Practice questions here, for every subject and every exam. Unlimited questions for unlimited attempts, given with answers and explanations.
$\mathrm{A}$ can do $25 \%$ of a work in 15 days. He worked for 15 days and then B alone completed the remaining work in 30 days. In how many days will $A$ and $B$ working together complete $50\%$ of that work?
A can do complete work in 15 × 4 = 60 days
Work done by A in 15 day is $25 \%$ of total work
$\Rightarrow 75 \%$ of the work is done by B in 30days
So, B can do complete work in $30 \times \frac{4}{3}=40$ days
Now , Let total work $=\operatorname{LCM}(60,40)=120$ units
$\Rightarrow$ A's 1 day work $\frac{120}{60}=2$ and B's 1 day work $\frac{120}{40}=3$
So, required number of days $=\frac{120}{2 \times(2+3)}$ $=12$ days
A and B working together can complete 45% of a work in 18 days. A alone can complete the same work in 60 days. A and B work together for 16 days, and then A leaves. B alone will complete the remaining work in:
Given:
$A$ and $B$ can complete $45 \%$ of a work $=18$ days
$A$ and $B$ can complete the whole work $=\frac{18}{45} \times 100=40$ Days
A alone can complete the whole work $=60$ days
Let total work $=\mathrm{LCM}$ of $40,60=120$ Units
Work done by B in 1 day $=3-2=1$ unit
Work done by $A$ and $B$ in 16 days $=16\times3 =48$ units
Remaining work $=120-48=72$ units
Therefore, the required time is taken by B to complete the remaining work $=\frac{72}{1}=72$ days
X is 60% more efficient than Y, and Y alone can do a work in 80 days. Working together, X and Y will complete 52% of the same work in
Let the efficiency of $Y=P$
Efficiency of $X=P \times \frac{(100+60)}{100}=\frac{8 P}{5}$
Total work $=P \times 80=80 P$
Therefore, required time to complete $52 \%$ of the same work by $A$ and $B=\frac{80 P \times\left(\frac{52}{100}\right)}{\left(P+\frac{8 P}{5}\right)}=\frac{\left(\frac{208 \mathrm{P}}{5}\right)}{\left(\frac{13 \mathrm{P}}{5}\right)}=$
$
\frac{208}{13}=16 \text { days }
$
Hence, option A is correct.
Let efficiency of A = 7x
Efficiency of B = 6x
Working together, they can complete the same work in 21 days.
Total work $=(6 x+7 x) \times 21=273 x$ unit
Work finished by B in 26 days $=(6 x) \times 26=156 x$ unit
Hence, Part of the work that is completed by B alone in 26 days $=\frac{156 x}{273 x}=\frac{4}{7}$
A and B can complete a work in 12 days and 20 days, respectively. They work together for 4 days. C alone completes the remaining work in 14 days. B and C together can complete the same work in:
Number of days taken by A to complete the work = 12 days
Number of days taken by B to complete the work= 20 days
Let total work = LCM (12,20) = 60 unit
Efficiency of A and B together = 5 + 3 = 8 unit/day
Both work together for 4 days only.
Work finished by them in 4 days =$4 \times 8=32$ unit
Remaining work = 60 unit – 32 unit = 28 unit
C completed the remaining work in 14 days.
Efficiency of C =$\frac{28}{14}=2$ unit/day
Efficiency of B and C together = 3 + 2 = 5 unit/day
Number of days taken by B and C to complete the whole work =$\frac{60}{5}=12$ days
Five men and 2 boys can do as much work in 30 days as 7 men and 10 boys can do in 15 days. How many boys should be employed along with 40 men to do the same work in 4 days?
$5 M+2 B=30$ days
$\Rightarrow 10 M+4 B=15$ days $\ldots .(i)$
also,
$7 \mathrm{M}+10 \mathrm{~B}=15$ days $\ldots \ldots$ (ii)
so, from (i) and (ii)
$10 M+4 B=7 M+10 B$
$\Rightarrow 3 M=6 B$
$\Rightarrow$ Efficiency ratio, $\frac{\mathrm{M}}{\mathrm{B}}=\frac{2}{1}$
So, total work $=(5 \times 2+2 \times 1) \times 30$ or $(7 \times 2+10 \times 1) \times 15$
$=360$ units
If 360 units is to be done in 4 days,
then work done each day $=90$ units
Now, $(40 \times 2+x \times1)=90 ------(x \rightarrow$ number of boys)
$x=10$
Hence, required number of boys is 10 .
$A$ can do a piece of work in $48$ days and $B$ can do it in $60$ days. They work together for $12$ days and then $A$ leaves. In what time (in days and hours) $B$ will complete $25\%$ of the remaining work?
Let total work $=$ LCM $(48,60)=240$ units
A's 1 Day work $=\frac{240}{48}=5$ units
B's 1 Day work $=\frac{240}{60}=4$ units
Work done by A and B together is 12 days
$
\begin{aligned}
&=(5+4) \times 12 \\
&=108 \text { units }
\end{aligned}
$
$25 \%$ of remaining work $=\frac{1}{4} \times(240-108)$
=33 units
Time required by B to complete $25 \%$ of the remaining work $=\frac{33}{4}=8 \frac{1}{4}$ days
$=8$ days 6 hours
4M + 6B = 8 days
6M + 4B = 7days
so, efficiency ratio
$\frac{\mathrm{M}}{\mathrm{B}}=\frac{8 \times 6-7 \times 4}{7 \times 6-8 \times 4}=\frac{20}{10}=\frac{2}{1}$
$\Rightarrow$ Total work $=(4 \times 2+6 \times 1) \times 8$ or $(6 \times 2+4$ $\times 1) \times 7$
$=112$ units
Required time $=\frac{112}{(5 \times 2+4 \times 1)}=\frac{112}{14}=8$
$\mathrm{A}$ can do a piece of work in 16 days and $\mathrm{B}$ can do it in 12 days. They worked together for 4 days and then $A$ left. $B$ did the rest of the work alone. For how many days did $B$ work in total to complete the work?
Let total work = LCM(16, 12) = 48units
A's 1 Day work $=\frac{48}{16}=3$ units and B's 1 Day work $\frac{48}{12}=4$ units
work done by A and B in 4 days = (3 + 4)×4 = 28 units
Remaining work will be done by B in $\frac{(48-28)}{4}$ day
= 5 days
So, B worked for total of 4 + 5 = 9days.
A' can complete $40 \%$ of a work in 6 days, while 'B' can complete the work in $7 \frac{1}{2}$ days. If they work together, then in how many days will the work be completed?
'A' can complete $40 \%$ of a work in 6 days.
Number of days taken by A to complete the work $=\frac{100}{40} \times 6=15$ days
'B' can complete the work in $7 \frac{1}{2}$ days.
$
\Rightarrow 7 \frac{1}{2}=\frac{15}{2} \text { days }
$
Let total work $=\operatorname{LCM}(15,\frac{15}{2})=15$ unit
Efficiency of $A$ and $B$ together $=1+2=3$ unit/day
Number of days taken by $A$ and $B$ together to complete the whole work $=\frac{15}{3}=5$ days
'A' can complete a work in 15 days and 'B' can complete the same work in 20 days. Working together, in how many days will they complete $70 \%$ of the same work?
'A' can complete a work in 15 days and 'B' can complete the same work in 20 days.
Let total work = LCM (15 , 20) = 60 unit
Number of days taken by A and B to complete the whole work together $=\frac{60}{7}$ days
Number of days taken by A and B to complete $70 \%$ of work together $=\frac{70}{100} \times \frac{60}{7}=6$ days
A tank is $25 \mathrm{~m}$ long, $12 \mathrm{~m}$ wide and $6 \mathrm{~m}$ deep. The cost (in ₹) of plastering its walls and bottom at ₹ $10$ per $\mathrm{m}^{2}$ is:
A tank is $25 \mathrm{~m}$ long, $12 \mathrm{~m}$ wide and $6 \mathrm{~m}$ deep.
Surface area of its walls and bottom $=2(l+b) h+l b=2(25+12) 6+25 \times 12=444+300=744 \mathrm{~m}^{2}$
Cost (in ₹) of plastering its walls and bottom at ₹ 10 per $\mathrm{m}^{2}=744 \times 10=$ Rs. 7440
X, Y and Z can do a piece of work in 46 days, 92 days and 23 days respectively. X started work. Y joined him after 2 days. If Z joins them after 8 days from the start, then for how many days did X work?
Let total work be LCM (46, 92, 23) = 92units
X's 1 day work $=\frac{92}{46}=2$ units
Y's 1 day work $=\frac{92}{92}=1$ unit
Z's 1 day work $=\frac{92}{23}=4$ units
Let $X$ worked for $a$ days, then,
$2\times a +1 \times(a-2)+4 \times(a-8)=92$ units
$\Rightarrow 7 a-34=92$
$\Rightarrow a=\frac{92+34}{7}=18$ days.
There are 3 punctures in a tire. The first puncture alone deflates the tire in 9 minutes, the second puncture alone deflates the tire in 18 minutes and the third puncture alone deflates the tire in 6 minutes. If the air leaks out at a constant rate, then how long (in minutes) does it take for all the punctures together to deflate the tire?
Let A, B, C be the three punctures.
also, let total work = LCM (9, 18, 6) =18 units
$\Rightarrow$ A's efficiency $=\frac{18}{9}=2$ units $/ \mathrm{min}$
$\Rightarrow$ B's efficiency $=\frac{18}{18}=1$ unit $/ \mathrm{min}$
$\Rightarrow$$C^{\prime}$ s efficiency $=\frac{18}{6}=3$ units $/ \mathrm{min}$
So, all punctures together will make the tire flat in $\frac{18}{(2+1+3)}=3$ minutes.
In a press there are three types of printing machines $\mathrm{A}, \mathrm{B}$, and C. Machine $\mathrm{A}$ can print ten thousand pages in 8 hours, machine B can do the same task in 10 hours, and machine C can do the same task in 12 hours. All three machines start the task at 9:00 a.m. Machine A breaks down at $11: 00$ a.m. and machines $B$ and C continue working. At 12:00 noon machine B also breaks down and machine $C$ alone has to complete the remaining task. What is the approximate time of completion of the task?
A's 1 hour work = 1/8
B's 1 hour work = 1/10
C's 1 hour work = 1/12
(A + B + C)'s 1 hour work = (1/8) + (1/10) + (1/12) = 37/120
Work done by A, B and C in 2 hour = 2 × (37/120) = 37/60
(B + C)'s 1 hour work = (1/10) + (1/12) = 11/60
Remaining work = 1 - (37/60) - (11/60)
⇒ (60 - 37 - 11)/60
⇒ 12/60 = 1/5
Remaining work complete by C = (1/5) × 12 = 2.4
Total time = 2 hour + 1 hour + 2:24 hour = 5:24 hour
Complete the task = 9 + 5:24 = 14:24 = 2:24
∴ Complete the task on 2:24 p.m.
MATHS MIRROR METHOD
LCM of 8, 10, and 12 is 120.
(A + B + C)'s work in 2 hour 9 a.m. to 11 a.m. = (15 + 12 + 10) × 2= 37 × 2= 74
(B + C)'s work in 1 hour 11 a.m. to 12 a.m.= (12 + 10) × 1= 22
Remaining work = 120 - (74 + 22) = 120 - 96 = 24
24 work complete by only C = 24/10 = 2.4
So, Total time taken = 2 hour + 1 hour + 2 hour 24 minutes = 5 hour 24 minutes
Then, 9 a.m. + 5 hour 24 minutes = 2:24 p.m.
∴ Complete the task on 2:24 am.
Person $A$ can complete $1/5$ of work in 3 days, while the efficiency of $B$ is half that of $A$. In how many days can $A$ and $B$ working together do half of the work?
A will do the complete work in 3 × 5 = 15days
Let A's efficiency be 2 units /day
B's efficiency will be $\frac{2}{2}=1$ unit/day
Total work = 15 × 2 = 30 units
So, required time $=\frac{30}{2 \times(2+1)}=5$ days
10 men can finish a piece of work in 10 days, whereas it takes 12 women to finish it in 10 days. If 8 men and 6 women undertake to complete the work, then in how many days will they complete it?
Total work $=10 \mathrm{M} \times 10=12 \mathrm{~W} \times 10$
$M: W=6: 5$
Total work $=10\times6\times10 =100 \times 6=600$ units
efficiency of 8 men and 6 women $=8 \mathrm{M}+6 \mathrm{~W}$
$= 8 \times 6+6 \times 5=78$ unit/day
Required Time $= \frac{600} { 78}$ days $= \frac{100} { 13}$ days
25 women can do a piece of work in 60 days. After how many days from the start of the work, should 15 more women join them so that the work is done in 45 days?
25 women can do a piece of work in 60 days
Let 15 more women join the work after x days
= 25 × 60 = 25 × x + 40 × (45 - x)
= 1500 = 25x + 40 × 45 - 40x
= 15x = 1800 - 1500
= 15x = 300
= x = 20 days
A man, a woman and a boy can do a piece of work in 3, 4 and 12 days respectively. How many boys should help one man and one woman to complete the same work in one day?
Let total work be LCM (3, 4, 12) = 12 units
1 Man's one day work = $\frac{12}{3}=4$ units
1 Woman's one day work $=\frac{12}{4}=3$ units
and 1 Boy's one day work $=\frac{12}{12}=1$ unit
Now, According to question
work to be done in 1 day $=12$ units
Let required number of boys be $x$ then,
1 × 4 + 1 × 3 + x × 1 = 12
x = 12 – 7 = 5
So, required number of boys = 5