Arithmetic Applications (Level 2): MCQ test, Set 10

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A can finish a piece of work in $12$ days. What part of the work does A finish in $2$ days?
A $\displaystyle \frac{1}{6}$ B $6$ C $\displaystyle \frac{5}{6}$ D $\displaystyle \frac{1}{12}$

Answer and solution

Answer: (A) $\displaystyle \frac{1}{6}$

A's 1 day's work $\displaystyle = \frac{1}{12}$

A's 2 days' work $\displaystyle = 2 \times \frac{1}{12} = \frac{2}{12} = \frac{1}{6}$
If the number of men to do a work is changed in the ratio $3 : 1$ , the time taken by them to finish the work changes in the ratio
A $3 : 1$ B $1 : 9$ C $9 : 1$ D $1 : 3$

Answer and solution

Answer: (D) $1 : 3$

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

$\displaystyle \frac{T_1}{T_2} = \frac{M_2}{M_1} = \frac{1}{3}$

∴ $T_1 : T_2 = 1 : 3$
A can finish a piece of work in $54$ days and B can finish it in $36$ days. In how many days can they finish it working together?
A $45$ days B $90$ days C $\displaystyle 21\frac{3}{5}$ days D $18$ days

Answer and solution

Answer: (C) $\displaystyle 21\frac{3}{5}$ days

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

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

∴ (A + B)'s 1 day's work $\displaystyle = \frac{1}{54} + \frac{1}{36} = \frac{2 + 3}{108} = \frac{5}{108}$

∴ Both together can finish the work in $\displaystyle \frac{108}{5}$ days $\displaystyle = 21\frac{3}{5}$ days
A and B together can finish a piece of work in $34$ days. A alone can do it in $51$ days. In how many days can B alone do it?
A $\displaystyle 20\frac{2}{5}$ days B $102$ days C $\displaystyle 42\frac{1}{2}$ days D $17$ days

Answer and solution

Answer: (B) $102$ days

(A + B)'s 1 day's work $\displaystyle = \frac{1}{34}$

A's 1 day's work $\displaystyle = \frac{1}{51}$

⇒ B's 1 day's work $\displaystyle = \frac{1}{34} - \frac{1}{51} = \frac{3 - 2}{102} = \frac{1}{102}$

∴ B alone can finish the work in $102$ days.
A, B and C together can finish a piece of work in $9$ days. A and B together can finish it in $12$ days. In how many days can C alone finish it?
A $\displaystyle 5\frac{1}{7}$ days B $3$ days C $\displaystyle 10\frac{1}{2}$ days D $36$ days

Answer and solution

Answer: (D) $36$ days

(A + B + C)'s 1 day's work $\displaystyle = \frac{1}{9}$

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

C's 1 day's work $=$ (A + B + C)'s 1 day's work $-$ (A + B)'s 1 day's work

$\displaystyle = \frac{1}{9} - \frac{1}{12} = \frac{4 - 3}{36} = \frac{1}{36}$

∴ C alone can finish the work in $36$ days
A can do a piece of work in $44$ days and B can finish it in $24$ days. They work together for $12$ days and then B goes away. In how many days will A finish the remaining work?
A $\displaystyle 5\frac{5}{11}$ days B $32$ days C $\displaystyle 41\frac{1}{6}$ days D $10$ days

Answer and solution

Answer: (D) $10$ days

A's 1 day's work $\displaystyle = \frac{1}{44}$

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

(A + B)'s 1 day's work $\displaystyle = \frac{1}{44} + \frac{1}{24} = \frac{6 + 11}{264} = \frac{17}{264}$

(A + B)'s 12 days' work $\displaystyle = 12 \times \frac{17}{264} = \frac{17}{22}$

Remaining work $\displaystyle = 1 - \frac{17}{22} = \frac{5}{22}$

This work is done by A in $\displaystyle \frac{5}{22} \times 44 = 10$ days
A can do a piece of work in $51$ days and B can finish it in $15$ days. They work together for $5$ days and then B goes away. In how many days will A finish the remaining work?
A $\displaystyle 46\frac{3}{5}$ days B $46$ days C $29$ days D $23$ days

Answer and solution

Answer: (C) $29$ days

A's 1 day's work $\displaystyle = \frac{1}{51}$

B's 1 day's work $\displaystyle = \frac{1}{15}$

(A + B)'s 1 day's work $\displaystyle = \frac{1}{51} + \frac{1}{15} = \frac{5 + 17}{255} = \frac{22}{255}$

(A + B)'s 5 days' work $\displaystyle = 5 \times \frac{22}{255} = \frac{22}{51}$

Remaining work $\displaystyle = 1 - \frac{22}{51} = \frac{29}{51}$

This work is done by A in $\displaystyle \frac{29}{51} \times 51 = 29$ days
A can do a piece of work in $20$ days and B in $15$ days. They work together for $6$ days. The rest of the work is finished by C in $7$ days. If they get ₹2100 for the whole work, what is A's share?
A ₹840 B ₹700 C ₹630 D ₹105

Answer and solution

Answer: (C) ₹630

A's 1 day's work $\displaystyle = \frac{1}{20}$ , B's 1 day's work $\displaystyle = \frac{1}{15}$

(A + B)'s 1 day's work $\displaystyle = \frac{1}{20} + \frac{1}{15} = \frac{3 + 4}{60} = \frac{7}{60}$

(A + B)'s 6 days' work $\displaystyle = \frac{7}{10}$

Remaining work $\displaystyle = 1 - \frac{7}{10} = \frac{3}{10}$ , done by C in $7$ days

A's 6 days' work $\displaystyle = \frac{6}{20} = \frac{3}{10}$ , B's 6 days' work $\displaystyle = \frac{6}{15} = \frac{2}{5}$ , C's 7 days' work $\displaystyle = \frac{3}{10}$

Consider $\displaystyle \frac{3}{10} : \frac{2}{5} : \frac{3}{10} = 3 : 4 : 3$

A's share $\displaystyle = 3 \times \frac{2100}{10} = 630$ , i.e. ₹630
A is thrice as good a workman as B and is therefore able to finish a piece of work in $52$ days less than B. How many days does A take to finish it alone?
A $156$ days B $78$ days C $\displaystyle 17\frac{1}{3}$ days D $26$ days

Answer and solution

Answer: (D) $26$ days

Suppose B finishes the work in $x$ days

⇒ A is able to finish it in $(x - 52)$ days

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

∴ $\displaystyle x - 52 = \frac{1}{3}x$

⇒ $3x - 156 = x$ ⇒ $2x = 156$ ⇒ $x = 78$

∴ Time taken by B $= 78$ days, time taken by A $= 26$ days
A can do a piece of work in $30$ days, B in $10$ days and C in $27$ days. All begin together, but A leaves the work after $2$ days and B leaves $3$ days before the work is finished. How long did the work last?
A $6$ days B $9$ days C $12$ days D $4$ days

Answer and solution

Answer: (B) $9$ days

A's 1 day's work $\displaystyle = \frac{1}{30}$ , B's $\displaystyle = \frac{1}{10}$ , C's $\displaystyle = \frac{1}{27}$

A, B, C together work for $2$ days; C alone works for the last $3$ days; B and C work together in between.

(A + B + C)'s 1 day's work $\displaystyle = \frac{1}{30} + \frac{1}{10} + \frac{1}{27} = \frac{9 + 27 + 10}{270} = \frac{46}{270} = \frac{23}{135}$

(A + B + C)'s 2 days' work $\displaystyle = 2 \times \frac{23}{135} = \frac{46}{135}$

C's 3 days' work $\displaystyle = 3 \times \frac{1}{27} = \frac{1}{9}$

Remaining work $\displaystyle = 1 - \left(\frac{46}{135} + \frac{1}{9}\right) = \frac{74}{135}$

(B + C)'s 1 day's work $\displaystyle = \frac{1}{10} + \frac{1}{27} = \frac{27 + 10}{270} = \frac{37}{270}$

Remaining work is done by (B + C) in $\displaystyle \frac{74}{135} \div \frac{37}{270} = 4$ days

Total time $= 2 + 4 + 3 = 9$ days
A can do a piece of work in $80$ days and B can do it in $96$ days. They work together for $4$ days. Then B leaves and A alone continues the work. $10$ days after that, C joins A and the work is completed in $36$ days more. In how many days can C do the work alone?
A $188$ days B $216$ days C $108$ days D $72$ days

Answer and solution

Answer: (C) $108$ days

A's 1 day's work $\displaystyle = \frac{1}{80}$ , B's 1 day's work $\displaystyle = \frac{1}{96}$

(A + B)'s 1 day's work $\displaystyle = \frac{1}{80} + \frac{1}{96} = \frac{6 + 5}{480} = \frac{11}{480}$

(A + B)'s 4 days' work $\displaystyle = 4 \times \frac{11}{480} = \frac{11}{120}$

A's 10 days' work $\displaystyle = \frac{10}{80} = \frac{1}{8}$

(A + C)'s 36 days' work $\displaystyle = 1 - \left(\frac{11}{120} + \frac{1}{8}\right) = \frac{47}{60}$

⇒ (A + C)'s 1 day's work $\displaystyle = \frac{47}{60} \div 36 = \frac{47}{2160}$

C's 1 day's work $\displaystyle = \frac{47}{2160} - \frac{1}{80} = \frac{47 - 27}{2160} = \frac{20}{2160} = \frac{1}{108}$

C can finish the work in $108$ days
A and B together can do a piece of work in $33$ days. B and C together can do it in $44$ days. If A is twice as good a workman as C, C alone can do it in
A $264$ days B $132$ days C $11$ days D $66$ days

Answer and solution

Answer: (B) $132$ days

(A + B)'s 1 day's work $\displaystyle = \frac{1}{33}$ , (B + C)'s 1 day's work $\displaystyle = \frac{1}{44}$

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

C's 2 days' work + B's 1 day's work $\displaystyle = \frac{1}{33}$

⇒ C's 1 day's work + (B + C)'s 1 day's work $\displaystyle = \frac{1}{33}$

⇒ C's 1 day's work $\displaystyle + \frac{1}{44} = \frac{1}{33}$

C's 1 day's work $\displaystyle = \frac{1}{33} - \frac{1}{44} = \frac{4 - 3}{132} = \frac{1}{132}$

∴ C alone can finish it in $132$ days
A and B working alone can finish a piece of work in $14$ and $17$ days respectively. If they work at it alternately for a day, A beginning, in how many days will the work be finished?
A $\displaystyle 15\frac{2}{7}$ days B $\displaystyle 16\frac{2}{7}$ days C $15$ days D $16$ days

Answer and solution

Answer: (A) $\displaystyle 15\frac{2}{7}$ days

A's 1 day's work $\displaystyle = \frac{1}{14}$ , B's 1 day's work $\displaystyle = \frac{1}{17}$

Work done in the first 2 days $\displaystyle = \frac{1}{14} + \frac{1}{17} = \frac{17 + 14}{238} = \frac{31}{238}$

Work done in the first 14 days $\displaystyle = 7 \times \frac{31}{238} = \frac{31}{34}$

On day 15 it is A's turn: work done in the first 15 days $\displaystyle = \frac{31}{34} + \frac{1}{14} = \frac{117}{119}$

Remaining work $\displaystyle = 1 - \frac{117}{119} = \frac{2}{119}$

B does it in $\displaystyle \frac{2}{119} \div \frac{1}{17} = \frac{2}{7}$ day

∴ Total no. of days $\displaystyle = 15 + \frac{2}{7} = 15\frac{2}{7}$ days
A certain number of men complete a piece of work in $60$ days. If there were $22$ men more, the work could be finished in $24$ days less. How many men were there originally?
A $11$ B $33$ C $55$ D $77$

Answer and solution

Answer: (B) $33$

Let the original number of men be $x$

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

$x \times 60 = (x + 22) \times 36$

$60x = 36x + 792$

$24x = 792$ ⇒ $x = 33$
If $7$ men or $15$ boys can finish a piece of work in $12$ days, in how many days can $7$ men and $3$ boys finish it?
A $18$ days B $\displaystyle 14\frac{2}{5}$ days C $\displaystyle 8\frac{2}{5}$ days D $10$ days

Answer and solution

Answer: (D) $10$ days

$7$ men $= 15$ boys

$7$ men $\displaystyle = \frac{15}{7} \times 7 = 15$ boys

$7$ men + $3$ boys $= 15$ boys + $3$ boys $= 18$ boys

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

$15 \times 12 = 18 \times x$ ⇒ $x = 10$ days
$7$ men and $6$ women working $10$ hours a day finish a piece of work in $3$ days; $8$ men and $8$ women working $6$ hours a day do the same work in $4$ days. Find the number of days in which the work is done by $4$ men only, working $6$ hours a day.
A $\displaystyle 16\frac{1}{4}$ days B $20$ days C $\displaystyle 13\frac{3}{4}$ days D $12$ days

Answer and solution

Answer: (B) $20$ days

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

Case (i): $(7m + 6w) \times 10 \times 3 = 210m + 180w$

Case (ii): $(8m + 8w) \times 6 \times 4 = 192m + 192w$

Same work: $210m + 180w = 192m + 192w$

⇒ $12w = 18m$ ⇒ $2w = 3m$

So $6$ women $= 9$ men, and $7m + 6w = 16$ men

$16$ men working $10$ hours a day for $3$ days do the same work as $4$ men working $6$ hours a day for $x$ days

$16 \times 10 \times 3 = 4 \times 6 \times x$ ⇒ $x = 20$ days
If $6$ men with $5$ boys can earn ₹4680 in $4$ days and $12$ men with $6$ boys can earn ₹16800 in $8$ days, in what time will $9$ men with $9$ boys earn ₹16605?
A $\displaystyle 4\frac{1}{2}$ days B $9$ days C $11$ days D $7$ days

Answer and solution

Answer: (B) $9$ days

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

$(6x + 5y) \times 4 = 4680$ ⇒ $6x + 5y = 1170 \quad \cdots (1)$

$(12x + 6y) \times 8 = 16800$ ⇒ $12x + 6y = 2100 \quad \cdots (2)$

(1) $\times 2$ ⇒ $12x + 10y = 2340$

(1) $\times 2$ $-$ (2) ⇒ $4y = 240$ ⇒ $y = 60$

From (1): $6x = 1170 - 300 = 870$ ⇒ $x = 145$

$(9x + 9y)n = 16605$

⇒ $(1305 + 540)n = 16605$ ⇒ $1845n = 16605$

⇒ $n = 9$ days
$15$ men are employed to do a piece of work which they could finish in $28$ days, but the men drop off by $3$ at the end of every $15$ days. In what time will the work be completed?
A $45$ days B $35$ days C $\displaystyle 31\frac{2}{3}$ days D $28$ days

Answer and solution

Answer: (C) $\displaystyle 31\frac{2}{3}$ days

Work done by 15 men in 15 days $\displaystyle = 15 \times \frac{1}{28} = \frac{15}{28}$

So 1 man's work in 15 days $\displaystyle = \frac{15}{28} \div 15 = \frac{1}{28}$

Remaining work $\displaystyle = \frac{13}{28}$ , remaining men $= 12$ ; work done by 12 men in 15 days $\displaystyle = 12 \times \frac{1}{28} = \frac{3}{7}$

Remaining work $\displaystyle = \frac{1}{28}$ , remaining men $= 9$ ; work done by 9 men in 15 days $\displaystyle = \frac{9}{28}$

$\displaystyle \frac{1}{28}$ work is done by 9 men in $\displaystyle \frac{1}{28} \div \frac{9}{28} \times 15 = 1\frac{2}{3}$ days

Total time $\displaystyle = 15 + 15 + 1\frac{2}{3} = 31\frac{2}{3}$ days
A can do half as much again as B can do in the same time. A alone can do a piece of work in $40$ days. In how many days can both together finish it?
A $24$ days B $36$ days C $30$ days D $50$ days

Answer and solution

Answer: (A) $24$ 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 $= 2k = 40$ ⇒ $k = 20$

Time taken by B $= 3k = 60$ days

A's 1 day's work $\displaystyle = \frac{1}{40}$ , B's 1 day's work $\displaystyle = \frac{1}{60}$

(A + B)'s 1 day's work $\displaystyle = \frac{1}{40} + \frac{1}{60} = \frac{3 + 2}{120} = \frac{5}{120} = \frac{1}{24}$

Time taken by (A + B) to finish the work $= 24$ days
The speed of a boat in still water is $5.5$ km/hr and the speed of the stream is $2.5$ km/hr. The speed of the boat upstream is
A $3$ km/hr B $5.5$ km/hr C $8$ km/hr D $4$ km/hr

Answer and solution

Answer: (A) $3$ km/hr

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

$= 5.5 - 2.5 = 3$ km/hr
A man rows upstream $23$ km and downstream $39$ km, taking $5$ hours each time. The velocity of the current is
A $6.2$ km/hr B $3.2$ km/hr C $1.6$ km/hr D $4.2$ km/hr

Answer and solution

Answer: (C) $1.6$ km/hr

Speed downstream: $\displaystyle x + y = \frac{39}{5}$ km/hr

Speed upstream: $\displaystyle x - y = \frac{23}{5}$ km/hr

Subtracting, $\displaystyle 2y = \frac{39}{5} - \frac{23}{5} = \frac{16}{5}$

$\displaystyle y = \frac{16}{10} = 1.6$ km/hr
A man can row $12$ km/hr in still water and he finds that it takes him twice as long to row up as to row down the river. The rate of the stream is
A $8$ km/hr B $10$ km/hr C $4$ km/hr D $6$ km/hr

Answer and solution

Answer: (C) $4$ 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 $2(12 - y) = 12 + y$

⇒ $24 - 2y = 12 + y$ ⇒ $3y = 12$

⇒ $y = 4$ km/hr
A man can row $12.5$ km/hr in still water. If the river is running at $6$ km/hr, it takes him 1 hour 30 minutes to row to a place and back. The place is at a distance of
A $14.43$ km B $9.75$ km C $9.375$ km D $7.215$ km

Answer and solution

Answer: (D) $7.215$ km

Let the required distance be $d$ km

Speed downstream $= 12.5 + 6 = 18.5$ km/hr

Speed upstream $= 12.5 - 6 = 6.5$ km/hr

Total time $=$ 1 hour 30 minutes $= 1.5$ hours

Time $\displaystyle = \frac{\text{Distance}}{\text{Speed}}$ , so $\displaystyle \frac{d}{18.5} + \frac{d}{6.5} = 1.5$

⇒ $\displaystyle d\left(\frac{2}{37} + \frac{2}{13}\right) = 1.5$

⇒ $\displaystyle d \times \frac{100}{481} = 1.5$

⇒ $d = 7.215$ km
The current of a stream runs at the rate of $1$ km/hr. A motor boat goes $8$ km upstream and back again to the starting point in 2 hours 20 minutes. The speed of the motor boat in still water is
A $7$ km/hr B $6$ km/hr C $8$ km/hr D $9$ km/hr

Answer and solution

Answer: (A) $7$ km/hr

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 $= 140$ min $\displaystyle = \frac{140}{60}$ hr $\displaystyle = \frac{7}{3}$ hr

$\displaystyle \frac{8}{x - 1} + \frac{8}{x + 1} = \frac{7}{3}$

⇒ $\displaystyle 8\left(\frac{(x + 1) + (x - 1)}{(x - 1)(x + 1)}\right) = \frac{7}{3}$

⇒ $\displaystyle \frac{16x}{x^2 - 1} = \frac{7}{3}$

⇒ $48x = 7x^2 - 7$

⇒ $7 x^{2} - 48 x - 7 = 0$

⇒ $\left(x - 7\right) \left(7 x + 1\right) = 0$

⇒ $x = 7$ (or) $\displaystyle x = -\frac{1}{7}$ (rejected, a speed cannot be negative)

∴ Speed of the boat in still water $= 7$ km/hr
A man can row $6$ km upstream and $24$ km downstream in $9$ hours, while he can row $12$ km upstream and $12$ km downstream in $9$ hours. The rate of the current is
A $1$ km/hr B $4$ km/hr C $3$ km/hr D $2$ km/hr

Answer and solution

Answer: (A) $1$ km/hr

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 $\displaystyle p = \frac{1}{x - y}$ , $\displaystyle q = \frac{1}{x + y}$

$\displaystyle \frac{6}{x - y} + \frac{24}{x + y} = 9$ ⇒ $6p + 24q = 9 \quad \cdots (1)$

$\displaystyle \frac{12}{x - y} + \frac{12}{x + y} = 9$ ⇒ $12p + 12q = 9 \quad \cdots (2)$

(1) $\times 2$ ⇒ $12p + 48q = 18$

(1) $\times 2$ $-$ (2) ⇒ $36q = 9$ ⇒ $\displaystyle q = \frac{1}{4}$

From (1): $6p + 6 = 9$ ⇒ $6p = 3$ ⇒ $\displaystyle p = \frac{1}{2}$

So $x - y = 2$ and $x + y = 4$

Adding, $2x = 6$ ⇒ $x = 3$ ; then $y = 4 - 3 = 1$

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MCQ test, Set 10