## Sunday, April 17, 2011

### Problems on Clocks

Clocks
Problems on Clocks

General Concepts
The face or dail of a watch is a circle whose circumference is divided into 60 equal parts , called minute spaces.

A clock has two hands , the smaller one is called the hour hand or short hand while the larger hand is called the minute hand or long hand.

In 60 minutes , the minute hand gains 55 minutes on the hour hand.
In every hour , both the hands coincide once.
The hands are in the same straight line when they are coincident or opposite to each other
When the two hands are at right angles, they are 15 minute spaces apart.
When the hands are in opposite directions, they are 30 minute spaces apart.
Too Fast and Too Slow :If a watch or a clock indicates 8.15, when the correct time is 8, it is said to be 15 minutes too fast

On the other hand , if it indicates 7.45, when the correct time is 8, it is said to be too slow.

Solved Problems

1. Find the angle between the minute hand and hour hand of a clock when the time is 7.20

Sol : Angle traced by hour hand in 12 hours = 360 Degrees

Angle traced by it in 7 hrs 20 min i.e 22/3 hrs = (360/12) x (22/3)= 2200

Angle traced by minute hand in 60 min = 3600

Angle traced by it in 20 minutes = (360/60) x 20 = 1200

Required angle = (2200 - 1200) = 1000

2. At what time between 2 and 3 o' clock will the hands of a clock together ?

Sol : At o' clock, the hour hand is at 2 and minute hand at 12, i.e they are 10 minute spaces apart.

To be together , the minute hand must gain 10 minutes over the hour hand.

Now, 55 minutes are gaines by it in 60 minutes

10 minutes will be gained in {(60/55) x 10} min =10 10/11 min

The hands will coincide at 10 10/11 min past 2

3. At what time between 4 and 5 o' clock will the hands of a clock be at right angle ?

Sol : At 4 o' clock, the minute hand will be 20 min spaces behind the hour hand.

Now , when the two hands are at right angles , they are 15 min. spaces apart.

So, they are at right angles in following two cases

Case I : When minute hand is 15 minute spaces behind the hour hand

In this case min hand will have to gain (20 -15) = 5 minute spaces

55 min. spaces will be gained by it in (60 x 5)/55 min = 5 5/11 min

They are at right angles at 5 5/1 min past 4

Case II : When the minute hand is 15 minute spaces ahead of the hour hand

They are at right angles at 38 2/11 min past 4

### Pipes and Cisterns

Important Facts:

1.INLET:A pipe connected with a tank or cistern or a reservoir,
that fills it, it is known as Inlet.

OUTLET:A pipe connected with a tank or a cistern or a reservoir,
emptying it, is known as Outlet.

2. i) If a pipe can fill a tank in x hours, then :
part filled in 1 hour=1/x.

ii)If a pipe can empty a tank in y hours, then :
part emptied in 1 hour=1/y.

iii)If a pipe can fill a tank in x hours and another pipe can
empty the full tank in y hours( where y>x), then on
opening both the pipes, the net part filled in
1 hour=(1/x -1/y).

iv)If a pipe can fill a tank in x hours and another pipe can
empty the full tank in y hours( where x>y), then on opening
both the pipes, the net part filled in 1 hour=(1/y -1/x).

v) If two pipes A and B can fill a tank in x hours and y hours
respectively. If both the pipes are opened simultaneously, part
filled by A+B in 1 hour= 1/x +1/y.

Simple Problems

1)Two pipes A& B can fill a tank in 36 hours and 45 hours respectively.
If both the pipes are opened simultaneously, how much time will be
taken to fill the tank?

Sol: Part filled by A in 1 hour=1/36
Part filled by B in 1 hour= 1/45;
Part filled by (A+B)’s in 1 hour=1/36 +1/45= 9/180 =1/20
Hence, both the pipes together will fill the tank in 20 hours.

2)Two pipes can fill a tank in 10 hours & 12 hours respectively. While
3rd pipe empties the full tank n 20 hours. If all the three pipes
operate simultaneously,in how much time will the tank be filled?

Sol: Net part filled in 1 hour=1/10 +1/12 -1/20
=8/60=2/15
The tank be filled in 15/2hours= 7 hrs 30 min

3)A cistern can be filled by a tap in 4 hours while it can be emptied
by another tap in 9 hours. If both the taps are opened simultaneously,
then after how much time will the cistern get filled?

Sol: Net part filled in 1 hour= 1/4 -1/9= 5/36
Therefore the cistern will be filled in 36/5 hours or 7.2 hours.

4)If two pipes function simultaneously, the reservoir will be filled in
12 days.One pipe fills the reservoir 10 hours faster than the other.
How many hours does it take the second pipe to fill the reservoir.

Sol: Let the reservoir be filled by the 1st pipe in x hours.
The second pipe will fill it in (x+10) hours
1/x + (1/(x+10))= 1/12
=> (2x+10)/((x)*(x+10))= 1/12
=> x=20
So, the second pipe will take 30 hours to fill the reservoir.

5)A cistern has two taps which fill it in 12 min and 15 min respectively.
There is also a waste pipe in the cistern. When all the three are opened,
the empty cistern is full in 20 min. How long will the waste pipe take to
empty the full cistern?

Sol: Work done by a waste pipe in 1 min
=1/20 -(1/12+1/15)= -1/10 (-ve means emptying)

6)A tap can fill a tank in 6 hours. After half the tank is filled, three
more similar taps are opened. What is the total time taken to fill the
tank completely?

Sol: Time taken by one tap to fill the half of the tank =3 hours
Part filled by the four taps in 1 hour=4/6=2/3
Remaining part=1 -1/2=1/2
Therefore, 2/3:1/2::1:x
or x=(1/2)*1*(3/2)=3/4 hours.
i.e 45 min
So, total time taken= 3hrs 45min.

7)A water tank is two-fifth full. Pipe A can fill a tank in 10 min. And B
can empty it in 6 min. If both pipes are open, how long will it take to
empty or fill the tank completely ?

Sol: Clearly, pipe B is faster than A and So, the tank will be emptied.
Part to be emptied=2/5.
Part emptied by (A+B) in 1 min= 1/6 -1/10=1/15
Therefore, 1/15:2/5::1:x or x=((2/5)*1*15)=6 min.
So, the tank be emptied in 6 min.

8)Bucket P has thrice the capacity as Bucket Q. It takes 60 turns for
Bucket P to fill the empty drum. How many turns it will take for both the
buckets P&Q, having each turn together to fill the empty drum?

Sol: Let the capacity of P be x lit.
Then capacity of Q=x/3 lit
Capacity of the drum=60x lit
Required number of turns= 60x/(x+(x/3))= 60x*3/4x=45

Complex Problems

1)Two pipes can fill a cistern in 14 hours and 16 hours respectively. The
pipes are opened simultaneously and it is found that due to leakage in the
bottom it took 32min more to fill the cistern. When the cistern is full,
in what time will the leak empty it?

Sol: Work done by the two pipes in 1 hour= 1/14+1/16=15/112
Time taken by these two pipes to fill the tank=112/15 hrs.
Due to leakage, time taken = 7 hrs 28 min+ 32 min= 8 hours
Therefore, work done by (two pipes + leak) in 1 hr= 1/8
work done by leak n 1 hour=15/112 -1/8=1/112
Leak will empty full cistern n 112 hours.

2)Two pipes A&B can fill a tank in 30 min. First, A&B are opened. After
7 min, C also opened. In how much time, the tank s full.

Sol: Part filled n 7 min = 7*(1/36+1/45)=7/20
Remaining part= 1-7/20=13/20
Net part filled in 1 min when A,B and C are opened=1/36 +1/45- 1/30=1/60
Now, 1/60 part is filled in 1 min.
13/20 part is filled n (60*13/20)=39 min
Total time taken to fill the tank=39+7=46 min

3)Two pipes A&B can fill a tank in 24 min and 32 min respectively. If
both the pipes are opened simultaneously, after how much time B should
be closed so that the tank is full in 18 min.

Sol: Let B be closed after x min, then part filled by (A+B) in x min+
part filled by A in (18-x) min=1
x(1/24+1/32) +(18-x)1/24 =1
=> x=8
Hence B must be closed after 8 min.

4)Two pipes A& B together can fill a cistern in 4 hours. Had they been
opened separately, then B would have taken 6 hours more than A to fill
the cistern. How much time will be taken by A to fill the cistern
separately?

Sol: Let the cistern be filled by pipe A alone in x hours.
Pipe B will fill it in x+6 hours
1/x + 1/x+6=1/4
Solving this we get x=6.
Hence, A takes 6 hours to fill the cistern separately.

5)A tank is filled by 3 pipes with uniform flow. The first two pipes
operating simultaneously fill the tan in the same time during which
the tank is filled by the third pipe alone. The 2nd pipe fills the tank
5 hours faster than first pipe and 4 hours slower than third pipe. The
time required by first pipe is :

Sol: Suppose, first pipe take x hours to fill the tank then
B & C will take (x-5) and (x-9) hours respectively.
Therefore, 1/x +1/(x-5) =1/(x-9)
On solving, x=15
Hence, time required by first pipe is 15 hours.

6)A large tanker can be filled by two pipes A& B in 60min and 40 min
respectively. How many minutes will it take to fill the tanker from
empty state if B is used for half the time & A and B fill it together for
the other half?

Sol: Part filled by (A+B) n 1 min=(1/60 +1/40)=1/24
Suppose the tank is filled in x minutes
Then, x/2(1/24+1/40)=1
=> (x/2)*(1/15)=1
=> x=30 min.

7)Two pipes A and B can fill a tank in 6 hours and 4 hours respectively.
If they are opened on alternate hours and if pipe A s opened first, in
how many hours, the tank shall be full.

Sol: (A+B)’s 2 hours work when opened alternatively =1/6+1/4 =5/12
(A+B)’s 4 hours work when opened alternatively=10/12=5/6
Remaining part=1 -5/6=1/6.
Now, it is A’s turn and 1/6 part is filled by A in 1 hour.
So, total time taken to fill the tank=(4+1)= 5 hours.

8)Three taps A,B and C can fill a tank in 12, 15 and 20 hours respectively.
If A is open all the time and B and C are open for one hour each
alternatively, the tank will be full in.

Sol: (A+B)’s 1 hour’s work=1/12+1/15=9/60=3/20
(A+C)’s 1 hour’s work=1/20+1/12=8/60=2/15
Part filled in 2 hours=3/20+2/15=17/60
Part filled in 2 hours=3/20+2/15= 17/60
Part filled in 6 hours=3*17/60 =17/20
Remaining part=1 -17/20=3/20
Now, it is the turn of A & B and 3/20 part is filled by A& B in 1 hour.
Therefore, total time taken to fill the tank=6+1=7 hours

9)A Booster pump can be used for filling as well as for emptying a tank.
The capacity of the tank is 2400 m3. The emptying capacity of the tank is
10 m3 per minute higher than its filling capacity and the pump needs 8
minutes lesser to empty the tank than it needs to fill it. What is the
filling capacity of the pump?

Sol: Let, the filling capacity of the pump be x m3/min
Then, emptying capacity of the pump=(x+10) m3/min.
So,2400/x – 2400/(x+10) = 8
on solving x=50.

### Idioms

A Bird In The Hand Is Worth Two In The Bush:
Having something that is certain is much better than taking a risk for more, because chances are you might lose everything.
A Blessing In Disguise: Something good that isn't recognized at first.
A Chip On Your Shoulder: Being upset for something that happened in the past.
A Dime A Dozen: Anything that is common and easy to get.

For More Idioms Visit http://www.idiomsite.com/

## Monday, April 4, 2011

### Compound Interest

Compound Interest

Compound Interest: sometimes it so happens that the borrower and the lender agree to fix up a certain unit of time, say yearly or half-yearly or quarterly to settle the previous amount.
In such case, the amount after first unit of time becomes the principal for the second unit, the amount after second unit becomes the principal for the third unit and so on.
After a specified period, the difference between the amount and the money borrowed is called the compound Interest (abbreviated as C.I.) for that period.
FORMULAE:
Let Principal =p, Rate = R% per annum, Time=n years
I.                   When interest is compounded annually:
Amount = P[ 1 + (R/100)]n
II.                When the interest is compounded half-yearly :
Amount = p[ 1 + (R/2)/100]2n
III.             When the interest is compounded quarterly :
Amount = P[ 1 + (R/4)/100 ]4n
IV.              When the interest is compounded annually but time is in                    fraction, say3 2/5 Years.
Amount = P{ [1 + (R/100)]3  x [ 1 + (2R/5)/100] }
V.                 When rates are different for different years, say R1%, R2%,               R3% for 1st,   2nd  and 3rd year respectively.
Then, Amount= P [ 1 + R1/100][ 1 + R2/100][ 1+ R3/100]
VI.              Present worth of Rs. X Due n years hence is given by :
Present worth = X/ (1 + R/100)n

Solved Problem
Ex.1. Find compound interest on Rs.6250 at 16% per annum for 2 years, compounded annually.
Sol.  Amount = Rs. [ 6250 x ( 1 + 16/100)2 ]
=  Rs.[ 6250 x 29/25 x 29/25 ] = Rs.8410.
.. C.I = Rs.(8410 – 6250) = Rs. 2160.
Ex.2. Find compound interest on Rs. 5000 at 12% per annum for 1 year,  compounded  half-yearly.
Sol.   Principal = Rs.5000, Rate=6% per half year
Time = 1 year   = 2 half-years
Amount = Rs. {5000 x ( 1 + 6/100)2] = Rs. [ 5000 x 53/50 x 53/50 ]
= Rs. 5618
.. C.I. = Rs. (5618 – 5000) = Rs. 618
Ex.3. Find compound interest on Rs. 16000 at 20% per annum for 9 months,compounded quarterly.
Sol.  Principal = Rs. 16000,  Time = 9 months = 3 quarters
Rate = 20% per annum = 5% per quarter.
.. Amount = Rs. [ 16000 x { 1 + 5/100 )3 ]
= Rs. [ 16000 x 21/20 x 21/20 x 21/20 ] = Rs. 18522.
.. C.I. = Rs. (18522 – 16000) = Rs. 2522.
Ex.4.  The difference between the compounded interest and simple interest on a certain sum at 10% per annum for 2 years is Rs.631. Find the sum.
Sol. Let the sum be Rs. X .Then,
C.I.. = x[ 1 + 10/100 ]2  - x ] = 21x/100
S.I.  = [ (Xx10x2)/100] = X/5
.. (C.I.) – (S.I.) = [ (21x)/100 – (x/5) ] = x/100
.. x/100 = 631     x= 63100
Hence, the sum is Rs. 63100
Ex.5. If the compound interest on a certain sum for 2 years at 12% per annum is  Rs. 1590, what would be the simple interest?
Sol. Let the sum be Rs. X. Then
X{ 1 + 12/100 }2 – X = 1590  or  (784X/625) – X = 1590
Or 159X/625 = 1590 or x = {1590 x 625}/159 =6250
.. Sum = Rs. 6250
.. S.I. = Rs. [ (6250 x 12 x 2)/100] = Rs. 1500
Ex.6. A sun of money amounts to Rs.6690 after 3 years and to Rs 10035 after 6  years on compound interest. Find the sum.
P{ 1 + R/100 }3 = 6690 ….(i)  and  P{ 1 + R/100 }6  = 10035  … (ii)
On Dividing, we get [ 1 + R/100 ]3 = 10035/6690 = 3/2
Substituting this value in (i), we get;
P x 3/2 =6690  or  P=[6690 x 2/3] = 4460
Hence, the sum is Rs. 4460
Ex.7. A sum of money doubles itself at compound interest in 15 years. In how many  years will it become eight times?
Sol.       P { 1 + R/100 }15 = 2P   ie 2P= [ 1 + R/100]15 = 2    … (i
Let  P[ 1 + R/100]n = 8P      [ 1 + R/100]n =8 = 23 = { [1 + R/100]15 }3
[ 1 + R/100]n = [ 1 + R/100]45
i.e      n=45
Thus required time = 45 years
Ex.8. A certain sum amounts to Rs.7350 in 2 years and to Rs.8575 in 3 years . Find  the sum and rate per cent.
Sol.  S.I. on Rs.7350 for 1 year = Rs.(8575 – 7350) = Rs.1225.
.. Rate = [(100 x 1225)/(7350 x 1)]% = 16 2/3%
Let the sum be Rs. X .Then
X[1 + 50/(3 x100)]2 = 7350
i.e X x 7/6 x 7/6 =7350
X= {7350 x 36/49]
.. Sum = Rs. 5400