Solutions to Practice Sets 1.1 to 1.5 and Problem Set 1, with step-by-step working, graphs and Cramer’s rule. Maharashtra State Board (Balbharati), Std 10 Mathematics Part 1 (Algebra).
1Complete the following activity to solve the simultaneous equations: 5x + 3y = 9 … (I); 2x − 3y = 12 … (II)
Adding equations (I) and (II):
(5x + 3y) + (2x − 3y) = 9 + 12
7x = 21
x = 217 ∴ x = 3
Placing x = 3 in equation (I):
5 × 3 + 3y = 9
3y = 9 − 15
3y = −6
y = −63 ∴ y = −2
∴ Solution is (x, y) = (3, −2)
2Solve the following simultaneous equations.
(1) 3a + 5b = 26; a + 5b = 22
3a + 5b = 26 … (I)
a + 5b = 22 … (II)
Subtracting (II) from (I):
2a = 4 ∴ a = 2
Substituting a = 2 in (II): 2 + 5b = 22
5b = 20 ∴ b = 4
(a, b) = (2, 4)
(2) x + 7y = 10; 3x − 2y = 7
x + 7y = 10 … (I)
3x − 2y = 7 … (II)
From (I), x = 10 − 7y … (III)
Substituting in (II):
3(10 − 7y) − 2y = 7
30 − 21y − 2y = 7
−23y = −23 ∴ y = 1
From (III), x = 10 − 7 × 1 = 3
(x, y) = (3, 1)
(3) 2x − 3y = 9; 2x + y = 13
2x − 3y = 9 … (I)
2x + y = 13 … (II)
Subtracting (I) from (II):
4y = 4 ∴ y = 1
Substituting y = 1 in (II): 2x + 1 = 13
2x = 12 ∴ x = 6
(x, y) = (6, 1)
(4) 5m − 3n = 19; m − 6n = −7
5m − 3n = 19 … (I)
m − 6n = −7 … (II)
Multiplying (I) by 2:
10m − 6n = 38 … (III)
Subtracting (II) from (III):
9m = 45 ∴ m = 5
Substituting m = 5 in (II): 5 − 6n = −7
−6n = −12 ∴ n = 2
(m, n) = (5, 2)
(5) 5x + 2y = −3; x + 5y = 4
5x + 2y = −3 … (I)
x + 5y = 4 … (II)
Multiplying (II) by 5:
5x + 25y = 20 … (III)
Subtracting (I) from (III):
23y = 23 ∴ y = 1
Substituting y = 1 in (II): x + 5 = 4 ∴ x = −1
(x, y) = (−1, 1)
(6) 13x + y = 103; 2x + 14y = 114
Multiplying the first equation by 3:
x + 3y = 10 … (I)
Multiplying the second equation by 4:
8x + y = 11 … (II)
Multiplying (II) by 3:
24x + 3y = 33 … (III)
Subtracting (I) from (III):
23x = 23 ∴ x = 1
Substituting x = 1 in (I): 1 + 3y = 10
3y = 9 ∴ y = 3
(x, y) = (1, 3)
(7) 99x + 101y = 499; 101x + 99y = 501
99x + 101y = 499 … (I)
101x + 99y = 501 … (II)
The coefficients of x and y are interchanged, so we add and subtract the equations.
Adding (I) and (II):
200x + 200y = 1000
∴ x + y = 5 … (III)
Subtracting (I) from (II):
2x − 2y = 2
∴ x − y = 1 … (IV)
Adding (III) and (IV):
2x = 6 ∴ x = 3
Substituting x = 3 in (III): 3 + y = 5 ∴ y = 2
(x, y) = (3, 2)
(8) 49x − 57y = 172; 57x − 49y = 252
49x − 57y = 172 … (I)
57x − 49y = 252 … (II)
Adding (I) and (II):
106x − 106y = 424
∴ x − y = 4 … (III)
Subtracting (I) from (II):
8x + 8y = 80
∴ x + y = 10 … (IV)
Adding (III) and (IV):
2x = 14 ∴ x = 7
Substituting x = 7 in (IV): 7 + y = 10 ∴ y = 3
(x, y) = (7, 3)
Practice Set 1.2
1Complete the following tables to draw graphs of the equations (I) x + y = 3 and (II) x − y = 4.
x + y = 3
x
3
−2
0
y
0
5
3
(x, y)
(3, 0)
(−2, 5)
(0, 3)
x − y = 4
x
4
−1
0
y
0
−5
−4
(x, y)
(4, 0)
(−1, −5)
(0, −4)
Working: For x + y = 3: when x = 3, y = 0; when y = 5, x = −2. For x − y = 4: when y = 0, x = 4; when x = −1, −1 − y = 4, so y = −5.
2Solve the following simultaneous equations graphically.
(1) x + y = 6; x − y = 4
x + y = 6
x
0
2
6
y
6
4
0
(x, y)
(0, 6)
(2, 4)
(6, 0)
x − y = 4
x
4
0
6
y
0
−4
2
(x, y)
(4, 0)
(0, −4)
(6, 2)
The lines intersect at (5, 1). Scale: 1 cm = 1 unit on both axes.
(x, y) = (5, 1)
(2) x + y = 5; x − y = 3
x + y = 5
x
0
2
5
y
5
3
0
(x, y)
(0, 5)
(2, 3)
(5, 0)
x − y = 3
x
3
0
5
y
0
−3
2
(x, y)
(3, 0)
(0, −3)
(5, 2)
The lines intersect at (4, 1).
(x, y) = (4, 1)
(3) x + y = 0; 2x − y = 9
x + y = 0
x
0
2
−2
y
0
−2
2
(x, y)
(0, 0)
(2, −2)
(−2, 2)
2x − y = 9
x
4
5
2
y
−1
1
−5
(x, y)
(4, −1)
(5, 1)
(2, −5)
The lines intersect at (3, −3).
(x, y) = (3, −3)
(4) 3x − y = 2; 2x − y = 3
3x − y = 2
x
0
1
2
y
−2
1
4
(x, y)
(0, −2)
(1, 1)
(2, 4)
2x − y = 3
x
0
2
1
y
−3
1
−1
(x, y)
(0, −3)
(2, 1)
(1, −1)
The lines intersect at (−1, −5).
(x, y) = (−1, −5)
(5) 3x − 4y = −7; 5x − 2y = 0
3x − 4y = −7
x
−1
3
−5
y
1
4
−2
(x, y)
(−1, 1)
(3, 4)
(−5, −2)
5x − 2y = 0
x
0
2
−2
y
0
5
−5
(x, y)
(0, 0)
(2, 5)
(−2, −5)
The lines intersect at (1, 2.5).
(x, y) = (1, 2.5)
The point of intersection has y = 2.5, which lies halfway between 2 and 3 on the Y-axis. Reading it accurately needs careful plotting; check it by substituting: 3(1) − 4(2.5) = −7 and 5(1) − 2(2.5) = 0.
(6) 2x − 3y = 4; 3y − x = 4
2x − 3y = 4
x
2
5
−1
y
0
2
−2
(x, y)
(2, 0)
(5, 2)
(−1, −2)
3y − x = 4
x
−4
2
−1
y
0
2
1
(x, y)
(−4, 0)
(2, 2)
(−1, 1)
The lines intersect at (8, 4).
(x, y) = (8, 4)
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Practice Set 1.3
1Fill in the blanks with correct number.
3245 = 3 × 5 − 2 × 4 = 15 − 8 = 7
2Find the values of following determinants.
(1) −1724
= (−1) × 4 − 7 × 2 = −4 − 14 = −18
−18
(2) 53−70
= 5 × 0 − 3 × (−7) = 0 + 21 = 21
21
(3) 73533212
= 73 × 12 − 53 × 32
= 76 − 156
= −86 = −43
−4/3
3Solve the following simultaneous equations using Cramer’s rule.
(4) 13x + y + 13x − y = 34; 12(3x + y) − 12(3x − y) = −18
Replacing 13x + y by p and 13x − y by q:
p + q = 34 … (I)
p2 − q2 = −18; multiplying by 2: p − q = −14 … (II)
Adding (I) and (II):
2p = 24 ∴ p = 14
Substituting in (I): 14 + q = 34 ∴ q = 12
13x + y = 14 ∴ 3x + y = 4 … (III)
13x − y = 12 ∴ 3x − y = 2 … (IV)
Adding (III) and (IV):
6x = 6 ∴ x = 1
Substituting in (III): 3 + y = 4 ∴ y = 1
(x, y) = (1, 1)
Practice Set 1.5
1Two numbers differ by 3. The sum of twice the smaller number and thrice the greater number is 19. Find the numbers.
Let the greater number be x and the smaller number be y.
From the first condition: x − y = 3 … (I)
From the second condition: 3x + 2y = 19 … (II)
Multiplying (I) by 2:
2x − 2y = 6 … (III)
Adding (II) and (III):
5x = 25 ∴ x = 5
Substituting in (I): 5 − y = 3 ∴ y = 2
The numbers are 5 and 2.
2Complete the following. The sides of a rectangle are 2x + y + 8 and 4x − y (opposite sides), and 2y and x + 4 (opposite sides). Find the values of x and y, and find the perimeter and area of the rectangle.
x = 12, y = 8; Perimeter = 112 units, Area = 640 sq. units
The textbook does not give a unit of length, so the perimeter is in units and the area in square units.
3The sum of father’s age and twice the age of his son is 70. If we double the age of the father and add it to the age of his son the sum is 95. Find their present ages.
Let the father’s present age be x years and the son’s present age be y years.
From the first condition: x + 2y = 70 … (I)
From the second condition: 2x + y = 95 … (II)
Adding (I) and (II):
3x + 3y = 165 ∴ x + y = 55 … (III)
Subtracting (I) from (II):
x − y = 25 … (IV)
Adding (III) and (IV):
2x = 80 ∴ x = 40
Substituting in (III): 40 + y = 55 ∴ y = 15
Father’s age = 40 years, son’s age = 15 years
4The denominator of a fraction is 4 more than twice its numerator. Denominator becomes 12 times the numerator, if both the numerator and the denominator are reduced by 6. Find the fraction.
Let the numerator be x and the denominator be y. The fraction is xy.
From the first condition: y = 2x + 4 … (I)
From the second condition: y − 6 = 12(x − 6)
∴ y − 6 = 12x − 72 ∴ y = 12x − 66 … (II)
From (I) and (II):
2x + 4 = 12x − 66
70 = 10x ∴ x = 7
Substituting in (I): y = 2 × 7 + 4 = 18
Check: 7 − 618 − 6 = 112, so the new denominator is 12 times the new numerator.
The fraction is 7/18.
5Two types of boxes A, B are to be placed in a truck having capacity of 10 tons. When 150 boxes of type A and 100 boxes of type B are loaded in the truck, it weighs 10 tons. But when 260 boxes of type A are loaded in the truck, it can still accommodate 40 boxes of type B, so that it is fully loaded. Find the weight of each type of box.
Let the weight of a box of type A be x kg and that of type B be y kg.
10 tons = 10 × 1000 kg = 10000 kg
From the first condition: 150x + 100y = 10000
Dividing by 50: 3x + 2y = 200 … (I)
From the second condition: 260x + 40y = 10000
Dividing by 20: 13x + 2y = 500 … (II)
Subtracting (I) from (II):
10x = 300 ∴ x = 30
Substituting in (I): 90 + 2y = 200
2y = 110 ∴ y = 55
Box A weighs 30 kg and box B weighs 55 kg.
6Out of 1900 km, Vishal travelled some distance by bus and some by aeroplane. Bus travels with average speed 60 km/hr and the average speed of aeroplane is 700 km/hr. It takes 5 hours to complete the journey. Find the distance Vishal travelled by bus.
Let the distance travelled by bus be x km and by aeroplane be y km.
From the first condition: x + y = 1900 … (I)
Time = distancespeed, so from the second condition:
x60 + y700 = 5
Multiplying by 2100 (the LCM of 60 and 700): 35x + 3y = 10500 … (II)
Multiplying (I) by 3:
3x + 3y = 5700 … (III)
Subtracting (III) from (II):
32x = 4800 ∴ x = 150
Then y = 1900 − 150 = 1750
Check: 15060 + 1750700 = 2.5 + 2.5 = 5 hours
Vishal travelled 150 km by bus.
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Problem Set 1
1Choose correct alternative for each of the following questions.
(1) To draw graph of 4x + 5y = 19, find y when x = 1. (A) 4 (B) 3 (C) 2 (D) −3
4(1) + 5y = 19 ∴ 5y = 15 ∴ y = 3
(B) 3
(2) For simultaneous equations in variables x and y, Dx = 49, Dy = −63, D = 7, then what is x? (A) 7 (B) −7 (C) 1/7 (D) −1/7
x = DxD = 497 = 7
(A) 7
(3) Find the value of 53−7−4. (A) −1 (B) −41 (C) 41 (D) 1
= 5 × (−4) − 3 × (−7) = −20 + 21 = 1
(D) 1
(4) To solve x + y = 3; 3x − 2y − 4 = 0 by determinant method, find D. (A) 5 (B) 1 (C) −5 (D) −1
The equations are x + y = 3 and 3x − 2y = 4.
D = 113−2 = (1 × −2) − (1 × 3) = −2 − 3 = −5
(C) −5
(5) ax + by = c and mx + ny = d and an ≠ bm, then these simultaneous equations have – (A) Only one common solution. (B) No solution. (C) Infinite number of solutions. (D) Only two solutions.
D = abmn = an − bm. Since an ≠ bm, D ≠ 0, so the equations have exactly one solution.
(A) Only one common solution.
2Complete the following table to draw the graph of 2x − 6y = 3.
2x − 6y = 3
x
−5
1.5
y
−136
0
(x, y)
(−5, −136)
(1.5, 0)
When x = −5: 2(−5) − 6y = 3 ∴ −6y = 13 ∴ y = −136
When y = 0: 2x = 3 ∴ x = 32 = 1.5
3Solve the following simultaneous equations graphically.
(1) 2x + 3y = 12; x − y = 1
2x + 3y = 12
x
0
6
−3
y
4
0
6
(x, y)
(0, 4)
(6, 0)
(−3, 6)
x − y = 1
x
1
0
5
y
0
−1
4
(x, y)
(1, 0)
(0, −1)
(5, 4)
The lines intersect at (3, 2).
(x, y) = (3, 2)
(2) x − 3y = 1; 3x − 2y + 4 = 0
x − 3y = 1
x
1
4
−5
y
0
1
−2
(x, y)
(1, 0)
(4, 1)
(−5, −2)
3x − 2y = −4
x
0
2
−4
y
2
5
−4
(x, y)
(0, 2)
(2, 5)
(−4, −4)
The lines intersect at (−2, −1).
(x, y) = (−2, −1)
(3) 5x − 6y + 30 = 0; 5x + 4y − 20 = 0
5x − 6y = −30
x
−6
6
−3
y
0
10
2.5
(x, y)
(−6, 0)
(6, 10)
(−3, 2.5)
5x + 4y = 20
x
4
8
−4
y
0
−5
10
(x, y)
(4, 0)
(8, −5)
(−4, 10)
The lines intersect at (0, 5).
(x, y) = (0, 5)
(4) 3x − y − 2 = 0; 2x + y = 8
3x − y = 2
x
0
1
3
y
−2
1
7
(x, y)
(0, −2)
(1, 1)
(3, 7)
2x + y = 8
x
0
4
3
y
8
0
2
(x, y)
(0, 8)
(4, 0)
(3, 2)
The lines intersect at (2, 4).
(x, y) = (2, 4)
(5) 3x + y = 10; x − y = 2
3x + y = 10
x
1
2
4
y
7
4
−2
(x, y)
(1, 7)
(2, 4)
(4, −2)
x − y = 2
x
2
0
5
y
0
−2
3
(x, y)
(2, 0)
(0, −2)
(5, 3)
The lines intersect at (3, 1).
(x, y) = (3, 1)
4Find the values of each of the following determinants.
(1) 4327
= 4 × 7 − 3 × 2 = 28 − 6 = 22
22
(2) 5−2−31
= 5 × 1 − (−2) × (−3) = 5 − 6 = −1
−1
(3) 3−114
= 3 × 4 − (−1) × 1 = 12 + 1 = 13
13
5Solve the following equations by Cramer’s method.
5p − 2q = −32; multiplying by 2: 10p − 4q = −3 … (II)
Adding (I) and (II):
20p = 2 ∴ p = 110
Substituting in (I): 1 + 4q = 5 ∴ q = 1
∴ 3x + 4y = 10 … (III) and 2x − 3y = 1 … (IV)
Multiplying (III) by 3 and (IV) by 4:
9x + 12y = 30 and 8x − 12y = 4
Adding:
17x = 34 ∴ x = 2
Substituting in (IV): 4 − 3y = 1 ∴ y = 1
(x, y) = (2, 1)
7Solve the following word problems.
(1) A two digit number and the number with digits interchanged add up to 143. In the given number the digit in unit’s place is 3 more than the digit in the ten’s place. Find the original number.
Let the digit in unit’s place be x and that in the ten’s place be y.
∴ the number = 10y + x
The number obtained by interchanging the digits is 10x + y
According to the first condition:
(10y + x) + (10x + y) = 143
∴ 11x + 11y = 143
x + y = 13 … (I)
From the second condition, digit in unit’s place = digit in ten’s place + 3
∴ x = y + 3 ∴ x − y = 3 … (II)
Adding equations (I) and (II):
2x = 16 ∴ x = 8
Putting this value of x in equation (I): 8 + y = 13 ∴ y = 5
The original number = 10y + x = 50 + 8 = 58
The original number is 58.
(2) Kantabai bought 1½ kg tea and 5 kg sugar from a shop. She paid ₹ 50 as return fare for rickshaw. Total expense was ₹ 700. Then she realised that by ordering online the goods can be bought with free home delivery at the same price. So next month she placed the order online for 2 kg tea and 7 kg sugar. She paid ₹ 880 for that. Find the rate of sugar and tea per kg.
Let the rate of tea be ₹ x per kg and that of sugar be ₹ y per kg.
Substituting in (I): 3x + 400 = 1300 ∴ 3x = 900 ∴ x = 300
Tea: ₹ 300 per kg; Sugar: ₹ 40 per kg
(3) To find number of notes that Anushka had, complete the following activity. Suppose that Anushka had x notes of ₹ 100 and y notes of ₹ 50 each. Anushka got ₹ 2500 from Anand in these denominations. If Anand had given her the amount by interchanging the number of notes, Anushka would have received ₹ 500 less than the previous amount.
From the first condition:
100x + 50y = 2500 ∴ 2x + y = 50 … equation I
If the numbers of notes are interchanged, she gets ₹ 2500 − ₹ 500 = ₹ 2000:
50x + 100y = 2000 ∴ x + 2y = 40 … equation II
Adding I and II:
3x + 3y = 90 ∴ x + y = 30 … (III)
Subtracting II from I:
x − y = 10 … (IV)
Adding (III) and (IV):
2x = 40 ∴ x = 20, and so y = 10
∴ The No. of notes (x, y) = (20, 10)
Anushka had 20 notes of ₹ 100 and 10 notes of ₹ 50.
(4) Sum of the present ages of Manish and Savita is 31. Manish’s age 3 years ago was 4 times the age of Savita. Find their present ages.
Let Manish’s present age be x years and Savita’s present age be y years.
From the first condition: x + y = 31 … (I)
3 years ago, their ages were (x − 3) and (y − 3) years.
From the second condition: x − 3 = 4(y − 3)
∴ x − 3 = 4y − 12 ∴ x − 4y = −9 … (II)
Subtracting (II) from (I):
5y = 40 ∴ y = 8
Substituting in (I): x + 8 = 31 ∴ x = 23
Manish: 23 years; Savita: 8 years
(5) In a factory the ratio of salary of skilled and unskilled workers is 5 : 3. Total salary of one day of both of them is ₹ 720. Find daily wages of skilled and unskilled workers.
Let the daily wage of a skilled worker be ₹ x and of an unskilled worker be ₹ y.
From the first condition: xy = 53 ∴ 3x = 5y ∴ 3x − 5y = 0 … (I)
From the second condition: x + y = 720 … (II)
Multiplying (II) by 5:
5x + 5y = 3600 … (III)
Adding (I) and (III):
8x = 3600 ∴ x = 450
Substituting in (II): 450 + y = 720 ∴ y = 270
Skilled worker: ₹ 450 per day; Unskilled worker: ₹ 270 per day
(6) Places A and B are 30 km apart and they are on a straight road. Hamid travels from A to B on bike. At the same time Joseph starts from B on bike, travels towards A. They meet each other after 20 minutes. If Joseph would have started from B at the same time but in the opposite direction (instead of towards A) Hamid would have caught him after 3 hours. Find the speed of Hamid and Joseph.
Let Hamid’s speed be x km/hr and Joseph’s speed be y km/hr.
20 minutes = 13 hour.
When they travel towards each other, together they cover 30 km in 1/3 hour:
13x + 13y = 30 ∴ x + y = 90 … (I)
When Joseph moves away in the same direction, Hamid gains the 30 km gap in 3 hours:
