Solving Problems (Based on Quadratic Equations) — Exercise 6(C)

A and B together can do a piece of work in 6 days. Whereas A alone can do the same work in 9 days. Then B alone will do same work in:

1. 15 days

2. 6 days

3. 18 days

4. 54 days

Answer

18 days

Reason

Given,

A alone can do the same work in 9 days.

So, one day work of A = $\dfrac{1}{9}$

Let B alone can do the same work in y days.

So, one day work of B = $\dfrac{1}{y}$

As together they can do work in 6 days.

∴ In one day, both of them do $\dfrac{1}{6}$th part of work.

$∴ \dfrac{1}{9} + \dfrac{1}{y} = \dfrac{1}{6}\\[1em] \Rightarrow \dfrac{1 \times y}{9\times y} + \dfrac{1\times 9}{y\times 9} = \dfrac{1}{6}\\[1em] \Rightarrow \dfrac{y}{9y} + \dfrac{9}{9y} = \dfrac{1}{6}\\[1em] \Rightarrow \dfrac{y + 9}{9y} = \dfrac{1}{6}\\[1em] \Rightarrow 6(y + 9) = 9y\\[1em] \Rightarrow 6y + 54 = 9y\\[1em] \Rightarrow 9y - 6y = 54\\[1em] \Rightarrow 3y = 54\\[1em] \Rightarrow y = \dfrac{54}{3}\\[1em] \Rightarrow y = 18$

∴ B alone can complete the work in 18 days.

Hence, option 3 is the correct option.

A can do a piece of work in 5 days and B can do the same work in (x + 5) days. The total number of days taken by A and B, working together, $3\dfrac{1}{3}$. The value of x is :

1. 5

2. 10

3. 15

4. $2\dfrac{1}{2}$

Answer

A can do a piece of work in 5 days.

So, in 1 day

A will do $\dfrac{1}{5}$ th of the work.

B can do a piece of work in (x + 5) days.

So, in 1 day

B will do $\dfrac{1}{x + 5}$ of the work.

Work done by A and B in 1 day = $\dfrac{1}{5} + \dfrac{1}{x + 5}$

Days taken to complete work = $\dfrac{1}{\text{Work done by A and B in 1 day}}$

$\Rightarrow 3\dfrac{1}{3} = \dfrac{1}{\dfrac{1}{5} + \dfrac{1}{x + 5}} \\[1em] \Rightarrow \dfrac{10}{3} = \dfrac{1}{\dfrac{x + 5 + 5}{5(x + 5)}} \\[1em] \Rightarrow \dfrac{10}{3} = \dfrac{5(x + 5)}{x + 10} \\[1em] \Rightarrow \dfrac{10}{3} = \dfrac{5x + 25}{x + 10} \\[1em] \Rightarrow 10(x + 10) = 3(5x + 25) \\[1em] \Rightarrow 10x + 100 = 15x + 75 \\[1em] \Rightarrow 15x - 10x = 100 -75 \\[1em] \Rightarrow 5x = 25 \\[1em] \Rightarrow x = \dfrac{25}{5} = 5.$

Hence, Option 1 is the correct option.

An article is bought for ₹ x and is sold at the profit of x %. If its selling price is ₹ 56, the cost price is :

1. ₹ 60

2. ₹ 50

3. ₹ 40

4. ₹ 28

Answer

Given,

C.P. = ₹ x

Profit = x %

S.P. = C.P. + Profit

$\Rightarrow 56 = x + \dfrac{x}{100} \times x \\[1em] \Rightarrow 56 = x + \dfrac{x^2}{100} \\[1em] \Rightarrow \dfrac{x^2 + 100x}{100} = 56 \\[1em] \Rightarrow x^2 + 100x = 5600 \\[1em] \Rightarrow x^2 + 100x - 5600 = 0\\[1em] \Rightarrow x^2 + 140x - 40x - 5600 = 0 \\[1em] \Rightarrow x(x + 140) - 40(x + 140) = 0 \\[1em] \Rightarrow (x - 40)(x + 140) = 0 \\[1em] \Rightarrow x - 40 = 0 \text{ or }x + 140 =0 \\[1em] \Rightarrow x = 40 \text{ or } x = -140.$

Since, cost cannot be negative.

∴ x ≠ -140.

Hence, Option 3 is the correct option.

An empty tank is filled by a pipe in 2 hours, whereas an another pipe empties the full tank in 3 hours. If the tank is empty and both the pipes are opened together the tank will be filled in:

1. 5 hours

2. 1 hour

3. 6 hours

4. none of these

Answer

6 hours

Reason

Given,

An empty tank is filled by a pipe in 2 hours.

So, one hour work of the pipe = $\dfrac{1}{2}$

Another pipe empties the full tank in 3 hours.

So, one hour work of other pipe = -$\dfrac{1}{3}$

When both pipes are opened together

$= \dfrac{1}{2} + \Big(-\dfrac{1}{3}\Big)\\[1em] = \dfrac{1}{2} - \dfrac{1}{3}\\[1em] = \dfrac{1 \times 3}{2 \times 3} - \dfrac{1 \times 2}{3 \times 2}\\[1em] = \dfrac{3}{6} - \dfrac{2}{6}\\[1em] = \dfrac{1}{6}$

Total time taken to fill the tank = $\dfrac{1}{\dfrac{1}{6}}$ = 6 hours.

Hence, option 3 is the correct option.

A can do a piece of work in 'x' days and B can do the same work in (x + 16) days. If both working together can do it in 15 days; calculate 'x'.

Answer

A can do work in x days

B can do work in (x + 16) days

In one day, A completes $\dfrac{1}{x}$ part of work

In one day, B completes $\dfrac{1}{x + 16}$ part of work

Given, both can do work in 15 days,

$\therefore \dfrac{1}{x} + \dfrac{1}{x + 16} = \dfrac{1}{15} \\[1em] \dfrac{x + 16 + x}{x(x + 16)} = \dfrac{1}{15} \\[1em] \dfrac{2x + 16}{x^2 + 16x} = \dfrac{1}{15} \\[1em] 15(2x + 16) = x^2 + 16x \\[1em] 30x + 240 = x^2+ 16x \\[1em] x^2 + 16x - 30x - 240 = 0 \\[1em] x^2 - 14x - 240 = 0 \\[1em] x^2 - 24x + 10x - 240 = 0 \\[1em] x(x - 24) + 10(x - 24) = 0 \\[1em] (x + 10)(x - 24) = 0 \\[1em] x + 10 = 0 \text{ or } x - 24 = 0 \\[1em] x = -10 \text{ or } x = 24.$

Since, no. of days cannot be negative,

∴ x ≠ -10.

Hence, x = 24 days.

One pipe can fill a cistern in 3 hours less than the other. The two pipes together can fill the cistern in 6 hours 40 minutes. Find the time that each pipe will take to fill the cistern.

Answer

Let second pipe fill in x hours, and first pipe in (x - 3) hours.

In one hour second pipe will fill $\dfrac{1}{x}$ and first pipe $\dfrac{1}{x - 3}$.

Given, together pipes can fill the cistern in 6 hours 40 minutes i.e. $\dfrac{400}{60}$ hours.

Hence, in one hour they will fill $\dfrac{1}{\dfrac{400}{60}} = \dfrac{60}{400}$.

$\therefore \dfrac{1}{x} + \dfrac{1}{x - 3} = \dfrac{60}{400} \\[1em] \Rightarrow \dfrac{x - 3 + x}{x(x - 3)} = \dfrac{3}{20} \\[1em] \Rightarrow \dfrac{2x - 3}{x^2 - 3x} = \dfrac{3}{20} \\[1em] \Rightarrow 20(2x - 3) = 3(x^2 - 3x) \\[1em] \Rightarrow 40x - 60 = 3x^2 - 9x \\[1em] \Rightarrow 3x^2 - 9x - 40x - 60 = 0 \\[1em] \Rightarrow 3x^2 - 49x - 60 = 0 \\[1em] \Rightarrow 3x^2 - 45x - 4x - 60 = 0 \\[1em] \Rightarrow 3x(x - 15) - 4(x - 15) = 0 \\[1em] \Rightarrow (3x - 4)(x - 15) = 0 \\[1em] \Rightarrow 3x - 4 = 0 \text{ or } x - 15 = 0 \\[1em] \Rightarrow x = \dfrac{4}{3} \text{ or } x = 15.$

Since first pipe takes 3 hours less than second pipe,

∴ x ≠ $\dfrac{4}{3}$ as in this case time will be negative which is not possible.

∴ x = 15, x - 3 = 12.

Hence, pipes will take 12 hours and 15 hours to fill cistern separately.

A trader bought an article for ₹ x and sold it for ₹ 52, thereby making a profit of (x - 10) percent on his outlay. Calculate the cost price.

Answer

C.P. = ₹ x

S.P. = ₹ 52

Profit = S.P. - C.P. = ₹(52 - x)

Profit % = $\dfrac{\text{Profit}}{\text{C.P.}} \times 100$

Substituting value in above equation we get,

$\Rightarrow x - 10 = \dfrac{52 - x}{x} \times 100 \\[1em] \Rightarrow x(x - 10) = 100(52 - x) \\[1em] \Rightarrow x^2 - 10x = 5200 - 100x \\[1em] \Rightarrow x^2 - 10x + 100x - 5200 = 0 \\[1em] \Rightarrow x^2 + 90x - 5200 = 0 \\[1em] \Rightarrow x^2 + 130x - 40x - 5200 = 0 \\[1em] \Rightarrow x(x + 130) - 40(x + 130) = 0 \\[1em] \Rightarrow (x - 40)(x + 130) = 0 \\[1em] \Rightarrow (x - 40) = 0 \text{ or } (x + 130) = 0 \\[1em] \Rightarrow x = 40 \text{ or } x = -130.$

Since, cost cannot be negative

∴ x ≠ -130.

Hence, C.P. = ₹ 40

The C.P. of an article is ₹ x which is sold for ₹ 27 at a loss of (x - 5) percent; find the value of x.

Answer

Given,

C.P. = ₹ x

S.P. = ₹ 27

Loss = (x - 5) %

Loss = C.P. - S.P.

= ₹ (x - 27)

As we know that Loss % = $\dfrac{\text{Loss}}{\text{C.P.}} \times 100$

Substituting the values, we get

$\Rightarrow (x - 5) = \dfrac{(x - 27)}{x} \times 100 \\[1em] \Rightarrow x(x - 5) = (x - 27) \times 100 \\[1em] \Rightarrow x^2 - 5x = 100x - 2700 \\[1em] \Rightarrow x^2 - 5x - 100x + 2700 = 0\\[1em] \Rightarrow x^2 - 105x + 2700 = 0\\[1em] \Rightarrow x^2 - 45x - 60x + 2700 = 0\\[1em] \Rightarrow x(x - 45) - 60(x - 45) = 0\\[1em] \Rightarrow (x - 45)(x - 60) = 0\\[1em] \Rightarrow (x - 45) = 0 \text{ or } (x - 60) = 0\\[1em] \Rightarrow x = 45 \text{ or } x = 60\\[1em]$

Hence, the value of C.P. = ₹ 45 or ₹ 60.

The C.P. of an article is ₹ x which is sold for ₹ 152 at a profit of (x + 10)%. Find the value of x.

Answer

Given,

C.P. = ₹ x

S.P. = ₹ 152

Profit = (x + 10) %

Profit = S.P. - C.P.

= ₹ (152 - x)

As we know that Profit % = $\dfrac{\text{Profit}}{\text{C.P.}} \times 100$

Substituting the values, we get

$\Rightarrow (x + 10) = \dfrac{(152 - x)}{x} \times 100 \\[1em] \Rightarrow x(x + 10) = (152 - x) \times 100 \\[1em] \Rightarrow x^2 + 10x = 15200 - 100x \\[1em] \Rightarrow x^2 + 10x - 15200 + 100x = 0\\[1em] \Rightarrow x^2 + 110x - 15200 = 0\\[1em] \Rightarrow x^2 + 190x - 80x - 15200 = 0\\[1em] \Rightarrow x(x + 190) - 80(x + 190) = 0\\[1em] \Rightarrow (x + 190)(x - 80) = 0\\[1em] \Rightarrow (x + 190) = 0 \text{ or } (x - 80) = 0\\[1em] \Rightarrow x = -190 \text{ or } x = 80\\[1em]$

The cost price of the article cannot be negative.

Hence, the value of C.P. = ₹ 80.

Two pipes running together can fill an empty cistern in $4\dfrac{20}{21}$ minutes. If one pipe takes 5 minutes more than the other pipe to fill the empty cistern. Find the time in which each pipe would fill the cistern.

Answer

Let the time taken by each pipe be x minutes and y minutes.

It is given that one pipe takes 5 minutes more than the other pipe to fill the empty cistern.

⇒ x = y + 5

So, one minute work of the pipe = $\dfrac{1}{x} = \dfrac{1}{y + 5}$

And, one minute work of the other pipe = $\dfrac{1}{y}$

Both pipes running together can fill the empty cistern in $4\dfrac{20}{21} = \dfrac{104}{21}$ minutes.

$\Rightarrow \dfrac{1}{x} + \dfrac{1}{y} = \dfrac{21}{104}\\[1em] \Rightarrow \dfrac{1}{y + 5} + \dfrac{1}{y} = \dfrac{21}{104}\\[1em] \Rightarrow \dfrac{1 \times y}{y(y + 5)} + \dfrac{1 \times (y + 5)}{y(y + 5)} = \dfrac{21}{104}\\[1em] \Rightarrow \dfrac{y}{y(y + 5)} + \dfrac{y + 5}{y(y + 5)} = \dfrac{21}{104}\\[1em] \Rightarrow \dfrac{y + y + 5}{y(y + 5)} = \dfrac{21}{104}\\[1em] \Rightarrow \dfrac{2y + 5}{y^2 + 5y} = \dfrac{21}{104}\\[1em] \Rightarrow 104(2y + 5) = 21(y^2 + 5y)\\[1em] \Rightarrow 208y + 520 = 21y^2 + 105y\\[1em] \Rightarrow 21y^2 + 105y - 208y - 520 = 0\\[1em] \Rightarrow 21y^2 - 103y - 520 = 0\\[1em] \Rightarrow 21y^2 + 65y - 168y - 105 = 0\\[1em] \Rightarrow y(21y + 65) - 8(21y + 65) = 0\\[1em] \Rightarrow (21y + 65)(y - 8) = 0\\[1em] \Rightarrow (21y + 65) = 0 \text{ or }(y - 8) = 0\\[1em] \Rightarrow y = -\dfrac{65}{21} \text{ or }y = 8\\[1em]$

As minutes cannot be negative.

So, one pipe takes 8 minutes and the other pipe takes 8 + 5 = 13 minutes.

Hence, the time taken by pipes = 8 minutes and 13 minutes.

In order to fill an empty swimming pool completely, a pipe of larger diameter alone takes 10 hour less than the time taken by the pipe of the smaller diameter alone. If the pipe of the larger diameter is used for 4 hours and the pipe of the smaller diameter is used for 9 hours, half of the pool is filled. In how many hours will the pipe of the larger diameter alone fill the pool.

Answer

Let the time taken by the larger diameter pipe to fill the pool alone be L hours and the time taken by the smaller diameter pipe to fill the pool alone be S hours.

From the problem, we know that the larger diameter pipe takes 10 hours less than the smaller diameter pipe. Hence, we can write the relationship as:

⇒ L = S - 10

The rate of the larger pipe is $\dfrac{1}{\text{L}}$ of the pool per hour.

The rate of the smaller pipe is $\dfrac{1}{\text{S}}$ of the pool per hour.

It is given that when the larger pipe is used for 4 hours and the smaller pipe is used for 9 hours, half of the pool is filled. So, the total work done in this scenario can be expressed as:

⇒ $4 \times \dfrac{1}{\text{L}} + 9 \times \dfrac{1}{\text{S}} = \dfrac{1}{2}$

Substituting the value of L = S - 10, we get

$\Rightarrow 4 \times \dfrac{1}{S - 10} + 9 \times \dfrac{1}{\text{S}} = \dfrac{1}{2}\\[1em] \Rightarrow \dfrac{4}{S - 10} + \dfrac{9}{\text{S}} = \dfrac{1}{2}\\[1em] \Rightarrow \dfrac{4S}{S(S - 10)} + \dfrac{9(S - 10)}{S(S - 10)} = \dfrac{1}{2}\\[1em] \Rightarrow \dfrac{4S}{S(S - 10)} + \dfrac{9S - 90}{S(S - 10)} = \dfrac{1}{2}\\[1em] \Rightarrow \dfrac{4S + 9S - 90}{S(S - 10)} = \dfrac{1}{2}\\[1em] \Rightarrow \dfrac{13S - 90}{S^2 - 10S} = \dfrac{1}{2}\\[1em] \Rightarrow 2(13S - 90) = S^2 - 10S\\[1em] \Rightarrow 26S - 180 = S^2 - 10S\\[1em] \Rightarrow S^2 - 10S - 26S + 180 = 0\\[1em] \Rightarrow S^2 - 36S + 180 = 0\\[1em] \Rightarrow S^2 - 30S - 6S + 180 = 0\\[1em] \Rightarrow S(S - 30) - 6(S - 30) = 0\\[1em] \Rightarrow (S - 30)(S - 6) = 0\\[1em] \Rightarrow (S - 30) = 0 \text{ or } (S - 6) = 0\\[1em] \Rightarrow S = 30 \text{ or } S = 6\\[1em]$

Substituting the value of S in equation, L = S - 10

⇒ L = 30 - 10 = 20 or 6 - 10 = -4

As time cannot be negative.

Hence, pipe of the larger diameter alone will take 20 hours to fill the pool.