Practice Man leaves and joins - quantitative aptitude Online Quiz (set-1) For All Competitive Exams
Q-1) 8 men can do a work in 12 days. After 6 days of work, 4 more men were engaged to finish the work. In how many days would the remaining work be completed?
(a)
(b)
(c)
(d)
Using Rule 1If $M_1$ men can finish $W_1$ work in $D_1$ days and $M_2$ men can finish $W_2$ work in $D_2$ days then, Relation is${M_1D_1}/{W_1} = {M_2D_2}/{W_2}$ andIf $M_1$ men finish $W_1$ work in $D_1$ days, working $T_1$ time each day and $M_2$ men finish $W_2$ work in $D_2$ days, working $T_2$ time each day, then${M_1D_1T_1}/{W_1} = {M_2D_2T_2}/{W_2}$
Work done by 8 men in 6 days
= $6/12 = 1/2$
Remaining work
= $1 - 1/2 = 1/2$
4 more men are engaged.
Total number of men = 8 + 4 = 12
By work and time formula
$W_1/{M_1D_1} = W_2/{M_2D_2}$, we have
$1/{8 × 12} = {1/2}/{12 × D_2}$
$D_2 = 1/2 × {8 × 12}/12$ = 4 days.
Q-2) A certain number of persons can complete a piece of work in 55 days. If there were 6 persons more, the work could be finished in 11 days less. How many persons were originally there ?
(a)
(b)
(c)
(d)
Originally, let there be x men
Now, more men, less days
(x + 6) : x : : 55 : 44
So, ${x + 6}/x = 55/44 = 5/4$
or 5x = 4x + 24 ⇒ x = 24
Using Rule 23Some people finish a certain work in 'D' days. If there were 'a' less people, then the work would be completed in 'd'days more, what was the number of people initially?Required number = $\text"a(D - d)"/\text"d"$ people
Here, D = 55, a = 6, d = 11
No of people initially = $\text"a(D - d)"/\text"d"$
= ${6(55 - 11)}/11$ = 24
Q-3) A and B can complete a work in 15 days and 10 days respectively. They started doing the work together but after 2 days, B had to leave and A alone completed the remaining work. The whole work was completed in :
(a)
(b)
(c)
(d)
Work done by (A + B) in 1 day
= $1/15 + 1/10 = {2 + 3}/30 = 5/30 = 1/6$
(A + B)’s 2 days’ work = $2/6 = 1/3$
Remaining work = $1 - 1/3 = 2/3$
This part is done by A alone.
Since, one work is done by A in 15 days.
$2/3$ work is done in $15 × 2/3$ = 10 days.
Total number of days = 10 + 2 = 12 days
Using Rule 20,
Here, m = 15, n = 10, p =2
A alone completed the work in
= $\text"mn - p(m+n)"/ \text"n"$days
= ${15 × 10 - 2(15 + 10)}/10$
= ${150 - 50}/10$ = 10 days
Total time taken= 10 + 2 = 12 days
Q-4) A can do a piece of work in 20 days which B can do in 12 days. B worked at it for 9 days. A can finish the remaining work in
(a)
(b)
(c)
(d)
Work done by B in 9 days = $9/12 = 3/4$ part
Remaining work
= $1 - 3/4 = 1/4$ which is done by A
Time taken by A = $1/4 × 20$ = 5 days
Q-5) A and B alone can complete work in 9 days and18 days respectively. They worked together; however 3 days before the completion of the work A left. In how many days was the work completed ?
(a)
(b)
(c)
(d)
Let the work be completed in x days.
According to the question,
A worked for (x –3) days, while B worked for x days.
${x - 3}/9 +x/18$ = 1
${2x - 6 + x}/18$ = 1
3x–6 = 18
3x = 18 + 6 = 24 ⇒ x = $24/3$ = 8 days
Using Rule 8,
Here, x = 9, y = 18, m = 3
Total time taken = ${(x + m)y}/{x + y}$
=${(9 + 3) × 18}/{9 + 18}$
= ${12 × 18}/27$ = 8 days
Q-6) A can finish a work in 24 days, B in 9 days and C in 12 days. B and C start the work but are forced to leave after 3 days. The remaining work was done by A in :
(a)
(b)
(c)
(d)
Work done by (B + C) in 3 days.
= $3 × (1/9 + 1/12)$
= $1/3 + 1/4 = {4 + 3}/12 = 7/12$
Remaining work = $1 - 7/12 = 5/12$
This part of work is done by A alone.
Now, $1/24$ part of work is done by A in 1 day.
∴ $5/12$ part of work will be done by A in
= $24 × 5/12$ = 10 days.
Q-7) A, B and C can complete a work in 10, 12 and 15 days respectively. They started the work together. But A left the work before 5 days of its completion. B also left the work 2 days after A left. In how many days was the work completed?
(a)
(b)
(c)
(d)
Let the work be completed in x days.
According to the question,
${x - 5}/10 + {x - 3}/12 + x/15$ = 1
${6x - 30 + 5x - 15 + 4x}/60$ = 1
15x - 45 = 60
15x = 105 ⇒ x = $105/15$ = 7
Hence, the work will be completed in 7 days.
Q-8) A and B can do a work in 45 days and 40 days respectively. They began the work together but A left after some time and B completed the remaining work in 23 days. After how many days of the start of the work did A leave ?
(a)
(b)
(c)
(d)
(A + B)’s 1 day’s work
= $(1/45 + 1/40) = {8 + 9}/360 = 17/360$
Work done by B in 23 days
= $1/40 × 23 = 23/40$
Remaining work = $1 - 23/40 = 17/40$
Now, $17/360$ work was done by (A + B) in 1 day
$17/40$ work was done by (A + B) in $1 × 360/17 × 17/40$ = 9 days.
Hence, A left after 9 days.
Using Rule 26A and B can do a piece of work in x and y days, respectively. Both begin together but after some days.A leaves the job and B completed the remaining work in a days. After how many days did A leave?Required time, t = $\text"(y - a)"/\text"(x + y)" × x$
Here, x = 45, y = 40, a = 23
Required time t= $\text"(y - a)"/\text"(x + y)" × x$
t = ${(40 - 23) × 45}/{45 + 40} = {17 × 45}/85$
t = 9 days
Q-9) A and B can separately complete a piece of work in 20 days and 30 days respectively. They worked together for some time, then B left the work. If A completed the rest of the work in 10 days, then B worked for
(a)
(b)
(c)
(d)
Let A and B worked together for x days
According to the question,
Part of work done by A for (x + 10) days
+ part of work done by B for x days = 1
${x + 10}/20 + x/30$ = 1
${3x + 30 + 2x}/60 = 1$
5x + 30 = 60
5x = 30 ⇒ x = $30/5$ = 6 days
Using Rule 20A can do a certain work in 'm' days and B can do the same work in 'n' days. They worked together for 'P' days and after this A left the work, then in how many days did B alone do the rest of work ?Required time = ${mn - P(m + n)}/m$ dayswhen after 'P' days B left the work, then in how many days did A alone do the rest of work?Required time = ${mn - P(m + n)}/n$ days
Here, m =20, n= 30, p = x and time taken by A alone = 10
10 = ${mn - p(m + n)}/n$
10 = ${30 × 20 - x(30 + 20)}/30$
300 = 600 - x 50
50x = 300 x = 6
B worked for 6 days
Q-10) A and B can do a piece of work in 28 and 35 days respectively. They began to work together but A leaves after sometime and B completed remaining work in 17 days. After how many days did A leave ?
(a)
(b)
(c)
(d)
Let A worked for x days.
According to question
$x/28 + (x + 17)/35 = 1$
${5x + 4(x + 17)}/140 = 1$
5x + 4x + 68 = 140
9x = 140 - 68 = 72
x = 8
∴ A worked for 8 days