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Question 1 :  Find the roots of the following quadratic equations by factorisation

(i) x2 –3x – 10 = 0

(ii) 2x2 + x – 6 = 0
(iii) √2 x2 + 7x + 5√2 = 0
(iv) 2x2 – x +1/8 = 0
(v) 100x2 – 20x + 1 = 0

Solutions:

(i) Given, x2 – 3x –10 =0

Taking LHS,

=>x2 – 5x +2x – 10

=>x(– 5) + 2(x –5)

=>(x – 5)(x + 2)

The roots of this equation, x2 –3x – 10 = 0 are the values of x for which (x – 5)(x + 2)= 0

Therefore, x – 5 = 0 or x +2 = 0

=> x = 5 or x =-2

(ii) Given, 2x2 + x –6 = 0

Taking LHS,

=> 2x2 + 4x –3x – 6

=> 2x(x + 2) – 3(x +2)

=> (x + 2)(2x – 3)

The roots of this equation, 2x2 + x –6=0 are the values of x for which (x – 5)(x + 2)= 0

Therefore, x + 2 = 0or 2x – 3 = 0

=> x = -2 or x =3/2

(iii) √2 x2 + 7x +5√2=0

Taking LHS,

=> √2 x+ 5x +2x + 5√2

=> x (√2x +5) + √2(√2x + 5)= (√2x + 5)(+ √2)

The roots of this equation, √2 x2 +7x + 5√2=0 are the values of x for which (x – 5)(x + 2)= 0

Therefore, √2x + 5 = 0or x + √2 = 0

=> x = -5/√2 or x =-√2

(iv) 2x2 – x +1/8= 0

Taking LHS,

=1/8 (16x2  – 8x +1)

= 1/8 (16x2  – 4x -4x +1)

= 1/8 (4x(4x  –1) -1(4x – 1))

= 1/8 (4– 1)2

The roots of this equation, 2x2 – x +1/8 = 0, are the values of x for which (4– 1)2= 0

Therefore, (4x – 1) = 0 or (4x –1) = 0

x = 1/4 or x = 1/4

(v) Given, 100x2 – 20x +1=0

Taking LHS,

= 100x2 – 10x –10x + 1

= 10x(10x – 1) -1(10x – 1)

= (10x – 1)2

The roots of this equation, 100x2 –20x + 1=0, are the values of x for which (10x – 1)2=0

(10x – 1) = 0 or (10x – 1) = 0

x = 1/10 or x = 1/10

Question 2 :

Solve the problems given in Example 1.

Represent the following situations mathematically:

(i) John and Jivantitogether have 45 marbles. Both of them lost 5 marbles each, and the product ofthe number of marbles they now have is 124. We would like to find out how manymarbles they had to start with.

(ii) A cottageindustry produces a certain number of toys in a day. The cost of production ofeach toy (in rupees) was found to be 55 minus the number of toys produced in aday. On a particular day, the total cost of production was ` 750. We would liketo find out the number of toys produced on that day.

Solutions:

(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.

Soiution

(i) Let us say, the number of marbles Johnhave = x.

Therefore, number of marble Jivanti have = 45– x

After losing 5 marbles each,

Number of marbles John have = x –5

Number of marble Jivanti have = 45 – x –5 = 40 – x

Given that the product of their marbles is124.

(– 5)(40 – x) = 124

x2 – 45x + 324 = 0

x2 – 36x – 9x +324 = 0

x(x – 36) -9(x – 36) = 0

(x – 36)(x – 9) = 0

Thus, we can say,

x – 36 = 0or x – 9 = 0

x = 36 or x = 9

Therefore,

If, John’s marbles = 36,

Then, Jivanti’s marbles = 45 – 36 = 9

And if John’s marbles = 9,

Then, Jivanti’s marbles = 45 – 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 ` 750. We would like to find out the number of toys produced on that day.

Solution

(ii) Let us say, number of toys produced in aday be x.

Therefore, cost of production of each toy =Rs(55 – x)

Given, total cost of production of the toys =Rs 750

x(55 – x) = 750

x2 – 55x + 750 = 0

x2 – 25x – 30x +750 = 0

x(x – 25)-30(x – 25) = 0

(x – 25)(x – 30) = 0

Thus, either x -25 = 0or x – 30 = 0

x = 25 or x = 30

Hence, the number of toys produced in a day,will be either 25 or 30.

Question 3 :

Find two numberswhose sum is 27 and product is 182.

Let us say, first number be x andthe second number is 27 – x.

Therefore, the product of two numbers

x(27 – x) = 182

x2 – 27x – 182 = 0

x2 – 13x – 14x + 182 = 0

x(x – 13) -14(x – 13) = 0

(x – 13)(x -14) = 0

Thus, either, x = -13 = 0or x – 14 = 0

x = 13 or x = 14

Therefore, if first number = 13, then secondnumber = 27 – 13 = 14

And if first number = 14, then second number =27 – 14 = 13

Hence, the numbers are 13 and 14.

Question 4 :

Find two consecutivepositive integers, sum of whose squares is 365.

Let us say, the two consecutive positiveintegers be x and x + 1.

Therefore, as per the given questions,

x2 + (x + 1)2 = 365

xx+1 + 2x = 365

2x2 + 2x – 364 = 0

x– 182 = 0

x+ 14x – 13x –182 = 0

x(x + 14) -13(x + 14)= 0

(x + 14)(x – 13) = 0

Thus, either, x + 14 = 0or x – 13 = 0,

x = – 14 or x = 13

since, the integers arepositive, so x can be 13, only.

x + 1 = 13 + 1 = 14

Therefore, two consecutive positive integerswill be 13 and 14.

Question 5 :

The altitude of aright triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find theother two sides.

Let us say, the base of the right trianglebe x cm.

Given, the altitude of right triangle =(x – 7) cm

From Pythagoras theorem, we know,

Base2 + Altitude2 =Hypotenuse2

x+ (x – 7)2 = 132

x+ x+ 49 –14x = 169

2x– 14x – 120 = 0

x– 7x – 60 = 0

x– 12x + 5x – 60 = 0

x(x – 12) + 5(x – 12) = 0

(x – 12)(x + 5) = 0

Thus, either x – 12 = 0or x + 5 = 0,

x = 12 or x = – 5

Since sides cannot benegative, x can only be 12.

Therefore, the base of thegiven triangle is 12 cm and the altitude of this triangle will be (12 – 7) cm =5 cm.

Question 6 :

A cottage industryproduces a certain number of pottery articles in a day. It was observed on aparticular day that the cost of production of each article (in rupees) was 3more than twice the number of articles produced on that day. If the total costof production on that day was Rs.90, find the number of articles produced andthe cost of each article.

Let us say, the number of articles producedbe x.

Therefore, cost of production of each article= Rs (2x + 3)

Given, total cost of production is Rs.90

x(2x + 3) = 90

2x+ 3x – 90 = 0

2x+ 15x -12x –90 = 0

x(2x + 15) -6(2x + 15) = 0

(2x + 15)(x – 6) = 0

Thus, either 2x + 15 = 0 or x –6 = 0

x = -15/2 or x = 6

As the number of articles produced can only bea positive integer, therefore, x can only be 6.

Hence, number of articles produced = 6

Cost of each article = 2 × 6 + 3 = Rs 15.

krishan