# Exercise : 5

(b) 6 days

(c) 18 days

(d) Data insufficient

(e) None of these

**Ans.b**

X’s one day’s work = ^{1}⁄_{15} th work.

Y’s one day’s work = ^{1}⁄_{15} + 50% of ^{1}⁄_{15} = ^{1}⁄_{10} th work.

∴ (X + Y)’s one day’s work = ^{1}⁄_{15} + ^{1}⁄_{10} = ^{1}⁄_{6} th work.

Hence, they together finish the work in 6 days.

^{1}⁄

_{3}days

(b) 7

^{1}⁄

_{3}days

(c) 6

^{1}⁄

_{3}days

(d) 7 days

(e) None of these

**Ans.b**

A’s one day’s work = ^{1}⁄_{8}th work.

B’s one day’s work = ^{1}⁄_{3}rd work.

∴ A’s 4 day’s work = 4 × ^{1}⁄_{8} = ^{1}⁄_{2}nd work

∴ In next two days, total wall = ^{1}⁄_{2} + 2(^{1}⁄_{8}) - 2(^{1}⁄_{3}) = ^{1}⁄_{12}th wall

Remaining wall = 1 - ^{1}⁄_{12} = ^{11}⁄_{12}th

Now, ^{1}⁄_{8}th wall is built up by A in one day.

∴ ^{11}⁄_{12}th wall is built up by A in 8 × ^{11}⁄_{12} = 7^{1}⁄_{3} days.

(b) 16

(c) 18

(d) 25

(e) None of these

**Ans.b**

Sakshi’s one day’s work = ^{1}⁄_{20}th work.

Tanya’s one day’s work

= ^{1}⁄_{20} + 25% of ^{1}⁄_{20} = ^{1}⁄_{16}th work

Hence, Tanya takes 16 days to complete the work.

(b) 20 min.

(c) 11 min.

(d) 13 min.

(e) None of these

**Ans.d**

Part filled by first tap in one min = ^{1}⁄_{12}th

Part filled by second tap in one min = ^{1}⁄_{18}th

Now, 2[^{1}⁄_{12} + ^{1}⁄_{18}] + unfilled part = 1

⇒ unfilled part = ^{13}⁄_{18}th

∵ ^{1}⁄_{18}th part of tank is filled by second tap in 1 min.

∴ ^{13}⁄_{18}th part of tank is filled by second tap in 1 min.

= 18 × ^{13}⁄_{18} min = 13 min.

(b) 24 hours

(c) 26 hours

(d) 18 hours

(e) None of these

**Ans.b**

∵ cistern fill in 6 hours.

∴ in 1 hour, filled part = ^{1}⁄_{6}th

Now, due to leakage, filled part in 1 hour = ^{1}⁄_{8}th

Part of the cistern emptied, due to leakage in 1 hour

= ^{1}⁄_{6} - ^{1}⁄_{8} = ^{1}⁄_{24}th

∴ The leakage will empty the full cistern in 24 hrs.

(b) 18 min

(c) 15 min

(d) 10 min

(e) None of these

**Ans.c**

Let B can fill the cistern in x min. Then,

then A can fill the cistern in ^{x}⁄_{3} min

Given x - ^{x}⁄_{3} = 10 ⇒ x = 15 min.

(b) 112 hours

(c) 115 hours

(d) 100 hours

(e) None of these

**Ans.b**

Cistern filled by both pipes in one hour

= ^{1}⁄_{14} + ^{1}⁄_{16} = ^{15}⁄_{112}th

∴ Both pipes filled the cistern in ^{112}⁄_{15} hrs.

Now, due to leakage both pipes filled the cistern in

^{112}⁄_{15} + ^{32}⁄_{60} = 8 hrs.

∴ due to leakage, filled part in one hour ^{1}⁄_{8}

∴ part of cistern emptied, due to leakage in one hour

= ^{15}⁄_{112} - ^{1}⁄_{8} = ^{1}⁄_{112}th

∴ in 112 hr, the leakage would empty the cistern.

(b) 8 min

(c) 10 min

(d) 12 min

(e) None of these

**Ans.b**

In one min, (A + B) fill the cistern = ^{1}⁄_{10} + ^{1}⁄_{15} = ^{1}⁄_{6}th

In 3 min, (A + B) fill the cistern = ^{3}⁄_{6} = ^{1}⁄_{2}th

Remaining part = 1 - ^{1}⁄_{2} = ^{1}⁄_{2}

∵ ^{1}⁄_{10}th part filled by A in one min.

∴ ^{1}⁄_{2}nd part filled by A in 10 × ^{1}⁄_{2} = 5 min.

∴ Total time = 3 + 5 = 8 min.

(b) 12 min

(c) 15 min

(d) 9 min

(e) None of these

**Ans.a**

Work done by the waste pipe in 1 minutes

= ^{1}⁄_{20} - (^{1}⁄_{12} + ^{1}⁄_{15}) = -^{1}⁄_{10} [-ve sign means emptying]

∴ Waste pipe will empty the full cistern in 10 minutes.

(b) 9 min.

(c) 10 min.

(d) 7 min.

(e) None of these

**Ans.a**

In one min, (A + B) fill the cistern = ^{1}⁄_{12} + ^{1}⁄_{18} = ^{5}⁄_{36}th

In 4 min, (A + B) fill the cistern = ^{5}⁄_{36} × 4 = ^{5}⁄_{9}th

Rest part = 1 - ^{5}⁄_{9} = ^{4}⁄_{9}th

∵ ^{1}⁄_{18}th part is filled by B in one min.

∴ ^{4}⁄_{9}th part is filled by B in 18 × ^{4}⁄_{9} = 8 min.