CBSE Class 10 Mathematics Quadratic Equation Worksheet Set 01

Quadratic Equation MCQ Questions with Answers Class 10 

Question. Solve: 6x² + 40 = 31x.
Answer : 8/3, 5/2

Question. Show that the equation 2x² + 5√3 x + 6 = 0 has real roots and solve it.
Answer : 

Question. Solve the equation: 1/(x + 4) - 1(x - 7) = 11/30, x ≠ -4.7
Answer : 2, 1

Question. The sum of the squares of two consecutive multiples of 7 is 637. Find the multiples.
Answer : 14 and 21

Question. Solve the equation: 4x² – 4ax + (a² – b²) = 0.
Answer : a + b / 2,  a - b /2

Question. Using quadratic formula, solve for x: abx² + (b² – ac)x – bc = 0.
Answer : c/a, b/a

Question. The sum of the squares of two consecutive odd numbers is 394. Find the numbers.
Answer : 13 and 15.

Question. Solve: 4√3x² + 5x – 2√3 = 0.
Answer : - (2/√3), √3/4

Question. 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.
Answer : (18 and 12) or (18 and –12).

Question. The total cost of a certain length of a piece of wire is ₹200. If the piece was 5 metres longer and each metre of wire costs ₹2 less, the cost of the piece would have remained unchanged. How long is the piece and what is its original rate per metre?
Answer : ₹ 10 per m.

Question. Solve the equation: x² – 2ax – (4b² – a²) = 0
Answer : x = (a – 2b) or x = (a + 2b)

Question. A girl is twice as old as her sister. Four years hence, the product of their ages (in years) will be 160. Find their present ages.
Answer : Sister’s present age = 6 years and girl’s present age = 12 years.

Applications of quadratic equations:

i. Read the statement of the problem carefully and determine what quantity (or quantities) must be found.

ii. Represent the unknown quantity (or quantities) by a variable (or variables).

iii. Identify the relationships existing in the problem and determine which expressions are equal and write an equation (or equations).

Note: Check the answers obtained by determining whether they fulfil the conditions of the original problem. It may happen that out of the two roots of the quadratic equations, only one satisfies the conditions of the problem, reject the other root.

1. A train travels at a certain average speed for a distance of 63 km and then travels a distance of 72km at an average speed of 6km/hr more than its original speed. If it takes 3 hours to complete the total journey, what is its original speed?

2. Find the two consecutive odd integers, the sum of whose squares is 202.

3. A natural number, when increased by 12, becomes equal to 160 times of its reciprocal. Find the number.

4. A motor boat whose speed is 18km/hr in still water takes 1 hour more to go 24 km upstream than to return to the same spot. Find the speed of the stream.

5. Find the roots of the quadratic equation:
1/3x² -√11x + 1 = 0

6. Solve for x:
x+1/x-1 + x-2/x+1 = 3

7. A shopkeeper buys a number of packets of biscuits for ` 80.x If he had bought 4 more packets for the same amount, each packet would have cost ` 1 less. How many packets did he buy?If

8. Solve the following equations using quadratic formula:

(i) 4x² – 4a²x + (a⁴ – b⁴) = 0

(ii) 12abx² –(9a² - 8b²)x – 6ab = 0II

9. Three consecutive natural numbers are such that the square of the middle number exceeds the difference of the square of the other two by 60, find the numbers.

10. If the equation (1+m²)x² + 2mcx + (c² - a²) = 0 has equal roots, prove that c² = a²(1 + m²).

11. A trader bought a number of articles for ` 900, ffive were damaged and he sold each of the rest at `2 more than what he paid for it, thus getting a profit of ` 80 on the whole transaction. Find the number of articles he bought.

Applications of quadratic equations:

i. Read the statement of the problem carefully and determine what quantity (or quantities) must be found.
ii. Represent the unknown quantity (or quantities) by a variable (or variables).
iii. Identify the relationships existing in the problem and determine which expressions are equal and write an equation (or equations).

Note: Check the answers obtained by determining whether they fulfil the conditions of the original problem. It may happen that out of the two roots of the quadratic equations, only one satisfies the conditions of the problem, reject the other root.

1. A train travels at a certain average speed for a distance of 63 km and then travels a distance of 72km at an average speed of 6km/hr more than its original speed. If it takes 3 hours to complete the total journey, what is its original speed?

2. Find the two consecutive odd integers, the sum of whose squares is 202.

3. A natural number, when increased by 12, becomes equal to 160 times of its reciprocal. Find the number.

4. A motor boat whose speed is 18km/hr in still water takes 1 hour more to go 24 km upstream than to return to the same spot. Find the speed of the stream.

5. Find the roots of the quadratic equation:
1/3x² -√11x + 1 =0

6. Solve for x:
x+1/x-1 + x-2/x+1 = 3

7. A shopkeeper buys a number of packets of biscuits for ` 80.x If he had bought 4 more packets for the same amount, each packet would have cost ` 1 less. How many packets did he buy?If

8. Solve the following equations using quadratic formula:
(i) 4x² – 4a²x + (a⁴ – b⁴) = 0
(ii) 12abx² – (9a² - 8b²)x – 6ab = 0II

9. Three consecutive natural numbers are such that the square of the middle number exceeds the difference of the square of the other two by 60, find the numbers.

10. If the equation (1+m²)x² + 2mcx + (c² - a²) = 0 has equal roots, prove that c² = a²(1 + m²).

11. A trader bought a number of articles for ` 900, ffive were damaged and he sold each of the rest at `2 more than what he paid for it, thus getting a profit of ` 80 on the whole transaction. Find the number of articles he bought.

Introduction

Euclid was a mathematician who lived around 300 BC in Greece. He is famous for his monumental work: Elements which consists of 13 books dealing in numbers, elementary algebra and geometry. He organized all the mathematical knowledge that had been compiled since the days of Thales (Greek Mathematician) who lived 300 years before Euclid. Euclid’s division algorithm is a technique to find the HCF/GCD/GCF of two numbers. This algorithm is one of the oldest algorithms (step by step procedure to solve a particular problem) in the history of Maths.

What is a dividend? Let us understand it with the help of a simple example. Can you divide 14 by 6?

    2  
6)14
  -12 
     2 

After division, we get 2 as the quotient and 2 as the remainde

Thus, we can also write 14 as 6 × 2 + 2. A dividend can thus be written as: 

Dividend = Divisor × Quotient + Remainder

Can you think of any other number which, when multiplied with 6, gives 14 as the dividend and 2 as the remainder?

Let us try it out with some other sets of dividends and divisors.
(1) Divide 100 by 20: 100 = 20 × 5 + 0
(2) Divide 117 by 15: 117 = 15 × 7 + 12
(3) Divide 67 by 17: 67 = 17 × 3 + 16

Thus, if we have a dividend and a divisor, then there will be a unique pair of a quotient and a remainder that will fit into the above equation.

This brings us to Euclid’s division lemma.

If a and b are positive integers, then there exist two unique integers, q and r, such that a = bq + r

This lemma is very useful for finding the H.C.F. of large numbers where breaking them into factors is difficult. This method is known as Euclid’s Division Algorithm.

SOLVED EXAMPLES

Example 1: Find the H.C.F. of 4032 and 262 using Euclid’s division algorithm.

Solution: Step 1:
First, apply Euclid’s division lemma on 4032 and 262.
4032 = 262 × 15 + 102

Step 2:
As the remainder is non-zero, we apply Euclid’s division lemma on 262 and 102.
262 = 102 × 2 + 58

Step 3:
Apply Euclid’s division lemma on 102 and 58.
102 = 58 × 1 + 44

Step 4:
Apply Euclid’s division lemma on 58 and 44.
58 = 44 × 1 + 14

Step 5:
Apply Euclid’s division lemma on 44 and 14.
44 = 14 × 3 + 2

Step 6:
Apply Euclid’s division lemma on 14 and 2.
14 = 2 × 7 + 0
In the problem given above, to obtain 0 as the remainder, the divisor has to be taken as 2.
Hence, 2 is the H.C.F. of 4032 and 262.

Example 2: Find the H.C.F. of 336 and 90 using Euclid’s division algorithm.

Solution:

As 336 > 90, we apply the division lemma to 336 and 90.

336 = 90 × 3 + 66

Applying Euclid’s division lemma to 90 and 66:

90 = 66 × 1 + 24

Applying Euclid’s division lemma to 66 and 24:

66 = 24 × 2 + 18

Applying Euclid’s division lemma to 24 and 18:

24 = 18 × 1 + 6

Applying Euclid’s division lemma to 18 and 6:

18 = 6 × 3 + 0

As the remainder is zero, we need not apply Euclid’s division lemma anymore. The divisor (6) is the required H.C.F.

ASSIGNMENT

Q1. Use Euclid’s division Lemma to find the HCF of:
1. 135 and 225
2. 196 and 38220
3. 867 and 255

Q2. Find more information on the contribution and life of Euclid from the internet.