# Model 1 Basics On Time & Work Practice Questions Answers Test With Solutions & More Shortcuts

#### TIME & WORK PRACTICE TEST [7 - EXERCISES]

Question : 16 [SSC CGL Prelim 2005]

A and B can complete a piece of work in 30 days, B and C in 20 days, while C and A in 15 days. If all of them work together, the time taken in completing the work will be

a) 13\$1/3\$ days

b) 12\$2/3\$ days

c) 10 days

d) 12 days

Work done by (A + B) in 1 day = \$1/30\$

Work done by (B + C) in 1 day = \$1/20\$

Work done by (C + A) in 1 day = \$1/15\$

Work done by 2 (A +B + C) in 1 day

= \$1/30 + 1/20 + 1/15 = {2 + 3 + 4}/60\$

= \$9/60 = 3/20\$

Work done by (A + B + C) in 1 day = \$3/40\$

(A + B + C) will do the work in \$40/3 = 13{1}/3\$ days

Using Rule 5,

Time taken = \${2 × 30 × 20 × 15}/{30 × 20 + 20 × 15 + 15 × 30}\$

= \$18000/{600 + 300 + 450}\$

= \$18000/1350 = 13{1}/3\$ days

Question : 17 [SSC CHSL 2011]

A and B can do a piece of work in 10 days. B and C can do it in 12 days. A and C can do it in 15 days. How long will A take to do it alone ?

a) 30 days

b) 40 days

c) 24 days

d) 20 days

(A+B)’s 1 day’s work = \$1/10\$

(B + C)’s 1 day’s work = \$1/12\$

(C + A)’s 1 day’s work = \$1/15\$

2(A + B + C)’s 1 day’s work = \$1/10 + 1/12 + 1/15\$

= \${6 + 5 + 4}/60 = 15/60 = 1/44\$

(A + B + C)’s 1 day’s work = \$1/8\$

A’s 1 day’s work = \$1/8 - 1/12\$

= \${3 - 2}/24 = 1/24\$

A will complete the work in 24 days.

Using Rule 19,

A alone can do in = \${2 × x × y × z}/{xy + yz - zx}\$

= \${2 × 10 × 12 × 15}/{10 × 12 + 12 × 15 - 15 × 10}\$

= \$3600/{120 + 180 - 150}\$

= \$3600/150\$ = 24 days

Question : 18 [SSC CPO S.I.2003]

A, B and C can complete a piece of work in 24, 6 and 12 days respectively. Working together, they will complete the same work in

a) 4 days

b) 3\${3}/7\$ days

c) \$1/4\$ day

d) \$7/24\$ day

A’s 1 day’s work = \$1/24\$

B’s 1 day’s work = \$1/6\$

C’s 1 day’s work = \$1/12\$

(A + B + C)’s 1 day’s work

=\$1/24 + 1/6 + 1/12 = {1 + 4 + 2}/24 = 7/24\$

The work will be completed by them in \$24/7\$ i.e., 3\$3/7\$ days

Using Rule 3
If A can do a work in 'x' days, B can do the same work in 'y' days, C can do the same work in 'z' days then, total time taken by A, B and C to complete the work together = \$1/{1/x + 1/y + 1/z} = {xyz}/{xy + yz + zx}\$and
If workers are more than 3 then total time taken by A, B, C ...... so on to complete the work together = \$1/{1/x + 1/y + 1/z + ...}\$

Time taken = \${24 × 6 × 12}/{24 × 6 + 6 × 12 + 24 × 12}\$

= \$1728/{144 + 72 + 288}\$

= \$1728/504 = 24/7 = 3{3}/7\$ days

Question : 19 [SSC MTS 2011]

A and B can do a piece of work in 72 days. B and C can do it in 120 days, A and C can do it in 90 days. In how many days all the three together can do the work ?

a) 150 days

b) 60 days

c) 80 days

d) 100 days

(A+B)’s 1 day’s work = \$1/72\$,

(B+C)’s 1 day’s work = \$1/120\$,

(C+A)’s 1 day’s work = \$1/90\$

2 (A + B + C)'s 1 days work = \$1/72 + 1/120 + 1/90\$

= \${5 + 3 + 4}/360 = 1/30\$

(A+B+C)’s 1 day’s work = \$1/60\$

(A+B+C) will do the work in 60 days.

Using Rule 5,

Time taken = \${2 × 72 × 120 × 90}/{72 × 120 + 120 × 90 + 72 × 90}\$

= \$1555200/{8640 + 10800 + 6480}\$

= \$1555200/25920\$ = 60 days

Question : 20 [SSC SO 2007]

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) 40 days

b) 30 days

c) 18 days

d) 24 days

(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\$

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

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