Simultaneous (Linear) Equations — Test Yourself

If 3y - 2x = 1 and 3x + 4y = 24, the value of x and y are :

1. 3 and 4

2. 3 and 3

3. 4 and 4

4. 4 and 3

Answer

Given,

First equation :

⇒ 3y - 2x = 1

⇒ 2x - 3y + 1 = 0 ..............(1)

Second equation :

⇒ 3x + 4y = 24

⇒ 3x + 4y - 24 = 0 ................(2)

Multiplying eq. (1) by 3, we get :

⇒ 6x - 9y + 3 = 0 ................(3)

Multiplying eq. (2) by 2, we get :

⇒ 6x + 8y - 48 = 0 ................(4)

Subtracting eq. (3) from (4) we get,

⇒ (6x + 8y - 48) - (6x - 9y + 3) = 0 - 0

⇒ 6x + 8y - 48 - 6x + 9y - 3 = 0

⇒ 17y - 51 = 0

⇒ 17y = 51

⇒ y = $\dfrac{51}{17}$

⇒ y = 3.

Substituting value of y in eq. (2) we get

⇒ 3x + 4(3) - 24 = 0

⇒ 3x + 12 - 24 = 0

⇒ 3x - 12 = 0

⇒ 3x = 12

⇒ x = $\dfrac{12}{3}$

⇒ x = 4

∴ x = 4 and y = 3

Hence, option 4 is the correct option.

On simplifying the equation 7(x - 1) - 6y = 5(x - y), we get:

1. 2x - y = 7

2. 2x + y = 7

3. x - 2y = 7

4. 3x - 2y = 7

Answer

Given,

⇒ 7(x - 1) - 6y = 5(x - y)

⇒ 7x - 7 - 6y = 5x - 5y

⇒ 7x - 7 - 6y - 5x + 5y = 0

⇒ 2x - 7 - y = 0

⇒ 2x - y = 7

Hence, option 1 is the correct option.

Solution of equation $\dfrac{2}{x} + \dfrac{3}{y}$ + 1 = 0 and $\dfrac{3}{x} + \dfrac{5}{y}$ + 2 = 0 is:

1. -1 and -1

2. -1 and -2

3. 1 and -1

4. none of these

Answer

Given, $\dfrac{2}{x} + \dfrac{3}{y}$ + 1 = 0 and $\dfrac{3}{x} + \dfrac{5}{y}$ + 2 = 0

Let $\dfrac{1}{x}$ be u and $\dfrac{1}{y}$ be v.

⇒ 2u + 3v + 1 = 0 ...................(1)

⇒ 3u + 5v + 2 = 0 ...................(2)

Multiplying eq. (1) by 3, we get,

⇒ 6u + 9v + 3 = 0 ................(3)

Multiplying eq. (2) by 2, we get,

⇒ 6u + 10v + 4 = 0 ................(4)

Subtracting eq. (3) from (4) we get,

⇒ 6u + 10v + 4 - (6u + 9v + 3) = 0 - 0

⇒ 6u + 10v + 4 - 6u - 9v - 3 = 0

⇒ v + 1 = 0

⇒ v = -1

⇒ $\dfrac{1}{y}$ = -1

⇒ y = -1

Substituting value of v in eq. (2) we get

⇒ 3u + 5(-1) + 2 = 0

⇒ 3u - 5 + 2 = 0

⇒ 3u - 3 = 0

⇒ 3u = 3

⇒ u = $\dfrac{3}{3}$

⇒ u = 1

⇒ $\dfrac{1}{x}$ = 1

⇒ x = 1

∴ x = 1 and y = -1

Hence, option 3 is the correct option.

In a two digit number, the sum of the digit is 11 and the tens digit minus unit digit is 5. The number is :

1. 38

2. 83

3. 29

4. 92

Answer

Let digit at ten's place be x.

It is given that the sum of the digit is 11.

So, digit at one's place be 11 - x.

And, the tens digit minus unit digit is 5.

⇒ x - (11 - x) = 5

⇒ x - 11 + x = 5

⇒ 2x - 11 = 5

⇒ 2x = 5 + 11

⇒ 2x = 16

⇒ x = $\dfrac{16}{2}$

⇒ x = 8.

The digit at one's place = 11 - x = 11 - 8 = 3.

∴ Number = 10 × 8 + 3 = 80 + 3 = 83.

Hence, option 2 is the correct option.

Based on the given information, find the height of the glass.

1. 16 cm

2. 14 cm

3. 12 cm

4. 10 cm

Answer

Total stack of 5 glasses = 34 cm

Bottom 2 glasses = 19 cm

∴ Top 3 uniform glasses = 34 - 19 = 15

3 uniform glasses = 15

∴ 1 uniform glass = $\dfrac{15}{3}$ = 5 cm

Bottom 2 glasses = 19 cm

∴ Bottom glass = 19 - 5 = 14 cm.

Hence, option 2 is the correct option.

Statement 1: x = 5 and y = 2 are the solution of equations x - y = 3 and 2x + y = 11.

Statement 2: x = 5 and y = 2 will be the solutions of the given equations if for each equation, the values on left hand side and right hand side are the same.

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

Given,

x = 5

y = 2

First equation :

⇒ x - y = 3

Substituting x = 5 and y = 2 in L.H.S. of first equation

⇒ 5 - 2

⇒ 3.

L.H.S. = R.H.S.

Second equation :

⇒ 2x + y = 11

Substituting x = 5 and y = 2 in L.H.S. of second equation

⇒ 2(5) + 2

⇒ 10 + 2

⇒ 12

L.H.S. ≠ R.H.S.

So, statement 1 is false.

A pair of values (x, y) is called a solution only if it makes both equations true, i.e., LHS = RHS for each equation.

So, statement 2 is true.

∴ Statement 1 is false , and statement 2 is true.

Hence, option 4 is the correct option.

Statement 1: The sum of two numbers x and y is 11. Twice the first number plus three times the second number equals to 25.

⇒ x + y = 11 and 2x + 3y = 25

Statement 2: The numbers are 7 and 4.

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

Given, The sum of two numbers x and y is 11.

⇒ x + y = 11

⇒ x = 11 - y ....................(1)

Twice the first number plus three times the second number equals to 25.

⇒ 2x + 3y = 25 .........(2)

So, statement 1 is true.

Substituting the value of x from equation (1) in (2), we get :

⇒ 2(11 - y) + 3y = 25

⇒ 22 - 2y + 3y = 25

⇒ 22 + y = 25

⇒ y = 25 - 22

⇒ y = 3.

Substitute the value of y in equation (1), we get

⇒ x = 11 - 3 = 8

The numbers are 8 and 3.

So, statement 2 is false.

∴ Statement 1 is true, and statement 2 is false.

Hence, option 3 is the correct option.

Assertion (A): Solutions of equations 3y - 2x = 1 and 3x + 4y = 24 is x = 4 and y = 3.

Reason (R): ∵ 3y - 2x = 3 x 3 - 2 x 4 = 1 and, 3x + 4y = 3 x 4 + 4 x 3 = 24

1. A is true, but R is false.

2. A is false, but R is true.

3. Both A and R are correct, and R is the correct explanation for A.

4. Both A and R are correct, and R is correct explanation for A.

Answer

If x = 4 and y = 3 are the solution of equation 3y - 2x = 1, then substitute x = 4 and y = 3 in L.H.S. and R.H.S., both values should be same.

Given,

3y - 2x = 1

Substituting value of x and y in L.H.S. of the above equation, we get :

⇒ 3y - 2x

⇒ 3 x 3 - 2 x 4

⇒ 9 - 8

⇒ 1

As, L.H.S. = R.H.S.

So, x = 4 and y = 3 are the solution of equation 3y - 2x = 1.

3x + 4y = 24

Substituting value of x and y in L.H.S. of the equation :

⇒ 3x + 4y

⇒ 3 x 4 + 4 x 3

⇒ 12 + 12

⇒ 24

As, L.H.S. = R.H.S.

So, x = 4 and y = 3 are the solution of equation 3x + 4y = 24.

∴ Both A and R are correct, and R is the correct explanation for A.

Hence, option 3 is the correct option.

Assertion (A): In a two digit number, three times the digit at ten's place is equals to two times the digit at the unit place and the sum of the digit is 5 then for number 10x + y, 3x = 2y and x + y = 5.

Reason (R): The number is 32.

1. A is true, but R is false.

2. A is false, but R is true.

3. Both A and R are correct, and R is the correct explanation for A.

4. Both A and R are correct, and R is not the correct explanation for A.

Answer

Let the two digit number be 10x + y.

It is given that three times the digit at ten's place is equals to two times the digit at the unit place.

⇒ 3x = 2y

⇒ x = $\dfrac{2}{3}$ y ....................(1)

And, the sum of the digit is 5.

⇒ x + y = 5 ........................(2)

So, assertion (A) is true.

Substitute the value of x from equation (1) in equation (2), we get :

$\Rightarrow \dfrac{2}{3}y + y = 5\\[1em] \Rightarrow \dfrac{2y + 3y}{3} = 5\\[1em] \Rightarrow \dfrac{5y}{3} = 5\\[1em] \Rightarrow y = \dfrac{5 \times 3}{5}\\[1em] \Rightarrow y = 3$

Substituting the value of y in equation (1),

⇒ x = $\dfrac{2}{3} \times 3$

⇒ x = 2

Thus, the number is 10x + y = 10(2) + 3 = 23.

So, reason (R) is false.

∴ A is true, but R is false.

Hence, option 1 is the correct option.

Solve the following pair of (simultaneous) equations using method of elimination by substitution :

3x + 2y = 11

2x - 3y + 10 = 0

Answer

Given,

Equations : 3x + 2y = 11 and 2x - 3y + 10 = 0

⇒ 3x + 2y = 11

⇒ 3x = 11 - 2y

⇒ x = $\dfrac{11 - 2y}{3}$ ........(1)

Substituting value of x from equation (1) in 2x - 3y + 10 = 0,

$\Rightarrow 2 \times \Big(\dfrac{11 - 2y}{3}\Big) - 3y + 10 = 0 \\[1em] \Rightarrow \dfrac{22 - 4y}{3} - 3y + 10 = 0 \\[1em] \Rightarrow \dfrac{22 - 4y - 9y + 30}{3} = 0 \\[1em] \Rightarrow \dfrac{52 - 13y}{3} = 0 \\[1em] \Rightarrow 52 - 13y = 0 \\[1em] \Rightarrow 13y = 52 \\[1em] \Rightarrow y = \dfrac{52}{13} = 4.$

Substituting value of y in equation (1), we get :

$\Rightarrow x = \dfrac{11 - 2 \times 4}{3} \\[1em] = \dfrac{11 - 8}{3} \\[1em] = \dfrac{3}{3} \\[1em] = 1.$

Hence, x = 1 and y = 4.

Solve the following pair of (simultaneous) equations using method of elimination by substitution :

2x - 3y + 6 = 0

2x + 3y - 18 = 0

Answer

Given,

Equations : 2x - 3y + 6 = 0 and 2x + 3y - 18 = 0

⇒ 2x - 3y + 6 = 0

⇒ 2x = 3y - 6

⇒ x = $\dfrac{3y - 6}{2}$ .......(1)

Substituting value of x from equation (1) in 2x + 3y - 18 = 0, we get :

$\Rightarrow 2 \times \Big(\dfrac{3y - 6}{2}\Big) + 3y - 18 = 0 \\[1em] \Rightarrow 3y - 6 + 3y - 18 = 0 \\[1em] \Rightarrow 6y - 24 = 0 \\[1em] \Rightarrow 6y = 24 \\[1em] \Rightarrow y = \dfrac{24}{6} = 4.$

Substituting value of y in equation (1), we get :

$\Rightarrow x = \dfrac{3 \times 4 - 6}{2} \\[1em] = \dfrac{12 - 6}{2} \\[1em] = \dfrac{6}{2} \\[1em] = 3.$

Hence, x = 3 and y = 4.

Solve the following pair of (simultaneous) equations using method of elimination by substitution :

$\dfrac{3x}{2} - \dfrac{5y}{3} + 2 = 0$

$\dfrac{x}{3} + \dfrac{y}{2} = 2\dfrac{1}{6}$

Answer

Given,

Equations : $\dfrac{3x}{2} - \dfrac{5y}{3} + 2 = 0, \dfrac{x}{3} + \dfrac{y}{2} = 2\dfrac{1}{6}$

$\Rightarrow \dfrac{3x}{2} - \dfrac{5y}{3} + 2 = 0 \\[1em] \Rightarrow \dfrac{9x - 10y + 12}{6} = 0 \\[1em] \Rightarrow 9x - 10y + 12 = 0 \\[1em] \Rightarrow 10y = 9x + 12 \\[1em] \Rightarrow y = \dfrac{9x + 12}{10} .......(1)$

Substituting value of y from equation (1) in $\dfrac{x}{3} + \dfrac{y}{2} = 2\dfrac{1}{6}$, we get :

$\Rightarrow \dfrac{x}{3} + \dfrac{\dfrac{9x + 12}{10}}{2} = 2\dfrac{1}{6} \\[1em] \Rightarrow \dfrac{x}{3} + \dfrac{9x + 12}{20} = \dfrac{13}{6} \\[1em] \Rightarrow \dfrac{20x + 3(9x + 12)}{60} = \dfrac{13}{6} \\[1em] \Rightarrow \dfrac{20x + 27x + 36}{60} = \dfrac{13}{6} \\[1em] \Rightarrow \dfrac{47x + 36}{60} = \dfrac{13}{6} \\[1em] \Rightarrow 47x + 36 = \dfrac{13}{6} \times 60 \\[1em] \Rightarrow 47x + 36 = 130 \\[1em] \Rightarrow 47x = 94 \\[1em] \Rightarrow x = \dfrac{94}{47} = 2.$

Substituting value of x in equation (1), we get :

$\Rightarrow y = \dfrac{9x + 12}{10} \\[1em] \Rightarrow y = \dfrac{9 \times 2 + 12}{10} \\[1em] \Rightarrow y = \dfrac{18 + 12}{10} \\[1em] \Rightarrow y = \dfrac{30}{10} = 3.$

Hence, x = 2 and y = 3.

Solve the following pair of (simultaneous) equations using method of elimination by substitution :

$\dfrac{x}{6} + \dfrac{y}{15} = 4$

$\dfrac{x}{3} - \dfrac{y}{12} = 4\dfrac{3}{4}$

Answer

Given,

Equations : $\dfrac{x}{6} + \dfrac{y}{15} = 4, \dfrac{x}{3} - \dfrac{y}{12} = 4\dfrac{3}{4}$

$\Rightarrow \dfrac{x}{6} + \dfrac{y}{15} = 4 \\[1em] \Rightarrow \dfrac{x}{6} = 4 - \dfrac{y}{15} \\[1em] \Rightarrow \dfrac{x}{6} = \dfrac{60 - y}{15} \\[1em] \Rightarrow x = \dfrac{6(60 - y)}{15} \\[1em] \Rightarrow x = \dfrac{2(60 - y)}{5} \\[1em] \Rightarrow x = \dfrac{120 - 2y}{5} ......(1)$

Substituting value of x from equation (1) in $\dfrac{x}{3} - \dfrac{y}{12} = 4\dfrac{3}{4}$, we get :

$\Rightarrow \dfrac{\dfrac{120 - 2y}{5}}{3} - \dfrac{y}{12} = 4\dfrac{3}{4} \\[1em] \Rightarrow \dfrac{120 - 2y}{15} - \dfrac{y}{12} = \dfrac{19}{4} \\[1em] \Rightarrow \dfrac{4(120 - 2y) - 5y}{60} = \dfrac{19}{4} \\[1em] \Rightarrow 480 - 8y - 5y = \dfrac{19}{4} \times 60 \\[1em] \Rightarrow 480 - 13y = 19 \times 15 \\[1em] \Rightarrow 480 - 13y = 285 \\[1em] \Rightarrow 13y = 480 - 285 \\[1em] \Rightarrow 13y = 195 \\[1em] \Rightarrow y = \dfrac{195}{13} = 15.$

Substituting value of y in equation (1), we get :

$\Rightarrow x = \dfrac{120 - 2 \times 15}{5} \\[1em] = \dfrac{120 - 30}{5} \\[1em] = \dfrac{90}{5} \\[1em] = 18.$

Hence, x = 18 and y = 15.

Find the value of m, if x = 2, y = 1 is a solution of the equation 2x + 3y = m.

Answer

Substituting value of x and y in 2x + 3y = m, we get :

⇒ 2(2) + 3(1) = m

⇒ 4 + 3 = m

⇒ m = 7.

Hence, m = 7.

Solve :

10% of x + 20% of y = 24, 3x - y = 20

Answer

Given,

Equations : 10% of x + 20% of y = 24, 3x - y = 20

$\Rightarrow 10$

Substituting value of x from equation (1) in 3x - y = 20, we get :

⇒ 3(240 - 2y) - y = 20

⇒ 720 - 6y - y = 20

⇒ 720 - 7y = 20

⇒ 7y = 720 - 20

⇒ 7y = 700

⇒ y = $\dfrac{700}{7}$ = 100.

Substituting value of y in equation (1), we get :

⇒ x = 240 - 2(100) = 240 - 200 = 40.

Hence, x = 40 and y = 100.

The value of expression mx - ny is 3 when x = 5 and y = 6 and its value is 8 when x = 6 and y = 5. Find the values of m and n.

Answer

Given,

Value of expression mx - ny is 3 when x = 5 and y = 6.

∴ 5m - 6n = 3

⇒ 5m = 3 + 6n

⇒ m = $\dfrac{3 + 6n}{5}$ ............(1)

Value of expression mx - ny is 8 when x = 6 and y = 5.

∴ 6m - 5n = 8

Substituting value of m from equation (1) in above equation.

$\Rightarrow 6\Big(\dfrac{3 + 6n}{5}\Big) - 5n = 8 \\[1em] \Rightarrow \dfrac{18 + 36n}{5} - 5n = 8 \\[1em] \Rightarrow \dfrac{18 + 36n - 25n}{5} = 8 \\[1em] \Rightarrow 18 + 11n = 5 \times 8 \\[1em] \Rightarrow 18 + 11n = 40 \\[1em] \Rightarrow 11n = 40 - 18 \\[1em] \Rightarrow 11n = 22 \\[1em] \Rightarrow n = \dfrac{22}{11} \\[1em] \Rightarrow n = 2.$

Substituting value of n in equation (1), we get :

$\Rightarrow m = \dfrac{3 + 6 \times 2}{5} \\[1em] = \dfrac{3 + 12}{5} \\[1em] = \dfrac{15}{5} \\[1em] = 3.$

Hence, m = 3 and n = 2.

Solve :

11(x - 5) + 10(y - 2) + 54 = 0

7(2x - 1) + 9(3y - 1) = 25

Answer

Given,

Equations : 11(x - 5) + 10(y - 2) + 54 = 0, 7(2x - 1) + 9(3y - 1) = 25

1st equation :

⇒ 11(x - 5) + 10(y - 2) + 54 = 0

⇒ 11x - 55 + 10y - 20 + 54 = 0

⇒ 11x + 10y - 21 = 0

⇒ 11x = 21 - 10y

⇒ x = $\dfrac{21 - 10y}{11}$ .........(1)

2nd equation :

⇒ 7(2x - 1) + 9(3y - 1) = 25

⇒ 14x - 7 + 27y - 9 = 25

⇒ 14x + 27y - 16 = 25

⇒ 14x + 27y = 25 + 16

⇒ 14x + 27y = 41

Substituting value of x from equation (1) in above equation :

$\Rightarrow 14\Big(\dfrac{21 - 10y}{11}\Big) + 27y = 41 \\[1em] \Rightarrow \dfrac{294 - 140y}{11} + 27y = 41 \\[1em] \Rightarrow \dfrac{294 - 140y + 297y}{11} = 41 \\[1em] \Rightarrow 294 + 157y = 41 \times 11 \\[1em] \Rightarrow 294 + 157y = 451 \\[1em] \Rightarrow 157y = 157 \\[1em] \Rightarrow y = 1.$

Substituting value of y in equation (1), we get :

$\Rightarrow x = \dfrac{21 - 10 \times 1}{11} \\[1em] = \dfrac{21 - 10}{11} \\[1em] = \dfrac{11}{11} \\[1em] = 1.$

Hence, x = 1 and y = 1.

Solve :

$\dfrac{7 + x}{5} - \dfrac{2x - y}{4} = 3y - 5$

$\dfrac{5y - 7}{2} + \dfrac{4x - 3}{6} = 18 - 5x$

Answer

Given,

1st equation :

$\Rightarrow \dfrac{7 + x}{5} - \dfrac{2x - y}{4} = 3y - 5 \\[1em] \Rightarrow \dfrac{4(7 + x) - 5(2x - y)}{20} = 3y - 5 \\[1em] \Rightarrow 28 + 4x - 10x + 5y = 20(3y - 5) \\[1em] \Rightarrow 28 - 6x + 5y = 60y - 100 \\[1em] \Rightarrow 6x = 28 + 100 + 5y - 60y \\[1em] \Rightarrow 6x = 128 - 55y \\[1em] \Rightarrow x = \dfrac{128 - 55y}{6} ......(1)$

2nd equation :

$\Rightarrow \dfrac{5y - 7}{2} + \dfrac{4x - 3}{6} = 18 - 5x \\[1em] \Rightarrow \dfrac{3(5y - 7) + 4x - 3}{6} = 18 - 5x \\[1em] \Rightarrow 15y - 21 + 4x - 3 = 6(18 - 5x) \\[1em] \Rightarrow 15y + 4x - 24 = 108 - 30x \\[1em] \Rightarrow 4x + 30x + 15y = 108 + 24 \\[1em] \Rightarrow 34x + 15y = 132.$

Substituting value of x from equation (1) in above equation :

$\Rightarrow 34\Big(\dfrac{128 - 55y}{6}\Big) + 15y = 132 \\[1em] \Rightarrow \dfrac{4352 - 1870y}{6} + 15y = 132 \\[1em] \Rightarrow \dfrac{4352 - 1870y + 90y}{6} = 132 \\[1em] \Rightarrow 4352 - 1780y = 792 \\[1em] \Rightarrow 1780y = 4352 - 792 \\[1em] \Rightarrow 1780y = 3560 \\[1em] \Rightarrow y = \dfrac{3560}{1780} = 2.$

Substituting value of y in equation (1), we get :

$\Rightarrow \dfrac{128 - 55 \times 2}{6}\\[1em] = \dfrac{128 - 110}{6} \\[1em] = \dfrac{18}{6} \\[1em] = 3.$

Hence, x = 3 and y = 2.

Solve :

$4x + \dfrac{x - y}{8}$ = 17

$2y + x - \dfrac{5y + 2}{3}$ = 2

Answer

Given,

1st equation :

$\Rightarrow 4x + \dfrac{x - y}{8} = 17 \\[1em] \Rightarrow \dfrac{32x + x - y}{8} = 17 \\[1em] \Rightarrow 33x - y = 136 \\[1em] \Rightarrow y = 33x - 136 .......(1)$

2nd equation :

$\Rightarrow 2y + x - \dfrac{5y + 2}{3} = 2 \\[1em] \Rightarrow \dfrac{6y + 3x - 5y - 2}{3} = 2 \\[1em] \Rightarrow y + 3x - 2 = 6 \\[1em] \Rightarrow y = 8 - 3x ........(2)$

From (1) and (2), we get :

⇒ 33x - 136 = 8 - 3x

⇒ 33x + 3x = 8 + 136

⇒ 36x = 144

⇒ x = $\dfrac{144}{36}$ = 4.

Substituting value of x in equation (1), we get :

⇒ y = 33(4) - 136 = 132 - 136 = -4.

Hence, x = 4 and y = -4.

Solve using cross-multiplication :

0.4x - 1.5y = 6.5

0.3x + 0.2y = 0.9

Answer

Simplifying first equation, we get :

⇒ 0.4x - 1.5y = 6.5

Multiplying both sides of the above equation by 10, we get :

⇒ 10(0.4x - 1.5y) = 10 × 6.5

⇒ 4x - 15y = 65

⇒ 4x - 15y - 65 = 0

Simplifying second equation, we get :

⇒ 0.3x + 0.2y = 0.9

⇒ 10(0.3x + 0.2y) = 0.9 × 10

⇒ 3x + 2y = 9

⇒ 3x + 2y - 9 = 0

So, the equations are

⇒ 4x - 15y - 65 = 0 ........(1)

⇒ 3x + 2y - 9 = 0 ........(2)

By cross-multiplication method, we get :

$\Rightarrow \dfrac{x}{(-15) \times (-9) - 2 \times (-65)} = \dfrac{y}{(-65) \times 3 - (-9) \times 4} = \dfrac{1}{4 \times 2 - 3 \times (-15)} \\[1em] \Rightarrow \dfrac{x}{135 + 130} = \dfrac{y}{-195 + 36} = \dfrac{1}{8 + 45} \\[1em] \Rightarrow \dfrac{x}{265} = \dfrac{y}{-159} = \dfrac{1}{53} \\[1em] \Rightarrow \dfrac{x}{265} = \dfrac{1}{53} \text{ and } \dfrac{y}{-159} = \dfrac{1}{53} \\[1em] \Rightarrow x = \dfrac{265}{53} \text{ and } y = \dfrac{-159}{53} \\[1em] \Rightarrow x = 5 \text{ and } y = -3.$

Hence, x = 5 and y = -3.

Solve using cross-multiplication :

$\sqrt{2}x - \sqrt{3}y = 0$

$\sqrt{5}x + \sqrt{2}y = 0$

Answer

Given, equations :

$\sqrt{2}x - \sqrt{3}y = 0$ ........(1)

$\sqrt{5}x + \sqrt{2}y = 0$ ........(2)

Multiplying equation (1) by $\sqrt{2}$, we get :

$\Rightarrow \sqrt{2}(\sqrt{2}x - \sqrt{3}y) = \sqrt{2} \times 0 \\[1em] \Rightarrow 2x - \sqrt{6}y = 0 \text{ .....(1)}$

Multiplying equation (2) by $\sqrt{3}$, we get :

$\Rightarrow \sqrt{3}(\sqrt{5}x + \sqrt{2}y) = \sqrt{3} \times 0 \\[1em] \Rightarrow \sqrt{15}x + \sqrt{6}y = 0 \text{ .....(2)}$

Adding equations (1) and (2), we get :

$\Rightarrow 2x - \sqrt{6}y + \sqrt{15}x + \sqrt{6}y = 0 \\[1em] \Rightarrow 2x + \sqrt{15}x = 0 \\[1em] \Rightarrow x(2 + \sqrt{15}) = 0 \\[1em] \Rightarrow x = 0.$

Substituting value of x in equation (1), we get :

$\Rightarrow 2 \times 0 - \sqrt{6}y = 0 \\[1em] \Rightarrow 0 - \sqrt{6}y = 0 \\[1em] \Rightarrow \sqrt{6}y = 0 \\[1em] \Rightarrow y = 0.$

Hence, x = 0 and y = 0.

Solve :

$\dfrac{3}{2x} + \dfrac{2}{3y} = -\dfrac{1}{3}$

$\dfrac{3}{4x} + \dfrac{1}{2y} = -\dfrac{1}{8}$

Answer

Given, equations :

$\dfrac{3}{2x} + \dfrac{2}{3y} = -\dfrac{1}{3}$ .......(1)

$\dfrac{3}{4x} + \dfrac{1}{2y} = -\dfrac{1}{8}$ .......(2)

Multiplying equation (1) by $\dfrac{1}{2}$, we get :

$\Rightarrow \dfrac{1}{2} \times \Big(\dfrac{3}{2x} + \dfrac{2}{3y}\Big) = \dfrac{1}{2} \times -\dfrac{1}{3} \\[1em] \Rightarrow \dfrac{1}{2} \times \dfrac{3}{2x} + \dfrac{1}{2} \times \dfrac{2}{3y} = -\dfrac{1}{6} \\[1em] \Rightarrow \dfrac{3}{4x} + \dfrac{1}{3y} = -\dfrac{1}{6} \text{ ........(3)}$

Subtracting equation (3) from (2), we get :

$\Rightarrow \Big(\dfrac{3}{4x} + \dfrac{1}{2y}\Big) - \Big(\dfrac{3}{4x} + \dfrac{1}{3y}\Big) = -\dfrac{1}{8} - \Big(-\dfrac{1}{6}\Big) \\[1em] \Rightarrow \dfrac{3}{4x} - \dfrac{3}{4x} + \dfrac{1}{2y} - \dfrac{1}{3y} = -\dfrac{1}{8} + \dfrac{1}{6} \\[1em] \Rightarrow \dfrac{3 - 2}{6y} = \dfrac{-3 + 4}{24} \\[1em] \Rightarrow \dfrac{1}{6y} = \dfrac{1}{24} \\[1em] \Rightarrow y = \dfrac{24}{6} = 4.$

Substituting value of y in equation (1), we get :

$\Rightarrow \dfrac{3}{2x} + \dfrac{2}{3 \times 4} = -\dfrac{1}{3} \\[1em] \Rightarrow \dfrac{3}{2x} + \dfrac{1}{6} = -\dfrac{1}{3} \\[1em] \Rightarrow \dfrac{3}{2x} = -\dfrac{1}{3} - \dfrac{1}{6} \\[1em] \Rightarrow \dfrac{3}{2x} = \dfrac{-2 - 1}{6} \\[1em] \Rightarrow \dfrac{3}{2x} = \dfrac{-3}{6} \\[1em] \Rightarrow x = \dfrac{3 \times 6}{2 \times -3}\\[1em] \Rightarrow x = \dfrac{18}{-6} = -3.$

Substituting value of x in equation (1), we get :

$\Rightarrow \dfrac{3}{2 \times -3} + \dfrac{2}{3y} = -\dfrac{1}{3} \\[1em] \Rightarrow -\dfrac{1}{2} + \dfrac{2}{3y} = -\dfrac{1}{3} \\[1em] \Rightarrow \dfrac{2}{3y} = -\dfrac{1}{3} + \dfrac{1}{2} \\[1em] \Rightarrow \dfrac{2}{3y} = \dfrac{-2 + 3}{6} \\[1em] \Rightarrow \dfrac{2}{3y} = \dfrac{1}{6} \\[1em] \Rightarrow y = \dfrac{2 \times 6}{3} \\[1em] \Rightarrow y = \dfrac{12}{3} = 4.$

Hence, x = -3 and y = 4.

Four times a certain two digit number is seven times the number obtained on interchanging its digits. If the difference between the digits is 4; find the number.

Answer

Let x be the digit at ten's place and y be the digit at unit's place.

Number = 10(x) + y = 10x + y

On reversing digits, the number becomes = 10(y) + x = 10y + x

Given,

Difference between digits = 4

⇒ x - y = 4

⇒ x = y + 4 .....(1)

Given,

Four times a certain two digit number is seven times the number obtained on interchanging its digits.

⇒ 4(10x + y) = 7(10y + x)

⇒ 40x + 4y = 70y + 7x

⇒ 40x - 7x = 70y - 4y

⇒ 33x = 66y

⇒ x = $\dfrac{66\text{y}}{33}$

⇒ x = 2y .........(2)

From (1) and (2), we get :

⇒ 2y = y + 4

⇒ 2y - y = 4

⇒ y = 4

Substituting value of y in equation (1), we get :

⇒ x = y + 4 = 4 + 4 = 8.

Number = 10x + y = 10(8) + 4 = 80 + 4 = 84.

Hence, number = 84.

The sum of a two digit number and the number obtained by interchanging the digits of the number is 121. If the digits of the number differ by 3, find the number.

Answer

Let x be the digit at ten's place and y be the digit at unit's place.

Number = 10(x) + y = 10x + y

On reversing digits, the number becomes = 10(y) + x = 10y + x.

Given,

The sum of a two digit number and the number obtained by interchanging the digits of the number is 121.

∴ 10x + y + 10y + x = 121

⇒ 11x + 11y = 121

⇒ 11(x + y) = 121

⇒ x + y = 11 ..........(1)

Let digits of the number differ by 3.

⇒ x - y = 3 ........(2)

or

⇒ y - x = 3..........(3)

Considering x - y = 3,

Adding equation (1) and (2), we get :

⇒ x + y + x - y = 11 + 3

⇒ 2x = 14

⇒ x = 7.

Substituting value of x in equation (1), we get :

⇒ 7 + y = 11

⇒ y = 11 - 7

⇒ y = 4.

Number = 10x + y = 10(7) + 4 = 70 + 4 = 74.

Considering y - x = 3,

Adding equation (1) and (3), we get :

⇒ x + y + y - x = 11 + 3

⇒ 2y = 14

⇒ y = $\dfrac{14}{2}$

⇒ y = 7.

Substituting value of x in equation (1), we get :

⇒ x + 7 = 11

⇒ x = 11 - 7

⇒ x = 4.

Number = 10x + y = 10(4) + 7 = 40 + 7 = 47.

Hence, number = 47 or 74.

A two digit number is obtained by multiplying the sum of the digits by 8. Also, it is obtained by multiplying the difference of the digits by 14 and adding 2. Find the number.

Answer

Let x be the digit at ten's place and y be the digit at unit's place.

Number = 10(x) + y = 10x + y

Given,

Number is obtained by multiplying the sum of the digits by 8.

∴ 8(x + y) = 10x + y

⇒ 8x + 8y = 10x + y

⇒ 10x - 8x = 8y - y

⇒ 2x = 7y

⇒ x = $\dfrac{7}{2}$y ..........(1)

Number is obtained by multiplying the difference of the digits by 14 and adding 2.

⇒ 14(x - y) + 2 = 10x + y .............(2)

or,

⇒ 14(y - x) + 2 = 10x + y .............(3)

Solving equation (2),

⇒ 14x - 14y + 2 = 10x + y

⇒ 14x - 10x = y + 14y - 2

⇒ 4x = 15y - 2 ...........(4)

Substituting value of x from equation (1) in equation (4), we get :

$\Rightarrow 4 \times \dfrac{7}{2}y = 15y - 2 \\[1em] \Rightarrow 14y = 15y - 2 \\[1em] \Rightarrow 15y - 14y = 2 \\[1em] \Rightarrow y = 2.$

Substituting value of y in equation (1), we get :

$\Rightarrow x = \dfrac{7}{2}y = \dfrac{7}{2} \times 2$ = 7.

Number = 10x + y = 10(7) + 2 = 70 + 2 = 72.

Solving equation (3),

⇒ 14(y - x) + 2 = 10x + y

⇒ 14y - 14x + 2 = 10x + y

⇒ 14y - y = 10x + 14x - 2

⇒ 13y = 24x + 2 ............(5)

Substituting value of x from equation (1) in equation (5), we get :

$\Rightarrow 13y = 24 \times \dfrac{7}{2}y + 2 \\[1em] \Rightarrow 13y = 84y + 2 \\[1em] \Rightarrow 84y - 13y = -2 \\[1em] \Rightarrow 71y = -2 \\[1em] \Rightarrow y = -\dfrac{2}{71}.$

This is not possible as y cannot be a fraction.

Hence, number = 72.

Two articles A and B are sold for ₹ 1,167 making 5% profit on A and 7% profit on B. If the two articles are sold for ₹ 1,165 a profit of 7% is made on A and a profit of 5% is made on B. Find the cost price of each article.

Answer

Let cost price of article A be ₹ x and article B be ₹ y.

Given,

Two articles A and B are sold for ₹ 1,167 making 5% profit on A and 7% profit on B.

$\Rightarrow \dfrac{105}{100}x + \dfrac{107}{100}y = 1167 \\[1em] \Rightarrow \dfrac{105x + 107y}{100} = 1167 \\[1em] \Rightarrow 105x + 107y = 116700 \\[1em] \Rightarrow 105x = 116700 - 107y \\[1em] \Rightarrow x = \dfrac{116700 - 107y}{105} .....(1)$

Given,

Two articles A and B are sold for ₹ 1,165 making 7% profit on A and 5% profit on B.

$\Rightarrow \dfrac{107}{100}x + \dfrac{105}{100}y = 1165 \\[1em] \Rightarrow \dfrac{107x + 105y}{100} = 1165 \\[1em] \Rightarrow 107x + 105y = 116500 .....(2)$

Substituting value of x from equation (1) in equation (2),

$\Rightarrow 107 \times \dfrac{116700 - 107y}{105} + 105y = 116500 \\[1em] \Rightarrow \dfrac{12486900 - 11449y}{105} + 105y = 116500 \\[1em] \Rightarrow \dfrac{12486900 - 11449y + 11025y}{105} = 116500 \\[1em] \Rightarrow 12486900 - 424y = 116500 \times 105 \\[1em] \Rightarrow 12486900 - 424y = 12232500 \\[1em] \Rightarrow 424y = 254400 \\[1em] \Rightarrow y = \dfrac{254400}{424} \\[1em] \Rightarrow y = 600$

Substituting value of y in equation (1), we get :

$\Rightarrow x = \dfrac{116700 - 107 \times 600}{105} \\[1em] = \dfrac{116700 - 64200}{105} \\[1em] = \dfrac{52500}{105} \\[1em] = 500.$

Hence, cost of article A = ₹ 500 and cost of article B = ₹ 600.

Pooja and Ritu can do a piece of work in $17\dfrac{1}{7}$ days. If one day work of Pooja be three fourth of one day work of Ritu; find in how many days each will do the work alone.

Answer

Let Pooja alone can do work in x days and Ritu alone can do in y days.

So, in one day Pooja can do $\dfrac{1}{x}$ th part of work and Ritu can do $\dfrac{1}{y}$ th part of work.

Given,

One day work of Pooja equals three fourth of one day work of Ritu.

$\therefore \dfrac{1}{x} = \dfrac{3}{4} \times \dfrac{1}{y} \\[1em] \Rightarrow \dfrac{1}{x} = \dfrac{3}{4y} \\[1em] \Rightarrow x = \dfrac{4y}{3} .......(1)$

Given,

Pooja and Ritu can do a piece of work in $17\dfrac{1}{7} \text{ or } \dfrac{120}{7}$ days.

So, in one day both of them can do $\dfrac{7}{120}$ th part of work.

$\therefore \dfrac{1}{x} + \dfrac{1}{y} = \dfrac{7}{120}$

Substituting value of x from equation (1) in above equation :

$\Rightarrow \dfrac{1}{\dfrac{4y}{3}} + \dfrac{1}{y} = \dfrac{7}{120} \\[1em] \Rightarrow \dfrac{3}{4y} + \dfrac{1}{y} = \dfrac{7}{120} \\[1em] \Rightarrow \dfrac{3 + 4}{4y} = \dfrac{7}{120} \\[1em] \Rightarrow \dfrac{7}{4y} = \dfrac{7}{120} \\[1em] \Rightarrow y = \dfrac{120 \times 7}{7 \times 4} \\[1em] \Rightarrow y = 30.$

Substituting value of y in equation (1), we get :

$\Rightarrow x = \dfrac{4 \times 30}{3} \\[1em] = 40.$

Hence, Pooja can do the work alone in 40 days and Ritu in 30 days.

Mr. and Mrs. Ahuja weigh x kg and y kg respectively. They both take a dieting course, at the end of which Mr. Ahuja loses 5 kg and weighs as much as his wife weighed before the course. Mrs. Ahuja loses 4 kg and weighs $\dfrac{7}{8}$ th of what her husband weighed before the course. Form two equations in x and y to find their weights before taking the dieting course.

Answer

Let Mr. Ahuja and Mrs. Ahuja weigh x kg and y kg respectively.

After one month,

Mr. Ahuja weighs (x - 5) kg and Mrs. Ahuja weighs (y - 4) kg.

Given,

After one month Mr. Ahuja weighs as much as his wife weighed before the course.

∴ x - 5 = y

⇒ x = y + 5 .........(1)

Given,

Mrs. Ahuja after one month weighs $\dfrac{7}{8}$ th of what her husband weighed before the course.

$\therefore y - 4 = \dfrac{7}{8} \times x \\[1em] \Rightarrow 8(y - 4) = 7x \\[1em] \Rightarrow 8y - 32 = 7x \\[1em] \Rightarrow 8y = 7x + 32 \\[1em] \Rightarrow y = \dfrac{7x + 32}{8} .......(2)$

Substituting value of x from equation (1) in equation (2), we get :

$\Rightarrow y = \dfrac{7(y + 5) + 32}{8} \\[1em] \Rightarrow y = \dfrac{7y + 35 + 32}{8} \\[1em] \Rightarrow 8y = 7y + 67 \\[1em] \Rightarrow 8y - 7y = 67 \\[1em] \Rightarrow y = 67.$

Substituting value of y in equation (1), we get :

⇒ x = 67 + 5 = 72.

Hence, weight of Mr. Ahuja and Mrs. Ahuja are 72 kg and 67 kg respectively.

A part of monthly expenses of a family is constant and the remaining vary with the number of members in the family. For a family of 4 persons, the total monthly expenses are ₹ 10,400; whereas for a family of 7 persons, the total monthly expenses are ₹ 15,800. Find the constant expenses per month and the monthly expenses on each member of a family.

Answer

Let constant monthly expense be ₹ x and for each member of the family monthly expense be ₹ y.

Given,

For a family of 4 persons, the total monthly expenses are ₹ 10,400.

⇒ x + 4y = 10400 .......(1)

For a family of 7 persons, the total monthly expenses are ₹ 15,800.

⇒ x + 7y = 15800 .......(2)

Subtracting equation (1) from (2), we get :

⇒ (x + 7y) - (x + 4y) = 15800 - 10400

⇒ x - x + 7y - 4y = 5400

⇒ 3y = 5400

⇒ y = $\dfrac{5400}{3}$ = ₹ 1800.

Substituting value of y in equation (1), we get :

⇒ x + 4(1800) = 10400

⇒ x + 7200 = 10400

⇒ x = 10400 - 7200 = ₹ 3200.

Hence, constant expense = ₹ 3200 and monthly expense = ₹ 1800.

The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is ₹ 315 and for a journey of 15 km, the charge paid is ₹ 465. What are the fixed charges and the charge per kilometer ? How much does a person have to pay for travelling a distance of 32 km ?

Answer

Let fixed charge be ₹ x and charge per kilometer be ₹ y.

Given,

For a distance of 10 km, the charge paid is ₹ 315.

⇒ x + 10y = 315 ........(1)

For a distance of 15 km, the charge paid is ₹ 465.

⇒ x + 15y = 465 ........(2)

Subtracting equation (1) from (2), we get :

⇒ (x + 15y) - (x + 10y) = 465 - 315

⇒ x - x + 15y - 10y = 150

⇒ 5y = 150

⇒ y = $\dfrac{150}{5}$ = ₹ 30.

Substituting value of y in equation (1), we get :

⇒ x + 10 × 30 = 315

⇒ x + 300 = 315

⇒ x = 315 - 300 = ₹ 15.

For 32 km,

Charge = x + 32y = 15 + 32 × 30

= 15 + 960 = ₹ 975.

Hence, fixed charge = ₹ 15, charge per kilometer = ₹ 30 and for 32 km charge = ₹ 975.

A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Geeta paid ₹ 27 for a book kept for seven days, while Mohit paid ₹ 21 for the book he kept for five days. Find the fixed charges and the charge for each extra day.

Answer

Let for first three days charge be ₹ x and for the days after that charge be ₹ y per day.

Given,

Geeta paid ₹ 27 for a book kept for seven days.

∴ x + (7 - 3)y = 27

⇒ x + 4y = 27 ........(1)

Mohit paid ₹ 21 for a book kept for five days.

∴ x + (5 - 3)y = 21

⇒ x + 2y = 21 ........(2)

Subtracting equation (2) from (1), we get :

⇒ (x + 4y) - (x + 2y) = 27 - 21

⇒ x - x + 4y - 2y = 6

⇒ 2y = 6

⇒ y = $\dfrac{6}{2}$

⇒ y = ₹ 3.

Substituting value of y in equation (1), we get :

⇒ x + 4(3) = 27

⇒ x + 12 = 27

⇒ x = 27 - 12

⇒ x = ₹ 15.

Hence, fixed charge = ₹ 15 and charge for each extra day = ₹ 3.

The area of rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. However, if the length of this rectangle increases by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.

Answer

Let the length of rectangle be x units and breadth be y units.

Given,

Area of rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units.

⇒ (x - 5)(y + 3) = xy - 9

⇒ xy + 3x - 5y - 15 = xy - 9

⇒ xy - xy + 3x - 5y = -9 + 15

⇒ 3x - 5y = 6 ........(1)

Given,

If the length of this rectangle increases by 3 units and the breadth by 2 units, the area increases by 67 square units.

⇒ (x + 3)(y + 2) = xy + 67

⇒ xy + 2x + 3y + 6 = xy + 67

⇒ xy - xy + 2x + 3y = 67 - 6

⇒ 2x + 3y = 61 ........(2)

Multiplying equation (1) by 2, we get :

⇒ 2(3x - 5y) = 2 × 6

⇒ 6x - 10y = 12 .........(3)

Multiplying equation (2) by 3, we get :

⇒ 3(2x + 3y) = 3 × 61

⇒ 6x + 9y = 183 .........(4)

Subtracting equation (3) from (4), we get :

⇒ (6x + 9y) - (6x - 10y) = 183 - 12

⇒ 6x - 6x + 9y + 10y = 171

⇒ 19y = 171

⇒ y = $\dfrac{171}{19}$ = 9.

Substituting value of y in equation (1), we get :

⇒ 3x - 5(9) = 6

⇒ 3x - 45 = 6

⇒ 3x = 45 + 6

⇒ 3x = 51

⇒ x = $\dfrac{51}{3}$ = 17.

Hence, length = 17 units and breadth = 9 units.

It takes 12 hours to fill a swimming pool using two pipes. If the pipe of larger diameter is used for 4 hours and the pipe of smaller diameter is used for 9 hours, only half of the pool is filled. How long would each pipe take to fill the swimming pool ?

Answer

Let the time taken by the first pipe be x hours and the time taken by the second pipe be y hours.

In 1 hour,

The first pipe can fill $\dfrac{1}{x}$ th of the pool

The second pipe can fill $\dfrac{1}{y}$ th part of pool.

Given,

It takes 12 hours to fill the pool.

So, in 1 hour both of them will fill $\dfrac{1}{12}$ th part of the pool.

$\therefore \dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{12}$ ...........(1)

Given,

If the pipe of larger diameter is used for 4 hours and the pipe of smaller diameter is used for 9 hours, only half of the pool is filled.

$\therefore \dfrac{4}{x} + \dfrac{9}{y} = \dfrac{1}{2}$ .............(2)

Multiplying equation (1) by 4, we get :

$\Rightarrow \dfrac{4}{x} + \dfrac{4}{y} = \dfrac{4}{12} \\[1em] \Rightarrow \dfrac{4}{x} + \dfrac{4}{y} = \dfrac{1}{3} .......(3)$

Subtracting equation (3) from (2), we get :

$\Rightarrow \dfrac{4}{x} + \dfrac{9}{y} - \Big(\dfrac{4}{x} + \dfrac{4}{y}\Big)= \dfrac{1}{2} - \dfrac{1}{3} \\[1em] \Rightarrow \dfrac{4}{x} - \dfrac{4}{x} + \dfrac{9}{y} - \dfrac{4}{y} = \dfrac{3 - 2}{6} \\[1em] \Rightarrow \dfrac{5}{y} = \dfrac{1}{6} \\[1em] \Rightarrow y = 5 \times 6 \\[1em] \Rightarrow y = 30.$

Substituting value of y in equation (1), we get :

$\Rightarrow \dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{12} \\[1em] \Rightarrow \dfrac{1}{x} + \dfrac{1}{30} = \dfrac{1}{12} \\[1em] \Rightarrow \dfrac{1}{x} = \dfrac{1}{12} - \dfrac{1}{30} \\[1em] \Rightarrow \dfrac{1}{x} = \dfrac{5 - 2}{60} \\[1em] \Rightarrow \dfrac{1}{x} = \dfrac{3}{60} \\[1em] \Rightarrow \dfrac{1}{x} = \dfrac{1}{20} \\[1em] \Rightarrow x = 20.$

Hence, pipe will larger diameter will fill pool in 20 hours and with smaller diameter in 30 hours.