Question paper, Set 3

B does $\displaystyle \frac{1}{14}$ of the work in 1 day.

∴ B finishes the whole work in $14$ days.

Answer: $14$ days

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

$\displaystyle 4 = \frac{1}{2} \times$ (Time taken by B)

∴ Time taken by B $= 2 \times 4 = 8$ days

Answer: $8$ days

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

$\displaystyle \frac{T_1}{T_2} = \frac{M_2}{M_1} = \frac{8}{20} = \frac{2}{5}$

∴ $T_1 : T_2 = 2 : 5$

Answer: $2 : 5$

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

$33 \times 18 = x \times 54$

$\displaystyle x = \frac{33 \times 18}{54} = 11$ men

Answer: $11$ men

(A + B)'s 1 day's work $\displaystyle = \frac{1}{11} + \frac{1}{55} = \frac{5 + 1}{55} = \frac{6}{55}$

Answer: $\displaystyle \frac{6}{55}$

(A + B)'s 1 day's work $\displaystyle = \frac{1}{20}$

B's 1 day's work $\displaystyle = \frac{1}{36}$

⇒ A's 1 day's work $\displaystyle = \frac{1}{20} - \frac{1}{36} = \frac{9 - 5}{180} = \frac{4}{180} = \frac{1}{45}$

∴ A alone can finish the work in $45$ days.

Answer: $45$ days

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

B's 1 day's work $\displaystyle = \frac{1}{36}$

(A + B)'s 1 day's work $\displaystyle = \frac{1}{24} + \frac{1}{36} = \frac{3 + 2}{72} = \frac{5}{72}$

(A + B)'s 12 days' work $\displaystyle = 12 \times \frac{5}{72} = \frac{5}{6}$

Remaining work $\displaystyle = 1 - \frac{5}{6} = \frac{1}{6}$

This work is done by A in $\displaystyle \frac{1}{6} \times 24 = 4$ days

Answer: $4$ days

A's 1 day's work $\displaystyle = \frac{1}{18}$, B's 1 day's work $\displaystyle = \frac{1}{36}$

(A + B)'s 1 day's work $\displaystyle = \frac{1}{18} + \frac{1}{36} = \frac{2 + 1}{36} = \frac{3}{36} = \frac{1}{12}$

(A + B)'s 10 days' work $\displaystyle = \frac{5}{6}$

Remaining work $\displaystyle = 1 - \frac{5}{6} = \frac{1}{6}$, done by C in $7$ days

A's 10 days' work $\displaystyle = \frac{10}{18} = \frac{5}{9}$, B's 10 days' work $\displaystyle = \frac{10}{36} = \frac{5}{18}$, C's 7 days' work $\displaystyle = \frac{1}{6}$

Consider $\displaystyle \frac{5}{9} : \frac{5}{18} : \frac{1}{6} = 10 : 5 : 3$

A's share $\displaystyle = 10 \times \frac{1260}{18} = 700$, i.e. ₹700

B's share $\displaystyle = 5 \times \frac{1260}{18} = 350$, i.e. ₹350

C's share $\displaystyle = 3 \times \frac{1260}{18} = 210$, i.e. ₹210

A's daily wage $\displaystyle = \frac{700}{10} = 70$, i.e. ₹70

B's daily wage $\displaystyle = \frac{350}{10} = 35$, i.e. ₹35

C's daily wage $\displaystyle = \frac{210}{7} = 30$, i.e. ₹30

Answer: Shares: A ₹700, B ₹350, C ₹210; daily wages: A ₹70, B ₹35, C ₹30

A's 1 day's work $\displaystyle = \frac{1}{30}$, B's $\displaystyle = \frac{1}{24}$, C's $\displaystyle = \frac{1}{10}$

A, B, C together work for $1$ day; C alone works for the last $4$ days; B and C work together in between.

(A + B + C)'s 1 day's work $\displaystyle = \frac{1}{30} + \frac{1}{24} + \frac{1}{10} = \frac{4 + 5 + 12}{120} = \frac{21}{120} = \frac{7}{40}$

C's 4 days' work $\displaystyle = 4 \times \frac{1}{10} = \frac{2}{5}$

Remaining work $\displaystyle = 1 - \left(\frac{7}{40} + \frac{2}{5}\right) = \frac{17}{40}$

(B + C)'s 1 day's work $\displaystyle = \frac{1}{24} + \frac{1}{10} = \frac{5 + 12}{120} = \frac{17}{120}$

Remaining work is done by (B + C) in $\displaystyle \frac{17}{40} \div \frac{17}{120} = 3$ days

Total time $= 1 + 3 + 4 = 8$ days

Answer: $8$ days

A's 1 day's work $\displaystyle = \frac{1}{8}$, B's 1 day's work $\displaystyle = \frac{1}{22}$

Work done in the first 2 days $\displaystyle = \frac{1}{8} + \frac{1}{22} = \frac{11 + 4}{88} = \frac{15}{88}$

Work done in the first 10 days $\displaystyle = 5 \times \frac{15}{88} = \frac{75}{88}$

On day 11 it is A's turn: work done in the first 11 days $\displaystyle = \frac{75}{88} + \frac{1}{8} = \frac{43}{44}$

Remaining work $\displaystyle = 1 - \frac{43}{44} = \frac{1}{44}$

B does it in $\displaystyle \frac{1}{44} \div \frac{1}{22} = \frac{1}{2}$ day

∴ Total no. of days $\displaystyle = 11 + \frac{1}{2} = 11\frac{1}{2}$ days

Answer: $\displaystyle 11\frac{1}{2}$ days

Let the original number of men be $x$

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

$x \times 78 = (x + 16) \times 46$

$78x = 46x + 736$

$32x = 736$ ⇒ $x = 23$

Answer: $23$ men

$7$ men $= 11$ boys

$7$ men $\displaystyle = \frac{11}{7} \times 7 = 11$ boys

$7$ men + $4$ boys $= 11$ boys + $4$ boys $= 15$ boys

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

$11 \times 30 = 15 \times x$ ⇒ $x = 22$ days

Answer: $22$ days

Let 1 man do $m$ and 1 woman do $w$ units of work in 1 hour.

Case (i): $(4m + 8w) \times 5 \times 3 = 60m + 120w$

Case (ii): $(3m + 2w) \times 10 \times 4 = 120m + 80w$

Same work: $60m + 120w = 120m + 80w$

⇒ $40w = 60m$ ⇒ $2w = 3m$

So $8$ women $= 12$ men, and $4m + 8w = 16$ men

$16$ men working $5$ hours a day for $3$ days do the same work as $6$ men working $5$ hours a day for $x$ days

$16 \times 5 \times 3 = 6 \times 5 \times x$ ⇒ $x = 8$ days

Answer: $8$ 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 A $= 2k = 6$ ⇒ $k = 3$

Time taken by B $= 3k = 9$ days

A's 1 day's work $\displaystyle = \frac{1}{6}$, B's 1 day's work $\displaystyle = \frac{1}{9}$

(A + B)'s 1 day's work $\displaystyle = \frac{1}{6} + \frac{1}{9} = \frac{3 + 2}{18} = \frac{5}{18}$

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

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

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

$= 15 + 2 = 17$ km/hr

Answer: $17$ km/hr

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

$\displaystyle = \frac{28}{5} = 5.6$ km/hr

Answer: $5.6$ km/hr

Speed downstream: $\displaystyle x + y = \frac{7}{2}$ km/hr

Speed upstream: $\displaystyle x - y = \frac{5}{2}$ km/hr

Adding, $\displaystyle 2x = \frac{7}{2} + \frac{5}{2} = \frac{12}{2}$

$\displaystyle x = \frac{12}{4} = 3$ km/hr

Answer: $3$ km/hr

Let the rate of the stream be $y$ km/hr

Same distance, twice the time, so speed down $= 2 \times$ speed up

Given $2(6 - y) = 6 + y$

⇒ $12 - 2y = 6 + y$ ⇒ $3y = 6$

⇒ $y = 2$ km/hr

Answer: $2$ km/hr

Let the required distance be $d$ km

Speed downstream $= 12.5 + 1.5 = 14$ km/hr

Speed upstream $= 12.5 - 1.5 = 11$ km/hr

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

Time $\displaystyle = \frac{\text{Distance}}{\text{Speed}}$, so $\displaystyle \frac{d}{14} + \frac{d}{11} = 2.5$

⇒ $\displaystyle d\left(\frac{1}{14} + \frac{1}{11}\right) = 2.5$

⇒ $\displaystyle d \times \frac{25}{154} = 2.5$

⇒ $d = 15.4$ km

Answer: $15.4$ km

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

Speed upstream $= (x - y)$ km/hr, speed downstream $= (x + y)$ km/hr. Put $\displaystyle p = \frac{1}{x - y}$, $\displaystyle q = \frac{1}{x + y}$

$\displaystyle \frac{12}{x - y} + \frac{12}{x + y} = 3$ ⇒ $12p + 12q = 3 \quad \cdots (1)$

$\displaystyle \frac{42}{x - y} + \frac{84}{x + y} = 14$ ⇒ $42p + 84q = 14 \quad \cdots (2)$

(1) $\times 7$ ⇒ $84p + 84q = 21$; (2) $\times 2$ ⇒ $84p + 168q = 28$

(2) $\times 2$ $-$ (1) $\times 7$ ⇒ $84q = 7$ ⇒ $\displaystyle q = \frac{1}{12}$

From (1): $12p + 1 = 3$ ⇒ $12p = 2$ ⇒ $\displaystyle p = \frac{1}{6}$

So $x - y = 6$ and $x + y = 12$

Adding, $2x = 18$ ⇒ $x = 9$; then $y = 12 - 9 = 3$

∴ Rate of the current $= 3$ km/hr, speed in still water $= 9$ km/hr

Answer: Rate of current $3$ km/hr, speed in still water $9$ km/hr