Let the original speed of train is x km/h
Time taken to cover 63 km with speed x km/h,
$Time = \frac {distance}{speed} =\frac { 63}{x}$ hours
After 63 km, speed of train becomes(x + 6) km/h
Time taken to cover 72km with speed (x + 6) km/h
$Time = \frac {distance}{speed} =\frac { 72}{x +6}$ hours
Now as per the question
$\frac { 63}{x}+ \frac { 72}{x +6}= 3 $
$\frac { 21}{x}+ \frac { 24}{x +6}= 1$
$ \frac {(21x + 126 + 24x) }{x(x+6}= 1$
$45x + 126 = x^2+ 6x$
$x^2- 39x - 126 = 0$
(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
$n(n-2)=24$
$n^2 -2n-24=0$
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$
$3y^2- 2y- 16=0$
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
$x(x+2)=224$
$x^2+2x-224=0$
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
$\frac {1}{p} + \frac {1}{q} = \frac {1}{6}$
or
$\frac {1}{q-5} + \frac {1}{q} = \frac {1}{6}$
$\frac { q + q-5}{q(q-5)} = \frac {1}{6}$
$ 6(2q-5) = q^2 -5q$
$q^2 -17q +30=0$
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
$ \frac {24}{18-x} = 1 + \frac {24}{18+x}$
$\frac {24}{18-x} = \frac {42+x}{18+x}$
$24(18+x) = (42+x) (18-x)$
$x^2 + 48x-182=0$
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
$x^2 +(x-7)^2 = 13^2$
$x^2 -7x -69=0$
x=12 or -5
Since x cannot be negative
x=12
$kx(x-2 \sqrt {5}) + 10 =0$
$kx(x-2 \sqrt {5}) + 10 =0$
or
$kx^2 -2 \sqrt {5} kx +10 =0$
For equal roots, Discriminant should be zero
$b^2 -4ac=0$
here $b= -2 \sqrt {5} k$, a=k ,c=10
$( -2 \sqrt {5} k)^2 -4 \times k \times 10=0$
$20k^2 -40k=0$
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 = $\frac {x-3}{x}$
Now when 2 is added
New fraction= $\frac {x-3+2}{x+2}=\frac {x-1}{x+2} $
According to the question
$\frac {x-3}{x} +\frac {x-1}{x+2}= \frac {29}{20}$
$ \frac {(x-3)(x+2) + x(x-1)}{x(x+2)} =\frac {29}{20}$
$\frac {2x^2 -2x-6}{x(x+2)} =\frac {29}{20}$
$20(2x^2 -2x-6) = 29x(x+2)$
$40x^2 -40x -120 =29x^2 + 58x$
$11x^2 -98x -120=0$
Factoring the quadratics
x=10 or -12/11
So fraction is$\frac {x-3}{x} = \frac {7}{10}$
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 \times \frac {1}{x} + 9 \times \frac {1}{y} = \frac {1}{2}$
or
$ \frac {4}{x} + \frac {9}{y} = \frac {1}{2}$
Now putting y = x + 10 in this
$ \frac {4}{x} + \frac {9}{x+10} = \frac {1}{2}$
$ \frac {4x+ 40 + 9x}{x(x+10)} = \frac {1}{2}$
$26x + 80 =x^2+10x$
$ x^2 - 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
$5x= 2x^2 -3$
$2x^2 -5x-3=0$
$2x^2 -6x + x-3=0$
$2x(x-3) + 1(x-3)=0$
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.
$D=b^2 -4ac \geq 0$
For equation (1) $x^2+2px+64=0$
$
(2p)^2- 4 \times 64 \geq 0$
$
p�\geq 64
$
$
p\geq 8$ or $p \leq -8$
For equation (2)
$x^2-8x+2p=0$
$64 -8p \geq0$
$8 \geq p$
Answer is $( -\infty , -8) \cup {8}$
Let age of zeba be x years
As per the question
$
(x-5)^2=11+5x
$
$x^2+25-10x=11+5x$
$x^2-15x+14=0
$
$x^2-14x-x+14=0$
$(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
$x^2 + (x+7)^2 =637$
$x^2 + x^2 + 49 + 14x =637$
$x^2 + 7x -294=0$
x=-21 and 14
So answer is -21,-14 or 14,17
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