Exercise 4.2 (NCERT) – Quadratic Equations, Class 10
| Chapter 4 | Exercise 4.2 | Class 10 |
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Below are the Quick links for all questions of Exercise 4.2, Quadratic Equations, Class 10. Click a link to view the solution of corresponding question.
Below are the solutions of every question of exercise 4.2 of chapter 4, Quadratic Equations, NCERT of class 10. YouTube video tutorial of every question is also embedded along with the written solution.
Question 1
Find the roots of the following quadratic equations by factorisation:
(i) x2 – 3x – 10 = 0
Splitting the middle term
Answer: Roots: -2, 5
(ii) 2x2 + x – 6 = 0
Splitting the middle term
Answer: Roots: 3/2, -2
(iii) √2x2 + 7x + 5√2 = 0
Splitting the middle term
Answer: Roots: -√2, -5/√2
(iv) 2x2 – x + 1/8 = 0
Answer: Roots: 1/4, 1/4
(v) 100 x2 – 20x + 1 = 0
Splitting the middle term
Answer: Roots: 1/10, 1/10
Question 2
Solve the problems given in Example 1.
(i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with.
Let John’s marble count = x
Then Jivanti’s marble count = 45 – x
Condition
Both lost 5 marbles, this implies
John’s marble count = x – 5
Jivanti’s marble count = 45 – x – 5 = 40 – x
product of the remanding marbles = 124
Splitting the middle term
Taking x = 9
No. of marbles John has = x = 9
No. of marbles Jivanti has = 45 – 9 = 36
Taking x = 36
No. of marbles John has = x = 36
No. of marbles Jivanti has = 45 – 36 = 9
Answer: 9, 36
(ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was Rs. 750. We would like to find out the number of toys produced on that day.
Let the number of toys produced = x
Then cost per toy = 55 – x
Condition
Total cost = 750
=> Total no. of toys x Cost of one toy = 750
=> x(55 – x) = 750
Answer: 25, 30
Question 3
Find two numbers whose sum is 27 and product is 182.
Let one number be = x
Condition 1:
Sum of numbers = 27
=> Second number = 27 – x
Condition 2:
Product of numbers = 182
Taking x = 13
First number = x = 13
Second number = 27 – 13 = 14
Taking x = 14
First number = x = 14
Second number = 27 – 14 = 13
In both cases the two numbers are 13, 14
Answer: 13, 14
Question 4
Find two consecutive positive integers, sum of whose squares is 365.
Let smaller positive integer be = x
Then, next integer = x + 1
Condition
Sum of squares of these numbers = 365
Reject x = -14
(because it is not a positive integer)
⇒ First number = x = 13
=> Second number = x + 1 = 14
Answer: 13, 14
Question 5
The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
Let the base = x cm
Then the altitude = x – 7 cm
Hypotenuse = 13 cm
Applying Pythagoras Theorem
Reject x = -5
(becasue length of a side cannot be -ve)
⇒ Base = x = 12 cm
=> Altitude = x – 7 = 5 cm
Answer: Base: 12 cm, Altitude: 5 cm
Question 6
A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was Rs. 90, find the number of articles produced and the cost of each article.
Let the number of articles produced = x
Cost per article = 2x + 3
Condition
Total cost = 90
=> Total no. of articles x Cost per article = 90
=> x (2x + 3) = 90
Reject -ve value as number of articles cannot be -ve
⇒ Total no. of articles produced = x = 6
=> Cost per article = 2x + 3 = 12 + 3 = 15 Rs.
Answer: Number of articles = 6, Cost per article = Rs. 15