Quadratic Equation
Mathematics Class Tenth
NCERT Exercise 4.3 - Q 4 - 11
Question(4) The sum of the reciprocals of Rehman's ages, (in years) 3 years ago and 5 years from now is 1/3. Find his present age.
Solution:
Let the present age of Rehman = x
∴ Rehman's age 3 years ago = x – 3
And, Rehman's age 5 years from now = x + 5
According to question
1/x – 3 + 1/x + 5 = 1/3
⇒ x + 5 + x – 3/(x – 3)(x + 5) = 1/3
⇒ 2x + 2/x2 + 5x – 3x – 15 = 1/3
⇒ 2x + 2/x2 + 2x – 15 = 1/3
By cross multiplication
⇒ 3(2x + 2) = x2 + 2x – 15
⇒ 6x + 6 = x2 + 2x – 15
⇒ x2 + 2x – 15 – 6x – 6 = 0
⇒ x2 + 2x – 6x – 21 = 0
⇒ x2 – 4x – 21 = 0
By expanding middle term
⇒ x2 – 7x + 3x – 21 = 0
⇒ x(x – 7) + 3(x – 7) = 0
⇒ (x + 3)(x – 7) = 0
Now, if x + 3 = 0
∴ x = – 3
And if x – 7 = 0
∴ x = 7
After discarding the negative value of x, the root of the given quadratic equation is 7
Thus, Rehman's present age is 7 years
Question (5) In a class test, the sum of Shefali's marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.
Solution:
Let marks got by Shefali in Mathematics = x
As per question, the sum of marks in Mathematics and English = 30
∴ Shefali's marks in English = 30 – x
If she got 2 more marks in Mathematics, ∴ her marks in Mathematics = x + 2
And if she got 3 marks less in English, ∴ her marks in English = 30 – x – 3
= 27 – x
Now, According to question, (x + 2)(27 – x) = 210
⇒ 27x – x2 + 54 – 2x = 210
⇒ – x2 + 25x – 2x + 54 = 210
⇒ x2 – 25x + 54 = 210
⇒ x2 – 25x + 54 – 210 = 0
⇒ x2 – 25x – 156 = 0
By taking minus as common
⇒ – (x2 – 25x + 156) = 0
⇒ x2 – 25x + 156 = 0
By expanding midle term (– 25x)
⇒ x2 – 12x – 13x + 156 = 0
⇒ x(x – 12) – 13(x – 12) = 0
⇒ (x – 12)(x – 13) = 0
Now, case – I
If x – 12 = 0
∴ x = 12
Thus, marks in Mathematics = 12
∴ marks in English = 30 – 12 = 18
Case – II
If x – 13 = 0
∴ x = 13
Thus, marks in Mathematics = 13
∴ Marks in English = 30 – 13 = 17
Thus, Shefali's marks in Mathematics = 12 and in English = 18 or marks in Mathematics = 13 and in English = 17
Question (6) The diagonal of a rectangular field is 60 meters more than the shorter side. If the longer side is 30 meters more than the shorter side, find the sides of the field.
Solution:
Let the shorter side of the given field = x meter
∴ Diagonal = x + 60 meter
And, longer side = x + 30 meter
We know that, Diagonal2 = Height2 + Base 2
⇒ (x + 60)2 = (x + 30)2 + x2
⇒ x2 +(2 × x × 60) + (60)2
= x2 + (2 × x × 30) + (30)2 + x2
⇒ x2 + 120x + 3600 = x2 + 60x + 900 + x2
⇒ x2 + 120x + 3600 = 2x2 + 60x + 900
⇒ x2 + 120x + 3600 – (2x2 + 60x + 900 ) = 0
⇒ x2 120x + 3600 – 2x2 – 60x – 900 = 0
⇒ (x2 – 2x2) + (120x – 60x) + (3600 – 900) = 0
⇒ – x2 + 60x + 2700 = 0
By taking minus as common
⇒ (x2 – 60x – 2700) = 0
⇒ x2 – 60x – 2700 = 0
By expanding middle term (– 60x)
⇒ x2 – 90x + 30x – 2700 = 0
⇒ x(x – 90) + 30(x – 90) = 0
⇒ (x + 30)(x – 90) = 0
Now, if x + 30 = 0
⇒ x = – 30
Since this value is negative, thus it is discarded
And if x – 90 = 0
∴ x = 90
Thus, shorter side of the given field = 90 meter
And longer side of the given field = 90 + 30 = 120 meter
Thus, sides of the given field are 120 meter and 90 meter
Question (7) The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers
Solution:
Let the larger number = x
∴ square of smaller number = 8x
Since the difference of squares of the two given numbers is equal to 180 (As per quesiton)
∴ x2 – 8x = 180
⇒ x2 – 8x – 180 = 0
By expanding middle term (– 8x) which can be written as (– 18x + 10x)
⇒ x2 – 18x + 10x – 180 = 0
⇒ x(x – 18) + 10(x – 18) = 0
⇒ (x + 10)(x – 18) = 0
Now, if x + 10 = 0
∴ x = – 10
Since, this number is negative, thus it will be discarded
And if x – 18 = 0
∴ x = 18
Since square of smaller number = 8x
∴ (smaller number)2 = 8 × 18
⇒ (smaller number)2 = 144
smaller number = 144
= smaller number = 12
Thus, numbers are 12 and 18
Proof
Given, differene of squares of two numbers is 180
∴ 182 – 122 = 324 – 144 = 180
Question (8) A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.
Solution:
Given, Distance = 360 km
Let, the speed of train = x km /h
Let, time taken at this speed = t hour
Thus, enhanced speed = x + 5 km/h
∴ time taken at enhanced speed = t – 1 hour
Now, we know that, time = Distance/Speed
t = 360/x ----(i)
And, t – 1 = 360/x + 5
⇒ t = 360/x + 5 +1 -----(ii)
Now, from equation (i) and (ii)
360/x = 360/ x + 5 + 1
⇒ 360/x– 360/x + 5 = 1
⇒ (360(x + 5)) – 360x)/x(x + 5) = 1
⇒ 360x + 1800 – 360x/x2 + 5x = 1
⇒ 1800 = x2 + 5x
⇒ x2 + 5x – 1800 = 0
By expanding middle term (5x)
⇒ x2 + 45x – 40x – 1800 = 0
⇒ x(x + 45) – 40(x + 45) = 0
⇒ (x + 45)(x + 40) = 0
Now, if x + 45 = 0
∴ x = – 45
Since this value is negative, hence discarded
And, if x – 40 = 0
∴ x = 40
Thus, speed of the train = 40 km/h
Question (9) Two water taps together can fill a tank in 93/8 hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the itme in which each tap can separately fill the tank.
Solution:
Let, the smaller tap fills the tank in x hours
∴ Larger tank will fill the tank in x – 10 hours
there4; in 1 hour part of tank filled by smaller tap = 1/x
And in 1 hour part of tank filled by larger tap = 1/x – 10
Given, in 75/8 hours both the tap together fills the tank (1 tank)
∴ in 1 hour both the tank will fill 8/75 part of tank.
∴ 1/x + 1/x – 10 = 8/75
⇒ (x – 10) + x/x(x – 10) = 8/75
⇒ x – 10 + x/x2 – 10x = 8/75
⇒ 2x – 10/x2 – 10x = 8/75
By cross multiplication
⇒ 75(2x – 10) = 8(x2 – 10x)
⇒ 150x – 750 = 8x2 – 80x
⇒ 150x – 750 – (8x2 – 80x) = 0
⇒ 150x – 750 – 8x2 + 80x = 0
⇒ 8x2 + 80x + 150x – 750 = 0
⇒ (8x2 – 80x – 150x + 750) = 0
⇒ 8x2 – 80x – 150x + 750 = 0
⇒ 8x2 – 230x + 750 = 0
⇒ 2(4x2 – 115x + 375) = 0
⇒ 4x2 – 115x + 375 = 0
⇒ 4x2 – 100x – 15x + 375 = 0
⇒ 4x(x – 25) – 15(x – 25) = 0
⇒ (4x – 15)(x – 25)= 0
Now, if 4x – 15 = 0
⇒ 4x = 15
⇒ x = 15/4
As this time is less than the difference in timing given, thus, it will be discarded
NOw, if x – 25 = 0
∴ x = 25
Thus, timing to fill the tank by smaller tap alone = 25 hour
∴ timing taken by larger tap alone = 25 – 10 = 15 hour
Thus, smaller tap will take 25 hour and larger tap will take 15 hour to fill the tank
Question (10) An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km/h more than that of the passenger train, find the average speed of the two trains.
Solution:
Let speed of the passenger train = x km/h
Thus, speed of the passenger train will be = x + 11 km/h
Given, distance = 132 km
Let time taken by passenger train = t hour
And given difference in timing taken by express train = 1 hour less
∴ time taken by express train = t – 1 hour
We know that, (time) = Distance/speed
⇒ t = 132/x -----(i)
And t – 1 = 132/x + 11
⇒ t = 132/x + 11 + 1 -----(ii)
Thus from equation (i) and (ii)
⇒ 132/x = 132/x + 11 + 1
⇒ 132/x – 132/x + 11 = 1
⇒ 132(x + 11) – 132x/x(x + 11) = 1
-->⇒ 132x + 1452 – 132x/x2 + 11x = 1
⇒ 1452/x2 + 11x = 1
After cross multiplication, we get
1452 = x + 11x
⇒ x2 + 11x – 1452 = 0
By expanding middle term, we get
x2 + 55x – 33x – 1452 = 0
⇒ x(x + 55) – 33(x + 55) = 0
⇒ (x + 55)(x – 33) = 0
Now, if x + 55 = 0
∴ x = – 55
Since, speed cannot be in negative, thus this value is dicarded
Now, if x – 33 = 0
∴ x = 33 (speed of passenger train)
Thus, speed of passenger train = 33 km/h
Thus, speed of express train = 33 + 11 = 44 km/h
∴ Average speed of both the train =33 + 44/2
= 77/2 = 38.5 km/h Answer
Question (11) Sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, find the sides of the two squares.
Solution:
Let side of first square = a
And let side of second square = b
Given, sum of the areas of two squares = 468 m2
And, difference of their perimeter = 24m
We know that, perimeter of a square = 4 x sides
Thus, perimeter of first square = 4 × a = 4a
Similarly, perimeter of second square = 4× b = 4b
∵ Difference in their perimeter = 24m
∴ 4a – 4b = 24m
⇒ 4(a – b) = 24m
⇒ a – b = 24/4 m
⇒ a – b = 6m
⇒ a = 6 + b m ----(i)
Now, we know that, Area of a square = (side)2
Thus, area of first square = a2
And area of second square = b2
As per question,
a2 + b2 = 468
after putting the value of a = 6 + b from equation (i)
⇒ (6 + b)2 + b2 = 468
⇒ 62 + b2 + (2 × 6 × b) + b2 = 468
= 36 + b2 + b2 + 12b = 468
⇒ 2b2 + 12b + 36 = 468
⇒ 2b2 + 12b + 36 – 468 = 0
⇒ 2b2 + 12b – 432 = 0
⇒ 2(b2 + 6b – 216) = 0
⇒ b2 + 6b – 216 = 0
After expanding the middle term (6b), we get
b2 + 18b – 12b – 216 = 0
⇒ b(b + 18)– 12(b + 18) = 0
⇒ (b + 18)(b – 12) = 0
Now, if b + 18 = 0
∴ b = – 18
Since, sides cannot be negative, thus this value will be discarded
And, if b – 12 = 0
∴ b = 12
By putting this value in equation (i)
a = 6 + b = 6 + 12
⇒ a = 18
Thus, side of one square is 12 m and another square is 18 m
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