Question paper, Set 2

Ravi paints the whole wall in $28$ hours.

∴ Ravi's 1 hour's work $\displaystyle = \frac{1}{28}$

Answer: $\displaystyle \frac{1}{28}$ of the wall

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

$\displaystyle 21 = \frac{1}{2} \times$ (Time taken by B)

∴ Time taken by B $= 2 \times 21 = 42$ days

Answer: $42$ days

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

$\displaystyle \frac{T_1}{T_2} = \frac{M_2}{M_1} = \frac{10}{16} = \frac{5}{8}$

∴ $T_1 : T_2 = 5 : 8$

Answer: $5 : 8$

(A + B)'s 1 day's work $\displaystyle = \frac{1}{36} + \frac{1}{44} = \frac{11 + 9}{396} = \frac{20}{396} = \frac{5}{99}$

Answer: $\displaystyle \frac{5}{99}$

A's 1 day's work $\displaystyle = \frac{1}{18}$

B's 1 day's work $\displaystyle = \frac{1}{30}$

∴ (A + B)'s 1 day's work $\displaystyle = \frac{1}{18} + \frac{1}{30} = \frac{5 + 3}{90} = \frac{8}{90} = \frac{4}{45}$

∴ Both together can finish the work in $\displaystyle \frac{45}{4}$ days $\displaystyle = 11\frac{1}{4}$ days

Answer: $\displaystyle 11\frac{1}{4}$ days

(A + B)'s 1 day's work $\displaystyle = \frac{1}{30}$

B's 1 day's work $\displaystyle = \frac{1}{45}$

⇒ A's 1 day's work $\displaystyle = \frac{1}{30} - \frac{1}{45} = \frac{3 - 2}{90} = \frac{1}{90}$

∴ A alone can finish the work in $90$ days.

Answer: $90$ days

A's 1 day's work $\displaystyle = \frac{1}{28}$

B's 1 day's work $\displaystyle = \frac{1}{48}$

(A + B)'s 1 day's work $\displaystyle = \frac{1}{28} + \frac{1}{48} = \frac{12 + 7}{336} = \frac{19}{336}$

Given (A + B + C)'s 1 day's work $\displaystyle = \frac{1}{14}$

C's 1 day's work $\displaystyle = \frac{1}{14} - \frac{19}{336} = \frac{24 - 19}{336} = \frac{5}{336}$

Consider $\displaystyle \frac{1}{28} : \frac{1}{48} : \frac{5}{336} = \frac{12}{336} : \frac{7}{336} : \frac{5}{336} = 12 : 7 : 5$

∴ ₹6960 must be divided in the ratio $12 : 7 : 5$ ($12 + 7 + 5 = 24$)

A's share $\displaystyle = 6960 \times \frac{12}{24} = 3480$, i.e. ₹3480

B's share $\displaystyle = 6960 \times \frac{7}{24} = 2030$, i.e. ₹2030

C's share $\displaystyle = 6960 \times \frac{5}{24} = 1450$, i.e. ₹1450

Answer: A ₹3480, B ₹2030, C ₹1450

Suppose B finishes the work in $x$ days

⇒ A is able to finish it in $(x - 54)$ days

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

∴ $\displaystyle x - 54 = \frac{1}{2}x$

⇒ $2x - 108 = x$ ⇒ $x = 108$

∴ Time taken by B $= 108$ days, time taken by A $= 54$ days

A's 1 day's work $\displaystyle = \frac{1}{54}$, B's 1 day's work $\displaystyle = \frac{1}{108}$

(A + B)'s 1 day's work $\displaystyle = \frac{1}{54} + \frac{1}{108} = \frac{2 + 1}{108} = \frac{3}{108} = \frac{1}{36}$

A and B together can finish the work in $36$ days

Answer: $36$ days

(A + B)'s 1 day's work $\displaystyle = \frac{1}{60}$, (B + C)'s 1 day's work $\displaystyle = \frac{1}{84}$

A's 1 day's work $=$ C's 2 days' work

C's 2 days' work + B's 1 day's work $\displaystyle = \frac{1}{60}$

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

⇒ C's 1 day's work $\displaystyle + \frac{1}{84} = \frac{1}{60}$

C's 1 day's work $\displaystyle = \frac{1}{60} - \frac{1}{84} = \frac{7 - 5}{420} = \frac{2}{420} = \frac{1}{210}$

B's 1 day's work $\displaystyle = \frac{1}{84} - \frac{1}{210} = \frac{5 - 2}{420} = \frac{3}{420} = \frac{1}{140}$

∴ B alone can finish the work in $140$ days

Answer: $140$ days

Let the original number of men be $x$

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

$x \times 71 = (x + 24) \times 47$

$71x = 47x + 1128$

$24x = 1128$ ⇒ $x = 47$

Answer: $47$ men

$6$ men $= 12$ boys

$4$ men $\displaystyle = \frac{12}{6} \times 4 = 8$ boys

$4$ men + $2$ boys $= 8$ boys + $2$ boys $= 10$ boys

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

$12 \times 10 = 10 \times x$ ⇒ $x = 12$ days

Answer: $12$ days

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

$(6x + 5y) \times 6 = 5340$ ⇒ $6x + 5y = 890 \quad \cdots (1)$

$(12x + 8y) \times 8 = 13120$ ⇒ $12x + 8y = 1640 \quad \cdots (2)$

(1) $\times 2$ ⇒ $12x + 10y = 1780$

(1) $\times 2$ $-$ (2) ⇒ $2y = 140$ ⇒ $y = 70$

From (1): $6x = 890 - 350 = 540$ ⇒ $x = 90$

$(7x + 8y)n = 11900$

⇒ $(630 + 560)n = 11900$ ⇒ $1190n = 11900$

⇒ $n = 10$ days

Answer: $10$ days

Work done by 60 men in 4 days $\displaystyle = 4 \times \frac{1}{8} = \frac{1}{2}$

So 1 man's work in 4 days $\displaystyle = \frac{1}{2} \div 60 = \frac{1}{120}$

Remaining work $\displaystyle = \frac{1}{2}$, remaining men $= 57$; work done by 57 men in 4 days $\displaystyle = 57 \times \frac{1}{120} = \frac{19}{40}$

Remaining work $\displaystyle = \frac{1}{40}$, remaining men $= 54$; work done by 54 men in 4 days $\displaystyle = \frac{9}{20}$

$\displaystyle \frac{1}{40}$ work is done by 54 men in $\displaystyle \frac{1}{40} \div \frac{9}{20} \times 4 = \frac{2}{9}$ days

Total time $\displaystyle = 4 + 4 + \frac{2}{9} = 8\frac{2}{9}$ days

Answer: $\displaystyle 8\frac{2}{9}$ 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 $= 3k = 24$ ⇒ $k = 8$

Time taken by A $= 2k = 16$ days

A's 1 day's work $\displaystyle = \frac{1}{16}$, B's 1 day's work $\displaystyle = \frac{1}{24}$

(A + B)'s 1 day's work $\displaystyle = \frac{1}{16} + \frac{1}{24} = \frac{3 + 2}{48} = \frac{5}{48}$

Time taken by (A + B) to finish the work $\displaystyle = 9\frac{3}{5}$ days

Answer: $\displaystyle 9\frac{3}{5}$ days

Speed downstream $= (x + y)$ km/hr

$= 11 + 5 = 16$ km/hr

Answer: $16$ 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 $x + y = 18$ and $x - y = 6$

Adding, $2x = 24$ ⇒ $x = 12$ km/hr

Subtracting, $2y = 12$ ⇒ $y = 6$ km/hr

Answer: Rate in still water $12$ km/hr, rate of current $6$ km/hr

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

$\displaystyle = \frac{46}{5} = 9.2$ km/hr

Answer: $9.2$ km/hr

Speed downstream: $\displaystyle x + y = \frac{36}{5}$ km/hr

Speed upstream: $\displaystyle x - y = \frac{12}{5}$ km/hr

Subtracting, $\displaystyle 2y = \frac{36}{5} - \frac{12}{5} = \frac{24}{5}$

$\displaystyle y = \frac{24}{10} = 2.4$ km/hr

Answer: $2.4$ km/hr

Let the required distance be $d$ km

Speed downstream $= 12.5 + 2 = 14.5$ km/hr

Speed upstream $= 12.5 - 2 = 10.5$ km/hr

Time $\displaystyle = \frac{\text{Distance}}{\text{Speed}}$, so $\displaystyle \frac{d}{14.5} + \frac{d}{10.5} = 3$

⇒ $\displaystyle d\left(\frac{2}{29} + \frac{2}{21}\right) = 3$

⇒ $\displaystyle d \times \frac{100}{609} = 3$

⇒ $d = 18.27$ km

Answer: $18.27$ km

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

Speed upstream $= (x - 4)$ km/hr, speed downstream $= (x + 4)$ km/hr

Total time $= 80$ min $\displaystyle = \frac{80}{60}$ hr $\displaystyle = \frac{4}{3}$ hr

$\displaystyle \frac{21}{x - 4} + \frac{21}{x + 4} = \frac{4}{3}$

⇒ $\displaystyle 21\left(\frac{(x + 4) + (x - 4)}{(x - 4)(x + 4)}\right) = \frac{4}{3}$

⇒ $\displaystyle \frac{42x}{x^2 - 16} = \frac{4}{3}$

⇒ $126x = 4x^2 - 64$

⇒ $2 x^{2} - 63 x - 32 = 0$

⇒ $\left(x - 32\right) \left(2 x + 1\right) = 0$

⇒ $x = 32$ (or) $\displaystyle x = -\frac{1}{2}$ (rejected, a speed cannot be negative)

∴ Speed of the boat in still water $= 32$ km/hr

Answer: $32$ km/hr