Question paper, Set 4

Ravi paints the whole wall in $49$ hours.

∴ Ravi's 1 hour's work $\displaystyle = \frac{1}{49}$

Answer: $\displaystyle \frac{1}{49}$ of the wall

Time taken by A $\displaystyle = \frac{1}{3} \times$ (Time taken by B)

$\displaystyle = \frac{1}{3} \times 33 = 11$ days

Answer: $11$ days

No. of men $\displaystyle \propto \frac{1}{\text{Time taken}}$

$\displaystyle \frac{T_1}{T_2} = \frac{M_2}{M_1} = \frac{45}{35} = \frac{9}{7}$

∴ $T_1 : T_2 = 9 : 7$

Answer: $9 : 7$

No. of men $\displaystyle \propto \frac{1}{\text{Time taken}}$

$28 \times 7 = 4 \times x$

$\displaystyle x = \frac{28 \times 7}{4} = 49$ hours

Answer: $49$ hours

(A + B)'s 1 day's work $\displaystyle = \frac{1}{15} + \frac{1}{66} = \frac{22 + 5}{330} = \frac{27}{330} = \frac{9}{110}$

Answer: $\displaystyle \frac{9}{110}$

A's 1 day's work $\displaystyle = \frac{1}{46}$

B's 1 day's work $\displaystyle = \frac{1}{23}$

∴ (A + B)'s 1 day's work $\displaystyle = \frac{1}{46} + \frac{1}{23} = \frac{1 + 2}{46} = \frac{3}{46}$

∴ Both together can finish the work in $\displaystyle \frac{46}{3}$ days $\displaystyle = 15\frac{1}{3}$ days

Answer: $\displaystyle 15\frac{1}{3}$ days

(A + B)'s 1 day's work $\displaystyle = \frac{1}{90}$

(B + C)'s 1 day's work $\displaystyle = \frac{1}{45}$

(C + A)'s 1 day's work $\displaystyle = \frac{1}{60}$

Adding, 2(A + B + C)'s 1 day's work $\displaystyle = \frac{1}{90} + \frac{1}{45} + \frac{1}{60} = \frac{2 + 4 + 3}{180} = \frac{9}{180} = \frac{1}{20}$

⇒ (A + B + C)'s 1 day's work $\displaystyle = \frac{1}{40}$

∴ A, B, C together can finish the work in $40$ days.

A's 1 day's work $\displaystyle = \frac{1}{40} - \frac{1}{45} = \frac{1}{360}$, so A alone takes $360$ days

B's 1 day's work $\displaystyle = \frac{1}{40} - \frac{1}{60} = \frac{1}{120}$, so B alone takes $120$ days

C's 1 day's work $\displaystyle = \frac{1}{40} - \frac{1}{90} = \frac{1}{72}$, so C alone takes $72$ days

Answer: Together $40$ days; A alone $360$ days, B alone $120$ days, C alone $72$ days

Suppose B finishes the work in $x$ days

⇒ A is able to finish it in $(x - 31)$ days

Time taken by A $\displaystyle = \frac{1}{2}$ (Time taken by B)

∴ $\displaystyle x - 31 = \frac{1}{2}x$

⇒ $2x - 62 = x$ ⇒ $x = 62$

∴ Time taken by B $= 62$ days, time taken by A $= 31$ days

A's 1 day's work $\displaystyle = \frac{1}{31}$, B's 1 day's work $\displaystyle = \frac{1}{62}$

(A + B)'s 1 day's work $\displaystyle = \frac{1}{31} + \frac{1}{62} = \frac{2 + 1}{62} = \frac{3}{62}$

A and B together can finish the work in $\displaystyle 20\frac{2}{3}$ days

Answer: $\displaystyle 20\frac{2}{3}$ days

A's 1 day's work $\displaystyle = \frac{1}{35}$, B's 1 day's work $\displaystyle = \frac{1}{70}$

(A + B)'s 1 day's work $\displaystyle = \frac{1}{35} + \frac{1}{70} = \frac{2 + 1}{70} = \frac{3}{70}$

B's 22 days' work $\displaystyle = 22 \times \frac{1}{70} = \frac{11}{35}$

Remaining work $\displaystyle = 1 - \frac{11}{35} = \frac{24}{35}$

(A + B)'s $x$ days' work $\displaystyle = \frac{24}{35}$ ⇒ $\displaystyle \frac{3}{70} \times x = \frac{24}{35}$

$x = 16$, so A left after $16$ days

Answer: $16$ days

A's 1 day's work $\displaystyle = \frac{1}{175}$, B's 1 day's work $\displaystyle = \frac{1}{60}$

(A + B)'s 1 day's work $\displaystyle = \frac{1}{175} + \frac{1}{60} = \frac{12 + 35}{2100} = \frac{47}{2100}$

(A + B)'s 30 days' work $\displaystyle = 30 \times \frac{47}{2100} = \frac{47}{70}$

A's 8 days' work $\displaystyle = \frac{8}{175}$

(A + C)'s 18 days' work $\displaystyle = 1 - \left(\frac{47}{70} + \frac{8}{175}\right) = \frac{99}{350}$

⇒ (A + C)'s 1 day's work $\displaystyle = \frac{99}{350} \div 18 = \frac{11}{700}$

C's 1 day's work $\displaystyle = \frac{11}{700} - \frac{1}{175} = \frac{11 - 4}{700} = \frac{7}{700} = \frac{1}{100}$

C can finish the work in $100$ days

Answer: $100$ days

Let the original number of men be $x$

No. of men $\displaystyle \propto \frac{1}{\text{No. of days}}$

$x \times 70 = (x + 4) \times 65$

$70x = 65x + 260$

$5x = 260$ ⇒ $x = 52$

Answer: $52$ men

$6$ men $= 10$ boys

$3$ men $\displaystyle = \frac{10}{6} \times 3 = 5$ boys

$3$ men + $20$ boys $= 5$ boys + $20$ boys $= 25$ boys

No. of boys $\displaystyle \propto \frac{1}{\text{No. of days}}$

$10 \times 15 = 25 \times x$ ⇒ $x = 6$ days

Answer: $6$ days

Let 1 man earn ₹$x$ and 1 boy earn ₹$y$ in 1 day.

$(2x + 7y) \times 9 = 2025$ ⇒ $2x + 7y = 225 \quad \cdots (1)$

$(4x + 7y) \times 11 = 3795$ ⇒ $4x + 7y = 345 \quad \cdots (2)$

(1) $\times 2$ ⇒ $4x + 14y = 450$

(1) $\times 2$ $-$ (2) ⇒ $7y = 105$ ⇒ $y = 15$

From (1): $2x = 225 - 105 = 120$ ⇒ $x = 60$

$(3x + 3y)n = 2025$

⇒ $(180 + 45)n = 2025$ ⇒ $225n = 2025$

⇒ $n = 9$ days

Answer: $9$ days

If B does $x$, A does $\displaystyle \frac{x}{2} + x = \frac{3x}{2}$ in the same time

Ratio of working of A and B $\displaystyle = \frac{3}{2} : 1 = 3 : 2$

Ratio of time taken for the same work $= 2 : 3$

Time taken by B $= 3k = 90$ ⇒ $k = 30$

Time taken by A $= 2k = 60$ days

A's 1 day's work $\displaystyle = \frac{1}{60}$, B's 1 day's work $\displaystyle = \frac{1}{90}$

(A + B)'s 1 day's work $\displaystyle = \frac{1}{60} + \frac{1}{90} = \frac{3 + 2}{180} = \frac{5}{180} = \frac{1}{36}$

Time taken by (A + B) to finish the work $= 36$ days

Answer: $36$ days

Speed upstream $= (x - y)$ km/hr

$= 9.5 - 0.5 = 9$ km/hr

Answer: $9$ km/hr

Let the man's rate in still water be $x$ km/hr and the rate of the current be $y$ km/hr.

Given $x + y = 11$ and $x - y = 8$

Adding, $2x = 19$ ⇒ $x = 9.5$ km/hr

Subtracting, $2y = 3$ ⇒ $y = 1.5$ km/hr

Answer: Rate in still water $9.5$ km/hr, rate of current $1.5$ km/hr

Speed $\displaystyle = \frac{\text{Distance}}{\text{Time}}$

$\displaystyle = \frac{27}{6} = 4.5$ km/hr

Answer: $4.5$ km/hr

Let the man's rate in still water be $x$ km/hr

Same distance, thrice the time, so speed down $= 3 \times$ speed up

Given $3(x - 4) = x + 4$

⇒ $3x - 12 = x + 4$ ⇒ $2x = 16$

⇒ $x = 8$ km/hr

Answer: $8$ km/hr

Let the required distance be $d$ km

Speed downstream $= 10 + 3 = 13$ km/hr

Speed upstream $= 10 - 3 = 7$ km/hr

Total time $=$ 2 hours 30 minutes $= 2.5$ hours

Time $\displaystyle = \frac{\text{Distance}}{\text{Speed}}$, so $\displaystyle \frac{d}{13} + \frac{d}{7} = 2.5$

⇒ $\displaystyle d\left(\frac{1}{13} + \frac{1}{7}\right) = 2.5$

⇒ $\displaystyle d \times \frac{20}{91} = 2.5$

⇒ $d = 11.375$ km

Answer: $11.375$ km

Let the speed of the boat in still water be $x$ km/hr

Speed upstream $= (x - 1)$ km/hr, speed downstream $= (x + 1)$ km/hr

Total time $= 85$ min $\displaystyle = \frac{85}{60}$ hr $\displaystyle = \frac{17}{12}$ hr

$\displaystyle \frac{12}{x - 1} + \frac{12}{x + 1} = \frac{17}{12}$

⇒ $\displaystyle 12\left(\frac{(x + 1) + (x - 1)}{(x - 1)(x + 1)}\right) = \frac{17}{12}$

⇒ $\displaystyle \frac{24x}{x^2 - 1} = \frac{17}{12}$

⇒ $288x = 17x^2 - 17$

⇒ $17 x^{2} - 288 x - 17 = 0$

⇒ $\left(x - 17\right) \left(17 x + 1\right) = 0$

⇒ $x = 17$ (or) $\displaystyle x = -\frac{1}{17}$ (rejected, a speed cannot be negative)

∴ Speed of the boat in still water $= 17$ km/hr

Answer: $17$ km/hr