Arithmetic Applications (Level 2): MCQ test, Set 9

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Choose one answer for each question, then check your score at the end. Each correct answer carries 1 mark; there is no negative marking.

If A can finish a piece of work in $9$ days, then A's 1 day's work is
A $9$ B $\displaystyle \frac{1}{9}$ C $\displaystyle \frac{1}{8}$ D $\displaystyle \frac{8}{9}$

Answer and solution

Answer: (B) $\displaystyle \frac{1}{9}$

A finishes the whole work in $9$ days.

∴ A's 1 day's work $\displaystyle = \frac{1}{9}$
$6$ men can finish a piece of work in $40$ days. In how many days can $15$ men finish the same work?
A $100$ days B $31$ days C $49$ days D $16$ days

Answer and solution

Answer: (D) $16$ days

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

$6 \times 40 = 15 \times x$

$\displaystyle x = \frac{6 \times 40}{15} = 16$ days
A can finish a piece of work in $60$ days and B can finish it in $12$ days. In how many days can they finish it working together?
A $72$ days B $10$ days C $\displaystyle \frac{1}{10}$ days D $36$ days

Answer and solution

Answer: (B) $10$ days

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

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

∴ (A + B)'s 1 day's work $\displaystyle = \frac{1}{60} + \frac{1}{12} = \frac{1 + 5}{60} = \frac{6}{60} = \frac{1}{10}$

∴ Both together can finish the work in $10$ days
A and B together can finish a piece of work in $38$ days. A alone can do it in $114$ days. In how many days can B alone do it?
A $76$ days B $57$ days C $\displaystyle 28\frac{1}{2}$ days D $152$ days

Answer and solution

Answer: (B) $57$ days

(A + B)'s 1 day's work $\displaystyle = \frac{1}{38}$

A's 1 day's work $\displaystyle = \frac{1}{114}$

⇒ B's 1 day's work $\displaystyle = \frac{1}{38} - \frac{1}{114} = \frac{3 - 1}{114} = \frac{2}{114} = \frac{1}{57}$

∴ B alone can finish the work in $57$ days.
A and B can do a piece of work in $70$ days; B and C in $84$ days; C and A in $60$ days. In how many days can A alone finish it?
A $210$ days B $140$ days C $105$ days D $\displaystyle 46\frac{2}{3}$ days

Answer and solution

Answer: (C) $105$ days

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

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

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

Adding, 2(A + B + C)'s 1 day's work $\displaystyle = \frac{1}{70} + \frac{1}{84} + \frac{1}{60} = \frac{6 + 5 + 7}{420} = \frac{18}{420} = \frac{3}{70}$

⇒ (A + B + C)'s 1 day's work $\displaystyle = \frac{3}{140}$

∴ A, B, C together can finish the work in $\displaystyle 46\frac{2}{3}$ days.

A's 1 day's work $\displaystyle = \frac{3}{140} - \frac{1}{84} = \frac{1}{105}$ , so A alone takes $105$ days
A can do a piece of work in $50$ days and B can finish it in $16$ days. They work together for $8$ days and then B goes away. In how many days will A finish the remaining work?
A $42$ days B $\displaystyle 45\frac{7}{8}$ days C $17$ days D $21$ days

Answer and solution

Answer: (C) $17$ days

A's 1 day's work $\displaystyle = \frac{1}{50}$

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

(A + B)'s 1 day's work $\displaystyle = \frac{1}{50} + \frac{1}{16} = \frac{8 + 25}{400} = \frac{33}{400}$

(A + B)'s 8 days' work $\displaystyle = 8 \times \frac{33}{400} = \frac{33}{50}$

Remaining work $\displaystyle = 1 - \frac{33}{50} = \frac{17}{50}$

This work is done by A in $\displaystyle \frac{17}{50} \times 50 = 17$ days
A can do a piece of work in $22$ days while B alone can do it in $44$ days. With the help of C they can finish it in $12$ days. If they are paid ₹11550 for the whole work, what is B's share?
A ₹6300 B ₹2100 C ₹3850 D ₹3150

Answer and solution

Answer: (D) ₹3150

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

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

(A + B)'s 1 day's work $\displaystyle = \frac{1}{22} + \frac{1}{44} = \frac{2 + 1}{44} = \frac{3}{44}$

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

C's 1 day's work $\displaystyle = \frac{1}{12} - \frac{3}{44} = \frac{11 - 9}{132} = \frac{2}{132} = \frac{1}{66}$

Consider $\displaystyle \frac{1}{22} : \frac{1}{44} : \frac{1}{66} = \frac{6}{132} : \frac{3}{132} : \frac{2}{132} = 6 : 3 : 2$

∴ ₹11550 must be divided in the ratio $6 : 3 : 2$ ( $6 + 3 + 2 = 11$ )

B's share $\displaystyle = 11550 \times \frac{3}{11} = 3150$ , i.e. ₹3150
A is twice as good a workman as B and is therefore able to finish a piece of work in $30$ days less than B. In how many days can they do it working together?
A $45$ days B $30$ days C $60$ days D $20$ days

Answer and solution

Answer: (D) $20$ days

Suppose B finishes the work in $x$ days

⇒ A is able to finish it in $(x - 30)$ days

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

∴ $\displaystyle x - 30 = \frac{1}{2}x$

⇒ $2x - 60 = x$ ⇒ $x = 60$

∴ Time taken by B $= 60$ days, time taken by A $= 30$ days

A's 1 day's work $\displaystyle = \frac{1}{30}$ , B's 1 day's work $\displaystyle = \frac{1}{60}$

(A + B)'s 1 day's work $\displaystyle = \frac{1}{30} + \frac{1}{60} = \frac{2 + 1}{60} = \frac{3}{60} = \frac{1}{20}$

A and B together can finish the work in $20$ days
A and B can do a piece of work in $20$ and $80$ days respectively. They began the work together, but A leaves after some days and B finishes the remaining work in $40$ days. After how many days did A leave?
A $8$ days B $48$ days C $40$ days D $10$ days

Answer and solution

Answer: (A) $8$ days

A's 1 day's work $\displaystyle = \frac{1}{20}$ , B's 1 day's work $\displaystyle = \frac{1}{80}$

(A + B)'s 1 day's work $\displaystyle = \frac{1}{20} + \frac{1}{80} = \frac{4 + 1}{80} = \frac{5}{80} = \frac{1}{16}$

B's 40 days' work $\displaystyle = 40 \times \frac{1}{80} = \frac{1}{2}$

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

(A + B)'s $x$ days' work $\displaystyle = \frac{1}{2}$ ⇒ $\displaystyle \frac{1}{16} \times x = \frac{1}{2}$

$x = 8$ , so A left after $8$ days
A can do a piece of work in $208$ days and B can do it in $130$ days. They work together for $10$ days. Then B leaves and A alone continues the work. $29$ days after that, C joins A and the work is completed in $75$ days more. In how many days can C do the work alone?
A $200$ days B $125$ days C $408$ days D $\displaystyle 472\frac{8}{11}$ days

Answer and solution

Answer: (A) $200$ days

A's 1 day's work $\displaystyle = \frac{1}{208}$ , B's 1 day's work $\displaystyle = \frac{1}{130}$

(A + B)'s 1 day's work $\displaystyle = \frac{1}{208} + \frac{1}{130} = \frac{5 + 8}{1040} = \frac{13}{1040} = \frac{1}{80}$

(A + B)'s 10 days' work $\displaystyle = 10 \times \frac{1}{80} = \frac{1}{8}$

A's 29 days' work $\displaystyle = \frac{29}{208}$

(A + C)'s 75 days' work $\displaystyle = 1 - \left(\frac{1}{8} + \frac{29}{208}\right) = \frac{153}{208}$

⇒ (A + C)'s 1 day's work $\displaystyle = \frac{153}{208} \div 75 = \frac{51}{5200}$

C's 1 day's work $\displaystyle = \frac{51}{5200} - \frac{1}{208} = \frac{51 - 25}{5200} = \frac{26}{5200} = \frac{1}{200}$

C can finish the work in $200$ days
A and B together can do a piece of work in $4$ days. B and C together can do it in $6$ days. If A is twice as good a workman as C, B alone can do it in
A $6$ days B $24$ days C $2$ days D $12$ days

Answer and solution

Answer: (D) $12$ days

(A + B)'s 1 day's work $\displaystyle = \frac{1}{4}$ , (B + C)'s 1 day's work $\displaystyle = \frac{1}{6}$

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

C's 2 days' work + B's 1 day's work $\displaystyle = \frac{1}{4}$

⇒ C's 1 day's work + (B + C)'s 1 day's work $\displaystyle = \frac{1}{4}$

⇒ C's 1 day's work $\displaystyle + \frac{1}{6} = \frac{1}{4}$

C's 1 day's work $\displaystyle = \frac{1}{4} - \frac{1}{6} = \frac{3 - 2}{12} = \frac{1}{12}$

B's 1 day's work $\displaystyle = \frac{1}{6} - \frac{1}{12} = \frac{2 - 1}{12} = \frac{1}{12}$

∴ B alone can finish the work in $12$ days
A and B working alone can finish a piece of work in $24$ and $6$ days respectively. If they work at it alternately for a day, A beginning, in how many days will the work be finished?
A $\displaystyle 9\frac{3}{5}$ days B $\displaystyle 4\frac{4}{5}$ days C $10$ days D $\displaystyle 9\frac{3}{4}$ days

Answer and solution

Answer: (D) $\displaystyle 9\frac{3}{4}$ days

A's 1 day's work $\displaystyle = \frac{1}{24}$ , B's 1 day's work $\displaystyle = \frac{1}{6}$

Work done in the first 2 days $\displaystyle = \frac{1}{24} + \frac{1}{6} = \frac{1 + 4}{24} = \frac{5}{24}$

Work done in the first 8 days $\displaystyle = 4 \times \frac{5}{24} = \frac{5}{6}$

On day 9 it is A's turn: work done in the first 9 days $\displaystyle = \frac{5}{6} + \frac{1}{24} = \frac{7}{8}$

Remaining work $\displaystyle = 1 - \frac{7}{8} = \frac{1}{8}$

B does it in $\displaystyle \frac{1}{8} \div \frac{1}{6} = \frac{3}{4}$ day

∴ Total no. of days $\displaystyle = 9 + \frac{3}{4} = 9\frac{3}{4}$ days
A certain number of men complete a piece of work in $64$ days. If there were $12$ men more, the work could be finished in $16$ days less. How many men were there originally?
A $60$ B $36$ C $48$ D $24$

Answer and solution

Answer: (B) $36$

Let the original number of men be $x$

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

$x \times 64 = (x + 12) \times 48$

$64x = 48x + 576$

$16x = 576$ ⇒ $x = 36$
If $8$ men or $12$ boys can finish a piece of work in $20$ days, in how many days can $2$ men and $3$ boys finish it?
A $40$ days B $48$ days C $32$ days D $10$ days

Answer and solution

Answer: (A) $40$ days

$8$ men $= 12$ boys

$2$ men $\displaystyle = \frac{12}{8} \times 2 = 3$ boys

$2$ men + $3$ boys $= 3$ boys + $3$ boys $= 6$ boys

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

$12 \times 20 = 6 \times x$ ⇒ $x = 40$ days
$4$ men and $6$ women working $8$ hours a day finish a piece of work in $5$ days; $2$ men and $6$ women working $7$ hours a day do the same work in $8$ days. Find the number of days in which the work is done by $7$ men only, working $5$ hours a day.
A $\displaystyle 11\frac{3}{7}$ days B $\displaystyle 18\frac{2}{7}$ days C $8$ days D $5$ days

Answer and solution

Answer: (C) $8$ days

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

Case (i): $(4m + 6w) \times 8 \times 5 = 160m + 240w$

Case (ii): $(2m + 6w) \times 7 \times 8 = 112m + 336w$

Same work: $160m + 240w = 112m + 336w$

⇒ $96w = 48m$ ⇒ $2w = m$

So $6$ women $= 3$ men, and $4m + 6w = 7$ men

$7$ men working $8$ hours a day for $5$ days do the same work as $7$ men working $5$ hours a day for $x$ days

$7 \times 8 \times 5 = 7 \times 5 \times x$ ⇒ $x = 8$ days
If $5$ men with $8$ boys can earn ₹4455 in $3$ days and $10$ men with $11$ boys can earn ₹14970 in $6$ days, in what time will $4$ men with $9$ boys earn ₹27265?
A $17$ days B $21$ days C $\displaystyle 9\frac{1}{2}$ days D $19$ days

Answer and solution

Answer: (D) $19$ days

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

$(5x + 8y) \times 3 = 4455$ ⇒ $5x + 8y = 1485 \quad \cdots (1)$

$(10x + 11y) \times 6 = 14970$ ⇒ $10x + 11y = 2495 \quad \cdots (2)$

(1) $\times 2$ ⇒ $10x + 16y = 2970$

(1) $\times 2$ $-$ (2) ⇒ $5y = 475$ ⇒ $y = 95$

From (1): $5x = 1485 - 760 = 725$ ⇒ $x = 145$

$(4x + 9y)n = 27265$

⇒ $(580 + 855)n = 27265$ ⇒ $1435n = 27265$

⇒ $n = 19$ days
$40$ men are employed to do a piece of work which they could finish in $38$ days, but the men drop off by $4$ at the end of every $16$ days. In what time will the work be completed?
A $48$ days B $\displaystyle 42\frac{2}{9}$ days C $\displaystyle 41\frac{1}{2}$ days D $38$ days

Answer and solution

Answer: (C) $\displaystyle 41\frac{1}{2}$ days

Work done by 40 men in 16 days $\displaystyle = 16 \times \frac{1}{38} = \frac{8}{19}$

So 1 man's work in 16 days $\displaystyle = \frac{8}{19} \div 40 = \frac{1}{95}$

Remaining work $\displaystyle = \frac{11}{19}$ , remaining men $= 36$ ; work done by 36 men in 16 days $\displaystyle = 36 \times \frac{1}{95} = \frac{36}{95}$

Remaining work $\displaystyle = \frac{1}{5}$ , remaining men $= 32$ ; work done by 32 men in 16 days $\displaystyle = \frac{32}{95}$

$\displaystyle \frac{1}{5}$ work is done by 32 men in $\displaystyle \frac{1}{5} \div \frac{32}{95} \times 16 = 9\frac{1}{2}$ days

Total time $\displaystyle = 16 + 16 + 9\frac{1}{2} = 41\frac{1}{2}$ days
A can do half as much again as B can do in the same time. B alone can do a piece of work in $39$ days. In how many days can both together finish it?
A $\displaystyle 23\frac{2}{5}$ days B $\displaystyle 15\frac{3}{5}$ days C $\displaystyle 32\frac{1}{2}$ days D $\displaystyle 19\frac{1}{2}$ days

Answer and solution

Answer: (B) $\displaystyle 15\frac{3}{5}$ 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 $= 3k = 39$ ⇒ $k = 13$

Time taken by A $= 2k = 26$ days

A's 1 day's work $\displaystyle = \frac{1}{26}$ , B's 1 day's work $\displaystyle = \frac{1}{39}$

(A + B)'s 1 day's work $\displaystyle = \frac{1}{26} + \frac{1}{39} = \frac{3 + 2}{78} = \frac{5}{78}$

Time taken by (A + B) to finish the work $\displaystyle = 15\frac{3}{5}$ days
A man can row downstream at the rate of $16$ km/hr and upstream at $12$ km/hr. The man's rate in still water is
A $4$ km/hr B $14$ km/hr C $28$ km/hr D $2$ km/hr

Answer and solution

Answer: (B) $14$ km/hr

Given $x + y = 16$ and $x - y = 12$

Adding, $2x = 28$ ⇒ $x = 14$ km/hr
A man rows upstream $18$ km and downstream $23$ km, taking $5$ hours each time. The velocity of the current is
A $0.5$ km/hr B $2$ km/hr C $1$ km/hr D $4.1$ km/hr

Answer and solution

Answer: (A) $0.5$ km/hr

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

Speed upstream: $\displaystyle x - y = \frac{18}{5}$ km/hr

Subtracting, $\displaystyle 2y = \frac{23}{5} - \frac{18}{5} = \frac{5}{5}$

$\displaystyle y = \frac{5}{10} = 0.5$ km/hr
A man can row $13.5$ 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 $11.5$ km/hr B $9$ km/hr C $4.5$ km/hr D $6.75$ km/hr

Answer and solution

Answer: (C) $4.5$ 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 $2(13.5 - y) = 13.5 + y$

⇒ $27 - 2y = 13.5 + y$ ⇒ $3y = 13.5$

⇒ $y = 4.5$ km/hr
A man can row $14$ km/hr in still water and he finds that it takes him thrice as long to row up as to row down the river. The rate of the stream is
A $11$ km/hr B $10.5$ km/hr C $7$ km/hr D $3.5$ km/hr

Answer and solution

Answer: (C) $7$ km/hr

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

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

Given $3(14 - y) = 14 + y$

⇒ $42 - 3y = 14 + y$ ⇒ $4y = 28$

⇒ $y = 7$ km/hr
A man can row $9$ km/hr in still water. If the river is running at $3$ km/hr, it takes him 3 hours to row to a place and back. The place is at a distance of
A $18$ km B $13.5$ km C $24$ km D $12$ km

Answer and solution

Answer: (D) $12$ km

Let the required distance be $d$ km

Speed downstream $= 9 + 3 = 12$ km/hr

Speed upstream $= 9 - 3 = 6$ km/hr

Time $\displaystyle = \frac{\text{Distance}}{\text{Speed}}$ , so $\displaystyle \frac{d}{12} + \frac{d}{6} = 3$

⇒ $\displaystyle d\left(\frac{1}{12} + \frac{1}{6}\right) = 3$

⇒ $\displaystyle d \times \frac{1}{4} = 3$

⇒ $d = 12$ km
The current of a stream runs at the rate of $4$ km/hr. A motor boat goes $36$ km upstream and back again to the starting point in 3 hours 45 minutes. The speed of the motor boat in still water is
A $20$ km/hr B $16$ km/hr C $24$ km/hr D $28$ km/hr

Answer and solution

Answer: (A) $20$ km/hr

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 $= 225$ min $\displaystyle = \frac{225}{60}$ hr $\displaystyle = \frac{15}{4}$ hr

$\displaystyle \frac{36}{x - 4} + \frac{36}{x + 4} = \frac{15}{4}$

⇒ $\displaystyle 36\left(\frac{(x + 4) + (x - 4)}{(x - 4)(x + 4)}\right) = \frac{15}{4}$

⇒ $\displaystyle \frac{72x}{x^2 - 16} = \frac{15}{4}$

⇒ $288x = 15x^2 - 240$

⇒ $5 x^{2} - 96 x - 80 = 0$

⇒ $\left(x - 20\right) \left(5 x + 4\right) = 0$

⇒ $x = 20$ (or) $\displaystyle x = -\frac{4}{5}$ (rejected, a speed cannot be negative)

∴ Speed of the boat in still water $= 20$ km/hr
A man can row $8$ km upstream and $84$ km downstream in $8$ hours, while he can row $24$ km upstream and $72$ km downstream in $9$ hours. The speed of the man in still water is
A $10$ km/hr B $12$ km/hr C $2$ km/hr D $8$ km/hr

Answer and solution

Answer: (A) $10$ 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 $\displaystyle p = \frac{1}{x - y}$ , $\displaystyle q = \frac{1}{x + y}$

$\displaystyle \frac{8}{x - y} + \frac{84}{x + y} = 8$ ⇒ $8p + 84q = 8 \quad \cdots (1)$

$\displaystyle \frac{24}{x - y} + \frac{72}{x + y} = 9$ ⇒ $24p + 72q = 9 \quad \cdots (2)$

(1) $\times 3$ ⇒ $24p + 252q = 24$

(1) $\times 3$ $-$ (2) ⇒ $180q = 15$ ⇒ $\displaystyle q = \frac{1}{12}$

From (1): $8p + 7 = 8$ ⇒ $8p = 1$ ⇒ $\displaystyle p = \frac{1}{8}$

So $x - y = 8$ and $x + y = 12$

Adding, $2x = 20$ ⇒ $x = 10$ ; then $y = 12 - 10 = 2$

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MCQ test, Set 9