Problems on Simultaneous Linear Equations — Exercise 6

The sum of two numbers is 50 and their difference is 16. Find the numbers.

Answer

Let two numbers be x and y.

Given, sum = 50 and difference = 16.

x + y = 50 ........(i)

x - y = 16 ........(ii)

Adding (i) and (ii) we get,

⇒ (x + y) + (x - y) = 50 + 16
⇒ 2x = 66
⇒ x = 33.

Substituting value of x in (i) we get,

⇒ 33 + y = 50
⇒ y = 50 - 33 = 17.

Hence, two numbers are 33 and 17.

The sum of two numbers is 2. If their difference is 20, find the numbers.

Answer

Let two numbers be x and y.

Given, sum = 2 and difference = 20.

x + y = 2 ........(i)

x - y = 20 ........(ii)

Adding (i) and (ii) we get,

⇒ (x + y) + (x - y) = 2 + 20
⇒ 2x = 22
⇒ x = 11.

Substituting value of x in (i) we get,

⇒ 11 + y = 2
⇒ y = 2 - 11 = -9.

Hence, two numbers are 11 and -9.

The sum of two numbers is 43. If the larger is doubled and the smaller is tripled, the difference is 36. Find the two numbers.

Answer

Let the larger number be x and smaller number be y.

Given, sum = 43.

x + y = 43 .......(i)

According to question,

2x - 3y = 36 .......(ii)

Multiplying (i) by 3 we get,

3x + 3y = 129 ......(iii)

Adding (ii) and (iii) we get,

⇒ (2x - 3y) + (3x + 3y) = 36 + 129
⇒ 5x = 165
⇒ x = 33.

Substituting value of x in (i) we get,

⇒ 33 + y = 43
⇒ y = 43 - 33 = 10.

Hence, two numbers are 33 and 10.

The cost of 5 kg of sugar and 7 kg of rice is ₹765 and the cost of 7 kg of sugar and 5 kg of rice is ₹735. Find the cost of 6 kg of sugar and 10 kg of rice.

Answer

Let cost of sugar and rice per kg be ₹ x and ₹ y respectively.

So, cost of 5 kg of sugar and 7 kg of rice = 5x + 7y and,

cost of 7 kg of sugar and 5 kg of rice = 7x + 5y.

Given,

Cost of 5 kg of sugar and 7 kg of rice is ₹ 765.

⇒ 5x + 7y = 765 .....................(1)

Cost of 7 kg of sugar and 5 kg of rice is ₹ 735

⇒ 7x + 5y = 735 .....................(2)

Multiplying eq (1) by 7, we get :

⇒ 7(5x + 7y) = 7 × 765

⇒ 35x + 49y = 5355 .....................(3)

Multiplying eq (2) by 5, we get :

⇒ 5(7x + 5y) = 5 × 735

⇒ 35x + 25y = 3675 .....................(4)

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

⇒ 35x + 49y - (35x + 25y) = 5355 - 3675

⇒ 35x - 35x + 49y - 25y = 5355 - 3675

⇒ 24y = 1680

⇒ y = $\dfrac{1680}{24}$

⇒ y = 70.

Substituting the value of y in (1) we get,

⇒ 5x + 7(70) = 765

⇒ 5x + 490 = 765

⇒ 5x = 765 - 490

⇒ 5x = 275

⇒ x = $\dfrac{275}{5}$

⇒ x = 55.

∴ Cost of 6 kg of sugar and 10 kg of rice = 6x + 10y

= 6(55) + 10(70) = 330 + 700 = ₹ 1,030.

Hence, cost of 6kg of sugar and 10 kg of rice = ₹ 1,030.

The class IX students of a certain public School wanted to give a farewell party to the outgoing student of class X. They decided to purchase two kinds of sweets, one costing ₹350 per kg and the other costing ₹ 440 per kg. They estimated that 36 kg of sweets were needed. If the total money spent on sweet was ₹ 14,000, find how much sweets of each kind they purchased.

Answer

Let first sweet purchased be x kg and second be y kg.

Then, x + y = 36 or y = 36 - x ......................(1)

Cost of first sweet = ₹ 350 per kg

Cost of second sweet = ₹ 440 per kg

Total cost of sweets will be 350x + 440y.

Total money spent on sweet = ₹ 14,000

⇒ 350x + 440y = 14000

Substituting the value of y from (1) in above equation we get,

⇒ 350x + 440(36 - x) = 14000

⇒ 350x + 15840 - 440x = 14000

⇒ 15840 - 90x = 14000

⇒ 90x = 15840 - 14000

⇒ 90x = 1840

⇒ x = $\dfrac{1840}{90}$

⇒ x = 20.44 ≈ 20.

Substituting this value of x in eq (1)

⇒ y = 36 - x

⇒ y = 36 - 20

⇒ y = 16.

Hence, the sweet costing ₹350 per kg purchased was 20 kg and the one costing ₹440 per kg was 16 kg.

If from twice the greater of two numbers 16 is subtracted, the result is half the other number. If from half the greater number 1 is subtracted, the result is still half the other number. What are the numbers?

Answer

Let the greater number be x and smaller number be y.

According to question if from twice the greater of two numbers 16 is subtracted, the result is half the other number.

∴ 2x - 16 = $\dfrac{y}{2}$

⇒ 2(2x - 16) = y

⇒ 4x - y = 32 .......(i)

According to question, if from half the greater number 1 is subtracted, the result is still half the other number.

$\therefore \dfrac{x}{2} - 1 = \dfrac{y}{2} \\[1em] \Rightarrow 2\Big(\dfrac{x}{2} - 1\Big) = y \\[1em] \Rightarrow x - y = 2 .......(ii)$

Subtracting eq. (ii) from eq. (i) we get,

⇒ 4x - y - (x - y) = 32 - 2

⇒ 4x -y - x + y = 30

⇒ 3x = 30

⇒ x = $\dfrac{30}{3}$

⇒ x = 10

Substituting value of x in (ii) we get,

10 - y = 2

⇒ y = 10 - 2

⇒ y = 8

Hence, two numbers are 8 and 10.

There are 38 coins in a collection of 20 paise coins and 25 paise coins. If the total value of the collection is ₹8.50, how many of each are there?

Answer

Let x be the no. of 20 paise coins and y be the no. of 25 paise coins.

x + y = 38 .......(i)

Total value of coins = 0.2x + 0.25y.

Given, total value = ₹8.50

⇒ 0.2x + 0.25y = 8.50 .......(ii)

Multiplying (ii) by 4 we get,

⇒ 0.8x + y = 34 .......(iii)

Subtracting (iii) from (i) we get,

⇒ x + y - (0.8x + y) = 38 - 34

⇒ 0.2x = 4

⇒ x = $\dfrac{4}{0.2}$

⇒ x = $\dfrac{40}{2}$

⇒ x = 20.

Substituting value of x in (i) we get,

⇒ 20 + y = 38

⇒ y = 38 - 20

⇒ y = 18.

Hence, no. of 25 paise coins = 18 and no. of 20 paise coins = 20.

A man has certain notes of denominations ₹ 200 and ₹ 50 which amount to ₹ 3,800. If the number of notes of each kind is interchanged, they amount to ₹ 600 less as before. Find the number of notes of each denomination.

Answer

Let number of ₹ 200 notes be x and number of ₹ 50 notes be y.

So, total amount = 200x + 50y.

Given, total amount = ₹ 3,800.

∴ 200x + 50y = 3800 ....................(1)

On exchanging number of notes, number of ₹ 200 notes will be y and number of ₹ 50 notes will be x.

Given, total amount in this case = 3200 (3800 - 600).

∴ 200y + 50x = 3200 ....................(2)

Multiplying eq. (1) by 4 we get,

⇒ 4(200x + 50y) = 3800 x 4

⇒ 800x + 200y = 15200 ....................(3)

Subtracting eq. (2) from (3) we get,

⇒ 800x + 200y - (200y + 50x) = 15200 - 3200

⇒ 800x - 50x + 200y - 200y = 12000

⇒ 750x = 12000

⇒ x = $\dfrac{12000}{750}$

⇒ x = 16.

Substituting the value of x in equation 1 we get,

⇒ 200(16) + 50y = 3800

⇒ 3200 + 50y = 3800

⇒ 50y = 3800 - 3200

⇒ 50y = 600

⇒ y = $\dfrac{600}{50}$

⇒ y = 12.

Hence, number of ₹ 200 notes = 16 and number of ₹ 50 notes = 12.

The ratio of two numbers is $\dfrac{2}{3}$. If 2 is subtracted from the first and 8 from the second, the ratio becomes the reciprocal of the original ratio. Find the numbers.

Answer

Let two numbers be x and y.

Given, ratio of two numbers is $\dfrac{2}{3}$.

$\therefore \dfrac{x}{y} = \dfrac{2}{3} \\[1em] \Rightarrow x = \dfrac{2y}{3} .......(i)$

Given, if 2 is subtracted from the first and 8 from the second, the ratio becomes the reciprocal of the original ratio.

$\therefore \dfrac{x - 2}{y - 8} = \dfrac{3}{2}$

Substituting value of x from (i) in above equation,

$\Rightarrow \dfrac{\dfrac{2y}{3} - 2}{y - 8} = \dfrac{3}{2} \\[1em] \Rightarrow \dfrac{\dfrac{2y - 6}{3}}{y - 8} = \dfrac{3}{2} \\[1em] \Rightarrow \dfrac{2y - 6}{3(y - 8)} = \dfrac{3}{2} \\[1em] \Rightarrow 2(2y - 6) = 3(y - 8) \times 3 \\[1em] \Rightarrow 4y - 12 = 9y - 72 \\[1em] \Rightarrow 9y - 4y = 72 - 12 \\[1em] \Rightarrow 5y = 60 \\[1em] \Rightarrow y = \dfrac{60}{5} \\[1em] \Rightarrow y = 12.$

Substituting value of y in eq. (i),

$x = \dfrac{2y}{3} = \dfrac{2 \times 12}{3}$ = 8.

Hence, the numbers are 8 and 12.

If 1 is added to the numerator of a fraction, it becomes $\dfrac{1}{5}$; if 1 is taken from the denominator, it becomes $\dfrac{1}{7}$, find the fraction.

Answer

Let numerator be x and denominator be y.

Given, if 1 is added to the numerator of a fraction, it becomes $\dfrac{1}{5}$.

$\therefore \dfrac{x + 1}{y} = \dfrac{1}{5}$

y = 5(x + 1) .......(i)

Given, if 1 is taken from the denominator, it becomes $\dfrac{1}{7}$.

$\therefore \dfrac{x}{y - 1} = \dfrac{1}{7}$

⇒ y - 1 = 7x
⇒ y = 1 + 7x ........(ii)

Equating (i) and (ii) we get,

⇒ 5(x + 1) = 1 + 7x

⇒ 5x + 5 = 1 + 7x

⇒ 7x - 5x = 5 - 1

⇒ 2x = 4

⇒ x = 2.

Substituting value of x in (ii) we get,

y = 1 + 7x = 1 + 7(2) = 1 + 14 = 15.

Original fraction = $\dfrac{x}{y} = \dfrac{2}{15}$.

Hence, fraction = $\dfrac{2}{15}$.

If the numerator of a certain fraction is increased by 2 and the denominator by 1, the fraction becomes equal to $\dfrac{5}{8}$ and if the numerator and denominator are each diminished by 1, the fraction becomes equal to $\dfrac{1}{2}$; find the fraction.

Answer

Let numerator be x and denominator be y.

Given, if the numerator of a certain fraction is increased by 2 and the denominator by 1, the fraction becomes equal to $\dfrac{5}{8}$

$\therefore \dfrac{x + 2}{y + 1} = \dfrac{5}{8}$

⇒ 8(x + 2) = 5(y + 1)

⇒ 8x + 16 = 5y + 5

⇒ 5y - 8x = 11 .......(i)

Given, if the numerator and denominator are each diminished by 1, the fraction becomes equal to $\dfrac{1}{2}$

$\therefore \dfrac{x - 1}{y - 1} = \dfrac{1}{2}$

⇒ 2(x - 1) = y - 1

⇒ 2x - 2 = y - 1

⇒ y = 2x - 1 ......(ii)

Substituting value of y from (ii) in eq. (i) we get,

⇒ 5(2x - 1) - 8x = 11

⇒ 10x - 5 - 8x = 11

⇒ 2x = 16

⇒ x = 8.

Substituting value of x in eq. (ii) we get,

y = 2x - 1 = 2(8) - 1 = 16 - 1 = 15.

Original fraction = $\dfrac{x}{y} = \dfrac{8}{15}.$

Hence, fraction = $\dfrac{8}{15}$.

Find the fraction which becomes $\dfrac{1}{2}$ when the denominator is increased by 4 and is equal to $\dfrac{1}{8}$ when the numerator is diminished by 5.

Answer

Let numerator be x and denominator be y.

Given, fraction becomes $\dfrac{1}{2}$ when the denominator is increased by 4.

$\therefore \dfrac{x}{y + 4} = \dfrac{1}{2}$

⇒ 2x = y + 4

⇒ 2x - y = 4 .......(i)

Given, fraction becomes $\dfrac{1}{8}$ when the numerator is diminished by 5

$\therefore \dfrac{x - 5}{y} = \dfrac{1}{8}$

⇒ 8(x - 5) = y ......(ii)

Substituting value of y from eq. (ii) in (i) we get,

⇒ 2x - [8(x - 5)] = 4

⇒ 2x - 8x + 40 = 4

⇒ -6x = 4 - 40

⇒ 6x = 36

⇒ x = 6.

Substituting value of x in (ii) we get,

⇒ 8(6 - 5) = y

⇒ y = 8(1) = 8.

Original fraction = $\dfrac{x}{y} = \dfrac{6}{8}$

Hence, fraction = $\dfrac{6}{8}$.

In a two digit number the sum of the digits is 7. If the number with the order of the digits reversed is 28 greater than twice the unit's digits of the original number, find the number.

Answer

Let's consider the digits at ten's place as x and let the digit at unit's place be y.

Then according to first condition,

⇒ x + y = 7

⇒ x = 7 - y ......(i)

The number if digits are reversed = 10 × y + x.

Given, number with the order of the digits reversed is 28 greater than twice the unit's digits of the original number

∴ (10 × y + x) - 2y = 28

Substituting value of x from (i) we get,

⇒ 10y + 7 - y - 2y = 28

⇒ 7y = 21

⇒ y = 3.

⇒ x = 7 - y = 7 - 3 = 4.

Original number = 10 × x + y = 10 × 4 + 3 = 43.

Hence, original number = 43.

A number of two digits exceeds four times the sum of its digits by 6 and it is increased by 9 on reversing the digits. Find the number.

Answer

Let's consider the digits at ten's place as x and let the digit at unit's place be y.

Number = 10 × x + y = 10x + y,

On reversing digits the number is = 10 × y + x = 10y + x.

According to first condition, we have

⇒ 10x + y - [4(x + y)] = 6

⇒ 10x + y - 4x - 4y = 6

⇒ 10x - 4x + y - 4y = 6

⇒ 6x - 3y = 6

⇒ 2x - y = 2 ......(i)

According to second condition, we have

⇒ 10x + y + 9 = 10y + x

⇒ 10x - x + y - 10y = -9

⇒ 9x - 9y = -9

⇒ x - y = -1

⇒ y - x = 1 ......(ii)

Adding eq. (i) and (ii) we get,

⇒ 2x - y + (y - x) = 2 + 1

⇒ x = 3.

Substituting value of x in (ii) we get,

⇒ y - 3 = 1

⇒ y = 4.

Number = 10 × x + y = 10 × 3 + 4 = 30 + 4 = 34.

Hence, number = 34.

When a two digit number is divided by the sum of its digits the quotient is 8. If the ten's digit is diminished by three times the unit's digit, the remainder is 1. What is the number?

Answer

Let's consider the digits at ten's place as x and let the digit at unit's place be y.

Number = 10 × x + y = 10x + y,

According to first condition, we have

$\Rightarrow \dfrac{10x + y}{x + y} = 8$

⇒ 10x + y = 8(x + y)

⇒ 10x + y = 8x + 8y

⇒ 10x - 8x + y - 8y = 0

⇒ 2x - 7y = 0 .......(i)

According to second condition, we have

x - 3y = 1 ......(ii)

Multiplying eq. (ii) by 2 we get,

2x - 6y = 2 .......(iii)

Subtracting eq. (i) from (iii) we get,

⇒ 2x - 6y - (2x - 7y) = 2 - 0

⇒ 2x - 2x - 6y + 7y = 2

⇒ y = 2.

Substituting value of y in (ii) we get,

⇒ x - 3(2) = 1

⇒ x - 6 = 1

⇒ x = 7.

Number = 10 × x + y = 10 × 7 + 2 = 72.

Hence, number = 72.

The result of dividing a number of two digits by the number with digits reversed is $1\dfrac{3}{4}$. If the sum of digits is 12, find the number.

Answer

Let's consider the digits at ten's place as x and let the digit at unit's place be y.

Number = 10 × x + y = 10x + y,

On reversing digits the number is = 10 × y + x = 10y + x.

According to first condition, we have

$\Rightarrow \dfrac{10x + y}{10y + x} = 1\dfrac{3}{4} \\[1em] \Rightarrow \dfrac{10x + y}{10y + x} = \dfrac{7}{4} \\[1em] \Rightarrow 4(10x + y) = 7(10y + x) \\[1em] \Rightarrow 40x + 4y = 70y + 7x \\[1em] \Rightarrow 40x - 7x + 4y - 70y = 0 \\[1em] \Rightarrow 33x - 66y = 0 \\[1em] \Rightarrow 11(3x - 6y) = 0 \\[1em] \Rightarrow 3x - 6y = 0 ........(i)$

According to second condition, we have

⇒ x + y = 12

⇒ x = 12 - y .......(ii)

Substituting value of x from (ii) in (i) we get,

⇒ 3(12 - y) - 6y = 0

⇒ 36 - 3y - 6y = 0

⇒ 36 - 9y = 0

⇒ 9y = 36

⇒ y = 4.

Substituting value of y in (ii) we get,

⇒ x = 12 - 4 = 8.

Number = 10 × x + y = 10 × 8 + 4 = 84.

Hence, number = 84.

The result of dividing a number of two digits by the number with the digits reversed is $\dfrac{5}{6}$. If the difference of digits is 1, find the number.

Answer

Let's consider the digits at ten's place as x and let the digit at unit's place be y.

Number = 10 × x + y = 10x + y,

On reversing digits the number is = 10 × y + x = 10y + x.

According to first condition we have,

$\Rightarrow \dfrac{10x + y}{10y + x} = \dfrac{5}{6} \\[1em] \Rightarrow 6(10x + y) = 5(10y + x) \\[1em] \Rightarrow 60x + 6y = 50y + 5x \\[1em] \Rightarrow 60x - 5x + 6y - 50y = 0 \\[1em] \Rightarrow 55x - 44y = 0 \\[1em] \Rightarrow 11(5x - 4y) = 0 \\[1em] \Rightarrow 5x - 4y = 0 .......(i)$

According to second condition we have,

⇒ y - x = 1

⇒ x = y - 1 .......(ii)

Substituting value of x in (i) we get,

⇒ 5(y - 1) - 4y = 0

⇒ 5y - 5 - 4y = 0

⇒ y = 5.

Substituting value of y in (ii) we get,

⇒ x = y - 1 = 5 - 1 = 4.

Number = 10 × x + y = 10 × 4 + 5 = 45.

Hence, number = 45.

A number of three digits has the hundred digit 4 times the unit digit and the sum of three digits is 14. If the three digits are written in the reverse order, the value of the number is decreased by 594. Find the number.

Answer

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

Then according to question,

Digit at hundred's place = 4y.

Number = 100 × 4y + 10 × x + y = 400y + 10x + y = 401y + 10x.

Reversing the number becomes = 100 × y + 10 × x + 4y = 100y + 10x + 4y = 104y + 10x.

Given, sum of digits = 14.

∴ 4y + x + y = 14
⇒ x + 5y = 14 ......(i)

According to second condition,

⇒ 401y + 10x = 104y + 10x + 594

⇒ 401y - 104y + 10x - 10x = 594

⇒ 297y = 594

⇒ y = 2.

Substituting value of y in (i) we get,

⇒ x + 5(2) = 14

⇒ x + 10 = 14

⇒ x = 4.

Number = 100 × 4y + 10 × x + y = 100 × 4(2) + 10 × 4 + 2 = 800 + 40 + 2 = 842.

Hence, number = 842.

Four years ago Marina was three times old as her daughter. Six years from now the mother will be twice as old as her daughter. Find their present ages.

Answer

Let present age of Marina be x years and daughter's age be y years.

Four years ago,

Age of Marina = (x - 4) years

Age of Marina's daughter = (y - 4) years

According to first condition in the problem,

⇒ (x - 4) = 3(y - 4)

⇒ x - 4 = 3y - 12

⇒ 3y - x = 8 .......(i)

After 6 years,

Age of Marina = (x + 6) years

Age of Marina's daughter = (y + 6) years

According to second condition in the problem,

⇒ x + 6 = 2(y + 6)

⇒ x + 6 = 2y + 12

⇒ x - 2y = 6 .......(ii)

Adding eq. (i) and (ii) we get,

⇒ (3y - x) + (x - 2y) = 8 + 6

⇒ 3y - 2y - x + x = 14

⇒ y = 14.

On substituting value of y in (ii) we get,

⇒ x - 2(14) = 6

⇒ x - 28 = 6

⇒ x = 34.

Hence, present age of Marina = 34 years and daughter = 14 years.

On selling a tea set at 5% loss and a lemon set at 15% gain, a shopkeeper gains ₹70. If he sells the tea set at 5% gain and lemon set at 10% gain he gains ₹130. Find the cost price of lemon set.

Answer

Let C.P. of tea set = ₹x and C.P. of lemon set = ₹y.

Given, Loss on tea set = 5% and gain on lemon set = 15%, total gain = ₹70.

$\therefore \dfrac{15}{100}y - \dfrac{5}{100}x = 70 \\[1em] \Rightarrow \dfrac{15y - 5x}{100} = 70 \\[1em] \Rightarrow 15y - 5x = 7000 \\[1em] \Rightarrow 3y - x = 1400 ......(i)$

Given, Gain on tea set = 5% and gain on lemon set = 10%, total gain = ₹130.

$\therefore \dfrac{5}{100}x + \dfrac{10}{100}y = 130 \\[1em] \Rightarrow \dfrac{5x + 10y}{100} = 130 \\[1em] \Rightarrow 5x + 10y = 13000 \\[1em] \Rightarrow x + 2y = 2600 ......(ii)$

Adding eq. (i) and (ii) we get,

⇒ (3y - x) + (x + 2y) = 1400 + 2600

⇒ 5y = 4000

⇒ y = 800.

Hence, C.P. of lemon set = ₹800.

A person invested some money at 12% simple interest and some other amount at 10% simple interest. He received yearly interest of ₹1300. If he had interchanged the amounts, he would have received ₹40 more as yearly interest. How much did he invest at different rates?

Answer

Let the amount invested at 12% S.I. be ₹x and at 10% S.I. be ₹y.

According to first condition,

$\Rightarrow \dfrac{12}{100} \times x + \dfrac{10}{100} \times y = 1300 \\[1em] \Rightarrow \dfrac{12x + 10y}{100} = 1300 \\[1em] \Rightarrow 12x + 10y = 130000 \\[1em] \Rightarrow 6x + 5y = 65000 .....(i)$

Now, let the amount invested at 12% S.I. be ₹y and at 10% S.I. be ₹x.

According to second condition,

$\Rightarrow \dfrac{12}{100}y + \dfrac{10}{100}x = 1340 \\[1em] \Rightarrow \dfrac{12y + 10x}{100} = 1340 \\[1em] \Rightarrow 12y + 10x = 134000 \\[1em] \Rightarrow 6y + 5x = 67000 ......(ii)$

Multiplying (i) by 6 and (ii) by 5, we have

⇒ 36x + 30y = 390000 .......(iii)

⇒ 30y + 25x = 335000 .......(iv)

Subtracting eq. (iv) from (iii) we get,

⇒ (36x + 30y) - (30y + 25x) = 390000 - 335000

⇒ 36x - 25x + 30y - 30y = 55000

⇒ 11x = 55000

⇒ x = 5000.

On substituting the value of x in (i) we get,

⇒ 6(5000) + 5y = 65000

⇒ 30000 + 5y = 65000

⇒ 5y = 65000 - 30000

⇒ 5y = 35000

⇒ y = 7000.

Hence, the investment at 12% is ₹5000 and the investment at 10% is ₹7000.

A shopkeeper sells a table at 8% profit and a chair at 10% discount, thereby getting ₹1008. If he had sold the table at 10% profit and chair at 8% discount, he would have got ₹20 more. Find the cost price of the table and the list price of the chair.

Answer

Let the cost price of table be ₹x and list price of chair be ₹y.

According to first condition,

$\Rightarrow \dfrac{108}{100}x + \dfrac{90}{100}y = 1008 \\[1em] \Rightarrow \dfrac{108x + 90y}{100} = 1008 \\[1em] \Rightarrow 108x + 90y = 100800 \\[1em] \Rightarrow 18(6x + 5y) = 100800 \\[1em] \Rightarrow 6x + 5y = \dfrac{100800}{18} \\[1em] \Rightarrow 6x + 5y = 5600 ......(i)$

According to second condition,

$\Rightarrow \dfrac{110}{100}x + \dfrac{92}{100}y = 1028 \\[1em] \Rightarrow \dfrac{110x + 92y}{100} = 1028 \\[1em] \Rightarrow 110x + 92y = 102800 \\[1em] \Rightarrow 2(55x + 46y) = 102800 \\[1em] \Rightarrow 55x + 46y = 51400 ......(ii)$

Multiplying (i) by 55 and (ii) by 6, we have

⇒ 330x + 275y = 308000 ........(iii)

⇒ 330x + 276y = 308400 ........(iv)

Subtracting eq. (iii) from (iv) we get,

⇒ (330x + 276y) - (330x + 275y) = 308400 - 308000

⇒ 330x - 330x + 276y - 275y = 400

⇒ y = 400.

On substituting value of y in (i) we get,

⇒ 6x + 5(400) = 5600

⇒ 6x + 2000 = 5600

⇒ 6x = 3600

⇒ x = 600.

Hence, cost of table = ₹600 and cost of chair = ₹400.

A and B have some money with them. A said to B, 'if you give me ₹100, my money will become 75% of the money left with you'. "B said to A" instead if you give me ₹100, your money will become 40% of my money. How much money did A and B have originally?

Answer

Let A have ₹x and B have ₹y.

According to first condition,

$\Rightarrow x + 100 = \dfrac{75}{100}(y - 100)$

⇒ 100(x + 100) = 75(y - 100)

⇒ 100x + 10000 = 75y - 7500

⇒ 75y - 100x = 10000 + 7500

⇒ 75y - 100x = 17500

⇒ 25(3y - 4x) = 25 × 700

⇒ 3y - 4x = 700 .......(i)

According to second condition,

$\Rightarrow \dfrac{40}{100}(y + 100) = x - 100$

⇒ 40(y + 100) = 100(x - 100)

⇒ 40y + 4000 = 100x - 10000

⇒ 100x - 40y = 14000

⇒ 20(5x - 2y) = 20 × 700

⇒ 5x - 2y = 700 .......(ii)

Multiplying (i) by 2 and (ii) by 3 we get,

6y - 8x = 1400 ......(iii)

15x - 6y = 2100 .....(iv)

Adding (iii) and (iv) we get,

⇒ (6y - 8x) + (15x - 6y) = 1400 + 2100

⇒ 6y - 6y - 8x + 15x = 3500

⇒ 7x = 3500

⇒ x = 500.

Substituting value of x in (i) we get,

⇒ 3y - 4(500) = 700

⇒ 3y - 2000 = 700

⇒ 3y = 2700

⇒ y = 900.

Hence, A has ₹500 while B has ₹900.

The students of a class are made to stand in (complete) rows. If one student is extra in a row, there would be 2 rows less, and if one student is less in a row, there would be 3 rows more. Find the number of students in the class.

Answer

Let number of students in a row be x and no. of rows be y.

Total no. of students = xy.

According to first condition we have,

⇒ (x + 1)(y - 2) = xy

⇒ xy - 2x + y - 2 = xy

⇒ xy - xy + y - 2x = 2

⇒ y - 2x = 2 .........(i)

According to second condition,

⇒ (x - 1)(y + 3) = xy

⇒ xy + 3x - y - 3 = xy

⇒ 3x - y = 3 + xy - xy

⇒ 3x - y = 3 ........(ii)

Adding (i) and (ii) we get,

⇒ (y - 2x) + (3x - y) = 2 + 3

⇒ y - y - 2x + 3x = 5

⇒ x = 5.

Substituting value of x in (i) we get,

⇒ y - 2(5) = 2

⇒ y - 10 = 2

⇒ y = 12.

xy = 5 × 12 = 60.

Hence, there are 60 students in the class.

A jeweller has bars of 18-carat gold and 12-carat gold. How much of each must be melted together to obtain a bar of 16-carat gold weighing 120 grams? (Pure gold is 24-carat).

Answer

Let us assume the quantity of 18 carat gold as x gm and 12 carat gold as y gm.

Given, total weight of new bar = 120 gm.

x + y = 120 ......(i)

Pure gold is 24 carat.

So, the purity of 18 carat gold = $\dfrac{18}{24} \times 100$

Purity of 12 carat gold = $\dfrac{12}{24} \times 100$

Purity of 16 carat gold = $\dfrac{16}{24} \times 100$

According to condition,

$75$

Multiplying (i) by 2 we get,

⇒ 2x + 2y = 240 .......(iii)

Subtracting eq (iii) from eq (ii)

⇒ (3x + 2y) - (2x + 2y) = 320 - 240

⇒ 3x - 2x + 2y - 2y = 80

⇒ x = 80.

Substituting value of x in (i) we get,

⇒ 80 + y = 120

⇒ y = 40.

Hence, jeweller requires 80 gm of 18 carat gold and 40 gm of 12 carat gold to obtain a bar of 16 carat gold weighing 120 gm.

A and B together can do a piece of work in 15 days. If A's one day work is $1\dfrac{1}{2}$ times the one day's work of B, find in how many days can each do the work.

Answer

Let A's one day work be x and B's one day work be y.

According to first condition given in the problem,

x = $\dfrac{3}{2}$y

2x = 3y

2x - 3y = 0 .......(i)

Also given, A and B together can do a piece of work in 15 days.

∴ x + y = $\dfrac{1}{15}$

⇒ 15(x + y) = 1

⇒ 15x + 15y = 1 .......(ii)

Multiplying (i) by 5 we get,

10x - 15y = 0 ......(iii)

Adding (ii) and (iii) we get,

⇒ 15x + 15y + 10x - 15y = 1 + 0

⇒ 25x = 1

⇒ x = $\dfrac{1}{25}$.

Substituting x in (i) we get,

$\Rightarrow 2 \times \dfrac{1}{25} - 3y = 0 \\[1em] \Rightarrow 3y = \dfrac{2}{25} \\[1em] \Rightarrow y = \dfrac{2}{75}.$

Since, A's one day work is x and B's one day work is y, so A can do complete work in $\dfrac{1}{x}$ and B can do work in $\dfrac{1}{y}$ days.

$\dfrac{1}{x} = \dfrac{1}{\dfrac{1}{25}} = 25$ days.

$\dfrac{1}{y} = \dfrac{1}{\dfrac{2}{75}} = \dfrac{75}{2} = 37\dfrac{1}{2}$ days.

Hence, A will do the work in 25 days and B will do the work in $37\dfrac{1}{2}$ days.

2 men and 5 women can do a piece of work in 4 days, while one man and one woman can finish it in 12 days. How long would it take for 1 man to do the work?

Answer

Let's assume that 1 man takes x days to do the work and y days for a women.

So, the amount of work done by 1 man in 1 day = $\dfrac{1}{x}$ and,

the amount of work done by 1 woman in 1 day = $\dfrac{1}{y}$.

∴ The amount of work done by 2 men in 1 day = $\dfrac{2}{x}$ and,

The amount of work done by 5 woman in 1 day = $\dfrac{5}{y}$.

According to the given conditions,

$\dfrac{2}{x} + \dfrac{5}{y} = \dfrac{1}{4}$ .......(i)

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

Multiplying (ii) by 5 we get,

$\dfrac{5}{x} + \dfrac{5}{y} = \dfrac{5}{12}$ .......(iii)

Subtracting (i) from (iii) we get,

$\Rightarrow \dfrac{5}{x} + \dfrac{5}{y} - \Big(\dfrac{2}{x} + \dfrac{5}{y}\Big) = \dfrac{5}{12} - \dfrac{1}{4} \\[1em] \Rightarrow \dfrac{5}{x} - \dfrac{2}{x} = \dfrac{5 - 3}{12} \\[1em] \Rightarrow \dfrac{3}{x} = \dfrac{2}{12} \\[1em] \Rightarrow \dfrac{3}{x} = \dfrac{1}{6} \\[1em] \Rightarrow x = 18.$

Hence, one man can do the work in 18 days.

A train covered a certain distance at a uniform speed. If the train had been 30 km/h faster, it would have taken 2 hours less than scheduled time. If the train were slower by 15km/h , it would have taken 2 hours more than the scheduled time. Find the length of the journey.

Answer

Let the actual speed of the train be x km/h and scheduled time be y hours.

Distance = Speed × Distance = xy.

According to first condition,

⇒ (x + 30)(y - 2) = xy

⇒ xy - 2x + 30y - 60 = xy

⇒ -2x + 30y = 60 ......(i)

According to second condition,

⇒ (x - 15)(y + 2) = xy

⇒ xy - 15y + 2x - 30 = xy

⇒ 2x - 15y = 30 .......(ii)

Adding (i) and (ii) we get,

(-2x + 30y) + (2x - 15y) = 60 + 30

15y = 90

y = 6.

Substituting value of y in (i) we get,

⇒ -2x + 30(6) = 60

⇒ -2x + 180 = 60

⇒ -2x = -120

⇒ x = 60.

Distance = xy = 60 × 6 = 360 km.

Hence, distance of journey = 360 km.

A boat takes 2 hours to go 40 km down the stream and it returns in 4 hours. Find the speed of the boat in still water and the speed of the stream.

Answer

Let speed of boat in still water be x km/h and speed of stream be y km/h.

Speed of boat downstream = (x + y) km/h

Speed of boat upstream = (x - y) km/h

Given, boat takes 2 hours to go 40 km down the stream

$\therefore \dfrac{40}{x + y} = 2$

⇒ 2(x + y) = 40

⇒ x + y = 20 .......(i)

Given, boat returns in 4 hours

$\therefore \dfrac{40}{x - y} = 4$

⇒ 4(x - y) = 40

⇒ x - y = 10 .......(ii)

Adding (i) and (ii) we get,

⇒ (x + y) + (x - y) = 20 + 10

⇒ 2x = 30

⇒ x = 15.

Substituting value of x in (i) we get,

⇒ 15 + y = 20

⇒ y = 5.

Hence, speed of boat in still water = 15 km/h and speed of stream = 5 km/h.

A boat sails a distance of 44 km in 4 hours with the current. It takes 4 hours 48 minutes longer to cover the same distance against the current. Find the speed of the boat in still water and the speed of the current.

Answer

Let speed of boat in still water be x km/h and speed of current be y km/h.

Speed of boat in the current = (x + y) km/h

Speed of boat against current = (x - y) km/h

Given, boat takes 4 hours to go 44 km with the current

$\therefore \dfrac{44}{x + y} = 4$

⇒ 4(x + y) = 44

⇒ x + y = 11 .......(i)

Given, boat takes 4 hours 48 minutes longer

$\therefore \dfrac{44}{x - y} = 4 + 4 + \dfrac{48}{60} \\[1em] \Rightarrow \dfrac{44}{x - y} = 4 + 4 + \dfrac{4}{5} \\[1em] \Rightarrow \dfrac{44}{x - y} = \dfrac{20 + 20 + 4}{5} \\[1em] \Rightarrow \dfrac{44}{x - y} = \dfrac{44}{5} \\[1em] \Rightarrow x - y = \dfrac{44 \times 5}{44} \\[1em] \Rightarrow x - y = 5 .......(ii)$

Adding (i) and (ii) we get,

⇒ (x + y) + (x - y) = 11 + 5

⇒ 2x = 16

⇒ x = 8.

Substituting value of x in (i) we get,

⇒ 8 + y = 11

⇒ y = 3.

Hence, speed of boat in still water = 8 km/h and speed of current = 3 km/h.

An aeroplane flies 1680 km with a head wind in 3.5 hours. On the return trip with same wind blowing, the plane takes 3 hours. Find the plane's air speed and the wind speed.

Answer

Let speed of aeroplane be x km/h and speed of wind be y km/h.

Speed of aeroplane in direction of wind = (x + y) km/h

Speed of aeroplane against wind = (x - y) km/h

Given, aeroplane takes 3.5 hours to go 1680 km against the wind

$\therefore \dfrac{1680}{x - y} = 3.5$

⇒ 3.5(x - y) = 1680

⇒ x - y = 480 .......(i)

Given, aeroplane takes 3 hours to go 1680 km with the wind

$\therefore \dfrac{1680}{x + y} = 3$

⇒ 1680 = 3(x + y)

⇒ x + y = 560 .......(ii)

Adding (i) and (ii) we get,

⇒ (x - y) + (x + y) = 480 + 560

⇒ 2x = 1040

⇒ x = 520.

Substituting value of x in (i) we get,

⇒ 520 - y = 480

⇒ y = 40.

Hence, speed of aeroplane = 520 km/h and speed of wind = 40 km/h.

A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When Bhawana takes food for 20 days, she has to pay ₹2600 as hostel charges; whereas when Divya takes food for 26 days, she pays ₹3020 as hostel charges. Find the fixed charges and the cost of food per day.

Answer

Let fixed charge be ₹x and per day food charge be ₹y.

Given, Bhawana pays ₹2600 when she takes food for 20 days.

x + 20y = 2600 .......(i)

Given, Divya pays ₹3020 when she takes food for 26 days.

x + 26y = 3020 .......(ii)

Subtracting (i) from (ii) we get,

(x + 26y) - (x + 20y) = 3020 - 2600
6y = 420
y = 70.

Substituting value of y in (i) we get,

x + 20(70) = 2600

x + 1400 = 2600

x = 1200.

Hence, fixed charge = ₹1200 and the cost of food per day = ₹70.