Class10 NCRT Pair of Linear Equations in Two Variables Exercise – 3.5 pdf || UP Board
Pair of Linear Equations in Two VariablesExercise – 3.5
1. Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions. In case there is a unique solution, find it by using cross multiplication method.
(i) x – 3y – 3 = 0………………(1)
3x – 9y – 2 = 0………………(2)
Here, a1/a2 = 1/3, b1/b2 = -3/-9 = 1/3 and c1/c2 = -3/-2 = 3/2
a1/a2 = b1/b2 ≠ c1/c2 Hence the pair of equations has no solution.
(ii) 2x + y – 5 = 0…………..(1)
3x + 2y – 8 = 0………………(2)
Here a1/a2 = 2/3, b1/b2 = ½ and c1/c2 = -5/-8
a1/a2 ≠ b1/b2 Therefore the pair of equations has a unique solution.
By cross-multiplication method,
x/((1×-8)-(2×-5))=y/((-5×3)- (-8×2))=1/((2×2)- (3×1))
x/2 = y/1 = 1/1
x/2 = 1/1 and y/1 = 1/1
x = 2 and y = 1
Hence, x = 2, y = 1
(iii) 3x – 5y – 20 = 0…………….(1)
6x – 10y – 40 = 0……………..(2)
Here, a1/a2 = 3/6 = 1/2, b1/b2 = -5/-10 = 1/2 and c1/c2 = -20/-40 =1/2
a1/a2 = b1/b2 = c1/c2 Therefore, the pair of equations has infinitely many solutions.
(iv) x – 3y – 7 = 0…………(1)
3x – 3y – 15 = 0…………….(2)
यहाँ a1/a2 = 1/3, b1/b2 = -3/-3 = 1/1
a1/a2 ≠ b1/b2 Therefore the pair of equations has a unique solution.
By cross-multiplication method,
x/((-3×-15)-(-3×-7))=y/((-7×3)-(-15×1))=1/((1×-3)- (3×-3))
x/24 = y/-6 = 1/6
x/24 = 1/6 and y/-6 = 1/6
x = 24/6 = 4 and y = -6/6 = -1
Hence, x = 4, y = -1
2. (i) For which values of a and b does the following pair of linear equations have an infinite number of solution?
2x + 3y = 7
(a – b)x + (a + b)y = 3a + b – 2
Solution:
2x + 3y – 7 = 0…………………..(1)
(a – b)x + (a + b)y – (3a + b – 2) = 0………………(2)
Here, a1/a2 = 2/(a - b), b1/b2 = 3/(a + b) and c1/c2 = -7/-(3a + b – 2) = 7/(3a + b – 2)
A pair of linear equations will have infinitely many solutions, if
a1/a2 = b1/b2 = c1/c2
2/(a - b) = 3/(a + b) = 7/(3a + b – 2)
2/(a - b) = 3/(a + b) और 3/(a + b) = 7/(3a + b – 2)
2(a + b) = 3(a - b) और 3(3a + b – 2) = 7(a + b)
2a + 2b = 3a – 3b और 9a + 3b – 6 = 7a + 7b
a = 5b……..(3) और 2a = 4b + 6………(4)
Substituting the value of 'a' in equation (4),
2×5b = 4b + 6
10b = 4b + 6
6b = 6
b = 1
Substituting the value of 'b' in equation (3),
a = 5×1 = 5
Hence, a = 5, b = 1
(ii) For which value of k will the following pair of linear equation have no solution?
3x + y = 1
(2k – 1)x + (k – 1)y = 2k + 1
Solution:
3x + y – 1 = 0……………..(1)
(2k – 1)x + (k – 1)y – (2k + 1) = 0…………..(2)
Here, a1/a2 = 3/(2k - 1), b1/b2 = 1/(k - 1) and c1/c2 = -1/-(2k + 1) = 1/(2k + 1)
Pairs of linear equations will have no solution, if
a1/a2 = b1/b2 ≠ c1/c2
3/(2k - 1) = 1/(k - 1) ≠ 1/(2k + 1)
3/(2k - 1) = 1/(k - 1)
3(k - 1) = 1(2k - 1)
3k – 3 = 2k – 1
3k – 2k = 3 – 1
k = 2
Hence, k = 2
3. Solve the following pair of linear equations by the substitution and cross-multiplication methods:
8x + 5y = 9; 3x + 2y = 4
Solution:
8x + 5y = 9………………..(1)
3x + 2y = 4………………(2)
By Substitution method,
From equation (1),
y = (9 – 8x)/5 …………..(3)
Substituting the value of 'y' in equation (2),
3x + 2× (9 – 8x)/5 = 4
15x + 18 – 16x = 20
-x = 2
x = -2
Substituting the value of 'x' in equation (1),
8×-2 + 5y = 9
-16 + 5y = 9
5y = 25
y = 25
Hence, x = -2, y = 5
By cross-multiplication method,
x/((5×-4)-(2×-9))=y/((-9×3)-(-4×8))=1/((8×2)- (3×5))
x/-2 = y/5 = 1/1
x/-2 = 1/1 and y/5 = 1/1
x = -2 and y = 5
Hence, x = -2, y = 5
4. Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method:
i. A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she to pay ₹1000 as hostel charges whereas a student B, Who takes food for 26 days, pays ₹1180 as hostel charges. Find the fixed charges and the cost of food per day.
Solution:
Let fixed value of monthly expenditure = ₹x
and cost of food per day = ₹ y
When a student takes food for 20 days, ₹ 1000 has to be paid for hostel expenses.
Therefore,
x + 26y = 1000
x + 26y – 1000 = 0 …………..(1)
When a student takes food for 26 days, ₹1180 has to be paid for hostel expenses.
Therefore,
x + 26y = 1180
x + 26y – 1180 = 0 …………….(2)
By cross-multiplication method,
x/((20×-1180)-(26×-1000))=y/((-1000×1)-(-1180×1))=1/((1×26)- (1×20))
x/(-23600 + 26000) = y/(-1000 + 1180) = 1/6
x/2400 = y/180 = 1/6
x/2400 = 1/6 and y/180 = 1/6
x = 2400/6 and y = 180/6
x = 400 and y = 30
Hence, the fixed charges is ₹ 400 and the cost of food per day is ₹ 30.
ii. A fraction becomes 1/3 when 1 is subtracted from the numerator and it becomes ¼ when 8 is added to its denominator. Find the fraction.
Solution:
Let numerator = x
and denominator = y
so fraction = x/y
According to the first condition,
(x – 1)/y = 1/3
3x – 3 = y
3x – y – 3 = 0 ……………(1)
According to the second condition,
x/(y + 8) = ¼
4x = y + 8
4x – y – 8 = 0 …………….(2)
By cross-multiplication method,
x/((-1×-8)-(-1×-3))=y/((-3×4)-(-8×3))=1/((3×-1)- (4×-1))
x/(8 – 3) = y/(-12 + 24) = 1/(-3 + 4)
x/5 = y/12 = 1/1
x/5 = 1/1 and y/12 = 1/1
x = 5 and y = 12
Hence, fraction = x/y = 5/12
iii. Yash Scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?
Solution:
Let the number of correct answers = x
and number of wrong answers = y
According to the first condition,
3x – y = 40
3x – y – 40 = 0 ………..(1)
According to the second condition,
4x – 2y = 50
4x – 2y – 50 = 0 ………….(2)
By cross-multiplication method,
x/((-1×-50)-(-2×-40))=y/((-40×4)-(-50×3))=1/((3×-2)- (4×-1))
x/( 50 – 80) = y/(-160 + 150) = 1/(-6 + 4)
x/(-30) = y/(-10) = 1/(-2)
x/(-30) = 1/(-2) and y/(-10) = 1/(-2)
x = 15 and y = 5
Total Number questions = x + y = 15 + 5 = 20
Hence, the total number of questions in the test was 20.
iv. Places A and B are 100 km apart on a highway. One car starts from A and another form B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the tow cars?
Solution:
Let the speed of the car starting from A = x km/h
and speed of car starting from B = y km/h
five hours later,
Distance covered by car starting from A = 5x km
Distance covered by car starting from B = 5y km
therefore,
5x – 5y = 100
x - y – 20 = 0 …………….(1)
If these cars move towards each other, they meet after 1 hour.
1 hour later,
Distance covered by car starting from A = x km
Distance covered by car starting from B = y km
therefore,
x + y = 100
x + y – 100 = 0 …………..(2)
By cross-multiplication method,
x/((-1×-100)-(1×-20))=y/((-20×1)-(-100×1))=1/((1×1)- (1×-1))
x/(100 + 20) = y/(-20 + 100) = 1/2
x/120 = y/80 = 1/2
x/120 = 1/2 and y/80 = 1/2
x = 120/2 and y = 80/2
x = 60 and y = 40
Hence, speed of car starting from A = 60 km/h and speed of car starting from B = 40 km/h.
v. The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breath is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.
Solution:
Let the length of the rectangle = x units
and breadth of rectangle = y units
Therefore, area of the rectangle = xy square units
If its length is decreased by 5 units and breadth is increased by 3 units, then area = (x – 5)(y + 3) square units
According to Question,
(x – 5)(y + 3) = xy – 9
xy + 3x – 5y – 15 = xy – 9
3x – 5y – 6 = 0 …………..(1)
If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. so,
(x + 3)(y + 2) = xy + 67
xy + 2x + 3y + 6 = xy + 67
2x + 3y – 61 = 0 …………..(2)
By cross-multiplication method,
x/((-5×-61)-(3×-6))=y/((-6×2)-(-61×3))=1/((3×3)- (2×-5))
x/(305 + 18) = y/(-12 + 183) = 1/(9 + 10)
x/323 = y/171 = 1/19
x/323 = 1/19 and y/171 = 1/19
x = 323/19 and y = 171/19
x = 17 and y = 9
Hence, the length of the rectangle is 17 units and the breadth is 9 units.
Download Class10 NCRT Pair of Linear Equations in Two Variables Exercise – 3.5 pdf
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