Practice Time and work - quantitative aptitude Online Quiz (set-1) For All Competitive Exams
Q-1) A labourer was appointed by a contractor on the condition that he would be paid Rs.75 for each day of his work but would be fined at the rate of Rs.15 per day for his absence, apart from losing his wages, After 20 days, the contractor paid the labourer Rs.1140. The number of days the labourer abstained from work was
(a)
(b)
(c)
(d)
Total salary for 20 days
= Rs.(75 × 20) = Rs.1500
Actual salary received = Rs.1140
Difference = Rs.(1500 - 1140) = Rs.360
Money deducted for 1 day’s absence from work
=Rs.(15 + 75) = Rs.90
Number of days he was absent
= $360/90 = 4$ days
Q-2) 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-3) A man, a woman and a boy can complete a work in 20 days, 30 days and 60 days respectively. How many boys must assist 2 men and 8 women so as to complete the work in 2 days ?
(a)
(b)
(c)
(d)
Part of work done by 2 men and 2 women in 2 days.
= $2(2/20 + 8/30)$
= $2(1/10 + 8/30) = 2({3 + 8}/30)$
= $22/30 = 11/15$
= Remaining work =$1 - 11/15 = 4/15$
Work done by 1 boy in 2 days
= $2/60 = 1/30$
Number of boys required to assist = $4/15 × 30 = 8$
Using Rule 14,
Here, A = 1, B = 1, C = 1
a = 20, b = 30, c = 60
$A_1 = 2, B_1$ = 8
Required time = $1/{A_1/{A × a} + B_1/{B × b} + C_1/{C × c}$
2 = ${1/{2/{1 × 20} + 8/{1 × 30} + x/{1 × 60}$
2 = $10/{2/2 + 8/3 + x/6}$
2 = $10/{{6 + 16 + x}/6}$
22 + x = 30 ⇒ x = 8
∴ Number of boys = 8
Q-4) 18 men or 36 boys working 6 hours a day can plough a field in 24 days. In how many days will 24 men and 24 boys working 9 hours a day plough the same field ?
(a)
(b)
(c)
(d)
18 men ≡ 36 boys ⇒ 1 man ≡ 2 boys
24 men + 24 boys ≡ (24 + 12) men ≡ 36 men
$M_1D_1T_1 = M_2D_2T_2$
$18 × 24 × 6 = 36 × D_2$ × 9
$D_2 = {18 × 24 × 6}/{36 × 9}$ = 8 days
Q-5) A man is twice as fast as a woman and a woman is twice as fast as a boy in doing a work. If all of them, a man, a woman and a boy can finish the work in 7 days, in how many days a boy will do it alone ?
(a)
(b)
(c)
(d)
Using Rule 11,
According to the question,
1 man ≡ 2 women ≡ 4 boys
1 man + 1 woman + 1 boy
= (4 + 2 +1) boys = 7 boys
$M_1D_1 = M_2D_2$
7 × 7 = 1 × $D_2$
∴ $D_2$ = 49 days
Q-6) 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-7) 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-8) 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-9) 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-10) 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-11) A and B together can complete a work in 8 days. B alone can complete that work in 12 days. B alone worked for four days. After that how long will A alone take to complete the work ?
(a)
(b)
(c)
(d)
Time taken by A
= ${8 × 12}/128 = {8 × 12}/4 = 24$ days
Work done of by B = $4/12 = 1/3$
Remaining work = $1 - 1/3 = 2/3$
Since, A can complete a work in 24 days
A can complete $2/3$ part of work in 24 × $2/3$ = 16 days
Q-12) If 1 man or 2 women or 3 boys can do a piece of work in 44 days, then the same piece of work will be done by 1 man, 1 woman and 1 boy in
(a)
(b)
(c)
(d)
1 man ≡ 2 women ≡ 3 boys
1 man + 1 woman + 1 boy
≡ 3 boys + $3/2$ boys + 1 boy
≡ $(3 + 3/2 + 1)$ boys ≡ $11/2$ boys
By $M_1D_1 = M_2D_2$,
3 × 44 = $11/2 × D_2$
$D_2 = {2 × 3 × 44}/11$ = 24 days
Using Rule 13,
Here, A = 1, B= 2, C = 3, a = 44
$A_1 = 1, B_1 = 1, C_1$ = 1
Required time = $a/{A_1/A + B_1/B + C_1/C}$ days
= $44/{1/1 + 1/2 + 1/3} = {44 × 6}/11$ = 24 days
Q-13) 3 men and 5 women can do a work in 14 days while 5 men can do it in 14 days. 5 men and 5 women can complete the work in
(a)
(b)
(c)
(d)
5 men can do 1 work in 14 days.
3 men will do $3/5$ work in 14 days.
Remaining work = $1 - 3/5 = 2/5$
5 women do $2/5$ work in 14 days.
Time taken by 5 women in doing 1 work
= ${14 × 5}/2 = 35$ days
(5 men + 5 women)’s 1 day’s work
= $1/14 + 1/35 = {5 + 2}/70 = 7/70 = 1/10$
∴Required time = 10 days.
Q-14) One man and one woman together can complete a piece of work in 8 days. A man alone can complete the work in 10 days. In how many days can one woman alone complete the work ?
(a)
(b)
(c)
(d)
Work done by 1 woman in 1 day
= $1/8 - 1/10 = {5 - 4}/40 = 1/40$
One woman will complete the work in 40 days.
Using Rule 4If A alone can do a certain work in 'x' days and A and B together can do the same work in 'y' days, then B alone can do the same work in $({xy}/{x - y})$ days.
Here, x = 10 and y = 8
Woman can do work in = $({xy}/{x - y})$ days
= $({10 × 8}/{10 - 8})$ = 40 days
Q-15) 4 men and 6 women can complete a work in 8 days, while 3 men and 7 women can complete it in 10 days. In how many days will 10 women complete it ?
(a)
(b)
(c)
(d)
Let 1 man’s 1 day’s work = x and
1 woman’s 1 day’s work = y
Then, 4x + 6y = $1/8$ and
$3x + 7y = 1/10$
From both equations,
we get y = $1/400$
10 women’s 1 day’s work = $10/400 = 1/40$
10 women will finish the work in 40 days.
Using Rule 11,
$A_1 = 4, B_1 = 6, D_1$ = 8
$A_2 = 3, B_2 = 7, D_2$ = 10
$A_3 = 0, B_3$ = 10
Required time = ${D_1D_2(A_1B_2 - A_2B_1)}/{D_1(A_1B_3 - A_3B_1) - D_2(A_2B_3 - A_3 B_2)}$
= ${8 × 10(4 × 7 - 3 × 6)}/{8(4 × 10 - 0 × 6) - 10(3 × 10 - 0 × 7)}$
= ${80 × 10}/20$ = 40 days
Q-16) If A and B together can complete a work in 18 days, A and C together in 12 days and B and C together in 9 days, then B alone can do the work in
(a)
(b)
(c)
(d)
(A + B)’s 1 day’s work = $1/18$
(B + C)’s 1 day’s work = $1/9$
(A + C)’s 1 day’s work = $1/12$
Adding all the above three,
2 (A + B + C)’s 1 day’s work = $1/18 + 1/9 + 1/12$
= ${2 + 4 + 3}/36 = 9/36 = 1/4$
(A + B + C)’s 1 day’s work = $1/8$
B’s 1 day’s work = (A + B + C)’s 1 day’s work - (A + C)’s 1 day’s work
= $1/8 - 1/12 = {3 - 2}/24 = 1/24$
Hence, B alone can do the work in 24 days.
Using Rule 19,
B alone can do in = ${2 × 18 × 9 × 12}/{-18 × 9 + 12 × 9 + 12 × 18}$
= ${36 × 108}/{-162 + 108 + 216} = {36 × 108}/162$ = 24 days
Q-17) A and B together can do a piece of work in 10 days. A alone can do it in 30 days. The time in which B alone can do it is
(a)
(b)
(c)
(d)
(A + B)’s 1 day’s work = $1/10$
A’s 1 day’s work = $1/30$
B’s 1 day’s work = $1/10 - 1/30$
= ${3 - 1}/30 = 2/30 = 1/15$
Hence, B, alone can complete the work in 15 days.
Using Rule 4If A alone can do a certain work in 'x' days and A and B together can do the same work in 'y' days, then B alone can do the same work in $({xy}/{x - y})$ days.
Time taken by B = ${30 × 10}/{30 - 10}$ = 15 days
Q-18) A and B can do a piece of work in 8 days, B and C can do it in 24 days, while C and A can do it in 8$4/7$ days. In how many days can C do it alone?
(a)
(b)
(c)
(d)
(A + B)’s 1 day’s work = $1/8$
(B + C)’s 1 day’s work = $1/24$
(C + A)’s 1 day’s work = $7/60$
On adding all three,
2 (A + B + C)’s 1 day’s work = $1/8 + 1/24 + 7/60$
= ${15 + 5 + 14}/120 = 34/120$
(A + B + C)’s 1 day’s work = $17/120$
C’s 1 day’s work
= $17/120 - 1/8 = {17 - 15}/120 = 1/60$
C alone will complete the work in 60 days.
Using Rule 19,
C alone can do in= ${2xyz}/{xy - yz + zx}$
= ${2 × 8 × 24 × 60/7}/{8 × 24 - 24 × 60/7 + 60/7 × 8$
= ${23040/7}/{192 - 1440/7 + 480/7}$
= ${23040/7}/{{1344 - 1440 + 480}/7}$
= $23040/7 × 7/384$ = 60 days
Q-19) A work can be completed by P and Q in 12 days, Q and R in 15 days, R and P in 20 days. In how many days P alone can finish the work?
(a)
(b)
(c)
(d)
(P + Q)’s 1 day’s work = $1/12$ ...(i)
(Q + R)’s 1 day’s work = $1/15$ ...(ii)
(R + P)’s 1 day’s work = $1/20$ ...(iii)
Adding all three equations,
2 (P + Q + R)’s 1 day’s work = $1/12 + 1/15 + 1/20$
${5 + 4 + 3}/60 = 12/60 = 1/5$
(P + Q + R)’s 1 day’s work = $1/10$ ...(iv)
P’s 1 day’s work = $1/10 - 1/15 = {3 - 2}/30 = 1/30$
P alone will complete the work in 30 days.
Q-20) A particular job can be completed by a team of 10 men in 12 days. The same job can be completed by a team of 10 women in 6 days. How many days are needed to complete the job if the two teams work together?
(a)
(b)
(c)
(d)
According to question,
10 men’s one day’s work = $1/12$
1 man one day’s work =$1/{12 × 10} = 1/120$
Similarly, 1 woman one day’s work = $1/{6 × 10} = 1/60$
(1 man + 1 woman)’s one day’s work = $1/120 + 1/60$
= ${1 + 2}/120 = 3/120 = 1/40$
(10 men + 10 women)’s one day’s work = $10/40 = 1/4$
Therefore, both the teams can finish the whole work in 4 days.