Worksheet Question & Solutions: Quadratic Equations

# Worksheet Question and Solutions: Quadratic Equations Class 10 Worksheet Maths

Q.1. A natural number, when increased by 12, equals 160 times its reciprocal. Find the number.

Ans:
Given, a natural number increased by 12 equals to 160 times its reciprocal.
We have to find the natural number.
Let the natural number be x.
So, x + 12 = 160(1/x)
On rearranging,
x2  + 12x = 160
x2  + 12x - 160 = 0
On factoring,
x2 + 20x - 8x - 160 = 0
x(x + 20) - 8(x + 20) = 0
(x - 8)(x + 20) = 0
Now, x - 8 = 0 x = 8
Also, x + 20 = 0
x = -20
Since a negative number cannot be negative, x = -20 is neglected.
Therefore, the natural number is x = 8.

Q.2. By increasing the list price of a book by ₹ 10, a person can buy 10 books less for ₹ 1200. Find the original list price of the book.

Ans:  Let original book price = X
Number of books that can be bought in 1200 =1200/x
Suppose price increases by 10.
New price = x + 10
No of books that can be bought in new price = 1200/(x + 10)
Now, this number is 10 less than the no of books that can be bought in 1200 at ‘x’ price.
So, 1200/x= 1200/(x + 10) + 10
Simplifying the above we get X2 + 30x - 1200 = 0
This is a quadratic equation having two solutions, 30 & -40.
Ignoring negative value the answer is x = 30
So original book price is Rs 30

Q.3. The hypotenuse of a right-angled triangle is 1 cm more than twice the shortest side. If the third side is 2 cm less than the hypotenuse, find the sides of the triangle.

Ans: (2x - 1)2 + x2 = (2x + 1)2
4x2 - 4x + 1 + x2 = 4x+ 4x + 1
x2 - 8x = 0
x(x - 8) = 0
Obviously x = 0 cm is not a realistic answer in this case.
The only acceptable answer is x = 8 cm
The two shorter sides are 8, 15 and the hypotenuse is 17 cm.

Q.4. A passenger train takes 2 hours less for a journey of 300 km, if its speed is increased by 5 km/hr from its usual speed. Find its usual speed.

Ans:
Given: A train takes two hours less for a journey of 300 km
Formula Used: Distance = Speed × Time
Calculation: Suppose the usual speed of the train = x km/hr
According to the question
300/x – 300/(x + 5) = 2
⇒ 300x + 1500 – 300x = 2x2 + 10x
⇒ 2x+ 10x – 1500 = 0
⇒ 2x2 + 60x – 50x – 1500 = 0
⇒ 2x (x + 30) – 50(x + 30) = 0
⇒ (x + 30) (2x – 50) = 0
⇒ x = 25 (As negative value of x is not possible)
∴ Usual speed of train is 25 km/hr.

Q.5. The numerator of a fraction is one less than its denominator. If three is added to each of the numerator and denominator, the fraction is increased by 3/28. Find the fraction.

Ans:
Let the denominator = x
Given that numerator is one less than the denominator
⇒ numerator = x – 1
So, the fraction  = (x - 1)/x
According to the question,

⇒ 28{(x2 + 2x) – (x2 –x + 3x – 3)} = 3 (x2 + 3x)
⇒ 28x2 + 56x – 28x2 – 56x + 84 = 3x2 + 9x
⇒ 3x+ 9x – 84 = 0
⇒ x2 + 3x – 28 = 0
⇒ x2 + 7x – 4x – 28 = 0
⇒ x(x + 7) – 4(x + 7) = 0
⇒ (x – 4) (x + 7) = 0
⇒ x = 4 and – 7
But x is a natural number
Hence, x = 4

Q.6. The difference of squares of two natural numbers is 45. The square of the smaller number is four times the larger number. Find the numbers

Ans:
Let the smaller natural number be x and larger natural number be y
Hence x2 = 4y → (1)
Given y2 – x2 = 45
⇒ y– 4y = 45
⇒ y2 – 4y – 45 = 0
⇒ y2 – 9y + 5y – 45 = 0
⇒ y ( y – 9 ) + 5 ( y – 9 ) = 0
⇒ ( y – 9 ) ( y + 5 ) = 0
⇒ ( y – 9 ) = 0 o r ( y + 5 ) = 0
∴ y = 9 o r y = − 5
But y is natural number, hence y ≠ - 5
Therefore, y = 9 Equation (1) becomes, x2 = 4 (9) = 36
∴ x = 6
Thus the two natural numbers are 6 and 9.

Q.7. Solve: x2 + 5√5x - 70 = 0

Ans:
Given, the equation is x2 + 5√5x - 70 = 0
We have to find whether the equation has real roots or not.
A quadratic equation ax2 + bx + c = 0 has 2 distinct real roots when the discriminant of the equation is greater than zero.
Discriminant = b2 - 4ac
Here, a = 1, b = 5√5 and c = -70
So, b2 - 4ac = (5√5)2 - 4(1)(-70)
= 125 + 280
= 405 > 0
So, the equation has 2 distinct real roots.
x = [-b ± √b2 - 4ac]/2a
x = (-5√5 ± √405)/2(1)
= (-5√5 ± 9√5)/2
Now, x = (-5√5 + 9√5)/2 = 4√5/2 = 2√5
x = (-5√5 - 9√5)/2 = -14√5/2 = -7√5
Therefore, the roots of the equation are -7√5 and 2√5.

Q.8. A train travels a distance of 300 km at a uniform speed. If the speed of the train is increased by 5 km an hour, the journey would have taken two hours less. Find the original speed of the train.

Ans:
Total distance travelled = 300 km
Let the speed of train = x km/hr
We know that,

Hence, time taken by train = 300/x
According to the question,
Speed of the train is increased by 5km an hour
∴ the new speed of the train = (x + 5)km/hr
Time taken to cover 300km = 300/(x + 5)
Given that time taken is 2hrs less from the previous time

⇒ 300x + 1500 – 300x = 2x (x + 5)
⇒ 1500 = 2x2 + 10x
⇒ 750 = x+ 5x
⇒ x2 + 5x – 750 = 0
⇒ x2 + 30x – 25x – 750 = 0
⇒ x (x + 30) – 25 (x + 30) = 0
⇒ (x – 25) (x + 30) = 0
⇒ (x – 25) = 0 or (x + 30) = 0
∴ x = 25 or x = – 30
Since, speed can’t be negative.
Hence, the speed of the train is 25km/hr

Q.9. The speed of a boat in still water is 11 km/ hr. It can go 12 km upstream and returns downstream to the original point in 2 hours 45 minutes. Find the speed of the stream.

Ans:
Let the speed of the stream be x km/hr
Given that, the speed boat in still water is 11 km/hr.
⇒ speed of the boat upstream = (11 - x) km/hr
⇒ speed of the boat downstream = (11 + x) km/hr
It is mentioned that the boat can go 12 km upstream and return downstream to its original point in 2 hr 45 min.
⇒ One-way Distance traveled by boat (d) = 12 km
⇒ Tupstream + Tdownstream  = 2 hr 45 min = (2 + 3/4) hr = 11/4 hr
⇒ [distance / upstream speed ] + [distance / downstream speed]   = 11/4
⇒ [ 12/ (11-x) ] + [ 12/ (11+x) ] = 11/4
⇒ 12 [ 1/ (11-x) + 1/(11+x) ] = 11/4
⇒ 12 [ {11 - x + 11 + x} / {121 - x2} ] = 11/4
⇒ 12 [ {22} / {121 - x2} ] = 11/4
⇒ 12 [ 2 / {121 - x2} ] = 1/4
⇒ 24 / {121 - x2}  = 1/4
⇒ 24 (4) = {121 - x2}
⇒ 96 = 121 - x2
⇒ x2 = 121 - 96
⇒ x2 = 25
⇒ x = + 5 or -5
As speed to stream can never be negative, we consider the speed of the stream(x) as 5 km/hr. Thus, the speed of the stream is 5 km/hr.

Q.10. Determine the value of k for which the quadratic equation 4x2 - 3kx + 1 = 0 has equal roots.

Ans: The given equation 4x2 – 3kx + 1 = 0 is in the form of ax2 + bx + c = 0
Where a = 4, b = -3k, c = 1
For the equation to have real and equal roots, the condition is
D = b2 – 4ac = 0
⇒ (-3k)2 – 4(4)(1) = 0
⇒ 9k2 – 16 = 0

Q.11. Using quadratic formula, solve the following equation for ‘x’:
ab x2 + (b2 - ac) x - bc = 0

Ans:

Q.12. The sum of the numerator and the denominator of a fraction is 12. If the denominator is increased by 3, the fraction becomes 1/2. Find the fraction.

Ans: Let’s assume the numerator of the fraction to be x and the denominator of the fraction to be y.
So, the required fraction is x/y.
From the question it’s given as, The sum of the numerator and denominator of the fraction is 12.
Thus, the equation so formed is,
x + y = 12
⇒ x + y – 12 = 0
And also it’s given in the question as,
If the denominator is increased by 3, the fraction becomes 1/2.
Putting this as an equation, we get
x/ (y + 3) = 1/2
⇒ 2x = (y + 3)
⇒ 2x – y – 3 = 0
The two equations are,
x + y – 12 = 0…… (i)
2x – y – 3 = 0…….. (ii)
Adding (i) and (ii), we get
x + y – 12 + (2x – y – 3) = 0
⇒ 3x -15 = 0
⇒ x = 5
Using x = 5 in (i), we find y
5 + y – 12 = 0
⇒ y = 7
Therefore, the required fraction is 5/7.

Q.13. Rewrite the following as a quadratic equation in x and then solve for x:

Ans: Given expression is

Solve the above expression

Cross multiplying the above equation
(4 – 3x) (2x + 3) = 5x
8x + 12 – 6x2 – 9x = 5x
– 6x2 + 8x – 9x – 5x + 12 = 0
– 6x2 – 6x + 12 = 0
Divide the above equation by -6 we get
x2 + x – 2 = 0
By following factorization method
x2 + 2x – x – 2 = 0
x (x + 2) -1(x + 2) = 0
(x + 2)(x – 1) = 0
x + 2 = 0 or x – 1= 0 x = -2, 1
The solution of the given expression is x = -2 and x = 1

Q.14. A two digit number is such that the product of its digits is 18. When 63 is subtracted from the number, the digits interchange their places. Find the number.

Ans: Let us consider, one’s digit of a two digit number = x and
Ten’s digit = y
The number is x + 10y
After interchanging the digits One’s digit = y
Ten’s digit = x
The number is y + 10x
As per the statement, xy = 18 ………(1)
And, x + 10y -63 = y + 10x
9y – 9x – y – 10x = 63
y – x = 7 …..(2)
using algebraic identity: (x + y)2 = (x – y)2 + 4xy
(x + y)2 = (-7)2 + 4(18) = 121 (Using (1))
(x + y)2 = 112
or x + y = 11 …..(3)
2y = 18
or y = 9
From (1): xy = 18
9x = 18
x = 2
Answer: The number is: x + 10y = 2 + 10(9) = 92

Q.15. A train covers a distance of 90 km at a uniform speed. Had the speed been 15 km/ hr more, it would have taken 30 minutes less for the journey. Find the original speed of the train.

Ans: Let the original speed of the train be x km/hr.
According to the question:

x cannot be negative; therefore, the original speed of train is 45 km/hr.

Q.16. Solve for x:

Ans: On cross multiply
(x + 1)(x + 2) + (x − 2)(x − 1) = 3(x − 1)(x + 2)
x2 + 3x + 2 + x− 3x + 2 = 3[x2 + x − 2]
2x2 + 4 = 3x2 + 3x − 6
x+ 3x − 10 = 0
(x + 5)(x − 2) = 0
[x = 2, −5]

Q.17. Using quadratic formula, solve the following for x: 9x2 − 3 (a2 + b2) x + a2 b2 = 0

Ans:

Q.18. A number consisting of two digits is seven times the sum of its digits. When 27 is subtracted from the number, the digits are reversed. Find the number.

Ans: Let the tens and the units digits of the required number be x and y, respectively.
Required number = (10x + y)
10x + y = 7(x + y)
10x + 7y = 7x + 7y or 3x – 6y = 0 ……….(i)
Number obtained on reversing its digits = (10y + x)
(10x + y) - 27 = (10y + x)
⇒ 10x – x + y – 10y = 27
⇒ 9x – 9y = 27 ⇒9(x – y) = 27
⇒ x – y = 3 ……..(ii)
On multiplying (ii) by 6, we get:
6x – 6y = 18 ………(iii)
On subtracting (i) from (ii), we get:
3x = 18
⇒ x = 6
On substituting x = 6 in (i) we get
3 × 6 – 6y = 0
⇒ 18 – 6y = 0
⇒ 6y = 18
⇒ y = 3
Number = (10x + y) = 10 × 6 + 3 = 60 + 3 = 63
Hence, the required number is 63.

Q.19. The sum of the squares of two consecutive odd numbers is 394. Find the numbers.

Ans: Let the two consecutive odd numbers be x and x + 2.
∴ x2 + (x + 2)2 = 394
⇒ x2 + x2 + 4x + 4 = 394
⇒ 2x2 + 4x – 390 = 0
⇒ x2 + 2x – 195 = 0
⇒ x2 + 15x – 13x – 195 = 0
or
x(x + 15) – 13(x + 15) = 0
⇒ x= 13 or x = – 15
∴  For  x = 13, x + 2 = 13 + 2 = 15
Thus, the required numbers are 13 and 15.

Q.20. Find the number which exceeds its positive square root by 20.

Ans: Let the Number be x
According to the given Question
√x + 20 = x
√x = x - 20
Squaring both the sides
x = x2 + 400 - 40x
[(a - b)2 =  a- 2ab + b2]
x2 - 41x + 400 = 0
x2 - 16x - 25x + 400 = 0
x(x - 16) - 25(x - 16) = 0
(x - 16)(x - 25) = 0
Ans = x is equal to  25 or 16 both are correct answers

Q.21. A two digit number is such that the product of its digits is 14. If 45 is added to the number, the digits interchange their places. Find the number.

Ans: Let the no be 'xy' with x × y = 14
As xy + 45 = yx(x, y > 0)
⇒ (10x + y) + 45 = 10y + x
⇒ 9x − 9y + 45 = 0
⇒ x − y + 5 = 0 & x × y = 14
⇒ x− 14/x + 5 = 0
⇒ x − 14/x + 5 = 0
⇒ x2 + 5x − 14 = 0
⇒ x = 2,−7 & y = 7, −2
Ad x, y > 0 ⇒ no is 27

Q.22. A two digit number is such that the product of its digits is 20. If 9 is added to the number, the digits interchange their places. Find the number.

Ans: Let the one’s digit be ‘a’ and ten’s digit be ‘b’.
Given, two - digit number is such that the product of its digits is 20.
⇒ ab = 20 ----- (1)
Also, If 9 is added to the number, the digits interchange their places.
⇒ 10b + a + 9 = 10a + b
⇒ a – b = 1 ----- (2)
Substituting value of a from eq1 in to eq2
⇒ 20/b – b = 1
⇒ b2 + b – 20 = 0
⇒ (b + 5)(b – 4) = 0
Thus,
b = 4 and a = b + 1 = 5
Number is 45.

Q.23. A two digit number is such that the product of its digits is 15. If 8 is added to the number, the digits interchange their places. Find the number.

Ans: Let the ten's digit be x and one's digit be y.
The number will be 10x + y.
Given product of its digits is 15
xy = 15
y = 15/x  --- (1).
Given that when 18 is added to the number the digits get interchanged.
10x + y + 18 = 10y + x
9x - 9y + 18 = 0
x - y + 2 = 0   ------- (2)
Substitute (1) in (2), we get
x - (15/x) + 2 = 0
x2 + 2x - 15 = 0
x2 + 5x - 3x - 15 = 0
x(x + 5) - 3(x + 5) = 0
(x + 5)(x - 3) = 0 x = -5, 3.
Since the digits cannot be negative, x = 3.
Substitute x = 3 in (1)
y = 15/3 = 5.
The number = 10x + y
= 10 * 3 + 5
= 30 + 5
= 35.

Q.24. Solve for x: abx2 + (b2 − ac) x − bc = 0

Ans:

Q.25. The sum of two numbers is 18. The sum of their reciprocals is 1/4.  Find the numbers.

Ans: Let The numbers be x and 18 - x

Q.26. The sum of two numbers ‘a’ and ‘b’ is 15, and sum of their reciprocals 1/a and 1/b is 3/10. Find the numbers ‘a’ and ‘b’.

Ans: The two numbers are a and b
According to the question,
⇒ a + b = 15
⇒ a = 15 − b ---- ( 1 )
⇒ 1/a  + 1/b  = 3/10
⇒ 1/15−b + 1/b = 3/10
⇒ b + 15 − b/b(15 − b) = 3/10
⇒ 150 = 3b(15−b)
⇒ 50 = b(15 − b)
⇒ 50 = 15b − b2
⇒ b2 − 15b + 50 = 0
⇒ b2 − 10b − 5b + 50 = 0
⇒ b(b − 10) − 5(b − 10) = 0
⇒ (b − 10)(b − 5) = 0
∴ b = 5 or b = 10
Putting both values of b in equation ( 1),
⇒ a = 15 − 5 or a = 15 − 10
∴  a = 10 or a = 5

Q.27. Find the roots of the following quadratic equation:

Ans: Given, 2/5 x2 - x - 3/5 = 0
⇒ 2x2 - 5x – 3 = 0
By splitting the middle term,
⇒ 2x2 - 6x + x – 3 = 0
Taking common in the expression,
⇒ 2x (x - 3) + 1 (x - 3) = 0
⇒ (2x + 1) (x - 3) = 0
∴ 2x + 1 = 0 and ∴ x – 3 = 0
∴ x = -1/2 and ∴ x = 3

Q.28. A natural number when subtracted from 28, becomes equal to 160 times its reciprocal.

Find the number.

Ans: Let the number be x
Now,
(28 - x) = 160 / x
28x - x2 = 160
x2 - 28x + 160 = 0
x2 - 20x - 8x + 160 = 0
x(x - 20) - 8(x - 20) = 0
(x - 8)(x - 20) = 0
x = 8 , 20

Q.29. Find two consecutive odd positive integers, sum of whose squares is 290.

Ans: Let one of the odd positive integer be x
then the other odd positive integer is x + 2
their sum of squares = x2 + (x + 2)2
= x2 + x2 + 4x + 4
= 2x2 + 4x + 4
Given that their sum of squares = 290
2x+ 4x + 4 = 290
2x2 + 4x = 286
2x2 + 4x − 286 = 0
x+ 2x − 143 = 0
x2 + 13x − 11x − 143 = 0
x(x + 13) − 11(x + 13) = 0
(x − 11) = 0,(x + 13) = 0
Therefore, x = 11 or −13
We always take positive value of x
So, x = 11 and (x + 2) = 11 + 2 = 13
Therefore , the odd positive integers are 11 and 13

Q.30. Find the values of k for which the quadratic equation
(k + 4) x2 + (k + 1) x + 1 = 0 has equal roots.

Also find these roots.

Ans: (k + 4)x2 + (k + 1)x + 1 = 0
D = b2 -4ac
= (k + 1)2 - 4(k + 4)(1)
= k+ 2k + 1 - 4k - 16
= k2 - 2k - 15
For equal roots, D = 0
D = 0
K2 - 2k - 15 = 0
k2 - 5k + 3k - 15 = 0
k(k - 5) + 3(k - 5) = 0
(k + 3)(k - 5) = 0
k + 3 = 0
OR k - 5 = 0
k = -3, k = 5

Q.31. Solve for x:

Ans:

Q.32. Solve for x:

Ans:
In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ 3(x2 – 7x + 10) + 3(x2 – 7x + 12)
= 10(x2 – 8x + 15)
⇒ 4x2 – 38x + 84 = 0
⇒ 2x2 – 19x + 42 = 0
⇒ 2x2 – 12x – 7x + 42 = 0
⇒ 2x(x – 6) – 7(x – 6) = 0
⇒ (x – 6)(2x - 7) = 0
⇒ x = 7/2, 6

Q.33. Find the value of k, for which one root of quadratic equation kx2 – 14x + 8 = 0 is six times the other.

Ans: Let one root = α
Other root = 6α

Q.34. If x =  2/3 and x = –3 are roots of the quadratic equation ax2 + 7x + b = 0, find the value of a and b.

Ans: Let us assume the quadratic equation be, Ax2 + Bx + C = 0.
Sum of the roots = -B/A

Q.35. If –5 is a root of the quadratic equation 2x2 + px – 15 = 0 and the quadratic equation p (x2 + x) + k = 0 has equal roots, find the value of k.

Ans: Given: -5 is a root of the quadratic equation 2x2 + px – 15 = 0
Substitute the value of x = -5
2(-5)2 + p(-5) – 15 = 0
50 – 5p – 15 = 0
35 – 5p = 0 p = 7
Again, In quadratic equation p(x2 + x) + k = 0
7 (x2 + x) + k = 0 (put value of p = 7)
7x2 + 7x + k = 0
Compare given equation with the general form of quadratic equation, which is ax2 + bx + c = 0
a = 7, b = 7, c = k
Find Discriminant:
D = b2 – 4ac
= (7)2 – 4 x 7 x k
= 49 – 28k
Since roots are real and equal, put D = 0
49 – 28k = 0
28k = 49 k = 7 / 4
The value of k is 7/4.

The document Worksheet Question and Solutions: Quadratic Equations Class 10 Worksheet Maths is a part of the Class 10 Course Mathematics (Maths) Class 10.
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## Mathematics (Maths) Class 10

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## FAQs on Worksheet Question and Solutions: Quadratic Equations Class 10 Worksheet Maths

 1. What is a quadratic equation?
Ans. A quadratic equation is a polynomial equation of degree 2, which can be written in the form ax^2 + bx + c = 0, where a, b, and c are constants and 'x' is the unknown variable.
 2. How do you solve a quadratic equation by factoring?
Ans. To solve a quadratic equation by factoring, follow these steps: 1. Write the equation in the form ax^2 + bx + c = 0. 2. Factorize the quadratic expression on the left side of the equation. 3. Set each factor equal to zero and solve the resulting linear equations. 4. Obtain the values of 'x' that satisfy the equation.
 3. What is the quadratic formula?
Ans. The quadratic formula is a formula used to solve quadratic equations. It states that for any quadratic equation of the form ax^2 + bx + c = 0, the solutions can be found using the formula x = (-b ± √(b^2 - 4ac)) / (2a).
 4. How do you determine the nature of the roots of a quadratic equation?
Ans. The nature of the roots of a quadratic equation can be determined by analyzing the discriminant (b^2 - 4ac) of the equation. - If the discriminant is greater than zero, the equation has two distinct real roots. - If the discriminant is equal to zero, the equation has two identical real roots. - If the discriminant is less than zero, the equation has two complex conjugate roots.
 5. Can a quadratic equation have only one solution?
Ans. No, a quadratic equation can either have two distinct real solutions, two identical real solutions, or two complex conjugate solutions. It cannot have only one solution.

## Mathematics (Maths) Class 10

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