155 Suggestion to exclude certain letters and numbers from captchas

CAPTCHA (Completely Automated Public Turing test to tell Computers and Humans Apart)

1) The main function of captchas is to distinguish between real people and artificial

bots.

2) The design of captchas should be approached holistically, with an emphasis on

striking a balance between usability, accessibility, security, and efficacy. Combining many tactics, such as using warped letters, extra obstacles, or other strategies like picture recognition or mathematically straightforward operations like addition, subtraction, or multiplication, may result in a more effective strategy.

3) It's important to keep in mind that captcha technology is always developing, and

researchers and developers are continuously looking for substitute ways to confirm that users are human, including invisible captchas that depend on behavioural analysis or biometrics. With these developments, usability and security are being balanced.

4) Additionally, it's critical to continually review and adapt captcha tactics since new

developments in machine learning and artificial intelligence may eventually make some solutions less effective. 

 

5) Here, we're trying to make the point that certain characters might not be utilized

while creating captcha software since they are unclear.

6) Following characters are confusing:

a) Looks the same in both upper and lower case:

i) 'c', 'j', 'k', 'o', 'p', 's', 'u', 'v', 'w', 'x', 'y', and 'z'

ii) 'C', 'J', 'K', 'O', 'P', 'S', 'U', 'V', 'W', 'X', 'Y', and 'Z'

Using these characters for making captchas is not recommended since they are nearly identical in upper- and lower-case.

b) Certain capital and lowercase letters are difficult to distinguish:

i) ) In specific fonds, the uppercase letter 'I' (capital 'i') occasionally

appears as the lowercase letter 'l' (lower 'L'). The user is confused about which one has to be taken. Therefore, while designing captchas, upper case letter 'I' and lower case letter 'l' must be eliminated.

c) Uncertainty over the characters for the numbers '0' (zero) and the upper and

lower case 'O' ('o'):

i) When the letter 'o' is typed in lower case, higher case, or the number '0'

in the captcha, there is a lot of misunderstanding. As a result, they might not be included when making captchas.

 

7) Because of their closeness in appearance, several characters—including the

capital and lowercase versions of the letters 'c', 'k', 'o', 'p', 'u', 'v', 'w', 'x', 'y', and 'z' — are excluded. 

 

8) This tactic must be carefully implemented, striking a balance between user ease

and security. Before implementing any substantial modifications to captcha systems, it is usually a good idea to undertake user testing and evaluate the impact on security.

9) We want to make it simpler for users to understand the CAPTCHA and decrease

the probability of mistakes or annoyance by getting rid of these potentially perplexing characters. However, it's crucial to keep in mind that CAPTCHAs should strike a balance between being difficult enough to stave off automated bots while still being clear to the majority of human users. A CAPTCHA may lose its effectiveness if it gets too simple to solve.

Here are my opinions. Everyone is welcome to share their worthwhile thoughts.

ANIL SATPUTE

154-NCERT-10-3-Pair of Linear Equations in Two Variables - Ex-3.7

NCERT

10th Mathematics

Exercise 3.7

Topic: 3 Pair of Linear Equations in Two Variables

Click here for ⇨ NCERT-10-3-Pair of Linear Equations in Two Variables-Ex-3.6

EXERCISE 3.7

Explanation:

1) Here, let x and y be two variables.

2) Convert the variables suitably to get linear equations in two variables.

3) Then solve these equations to get the values of our variables. 

Solution:

Q1. The ages of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Ani, and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.

1) Let Ani's age be x and Biju's age be y.

2) The age of Ani's father Dharam is 2x.

3) The age of Biju's sister Cathy is y/2.

4) As the ages of Cathy and Dharam differ by 30 years, as Dharam is older than

Cathy, we have,

2x - y/2 = 30

4x - y = 60 ---------------------- equation 1

5) Age difference between Ani and Biju is 3 years. Here we need to take 2 cases.

6) Case 1: a) Ani is older than Biju.

x - y = 3 ---------------------- equation 2.

7) Subtract equation 2 from equation 1, and we get,

4x - y = 60

  x - y =   3

     (-)   (+)   (-)

    ----------------------- 

3x     =  57

x = 57/3

x = 19 ---------------------- equation 3.

8) Put the value of x = 19 from equation 3 in equation 2, and we get,

x - y = 3

19 - y = 3

y = 19 - 3

y = 16 ---------------------- equation 4

9) If Ani is older than Biju, Ani's age is 19 years and Biju's age is 16 years.

10) Case 2: b) Biju is older than Ani.

y - x = 3

- x + y = 3 ---------------------- equation 5.

11) Add equation 5 to equation 1, and we get,

4x - y = 60

- x + y =  3

    ----------------------- 

3x     =  63

x = 63/3

x = 21 ---------------------- equation 6.

12) Put the value of x = 21 from equation 6 in equation 5, and we get,

- x + y = 3

-21 + y = 3

y = 21 + 3

y = 24 ---------------------- equation 4

13) If Biju is older than Ani, Ani's age is 21 years and Biju's age is 24 years.

14) If Ani is older than Biju, Ani's age is 19 years and Biju's age is 16 years.

Q2. One says, “Give me a hundred, friend! I shall then become twice as rich as you”. The other replies, “If you give me ten, I shall be six times as rich as you”. Tell me what is the amount of their (respective) capital? [From the Bijaganita of Bhaskara II] [Hint : x + 100 = 2(y – 100), y + 10 = 6(x – 10)].

1) Let the first friend has Rs x and the other friend has Rs y.

2) If the first friend gets Rs 100 from the second friend, the new amount with them,

a) First friend has (x + 100) 

b) Second friend has (y - 100), so, according to first condition, we have,

x + 100 = 2(y - 100)

x + 100 = 2y - 200
x - 2y = - 200 - 100
x - 2y = - 300

x = 2y - 300 ------------ equation 1

3) If the second friend gets Rs 10 from the first friend, the new amount with them,

a) First friend has (x - 10) 

b) Second friend has (y + 10), so, according to second condition, we have,

y + 10 = 6(x - 10)

y + 10 = 6x - 60
6x - y = 10 + 60
6x - y = 70 ------------ equation 2

4) Put the value of x = 2y - 300 from equation 1 in equation 2, we get,

6x - y = 70

6(2y - 300) - y = 70

12y - 1800 - y = 70

12y - y = 70 + 1800

11y = 1870

y = 1870/11

y = 170 -------------------------- equation 3.

5) Put the value of y = 170 from equation 3 in equation 2, we get,

6x - y = 70

6x - 170 = 70

6x = 70 + 170

6x = 240 

x = 240/6 

x = 40 -------------------------- equation 4.

6) The first friend has Rs 40 and the second friend has Rs 170.

Q3. A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.

1) Let uniform speed be x km/h and the time taken to cover the distance be y hrs.

2) We know that distance = speed x time, so the total distance = xy km.

3) When the train travels 10 km/h faster, then the time will be 2 hrs less, so,

(x + 10)(y - 2) = xy

xy - 2x + 10y - 20 = xy
- 2x + 10y - 20 = 0
- 2x + 10y = 20 ------------ equation 1

4) When the train travels 10 km/h slower, then the time will be 3 hrs more, so,

(x - 10)(y + 3) = xy

xy + 3x - 10y - 30 = xy
3x - 10y - 30 = 0
3x - 10y = 30 ------------ equation 2

5) Add equation 1 to equation 2, and we get,

   3x - 10y = 30

- 2x + 10y = 20

     -------------------------- 

    x           = 50

x = 50 ---------------------- equation 3.

6) Put the value of x = 50 from equation 3 in equation 2, we get,

3x - 10y = 30

3(50) - 10y = 30

150 - 10y = 30

10y = 150 - 30

10y = 120

y = 120/10

y = 12 -------------------------- equation 4.

7) The uniform speed is 50 km/h and the time taken to cover the distance is 12 hrs.

8) So, the total distance traveled by the train is 50 x 12 = 600 km.

Q4. The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.

1) Let the number of rows be x and the number of students in a row is y.

2) Total number of students = Number of rows x Number of students in a row = xy.

3) When 3 students are more in a row, then the number of rows will be less by 1,

(x - 1)(y + 3) = xy

xy + 3x - y - 3 = xy
3x - y - 3 = 0
3x - y = 3 ------------ equation 1

4) When 3 students are less in a row, then the number of rows will be more by 2,

(x + 2)(y - 3) = xy

xy - 3x + 2y - 6 = xy
-3x + 2y - 6 = 0
-3x + 2y = 6 ------------ equation 2

5) Add equation 2 to equation 1, and we get,   

   3x -   y = 3

- 3x + 2y = 6

     -------------------------- 

            y = 9

y = 9 ---------------------- equation 3.

6) Put the value of y = 9 from equation 3 in equation 1, we get,

3x - y = 3

3x - 9 = 3

3x = 9 + 3

3x = 12

x = 12/3

x = 4 -------------------------- equation 4.

7) The number of rows is 4 and the number of students in a row is 9.

8) So, the number of students in the class = 9 x 4 = 36.

Q5. In a ∆ ABC, < C = 3 < B = 2 (< A + < B). Find the three angles.

1) Let the < A be x, and < B be y.

2) According to the first condition, 

< C = 3y -------------------------- equation 1.

3) We know that the sum of the angles of a triangle is 180 degrees, 

< A + < B + < C = 180

x + y + 3y = 180

x + 4y = 180 -------------------------- equation 2.

4) According to the given condition,

< C = 2 (< A + < B)

3y = 2(x + y)

3y = 2x + 2y
3y - 2y = 2x
y = 2x -------------------------- equation 3.  

5) Put the value of y = 2x from equation 3 in equation 2, we get,

x + 4y = 180

x + 4(2x) = 180

x + 8x = 180

9x = 180

x = 20 -------------------------- equation 4.

6) Put the value of x = 20 from equation 4 in equation 3, we get,

y = 2x

y = 2(20)

y = 40 -------------------------- equation 5.

7) The angles are < A = 20⁰, < B = 40⁰, and < C = 120⁰.

Q6. Draw the graphs of the equations 5x – y = 5 and 3x – y = 3. Determine the coordinates of the vertices of the triangle formed by these lines and the y-axis.

1) First equation: 5x - y = 5, so, y = 5x - 5.

2) Take ant two points as follows:

3) Second equation: 3x - y = 3, so, y = 3x - 3.

4) Take ant two points as follows:

5) Now we will draw the graph:

6) So, the coordinates of the vertices of a triangle ABC are A(1, 0), B(0, -3), and C(0, -5).

Q7. Solve the following pair of linear equations:

(i) px + qy = p – q;     qx – py = p + q

(ii) ax + by = c;         bx + ay = 1 + c

(iii) (x/a) - (y/b) = 0;     ax + by = a² + b²

(iv) (a – b)x + (a + b) y = a² – 2ab – b²;     (a + b)(x + y) = a² + b²

(v) 152x – 378y = – 74;     –378x + 152y = – 604

Solution:

(i) px + qy = p – q;     qx – py = p + q

1) The given equations are 

px + qy = p – q -------------------------- equation 1

qx – py = p + q -------------------------- equation 2

2) Multiply equation 1 by p, we get,

p(px + qy) = p(p – q)

p²x + pqy = p² – pq -------------------------- equation 3

3) Multiply equation 2 by q, we get,

q(qx – py) = q(p + q)

q²x - pqy = pq + q² -------------------------- equation 4

4) Add equation 4 to equation 3, we get, 

p²x + pqy = p² – pq

q²x -  pqy = pq + q²

 ---------------------------------------- 

 (p² + q²) x      =  (p² + q²)

x = (p² + q²)/(p² + q²)

x = 1 -------------------------- equation 5

5) Put the value of x = 1 from equation 5 in equation 1, and we get, 

px + qy = p – q

p(1) + qy = p – q

   p + qy = p – q

qy = – q

  y = – q/q

  y = – 1 -------------------------- equation 6

6) So, x = 1 and y = - 1. 

 

(ii) ax + by = c;         bx + ay = 1 + c

1) The given equations are 

ax + by = c -------------------------- equation 1

bx + ay = 1 + c -------------------------- equation 2

2) Multiply equation 1 by a, we get,

a(ax + by) = ac

a²x + aby = ac -------------------------- equation 3

3) Multiply equation 2 by b, we get,

b(bx + ay) = b(c + 1)

b²x + aby = bc + b -------------------------- equation 4

4) Subtract equation 4 from equation 3, we get, 

a²x + aby = ac 

b²x + aby = bc + b

   (-)       (-)        (-) 
 ---------------------------------------- 

 (a² - b²) x      =  (ac - bc - b)

x = [c(a - b) - b]/(a² - b²) -------------------------- equation 5

5) Put the value of x = [c(a - b) - b]/(a² - b²) from equation 5 in equation 1, we get, 

ax + by = c

a[c(a - b) - b]/(a² - b²) + by = c

[ac(a - b) - ab]/(a² - b²) + by = c

by = c - [ac(a - b) - ab]/(a² - b²)

by = {c(a² - b²) - [ac(a - b) - ab]}/(a² - b²)

by = [ca² - cb² - a²c + abc + ab]/(a² - b²)

by = [abc - cb² + ab]/(a² - b²)

by = [bc(a - b) + ab]/(a² - b²)

by = b[c(a - b) + a]/(a² - b²)
  y = [c(a - b) + a]/(a² - b²) -------------------------- equation 6

6) So, x = [c(a - b) - b]/(a² - b²) and y = [c(a - b) + a]/(a² - b²). 

(iii) (x/a) - (y/b) = 0;     ax + by = a² + b²

1) The given equations are 

(x/a) - (y/b) = 0 -------------------------- equation 1

ax + by = a² + b² -------------------------- equation 2 

2) So, from equation 1, we have

(x/a) - (y/b) = 0

(x/a) = (y/b)

x = (ay/b) -------------------------- equation 3

3) Put the value of x = (ay/b) from equation 3 in equation 2, we get, 

ax + by = a² + b²

a(ay/b) + by = a² + b²

(a²y/b) + by = a² + b²

(a²y + b²y)/b = a² + b²

(a² + b²)y/b = a² + b²

(a² + b²)y = b(a² + b²)

y = b(a² + b²)/(a² + b²)

y = b -------------------------- equation 4

4) Put the value of y = b from equation 4 in equation 3, we get,

x = (ay/b)

x = (ab/b)
x = a 

5) So, x = a and y = b. 

(iv) (a – b)x + (a + b) y = a² – 2ab – b²;     (a + b)(x + y) = a² + b²

1) The given equations are 

(a – b)x + (a + b) y = a² – 2ab – b² -------------------------- equation 1

(a + b)(x + y) = a² + b² -------------------------- equation 2 

2) So, from equation 2, we have,   

(a + b)(x + y) = a² + b²

(a + b)x + (a + b)y = a² + b² -------------------------- equation 3

3) Subtract equation 1 from equation 3, we get,

(a + b)x + (a + b)y = a² + b²

(a – b)x + (a + b)y = a² – 2ab – b²

     (-)           (-)               (-) 
 --------------------------------------------------------- 

       2bx                 =  2b² + 2ab

2bx = 2b(b + a)
x = (a + b) -------------------------- equation 4 

4) Put the value of x = (a + b) from equation 4 in equation 1, we get, 

(a – b)x + (a + b) y = a² – 2ab – b²

(a – b)(a + b) + (a + b) y = a² – 2ab – b²

(a² – b²) + (a + b) y = (a² – b²) – 2ab

(a + b) y = – 2ab

y = – 2ab/(a + b) -------------------------- equation 5.

5) So, x = (a + b) and y = – 2ab/(a + b). 

(v) 152x – 378y = – 74;     –378x + 152y = – 604 

1) The given equations are

152x – 378y = – 74 -------------------------- equation 1  

–378x + 152y = – 604 -------------------------- equation 2

2) Add equation 2 to equation 1, we get, 

152x – 378y = – 74

–378x + 152y = – 604

 ---------------------------------------------- 

      – 226 x – 226 y = – 678

x + y = (-678)/(-226)

x + y = 3 -------------------------- equation 3

3) Subtract equation 2 from equation 1, we get,

152x – 378y = – 74

–378x + 152y = – 604

     (+)        (-)            (+) 
 --------------------------------------------------------- 

530x – 530y = 530

x – y = 530/530

x – y = 1 

x = y + 1 -------------------------- equation 4 

4) Put the value of x = y + 1 from equation 4 in equation 3, we get,

x + y = 3

(y + 1) + y = 3

2y + 1 = 3

2y = 3 - 1

y = 2/2

y = 1 -------------------------- equation 5.

5) Put the value of y = 1 from equation 5 in equation 4, we get,

x = y + 1

x = 1 + 1

x = 2 -------------------------- equation 6.

6) So, x = 2 and y = 1.

Q8. ABCD is a cyclic quadrilateral. Find the angles of the cyclic quadrilateral.

1) We know that the sum of the opposite angles of a cyclic quadrilateral is 180⁰.

2) So according to the diagram, we have, 

(-7x + 5) + (3y - 5) = 180 -------------------------- equation 1.

(-4x) + (4y + 20) = 180 -------------------------- equation 2. 

3) From equation 1 we get, 

(-7x + 5) + (3y - 5) = 180

-7x + 3y = 180

7x - 3y = -180 -------------------------- equation 3.

4) From equation 2 we get, 

(-4x) + (4y + 20) = 180

(-x) + (y + 5) = 45
-x + y = 45 - 5
-x + y = 40
y = x + 40 -------------------------- equation 4.

5) Put the value of y = x + 40 from equation 4 in equation 3, we get,

7x - 3y = -180

7x - 3(x + 40) = -180

7x - 3x - 120 = -180

4x = 120 -180

4x = - 60 

x = - 60/4

x = - 15 -------------------------- equation 5.

6) Put the value of x = - 15 from equation 5 in equation 4, we get,

y = x + 40

y = - 15 + 40

y = 25 -------------------------- equation 6.

7) So, here, x = - 15 and y = 25, therefore we have,

< A = (4y + 20) = (4(25) + 20) = (100 + 20) = 120, so , < A =120.

< B = (3y - 5) = (3(25) - 5) = (75 - 5) = 70, so, < B = 70.

< C = (- 4x) = (- 4(- 15)) = 60, so, < C = 60.

< D = (- 7x + 5) = (- 7(- 15) + 5) = (105 + 5) = 110, < D = 110.

Click here for ⇨ NCERT-10-4-Quadratic Equations - Ex-4.1

ANIL SATPUTE

153-NCERT-10-3-Pair of Linear Equations in Two Variables - Ex-3.6

NCERT

10th Mathematics

Exercise 3.6

Topic: 3 Pair of Linear Equations in Two Variables

Click here for ⇨ NCERT-10-3-Pair of Linear Equations in Two Variables-Ex-3.5

EXERCISE 3.6

Q1. Solve the following pairs of equations by reducing them to a pair of linear equations: 

(i) 1/(2x) + 1/(3y) = 2;   1/(3x) + 1/(2y) = 13/6,

(ii) (2/√x) + (3/√y) = 2;  (4/√x) - (9/√y) = - 1,

(iii) 4/x + 3y = 14;  3/x - 4y = 23,

(iv) 5/(x - 1) + 1/(y - 2) = 2;  6/(x - 1) - 3/(y - 2) = 1,

(v) (7x - 2y)/xy = 5;  (8x + 7y)/xy = 15,

(vi) 6x + 3y = 6xy;  2x + 4y = 5xy,

(vii) 10/(x + y) + 2/(x - y) = 4;  15/(x + y) - 5/(x - y) = -2

(viii) 1/(3x + y) + 1/(3x - y) = 3/4;  1/[2(3x + y)] - 1/[2(3x - y)] = -1/8.

Explanation:

1) Here, let x and y be two variables.

2) Convert the variables suitably to get linear equations in two variables.

3) Then solve these equations to get the values of our variables. 

Solution:

(i) 1/(2x) + 1/(3y) = 2;   1/(3x) + 1/(2y) = 13/6.

1) Given equations are

1/(2x) + 1/(3y) = 2;  1/(3x) + 1/(2y) = 13/6

2) In the equation, 1/(2x) + 1/(3y) = 2, put 1/x = p, and 1/y = q, so we have,

1/(2x) + 1/(3y) = 2

p/2 + q/3 = 2

(3p + 2q)/6 = 2

3p + 2q = 2(6)

3p + 2q = 12 ---------------equation 1

3) In the equation, 1/(3x) + 1/(2y) = 13/6, put 1/x = p, and 1/y = q, so we have,

1/(3x) + 1/(2y) = 13/6 

p/3 + q/2 = 13/6

(2p + 3q)/6 = 13/6

2p + 3q = 13

2p = 13 - 3q

p = (13 - 3q)/2 ---------------equation 2

4) Substitute the value of p = (13 - 3q)/2 from equation 2 in equation 1, we get

3p + 2q = 12

3(13 - 3q)/2 + 2q = 12

(39 - 9q)/2 + 2q = 12

(39 - 9q + 4q)/2 = 12

(39 - 5q) = 12 (2)

(39 - 5q) = 24

5q = 39 - 24

q = 15/5

q = 3 ------------ equation 3

5) Put the value of q = 3 from equation 3 in equation 2, and we get

p = (13 - 3q)/2

p = (13 - 3(3))/2

p = (13 - 9)/2

p = (4)/2

p = 2 ------------ equation 4

6) The value of p = 2 and the value of q = 3.

7) As, p = 1/x and p = 2, we have x = 1/2

8) As, q = 1/y and q = 3, we have y = 1/3.

(ii) (2/√x) + (3/√y) = 2;  (4/√x) - (9/√y) = - 1

1) Given equations are

(2/√x) + (3/√y) = 2;  (4/√x) - (9/√y) = - 1

2) In the equation, (2/√x) + (3/√y) = 2, put 1/√x = p, and 1/√y = q, so we have,

(2/√x) + (3/√y) = 2

2p + 3q = 2 ---------------equation 1

3) In the equation, (4/√x) - (9/√y) = - 1, put 1/√x = p, and 1/√y = q, so we have,

(4/√x) - (9/√y) = - 1 

4p - 9q = - 1

4p = 9q - 1

p = (9q - 1)/4 ---------------equation 2

4) Substitute the value of p = (9q - 1)/4 from equation 2 in equation 1, we get,

2p + 3q = 2

2(9q - 1)/4 + 3q = 2

(9q - 1)/2 + 3q = 2

(9q - 1 + 6q)/2 = 2 

15q = 4 + 1

q = 5/15

q = 1/3 ------------ equation 3

5) Put the value of q = 1/3 from equation 3 in equation 2, and we get

p = (9q - 1)/4

p = (9(1/3) - 1)/4

p = (3 - 1)/4

p = (2)/4

p = 1/2 ------------ equation 4

6) The value of p = 1/2 and the value of q = 1/3.

7) As, p = 1/√x and p = 1/2, we have √x = 2, so x = 4.

8) As, q = 1/√y and q = 1/3, we have √y = 3, so y = 9.

(iii) 4/x + 3y = 14;  3/x - 4y = 23

1) Given equations are

4/x + 3y = 14;  3/x - 4y = 23

2) In the equation, 4/x + 3y = 14, put 1/x = p, so we have,

4/x + 3y = 14

4p + 3y = 14 ---------------equation 1

3) In the equation, 1/(3x) + 1/(2y) = 13/6, put 1/x = p, and 1/y = q, so we have,

3/x - 4y = 23 

3p - 4y = 23

3p = 23 + 4y

p = (23 + 4y)/3 ---------------equation 2

4) Substitute the value of p = (23 + 4y)/3 from equation 2 in equation 1, we get,

4p + 3y = 14

(4(23 + 4y)/3) + 3y = 14

(92 + 16y)/3 + 3y = 14

(92 + 16y + 9y)/3 = 14

(92 + 25y)/3 = 14 

(92 + 25y)    = 14 (3) 

(92 + 25y)    = 42

(92 + 25y)    = 42

25y = 42 - 92

25y = - 50

y = - 50/25

y = - 2 ------------ equation 3

5) Put the value of y = - 2 from equation 3 in equation 2, and we get

p = (23 + 4y)/3

p = (23 + 4(-2))/3

p = (23 - 8)/3
p = (15)/3
p = 5 ------------ equation 4

6) The value of p = 5 and the value of y = - 2. 

7) As, p = 1/x and p = 5, we have x = 1/5 and y = - 2.

(iv) 5/(x - 1) + 1/(y - 2) = 2;  6/(x - 1) - 3/(y - 2) = 1

1) Given equations are

5/(x - 1) + 1/(y - 2) = 2;  6/(x - 1) - 3/(y - 2) = 1

2) In the equation, 5/(x - 1) + 1/(y - 2) = 2, put 1/(x - 1) = p, and 1/(y - 2) = q, 

so we have,

5/(x - 1) + 1/(y - 2) = 2

5p + q = 2

q = 2 - 5p ---------------equation 1

3) In the equation, 6/(x - 1) - 3/(y - 2) = 1, put 1/(x - 1) = p, and 1/(y - 2) = q, 

so we have,

6/(x - 1) - 3/(y - 2) = 1 

6p - 3q = 1 ---------------equation 2

4) Substitute the value of q = (2 - 5p) from equation 1 in equation 2, we get,

6p - 3q = 1

6p - 3(2 - 5p) = 1

6p - 6 + 15p = 1

6p + 15p = 1 + 6

21p = 7

p = 7/21

p = 1/3 ---------------equation 3

5) Put the value of p = 1/3 from equation 3 in equation 1, and we get

q = 2 - 5p

q = 2 - 5(1/3)

q = (6 - 5)/3
q = 1/3 ------------ equation 4

6) The value of p = 1/3 and the value of q = 1/3.

7) As, p = 1/(x - 1) and p = 1/3, we have (x - 1) = 3, so x = 4.

8) As, q = 1/(y - 2) and q = 1/3, we have (y - 2) = 3, so y = 5.

(v) (7x - 2y)/xy = 5;  (8x + 7y)/xy = 15

1) First we will simplify the first equation (7x - 2y)/xy = 5:

(7x - 2y)/xy = 5

(7x)/xy - (2y)/xy = 5

7/y - 2/x = 5

- 2/x + 7/y = 5, here, put 1/x = p and 1/y = q, we get,

- 2p + 7q = 5
2p - 7q = - 5
2p = 7q - 5
  p = (7q - 5)/2 ---------------equation 1

2) First we will simplify the second equation (8x + 7y)/xy = 15:

(8x + 7y)/xy = 15

(8x)/xy + (7y)/xy = 15

8/y + 7/x = 15

7/x + 8/y = 15, here, put 1/x = p and 1/y = q, we get,
7p + 8q = 15 ---------------equation 2 

3) Substitute the value of p = (7q - 5)/2 from equation 1 in equation 2, we get,

7p + 8q = 15

7(7q - 5)/2 + 8q = 15

(49q - 35)/2 + 8q = 15

(49q - 35 + 16q)/2 = 15

(49q + 16q - 35) = 15 (2) 

(65q - 35) = 30

65q = 35 + 30

65q = 65

q = 1 ---------------equation 3

4) Put the value of q = 1 from equation 3 in equation 1, and we get

p = (7q - 5)/2

p = (7(1) - 5)/2

p = (7 - 5)/2

p = 2/2 

p = 1 ------------ equation 4

5) The value of p = 1 and the value of q = 1.

6) As, p = 1/x and p = 1, we have x = 1.

7) As, q = 1/y and q = 1, we have y = 1.

(vi) 6x + 3y = 6xy;  2x + 4y = 5xy

1) First we will simplify the first equation 6x + 3y = 6xy:

6x + 3y = 6xy

(6x)/xy + (3y)/xy = 6

6/y + 3/x = 6

3/x + 6/y = 6, here, put 1/x = p and 1/y = q, we get,

3p + 6q = 6
  p + 2q = 2
  p = 2 - 2q ---------------equation 1

2) First we will simplify the second equation 2x + 4y = 5xy:

2x + 4y = 5xy

(2x)/xy + (4y)/xy = 5

2/y + 4/x = 5

4/x + 2/y = 5, here, put 1/x = p and 1/y = q, we get,
4p + 2q = 5 ---------------equation 2

3) Substitute the value of p = 2 - 2q from equation 1 in equation 2, we get,

4p + 2q = 5

4(2 - 2q) + 2q = 5

8 - 8q + 2q = 5

8 - 6q = 5

 - 6q = 5 - 8

 - 6q = - 3

     q = (- 3)/(- 6) 

q = 1/2 ---------------equation 3

4) Put the value of q = 1/2 from equation 3 in equation 1, and we get

p = 2 - 2q

p = 2 - 2(1/2)

p = 2 - 1

p = 1 ------------ equation 4

5) The value of p = 1 and the value of q = 1/2.

6) As, p = 1/x and p = 1, we have x = 1.

7) As, q = 1/y and q = 1/2, we have y = 2.

(vii) 10/(x + y) + 2/(x - y) = 4;  15/(x + y) - 5/(x - y) = - 2

1) Given equations are

10/(x + y) + 2/(x - y) = 4;  15/(x + y) - 5/(x - y) = - 2

2) In the equation, 10/(x + y) + 2/(x - y) = 4, put 1/(x + y) = p, and 1/(x - y) = q, 

so we have,

10/(x + y) + 2/(x - y) = 4

10p + 2q = 4

5p + q = 2 

q = 2 - 5p ---------------equation 1

3) In the equation, 15/(x + y) - 5/(x - y) = - 2, put 1/(x + y) = p, and 1/(x - y) = q, 

so we have,

15/(x + y) - 5/(x - y) = - 2 

15p - 5q = - 2 ---------------equation 2

4) Substitute the value of q = (2 - 5p) from equation 1 in equation 2, we get,

15p - 5q = - 2

15p - 5(2 - 5p) = - 2

15p - 10 + 25p = - 2

15p + 25p = - 2 + 10

40p = 8

p = 8/40

p = 1/5 ---------------equation 3

5) Put the value of p = 1/5 from equation 3 in equation 1, and we get

q = 2 - 5p

q = 2 - 5(1/5)

q = 2 - 1
q = 1 ------------ equation 4

6) The value of p = 1/5 and the value of q = 1.

7) As, p = 1/(x + y) and p = 1/5, we have 

(x + y) = 5 ------------ equation 5 

8) As, q = 1/(x - y) and q = 1, we have 

(x - y) = 1 ------------ equation 6

9) Adding equation 5 and equation 6, we get,

x + y + x - y = 5 + 1

2x = 6, so x = 3 ------------ equation 7

10) put x = 3 from equation 7 in equation 5, and we get,

(x + y) = 5

(3 + y) = 5

y = 5 - 3

y = 2 ------------ equation 8.

11) So, x = 3, and y = 2.

 

(viii) 1/(3x + y) + 1/(3x - y) = 3/4;  1/[2(3x + y)] - 1/[2(3x - y)] = -1/8

1) Given equations are

1/(3x + y) + 1/(3x - y) = 3/4;  1/[2(3x + y)] - 1/[2(3x - y)] = -1/8

2) In the equation, 1/(3x + y) + 1/(3x - y) = 3/4, put 1/(3x + y) = p, and 1/(3x - y) = q, 

so we have,

1/(3x + y) + 1/(3x - y) = 3/4

p + q = 3/4 ---------------equation 1

3) In the equation, 1/[2(3x + y)] - 1/[2(3x - y)] = -1/8, put 1/(3x + y) = p, and 

1/(3x - y) = q, so we have,

1/[2(3x + y)] - 1/[2(3x - y)] = -1/8

p/2 - q/2 = -1/8 

p - q = -1/4 ---------------equation 2

4) Adding equation 1, and equation 2, we get,

p + q = 3/4

p - q = -1/4

    ----------------------- 

2p      =  2/4 

p  = 2/8

p = 1/4 ---------------equation 3

5) Put the value of p = 1/4 from equation 3 in equation 1, and we get

p + q = 3/4

1/4 + q = 3/4

q = 3/4 - 1/4

q = (3 - 1)/4

q = 2/4

q = 1/2 ------------ equation 4

6) The value of p = 1/4 and the value of q = 1/2.

7) As, p = 1/(3x + y) and p = 1/4, we have 

(3x + y) = 4 ------------ equation 5 

8) As, q = 1/(3x - y) and q = 1/2, we have 

(3x - y) = 2 ------------ equation 6

9) Adding equation 5 and equation 6, we get,

3x + y = 4

3x - y = 2

    ----------------------- 

6x      =  6 

x = 1 ---------------equation 7

10) put x = 1 from equation 7 in equation 5, and we get,

(3x + y) = 4

(3 + y) = 4

y = 4 - 3

y = 1 ------------ equation 8.

11) So, x = 1, and y = 1.

 

Q2. Formulate the following problems as a pair of equations, and hence find their solutions:

(i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.

 

(ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.

 

(iii) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.

Explanation:

1) Here, let x and y be two variables.

2) Apply the given conditions and frame the equations.

3) We will get two equations from the above two conditions, then solve these

equations to get the values of x and y. 

Solution:

(i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.

1) Let Ritu's speed in still water be x km/h.

2) Let the speed of the current be y km/h.

3) Ritu's speed of rowing downstream (x + y) km/h.

4) Ritu's speed of rowing upstream (x - y) km/h.

5) According to the first condition, as Ritu can row downstream 20 km in 2 hrs,  

2(x + y) = 20

(x + y) = 10 -------------------------- equation 1.

6) According to the first condition, as Ritu can row upstream 4 km in 2 hrs,  

2(x - y) = 4

(x - y) = 2-------------------------- equation 2.

7) Adding equation 1 and equation 2, we get,

(x + y) + (x - y) = 10 + 2

2x = 12

x = 12/2

x = 6 -------------------------- equation 3.

8) Put the value of x = 6 from equation 3 in equation 1, we get,

(x + y) = 10

(6 + y) = 10

y = 10 - 6

y = 4 -------------------------- equation 4.

9) Ritu's speed of rowing in still water is 6 km/h and the speed of the current is 4

km/h.

 

(ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.

1) Let the number of days taken by one woman to finish the work = x. So work done

by a woman in 1 day will be 1/x

2) Let the number of days taken by one man to finish the work = y. So work done

by a man in 1 day will be 1/y

3) As 2 women and 5 men finish the work in 4 days, so we have,

(2/x + 5/y) = 1/4, put 1/x = p and 1/y = q, we have,

2p + 5q = 1/4

4(2p + 5q) = 1

8p + 20q = 1 -------------------------- equation 1. 

4) As 3 women and 6 men finish the work in 3 days, so we have,

(3/x + 6/y) = 1/3, put 1/x = p and 1/y = q, we have,

3p + 6q = 1/3
3(p + 2q) = 1/3

(p + 2q) = 1/9

p = 1/9 - 2q -------------------------- equation 2.

5) Put the value of p = (1/9 - 2q) from equation 2 in equation 1, we get,

8p + 20q = 1

8(1/9 - 2q) + 20q = 1

8/9 - 16q + 20q = 1

8/9 + 4q = 1

4q = 1 - 8/9

4q = (9 - 8)/9

  q = 1/36 -------------------------- equation 3.

6) Put the value of q = 1/36 from equation 3 in equation 2, and we get,

p = 1/9 - 2q

p = 1/9 - 2(1/36)

p = 1/9 - 1/18

p = 2/18 - 1/18
p = (2 - 1)/18
p = 1/18 -------------------------- equation 4.

7) The value of p = 1/18 and the value of q = 1/36.

8) As, p = 1/x and p = 1/18, we have x = 18.

9) As, q = 1/y and q = 1/36, we have y = 36. 

10) Number of days taken by a woman to finish the work is 18 and by a man is 36.

(iii) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.

1) Let the speed of the train be x km/h.

2) Let the speed of the bus be y km/h.

3) Total distance traveled by Roohi is 300 km.

4) According to the first condition, she traveled 60 km by train and 240 km by bus in

4 hrs, so, we have,

60/x + 240/y = 4, put 1/x = p and 1/y = q, we have,

60p + 240q = 4

4(15p + 60q) = 4

15p + 60q = 1

15(p + 4q) = 1

p + 4q = 1/15
p = ((1/15) - 4q) -------------------------- equation 1.

5) According to the second condition, she traveled 100 km by train and 200 km by

bus in 4 hrs and 10 minutes i.e. 4 + (10/60) = 4 + (1/6) = 25/6 hrs, so, we have,

100/x + 200/y = 25/6, put 1/x = p and 1/y = q, we have,

100p + 200q = 25/6

25(4p + 8q) = 25/6

(4p + 8q) = 1/6 -------------------------- equation 2.

6) Put the value of p = ((1/15) - 4q) from equation 1 in equation 2, we get,

(4p + 8q) = 1/6

(4((1/15) - 4q) + 8q) = 1/6

(4/15 - 16q + 8q) = 1/6

(4/15 - 8q) = 1/6 

8q = 4/15 - 1/6 

8q = (8/30) - (5/30)

8q = 3/30

8q = 1/10

q = 1/(10(8)) 

q = 1/80 -------------------------- equation 3.

7) Put the value of q = 1/80 from equation 3 in equation 1, we get,

p = ((1/15) - 4q)

p = ((1/15) - 4(1/80))

p = ((1/15) - 1/20))

p = ((4/60) - 3/60))

p = (4 - 3)/60

p = 1/60  -------------------------- equation 4.

8) The value of p = 1/60 and the value of q = 1/80.

9) As, p = 1/x and p = 1/60, we have x = 60.

10) As, q = 1/y and q = 1/80, we have y = 80.

11) The speed of the train is 60 km/h. 

12) The speed of the bus is 80 km/h.

Click here for ⇨ NCERT-10-3-Pair of Linear Equations in Two Variables - Ex-3.7

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