Linear Equations and Inequalities in One Variable — Exercise 12.2

Three more than twice a number is equal to four less than the number. Find the number.

Answer

Let the number be $x$.

Three more than twice the number = $2x + 3$

Four less than the number = $x - 4$

Given,

Three more than twice a number is equal to four less than the number.

$\therefore 2x + 3 = x - 4 \\[1em] \Rightarrow 2x - x = - 4 - 3 \\[1em] \Rightarrow x = - 7$

Hence, the required number = -7.

When four consecutive integers are added, the sum is 46. Find the integers.

Answer

Let the four consecutive integers be $x, x + 1, x + 2$ and $x + 3$.

According to the given condition,

$\Rightarrow x + (x + 1) + (x + 2) + (x + 3) = 46 \\[1em] \Rightarrow 4x + 6 = 46 \\[1em] \Rightarrow 4x = 46 - 6 \\[1em] \Rightarrow 4x = 40 \\[1em] \Rightarrow x = \dfrac{40}{4} \\[1em] \Rightarrow x = 10$

So, the integers are 10, 10 + 1 = 11, 10 + 2 = 12 and 10 + 3 = 13.

Hence, the required integers are 10, 11, 12 and 13.

Manjula thinks of a number and subtracts $\dfrac{7}{3}$ from it. She multiplies the result by 6. The result now obtained is 2 less than twice the original number she thought of. What is the number?

Answer

Let the number thought of by Manjula be $x$.

On subtracting $\dfrac{7}{3}$ from it, we get $x - \dfrac{7}{3}$

On multiplying the result by 6, we get $6\left(x - \dfrac{7}{3}\right)$

2 less than twice the original number = $2x - 2$

According to the given condition,

$\Rightarrow 6\left(x - \dfrac{7}{3}\right) = 2x - 2 \\[1em] \Rightarrow 6x - 14 = 2x - 2 \\[1em] \Rightarrow 6x - 2x = - 2 + 14 \\[1em] \Rightarrow 4x = 12 \\[1em] \Rightarrow x = \dfrac{12}{4} \\[1em] \Rightarrow x = 3$

Hence, the required number = 3.

A positive number is 7 times another number. If 15 is added to both the numbers, then one of the new number becomes $\dfrac{5}{2}$ times the other new number. What are the numbers?

Answer

Let the smaller number be $x$.

Then, the other number = $7x$

On adding 15 to both the numbers, the new numbers are $x + 15$ and $7x + 15$.

According to the given condition,

⇒ $7x + 15 = \dfrac{5}{2}(x + 15)$

By cross multiplication, we get

$\Rightarrow 2(7x + 15) = 5(x + 15) \\[1em] \Rightarrow 14x + 30 = 5x + 75 \\[1em] \Rightarrow 14x - 5x = 75 - 30 \\[1em] \Rightarrow 9x = 45 \\[1em] \Rightarrow x = \dfrac{45}{9} \\[1em] \Rightarrow x = 5$

So, the other number = 7 × 5 = 35

Hence, the required numbers are 35 and 5.

When three consecutive even integers are added, the sum is zero. Find the integers.

Answer

Let the three consecutive even integers be $x, x + 2$ and $x + 4$.

The sum of three consecutive even integers = 0.

$\therefore x + (x + 2) + (x + 4) = 0 \\[1em] \Rightarrow 3x + 6 = 0 \\[1em] \Rightarrow 3x = - 6 \\[1em] \Rightarrow x = - \dfrac{6}{3} \\[1em] \Rightarrow x = - 2$

So, the integers are -2, -2 + 2 = 0 and -2 + 4 = 2.

Hence, the required integers are -2, 0 and 2.

Find two consecutive odd integers such that two-fifth of the smaller exceeds two-ninth of the greater by 4.

Answer

Let the smaller odd integer be $x$.

Then, the greater odd integer = $x + 2$

Given,

Two-fifth of the smaller exceeds two-ninth of the greater by 4.

⇒ $\dfrac{2}{5}x - \dfrac{2}{9}(x + 2) = 4$

L.C.M. of 5 and 9 = 45

Multiplying both sides by 45, we get:

$\Rightarrow \Big[\dfrac{2}{5}x - \dfrac{2}{9}(x + 2)\Big] \times 45 = 4 \times 45 \\[1em] \Rightarrow \dfrac{2}{5}x \times 45 - \dfrac{2}{9}(x + 2) \times 45 = 180 \\[1em] \Rightarrow 18x - 10(x + 2) = 180 \\[1em] \Rightarrow 18x - 10x - 20 = 180 \\[1em] \Rightarrow 8x = 180 + 20 \\[1em] \Rightarrow 8x = 200 \\[1em] \Rightarrow x = \dfrac{200}{8} \\[1em] \Rightarrow x = 25$

So, the greater odd integer = 25 + 2 = 27

Hence, the required integers are 25 and 27.

The denominator of a fraction is 1 more than twice its numerator. If the numerator and denominator are both increased by 5, it becomes $\dfrac{3}{5}$. Find the original fraction.

Answer

Let the numerator of the fraction be $x$.

Then, its denominator = $2x + 1$

So, the original fraction = $\dfrac{x}{2x + 1}$

On increasing the numerator and the denominator by 5, we get $\dfrac{x + 5}{2x + 1 + 5} = \dfrac{x + 5}{2x + 6}$

According to the given condition,

⇒ $\dfrac{x + 5}{2x + 6} = \dfrac{3}{5}$

By cross multiplication, we get

$\Rightarrow 5(x + 5) = 3(2x + 6) \\[1em] \Rightarrow 5x + 25 = 6x + 18 \\[1em] \Rightarrow 25 - 18 = 6x - 5x \\[1em] \Rightarrow x = 7$

So, the denominator = 2 × 7 + 1 = 15

Hence, the original fraction = $\mathbf{\dfrac{7}{15}}$.

Find two positive numbers in the ratio 2 : 5 such that their difference is 15.

Answer

Let the two positive numbers be $2x$ and $5x$.

According to the given condition,

$\Rightarrow 5x - 2x = 15 \\[1em] \Rightarrow 3x = 15 \\[1em] \Rightarrow x = \dfrac{15}{3} \\[1em] \Rightarrow x = 5$

So, the numbers are 2 × 5 = 10 and 5 × 5 = 25.

Hence, the required numbers are 10 and 25.

What number should be added to each of the numbers 12, 22, 42 and 72 so that the resulting numbers may be in proportion?

Answer

Let the number to be added be $x$.

On adding $x$ to each of the given numbers, the numbers become $12 + x, 22 + x, 42 + x$ and $72 + x$.

As the resulting numbers are in proportion,

⇒ $\dfrac{12 + x}{22 + x} = \dfrac{42 + x}{72 + x}$

By cross multiplication, we get

$\Rightarrow (12 + x)(72 + x) = (22 + x)(42 + x) \\[1em] \Rightarrow 864 + 12x + 72x + x^{2} = 924 + 22x + 42x + x^{2} \\[1em] \Rightarrow 864 + 84x + x^{2} = 924 + 64x + x^{2} \\[1em] \Rightarrow 84x - 64x = 924 - 864 \\[1em] \Rightarrow 20x = 60 \\[1em] \Rightarrow x = \dfrac{60}{20} \\[1em] \Rightarrow x = 3$

Hence, the required number to be added = 3.

The digits of a two-digit number differ by 3. If the digits are interchanged and the resulting number is added to the original number, we get 143. What can be the original number?

Answer

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

So, the original number = $10y + x$

On interchanging the digits, the new number = $10x + y$

According to the given condition,

$\Rightarrow (10y + x) + (10x + y) = 143 \\[1em] \Rightarrow 11x + 11y = 143 \\[1em] \Rightarrow 11(x + y) = 143 \\[1em] \Rightarrow x + y = \dfrac{143}{11} \\[1em] \Rightarrow x + y = 13 \quad \text{...(1)}$

Also given, the digits differ by 3.

Case 1 : Let $y - x = 3$ ...(2)

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

$\Rightarrow 2y = 16 \\[1em] \Rightarrow y = 8$

Substituting $y = 8$ in equation (1), we get $x = 13 - 8 = 5$

So, the number = 10 × 8 + 5 = 85

Case 2 : Let $x - y = 3$ ...(3)

Adding equations (1) and (3), we get

$\Rightarrow 2x = 16 \\[1em] \Rightarrow x = 8$

Substituting $x = 8$ in equation (1), we get $y = 13 - 8 = 5$

So, the number = 10 × 5 + 8 = 58

Hence, the original number can be 58 or 85.

Sum of the digits of a two digit number is 11. When we interchange the digits, it is found that the resulting new number is greater than the original number by 63. Find the two digit number.

Answer

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

So, the original number = $10y + x$

On interchanging the digits, the new number = $10x + y$

Given, sum of the digits = 11

⇒ $x + y = 11 \quad \text{...(1)}$

According to the given condition,

$\Rightarrow (10x + y) - (10y + x) = 63 \\[1em] \Rightarrow 9x - 9y = 63 \\[1em] \Rightarrow 9(x - y) = 63 \\[1em] \Rightarrow x - y = \dfrac{63}{9} \\[1em] \Rightarrow x - y = 7 \quad \text{...(2)}$

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

$\Rightarrow 2x = 18 \\[1em] \Rightarrow x = 9$

Substituting $x = 9$ in equation (1), we get $y = 11 - 9 = 2$

So, the number = 10 × 2 + 9 = 29

Hence, the required two digit number = 29.

Ritu is now four times as old as her brother Raju. In 4 years time, her age will be twice of Raju's age. What are their present ages?

Answer

Let the present age of Raju be $x$ years.

Then, the present age of Ritu = $4x$ years

After 4 years, Raju's age = $(x + 4)$ years and Ritu's age = $(4x + 4)$ years

According to the given condition,

$\Rightarrow 4x + 4 = 2(x + 4) \\[1em] \Rightarrow 4x + 4 = 2x + 8 \\[1em] \Rightarrow 4x - 2x = 8 - 4 \\[1em] \Rightarrow 2x = 4 \\[1em] \Rightarrow x = \dfrac{4}{2} \\[1em] \Rightarrow x = 2$

So, Ritu's present age = 4 × 2 = 8 years

Hence, the present ages of Ritu and Raju are 8 years and 2 years respectively.

A father is 7 times as old as his son. Two years ago, the father was 13 times as old as his son. How old are they now?

Answer

Let the present age of the son be $x$ years.

Then, the present age of the father = $7x$ years

Two years ago, son's age = $(x - 2)$ years and father's age = $(7x - 2)$ years

According to the given condition,

$\Rightarrow 7x - 2 = 13(x - 2) \\[1em] \Rightarrow 7x - 2 = 13x - 26 \\[1em] \Rightarrow 26 - 2 = 13x - 7x \\[1em] \Rightarrow 6x = 24 \\[1em] \Rightarrow x = \dfrac{24}{6} \\[1em] \Rightarrow x = 4$

So, father's present age = 7 × 4 = 28 years

Hence, the present ages of the father and the son are 28 years and 4 years respectively.

The ages of Sona and Sonali are in the ratio 5 : 3. Five years hence, the ratio of their ages will be 10 : 7. Find their present ages.

Answer

Let the present ages of Sona and Sonali be $5x$ years and $3x$ years respectively.

Five years hence, Sona's age = $(5x + 5)$ years and Sonali's age = $(3x + 5)$ years

According to the given condition,

⇒ $\dfrac{5x + 5}{3x + 5} = \dfrac{10}{7}$

By cross multiplication, we get

$\Rightarrow 7(5x + 5) = 10(3x + 5) \\[1em] \Rightarrow 35x + 35 = 30x + 50 \\[1em] \Rightarrow 35x - 30x = 50 - 35 \\[1em] \Rightarrow 5x = 15 \\[1em] \Rightarrow x = \dfrac{15}{5} \\[1em] \Rightarrow x = 3$

So, Sona's present age = 5 × 3 = 15 years

and Sonali's present age = 3 × 3 = 9 years

Hence, the present ages of Sona and Sonali are 15 years and 9 years respectively.

An employee works in a company on a contract of 30 days on the condition that he will receive ₹ 800 for each day he works and he will be fined ₹ 80 for each day he is absent. If he receives ₹ 15200 in all, for how many days did he remain absent?

Answer

Let the number of days for which the employee remained absent be $x$.

Then, the number of days for which he worked = $30 - x$

Amount received for the days he worked = ₹ $800(30 - x)$

Amount fined for the days he remained absent = ₹ $80x$

According to the given condition,

$\Rightarrow 800(30 - x) - 80x = 15200 \\[1em] \Rightarrow 24000 - 800x - 80x = 15200 \\[1em] \Rightarrow 24000 - 880x = 15200 \\[1em] \Rightarrow 24000 - 15200 = 880x \\[1em] \Rightarrow 880x = 8800 \\[1em] \Rightarrow x = \dfrac{8800}{880} \\[1em] \Rightarrow x = 10$

Hence, the employee remained absent for 10 days.

I have a total of ₹ 300 in coins of denomination ₹ 1, ₹ 2 and ₹ 5. The number of ₹ 2 coins is 3 times the number of ₹ 5 coins. The total number of coins is 160. How many coins of each denomination do I have?

Answer

Let the number of ₹ 5 coins be $x$.

Then, the number of ₹ 2 coins = $3x$

As the total number of coins is 160,

number of ₹ 1 coins = $160 - x - 3x = 160 - 4x$

Value of ₹ 5 coins = ₹ $5x$

Value of ₹ 2 coins = ₹ $2 \times 3x$ = ₹ $6x$

Value of ₹ 1 coins = ₹ $1 \times (160 - 4x)$ = ₹ $(160 - 4x)$

According to the given condition,

$\Rightarrow 5x + 6x + (160 - 4x) = 300 \\[1em] \Rightarrow 7x + 160 = 300 \\[1em] \Rightarrow 7x = 300 - 160 \\[1em] \Rightarrow 7x = 140 \\[1em] \Rightarrow x = \dfrac{140}{7} \\[1em] \Rightarrow x = 20$

So, the number of ₹ 5 coins = 20,

the number of ₹ 2 coins = 3 × 20 = 60

and the number of ₹ 1 coins = 160 - 4 × 20 = 160 - 80 = 80

Hence, there are 80 coins of ₹ 1, 60 coins of ₹ 2 and 20 coins of ₹ 5.

A local bus is carrying 40 passengers, some with ₹ 5 tickets and the remaining with ₹ 7.50 tickets. If the total receipts from these passengers is ₹ 230, find the number of passengers with ₹ 5 tickets.

Answer

Let the number of passengers with ₹ 5 tickets be $x$.

Then, the number of passengers with ₹ 7.50 tickets = $40 - x$

Amount received from ₹ 5 tickets = ₹ $5x$

Amount received from ₹ 7.50 tickets = ₹ $7.50(40 - x)$

According to the given condition,

$\Rightarrow 5x + 7.50(40 - x) = 230 \\[1em] \Rightarrow 5x + 300 - 7.50x = 230 \\[1em] \Rightarrow 300 - 2.50x = 230 \\[1em] \Rightarrow 300 - 230 = 2.50x \\[1em] \Rightarrow 2.50x = 70 \\[1em] \Rightarrow x = \dfrac{70}{2.50} \\[1em] \Rightarrow x = 28$

Hence, the number of passengers with ₹ 5 tickets = 28.

On a school picnic, a group of students agree to pay equally for the use of a full boat and pay ₹ 50 each. If there had been 3 more students in the group, each would have paid ₹ 10 less. How many students were there in the group?

Answer

Let the number of students in the group be $x$.

As each student pays ₹ 50, the total fare of the boat = ₹ $50x$

If there had been 3 more students, the number of students = $x + 3$

and the amount paid by each student = ₹ (50 - 10) = ₹ 40

So, the total fare of the boat = ₹ $40(x + 3)$

As the fare of the boat remains the same,

$\Rightarrow 40(x + 3) = 50x \\[1em] \Rightarrow 40x + 120 = 50x \\[1em] \Rightarrow 120 = 50x - 40x \\[1em] \Rightarrow 10x = 120 \\[1em] \Rightarrow x = \dfrac{120}{10} \\[1em] \Rightarrow x = 12$

Hence, there were 12 students in the group.

Half of a herd of deer are grazing in the field and three fourths of the remaining are playing nearby. The rest 9 are drinking water from the pond. Find the number of deer in the herd.

Answer

Let the number of deer in the herd be $x$.

Number of deer grazing in the field = $\dfrac{x}{2}$

Remaining deer = $x - \dfrac{x}{2} = \dfrac{x}{2}$

Number of deer playing nearby = $\dfrac{3}{4} \times \dfrac{x}{2} = \dfrac{3x}{8}$

According to the given condition,

⇒ $x - \dfrac{x}{2} - \dfrac{3x}{8} = 9$

L.C.M. of 2 and 8 = 8

Multiplying both sides by 8, we get:

$\Rightarrow \Big[x - \dfrac{x}{2} - \dfrac{3x}{8}\Big] \times 8 = 9 \times 8 \\[1em] \Rightarrow x \times 8 - \dfrac{x}{2} \times 8 - \dfrac{3x}{8} \times 8 = 72 \\[1em] \Rightarrow 8x - 4x - 3x = 72 \\[1em] \Rightarrow x = 72$

Hence, the number of deer in the herd = 72.

Sakshi takes some flowers in a basket and visits three temples one by one. At each temple, she offers one half of the flowers from the basket. If she is left with 6 flowers at the end, find the number of flowers she had in the beginning.

Answer

Let the number of flowers Sakshi had in the beginning be $x$.

At the first temple, she offers $\dfrac{x}{2}$ flowers, so flowers left = $x - \dfrac{x}{2} = \dfrac{x}{2}$

At the second temple, she offers half of $\dfrac{x}{2}$, so flowers left = $\dfrac{1}{2} \times \dfrac{x}{2} = \dfrac{x}{4}$

At the third temple, she offers half of $\dfrac{x}{4}$, so flowers left = $\dfrac{1}{2} \times \dfrac{x}{4} = \dfrac{x}{8}$

According to the given condition,

⇒ $\dfrac{x}{8} = 6$

By cross multiplication, we get

$\Rightarrow x = 6 \times 8 \\[1em] \Rightarrow x = 48$

Hence, Sakshi had 48 flowers in the beginning.

Two supplementary angles differ by 50°. Find the measure of each angle.

Answer

Let the smaller angle be $x$°.

As the angles are supplementary, their sum = 180°

So, the greater angle = $(180 - x)$°

According to the given condition,

$\Rightarrow (180 - x)\degree - x\degree = 50\degree \\[1em] \Rightarrow 180 - 2x = 50\degree \\[1em] \Rightarrow 180\degree - 50\degree = (2x)\degree \\[1em] \Rightarrow (2x)\degree = 130\degree \\[1em] \Rightarrow x\degree = \dfrac{130}{2} \\[1em] \Rightarrow x\degree = 65\degree$

So, the greater angle = 180° - 65° = 115°

Hence, the two angles are 65° and 115°.

If the angles of a triangle are in the ratio 5 : 6 : 7, find the angles.

Answer

Let the angles of the triangle be $(5x)\degree, (6x)\degree$ and $(7x)\degree$.

We know that,

The sum of the angles of a triangle is 180°.

$\Rightarrow (5x)\degree + (6x)\degree + (7x)\degree = 180\degree \\[1em] \Rightarrow (18x)\degree = 180\degree \\[1em] \Rightarrow x\degree = \dfrac{180}{18} \\[1em] \Rightarrow x\degree = 10\degree$

So, the angles are 5 × 10 = 50°, 6 × 10 = 60° and 7 × 10 = 70°.

Hence, the angles of the triangle are 50°, 60° and 70°.

Two equal sides of an isosceles triangle are $3x - 1$ and $2x + 2$ units. The third side is $2x$ units. Find $x$ and the perimeter of the triangle.

Answer

As the triangle is isosceles, its two equal sides are equal in length.

$\Rightarrow 3x - 1 = 2x + 2 \\[1em] \Rightarrow 3x - 2x = 2 + 1 \\[1em] \Rightarrow x = 3$

So, the equal sides = 3 × 3 - 1 = 8 units and 2 × 3 + 2 = 8 units

and the third side = 2 × 3 = 6 units

Perimeter of the triangle = Sum of all the sides

⇒ Perimeter = 8 + 8 + 6 = 22 units

Hence, $\bm{x = 3}$ and the perimeter of the triangle = 22 units.

If each side of a triangle is increased by 4 cm, the ratio of the perimeters of the new triangle and the given triangle is 7 : 5. Find the perimeter of the given triangle.

Answer

Let the perimeter of the given triangle be $x$ cm.

As each of the three sides is increased by 4 cm, the perimeter increases by (3 × 4) cm = 12 cm.

So, the perimeter of the new triangle = $(x + 12)$ cm

According to the given condition,

⇒ $\dfrac{x + 12}{x} = \dfrac{7}{5}$

By cross multiplication, we get

$\Rightarrow 5(x + 12) = 7x \\[1em] \Rightarrow 5x + 60 = 7x \\[1em] \Rightarrow 60 = 7x - 5x \\[1em] \Rightarrow 2x = 60 \\[1em] \Rightarrow x = \dfrac{60}{2} \\[1em] \Rightarrow x = 30$

Hence, the perimeter of the given triangle = 30 cm.

The length of a rectangle is 5 cm less than twice its breadth. If the length is decreased by 3 cm and breadth increased by 2 cm, the perimeter of the resulting rectangle is 72 cm. Find the area of the original rectangle.

Answer

Let the breadth of the original rectangle be $x$ cm.

Then, its length = $(2x - 5)$ cm

New length = $(2x - 5 - 3)$ cm = $(2x - 8)$ cm

New breadth = $(x + 2)$ cm

We know that,

Perimeter of a rectangle = 2(length + breadth)

According to the given condition,

$\Rightarrow 2[(2x - 8) + (x + 2)] = 72 \\[1em] \Rightarrow 2(3x - 6) = 72 \\[1em] \Rightarrow 3x - 6 = \dfrac{72}{2} \\[1em] \Rightarrow 3x - 6 = 36 \\[1em] \Rightarrow 3x = 36 + 6 \\[1em] \Rightarrow 3x = 42 \\[1em] \Rightarrow x = \dfrac{42}{3} \\[1em] \Rightarrow x = 14$

So, breadth of the original rectangle = 14 cm

and length of the original rectangle = 2 × 14 - 5 = 23 cm

Area of the original rectangle = length × breadth

⇒ Area = 23 × 14 = 322 cm²

Hence, the area of the original rectangle = 322 cm².

A rectangle is 10 cm long and 8 cm wide. When each side of the rectangle is increased by $x$ cm, its perimeter is doubled. Find the equation in $x$ and hence find the area of the new rectangle.

Answer

Length of the given rectangle = 10 cm and breadth = 8 cm

When each side is increased by $x$ cm,

new length = $(10 + x)$ cm and new breadth = $(8 + x)$ cm

We know that,

Perimeter of a rectangle = 2(length + breadth)

As the perimeter of the new rectangle is double the perimeter of the given rectangle,

⇒ $2[(x + 10) + (x + 8)] = 2 \times 2(10 + 8)$

Solving the above equation, we get

$\Rightarrow 2(2x + 18) = 2 \times 36 \\[1em] \Rightarrow 2x + 18 = 36 \\[1em] \Rightarrow 2x = 36 - 18 \\[1em] \Rightarrow 2x = 18 \\[1em] \Rightarrow x = \dfrac{18}{2} \\[1em] \Rightarrow x = 9$

So, new length = 10 + 9 = 19 cm and new breadth = 8 + 9 = 17 cm

Area of the new rectangle = length × breadth

⇒ Area = 19 × 17 = 323 cm²

Hence, the equation is $\bm{2[(x + 10) + (x + 8)] = 2 \times 2(10 + 8)}$ and the area of the new rectangle = 323 cm².

A steamer travels 90 km downstream in the same time as it takes to travel 60 km upstream. If the speed of the stream is 5 km/h, find the speed of the steamer in still water.

Answer

Let the speed of the steamer in still water be $x$ km/h.

Given, speed of the stream = 5 km/h

So, speed of the steamer downstream = $(x + 5)$ km/h

and speed of the steamer upstream = $(x - 5)$ km/h

We know that,

Time = $\dfrac{\text{Distance}}{\text{Speed}}$

Time taken to travel 90 km downstream = $\dfrac{90}{x + 5}$ hours

Time taken to travel 60 km upstream = $\dfrac{60}{x - 5}$ hours

According to the given condition,

⇒ $\dfrac{90}{x + 5} = \dfrac{60}{x - 5}$

By cross multiplication, we get

$\Rightarrow 90(x - 5) = 60(x + 5) \\[1em] \Rightarrow 90x - 450 = 60x + 300 \\[1em] \Rightarrow 90x - 60x = 300 + 450 \\[1em] \Rightarrow 30x = 750 \\[1em] \Rightarrow x = \dfrac{750}{30} \\[1em] \Rightarrow x = 25$

Hence, the speed of the steamer in still water = 25 km/h.

A streamer goes downstream and covers the distance between two ports in 5 hours while it covers the same distance upstream in 6 hours. If the speed of the stream is 1 km/h, find the speed of the streamer in still water and the distance between two ports.

Answer

Let the speed of the streamer in still water be $x$ km/h.

Given, speed of the stream = 1 km/h

So, speed of the streamer downstream = $(x + 1)$ km/h

and speed of the streamer upstream = $(x - 1)$ km/h

We know that,

Distance = Speed × Time

Distance covered downstream in 5 hours = $5(x + 1)$ km

Distance covered upstream in 6 hours = $6(x - 1)$ km

As the distance between the two ports is the same,

$\Rightarrow 5(x + 1) = 6(x - 1) \\[1em] \Rightarrow 5x + 5 = 6x - 6 \\[1em] \Rightarrow 5 + 6 = 6x - 5x \\[1em] \Rightarrow x = 11$

So, the distance between the two ports = 5(11 + 1) = 5 × 12 = 60 km

Hence, the speed of the streamer in still water = 11 km/h and the distance between the two ports = 60 km.

Distance between two places A and B is 350 km. Two cars start simultaneously from A and B towards each other and the distance between them after 4 hours is 62 km. If speed of one car is 8 km/h less than the speed of other car, find the speed of each car.

Answer

Let the speed of the faster car be $x$ km/h.

Then, the speed of the other car = $(x - 8)$ km/h

Let the faster car start from place A and the slower car start from place B.

Let P and Q be their positions after 4 hours.

We know that,

Distance = Speed × Time

AP = $4x$ km, BQ = $4(x - 8)$ km and PQ = 62 km

According to the given condition,

$\Rightarrow \text{AP} + \text{PQ} + \text{QB} = 350 \\[1em] \Rightarrow 4x + 62 + 4(x - 8) = 350 \\[1em] \Rightarrow 4x + 62 + 4x - 32 = 350 \\[1em] \Rightarrow 8x + 30 = 350 \\[1em] \Rightarrow 8x = 350 - 30 \\[1em] \Rightarrow 8x = 320 \\[1em] \Rightarrow x = \dfrac{320}{8} \\[1em] \Rightarrow x = 40$

So, the speed of the other car = 40 - 8 = 32 km/h

Hence, the speeds of the two cars are 40 km/h and 32 km/h.