Arithmetic Applications (Level 2): MCQ test, Set 4

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A can finish a piece of work in $60$ days. What part of the work does A finish in $18$ days?
A $\displaystyle \frac{1}{60}$ B $\displaystyle \frac{7}{10}$ C $\displaystyle \frac{3}{10}$ D $\displaystyle \frac{10}{3}$

Answer and solution

Answer: (C) $\displaystyle \frac{3}{10}$

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

A's 18 days' work $\displaystyle = 18 \times \frac{1}{60} = \frac{18}{60} = \frac{3}{10}$
If the number of men to do a work is changed in the ratio $3 : 7$ , the time taken by them to finish the work changes in the ratio
A $49 : 9$ B $9 : 49$ C $3 : 7$ D $7 : 3$

Answer and solution

Answer: (D) $7 : 3$

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

$\displaystyle \frac{T_1}{T_2} = \frac{M_2}{M_1} = \frac{7}{3}$

∴ $T_1 : T_2 = 7 : 3$
A can do a piece of work in $6$ days and B alone can do it in $8$ days. How much time will both take to finish the work?
A $7$ days B $14$ days C $\displaystyle 3\frac{3}{7}$ days D $2$ days

Answer and solution

Answer: (C) $\displaystyle 3\frac{3}{7}$ days

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

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

∴ (A + B)'s 1 day's work $\displaystyle = \frac{1}{6} + \frac{1}{8} = \frac{4 + 3}{24} = \frac{7}{24}$

∴ Both together can finish the work in $\displaystyle \frac{24}{7}$ days $\displaystyle = 3\frac{3}{7}$ days
A and B together can do a piece of work in $36$ days. B alone can finish it in $52$ days. In how many days can A alone finish the work?
A $16$ days B $\displaystyle 21\frac{3}{11}$ days C $44$ days D $117$ days

Answer and solution

Answer: (D) $117$ days

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

B's 1 day's work $\displaystyle = \frac{1}{52}$

⇒ A's 1 day's work $\displaystyle = \frac{1}{36} - \frac{1}{52} = \frac{13 - 9}{468} = \frac{4}{468} = \frac{1}{117}$

∴ A alone can finish the work in $117$ days.
A, B and C together can finish a piece of work in $26$ days. A and B together can finish it in $39$ days. In how many days can C alone finish it?
A $\displaystyle 32\frac{1}{2}$ days B $78$ days C $13$ days D $\displaystyle 15\frac{3}{5}$ days

Answer and solution

Answer: (B) $78$ days

(A + B + C)'s 1 day's work $\displaystyle = \frac{1}{26}$

(A + B)'s 1 day's work $\displaystyle = \frac{1}{39}$

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

$\displaystyle = \frac{1}{26} - \frac{1}{39} = \frac{3 - 2}{78} = \frac{1}{78}$

∴ C alone can finish the work in $78$ days
A can do a piece of work in $48$ days and B can finish it in $60$ days. They work together for $10$ days and then B goes away. In how many days will A finish the remaining work?
A $\displaystyle 37\frac{1}{2}$ days B $38$ days C $\displaystyle 46\frac{1}{5}$ days D $30$ days

Answer and solution

Answer: (D) $30$ days

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

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

(A + B)'s 1 day's work $\displaystyle = \frac{1}{48} + \frac{1}{60} = \frac{5 + 4}{240} = \frac{9}{240} = \frac{3}{80}$

(A + B)'s 10 days' work $\displaystyle = 10 \times \frac{3}{80} = \frac{3}{8}$

Remaining work $\displaystyle = 1 - \frac{3}{8} = \frac{5}{8}$

This work is done by A in $\displaystyle \frac{5}{8} \times 48 = 30$ days
A can do a piece of work in $13$ days and B in $26$ days. They work together for $7$ days. The rest of the work is finished by C in $6$ days. If they get ₹1560 for the whole work, what is C's daily wage?
A ₹50 B ₹120 C ₹300 D ₹60

Answer and solution

Answer: (A) ₹50

A's 1 day's work $\displaystyle = \frac{1}{13}$ , B's 1 day's work $\displaystyle = \frac{1}{26}$

(A + B)'s 1 day's work $\displaystyle = \frac{1}{13} + \frac{1}{26} = \frac{2 + 1}{26} = \frac{3}{26}$

(A + B)'s 7 days' work $\displaystyle = \frac{21}{26}$

Remaining work $\displaystyle = 1 - \frac{21}{26} = \frac{5}{26}$ , done by C in $6$ days

A's 7 days' work $\displaystyle = \frac{7}{13}$ , B's 7 days' work $\displaystyle = \frac{7}{26}$ , C's 6 days' work $\displaystyle = \frac{5}{26}$

Consider $\displaystyle \frac{7}{13} : \frac{7}{26} : \frac{5}{26} = 14 : 7 : 5$

A's share $\displaystyle = 14 \times \frac{1560}{26} = 840$ , i.e. ₹840

B's share $\displaystyle = 7 \times \frac{1560}{26} = 420$ , i.e. ₹420

C's share $\displaystyle = 5 \times \frac{1560}{26} = 300$ , i.e. ₹300

C's daily wage $\displaystyle = \frac{300}{6} = 50$ , i.e. ₹50
A is thrice as good a workman as B and is therefore able to finish a piece of work in $34$ days less than B. How many days does A take to finish it alone?
A $51$ days B $102$ days C $\displaystyle 11\frac{1}{3}$ days D $17$ days

Answer and solution

Answer: (D) $17$ days

Suppose B finishes the work in $x$ days

⇒ A is able to finish it in $(x - 34)$ days

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

∴ $\displaystyle x - 34 = \frac{1}{3}x$

⇒ $3x - 102 = x$ ⇒ $2x = 102$ ⇒ $x = 51$

∴ Time taken by B $= 51$ days, time taken by A $= 17$ days
A can do a piece of work in $9$ days, B in $12$ days and C in $36$ days. All begin together, but A leaves the work after $1$ day and B leaves $4$ days before the work is finished. How long did the work last?
A $11$ days B $6$ days C $15$ days D $7$ days

Answer and solution

Answer: (A) $11$ days

A's 1 day's work $\displaystyle = \frac{1}{9}$ , B's $\displaystyle = \frac{1}{12}$ , C's $\displaystyle = \frac{1}{36}$

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 $\displaystyle = \frac{1}{9} + \frac{1}{12} + \frac{1}{36} = \frac{4 + 3 + 1}{36} = \frac{8}{36} = \frac{2}{9}$

C's 4 days' work $\displaystyle = 4 \times \frac{1}{36} = \frac{1}{9}$

Remaining work $\displaystyle = 1 - \left(\frac{2}{9} + \frac{1}{9}\right) = \frac{2}{3}$

(B + C)'s 1 day's work $\displaystyle = \frac{1}{12} + \frac{1}{36} = \frac{3 + 1}{36} = \frac{4}{36} = \frac{1}{9}$

Remaining work is done by (B + C) in $\displaystyle \frac{2}{3} \div \frac{1}{9} = 6$ days

Total time $= 1 + 6 + 4 = 11$ days
A can do a piece of work in $120$ days and B can do it in $240$ days. They work together for $12$ days. Then B leaves and A alone continues the work. $27$ days after that, C joins A and the work is completed in $35$ days more. In how many days can C do the work alone?
A $225$ days B $70$ days C $56$ days D $105$ days

Answer and solution

Answer: (D) $105$ days

A's 1 day's work $\displaystyle = \frac{1}{120}$ , B's 1 day's work $\displaystyle = \frac{1}{240}$

(A + B)'s 1 day's work $\displaystyle = \frac{1}{120} + \frac{1}{240} = \frac{2 + 1}{240} = \frac{3}{240} = \frac{1}{80}$

(A + B)'s 12 days' work $\displaystyle = 12 \times \frac{1}{80} = \frac{3}{20}$

A's 27 days' work $\displaystyle = \frac{27}{120} = \frac{9}{40}$

(A + C)'s 35 days' work $\displaystyle = 1 - \left(\frac{3}{20} + \frac{9}{40}\right) = \frac{5}{8}$

⇒ (A + C)'s 1 day's work $\displaystyle = \frac{5}{8} \div 35 = \frac{1}{56}$

C's 1 day's work $\displaystyle = \frac{1}{56} - \frac{1}{120} = \frac{15 - 7}{840} = \frac{8}{840} = \frac{1}{105}$

C can finish the work in $105$ days
A and B together can do a piece of work in $30$ days. B and C together can do it in $45$ days. If A is twice as good a workman as C, C alone can do it in
A $15$ days B $90$ days C $45$ days D $180$ days

Answer and solution

Answer: (B) $90$ days

(A + B)'s 1 day's work $\displaystyle = \frac{1}{30}$ , (B + C)'s 1 day's work $\displaystyle = \frac{1}{45}$

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

C's 2 days' work + B's 1 day's work $\displaystyle = \frac{1}{30}$

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

⇒ C's 1 day's work $\displaystyle + \frac{1}{45} = \frac{1}{30}$

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

∴ C alone can finish it in $90$ days
A and B working alone can finish a piece of work in $6$ and $10$ days respectively. If they work at it alternately for a day, A beginning, in how many days will the work be finished?
A $\displaystyle 7\frac{1}{3}$ days B $\displaystyle 3\frac{3}{4}$ days C $8$ days D $\displaystyle 7\frac{1}{2}$ days

Answer and solution

Answer: (A) $\displaystyle 7\frac{1}{3}$ days

A's 1 day's work $\displaystyle = \frac{1}{6}$ , B's 1 day's work $\displaystyle = \frac{1}{10}$

Work done in the first 2 days $\displaystyle = \frac{1}{6} + \frac{1}{10} = \frac{5 + 3}{30} = \frac{8}{30} = \frac{4}{15}$

Work done in the first 6 days $\displaystyle = 3 \times \frac{4}{15} = \frac{4}{5}$

On day 7 it is A's turn: work done in the first 7 days $\displaystyle = \frac{4}{5} + \frac{1}{6} = \frac{29}{30}$

Remaining work $\displaystyle = 1 - \frac{29}{30} = \frac{1}{30}$

B does it in $\displaystyle \frac{1}{30} \div \frac{1}{10} = \frac{1}{3}$ day

∴ Total no. of days $\displaystyle = 7 + \frac{1}{3} = 7\frac{1}{3}$ days
A certain number of men complete a piece of work in $78$ days. If there were $5$ men more, the work could be finished in $10$ days less. How many men were there originally?
A $39$ B $34$ C $29$ D $44$

Answer and solution

Answer: (B) $34$

Let the original number of men be $x$

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

$x \times 78 = (x + 5) \times 68$

$78x = 68x + 340$

$10x = 340$ ⇒ $x = 34$
If $3$ men or $9$ boys can finish a piece of work in $8$ days, in how many days can $5$ men and $3$ boys finish it?
A $16$ days B $4$ days C $9$ days D $\displaystyle 15\frac{3}{7}$ days

Answer and solution

Answer: (B) $4$ days

$3$ men $= 9$ boys

$5$ men $\displaystyle = \frac{9}{3} \times 5 = 15$ boys

$5$ men + $3$ boys $= 15$ boys + $3$ boys $= 18$ boys

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

$9 \times 8 = 18 \times x$ ⇒ $x = 4$ days
$4$ men and $4$ women working $9$ hours a day finish a piece of work in $6$ days; $3$ men and $2$ women working $9$ hours a day do the same work in $9$ days. Find the number of days in which the work is done by $9$ men only, working $4$ hours a day.
A $4$ days B $12$ days C $18$ days D $9$ days

Answer and solution

Answer: (D) $9$ days

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

Case (i): $(4m + 4w) \times 9 \times 6 = 216m + 216w$

Case (ii): $(3m + 2w) \times 9 \times 9 = 243m + 162w$

Same work: $216m + 216w = 243m + 162w$

⇒ $54w = 27m$ ⇒ $2w = m$

So $4$ women $= 2$ men, and $4m + 4w = 6$ men

$6$ men working $9$ hours a day for $6$ days do the same work as $9$ men working $4$ hours a day for $x$ days

$6 \times 9 \times 6 = 9 \times 4 \times x$ ⇒ $x = 9$ days
If $6$ men with $2$ boys can earn ₹5520 in $8$ days and $12$ men with $11$ boys can earn ₹4455 in $3$ days, how much does 1 man earn in a day?
A ₹125 B ₹115 C ₹110 D ₹15

Answer and solution

Answer: (C) ₹110

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

$(6x + 2y) \times 8 = 5520$ ⇒ $6x + 2y = 690 \quad \cdots (1)$

$(12x + 11y) \times 3 = 4455$ ⇒ $12x + 11y = 1485 \quad \cdots (2)$

(1) $\times 2$ ⇒ $12x + 4y = 1380$

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

From (1): $6x = 690 - 30 = 660$ ⇒ $x = 110$
$36$ men are employed to do a piece of work which they could finish in $12$ days, but the men drop off by $3$ at the end of every $6$ days. In what time will the work be completed?
A $18$ days B $\displaystyle 13\frac{1}{11}$ days C $\displaystyle 12\frac{3}{5}$ days D $12$ days

Answer and solution

Answer: (C) $\displaystyle 12\frac{3}{5}$ days

Work done by 36 men in 6 days $\displaystyle = 6 \times \frac{1}{12} = \frac{1}{2}$

So 1 man's work in 6 days $\displaystyle = \frac{1}{2} \div 36 = \frac{1}{72}$

Remaining work $\displaystyle = \frac{1}{2}$ , remaining men $= 33$ ; work done by 33 men in 6 days $\displaystyle = 33 \times \frac{1}{72} = \frac{11}{24}$

Remaining work $\displaystyle = \frac{1}{24}$ , remaining men $= 30$ ; work done by 30 men in 6 days $\displaystyle = \frac{5}{12}$

$\displaystyle \frac{1}{24}$ work is done by 30 men in $\displaystyle \frac{1}{24} \div \frac{5}{12} \times 6 = \frac{3}{5}$ days

Total time $\displaystyle = 6 + 6 + \frac{3}{5} = 12\frac{3}{5}$ 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 $44$ days. In how many days can B alone finish it?
A $66$ days B $88$ days C $\displaystyle 29\frac{1}{3}$ days D $22$ days

Answer and solution

Answer: (A) $66$ 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 $= 2k = 44$ ⇒ $k = 22$

Time taken by B $= 3k = 66$ days
A man rows in still water at $12.5$ km/hr and the rate of the current is $1$ km/hr. The man's rate with the current is
A $13.5$ km/hr B $11.5$ km/hr C $12.5$ km/hr D $5.75$ km/hr

Answer and solution

Answer: (A) $13.5$ km/hr

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

$= 12.5 + 1 = 13.5$ km/hr
A man rows upstream $33$ km and downstream $36$ km, taking $6$ hours each time. The velocity of the current is
A $1.5$ km/hr B $0.25$ km/hr C $0.5$ km/hr D $5.75$ km/hr

Answer and solution

Answer: (B) $0.25$ km/hr

Speed downstream: $\displaystyle x + y = \frac{36}{6}$ km/hr

Speed upstream: $\displaystyle x - y = \frac{33}{6}$ km/hr

Subtracting, $\displaystyle 2y = \frac{36}{6} - \frac{33}{6} = \frac{3}{6}$

$\displaystyle y = \frac{3}{12} = 0.25$ km/hr
A boat goes $58$ km downstream in $5$ hours. Its speed downstream is
A $11.6$ km/hr B $53$ km/hr C $290$ km/hr D $14.5$ km/hr

Answer and solution

Answer: (A) $11.6$ km/hr

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

$\displaystyle = \frac{58}{5} = 11.6$ km/hr
A man can row $15$ 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 $5$ km/hr B $3.75$ km/hr C $7.5$ km/hr D $11.25$ km/hr

Answer and solution

Answer: (C) $7.5$ 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 $3(15 - y) = 15 + y$

⇒ $45 - 3y = 15 + y$ ⇒ $4y = 30$

⇒ $y = 7.5$ km/hr
A man can row $6$ km/hr in still water. If the river is running at $3$ km/hr, it takes him 2 hours to row to a place and back. The place is at a distance of
A $9$ km B $4.5$ km C $6$ km D $5.5$ km

Answer and solution

Answer: (B) $4.5$ km

Let the required distance be $d$ km

Speed downstream $= 6 + 3 = 9$ km/hr

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

Time $\displaystyle = \frac{\text{Distance}}{\text{Speed}}$ , so $\displaystyle \frac{d}{9} + \frac{d}{3} = 2$

⇒ $\displaystyle d\left(\frac{1}{9} + \frac{1}{3}\right) = 2$

⇒ $\displaystyle d \times \frac{4}{9} = 2$

⇒ $d = 4.5$ km
The current of a stream runs at the rate of $5$ km/hr. A motor boat goes $5$ km upstream and back again to the starting point in 45 minutes. The speed of the motor boat in still water is
A $20$ km/hr B $10$ km/hr C $25$ km/hr D $15$ km/hr

Answer and solution

Answer: (D) $15$ km/hr

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

Speed upstream $= (x - 5)$ km/hr, speed downstream $= (x + 5)$ km/hr

Total time $= 45$ min $\displaystyle = \frac{45}{60}$ hr $\displaystyle = \frac{3}{4}$ hr

$\displaystyle \frac{5}{x - 5} + \frac{5}{x + 5} = \frac{3}{4}$

⇒ $\displaystyle 5\left(\frac{(x + 5) + (x - 5)}{(x - 5)(x + 5)}\right) = \frac{3}{4}$

⇒ $\displaystyle \frac{10x}{x^2 - 25} = \frac{3}{4}$

⇒ $40x = 3x^2 - 75$

⇒ $3 x^{2} - 40 x - 75 = 0$

⇒ $\left(x - 15\right) \left(3 x + 5\right) = 0$

⇒ $x = 15$ (or) $\displaystyle x = -\frac{5}{3}$ (rejected, a speed cannot be negative)

∴ Speed of the boat in still water $= 15$ km/hr
A man can row $12$ km upstream and $35$ km downstream in $11$ hours, while he can row $18$ km upstream and $15$ km downstream in $9$ hours. The speed of the man in still water is
A $3$ km/hr B $1$ km/hr C $4$ km/hr D $5$ km/hr

Answer and solution

Answer: (C) $4$ 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 $\displaystyle p = \frac{1}{x - y}$ , $\displaystyle q = \frac{1}{x + y}$

$\displaystyle \frac{12}{x - y} + \frac{35}{x + y} = 11$ ⇒ $12p + 35q = 11 \quad \cdots (1)$

$\displaystyle \frac{18}{x - y} + \frac{15}{x + y} = 9$ ⇒ $18p + 15q = 9 \quad \cdots (2)$

(1) $\times 3$ ⇒ $36p + 105q = 33$ ; (2) $\times 2$ ⇒ $36p + 30q = 18$

(1) $\times 3$ $-$ (2) $\times 2$ ⇒ $75q = 15$ ⇒ $\displaystyle q = \frac{1}{5}$

From (1): $12p + 7 = 11$ ⇒ $12p = 4$ ⇒ $\displaystyle p = \frac{1}{3}$

So $x - y = 3$ and $x + y = 5$

Adding, $2x = 8$ ⇒ $x = 4$ ; then $y = 5 - 4 = 1$

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MCQ test, Set 4