Exercise 4.1
ANS:1
It is of the form
.
Hence, the given equation is a quadratic equation.
It is of the form.
Hence, the given equation is a quadratic equation.
It is not of the form.
Hence, the given equation is not a quadratic equation.
It is of the form.
Hence, the given equation is a quadratic equation.
It is of the form .
Hence, the given equation is a quadratic equation.
It is not of the form.
Hence, the given equation is not a quadratic equation.
It is not of the form .
Hence, the given equation is not a quadratic equation.
It is of the form .
Hence, the given equation is a quadratic equation.
ANS: 2
(i) Let the breadth of the plot be x m.
Hence, the length of the plot is (2x + 1) m.
Area of a rectangle = Length × Breadth
∴ 528 = x (2x + 1)
(ii) Let the consecutive integers be x and x + 1.
It is given that their product is 306.
∴
(iii) Let Rohan’s age be x.
Hence, his mother’s age = x + 26
3 years hence,
Rohan’s age = x + 3
Mother’s age = x + 26 + 3 = x + 29
It is given that the product of their ages after 3 years is 360.
(iv) Let the speed of train be x km/h.
Time taken to travel 480 km =
In second condition, let the speed of train =
km/h
It is also given that the train will take 3 hours to cover the same distance.
Therefore, time taken to travel 480 km =
hrs
Speed × Time = Distance
Roots of this equation are the values for which
= 0
∴
= 0 or 
= 0
i.e., x = 5 or x = −2
Roots of this equation are the values for which
= 0
∴= 0 or 
= 0
i.e., x = −2 or x =
Roots of this equation are the values for which
= 0
∴
= 0 or 
= 0
i.e., x =
or x = 
Roots of this equation are the values for which
= 0
Therefore,
i.e.,
Roots of this equation are the values for which
= 0
Therefore,
i.e.,
ANS: 2
(i) Let the number of John’s marbles be x.
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.
Either
= 0 or x − 9 = 0
i.e., x = 36 or x = 9
If the number of John’s marbles = 36,
Then, number of Jivanti’s marbles = 45 − 36 = 9
If number of John’s marbles = 9,
Then, number of Jivanti’s marbles = 45 − 9 = 36
(ii) Let the number of toys produced be x.
∴ Cost of production of each toy = Rs (55 − x)
It is given that, total production of the toys = Rs 750
Either
= 0 or x − 30 = 0
i.e., x = 25 or x = 30
Hence, the number of toys will be either 25 or 30.
ANS: 3
Let the first number be x and the second number is 27 − x.
Therefore, their product = x (27 − x)
It is given that the product of these numbers is 182.
Either
= 0 or x − 14 = 0
i.e., 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.
ANS: 4
Let the consecutive positive integers be x and x + 1.
Either x+ 14 = 0 or x − 13 = 0, i.e., x = −14 or x = 13
Since the integers are positive, x can only be 13.
∴ x+ 1 = 13 + 1 = 14
Therefore, two consecutive positive integers will be 13 and 14.
ANS: 5
Let the base of the right triangle be x cm.
Its altitude = (x − 7) cm
Either x− 12 = 0 or x + 5 = 0, i.e., x = 12 orx = −5
Since sides are positive, x can only be 12.
Therefore, the base of the given triangle is 12 cm and the altitude of this triangle will be (12 − 7) cm = 5 cm.
ANS: 6
Let the number of articles produced be x.
Therefore, cost of production of each article = Rs (2x + 3)
It is given that the total production is Rs 90.
Either 2x+ 15 = 0 or x − 6 = 0, i.e., x =
or x = 6
As the number of articles produced can only be a positive integer, therefore, x can only be 6.
Hence, number of articles produced = 6
Cost of each article = 2 × 6 + 3 = Rs 15
Exercise 4.3
ANS: 1
ANS: 2
ANS: 3
ANS: 4
Let the present age of Rehman be x years.
Three years ago, his age was (x − 3) years.
Five years hence, his age will be (x + 5) years.
It is given that the sum of the reciprocals of Rehman’s ages 3 years ago and 5 years from now is
.
However, age cannot be negative.
Therefore, Rehman’s present age is 7 years.
ANS: 5
Let the marks in Maths be x.
Then, the marks in English will be 30 − x.
According to the given question,
If the marks in Maths are 12, then marks in English will be 30 − 12 = 18
If the marks in Maths are 13, then marks in English will be 30 − 13 = 17
ANS: 6
Let the shorter side of the rectangle be x m.
Then, larger side of the rectangle = (x + 30) m
However, side cannot be negative. Therefore, the length of the shorter side will be
90 m.
Hence, length of the larger side will be (90 + 30) m = 120 m
ANS: 7
Let the larger and smaller number be x and y respectively.
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 larger number will be 18 only.
Therefore, the numbers are 18 and 12 or 18 and −12.
ANS: 8
Let the speed of the train be x km/hr.
Time taken to cover 360 km
hr
According to the given question,
However, speed cannot be negative.
Therefore, the speed of train is 40 km/h
ANS: 9
Let the time taken by the smaller pipe to fill the tank be x hr.
Time taken by the larger pipe = (x − 10) hr
Part of tank filled by smaller pipe in 1 hour =
Part of tank filled by larger pipe in 1 hour =
It is given that the tank can be filled in
hours by both the pipes together. Therefore,
Time taken by the smaller pipe cannot be
= 3.75 hours. As in this case, the time taken by the larger pipe will be negative, which is logically not possible.
Therefore, time taken individually by the smaller pipe and the larger pipe will be 25 and 25 − 10 =15 hours respectively.
ANS: 10
Let the average speed of passenger train be x km/h.
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.
Speed cannot be negative.
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.
ANS: 11
Let the sides of the two squares be x m and y m. Therefore, their perimeter will be 4x and 4y respectively and their areas will be x2 and y2respectively.
It is given that
4x − 4y = 24
x − y = 6
x = y + 6
However, side of a square cannot be negative.
Hence, the sides of the squares are 12 m and (12 + 6) m = 18 m
Exercise 4.4
ANS: 1
We know that for a quadratic equation ax2 + bx + c= 0, discriminant is b2 − 4ac.
(A) If b2− 4ac > 0 → two distinct real roots
(B) If b2− 4ac = 0 → two equal real roots
(C) 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.
(II)
Comparing this equation with ax2 + bx + c= 0, we obtain
Discriminant
= 48 − 48 = 0
As b2 − 4ac = 0,
Therefore, real roots exist for the given equation and they are equal to each other.
And the roots will be
and.
Therefore, the roots are
and.
(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.
Therefore, the roots are
or
.
ANS: 2
We know that if an equation ax2 + bx + c = 0 has two equal roots, its discriminant
(b2 −4ac) will be 0.
(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’.
Therefore, if this equation has two equal roots, k should be 6 only.
ANS: 3
Let the breadth of mango grove be l.
Length of mango grove will be 2l.
Area of mango grove = (2l) (l)
= 2l2
Comparing this equation with al2 + bl + c = 0, we obtain
a = 1 b = 0, c = 400
Discriminant = b2 − 4ac = (0)2 − 4 × (1) × (− 400) = 1600
Here, b2− 4ac > 0
Therefore, the equation will have real roots. And hence, the desired rectangular mango grove can be designed.
However, length cannot be negative.
Therefore, breadth of mango grove = 20 m
Length of mango grove = 2 × 20 = 40 m
ANS: 4
Let the age of one friend be x years.
Age of the other friend will be (20 − x) years.
4 years ago, age of 1st friend = (x − 4) years
And, age of 2nd friend = (20 − x − 4)
= (16 − x) years
Given that,
(x− 4) (16 − x) = 48
16x− 64 − x2 + 4x = 48
− x2+ 20x − 112 = 0
x2− 20x + 112 = 0
Comparing this equation with ax2 + bx + c = 0, we obtain
a = 1, b = −20, c = 112
Discriminant = b2 − 4ac = (− 20)2− 4 (1) (112)
= 400 − 448 = −48
As b2− 4ac < 0,
Therefore, no real root is possible for this equation and hence, this situation is not possible.
ANS: 5
Let the length and breadth of the park be l and b.
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
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. And hence, this situation is possible.
Root of this equation,
Therefore, length of park, l = 20 m
And breadth of park, b = 40 − l = 40 − 20 = 20 m
ANS:1
It is of the form
.Hence, the given equation is a quadratic equation.
It is of the form
.Hence, the given equation is a quadratic equation.
It is not of the form
.Hence, the given equation is not a quadratic equation.
It is of the form
.Hence, the given equation is a quadratic equation.
It is of the form .Hence, the given equation is a quadratic equation.
It is not of the form
.Hence, the given equation is not a quadratic equation.
It is not of the form .Hence, the given equation is not a quadratic equation.
It is of the form .Hence, the given equation is a quadratic equation.
ANS: 2
(i) Let the breadth of the plot be x m.
Hence, the length of the plot is (2x + 1) m.
Area of a rectangle = Length × Breadth
∴ 528 = x (2x + 1)
(ii) Let the consecutive integers be x and x + 1.
It is given that their product is 306.
∴
(iii) Let Rohan’s age be x.
Hence, his mother’s age = x + 26
3 years hence,
Rohan’s age = x + 3
Mother’s age = x + 26 + 3 = x + 29
It is given that the product of their ages after 3 years is 360.
(iv) Let the speed of train be x km/h.
Time taken to travel 480 km =
In second condition, let the speed of train =
km/hIt is also given that the train will take 3 hours to cover the same distance.
Therefore, time taken to travel 480 km =
hrsSpeed × Time = Distance
Exercise 4.2
ANS :1
Roots of this equation are the values for which
= 0∴
= 0 or = 0i.e., x = 5 or x = −2
Roots of this equation are the values for which
= 0∴
= 0 or = 0i.e., x = −2 or x =
Roots of this equation are the values for which
= 0∴
= 0 or = 0i.e., x =
or x = Roots of this equation are the values for which
= 0Therefore,
i.e.,
Roots of this equation are the values for which
= 0Therefore,
i.e.,
ANS: 2
(i) Let the number of John’s marbles be x.
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.
Either
= 0 or x − 9 = 0i.e., x = 36 or x = 9
If the number of John’s marbles = 36,
Then, number of Jivanti’s marbles = 45 − 36 = 9
If number of John’s marbles = 9,
Then, number of Jivanti’s marbles = 45 − 9 = 36
(ii) Let the number of toys produced be x.
∴ Cost of production of each toy = Rs (55 − x)
It is given that, total production of the toys = Rs 750
Either
= 0 or x − 30 = 0i.e., x = 25 or x = 30
Hence, the number of toys will be either 25 or 30.
ANS: 3
Let the first number be x and the second number is 27 − x.
Therefore, their product = x (27 − x)
It is given that the product of these numbers is 182.
Either
= 0 or x − 14 = 0i.e., 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.
ANS: 4
Let the consecutive positive integers be x and x + 1.
Either x+ 14 = 0 or x − 13 = 0, i.e., x = −14 or x = 13
Since the integers are positive, x can only be 13.
∴ x+ 1 = 13 + 1 = 14
Therefore, two consecutive positive integers will be 13 and 14.
ANS: 5
Let the base of the right triangle be x cm.
Its altitude = (x − 7) cm
Either x− 12 = 0 or x + 5 = 0, i.e., x = 12 orx = −5
Since sides are positive, x can only be 12.
Therefore, the base of the given triangle is 12 cm and the altitude of this triangle will be (12 − 7) cm = 5 cm.
ANS: 6
Let the number of articles produced be x.
Therefore, cost of production of each article = Rs (2x + 3)
It is given that the total production is Rs 90.
Either 2x+ 15 = 0 or x − 6 = 0, i.e., x =
or x = 6As the number of articles produced can only be a positive integer, therefore, x can only be 6.
Hence, number of articles produced = 6
Cost of each article = 2 × 6 + 3 = Rs 15
Exercise 4.3
ANS: 1
ANS: 2
ANS: 3
ANS: 4
Let the present age of Rehman be x years.
Three years ago, his age was (x − 3) years.
Five years hence, his age will be (x + 5) years.
It is given that the sum of the reciprocals of Rehman’s ages 3 years ago and 5 years from now is
.However, age cannot be negative.
Therefore, Rehman’s present age is 7 years.
ANS: 5
Let the marks in Maths be x.
Then, the marks in English will be 30 − x.
According to the given question,
If the marks in Maths are 12, then marks in English will be 30 − 12 = 18
If the marks in Maths are 13, then marks in English will be 30 − 13 = 17
ANS: 6
Let the shorter side of the rectangle be x m.
Then, larger side of the rectangle = (x + 30) m
However, side cannot be negative. Therefore, the length of the shorter side will be
90 m.
Hence, length of the larger side will be (90 + 30) m = 120 m
ANS: 7
Let the larger and smaller number be x and y respectively.
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 larger number will be 18 only.
Therefore, the numbers are 18 and 12 or 18 and −12.
ANS: 8
Let the speed of the train be x km/hr.
Time taken to cover 360 km
hrAccording to the given question,
However, speed cannot be negative.
Therefore, the speed of train is 40 km/h
ANS: 9
Let the time taken by the smaller pipe to fill the tank be x hr.
Time taken by the larger pipe = (x − 10) hr
Part of tank filled by smaller pipe in 1 hour =
Part of tank filled by larger pipe in 1 hour =
It is given that the tank can be filled in
hours by both the pipes together. Therefore,Time taken by the smaller pipe cannot be
= 3.75 hours. As in this case, the time taken by the larger pipe will be negative, which is logically not possible.Therefore, time taken individually by the smaller pipe and the larger pipe will be 25 and 25 − 10 =15 hours respectively.
ANS: 10
Let the average speed of passenger train be x km/h.
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.
Speed cannot be negative.
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.
ANS: 11
Let the sides of the two squares be x m and y m. Therefore, their perimeter will be 4x and 4y respectively and their areas will be x2 and y2respectively.
It is given that
4x − 4y = 24
x − y = 6
x = y + 6
However, side of a square cannot be negative.
Hence, the sides of the squares are 12 m and (12 + 6) m = 18 m
Exercise 4.4
ANS: 1
We know that for a quadratic equation ax2 + bx + c= 0, discriminant is b2 − 4ac.
(A) If b2− 4ac > 0 → two distinct real roots
(B) If b2− 4ac = 0 → two equal real roots
(C) 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.
(II)
Comparing this equation with ax2 + bx + c= 0, we obtain
Discriminant
= 48 − 48 = 0
As b2 − 4ac = 0,
Therefore, real roots exist for the given equation and they are equal to each other.
And the roots will be
and.Therefore, the roots are
and.(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.
Therefore, the roots are
or.ANS: 2
We know that if an equation ax2 + bx + c = 0 has two equal roots, its discriminant
(b2 −4ac) will be 0.
(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’.
Therefore, if this equation has two equal roots, k should be 6 only.
ANS: 3
Let the breadth of mango grove be l.
Length of mango grove will be 2l.
Area of mango grove = (2l) (l)
= 2l2
Comparing this equation with al2 + bl + c = 0, we obtain
a = 1 b = 0, c = 400
Discriminant = b2 − 4ac = (0)2 − 4 × (1) × (− 400) = 1600
Here, b2− 4ac > 0
Therefore, the equation will have real roots. And hence, the desired rectangular mango grove can be designed.
However, length cannot be negative.
Therefore, breadth of mango grove = 20 m
Length of mango grove = 2 × 20 = 40 m
ANS: 4
Let the age of one friend be x years.
Age of the other friend will be (20 − x) years.
4 years ago, age of 1st friend = (x − 4) years
And, age of 2nd friend = (20 − x − 4)
= (16 − x) years
Given that,
(x− 4) (16 − x) = 48
16x− 64 − x2 + 4x = 48
− x2+ 20x − 112 = 0
x2− 20x + 112 = 0
Comparing this equation with ax2 + bx + c = 0, we obtain
a = 1, b = −20, c = 112
Discriminant = b2 − 4ac = (− 20)2− 4 (1) (112)
= 400 − 448 = −48
As b2− 4ac < 0,
Therefore, no real root is possible for this equation and hence, this situation is not possible.
ANS: 5
Let the length and breadth of the park be l and b.
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
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. And hence, this situation is possible.
Root of this equation,
Therefore, length of park, l = 20 m
And breadth of park, b = 40 − l = 40 − 20 = 20 m
No comments:
Post a Comment