NCERT full solutions here you can find full detailed solutions for all the questions with proper format. class 10 maths ncert solutions chapter 3 exercise 3.3.
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
- NCERT Solutions For Class 10 Maths – Chapter 2 Exercise 2.1 Polynomials Real Number
- NCERT Class 10th – Chapter 2 Polynomial Solutions – Exercise 2.2
- NCERT Class 10th – Chapter 2 Polynomial Solutions – Exercise 2.3
EXERCISE:-3.3
QUE :-1 Solve the following pair of linear equations by the substitution method.
(i) x + y = 14 , x – y = 4 (ii) s – t = 3 ,s/3+t/2=6
(iii) 3x – y = 3 , 9x – 3y = 9 (iv) 0.2x + 0.3y = 1.3,0.4x + 0.5y = 2.3
(v) √2x +√3y =0 , √3x -√2y =0 (vi) 3x/2-5y/3=-2 , x/3 +y/2=13/6
SOL :-
(i) x + y = 14 ………………………(1)
x – y = 4 ………………………..(2)
from the equation (1) , we get
y = 14 -x …………………. (3)
putting the value of x in equation (3) ,we get
x – (14-x) =4
2x = 18
x = 9
putting the value of x in equation 3
y = 14 – 9 = 5
x = 9 , y= 5
(ii) s – t = 3 ……….(1)
s/3+t/2=6 …………..(2)
from equ (1)
s = 3 + t …………(3)
putting the value of s in equ (2)
(3+t)/3 +t/2 =6
= (6+2t +3t)/6 =6
6 + 5t = 36
t = (36 -6 )/5 =6
putting the value of t in equ.(3)
s = 3 + 6 = 9
then s =9 and t =6
(iii) 3x – y = 3 ………………….(1),
9x – 3y = 9 ………………….(2)
from the equation (1)
y = 3x -3 ………………….(3)
putting the value of y in equ. (2)
9x – 3 (3x – 3 )= 9
9x – 9x -9 = 9
0 = 0
then pair of linear equ. have infinite solution.
(iv) 0.2x + 0.3y = 1.3……………..(1)
0.4x + 0.5y = 2.3 …………………(2)
from te equ. (1) we get
0.2x + 0.3y = 1.3
y=( 1.3 -0.2x )/0.3………………….(3)
putting the value of y in equ. (2)
0.4x + 0.5[( 1.3 -0.2x )/0.3] = 2.3
0.4x + [(.65 -0.1x )/0.3] = 2.3
0.12x +.65 – 0.1x =.69
0.02x =0.04
x= 2
putting the value of x in equ. (3)
y=( 1.3 -0.2×2 )/0.3
y = (1.3 -0.4)/0.3
y= 0.9/0.3
y=3
so x = 2 and y =3
(v) √2x +√3y =0……………………….(1)
√3x -√2y =0 ………………………(2)
from the equ (1) we get
√2x +√3y =0
y = √2x /√3…………………(3)
putting the value of y in equ. (2)
√3x -√2 (√2x /√3)=0
3x -2x = 0
x = 0
putting the value of x in equ. (3)
y = √2x /√3
y= 0
hence x= 0 and y = 0
(vi) 3x/2-5y/3=-2 ……………….(1)
x/3 +y/2=13/6…………….(2)
from the equ. (1) we get
5/3y = 3x/2 +2
y = ( 3x +4 )/2 ×3 / 5
y = ( 9x +12 ) /10 …………………….(3)
putting the value of y in equ. (2)
x/3 +(( 9x +12 ) /10)/2=13/6
x/3 +( 9x +12 ) /20)=13/6
(20x +27x +36) /60 =13/6
47x + 36 = 780/6
47 x = 130 -36
x= 94/47 =2
putting the value x in equ . (3)
y = ( 9×2 +12 ) /10
y = 30/10 = 3
so x = 2 and y =3
QUE :-2 Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which
y = mx + 3.
SOL:-
2x + 3y = 11……………….(1)
2x – 4y = – 24 ………………(2)
from equ. (1)
2x + 3y = 11
y = (11 -2x )/3 ……………..(3)
putting the value of y in equ. (2)
2x – 4[(11 -2x )/3] = – 24
6x + 8x -44 = -72
14 x = -72 +44
x = -28/14 = -2
putting the value of x in equ, (3)
y = (11 -2x )/3
y = (11 -2(-2) )/3
y= 15/3 = 5
so x = -2 and y = 5
put the value of x and y in y = mx + 3 we get
5 = -2m +3
m = 2/-2=-2
QUE :-3. Form the pair of linear equations for the following problems and find their solution by
substitution method.
(i) The difference between two numbers is 26 and one number is three times the other.
Find them.
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find
them.
(iii) The coach of a cricket team buys 7 bats and 6 balls for ` 3800. Later, she buys 3
bats and 5 balls for ` 1750. Find the cost of each bat and each ball.
(iv) The taxi charges in a city consist of a fixed charge together with the charge for the
distance covered. For a distance of 10 km, the charge paid is ` 105 and for a
journey of 15 km, the charge paid is ` 155. What are the fixed charges and the
charge per km? How much does a person have to pay for travelling a distance of
25 km?
(v) A fraction becomes9 /11, if 2 is added to both the numerator and the denominator.
If, 3 is added to both the numerator and the denominator it becomes5/6. Find thefraction.
(vi) Five years hence, the age of Jacob will be three times that of his son. Five years
ago, Jacob’s age was seven times that of his son. What are their present ages?
SOL :-
(i) let the frist number = x
let the second number =y
according to question
x = 3y …………………..(1)
the diffrence between two numberis 26 therefore
x -y = 26 ………………………(2)
putting the value of x in equ. (2)
3y – y = 26
2y = 26
y = 13
x = 3y =3 × 13 =39
hence , one number is 13 and another number is 39
(ii) let the larger angel is = x
smaaler angel is = y
according to the question
x = y +18 ……………(1)
bothe angels are supplementary therefore
x + y = 180…………………..(2)
putting the value of x in equ. (2)
y +18+ y = 180
2 y = 180-18 =162
y = 162/2=81
putting the value of y in equ. (2)
x + y = 180
x +81 = 180
x = 180 -81 =99
hence x = 99 and y = 81
(iii) let the cost of bat =x
and the cost of baal is = y
accordind to que.
7x + 6y =3800……….(1)
3x + 5y = 1750 ……….(2)
from equ. (1) we get
y = (3800 -7x )/6……………(3)
putting the value of y in equ. (2)
3x + 5(3800 -7x )/6 = 1750
(18x +19000 -35x )/6 =1750
-17x +19000 = 1750 × 6
-17x = 10500 -19000
x =-8500 /-17 = 500
putting the value of x in equ. (3)
y = (3800 -7×500 )/6
y = 300/6 =50
therefore cost of 1 bat =500
and cost of 1 ball = 50
(iv) let the fix charge of taxi = x
and the charge for distance with the rate / km =y
x +10y = 105 ………….(1)
x + 15y = 155 ……………(2)
from the equ. (1) we get
x = 105 -10y ………………..(3)
putting the value x in equ. (2)
105 -10y+ 15y = 155
5y = 45
y =9
putting the value y in equ. (3)
x = 105 -10×9 =15
so the fix charge of taxi =15
and the charge for distance with the rate / km = 9 rs/km
(v) let the numretor is =x
and the denomenetor is =y
according to the que.
(x +2) /(y+2) =9/11 ……………….(1)
(x +3)/(y +3) = 5/6 …………………(2)
from equ. (1) we get
x +2 =( 9y +18 ) /11
x=( 9y +18 -22 ) /11
x = (9y -4)/11 ………..(3)
putting the value x in equ. (2)
(x +3)/(y +3) = 5/6
((9y -4)/11)+3)/(y +3) = 5/6
(9y -4 +33)/11 =( 5y +15) /6
54y -24 +198 = 55y +165
54y -55y = 165 -174
-y = -9
y= 9
putting the value of y in equ (3)
x = (9×9 -4)/11 =77/11=7
so the fraction is =7/9
(vi) let present age of jacob =x
and present age of his son =y
according to que.
x + 5 = 3(y +5) …………….(1) five year
x-5 = 7(y-5) ……………..(2)
from equ.(1) we get
x + 5 = 3(y +5)
x = 3y +15 -5 =3y +10 ………………(3)
putting the value of x in equ (2)
3y +10 -5 = 7(y-5)
3y + 5 = 7y -35
3y -7y = -40
-4y =-40
y =10
putting the value y in equ. (3)
x=3y +10
x = 30 +10 =40
so jacob’s present age is 40 year
and his son at present 10 year old.
You can see the solution for complete chapter here –
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