Question 1.
A train travels at a certain average speed for a distance of 63 km and then travels a distance of 72 km at an average speed of 6 km/h more than its original speed. If it takes 3 hours to complete the total journey, what is its original average speed? Solution
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
âˆ´x = 42 and -3 , but x â‰ -3 as speed cannot be negative
Hence, original speed of train = 42 km/h
Question 2.
Determine two consecutive negative even integers whose product is 24? Solution
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)
Question 3
A rectangle has a length that is 2 less than 3 times the width. If the area of this rectangle is 16, find the dimensions and the perimeter. Solution
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
Question 4
The difference of two number is 2 and product of them is 224.Find the Numbers Solution
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
Question 5
X and Y working together can complete a certain project is 6 days. If X alone works on the project, He will take 5 days less than Y to complete. Find days required for X and Y to complete the project alone? Solution
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
Question 6
A motor boat whose speed is 18 km/h in still water takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream. Solution
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
Question 7
The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides Solution
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
Question 8
Find the value of k for which the roots of the quadratic equation are equal Solution
$kx(x-2 \sqrt {5}) + 10 =0$
$kx(x-2 \sqrt {5}) + 10 =0$
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
Question 9
The numerator of a fraction is 3 less than its denominator . If 2 is added to both the numerator and its denominator, the sum of the new fraction and original fraction is 29/20. Find the original fraction Solution
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$\frac {x-3}{x} = \frac {7}{10}$
Question 10
To fill a swimming pool two pipes are used if the pipe of larger diameter used for 4 hour and the pipe of a smaller diameter for 9 hours only half of the pool can be filled .find how much long it would take for each pipe to fill the pool separately if the pool of a smaller diameter takes 10 hour more than the pool of larger diameter to fill the pool? Solution
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
Question 11
Five times of a positive integer is less than twice its square by 3. Find the integer. Solution
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
Question 12
Determine the positive values of P for which equation $x^2+2px+64=0$ and $x^2-8x+2p=0$ will both have real roots Solution
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 (2)
$x^2-8x+2p=0$
$64 -8p \geq0$
$8 \geq p$
Answer is $( -\infty , -8) \cup {8}$
Question 13
If zeba were younger by 5 years than what she really is then the square of her age would have been 11 more than five times her actually age. What is her age now? Solution
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
Question 14
The sum of the squares of the two consecutive multiples of 7 is 637, find the multiples? Solution
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