Solutions to Practice Sets 2.1 to 2.6 and Problem Set 2, with step-by-step working for factorisation, completing the square and the formula. Maharashtra State Board (Balbharati), Std 10 Mathematics Part 1 (Algebra).
7The roots of each of the following quadratic equations are real and equal, find k.
(1) 3y2 + ky + 12 = 0
a = 3, b = k, c = 12
Roots are real and equal, so Δ = 0.
k2 − 4 × 3 × 12 = 0
∴ k2 − 144 = 0 ∴ k2 = 144
∴ k = 12 or k = −12
k = 12 or k = −12
(2) kx(x − 2) + 6 = 0
kx2 − 2kx + 6 = 0
a = k, b = −2k, c = 6
Roots are real and equal, so Δ = 0.
(−2k)2 − 4 × k × 6 = 0
∴ 4k2 − 24k = 0
∴ 4k(k − 6) = 0
∴ k = 0 or k = 6
If k = 0, the equation becomes 6 = 0, which is not a quadratic equation. ∴ k ≠ 0.
k = 6
Practice Set 2.6
1Product of Pragati’s age 2 years ago and 3 years hence is 84. Find her present age.
Let Pragati’s present age be x years.
Age 2 years ago = (x − 2) years; age 3 years hence = (x + 3) years.
(x − 2)(x + 3) = 84
∴ x2 + 3x − 2x − 6 = 84
∴ x2 + x − 90 = 0
∴ x2 + 10x − 9x − 90 = 0
∴ x(x + 10) − 9(x + 10) = 0
∴ (x + 10)(x − 9) = 0
∴ x = −10 or x = 9
Age cannot be negative, so x ≠ −10. ∴ x = 9
Pragati’s present age is 9 years.
2Sum of squares of 2 consecutive natural even numbers is 244; find the numbers.
Let the numbers be x and x + 2.
x2 + (x + 2)2 = 244
∴ x2 + x2 + 4x + 4 = 244
∴ 2x2 + 4x − 240 = 0
Dividing by 2: x2 + 2x − 120 = 0
∴ x2 + 12x − 10x − 120 = 0
∴ (x + 12)(x − 10) = 0
∴ x = −12 or x = 10
The numbers are natural numbers, so x ≠ −12. ∴ x = 10 and x + 2 = 12
Check: 102 + 122 = 100 + 144 = 244
The numbers are 10 and 12.
3In the orange garden of Mr. Madhusudan there are 150 orange trees. The number of trees in each row is 5 more than that in each column. Find the number of trees in each row and each column with the help of the flow chart.
No. of trees in a column = x
No. of trees in a row = x + 5
Total no. of trees = x(x + 5) = 150
Frame quadratic equation:x2 + 5x − 150 = 0
Find x:x2 + 15x − 10x − 150 = 0 → (x + 15)(x − 10) = 0 → x = −15 or x = 10. Number of trees cannot be negative, so x = 10.
No. of trees in a column = 10
No. of trees in a row = 10 + 5 = 15
10 trees in each column and 15 trees in each row
4Vivek is older than Kishor by 5 years. The sum of the reciprocals of their ages is 1/6. Find their present ages.
Let Kishor’s present age be x years. Then Vivek’s age is (x + 5) years.
1x + 1x + 5 = 16
∴ x + 5 + xx(x + 5) = 16
∴ 6(2x + 5) = x(x + 5)
∴ 12x + 30 = x2 + 5x
∴ x2 − 7x − 30 = 0
∴ x2 − 10x + 3x − 30 = 0
∴ (x − 10)(x + 3) = 0
∴ x = 10 or x = −3
Age cannot be negative. ∴ x = 10 and x + 5 = 15
Kishor is 10 years old and Vivek is 15 years old.
5Suyash scored 10 marks more in second test than that in the first. 5 times the score of the second test is the same as square of the score in the first test. Find his score in the first test.
Let the score in the first test be x. Then the score in the second test is x + 10.
5(x + 10) = x2
∴ x2 − 5x − 50 = 0
∴ x2 − 10x + 5x − 50 = 0
∴ (x − 10)(x + 5) = 0
∴ x = 10 or x = −5
A score in a test cannot be negative here. ∴ x = 10
Check: second test = 20; 5 × 20 = 100 = 102
Suyash scored 10 marks in the first test.
6Mr. Kasam runs a small business of making earthen pots. Production cost of each pot is ₹ 40 more than 10 times the total number of pots he makes in one day. If production cost of all pots per day is ₹ 600, find production cost of one pot and number of pots he makes per day.
Let the number of pots made per day be x.
Production cost of one pot = ₹ (10x + 40)
x(10x + 40) = 600
∴ 10x2 + 40x − 600 = 0
Dividing by 10: x2 + 4x − 60 = 0
∴ x2 + 10x − 6x − 60 = 0
∴ (x + 10)(x − 6) = 0
∴ x = −10 or x = 6
Number of pots cannot be negative. ∴ x = 6
Cost of one pot = 10 × 6 + 40 = ₹ 100
Check: 6 × 100 = ₹ 600
He makes 6 pots per day; production cost of one pot is ₹ 100.
7Pratik takes 8 hours to travel 36 km downstream and return to the same spot. The speed of boat in still water is 12 km per hour. Find the speed of water current.
8Pintu takes 6 days more than those of Nishu to complete certain work. If they work together they finish it in 4 days. How many days would it take to complete the work if they work alone?
Let Nishu take x days alone. Then Pintu takes (x + 6) days.
Work done in one day: Nishu 1x, Pintu 1x + 6, together 14
1x + 1x + 6 = 14
∴ 4(2x + 6) = x(x + 6)
∴ 8x + 24 = x2 + 6x
∴ x2 − 2x − 24 = 0
∴ x2 − 6x + 4x − 24 = 0
∴ (x − 6)(x + 4) = 0
∴ x = 6 or x = −4
Number of days cannot be negative. ∴ x = 6 and x + 6 = 12
Nishu takes 6 days and Pintu takes 12 days.
9If 460 is divided by a natural number, quotient is 6 more than five times the divisor and remainder is 1. Find quotient and divisor.
Let the divisor be x. Then the quotient is 5x + 6.
Dividend = divisor × quotient + remainder
∴ 460 = x(5x + 6) + 1
∴ 5x2 + 6x − 459 = 0
5 × (−459) = −2295 = 51 × (−45) and 51 − 45 = 6
∴ 5x2 + 51x − 45x − 459 = 0
∴ x(5x + 51) − 9(5x + 51) = 0
∴ (5x + 51)(x − 9) = 0
∴ x = −515 or x = 9
The divisor is a natural number. ∴ x = 9
Quotient = 5 × 9 + 6 = 51
Check: 9 × 51 + 1 = 459 + 1 = 460
Divisor = 9, Quotient = 51
10In the figure, □ABCD is a trapezium, AB ∥ CD and its area is 33 cm2. AB = x, DC = 2x + 1, AM = x − 4, BC = x − 2 and AD = BC. Find the lengths of all sides of □ABCD. Fill in the empty boxes to get the solution.
Trapezium ABCD with AB ∥ CD; AM ⊥ DC is the height, AD = BC.
□ABCD is a trapezium, AB ∥ CD.
A(□ABCD) = 12(AB + CD) × AM
33 = 12(x + 2x + 1) × (x − 4)
∴ 66 = (3x + 1) × (x − 4)
∴ 66 = 3x2 − 12x + x − 4
∴ 3x2 − 11x − 70 = 0
∴ 3x2 − 21x + 10x − 70 = 0
∴ 3x(x − 7) + 10(x − 7) = 0
∴ (3x + 10)(x − 7) = 0
∴ (3x + 10) = 0 or x − 7 = 0
∴ x = −103 or x = 7
But length is never negative. ∴ x ≠ −103 ∴ x = 7
AB = x = 7 cm, CD = 2x + 1 = 15 cm, AD = BC = x − 2 = 5 cm
AB = 7 cm, CD = 15 cm, AD = BC = 5 cm
Check: the height AM = 7 − 4 = 3 cm, and ½ × (7 + 15) × 3 = 33 cm². Also each slanting side extends (15 − 7) ÷ 2 = 4 cm beyond AB, and √(4² + 3²) = 5 cm, which matches AD = BC = 5 cm.
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Problem Set 2
1Choose the correct answers for the following questions.
(1) Which one is the quadratic equation? (A) 5x − 3 = x2 (B) x(x + 5) = 2 (C) n − 1 = 2n (D) 1x2(x + 2) = x
(B) x(x + 5) = 2 gives x2 + 5x − 2 = 0, whose maximum index is 2.
(A) becomes x3 + 3x − 5 = 0 and (D) becomes x3 − x − 2 = 0 (index 3); (C) is linear.
(B) x(x + 5) = 2
(2) Out of the following equations which one is not a quadratic equation? (A) x2 + 4x = 11 + x2 (B) x2 = 4x (C) 5x2 = 90 (D) 2x − x2 = x2 + 5
In (A), x2 cancels on both sides, leaving 4x = 11, which is linear.
(A) x² + 4x = 11 + x²
(3) The roots of x2 + kx + k = 0 are real and equal, find k. (A) 0 (B) 4 (C) 0 or 4 (D) 2
Δ = k2 − 4 × 1 × k = 0
∴ k(k − 4) = 0 ∴ k = 0 or k = 4
(C) 0 or 4
(4) For √2x2 − 5x + √2 = 0 find the value of the discriminant. (A) −5 (B) 17 (C) √2 (D) 2√2 − 5
Δ = (−5)2 − 4 × √2 × √2 = 25 − 8 = 17
(B) 17
(5) Which of the following quadratic equations has roots 3, 5? (A) x2 − 15x + 8 = 0 (B) x2 − 8x + 15 = 0 (C) x2 + 3x + 5 = 0 (D) x2 + 8x − 15 = 0
Sum = 8, product = 15, so the equation is x2 − 8x + 15 = 0.
(B) x² − 8x + 15 = 0
(6) Out of the following equations, find the equation having the sum of its roots −5. (A) 3x2 − 15x + 3 = 0 (B) x2 − 5x + 3 = 0 (C) x2 + 3x − 5 = 0 (D) 3x2 + 15x + 3 = 0
For (D), sum of roots = −ba = −153 = −5. (A) and (B) give 5; (C) gives −3.
(D) 3x² + 15x + 3 = 0
(7) √5m2 − √5m + √5 = 0 — which of the following statements is true for this equation? (A) Real and unequal roots (B) Real and equal roots (C) Roots are not real (D) Three roots
Δ = (−√5)2 − 4 × √5 × √5 = 5 − 20 = −15 < 0
(C) Roots are not real
(8) One of the roots of equation x2 + mx − 5 = 0 is 2; find m. (A) −2 (B) −12 (C) 12 (D) 2
22 + 2m − 5 = 0 ∴ 2m = 1 ∴ m = 12
(C) 1/2
2Which of the following equations is quadratic?
(1) x2 + 2x + 11 = 0
The maximum index of the variable is 2.
Quadratic
(2) x2 − 2x + 5 = x2
x2 cancels on both sides: −2x + 5 = 0, which is linear.
Not quadratic
(3) (x + 2)2 = 2x2
x2 + 4x + 4 = 2x2
∴ x2 − 4x − 4 = 0 (maximum index 2)
Quadratic
3Find the value of discriminant for each of the following equations.
(1) 2y2 − y + 2 = 0
a = 2, b = −1, c = 2
Δ = (−1)2 − 4 × 2 × 2 = 1 − 16 = −15
Δ = −15
(2) 5m2 − m = 0
a = 5, b = −1, c = 0
Δ = (−1)2 − 4 × 5 × 0 = 1
Δ = 1
(3) √5x2 − x − √5 = 0
a = √5, b = −1, c = −√5
Δ = (−1)2 − 4 × √5 × (−√5) = 1 + 20 = 21
Δ = 21
4One of the roots of quadratic equation 2x2 + kx − 2 = 0 is −2, find k.
2(−2)2 + k(−2) − 2 = 0
∴ 8 − 2k − 2 = 0
∴ 6 = 2k ∴ k = 3
k = 3
5Two roots of quadratic equations are given; frame the equation.
(1) 10 and −10
Let α = 10 and β = −10.
α + β = 0
α × β = −100
The quadratic equation is x2 − (α + β)x + αβ = 0
∴ x2 − 0x − 100 = 0
x² − 100 = 0
(2) 1 − 3√5 and 1 + 3√5
Let α = 1 − 3√5 and β = 1 + 3√5.
α + β = 2
α × β = 12 − (3√5)2 = 1 − 45 = −44
The quadratic equation is x2 − (α + β)x + αβ = 0
x² − 2x − 44 = 0
(3) 0 and 7
Let α = 0 and β = 7.
α + β = 7
α × β = 0
The quadratic equation is x2 − (α + β)x + αβ = 0
x² − 7x = 0
6Determine the nature of roots for each of the quadratic equation.
(1) 3x2 − 5x + 7 = 0
Δ = (−5)2 − 4 × 3 × 7 = 25 − 84 = −59
Δ < 0, so the roots are not real.
(2) √3x2 + √2x − 2√3 = 0
Δ = (√2)2 − 4 × √3 × (−2√3) = 2 + 24 = 26
Δ > 0, so the roots are real and unequal.
(3) m2 − 2m + 1 = 0
Δ = (−2)2 − 4 × 1 × 1 = 4 − 4 = 0
Δ = 0, so the roots are real and equal.
7Solve the following quadratic equations.
(1) 1x + 5 = 1x2
x2 = x + 5 (x ≠ 0, x ≠ −5)
∴ x2 − x − 5 = 0
a = 1, b = −1, c = −5; Δ = 1 + 20 = 21
x = 1 ± √212
Roots: (1 + √21)/2 and (1 − √21)/2
(2) x2 − 3x10 − 110 = 0
Multiplying by 10: 10x2 − 3x − 1 = 0
∴ 10x2 − 5x + 2x − 1 = 0
∴ 5x(2x − 1) + 1(2x − 1) = 0
∴ (2x − 1)(5x + 1) = 0
∴ x = 12 or x = −15
Roots: 1/2 and −1/5
(3) (2x + 3)2 = 25
∴ 2x + 3 = 5 or 2x + 3 = −5
∴ 2x = 2 or 2x = −8
∴ x = 1 or x = −4
Roots: 1 and −4
(4) m2 + 5m + 5 = 0
a = 1, b = 5, c = 5; Δ = 25 − 20 = 5
m = −5 ± √52
Roots: (−5 + √5)/2 and (−5 − √5)/2
(5) 5m2 + 2m + 1 = 0
a = 5, b = 2, c = 1; Δ = 4 − 20 = −16
m = −2 ± √−1610
√−16 is not a real number.
The equation has no real roots.
(6) x2 − 4x − 3 = 0
a = 1, b = −4, c = −3; Δ = 16 + 12 = 28
x = 4 ± √282 = 4 ± 2√72 = 2 ± √7
Roots: 2 + √7 and 2 − √7
8Find m if (m − 12)x2 + 2(m − 12)x + 2 = 0 has real and equal roots.
a = m − 12, b = 2(m − 12), c = 2
Roots are real and equal, so Δ = 0.
[2(m − 12)]2 − 4(m − 12)(2) = 0
∴ 4(m − 12)2 − 8(m − 12) = 0
∴ 4(m − 12)[(m − 12) − 2] = 0
∴ 4(m − 12)(m − 14) = 0
∴ m = 12 or m = 14
If m = 12, the coefficient of x2 is 0 and the equation becomes 2 = 0, which is not quadratic. ∴ m ≠ 12.
m = 14
9The sum of two roots of a quadratic equation is 5 and sum of their cubes is 35, find the equation.
α + β = 5 and α3 + β3 = 35
α3 + β3 = (α + β)3 − 3αβ(α + β)
∴ 35 = 125 − 15αβ
∴ 15αβ = 90 ∴ αβ = 6
Equation: x2 − (α + β)x + αβ = 0
x² − 5x + 6 = 0
10Find quadratic equation such that its roots are square of sum of the roots and square of difference of the roots of equation 2x2 + 2(p + q)x + p2 + q2 = 0.
Let α, β be the roots of 2x2 + 2(p + q)x + p2 + q2 = 0.
Product of new roots = (p + q)2 × [−(p − q)2] = −(p2 − q2)2
Required equation: x2 − (sum)x + product = 0
x² − 4pqx − (p² − q²)² = 0
11Mukund possesses ₹ 50 more than what Sagar possesses. The product of the amount they have is 15,000. Find the amount each one has.
Let Sagar have ₹ x. Then Mukund has ₹ (x + 50).
x(x + 50) = 15000
∴ x2 + 50x − 15000 = 0
∴ x2 + 150x − 100x − 15000 = 0
∴ (x + 150)(x − 100) = 0
∴ x = −150 or x = 100
Amount cannot be negative. ∴ x = 100 and x + 50 = 150
Sagar has ₹ 100 and Mukund has ₹ 150.
12The difference between squares of two numbers is 120. The square of smaller number is twice the greater number. Find the numbers.
Let the greater number be x and the smaller number be y.
x2 − y2 = 120 … (I)
y2 = 2x … (II)
Substituting (II) in (I): x2 − 2x − 120 = 0
∴ x2 − 12x + 10x − 120 = 0
∴ (x − 12)(x + 10) = 0
∴ x = 12 or x = −10
If x = −10, then y2 = −20, which is impossible. ∴ x = 12
y2 = 2 × 12 = 24 ∴ y = √24 = 2√6 or y = −2√6
The numbers are 12 and 2√6 (or 12 and −2√6).
Both values of the smaller number satisfy the conditions: (2√6)² = (−2√6)² = 24 = 2 × 12, and 144 − 24 = 120. If only positive numbers are intended, the answer is 12 and 2√6.
13Ranjana wants to distribute 540 oranges among some students. If 30 students were more each would get 3 oranges less. Find the number of students.
Let the number of students be x. Each gets 540x oranges.
With 30 more students, each gets 540x + 30 oranges.
540x − 540x + 30 = 3
∴ 540 × 30x(x + 30) = 3
∴ 16200 = 3x(x + 30)
∴ x2 + 30x − 5400 = 0
∴ x2 + 90x − 60x − 5400 = 0
∴ (x + 90)(x − 60) = 0
∴ x = −90 or x = 60
Number of students cannot be negative. ∴ x = 60
Check: 540 ÷ 60 = 9 and 540 ÷ 90 = 6; 9 − 6 = 3
There are 60 students.
14Mr. Dinesh owns an agricultural farm at village Talvel. The length of the farm is 10 metre more than twice the breadth. He dug a square shaped pond inside the farm. The side of pond is 1/3 of the breadth of the farm. The area of the farm is 20 times the area of the pond. Find the length and breadth of the farm and of the pond.
Let the breadth of the farm be x m. Then length = (2x + 10) m and side of the pond = x3 m.
Area of farm = 20 × area of pond
x(2x + 10) = 20 × (x3)2 = 20x29
∴ 9x(2x + 10) = 20x2
∴ 18x2 + 90x = 20x2
∴ 2x2 − 90x = 0
∴ 2x(x − 45) = 0
∴ x = 0 or x = 45
Breadth cannot be 0. ∴ x = 45
Length = 2 × 45 + 10 = 100 m; side of pond = 45 ÷ 3 = 15 m
Check: 100 × 45 = 4500 m² and 20 × 15² = 4500 m²
Farm: length 100 m, breadth 45 m; Pond: side 15 m (length = breadth = 15 m)
15A tank fills completely in 2 hours if both the taps are open. If only one of the taps is open at the given time, the smaller tap takes 3 hours more than the larger one to fill the tank. How much time does each tap take to fill the tank completely?
Let the larger tap take x hours. Then the smaller tap takes (x + 3) hours.
