Call Now +91-1244985228

Arinjay Academy

Study Maths, Hindi, Accounts, Economics, for K-12 and International Taxation for CA Final Students. Visit Wesite or download app for updates.

You are here: Home / Archives for NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.2

NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.2

/

NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.2

Download NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.2

1. Find the roots of the following quadratic equations by factorization:

(i) x2-3x-10 = 0

Solution:

x2-3x-10 = 0
x2-5x+2x-10 = 0
x(x-5)+2(x-5) = 0
(x-5)(x+2) = 0
⇒ x-5 = 0 or, x+2 = 0
x = 5 or x =-2
Hence the required roots are 5, -2.

(ii) 2x2+x-6=0

Solution:

2x2+x-6 = 0
2x2+4x-3x-6 = 0
2x(x+2)-3(x+2) = 0
(2x-3)(x+2) = 0
⇒ 2x-3 = 0 or, x+2 = 0
x = \cfrac { 3 }{ 2 }  or x = -2
Hence the required roots are \cfrac { 3 }{ 2 } , -2.

(iii) √2x2+7x+5√2 = 0

Solution:

√2x2+7x+5√2 = 0
√2x2+2x+5x+ 5√2 = 0
√2x(x+√2)+5(x+√2) = 0
(x+√2)(√2x +5) = 0
⇒ x+√2 = 0 or, √2x +5 = 0
x = -√2 or x = –\cfrac { 5 }{ \sqrt { 2 } }

Hence the required roots are -√2, – \cfrac { 5 }{ \sqrt { 2 } } .

(iv) 2x2-x+\cfrac { 1 }{ 8 } =0

Solution:

2x2-x+ \cfrac { 1 }{ 8 }  =0
16x2-8x+1 = 0
(4x)2-2.4x.1+12 = 0
(4x-1)2= 0
⇒ 4x-1 = 0  or 4x-1 = 0
x = \cfrac { 1 }{ 4 }   or x = \cfrac { 1 }{ 4 }
Hence the required roots are \cfrac { 1 }{ 4 } , \cfrac { 1 }{ 4 }

(v) 100x2-20x+1 = 0

Solution:

100x2-20x+1 = 0
(10x)2-2×10x.1+12= 0
(10x-1)2 = 0
10x-1 = 0 or 10x-1 = 0
x = \cfrac { 1 }{ 10 } or x = \cfrac { 1 }{ 10 }
Hence the required roots are \cfrac { 1 }{ 10 } or \cfrac { 1 }{ 10 }

 

2. Solve the problem given in example 1.

(i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each others, and product of the marbles they now have is 124. We would like to find how many marbles they had to start with?

Solution: Let the number of marbles John had be x.
Then the number of marbles Jivanti had be 45-x
They lost 5 marbles to each others.
Hence now the number of marbles of John is x-5 and that of Jivanti is 45-x-5 = 40-x.
Then by given condition we have
(x-5)(40-x) = 124
40x-200-x2+5x = 124
x2-45x+324 = 0
x2-36x-9x+324 = 0
x(x-36)-9(x-36) = 0
(x-36)(x-9) = 0
x-36 = 0
x = 36
or x-9 = 0
x = 9
Therefore, they had 36 and 9 marvels respectively.

(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: Let the number of toys produced on that day is x.
Therefore, the cost of production of each toy on that day is (55-x)
So, the total cost of production on that day is = x(55-x)
Then using the given condition, we have,
x(55-x) = 750
55x-x2 =750
x2-55x+750 = 0
x2-25x-30x+750 = 0
x(x-25)-30(x-25) = 0
(x-25)(x-30) = 0
x-25 = 0
x =25
or x-30 =0
x = 30
Therefore, the number of toys produced on that day is 25 or 30.

 

3. Find the two numbers whose sum is 27 and product is 182.

Solution: Let one number is x. Then another number is 27-x.
Then by the given condition,
x(27-x) = 182
27x- x2-182 = 0
x2-27x+182 = 0
x2-14x-13x+182 = 0
x(x-14)-13(x-14) = 0
(x-14)(x-13) = 0
x-14 = 0
x =14
or, x-13 = 0
x =13
Hence the required two numbers are 13 and 14.

 

4. Find the two consecutive positive integers, sum of whose squares is 365.

Solution: Let the two consecutive positive integers be x and x+1
Then by the given condition,
x2+(x+1)2 = 365
x2+ x2+2x+1 = 365
2x2+2x-364 = 0
x2+x-182 = 0
x2+14x-13x-182 = 0
x(x+14)-13(x+14) = 0
(x+14)(x-13) = 0
x = -14 or 13
Since the numbers are positive so x = -14 not possible.
So, the numbers are 13 and 13+1 = 14.

 

5. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.

Solution: Let base of the right triangle is x cm. Then the altitude is x-7 cm.
Then by the property of right triangle and using the given property we have
x2+(x-7)2 = (13)2
x2+ x2-14x+49 = 169
2x2-14x-120 = 0
x2-7x-60 = 0
x2-12x+5x-60 = 0
x(x-12)+5(x-12) = 0
(x-12)(x+5) = 0
x-12 = 0
x =12
or, x+5 = 0
x = -5
Since length can’t be negative, so x = -5 is not possible.
Hence the base is 12 cm and the altitude is (12-7)cm = 5 cm

 

6. A cottage industry produces a certain number of pottery articles in a day. It was observed that on a particular day that the cost of production of each articles (in rupees) was 3 more than twice the number of articles produced on that day. If the total number of production on that day was ₹90, find the number of articles produced and the cost of each article.

Solution: Let the number of articles produced on that day is x. then the price on that day is 2x+3
Then using the given condition we have,
x(2x+3) = 90
2x2+3x-90 = 0
2x2+15x-12x-90 = 0
x(2x+15)-6(2x+15)=0
(2x+15)(x-6) = 0
2x+15 = 0
x = – \cfrac { 15 }{ 2 }
or, x-6 = 0
x = 6
Since Articles can’t be negative then x = –\cfrac { 15 }{ 2 }  is not possible.
Therefore, the number of articles produced on that day = 6
Cost of each article = 2×6+3 = ₹15

Download NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.2