In the previous sections we saw some advanced cases in the solution of two equations with two variables. We also saw some solved examples. In this section, we will see more solved examples.
Solved example 14.6
The distance travelled in t seconds by an object starting with a speed of u metres per second, and moving along a straight line, with speed increasing at the rate of a metres/second every second is given by d = ut + 1⁄2 at2 metres. An object moving in this manner travels 10 metres in 2 seconds and 28 metres in 4 seconds. With what speed did it start? At what rate does it's speed change?
Solution:
• The equation d = ut + 1⁄2 at2 gives us the distance d travelled by an object. But to use the equation, we must have three parameters: initial speed u, acceleration a, and time t.
• The object travelled with a certain initial speed. This speed is not given to us. But two readings were taken during it's travel
• According to the first reading, the object travelled 10 metres during a time interval of 2 seconds. That is., d= 10, and t = 2.
1. So we can write: 10 = (u × 2) + (1⁄2 × a × 22) ⇒ 10 = 2u + 2a
• According to the second reading, it travelled 28 metres in 4 seconds. That is., d = 28, and t = 4
2. So we can write: 28 = (u × 4) + (1⁄2 × a × 42) ⇒ 28 = 4u + 8a
3. Multiply (1) by 2 ⇒ (1) × 2 ⇒ 20 = 4u + 4a
4. Isolate 4u from (3): 4u = 20 - 4a
5. Substitute this 4u in (2): 28 = (20 - 4a) + 8a ⇒ 28 = 20 + 4a ⇒ 4a = 28 - 20 ⇒ 4a = 8
∴ a = 8/4 = 2
6. substitute this value of a in (1). We get: 10 = 2u + 2 × 2 ⇒ 10 = 2u + 4 ⇒ 2u = 6
∴ u = 6/2 = 3
7. So we can write: Initial speed = u = 3 meter/sec and, acceleration a = 2 metre/ sec2
8. Check: Use (2): 28 = 4u + 8a: 28 = 4 × 3 + 8 × 2 ⇒ 28 = 12 + 16 ⇒ 28 = 28
Solved example 14.7
Four times a number added to three times a second number gives 43. Two times the second number subtracted from three times the first number gives 11. What are the numbers?
Solution:
1. Let the first number = x
2. Let the second number = y
3. Use condition 1: 4 times first number + 3 times second number = 43: 4x + 3y = 43
4. Use condition 2: 3 times first number - 2 times second number = 11: 3x - 2y = 11
5. Multiply (3) by 3 (Here 3 is the coefficient of the x term in the 'other equation')
⇒ (3) × 3 ⇒ 12x + 9y = 129
6. Multiply (4) by 4 (Here 4 is the coefficient of the x term in the 'other equation')
⇒ (4) × 4 ⇒ 12x - 8y = 44
7. Isolate 12x from (6) ⇒ 12x = 44 + 8y
8. Substitute this 12x in (5)
⇒ (44 + 8y) + 9y = 129 ⇒ 44 + 17y = 129 ⇒ 17y = 129 - 44 ⇒ 17y = 85
∴ y = 85/17 = 5
9. Substitute this value of y in (3)
We get: 4x + 3 × 5 = 43 ⇒ 4x + 15 = 43 ⇒ 4x = 43 -15 ⇒ 4x = 28
∴ x = 28/4 = 7
10. Thus we can write: First number = x = 7, and second number = y = 5
11. Check: Use condition 1: 3 times first number - 2 times second number = 11: (3 × 7) - (2 × 5) = 21 - 10 = 11
Solved example 14.8
The sum of the digits of a two digit number is 11. The number got by interchanging the digits is 27 more than the original number. What is the number ?
Solution:
• We have a two digit number. Let the digits be x and y
• Let x be in the tens place and y be in the ones place
• Then the value of the number = 10x +y
1. Use condition 1: Sum of the digits = 11: x + y = 11
2. Use condition 2: The number got by interchanging the digits is 27 more than the original number
• The value of the number got by interchanging the digits will be 10y + x
• This value is 27 more than the value of the original number. So we can write:
10y + x = (10x +y) + 27 ⇒ 10y - 10x + x - y = 27 ⇒ 9y - 9x = 27 ⇒ y - x = 3 (dividing both sides by 9)
3. Isolate y from (2): y = 3 + x
4. Substitute this value of y in (1): x + (3 + x) = 11 ⇒ 2x + 3 = 11 ⇒ 2x = 8
∴ x = 8/2 = 4
5. Substitute this value of x in (1). We get: 4 + y = 11 ⇒ y = 11 - 4 = 7
6. So we can write: the original number = 47, and the new number = 74
7. Check: second number - first number = 74 - 47 = 27
Solved example 14.9
Four years ago, Mr.A's age was three times the age of Mr.B. After two years, Mr. A's age will be two times the age of Mr.B. What are their ages now?
Solution:
1. Let the present age of Mr.A = x
2. Let the present age of Mr.B = y
3. Use condition 1: Age 4 years ago:
• Age of Mr.A 4 years ago = x-4
• Age of Mr.B 4 years ago = y-4
• Applying the condition, we get: x - 4 = 3(y-4) ⇒ x - 4 = 3y - 12 ⇒ 3y - x = 8
4. Use condition 2: Age after 2 years:
• Age of Mr.A after 2 years = x +2
• Age of Mr.B after 2 years = y +2
• Applying the condition, we get: x + 2 = 2(y +2) ⇒ x + 2 = 2y + 4 ⇒ x - 2y = 2
5. Isolate x from (4): x = 2 + 2y
6. Substitute this value of x in (3). We get: 3y -(2 + 2y) = 8 ⇒ 3y -2 - 2y = 8 ⇒ y -2 = 8 ⇒ y = 10
7. Substitute this value of y in (3). We get: 3 × 10 - x = 8 ⇒ 30 - x = 8 ⇒ x = 30 - 8 = 22
8. So we can write: Present age of Mr.A = x = 22, and present age of Mr.B = y = 10
9. Check: Use (4): 22 - 2 × 10 = 2 ⇒ 22 - 20 = 2 ⇒ 2 = 2
Solved example 14.10
If the length of a rectangle is increased by 5 m, and the breadth decreased by 3 m, the area will decrease by 5 sq.m. If the length is increased by 3 m, and breadth increased by 2 m, the area will increase by 50 sq.m. What is the original length and breadth?
Solution:
1. Let the original length = x
2. Let the original breadth = y
3. So original area = xy
4. Use condition 1: (x+5) (y-3) = xy-5 ⇒ xy+5y-3x-15 = xy-5 ⇒ 5y-3x = 10
5. Use condition 2: (x+3) (y+2) = xy+50 ⇒ xy+3y+2x+6 = xy+50 ⇒ 3y+2x = 44
6. Multiply (4) by 3 (Here 3 is the coefficient of the y term in the 'other equation')
⇒ (4) × 3 ⇒ 15y - 9x = 30
7. Multiply (5) by 5 (Here 5 is the coefficient of the y term in the 'other equation')
⇒ (5) × 5 ⇒ 15y + 10x = 220
8. Isolate 15y from (7). We get: 15y = 220-10x
9. Substitute this value of 15y in (6). We get: (220-10x)-9x = 30 ⇒ 220-19x = 30 ⇒ 190 = 19x
∴ x = 190/19 = 10
10. Substitute this value of x in (4). We get: 5y-3 10 = 10 ⇒ 5y-30 = 10 ⇒ 5y = 40
∴ y = 40/5 = 8
11. So we can write: length and breadth of the original rectangle are x = 10 m and y = 8 m respectively
12. Check: Use (5): 3 8 + 2 10 = 44 ⇒ 24 + 20 = 44 ⇒ 44 = 44
In the next section we will see another type of equation.
Solved example 14.6
The distance travelled in t seconds by an object starting with a speed of u metres per second, and moving along a straight line, with speed increasing at the rate of a metres/second every second is given by d = ut + 1⁄2 at2 metres. An object moving in this manner travels 10 metres in 2 seconds and 28 metres in 4 seconds. With what speed did it start? At what rate does it's speed change?
Solution:
• The equation d = ut + 1⁄2 at2 gives us the distance d travelled by an object. But to use the equation, we must have three parameters: initial speed u, acceleration a, and time t.
• The object travelled with a certain initial speed. This speed is not given to us. But two readings were taken during it's travel
• According to the first reading, the object travelled 10 metres during a time interval of 2 seconds. That is., d= 10, and t = 2.
1. So we can write: 10 = (u × 2) + (1⁄2 × a × 22) ⇒ 10 = 2u + 2a
• According to the second reading, it travelled 28 metres in 4 seconds. That is., d = 28, and t = 4
2. So we can write: 28 = (u × 4) + (1⁄2 × a × 42) ⇒ 28 = 4u + 8a
3. Multiply (1) by 2 ⇒ (1) × 2 ⇒ 20 = 4u + 4a
4. Isolate 4u from (3): 4u = 20 - 4a
5. Substitute this 4u in (2): 28 = (20 - 4a) + 8a ⇒ 28 = 20 + 4a ⇒ 4a = 28 - 20 ⇒ 4a = 8
∴ a = 8/4 = 2
6. substitute this value of a in (1). We get: 10 = 2u + 2 × 2 ⇒ 10 = 2u + 4 ⇒ 2u = 6
∴ u = 6/2 = 3
7. So we can write: Initial speed = u = 3 meter/sec and, acceleration a = 2 metre/ sec2
8. Check: Use (2): 28 = 4u + 8a: 28 = 4 × 3 + 8 × 2 ⇒ 28 = 12 + 16 ⇒ 28 = 28
Solved example 14.7
Four times a number added to three times a second number gives 43. Two times the second number subtracted from three times the first number gives 11. What are the numbers?
Solution:
1. Let the first number = x
2. Let the second number = y
3. Use condition 1: 4 times first number + 3 times second number = 43: 4x + 3y = 43
4. Use condition 2: 3 times first number - 2 times second number = 11: 3x - 2y = 11
5. Multiply (3) by 3 (Here 3 is the coefficient of the x term in the 'other equation')
⇒ (3) × 3 ⇒ 12x + 9y = 129
6. Multiply (4) by 4 (Here 4 is the coefficient of the x term in the 'other equation')
⇒ (4) × 4 ⇒ 12x - 8y = 44
7. Isolate 12x from (6) ⇒ 12x = 44 + 8y
8. Substitute this 12x in (5)
⇒ (44 + 8y) + 9y = 129 ⇒ 44 + 17y = 129 ⇒ 17y = 129 - 44 ⇒ 17y = 85
∴ y = 85/17 = 5
9. Substitute this value of y in (3)
We get: 4x + 3 × 5 = 43 ⇒ 4x + 15 = 43 ⇒ 4x = 43 -15 ⇒ 4x = 28
∴ x = 28/4 = 7
10. Thus we can write: First number = x = 7, and second number = y = 5
11. Check: Use condition 1: 3 times first number - 2 times second number = 11: (3 × 7) - (2 × 5) = 21 - 10 = 11
Solved example 14.8
The sum of the digits of a two digit number is 11. The number got by interchanging the digits is 27 more than the original number. What is the number ?
Solution:
• We have a two digit number. Let the digits be x and y
• Let x be in the tens place and y be in the ones place
• Then the value of the number = 10x +y
1. Use condition 1: Sum of the digits = 11: x + y = 11
2. Use condition 2: The number got by interchanging the digits is 27 more than the original number
• The value of the number got by interchanging the digits will be 10y + x
• This value is 27 more than the value of the original number. So we can write:
10y + x = (10x +y) + 27 ⇒ 10y - 10x + x - y = 27 ⇒ 9y - 9x = 27 ⇒ y - x = 3 (dividing both sides by 9)
3. Isolate y from (2): y = 3 + x
4. Substitute this value of y in (1): x + (3 + x) = 11 ⇒ 2x + 3 = 11 ⇒ 2x = 8
∴ x = 8/2 = 4
5. Substitute this value of x in (1). We get: 4 + y = 11 ⇒ y = 11 - 4 = 7
6. So we can write: the original number = 47, and the new number = 74
7. Check: second number - first number = 74 - 47 = 27
Solved example 14.9
Four years ago, Mr.A's age was three times the age of Mr.B. After two years, Mr. A's age will be two times the age of Mr.B. What are their ages now?
Solution:
1. Let the present age of Mr.A = x
2. Let the present age of Mr.B = y
3. Use condition 1: Age 4 years ago:
• Age of Mr.A 4 years ago = x-4
• Age of Mr.B 4 years ago = y-4
• Applying the condition, we get: x - 4 = 3(y-4) ⇒ x - 4 = 3y - 12 ⇒ 3y - x = 8
4. Use condition 2: Age after 2 years:
• Age of Mr.A after 2 years = x +2
• Age of Mr.B after 2 years = y +2
• Applying the condition, we get: x + 2 = 2(y +2) ⇒ x + 2 = 2y + 4 ⇒ x - 2y = 2
5. Isolate x from (4): x = 2 + 2y
6. Substitute this value of x in (3). We get: 3y -(2 + 2y) = 8 ⇒ 3y -2 - 2y = 8 ⇒ y -2 = 8 ⇒ y = 10
7. Substitute this value of y in (3). We get: 3 × 10 - x = 8 ⇒ 30 - x = 8 ⇒ x = 30 - 8 = 22
8. So we can write: Present age of Mr.A = x = 22, and present age of Mr.B = y = 10
9. Check: Use (4): 22 - 2 × 10 = 2 ⇒ 22 - 20 = 2 ⇒ 2 = 2
Solved example 14.10
If the length of a rectangle is increased by 5 m, and the breadth decreased by 3 m, the area will decrease by 5 sq.m. If the length is increased by 3 m, and breadth increased by 2 m, the area will increase by 50 sq.m. What is the original length and breadth?
Solution:
1. Let the original length = x
2. Let the original breadth = y
3. So original area = xy
4. Use condition 1: (x+5) (y-3) = xy-5 ⇒ xy+5y-3x-15 = xy-5 ⇒ 5y-3x = 10
5. Use condition 2: (x+3) (y+2) = xy+50 ⇒ xy+3y+2x+6 = xy+50 ⇒ 3y+2x = 44
6. Multiply (4) by 3 (Here 3 is the coefficient of the y term in the 'other equation')
⇒ (4) × 3 ⇒ 15y - 9x = 30
7. Multiply (5) by 5 (Here 5 is the coefficient of the y term in the 'other equation')
⇒ (5) × 5 ⇒ 15y + 10x = 220
8. Isolate 15y from (7). We get: 15y = 220-10x
9. Substitute this value of 15y in (6). We get: (220-10x)-9x = 30 ⇒ 220-19x = 30 ⇒ 190 = 19x
∴ x = 190/19 = 10
10. Substitute this value of x in (4). We get: 5y-3 10 = 10 ⇒ 5y-30 = 10 ⇒ 5y = 40
∴ y = 40/5 = 8
11. So we can write: length and breadth of the original rectangle are x = 10 m and y = 8 m respectively
12. Check: Use (5): 3 8 + 2 10 = 44 ⇒ 24 + 20 = 44 ⇒ 44 = 44
In the next section we will see another type of equation.
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