Time and Work Questions and Answers for SSC Exams - 1

Ans: C

B’s 3 day’s work = $$({1}/{21} × 3)$$ = $${1}/{7}$$.

Remaining work = $$(1 - {1}/{7})$$ = $${6}/{7}$$.

(A + B)’s 1 day’s work = $$({1}/{14} + {1}/{21})$$ = $${5}/{42}$$

Now, $${5}/{42}$$ work is done by A and B in 1 day.

∴ $${6}/{7}$$ work is done by A and B in $$({42}/{5} × {6}/{7})$$ = $${36}/{5}$$ days.

Hence, total time taken = $$(3 + {36}/{5})$$ days = $$10{1}/{5}$$ days

Ans: C

(A + B + C)’s 1 day’s work = $${1}/{4}$$

A’s 1 day’s work = $${1}/{16}$$

B’s 1 day’s work = $${1}/{12}$$

C’s 1 day’s work = $${1}/{4}$$ - ($${1}/{16}$$ + $${1}/{12}$$) = $$({1/4} - {7/48})$$ = $${5}/{48}$$

So, C alone can do the work in $${48}/{5}$$ = $$9{3}/{5}$$ days.

Ans: D

5 men’s 15 day’s work = 5 men’s 10 day’s work + 10 women’s 5 day’s work.

⇒ 5 men’s 5 day’s work = 10 women’s 5 day’s work.

⇒ 10 men’s 5’s day work = $$({1}/{15} × 5)$$ = $${1}/{3}$$

⇒ 10 women’s 1 day’s work = $${1}/{15}$$.

∴ 10 women can complete the work in 15 days.

Ans: C

A’s 1 hour’s work = $${1} /{4}$$

(B + C)’s 1’s hour’s work = $${1} / {3}$$

(A + C)’s 1 hour’s work = $${1} / {2}$$.

(A + B + C)’s 1 hour’s work = $${1} / {4} + {1} / {3}$$ = $${7} / {12}$$.

∴ B’s 1 hour’s work = (A + B+ C)’s 1 hour’s work - (A + C)’s 1 hour’s work

                      = $$7 / 12$$ - $${1} / {2}$$ = $${1} / {12}$$.

So, B alone can complete the work in 12 hours.

Ans: B

Total time taken by David and Michael together

= 4 days 19 hrs 12 min = 4 days $$19{1}/{5}$$ hrs

= 4 days + $$({96}/{5} × {1}/{24})$$days = $$4{4}/{5}$$days = $${24}/{5}$$days.

(David + Michael)’s 1 day’s work = $${5}/{24}$$

(David’s 1 day’s work) : (Michael’s 1 day’s work) = $${2}/{3} : 1$$ = 2 : 3.

∴ Michael’s 1 day’s work = $$({5}/{24} × {3}/{5})$$ = $${1}/{8}$$.

Hence, Michael alone can finish the job in 8 days.