Practice Time and work - quantitative aptitude Online Quiz (set-2) For All Competitive Exams
Q-1) If 12 men working 8 hours a day complete the work in 10 days, how long would 16 men working 7$1/2$ hours a day take to complete the same work?
(a)
(b)
(c)
(d)
Using Rule 1,
| Men | Working hours | Days |
| 12 | 8 | 10 |
| ↑ | ↑ | ↓ |
| 16 | 7$1/2$ | x |
∴ ${16 : 12}/{15/2 : 8}]$ : : 10 : x
16 × $15/2$ × x = 12 × 8 × 10
8 × 15 × x = 12 × 8 × 10
$x = {12 × 8 × 10}/{8 × 15}$ = 8 days
Q-2) P can complete $1/4$ of a work in 10 days, Q can complete 40% of the same work in 15 days, R, completes $1/3$ of the work in 13 days and S, $1/6$ of the work in 7 days. Who will be able to complete the work first ?
(a)
(b)
(c)
(d)
Time taken by P in completing 1 work
= 10 × 4 = 40 days
Time taken by Q in completing 1 work
= ${15 × 5}/2 = 75/2$ days
Time taken by R in completing 1 work
= 13 × 3 = 39 days
Time taken by S in completing 1 work
= 7 × 6 = 42 days
Clearly, Q took the least time i.e. $75/2$ or 37${1}/2$ days.
Q-3) If 28 men complete $7/8$ of a piece of work in a week, then the number of men, who must be engaged to get the remaining work completed in another week, is
(a)
(b)
(c)
(d)
Using Rule1,
| Work | Days | Men |
| $7/8$ | 7 | 28 |
| ↓ | ↓ | ↓ |
| $1/8$ | 7 | x |
$7/8 : 1 8 : : 28 : x$
where x is no. of men
$7/8 × x = 1/8 × 28$
$x = {28 × 8}/{7 × 8}$ = 4
Q-4) P can do $(1/4)$th of work in 10 days, Q can do 40% of work in 40 days and R can do $(1/3)$rd of work in 13 days. Who will complete the work first?
(a)
(b)
(c)
(d)
Since, P does $1/4$ th work in 10 days.
P will do 1 work in 10 × 4 = 40 days
Since, Q, does 40% part of work in 40 days
Q will do 100% work in
${40 × 100}/40$ = 100 days
Since, R, does $1/3$ rd work in 13 days.
R will do 1 work in 13 × 3 = 39 days
Q-5) A can complete $2/3$ of a work in 4 days and B can complete $3/5$ of the work in 6 days. In how many days can both A and B together complete the work ?
(a)
(b)
(c)
(d)
Using basics of Rule 2,
Time taken by A to complete the work
= ${4 × 3}/2$ = 6 days
Time taken by B to complete the work
= ${6 × 5}/3$ = 10 days
(A + B)’s 1 day’s work
= $1/6 + 1/10 = {5 + 3}/30 = 8/30 = 4/15$
A and B together will complete the work in
$15/4 = 3{3}/4$ days.
Q-6) 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-7) 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
Q-8) A and B can do a job in 6 and 12 days respectively. They began the work together but A leaves after 3 days. Then the total number of days needed for the completion of the work is :
(a)
(b)
(c)
(d)
A’s one day’s work = $1/6$
B’s one day’s work = $1/12$
(A + B)’s one day’s work = $1/6 + 1/12 = {2 + 1}/12 = 1/4$
(A + B)’s three day’s work = $3/4$
Remaining work = $1 - 3/4 = 1/4$
Total required number of days
= $1/4 × 12/1 + 3$ = 3 + 3 = 6 days
Using Rule 20,
Here, m = 6, n= 12, and p = 3
Time taken by B = $\text"mn - p(m+n)"/ \text"m"$
= ${6 × 12 - (6 + 12) × 3}/6 = {72 - 54}/6$ = 3 days
Total number of days taken to finish the works = 6 days
Q-9) A can do a piece of work in 12 days and B can do it in 18 days. They work together for 2 days and then A leaves. How long will B take to finish the remaining work ?
(a)
(b)
(c)
(d)
(A+B)’s 2 days’ work = $2(1/12 + 1/18) = 10/36$
Remaining work = $1 - 10/36 = 26/36$
Time taken by B to complete $26/36$ part of work
= $26/36$ × 18 = 13 days
Using Rule 20,
Here, m = 12, n = 18, p = 2
Time taken by B = $\text"mn - p(m+n)"/ \text"m"$
= ${12 × 18 - 2(12 + 18)}/12$
= ${216 - 60}/12$ = 13 days
Q-10) If 3 men or 4 women can plough a field in 43 days, how long will 7 men and 5 women take to plough it ?
(a)
(b)
(c)
(d)
3 men = 4 women
1 man = $4/3$ women
7 men = ${7 × 4}/3 = 28/3$ women
7 men + 5 women = $28/3 + 5$
= ${28 + 15}/3 = 43/3$ Women
Now, $M_1D_1 = M_2D_2$
4 × 43 = $43/3 × D_2$ ,
where $D_2$ = number of days
$D_2 = {4 × 3 × 43}/43$ = 12 days.
Using Rule 12,
Here, A = 3, B = 4, a = 43, $A_1$ = 7 and $B_1$ = 5
Time taken = ${a(A × B)}/{A_1B + B_1A}$
= ${43(3 × 4)}/{7 × 4 + 5 × 3}$
= ${43 × 12}/43$ = 12 days
Q-11) 6 men and 8 women can do a work in 10 days, Then 3 men and 4 women can do the same work in
(a)
(b)
(c)
(d)
6m + 8w ≡ 10 days
2 (3m + 4w) ≡ 10 days
3m + 4w ≡ 20 days
[Since the workforce has become half of the original force, so number of days must be double].
Using Rule 14,
Let us assume efficiency of 6 men = efficiency of 8 men.
A = 6, a = 20, B = 8, b = 20
$A_1 = 3, B_1$ = 4
Required time = $1/{A_1/{A × a} + B_1/{B × b}}$
= $1/{3/{6 × 20} + 4/{8 × 20}}$
= $1/{1/40 + 1/40} = 40/2$ = 20 days
Q-12) One man or two women or three boys can do a piece of work in 88 days. One man, one woman and one boy will do it in
(a)
(b)
(c)
(d)
1 man ≡ 2 women ≡ 3 boys
1 man + 1 woman + 1 boy
= $(3 + 3/2 + 1)$ boys
= $({6 + 3 + 2}/2)$ boys= $11/2$ boys
$M_1D_1 = M_2D_2$
3 × 88 = $11/2 × D_2$
$D_2 = {3 × 2 × 88}/11$ = 48 days
Using Rule 13,
Here, A = 1, B = 2, C = 3, a= 88
$A_1 = 1, B_1 = 1, C_1$ = 1
Required time = $a/{A_1/A + B_1/B + C_1/C}$
= $88/{1/1 + 1/2 + 1/3} = 88/{{6 + 3 + 2}/6}$ = 48 days
Q-13) 8 children and 12 men complete a certain piece of work in 9 days. Each child takes twice the time taken by a man to finish the work. In how many days will 12 men finish the same work ?
(a)
(b)
(c)
(d)
Using Rule 1,
2 children ≡ 1 man
8 children + 12 men ≡ 16 men
$M_1D_1 = M_2D_2$
16 × 9 = 12 × $D_2$
$D_2 = {16 × 9}/12$ = 12 days.
Q-14) 2 men and 3 women can do a piece of work in 10 days while 3 men and 2 women can do the same work in 8 days. Then, 2 men and 1 woman can do the same work in
(a)
(b)
(c)
(d)
2 × 10 men + 3 × 10 women = 3 × 8 men + 2 × 8 women
20 men + 30 women = 24 men + 16 women
4 men = 14 women or 2 men = 7 women
2 men + 3 women = 10 women
2 men + 1 woman = 8 women
$M_1D_1 = M_2D_2$
10 × 10 = 8 × $D_2$
$D_2 = 25/2 = 12{1}/2$ days
Using Rule 11
Here, $A_1 = 2, B_1 = 3, D_1$ = 10
$A_2 = 3, B_2 = 2, D_2$ = 8
$A_3 = 2, B_3$ = 1
Required time = ${D_1D_2(A_1B_2 - A_2B_1)}/{D_1(A_1B_3 - A_3B_1) - D_2(A_2B_3 - A_3 B_2)}$ days.
= ${10 × 8(2 × 2 - 3 × 3)}/{10(2 × 1 - 2 × 3) - 8(3 × 1 - 2 × 2)}$
= ${80(4 - 9)}/{10(2 - 6) - 8(3 - 4)}$
= ${-400}/{-40 + 8} = {-400}/{-32} = 25/2 = 12{1}/2$ days
Q-15) If 1 man or 2 women or 3 boys can complete a piece of work in 88 days, then 1 man, 1 woman and 1 boy together will complete it in
(a)
(b)
(c)
(d)
1 man = 2 women ≡ 3 boys 1 man + 1 woman + 1 boy
= $(3 + 3/2 + 1)$ boys = $11/2$ boys
$M_1D_1 = M_2D_2$
$3 × 88 = 11/2 × D_2$
$D_2 = {2 × 3 × 88}/11$ = 48 days
Using Rule 13If A men or B boys or C women can do a certain work in 'a' days, then $A_1$ men, $B_1$ boys and $C_1$ women can do the same work inTime taken = $a/{A_1/A + B_1/B + C_1/C}$
Here, A = 1, B= 2, C = 3, a = 88
$A_1 = 1, B_1 = 1, C_1$ = 1
Time taken = $a/{A_1/A + B_1/B + C_1/C}$
= $88/{1/1 + 1/2 + 1/3}$
= ${88 × 6}/{6 + 3 + 2}$ = 48 days
Q-16) A and B can do a piece of work in 10 days. B and C can do it in 12 days. C and A in 15 days. In how many days will C finish it alone ?
(a)
(b)
(c)
(d)
(A + B)'s 1 day's work = $1/10$ ... (i)
(B + C)'s 1 day’s work = $1/12$ ... (ii)
(C + A)’s 1 day’s work = $1/15$ ... (iii)
On adding all these,
2(A + B + C)’s 1 day’s work = $1/10 + 1/12 + 1/15$
= ${6 + 5 + 4}/60 = 1/4$
(A + B + C)’s 1 day’s work = $1/8$ … (iv)
C’s 1 day’s work = $1/8 - 1/10 = {5 - 4}/40 = 1/40$
C will finish the work in 40 days.
Using Rule 19,
C alone can do in = ${2 × 10 × 12 × 15}/{10 × 12 - 12 × 15 + 10 × 15}$
= ${240 × 15}/{120 - 180 + 150} = {240 × 15}/90$ = 40 days
Q-17) A and B undertook to do a piece of work for Rs.4500. A alone could do it in 8 days and B alone in 12 days. With the assistance of C they finished the work in 4 days. Then C’s share of the money is
(a)
(b)
(c)
(d)
Using Rule 25If A, B and C finish the work in m, n and p days respectively and they receive the total wages R, then the ratio of their wages is $1/m : 1/n : 1/p$
C’s 1 day’s work
= $1/4 - (1/8 + 1/12) = 1/4 - ({3 + 2}/24)$
= $1/4 - 5/24 = {6 - 5}/24 = 1/24$
A : B : C = $1/8 : 1/12 : 1/24$ = 3 : 2 : 1
C’s share = Rs.$(1/6 × 4500)$ = Rs.750
Q-18) The daily wages of A and B respectively are Rs.3.50 and Rs.2.50. When A finishes a certain work, he gets a total wage of Rs.63. When B does the same work, he gets a total wage of Rs.75. If both of them do it together what is the cost of the work ?
(a)
(b)
(c)
(d)
Time taken by A
= $63/{3.50}$ = 18 days
Time taken by B
= $75/{2.5}$ = 30 days
(A + B)’s 1 day’s work
= $1/18 + 1/30$
= ${5 + 3}/90 = 8/90 = 4/45$
Required time = $45/4$ days
Total wages
= Rs.$45/4$ × (3.50 + 2.50)
= Rs.$(45/4 × 6)$ = Rs.67.5
Q-19) 5 men can do a piece of work in 6 days while 10 women can do it in 5 days. In how many days can 5 women and 3 men do it ?
(a)
(b)
(c)
(d)
5 × 6 men = 10 × 5 women
3 men = 5 women
5 women + 3 men = 6 men
5 men complete the work in 6 days
6 men will complete the work in ${5 × 6}/6$ = 5 days
Using Rule 14
If 'A' men can do a certain work in 'a' days and 'B' women can do the same work in 'b' days, then the total time is taken when $A_1$ men and $B_1$ women work together isTime taken = $1/{A_1/{A . a} + B_1/{B . b}}$If A men do a certain work in 'a' days, B women do the same work in 'b' days and C boys do the same work in 'c' days then the total time taken when $A_1$ men, $B_1$ women and $C_1$ boys can work together is,
Total time taken = $1/{(A_1/{A . a} + B_1/{B . b} + C_1/{C . c})}$
Here, A = 5, a = 6, B = 10, b = 5, $A_1$ = 3, $B_1$ = 5
Time taken = $1/{A_1/{A × a} + B_1/{B × b}}$
= $1/{3/{5 × 6} + 5/{10 × 5}}$
= $1/{1/10 + 1/10}$ = 5 days
Q-20) If 10 men or 20 women or 40 children can do a piece of work in 7 months, then 5 men, 5 women and 5 children together can do half of the work in :
(a)
(b)
(c)
(d)
10 men ≡ 20 women
1 man = 2 women = 5 children
1 woman = 2 children
5 men + 5 women + 5 children
= 20 + 10 + 5 = 35 children
$M_1D_1 = M_2D_2$
40 × 7 = 35 × $D_2$
$D_2 = {40 ×7}/35$ = 8 months
5 men, 5 women and 5 children can do half of the work in 8 months
Required time = 4 months.
Using Rule 13,
Here, A = 10, B= 20, C = 40, a = 7
$A_1 = 5, B_1 = 5, C_1$ = 5
Time taken to do same work = $a/{A_1/A + B_1/B + C_1/C}$
= $7/{5/10 + 5/20 + 5/40}$
= $7/{1/2 + 1/4 + 1/8}$
= $7/{{4 + 2 + 1}/8}$ = 8 months
Half of the work they do in 4 months.