 hour
SimilaryIf larger pipe takes y hours to fill the pool, it will fill 1/y part in 1 hour
According to question
4×1x+9×1y=12
or
4x+9y=12
Now putting y = x + 10 in this
4x+9x+10=12
4x+40+9xx(x+10)=12
26x+80=x2+10x
x2−16x−80=0
OR (x -20)(x+4) = 0
Neglecting negative root,
x = 20 h
And y = 30 h
Let x be the number,then
5x=2x2−3
2x2−5x−3=0
2x2−6x+x−3=0
2x(x−3)+1(x−3)=0
or
(2x+1)(x-3)=0
or x=3 as x is a positive integer
For an equation to have real roots, its discriminant should be greater than or equal to 0.
D=b2−4ac≥0
For equation (1) x2+2px+64=0
(2p)2−4×64≥0
p�≥64
p≥8 or p≤−8
For equation (2)
x2−8x+2p=0
64−8p≥0
8≥p
Answer is (−∞,−8)∪8
Let age of zeba be x years
As per the question
(x−5)2=11+5x
x2+25−10x=11+5x
x2−15x+14=0
x2−14x−x+14=0
$(x-1)(x-14)
So Zeba age will be 14 yrs because if she was 1 year old the 5 years younger cannot happen
Let x and x+7 are the two consecutive multiples of 7
Now as per question
x2+(x+7)2=637
x2+x2+49+14x=637
x2+7x−294=0
x=-21 and 14
So answer is -21,-14 or 14,17
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