# Class 10 RD Sharma Solutions – Chapter 8 Quadratic Equations – Exercise 8.2

### Question 1: **The product of two consecutive positive integers is 306. Form the quadratic equation to find the integers, if x denotes the smaller integer.**

**Solution:**

Given that the smallest integer of the two consecutive positive integer is denoted by x. Hence, the two integers are x and (x+1). Now its also given that the product of the two consecutive positive integers is 306.

Hence, x*(x+1)= 306

or x

^{2}+ x=306or x

^{2}+x-306=0hence, the required quadratic equation is x

^{2}+x-306=0 .

### Question 2: **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 128. Form the quadratic equation to find how many marbles they had to start with, if John had x marbles.**

**Solution:**

Given that the John and Jivanti together have 45 marbles and john had x marbles. Now , let us assume that the number of marbles Jivanti had is y. Hence, according to question, x+y=45

or y=45-x

So, Jivanti had (45-x) marbles. Now , it is given in the question that both of them lost 5 marbles.

Hence, the number of marbles John has after losing 5 marbles is = x-45

and, the number of marbles Jivanti has after losing 5 marbles is = 45-x-5= 40-x

Now, given that the product of the number of marbles they have now is = 128

Hence, (x-5)*(40-x)=128

or 40x – x

^{2}– 200 + 5x – 128 = 0or -x

^{2}+ 45x – 328 = 0or x

^{2}– 45x + 328 = 0Therefore, the required quadratic equation is x

^{2}– 45x + 328 = 0.

### Question 3: **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 articles produced in a day. On a particular day, the total cost of production was Rs. 750. If x denotes the number of toys produced that day, form the quadratic equation of find x.**

**Solution:**

Given that x denotes the number of toys produced that day.

Hence, according to question the cost of production of each toy is= 55- Number of toys produced in a day = 55-x

now, we can say that the total cost of production on a particular day is Rs x * (55-x)

Given that the on a particular day, the total cost of production was Rs. 750

hence, x * (55 – x) = 750

or x

^{2}– 55x + 750 = 0Therefore, the required quadratic a equation is x

^{2}– 55x + 750 = 0 .

### Question 4: **The height of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, form the quadratic equation to find the base of the triangle.**

**Solution:**

Let us assume that the base of the right triangle is x. Now, given that the height of the right triangle is 7 cm less than its base.

Hence, the height of the right triangle is = (x-7) cm

Given that the hypotenuse of the right triangle is 13 cm.

Hence, according to

Pythagoras Theorem,(Hypotenuse)

^{2}= (Base)^{2}+ (Height)^{2}or (13 )

^{2}= x^{2}+ (x – 7)^{2}or 169 = x

^{2}+ x^{2}– 14x + 49or 2x

^{2}– 14x + 49 – 169 = 0or 2x

^{2}– 14x – 120 = 0or x

^{2}– 7x – 60 = 0Therefore, the required quadratic equation is x

^{2}– 7x – 60 = 0.

### Question 5: **An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore. If the average speed of the express train is 11 km/hr more than that of the passenger train, form the quadratic equation to find the average speed of express train.**

**Solution:**

Let us assume that the average speed of the express train is x km/hr. Now, given that the average speed of the express train is 11 km/hr more than that of the passenger train. We can say that,

The average speed of the passengers train is = (x-11) Km/hr

Now, according to the question the total distance travelled by each train is 132km.

We know that,

Time taken to travel= Distance travelled/average speedTime taken by the express train (t1) = distance travelled/average speed of the express train

= hr

Time taken by the passengers train (t2) = distance travelled/average speed of the passengers train

= hr

Given that the time taken by the express train is 1 h less than the passengers train,

Hence, t2-t1=1

or

or

or 132*(x-x+11)=x*(x-11)

or 132*11=x

^{2}-11xor x

^{2}-11x=1452or x

^{2}-11x-1452=0Therefore, the required quadratic equation is x

^{2}-11x-1452=0.

### Question 6: **A train travels 360 km at a uniform speed. If the speed had been 5 km/hr more, it would have taken 1 hour less for the same journey. Form the quadratic equation to find the speed of the train.**

**Solution:**

Let us assume that the speed of the train is x km/hr. Given that the distance covered by the train at a uniform speed is 360 km.

hence, the time travelled by the train is = Distance covered/speed of the train (t1)= hr

now if the speed of the train is 5 km/hr more then the speed of the should be = (x+5) km/hr

hence, the time travelled by the train when it’s speed is 5 km/h more is (t2)= hr

Now , given that the train would have taken 1 hour less if it’s speed was 5 km/hr more.

hence, t1-t2=1

or

or

or 360*5=x*(x+5)

or x

^{2}+5x=1800or x

^{2}+5x=1800Therefore, the required quadratic equation is x

^{2}+5x=1800.

Attention reader! Don’t stop learning now. Join the **First-Step-to-DSA Course for Class 9 to 12 students ****, **specifically designed to introduce data structures and algorithms to the class 9 to 12 students