 Quadratic Equation NCERT Exercise 4.3 part four Class Ten Mathematics

## Exercise 4.3 Part 4

Question 5: In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.

Solution: Let us assume, marks in Mathematics = x

Therefore, marks in English = 30 – x

If she scores 2 marks more in Mathematics; then marks in mathematics = x +2

And if she scores 3 marks less in English, the marks in English = 30 – x – 3 = 27 – x

As per question;

(x + 2)(27 – x) = 210

Or, 27x – x^2 + 54 – 2x = 210

Or, 25x – x^2 + 54 – 210 = 0

Or, 25x – x^2 – 156 = 0

Or, x^2 – 25x + 156 = 0

Or, x^2 – 12x – 13x + 156 = 0

Or, x(x – 12) – 13(x – 12) = 0

Or, (x – 12)(x – 13) = 0

Hence, x = 12 and x = 13

Case 1: If x = 13, then marks in English = 30 – 13 = 17

Case 2: If x = 12, then marks in English = 30 – 12 = 18

In both the cases; after adding 2 marks to mathematics and deducting 3 marks from English; the product of resultants is 210

Question 6: The diagonal of a rectangular field is 60 meters more than the shorter side. If the longer side is 30 meters more than the shorter side, find the sides of the field.

Solution: Let us assume, the shorter side = x

Then; longer side = x + 30 and diagonal = x + 60

Using Pythagoras Theorem, we get following equation:

(x + 60)^2 = x^2 + (x + 30)^2

Or, x^2 + 120x + 3600

= x2 + x2 + 60x + 900

Or, -x^2 + 60x + 2700 = 0

Or, x^2 - 60x - 2700 = 0

Or, x^2 - 90x + 30x - 2700 = 0

Or, x(x - 90) + 30(x - 90) = 0

Or,(x + 30)(x - 90) = 0

Or, x = -30 and x = 90

Discarding the negative value, we have x = 90 m, longer side = 120 m and diagonal = 150 m

Question 7: The difference of squares of two numbers is 180. The square of the smaller number is 8 times the large number. Find the two numbers.

Solution: Let us assume, larger number = x

Hence, square of smaller number = 8x

As per question;

x^2 – 8x = 180

Or, x^2 – 8x – 180 = 0

Or, x^2 – 18x + 10x – 180 = 0

Or, x(x – 18) + 10(x – 18) = 0

Or, (x + 10)(x – 18) = 0

Hence, x = - 10 and x = 18

Discarding the negative value; x = 18

Smaller number

=sqrt(8xx18)=sqrt(144)=12

Hence, the numbers are; 12 and 18