Ncert Solutions For Class 10 Maths Ex 3.5

Ncert Solutions For Class 10 Maths Chapter 3 Ex 3.5

Question 1:

 From the given pairs of linear equations find the ones with unique solution, no solution or infinitely many solutions. If there is a unique solution, find it by using cross multiplication method.

(i) x-3y-3 =0 and  3x-9y-2=0               (ii) 2x+y =5 and 3x +2y =8

(iii)6x – 10y =40 and 3x-5y = 20    (iv)x -3y – 7 =0  and 3x -3y -15=0

 

Solution:

(i)      3x – 9y -2 =0

x – 3y – 3 =0

a1/a1=1/3 ,         b1/b2  = -3/-9 =1/3,     c1/c2=-3/-2 = 3/2

a1a2=b1b2c1c2

Since, the given set of lines are parallel to each other they will not intersect each other and     therefore there is no solution for these equation.

 

(ii) 2x + y = 5

3x +2y = 8

a1a2=23,b1b2=12,c1c2=58 a1a2b1b2

Since they intersect at a  unique point these equations will have a unique solution by cross multiplication method:

xb1c2c1b2=yc1a2c2a1=1a1b2b1a2 x8+10=y15+16=143

x/ 2 = y/1 = 1

x =2 , y =1.

 

(iii) 6x -10y = 40

3x – 5y = 20

a/ a2 = 3/6 = 1/2 ,               b1/b2 = -5/-10 =1/2,     c1/c2 =-20/-40 = ½

a1/a= b1/b2 = c1/c2

Since the given sets of lines are overlapping each other there will be infinite number of solutions for this pair of equation.

 

(iv) 3x – 3y – 15 =0

x – 3y – 7 = 0

a1/a= 1/3,     b1/b2 = 1 ,   c1/c= -7/-15

a1a2b1b2

Since this pair of lines are intersecting each other at a unique point, there will be  a unique solution.

By cross multiplication,

x/(45-21) = y/(-21+15) = 1/(-3+9))

x/24 = y/-6 =1/6

x/24 = 1/6 and y/-6 = 1/6

∴  x = 4 and y =1.

 

Question 2:

(i) Find the values of a and b at which the following pair of linear equations will have infinite solution.

3y + 2x = 7

(a+b)y +(a-b)x = 3a + b -2

(ii) Find the value of k at which the following set of linear equations won’t have a solution.

3x + y =1

(2k – 1)x + (k-1)y = 2k + 1

Solution:

(i) 3y + 2x -7 =0

(a + b) y + (a-b)y – (3a + b -2) = 0

a1/a2=2/a-b ,               b1/b2=3/a+b ,               c1/c2=-7/-(3a + b -2)

For infinitely many solutions,

a1/a= b1/b= c1/c2

2/a-b = 7/3a + b – 2

6a + 2b – 4 = 7a – 7b

a – 9b = -4  ……(i)

2/a-b = 3/a+b

2a + 2b = 3a – 3b

a -5b = 0 …….(ii)

Subtracting (i) from (ii), we get

4b = 4

b =1

Substituting this eqn in (ii), we get

a -5 x 1= 0

a = 5

Thus at a = 5 and b = 1 the given equations will have infinite solutions.

 

(ii) 3x + y -1 = 0

(2k -1)x  +  (k-1)y – 2k -1 = 0

a1/a2 = 3/2k -1 ,           b1/b2=1/k-1 ,c1/c2 = -1/-2k -1 = 1/ 2k +1

For no solutions

a1/a2  = b1/b2 ≠ c1/c2

3/2k-1 = 1/k -1   ≠ 1/2k +1

3/2k – 1 = 1/k -1

3k -3 = 2k -1

k =2

Therefore, for k = 2 the given pair of linear equations will have no solution.

 

Question 3:

Solve the given pair of linear equations using cross multiplication method and substitution method:

3x + 2y = 4

8x + 5y =9

Solution:

  1. 8x + 5y = 9 ….(1)

3x   +  2y = 4 ….(2)

From equation (2) we get

x = 4 – 2y / 3  …. (3)

Using this value in equation 1 we get

8(4-2y)/3 + 5y = 9

32 – 16y +15y = 27

-y = -5

y = 5 ……(4)

Using this value in equation (2), we get

3x + 10 = 4

x = -2

Thus , x = -2 and y = 5.

 

Now, Using Cross Multiplication method:

8x +5y – 9 = 0

3x + 2y – 4 = 0

x/(-20+18) = y/(-27 + 32 ) = 1/(16-15)

-x/2 = y/5 =1/1

Thus , x = -2 and y =5.

 

Question 4:

For the problems given below, form a pair of linear equations and also solve them (if a solution exists).

(i) When 1 is subtracted from the numerator of a fraction it becomes 1/3 and it becomes 1/4 when 8 is added to the denominator. Find the fraction.

(ii) Gah and Hag are 100kms apart on a highway. A car starts from Gah and another from Hag at the same time. If the cars move towards each other, they meet in one hour and if they move in the same direction, they meet after 5 hours. Find the speed of the two cars.

(iii) Monthly hostel charges are fixed, except for the mess charge which is included in it. Dom a eats for 20 days in the mess and pays Rs1000 as her monthly hostel charge. Nimi eats for 26 days and pays Rs.1180 as hostel charges. What is the fixed charge and the cost of food per day?

(iv) Lama scored 40 in an exam, acquiring 3 marks higher for each right answer and losing 1 mark for every wrong answer. If 4 marks had been awarded for every correct answer and 2 marks reduced for every wrong answer, Lama would have scored 50 marks. Find the number of questions in the exam.

Solution:

(i)    let the fraction be x/y

So, according to question :

(x-1)/y = 1/3 ­­­ =>  3x – y = 3…..(1)

x/(y + 8) = 1/4  => 4x –y =8 ….(2)

Subtracting equation (1) from (2) , we get

x = 5 ……(3)

Using this value in equation (2) we get

4×5 – y = 8

y= 12.

Therefore, the fraction is 5/12.

 

(ii)Let the speed of the first and second cars be x km/h and y km/h respectively.

Respective speed of the two cars when they are moving in the same direction = (x -y) km/h

Respective speed of the two cars when they are headed towards each other = (x + y)km/h

According to the question;

5(x-y) =100

    x – y =20 ….. (1)

Also,

1(x +y )=100 …..(2)

Adding equations (1) and (2) we get

2x =120

x = 60 km/h ……(3)

Using this in equation (1) we get

60-20= y

y= 40 km/h.

 

(iii) let x be the fixed charge and y be the charge of food per day.

According to the question

x + 20y = 1000…. (i)

x +  26y = 1180….(ii)

Subtracting (i) from  (ii) we get

6y=180

y= Rs.30

Using this value in equation (ii) we get

x = 1180 -26 x 30

= Rs.400.

 

(iv) Let x be the number of correct numbers and y be the number of incorrect answers.

According to the question,

3x – y = 40 …..(i)

4x – 2y = 50

    2x-y = 25….(ii)

Subtracting equation (ii) from (i), we get

x=15…………(iii)

Using this in equation (i) we get

3(15) – 40 =y

y= 5

Therefore, the number of correct answers = 15

the number of incorrect answers= 5

the number of questions              =20