Practice Set 5.1
1By using variables x and y form any five linear equations in two variables.
Sample answers (any equation of the form ax + by + c = 0, with a and b not both zero, is correct):
(1) x + y = 10 (2) 2x − y = 3 (3) 3x + 4y = 12 (4) x − 5y + 2 = 0 (5) x2 + y = 7
2Write five solutions of the equation x + y = 7.
Any pair of numbers whose sum is 7 is a solution, for example:
(0, 7), (1, 6), (2, 5), (3, 4), (−1, 8)
3Solve the following sets of simultaneous equations.
(i) x + y = 4 ; 2x − 5y = 1
x + y = 4 …(I) 2x − 5y = 1 …(II)
Multiply (I) by 5: 5x + 5y = 20 …(III)
Add (II) and (III): 7x = 21 ⇒ x = 3
Put x = 3 in (I): 3 + y = 4 ⇒ y = 1
x = 3, y = 1
(ii) 2x + y = 5 ; 3x − y = 5
Adding the two equations: 5x = 10 ⇒ x = 2
Put x = 2 in 2x + y = 5: 4 + y = 5 ⇒ y = 1
x = 2, y = 1
(iii) 3x − 5y = 16 ; x − 3y = 8
3x − 5y = 16 …(I) x − 3y = 8 …(II)
Multiply (II) by 3: 3x − 9y = 24 …(III)
(I) − (III): 4y = −8 ⇒ y = −2
Put y = −2 in (II): x + 6 = 8 ⇒ x = 2
x = 2, y = −2
(iv) 2y − x = 0 ; 10x + 15y = 105
From the first equation, x = 2y
Substitute in the second: 10(2y) + 15y = 105 ⇒ 35y = 105 ⇒ y = 3
x = 2 × 3 = 6
x = 6, y = 3
(v) 2x + 3y + 4 = 0 ; x − 5y = 11
2x + 3y = −4 …(I) x − 5y = 11 …(II)
Multiply (II) by 2: 2x − 10y = 22 …(III)
(I) − (III): 13y = −26 ⇒ y = −2
Put y = −2 in (II): x + 10 = 11 ⇒ x = 1
x = 1, y = −2
(vi) 2x − 7y = 7 ; 3x + y = 22
2x − 7y = 7 …(I) 3x + y = 22 …(II)
Multiply (II) by 7: 21x + 7y = 154 …(III)
(I) + (III): 23x = 161 ⇒ x = 7
Put x = 7 in (II): 21 + y = 22 ⇒ y = 1
x = 7, y = 1
Practice Set 5.2
1In an envelope there are some 5 rupee notes and some 10 rupee notes. Total amount of these notes together is 350 rupees. Number of 5 rupee notes are less by 10 than number of 10 rupee notes. Then find the number of 5 rupee and 10 rupee notes.
Let the number of 5 rupee notes = x and 10 rupee notes = y.
Total amount: 5x + 10y = 350 ⇒ x + 2y = 70 …(I)
5 rupee notes are 10 fewer: x = y − 10 …(II)
Substitute (II) in (I): y − 10 + 2y = 70 ⇒ 3y = 80 ⇒ y = 803
This is not a whole number, so the data as printed cannot describe a real number of notes.
With the data as printed there is no whole-number answer. If the 5 rupee notes were fewer by 20, the answer would be 10 notes of 5 rupees and 30 notes of 10 rupees (5 × 10 + 10 × 30 = 350).
2The denominator of a fraction is 1 less than twice its numerator. If 1 is added to numerator and denominator respectively, the ratio of numerator to denominator is 3 : 5. Find the fraction.
Let the fraction be xy. Then y = 2x − 1 …(I)
x + 1y + 1 = 35 ⇒ 5x + 5 = 3y + 3 ⇒ 5x − 3y = −2 …(II)
Substitute (I) in (II): 5x − 3(2x − 1) = −2 ⇒ −x + 3 = −2 ⇒ x = 5
y = 2 × 5 − 1 = 9
The fraction is 59 (check: 610 = 35)
3The sum of ages of Priyanka and Deepika is 34 years. Priyanka is elder to Deepika by 6 years. Then find their today’s ages.
Let Priyanka’s age = x, Deepika’s age = y. x + y = 34 …(I), x − y = 6 …(II)
Adding: 2x = 40 ⇒ x = 20; then y = 34 − 20 = 14
Priyanka: 20 years Deepika: 14 years
4The total number of lions and peacocks in a certain zoo is 50. The total number of their legs is 140. Then find the number of lions and peacocks in the zoo.
Let lions = x, peacocks = y. x + y = 50 …(I)
A lion has 4 legs, a peacock 2: 4x + 2y = 140 ⇒ 2x + y = 70 …(II)
(II) − (I): x = 20; then y = 50 − 20 = 30
Lions: 20 Peacocks: 30
5Sanjay gets fixed monthly income. Every year there is a certain increment in his salary. After 4 years, his monthly salary was Rs. 4500 and after 10 years his monthly salary became 5400 rupees, then find his original salary and yearly increment.
Let the original salary = x rupees and yearly increment = y rupees.
x + 4y = 4500 …(I) x + 10y = 5400 …(II)
(II) − (I): 6y = 900 ⇒ y = 150; then x = 4500 − 600 = 3900
Original salary: Rs. 3900 Yearly increment: Rs. 150
6The price of 3 chairs and 2 tables is 4500 rupees and price of 5 chairs and 3 tables is 7000 rupees, then find the total price of 2 chairs and 2 tables.
Let price of a chair = x, a table = y. 3x + 2y = 4500 …(I), 5x + 3y = 7000 …(II)
(I) × 3: 9x + 6y = 13500; (II) × 2: 10x + 6y = 14000
Subtracting: x = 500; from (I): 1500 + 2y = 4500 ⇒ y = 1500
Price of 2 chairs and 2 tables = 2(500) + 2(1500) = 4000
Rs. 4000
7The sum of the digits in a two-digit number is 9. The number obtained by interchanging the digits exceeds the original number by 27. Find the two-digit number.
Let the units digit = x and tens digit = y. Original number = 10y + x
x + y = 9 …(I)
(10x + y) − (10y + x) = 27 ⇒ 9x − 9y = 27 ⇒ x − y = 3 …(II)
Adding: 2x = 12 ⇒ x = 6; y = 3
The number is 36 (63 − 36 = 27)
8*In △ABC, the measure of ∠A is equal to the sum of the measures of ∠B and ∠C. Also the ratio of measures of ∠B and ∠C is 4 : 5. Then find the measures of angles of the triangle.
∠A + ∠B + ∠C = 180° and ∠A = ∠B + ∠C ⇒ 2∠A = 180° ⇒ ∠A = 90°
Let ∠B = x, ∠C = y. x + y = 90 …(I) and xy = 45 ⇒ 5x − 4y = 0 …(II)
(I) × 4 + (II): 9x = 360 ⇒ x = 40; y = 50
∠A = 90°, ∠B = 40°, ∠C = 50°
9*Divide a rope of length 560 cm into 2 parts such that twice the length of the smaller part is equal to 13 of the larger part. Then find the length of the larger part.
Let the smaller part = x cm and larger part = y cm. x + y = 560 …(I)
2x = y3 ⇒ y = 6x …(II)
Substitute: x + 6x = 560 ⇒ x = 80; y = 480
Larger part = 480 cm (smaller part 80 cm)
10In a competitive examination, there were 60 questions. The correct answer would carry 2 marks, and for incorrect answer 1 mark would be subtracted. Yashwant had attempted all the questions and he got total 90 marks. Then how many questions he got wrong?
Let correct answers = x and wrong answers = y. x + y = 60 …(I)
2x − y = 90 …(II)
Adding: 3x = 150 ⇒ x = 50; y = 10
He got 10 questions wrong (50 correct: 100 − 10 = 90)
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Problem Set 5
1Choose the correct alternative answers for the following questions.
- (i) If 3x + 5y = 9 and 5x + 3y = 7 then what is the value of x + y? — (A) 2
Adding: 8x + 8y = 16, so x + y = 2.
- (ii) ‘When 5 is subtracted from length and breadth of the rectangle, the perimeter becomes 26.’ Mathematical form? — (C) x + y = 23
2[(x − 5) + (y − 5)] = 26 ⇒ x + y − 10 = 13 ⇒ x + y = 23
- (iii) Ajay is younger than Vijay by 5 years. Sum of their ages is 25 years. What is Ajay’s age? — (C) 10
a + (a + 5) = 25 ⇒ a = 10
2Solve the following simultaneous equations.
(i) 2x + y = 5 ; 3x − y = 5
Adding: 5x = 10 ⇒ x = 2; 4 + y = 5 ⇒ y = 1. x = 2, y = 1
(ii) x − 2y = −1 ; 2x − y = 7
From the first, x = 2y − 1. Then 2(2y − 1) − y = 7 ⇒ 3y = 9 ⇒ y = 3; x = 5. x = 5, y = 3
(iii) x + y = 11 ; 2x − 3y = 7
x = 11 − y. Then 2(11 − y) − 3y = 7 ⇒ 22 − 5y = 7 ⇒ y = 3; x = 8. x = 8, y = 3
(iv) 2x + y = −2 ; 3x − y = 7
Adding: 5x = 5 ⇒ x = 1; 2 + y = −2 ⇒ y = −4. x = 1, y = −4
(v) 2x − y = 5 ; 3x + 2y = 11
First × 2: 4x − 2y = 10. Adding to the second: 7x = 21 ⇒ x = 3; 6 − y = 5 ⇒ y = 1. x = 3, y = 1
(vi) x − 2y = −2 ; x + 2y = 10
Adding: 2x = 8 ⇒ x = 4; 4 + 2y = 10 ⇒ y = 3. x = 4, y = 3
3By equating coefficients of variables, solve the following equations.
(i) 3x − 4y = 7 ; 5x + 2y = 3
Second × 2: 10x + 4y = 6. Add to the first: 13x = 13 ⇒ x = 1
5(1) + 2y = 3 ⇒ 2y = −2 ⇒ y = −1
x = 1, y = −1
(ii) 5x + 7y = 17 ; 3x − 2y = 4
First × 2: 10x + 14y = 34; second × 7: 21x − 14y = 28
Adding: 31x = 62 ⇒ x = 2; 6 − 2y = 4 ⇒ y = 1
x = 2, y = 1
(iii) x − 2y = −10 ; 3x − 5y = −12
First × 3: 3x − 6y = −30
(3x − 5y) − (3x − 6y) = −12 − (−30) ⇒ y = 18
x − 36 = −10 ⇒ x = 26
x = 26, y = 18
(iv) 4x + y = 34 ; x + 4y = 16
Second × 4: 4x + 16y = 64. Subtract the first: 15y = 30 ⇒ y = 2
x + 8 = 16 ⇒ x = 8
x = 8, y = 2
4Solve the following simultaneous equations.
(i) x3 + y4 = 4 ; x2 − y4 = 1
Multiply the first by 12: 4x + 3y = 48 …(I); the second by 4: 2x − y = 4 …(II)
(II) × 3: 6x − 3y = 12. Add to (I): 10x = 60 ⇒ x = 6
12 − y = 4 ⇒ y = 8
x = 6, y = 8
(ii) x3 + 5y = 13 ; 2x + y2 = 19
Multiply the first by 3: x + 15y = 39 …(I); the second by 2: 4x + y = 38 …(II)
(I) × 4: 4x + 60y = 156. Subtract (II): 59y = 118 ⇒ y = 2
x + 30 = 39 ⇒ x = 9
x = 9, y = 2
(iii) 2x + 3y = 13 ; 5x − 4y = −2
Let 1x = p and 1y = q: 2p + 3q = 13 …(I), 5p − 4q = −2 …(II)
(I) × 4: 8p + 12q = 52; (II) × 3: 15p − 12q = −6
Adding: 23p = 46 ⇒ p = 2; 4 + 3q = 13 ⇒ q = 3
1x = 2 ⇒ x = 12; 1y = 3 ⇒ y = 13
x = 12, y = 13
5*A two digit number is 3 more than 4 times the sum of its digits. If 18 is added to this number, the sum is equal to the number obtained by interchanging the digits. Find the number.
Let the units digit = x and tens digit = y. Number = 10y + x
10y + x = 4(x + y) + 3 ⇒ 6y − 3x = 3 ⇒ 2y − x = 1 …(I)
10y + x + 18 = 10x + y ⇒ 9x − 9y = 18 ⇒ x − y = 2 …(II)
Adding (I) and (II): y = 3; then x = 5
The number is 35 (4 × 8 + 3 = 35 and 35 + 18 = 53)
6The total cost of 8 books and 5 pens is 420 rupees and the total cost of 5 books and 8 pens is 321 rupees. Find the cost of 1 book and 2 pens.
Let a book cost x and a pen cost y rupees. 8x + 5y = 420 …(I), 5x + 8y = 321 …(II)
Adding: 13x + 13y = 741 ⇒ x + y = 57 …(III)
Subtracting: 3x − 3y = 99 ⇒ x − y = 33 …(IV)
From (III) and (IV): x = 45, y = 12
Cost of 1 book and 2 pens = 45 + 2 × 12 = 69
Rs. 69
7*The ratio of incomes of two persons is 9 : 7. The ratio of their expenses is 4 : 3. Every person saves rupees 200, find the income of each.
Let the incomes be 9x and 7x, and expenses 4y and 3y.
Saving = income − expense: 9x − 4y = 200 …(I), 7x − 3y = 200 …(II)
(I) × 3: 27x − 12y = 600; (II) × 4: 28x − 12y = 800
Subtracting: x = 200
Incomes: Rs. 1800 and Rs. 1400
8*If the length of a rectangle is reduced by 5 units and its breadth is increased by 3 units, then the area is reduced by 8 square units. If length is reduced by 3 units and breadth is increased by 2 units, then the area will increase by 67 square units. Find the length and breadth of the rectangle.
Let length = x and breadth = y units.
(x − 5)(y + 3) = xy − 8 ⇒ 3x − 5y − 15 = −8 ⇒ 3x − 5y = 7 …(I)
(x − 3)(y + 2) = xy + 67 ⇒ 2x − 3y − 6 = 67 ⇒ 2x − 3y = 73 …(II)
(I) × 2: 6x − 10y = 14; (II) × 3: 6x − 9y = 219. Subtracting: y = 205
3x = 7 + 5 × 205 = 1032 ⇒ x = 344
Length = 344 units, Breadth = 205 units
9*The distance between two places A and B on road is 70 kilometers. A car starts from A and the other from B. If they travel in the same direction, they will meet after 7 hours. If they travel towards each other they will meet after 1 hour, then find their speeds.
Let the speed of the car from A = x km/hr and from B = y km/hr (x > y, since the car from A catches up).
Same direction: the gap of 70 km closes at (x − y) km/hr in 7 hours: 7(x − y) = 70 ⇒ x − y = 10 …(I)
Towards each other: (x + y) × 1 = 70 ⇒ x + y = 70 …(II)
Adding: 2x = 80 ⇒ x = 40; y = 30
Car from A: 40 km/hr Car from B: 30 km/hr
10*The sum of a two digit number and the number obtained by interchanging its digits is 99. Find the number.
Let the units digit = x and tens digit = y.
(10y + x) + (10x + y) = 99 ⇒ 11x + 11y = 99 ⇒ x + y = 9
Only one condition is given, so there is no single answer: any two-digit number whose digits add up to 9 works.
18, 27, 36, 45, 54, 63, 72, 81 or 90
For example, 36 + 63 = 99 and 45 + 54 = 99. For 90 the interchanged number is 09 = 9, and 90 + 9 = 99.
