# S Chand ICSE Maths Solutions Class 10 Chapter 5 Quadratic Equations Revision Exercise

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**Quadratic Equations***.*

**ICSE Class 10 Maths syllabus**## S Chand ICSE Maths Solutions for Class 10 Chapter 5 Revision Exercise

**S Chand ICSE Solutions Class 10 Maths Quadratic Equation Revision Exercise: Ques No 1**

Solve for x and give your answer correct to 2 decimal places: x^{2} – 10x + 6 = 0.

**S Chand ICSE Maths Solutions:**

Comparing with ax^{2} + bx + c = 0, we have a = 1, b = -10, c = 6.

On using quadratic formula, we get

Thus, x = 9.36 or 0.64

**ICSE Maths Solutions Class 10 Quadratic Equations: Ques No 2**

Solve using the quadratic formula : x^{2} – 4x + 1 = 0.

**Solutions:**

Comparing with ax^{2} + bx + c = 0, we have a = 1, b = -4, c = 1.

On using quadratic formula, we get

Thus, x = 3.732 or 0.268

**Ques No 3**

Solve the equation 3x^{2} – x – 7 = 0 and give your answer correct to two decimal places.

**Solutions:**

Comparing with ax^{2} + bx + c = 0, we have a = 3, b = -1, c = -7.

On using quadratic formula, we get

Thus, x = 1.70 or -1.37

**ICSE Maths Solutions Class 10 Quadratic Equations: Ques No 4**

Solve the following equation and give your answer up to two decimal places : x^{2} – 5x – 10 = 0.

**Solutions:**

Comparing with ax^{2} + bx + c = 0, we have a = 1, b = -5, c = -10.

On using quadratic formula, we get

Thus, x = 6.53 or -1.53

**Ques No 5**

Solve the equation 2x – 1/x = 7. Write your answer correct to two decimal places.

**Solutions:**

Given that 2x – 1/x = 7 ⇒ 2x^{2} – 7x – 1 = 0.

Comparing with ax^{2} + bx + c = 0, we have a = 2, b = -7, c = -1.

On using quadratic formula, we get

Thus, x = 3.64 or -0.14

**Ques No 6**

The bill for a number of people for overnight stay in Rs. 4800. If there were 4 more, bill each person had to pay would have reduced by Rs. 200. Find the number of people staying overnight.

**Solutions:**

Let the number of people staying overnight is x. Then, bill for each person is Rs. 4800/x.

According to question, we have

4800/(x + 4) = 4800/x – 200

⇒ x^{2} + 4x – 96 = 0

Comparing with ax^{2} + bx + c = 0, we have a = 1, b = 4, c = -96.

On using quadratic formula, we get

Thus, number of people is x = 8.

**Ques No 7**

An aeroplane travelled a distance of 400 km at an average speed of x km/hr. On the return journey, the speed was increased by 40 km/hr. Write down an expression for the time taken by

(i) the onward journey

(ii) the return journey.

If the return journey took 30 minutes less than onward journey, write down an equation in x and find its value.

**Solutions:**

(i) Time taken in onward journey = 400/x hr

(i) Time taken in return journey = 400/(x + 40) hr

Given that 400/(x + 40) – 400/x = 1/2

⇒ x^{2} + 40x – 32000 = 0

Comparing with ax^{2} + bx + c = 0, we have a = 1, b = 40, c = -32000.

On using quadratic formula, we get

Thus, we get x = 160 km/hr

**Ques No 8**

In an auditorium, seats were arranged in rows and columns. The number of rows were equal to the number of seats in each row. When the number of rows were doubled and the number of seats of seats in each row were reduced by 10, the total number of seats increased by 300. Find

(i) the number of rows in the original arrangement

(ii) the number of seats in the auditorium after rearrangement

**Solutions:**

Let the number of rows is x.

Then, the total number of seats in auditorium is x^{2}.

According to question, 2x(x – 10) = x^{2} + 300

⇒ x^{2} – 20x – 300 = 0

Comparing with ax^{2} + bx + c = 0, we have a = 1, b = -20, c = -300.

Thus, the number of seats in rows in the original arrangement is x = 30

Thus, number of seats in the auditorium after rearrangement is x^{2} + 300 = 1200

**Ques No 9**

P and Q are centres of circles of radius 9 cm and 2 cm respectively. PQ = 17 cm. R is the centre of a circle of radius x cm, which touches the above circles externally. Given that angle PRQ = 90 degree, write an equation in x and solve it.

**Solutions:**

According to the question, we have

(x + 9)^{2} + (x + 2)^{2} = 17^{2}

⇒ x^{2} + 11x – 102 = 0

Comparing with ax^{2} + bx + c = 0, we have a = 1, b = 11, c = -102.

Thus, the radius of the circle having center R is x = 6.

**Ques No 10**

By increasing the speed of the car by 10 km/hr, the time of journey for a distance of 72 km is reduced by 36 min. Find the original speed of the car.

**Solutions:**

Let the original speed of the car is x km/hr. Then, time taken for 72 km is 72/x hr.

According to question,

72/(x + 10) = 72/x – 36/60

⇒ x^{2} + 10x – 1200 = 0

Comparing with ax^{2} + bx + c = 0, we have a = 1, b = 1, c = -1200.

Thus, the the original speed of the car is x = 30 km/hr.

**Ques No 11**

A shopkeeper buys a certain number of books for Rs. 720. If the cost per book was Rs. 5 less the number of books that could bought for Rs. 720 would be 2 more. Taking the original cost of each book to be Rs x, write an equation in x and solve it.

**Solutions:**

Given that the original cost of each book to be Rs x. Then, the number of books is 720/x.

According to question,

720/(x – 5) = 720/x + 2

⇒ x^{2} – 5x – 1800 = 0

Comparing with ax^{2} + bx + c = 0, we have a = 1, b = -5, c = -1800.

Thus, x = 45.

**Ques No 12**

Solve the quadratic equation for x and give your answer correct to 2 decimal places : x^{2} – 3x – 9 = 0

**Solutions:**

Comparing with ax^{2} + bx + c = 0, we have a = 1, b = -3, c = -9.

On using quadratic formula, we get

Thus, x = 4.85 or -1.85

**Ques No 13**

Five years ago, a women age was the square of her son’s age. Ten years hence her age will be twice that of her son’s age. Find : (i) the age of her son five years ago (ii) the present age of the women.

**Solutions:**

Let, five years ago, son age was x years and his mother’s age was x^{2} years.

Then, the current age of son is (x + 5) and his mother’s age is (x^{2} + 5) years.

Now, after 10 years, we have

(x^{2} + 5 + 10) = 2(x + 5 + 10)

⇒ x^{2} – 2x – 15 = 0

Comparing with ax^{2} + bx + c = 0, we have a = 1, b = -2, c = -15.

Thus,

(i) the age of her son five years ago was x = 5 years

(ii) the present age of the women is (x^{2} + 5) = 30 years.

**Ques No 14**

Solve the following quadratic equation for x and give your answer correct to decimal places 5x(x + 2) = 3.

**Solutions:**

5x(x + 2) = 3

⇒ 5x^{2} + 10x – 3 = 0

Comparing with ax^{2} + bx + c = 0, we have a = 5, b = 10, c = -3.

Thus, x = 0.26 or -2.26

**Ques No 15**

Some students planned a picnic. The budget for the food was Rs 480. As eight of them failed to join the party, the cost of the food for each member increased by Rs 10. Find how many students went for the picnic.

**Solutions:**

Let the number of students went for picnic is x. Then, cost of food for each member is Rs 480/x.

According to question, we have

480/(x – 8) = 480/x + 10

⇒ x^{2} – 8x – 384 = 0

Comparing with ax^{2} + bx + c = 0, we have a = 1, b = -8, c = -384.

Thus, required number of students is x = 24.

**Ques No 16**

Solve the following quadratic equation and give the answer correct to two significant figures 4x^{2} – 7x + 2 = 0.

**Solutions:**

4x^{2} – 7x + 2 = 0

Comparing with ax^{2} + bx + c = 0, we have a = 4, b = -7, c = 2.

Thus, x = 1.40 or 0.36

**Ques No 17**

The speed of an express train is x km/hr and the speed of an ordinary train is 12 km/hr less than that of the express train. If the ordinary train takes one hour longer than the express train to cover a distance of 240 km, find the speed of the express train.

**Solutions:**

Given that the speed of an express train is x km/hr, then the speed of the ordinary train is (x – 12) km/hr.

According to question, we have

⇒ 240/(x – 12) = 240/x + 1

⇒ x^{2} – 12x – 2880 = 0

Comparing with ax^{2} + bx + c = 0, we have a = 1, b = -12, c = -2880.

On using quadratic formula, we get

Thus, speed of express train is 60 km/hr.

**Ques No 18**

Without solving the following quadratic equation, find the value of p for which the roots are equal : px^{2} – 4x + 3 = 0.

**Solutions:**

px^{2} – 4x + 3 = 0

Comparing with ax^{2} + bx + c = 0, we have a = p, b = -4, c = 3.

The roots of the equation are equal.

⇒ D = 0

⇒ b^{2} – 4ac = 0

⇒ (-4)^{2} – 4(p)(3) = 0

⇒ 16 – 12p = 0

⇒ p = 16/12 = 4/3

**Ques No 19**

Solve the following equation x – 18/x = 6. Give your answer correct to 2 significant figures.

**Solutions:**

x – 18/x = 6

⇒ x^{2} – 6x – 18 = 0

Comparing with ax^{2} + bx + c = 0, we have a = 1, b = -6, c = -18.

On using quadratic formula, we get

Thus, x = 8.20 or -2.20

**Ques No 20**

Rs 480 is divided equally among x children. If the number of children were 20 more, then each would have got Rs. 12 less. Find x.

**Solutions:**

Each child got Rs (480/x).

According to question, we have

480/(x + 20) = 480/x – 12

⇒ x^{2} + 20x – 800 = 0

⇒ x^{2} + 40x – 20x – 800 = 0

⇒ x(x + 40) – 20(x + 40) = 0

⇒ (x – 20)(x + 40) = 0

⇒ x = 20 or -40.

Thus, x = 20

**Ques No 21**

Without solving the following quadratic equation, find the value of m for which the given equation has real and equal roots : x^{2} + 2(m – 1)x + (m + 5) = 0.

**Solutions:**

x^{2} + 2(m – 1)x + (m + 5) = 0

Comparing with ax^{2} + bx + c = 0, we have a = 1, b = 2(m – 1), c = (m + 5).

The roots of the equation are equal.

⇒ D = 0

⇒ b^{2} – 4ac = 0

⇒ [2(m – 1)]^{2} – 4(1)(m + 5) = 0

⇒ 4(m – 1)^{2} – 4(m + 5) = 0

⇒ m^{2} – 2m + 1 – m – 5 = 0

⇒ m^{2} – 3m – 4 = 0

⇒ m^{2} – 4m + m – 4 = 0

⇒ m(m – 4) + 1(m – 4) = 0

⇒ (m + 1)(m – 4) = 0

⇒ m = -1 or 4

**Ques No 22**

A car covers a distance of 400 km at a certain speed. Had the speed been 12 km/hr more, the time taken for the journey would have been 1 hour 40 minutes less. Find the original speed of the car.

**Solutions:**

Let original speed of the car is x km/hr.

According to question, we have

400/(x + 12) = 400/x – 100/60

⇒ x^{2} + 12x – 2880 = 0

⇒ x^{2} + 60x – 48x – 2880 = 0

⇒ x(x + 60) – 48(x + 60) = 0

⇒ (x – 48)(x + 60) = 0

⇒ x = 48 or -60

Thus, the original speed of the car is 48 km/hr.

**Ques No 23**

(i) Solve the following equation and calculate the answer correct to two decimal places : x^{2} – 5x – 10 = 0.

(ii) without solving the following quadratic equation has real and equal roots : x^{2} + (p – 3)x + p = 0

**Solutions:**

(i) Given that x^{2} – 5x – 10 = 0

Comparing with ax^{2} + bx + c = 0, we have a = 1, b = -5, c = -10.

On using quadratic formula, we get

Thus, we get x = 6.53 or -1.53

(ii) x^{2} + (p – 3)x + p = 0

Comparing with ax^{2} + bx + c = 0, we have a = 1, b = (p – 3), c = p.

The roots of the equation are real and equal.

⇒ D = 0

⇒ b^{2} – 4ac = 0

⇒ (p – 3)^{2} – 4(1)(p) = 0

⇒ p^{2} – 6p + 9 – 4p = 0

⇒ p^{2} – 10p + 9 = 0

⇒ p^{2} – 9p – p + 9 = 0

⇒ p(p – 9) – 1(p – 9) = 0

⇒ (p – 1)(p – 9) = 0

⇒ p = 1 or 9

**Ques No 24**

A shopkeeper purchases a certain number of books for Rs. 960. If the cost per book was Rs. 8 less, the number of books that could be purchased for Rs 960 would be 4 more. Find the number of books.

**Solutions:**

Let the number of books is x.

According to question, we have

960/(x + 4) = 960/x – 8

⇒ x^{2} + 4x – 480 = 0

⇒ x^{2} + 24x – 20x – 480 = 0

⇒ x(x + 24) – 20(x + 24) = 0

⇒ (x – 20)(x + 24) = 0

⇒ x = 20 or -24

Thus, the number of books is x = 20.

**Ques No 25**

Solve for x using the quadratic formula. Write your answer correct to two significant figures (x – 1)^{2} – 3x + 4 = 0.

**Solutions:**

Given that (x – 1)^{2} – 3x + 4 = 0

⇒ x^{2} – 2x + 1 – 3x + 4 = 0

⇒ x^{2} – 5x + 5 = 0

Comparing with ax^{2} + bx + c = 0, we have a = 1, b = -5, c = 5.

On using quadratic formula, we get

Thus, we get x = 3.62 or 1.38

**Ques No 26**

A two digit positive number is such that the product of its digit is 6. If 9 is added to the number, the digits interchange their places. Find the number.

**Solutions:**

Let the unit digit of the number is x, then the tens digit is 6/x.

Thus, the number is 10(6/x) + x.

According to question, we have

10(6/x) + x + 9 = 10x + 6/x

⇒ x^{2} – x – 6 = 0

⇒ x^{2} – 3x + 2x – 6 = 0

⇒ x(x – 3) + 2(x – 3) = 0

⇒ (x + 2)(x – 3) = 0

⇒ x = 3 or -2

Thus, the unit digit is x = 3 and the tens digit is 6/x = 2 and hence the number is 23.

**ICSE Maths Solutions Class 10 Quadratic Equations: Ques No 27**

Find the value of k for which x = 3 is the solution of the quadratic equation, (k + 2)x^{2} – kx + 6 = 0. Thus, find the roots of the equation.

**Solutions:**

Since x = 3 is the solution of (k + 2)x^{2} – kx + 6 = 0.

Put x = 3 in (k + 2)x^{2} – kx + 6 = 0, we get

9(k + 2) – 3k + 6 = 0

⇒ k = -4

Now, put k = -4 in (k + 2)x^{2} – kx + 6 = 0.

(-4 + 2)x^{2} – (-4)x + 6 = 0

⇒ -2x^{2} + 4x + 6 = 0

⇒ x^{2} – 2x – 3 = 0

⇒ x^{2} – 3x + x – 3 = 0

⇒ x(x – 3) + 1(x – 3) = 0

⇒ (x + 1)(x – 3) = 0

⇒ x = -1 or 3

**Ques No 28**

Sum of two natural numbers is 8 and the difference of their reciprocal is 2/15. Find the numbers.

**Solutions:**

Let first number is x and second number is (8 – x).

According to question, we have

1/x – 1/(8 – x) = 2/15

⇒(8 – x – x) / x(8 – x) = 2/15

⇒ 15(8 – 2x) = 2x(8 – x)

⇒ 120 – 30x = 16x – 2x^{2}

⇒ 2x^{2} – 46x + 120 = 0

⇒ x^{2} – 23x + 60 = 0

⇒ x^{2} – 20x – 3x + 60 = 0

⇒ x(x – 20) – 3(x – 20) = 0

⇒ (x – 3)(x – 20) = 0

⇒ x = 3 or 20 and (8 – x) = 5 or -17

Since both numbers are natural, hence the numbers are 3 and 5.

**Ques No 29**

Solve the quadratic equation x^{2} – 3(x + 3) = 0. Give your answer correct to two significant figures.

**Solutions:**

x^{2} – 3(x + 3) = 0

⇒ x^{2} – 3x – 9 = 0

Comparing with ax^{2} + bx + c = 0, we have a = 1, b = -3, c = -9.

On using quadratic formula, we get

Thus, x = 4.85 or -1.85

**S Chand ICSE Solutions Class 10 Maths Quadratic Equation Revision Exercise: Ques No 30**

A bus covers a distance of 240 km at uniform speed. Due to heavy rain its speed gets reduced by 10 km/h and as such it takes two hours longer to cover the total distance. Assuming the uniform speed to be x km/hr from an equation and solve it to evaluate x.

**S Chand ICSE Maths Solutions:**

According to question, we have

240/(x – 10) = 240/x + 2

⇒ x^{2} – 10x – 1200 = 0

⇒ x^{2} – 40x + 30x – 1200 = 0

⇒ x(x – 40) + 30(x – 40) = 0

⇒ (x + 30)(x – 40) = 0

⇒ x = -30 or 40

Thus, the uniform speed of the bus is x = 40 km/hr.

S Chand ICSE Class 10 Maths Chapterwise Notes & Solutions |
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Chapter 1 : GST Notes & Solutions | Notes | Exercise 1 |

Chapter 2 : Banking Notes & Solutions| Notes | Exercise 2 | Revision Exercise |

Chapter 3 : Shares & Dividends Notes & Solutions| Notes | Exercise 3A | Exercise 3B | Revision Exercise |

Chapter 4 : Linear Inequations in One Variable Notes & Solutions| Notes | Exercise 4 | Revision Exercise |

Chapter 5 : Quadratic Equations Notes & Solutions| Notes | Exercise 5A | Exercise 5B | Exercise 5C| Exercise 5D| Exercise 5E| Revision Exercise |

Chapter 6 : Ratios & Proportions Notes & Solutions| Notes | Exercise 6A | Exercise 6B | Exercise 6C| Revision Exercise |

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