Class 10 Mathematics

Quadratic Equation

Exercise 4.3 (NCERT) Part 5

Question 8: A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.

Solution: Let us assume, speed of train = x

We know; time = distance/speed

In case of normal speed;

`t=(360)/(x)`

In case of increased speed;

`t-1=(360)/(x+5)`

Or, `t=(360)/(x+5)+1`


From above equations;

`(360)/(x)=(360)/(x+5)+1`

Or, `(360)/(x)=(360+x+5)/(x+5)`

Or, `360x+1800=x^2+365x`

Or, `x^2+5x-1800=0`

Or, `x^2+45x-40x-1800=0`

Or, `x(x+45)-40(x+45)=0`

Or, `(x-40(x+45)=0`

x = 40 and x = - 45

Discarding the negative value; we have speed of train = 40 km/h

Question 9: Two water taps together can fill a tank 9 and 3/8 hours. The tap of larger diameter takes 10 hour less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

Solution: Let us assume, smaller tap takes x hours to fill the tank

Then, time taken by larger tap `= x – 10`

In 1 hour, the smaller tap will fill `1/x` of tank

In 1 hour, the larger tap will fill `(1)/(x-10)` of tank.

As per question;

`1/x+(1)/(x-10)=(8)/(75)`

Or, `(x-10+x)/(x^2-10x)=(8)/(75)`

Or, `150x-750=8x^2-80x`

Or, `8x^2-80x-150x+750=0`

Or, `8x^2-230x+750=0`

Or, `4x^2-115x+375=0`

Or, `4x^2-100x-15x+375=0`

Or, `4x(x-25)-15(x-25)=0`

Or, `(4x-15)(x-25)=0`

Or, `x=(15)/(4)` and `x=25`

Since 15/4 is less than the difference in their individual timings hence time taken by smaller tap = 25 hours and that by larger tap = 15 hours

Question: 10: An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km /h more than that of the passenger train, find the average speed of the two trains.

Solution: Let us assume, speed of passenger train = x, then speed of express train `= x + 11`

Using time = distance/speed, we get following equations;

`t=(132)/(x)`

And, `t-1=(132)/(x+11)`

Or, `t=(132)/(x+11)+1`


From above equations;

`(132)/(x)=(132+x+11)/(x+11)`

Or,`132x + 1452 = x^2 + 143x`

Or,`x^2 + 143x - 132x - 1452 = 0`

Or, `x^2 + 11x - 1452 = 0`

Or, `x^2 + 44x - 33x - 1452 = 0`

Or, `x(x + 44) - 33(x + 44) = 0`

Or,`(x - 33)(x + 44) = 0`

Hence, `x = 33` and `x = - 44`

Discarding the negative value, speed of passenger train = 33 km/h and speed of express train = 44 km/h

Average speed can be calculated as follows:

`(33+44)/(2)=38.5 km//hr`


Question 11: Sum of the areas of two squares is 468 square meter. If the difference of the perimeters is 24 m, find the sides of the two squares.

Solution: We know perimeter `= 4 xx \text(side)`

If x and y are the sides of two squares, then;

`4x – 4y = 24`

Or, `x – y = 6`

Or, `y = x – 6`

Now sum of areas can be given by following equation:

`x^2 + (x-6)^2 = 468`

Or, `x^2 + x^2 - 12x + 36 = 468`

Or, `2x^2 - 12x + 36 - 468 = 0`

Or, `2x^2 - 12x - 432 = 0`

Or, `x^2 - 6x - 216 = 0`

Or, `x^2 - 18x + 12x - 216 = 0`

Or, `x(x - 18) + 12(x - 18) = 0`

Or,`(x + 12)(x - 18) = 0`

Hence; `x = -12` and `x = 18`

Hence, sides of squares are; 12 m and 18 m


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Exercise 4.1(1)

Exercise 4.1(2)

Exercise 4.2(1)

Exercise 4.2(2)

Exercise 4.2(3)

Exercise 4.3(part1)

Exercise 4.3(part2)

Exercise 4.3(part3)

Exercise 4.3(part4)

Exercise 4.4(part1)

Exercise 4.4(part2)