NCERT full solutions here you can find full detailed solutions for all the questions with proper format. class 10 maths ncert solutions chapter 4 exercise 4.2.
You can see the solution for complete chapter here –
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Complete 10th Class Maths Solution
- NCERT Solutions For Class 10 Maths – Chapter 1 Exercise 1.2 Real Number
- NCERT Solutions For Class 10 Maths – Chapter 1 Exercise 1.3 Real Number
- NCERT Solutions For Class 10 Maths – Chapter 1 Exercise 1.4 Real Number
EXERCISE -4.2
QUE:-1. Find the roots of the following quadratic equations by factorisation:
(i) x² -3x-10=0 (ii) 2x² +x -6 =0
(iii)√2x² +7x +5√2 (iv) 2x² -x +1/8=0
(v) 100x² -20x +1=0
SOL:-
(i) x² -3x-10=0
= x² -5x +2x -10
= x(x-5) +2(x-5)
= (x-5)(x+2)
roots of the quadratic equation is (x-5)(x+2)=0
x-5 =0 ; x=5
x+2=0 ; x=-2
(ii) 2x² +x -6 =0
= 2x² -3x -4x -6
= x (2x -3) +2(2x-3)
= (2x-3)(x-2)
roots of the quadratic equation is (2x-3)(x+2)=0
x=3/2 and x=-2
(iii)√2x² +7x +5√2 =0
=√2x² +5x +2 +5√2
= x(√2x +5) +√2(√2x+5)
= (√2x +5) (x+√2)
roots of the quadratic equation is (√2x +5) (x+√2)=0
x=-5/√2 ; x=-√2
(iv) 2x² -x +1/8=0
=8/8(2x² -x +1/8)
= 1/8(16x² -8x +1)
= 1/8(4x(4x -1)-1(4x-1))
=1/8 (4x -1)(4x -1)
roots of the quadratic equation is 1/8 (4x -1)(4x -1)=0
x=1/4 ;x=1/4
(v) 100x² -20x +1=0
100x² -10x -10x +1 =0
=10x(10x -1)-1(10x-1)
= (10x-1) (10x -1)
roots of the quadratic equation is (10x-1) (10x -1)=0
x =1/10 ;x =1/10
QUE:-2. Solve the problems given in Example 1.
(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.
(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.
SOL:-
(i) Let the number of johns marbles be x.
Therefore ,number of jivantis marble=45 – x
after losing 5 marbles
number of john’s marbles =x-5
and number of jaynti’s marbles = 45-x-5 =40-x
according to the question
product of their marbles is 124
therefore( x-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
x =36 ;x=9
if the number marble have john =9
then jivantis marbles =45 -9=36
if john have =36 marbles
then jivanti ‘s marbles =45-36=9
(ii) let the number of toys produce =x
cost of each toy =55-x
total production of toys =750
x (55-x) =750
55x -x² =750
x² -55x +750=0
x²-25x -30x +750
x (x-25) -30(x-25)=0
(x-25)(x-30)=0
x-25 =0 ; x=25
or x-30=0 ; x=30
so the number of toys either 25 or 30.
QUE:-3. Find two numbers whose sum is 27 and product is 182.
SOL:-
let the the frist number is x and the second number is 27-x
according to the question
product of number x(27-x) =182
27x -x² -182 =0
x² -27x +182 =0
x² – 13x +14x +182 =0
x(x -13) -14(x -13) =0
(x-13)(x-14) =0
x -13 =0 ; x=13
x-14=0 ; x=14
QUE:-4. Find two consecutive positive integers, sum of whose squares is 365.
SOL:-
let the first number is x and second number x+1
according to the question
x² +(x+1)² =365
x² +x² +2x +1 =365
2x² +2x +1 -365 =0
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) =0 ;x=-14
x-13 =0 ; x=13
so the numbers are 13 and 14
QUE:-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.
SOL:-
let the base of right angel triangle x
according to the quetion altitude is x -7
by pythagoras theorem
x² + ( x -7)² =13²
x ² + x² +49 -14x =169
2x² -14x -120=0
x ² -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
so base of triangle is 12 and altitude of triangle is 12-7 =5
QUE:-6. A cottage industry produces a certain number of pottery articles in a day. It was observed
on a particular day that the cost of production of each article (in rupees) was 3 more than
twice the number of articles produced on that day. If the total cost of production on that
day was ` 90, find the number of articles produced and the cost of each article.
SOL:-
let the article produce is x
cost of article produced =2x +3
∴ 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 =-15/2
(x-6) =0
x=6
so the numer of article produce is 6
cost of each article is 2×15+3 =15
You can see the solution for complete chapter here –
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