Ncert Solutions For Class 10 Maths Ex 4.1

Ncert Solutions For Class 10 Maths Chapter 4 Ex 4.1

Polynomials with degree 2 are called quadratic polynomials. When this polynomial is equated to zero, we get a quadratic equation.

Its general form is a2 + bx + c = 0, where a, b and c are real numbers and a ≠ 0.

In many real life situations, we deal with quadratic equations.

Suppose, we have to make a table of 50m2 area with its length twice as its breadth then,

Capture2

Let x be the breadth of the table Therefore, its length will be 2x

Since, length × breadth = area

Therefore,                x.2x = 50

So,                               2x2 = 50 (quadratic equation)

x2 =25   that gives,    x = 5

Thus the length of that table will be = 2x = 10m and breadth will be 5m.

 

Represent the following statements mathematically:

Q.1 Rahul and Nikhil together have 50 chocolates. Both of them lost 5 chocolates each and the product of number of chocolates they have now is 300.Find how many chocolates they actually had?

Sol.

Let,      x be the number of chocolates Rahul had Then, the total number of chocolates Nikhil had = (50 – x) chocolates.

After loosing 5 chocolates Rahul had (x – 5) chocolates and Nikhil had (50 – x – 5) chocolates.

Now, according to the given condition:

(x – 5) (45 – x) = 300

45x – x2 – 225 + 5x=300

x2 – 50x + 525 = 0

This is the required quadratic equation.

 

Q2. Check whether (x – 7)2 +4 = 2x – 9 is a quadratic equation.

Sol.

On, simplifying the above equation we get:

x2 + 49 – 14x + 4 = 2x – 9

x2 – 16x + 62=0

General form of quadratic equation is ax2 + bx + c = 0

Therefore, the given equation is a quadratic equation.

 

Q.3 Check whether x(x-7) + 3 = (x + 6) (x – 6) is a quadratic equation.

Sol.

On, simplifying the above equation we get:

x2 – 7x + 3 = x2 – 36

    7x – 33 = 0

This is not a quadratic equation because general form of a quadratic equation is ax2+bx+c=0 with a ≠ 0 and in this equation a = 0.

 

Q.4 Check whether (x – 2)3 = 0 is a quadratic equation.

Sol.

On, simplifying the above equation we get:

x3 – 8 – 6x2+12x = 0.

Degree of this polynomial equation is 3,

Thus, it is a cubic polynomial equation not a quadratic equation.

 

Q.5 Rahul’s father is 27 years older than him. 3 years from now the product of their ages will be 1224. We would like to find Rohan’s present age.

Sol.

Let the present age of Rahul be x years

Then, present age of Rahul’s father will be = (x + 27) years

Now, their ages after 3 years:

Rahul’s age = (x + 3) years

Rahul’s father age = x + 27 + 3 = (x + 30) years

According to given condition:

(x+3) (x+30) = 1224

x2 + 30x +3x +90 –1224 = 0

               x2 + 33x – 1134 = 0. (Where x is the present age of Rahul in years)

This is the required quadratic equation.

 

Q.6 There is a rectangular park of area 860m 2. The length (meters) of the park is three more than twice its breadth (meters). Form a quadratic equation to solve this problem.

Sol.

Let, x be length (meters) and y be breadth (meter) of the rectangular plot

Now, according to given condition:-

x = 3 + 2y ……… (1)

And

x.y = 860m2…… (2)

Now, substituting equation (1) in (2) we get

(3 + 2y) y = 860

2y2 + 3y = 860

2y2 + 3y – 860 = 0 where y = breadth (in meters)

This is the required quadratic equation.

 

Q.7 A train is covering a distance of 540 km from one city to another city at a uniform speed. If the speed of train had been reduced by 6 km/h, then to cover the same distance the train would have taken 1 hour more. Form a quadratic equation of this situation.

Sol.

Let the train travels at the uniform speed of x km/hr.

Therefore,

Time taken to cover 540km = 540xhours

Now,

When speed is reduced by 6km/h then time taken to cover the same distance =  540x6hours

Now, according to given condition:

540x6540x=1

540x – 540 ( x – 6 )  = x ( x – 6 )

x2 – 6x – 3240  = 0.   Where x= speed of train in km/h

This is the required quadratic equation.

 Finding roots of quadratic equation by factorization