Notesale: Turn your study into money

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Question 1:
Check whether the following are quadratic equations:

Answer 1:

It is of the form


...


It is of the form


...


It is not of the form


...


It is of the form


...


It is of the form


...


It is not of the form


...


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...


It is of the form


...


Question 2:
Represent the following situations in the form of quadratic equations
...
The length of the plot (in
metres) is one more than twice its breadth
...

(ii)

The product of two consecutive positive integers is 306
...

(iii)

Rohan’s mother is 26 years older than him
...
We would like to find
Rohan’s present age
...
If the

speed had been 8 km/h less, then it would have taken 3 hours more to
cover the same distance
...

Answer 2:
(i) Let the breadth of the plot be x m
...

Area of a rectangle = Length × Breadth
∴ 528 = x (2x + 1)
(ii) Let the consecutive integers be x and x + 1
...

∴
(iii)Let Rohan’s age be x
...


(iv) Let the speed of train be x km/h
...

Therefore, time taken to travel 480 km =
Speed × Time = Distance

3

hrs

Question 1:
Find the roots of the following quadratic equations by factorisation:

Answer 1:

Roots of this equation are the values for which
∴

= 0 or

=0

=0

i
...
, x = 5 or x = −2

Roots of this equation are the values for which
∴ x + 2 = 0 or 2x – 3 = 0
i
...
, x = −2 or x = 3/2

1

=0

Roots of this equation are the values for which

= 0 or

i
...
, x =

=0

=0

or x =

Roots of this equation are the values for which

=0

Therefore,

Roots of this equation are the values for which

=0

Therefore,

Question 2:
(i) John and Jivanti together have 45 marbles
...
Find out how many marbles they had to start with
...
The cost of
production of each toy (in rupees) was found to be 55 minus the number
of toys produced in a day
...
Find out the number of toys produced on that day
...

Therefore, number of Jivanti’s marble = 45 − x
After losing 5 marbles,
Number of John’s marbles = x − 5
Number of Jivanti’s marbles = 45 − x − 5 = 40 − x
It is given that the product of their marbles is 124
...
e
...

∴ Cost of production of each toy = Rs (55 − x)
It is given that, total production of the toys = Rs 750

3

x – 25 = 0 or x − 30 = 0
i
...
, x = 25 or x = 30
Hence, the number of toys will be either 25 or 30
...

Answer 3:
Let the first number be x and the second number is 27 − x
...


x – 13 = 0 or x − 14 = 0
i
...
, x = 13 or x = 14
If first number = 13, then
Other number = 27 − 13 = 14
If first number = 14, then
Other number = 27 − 14 = 13 Therefore,
the numbers are 13 and 14
...

Answer 4:
Let the consecutive positive integers be x and x + 1
...
e
...

∴ x + 1 = 13 + 1 = 14
Therefore, two consecutive positive integers will be 13 and 14
...
If the hypotenuse
is 13 cm, find the other two sides
...

Its altitude = (x − 7) cm

5

Either x − 12 = 0 or x + 5 = 0, i
...
, x = 12 or x = −5
Since sides are positive, x can only be 12
...


Question 6:
A cottage industry produces a certain number of pottery articles in a day
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If the total cost of production on that day was Rs 90, find
the number of articles produced and the cost of each article
...

Therefore, cost of production of each article = Rs (2x + 3)
It is given that the total production is Rs 90
...
e
...

Hence, number of articles produced = 6
Cost of each article = 2 × 6 + 3 = Rs 15

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Question 1:
Find the roots of the following quadratic equations, if they exist, by the
method of completing the square:

Answer 1:

1

2

Question 2:
Find the roots of the quadratic equations given in Q
...

Answer 2:

3

4

Question 3:
Find the roots of the following equations:

Answer 3:

5

Question 4:
The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and
5 years from now is 1/3
...

Answer 4:
Let the present age of Rehman be x years
...

Five years hence, his age will be (x + 5) years
...


However, age cannot be negative
...


Question 5:
In a class test, the sum of Shefali’s marks in Mathematics and English is
30
...
Find her marks
in the two subjects
...

Then, the marks in English will be 30 − x
...
If the longer side is 30 metres more than the shorter side, find the
sides of the field
...

Then, larger side of the rectangle = (x + 30) m

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However, side cannot be negative
...

Hence, length of the larger side will be (90 + 30) m = 120 m

Question 7:
The difference of squares of two numbers is 180
...
Find the two numbers
...

According to the given question,

However, the larger number cannot be negative as 8 times of the larger
number will be negative and hence, the square of the smaller number
will be negative which is not possible
...


Therefore, the numbers are 18 and 12 or 18 and −12
...
If the speed had been 5 km/h
more, it would have taken 1 hour less for the same journey
...

Answer 8:
Let the speed of the train be x km/hr
...
Therefore,
the speed of train is 40 km/h

Question 9:
3

Two water taps together can fill a tank in 9 8 hours
...
Find the time in which each tap can separately fill the tank
...

Time taken by the larger pipe = (x − 10) hr

9

Part of tank filled by smaller pipe in 1 hour =
Part of tank filled by larger pipe in 1 hour =
3

It is given that the tank can be filled in 9 8 =

75
8

hours by both the pipes

together
...
75 hours
...

Therefore, time taken individually by the smaller pipe and the larger pipe
will be 25 and 25 − 10 =15 hours respectively
...
If the average speeds of the
express train is 11 km/h more than that of the passenger train, find the
average speed of the two trains
...

Average speed of express train = (x + 11) km/h
It is given that the time taken by the express train to cover 132 km is 1
hour less than the passenger train to cover the same distance
...

Therefore, the speed of the passenger train will be 33 km/h and thus,
the speed of the express train will be 33 + 11 = 44 km/h
...
If the difference of their
perimeters is 24 m, find the sides of the two squares
...
Therefore, their
perimeter will be 4x and 4y respectively and their areas will be x2 and
y2 respectively
...

Hence, the sides of the squares are 12 m and (12 + 6) m = 18 m

12

Question 1:
Find the nature of the roots of the following quadratic equations
...



If b2 − 4ac > 0 → two distinct real roots



If b2 − 4ac = 0 → two equal real roots



If b2 − 4ac < 0 → no real roots

(i)

2x2 −3x + 5 = 0

Comparing this equation with ax2 + bx + c = 0,
we obtain a = 2, b = −3, c = 5
Discriminant = b2 − 4ac = (− 3)2 − 4 (2) (5) = 9 – 40 = −31
As b2 − 4ac < 0,
Therefore, no real root is possible for the given equation
...


1

And the roots will be

−𝑏
2𝑎

and

Therefore, the roots are

2
√3

−𝑏
2𝑎


...


√3

(iii) 2x2 − 6x + 3 = 0
Comparing this equation with ax2 + bx + c = 0, we obtain a
= 2, b = −6, c = 3
Discriminant = b2 − 4ac = (− 6)2 − 4 (2) (3)
= 36 − 24 = 12
As b2 − 4ac > 0,
Therefore, distinct real roots exist for this equation as follows
...


Question 2:
Find the values of k for each of the following quadratic equations, so that
they have two equal roots
...

(i) 2x2 + kx + 3 = 0
Comparing equation with ax2 + bx + c = 0,
we obtain a = 2, b = k, c = 3
Discriminant = b2 − 4ac = (k)2− 4(2) (3) = k2 − 24
For equal roots,
Discriminant = 0
k2 − 24 = 0
k2 = 24

(ii)

kx (x − 2) + 6 = 0

or kx2 − 2kx + 6 = 0
Comparing this equation with ax2 + bx + c = 0,
we obtain a = k, b = −2k, c = 6
Discriminant = b2 − 4ac = (− 2k)2 − 4 (k) (6) = 4k2 − 24k
For equal roots, b2 − 4ac = 0
4k2 − 24k = 0
4k (k − 6) = 0
Either 4k = 0 or k = 6 = 0
k = 0 or k = 6
However, if k = 0, then the equation will not have the terms ‘x2’ and
‘x’
...


3

Question 3:
Is it possible to design a rectangular mango grove whose length is twice
its breadth, and the area is 800 m2?
If so, find its length and breadth
...

Length of mango grove will be 2l
...
And hence, the desired
rectangular mango grove can be designed
...

Therefore, breadth of mango grove = 20 m
Length of mango grove = 2 × 20 = 40 m

Question 4:
Is the following situation possible? If so, determine their present ages
...
Four years ago, the
product of their ages in years was 48
...

Age of the other friend will be (20 − x) years
...


Question 5:
Is it possible to design a rectangular park of perimeter 80 and area 400
m2? If so find its length and breadth
...

Perimeter = 2 (l + b) = 80 l + b = 40
Or, b = 40 − l
Area = l × b = l (40 − l) = 40l − l2
40l − l2 = 400
l2 − 40l + 400 = 0

5

Comparing this equation with al2 + bl + c = 0,
we obtain a = 1, b = −40, c = 400
Discriminate = b2 − 4ac = (− 40)2 −4 (1) (400) = 1600 − 1600 = 0
As b2 − 4ac = 0,
Therefore, this equation has equal real roots
...

Root of this equation,

Therefore, length of park, l = 20 m
And breadth of park, b = 40 − l = 40 − 20 = 20 m

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