Programmers interview

1. Principal: The money borrowed or lent out for a certain period is called the
principalor the sum.
2. Interest: Extra money paid for using other's money is called interest.
3. Simple Interest (S.I.) : If the interest on a sum borrowed for a certain period is reckoned uniformly, then it is called simple interest.
Let Principal = P, Rate = R% per annum (p.a.) and Time = T years. Then,
    i). S.I. =  (P*R*T)/100
   ii). P = (100*S.I)/(R*T)
       R = (100*S.I)/(P*T)
       T = (100*S.I)/(P*R)

1. Find the simple interest on Rs. 68,000 at 16 2/3% per annum for 9 months.

P = Rs.68000 

R = 50/3% p.a
T = 9/12 years  = 3/4 years
S.I. = (P*T*R*)/100 = (68,000*(50/3)*(3/4)*(1/100)) = 8500 R.s

2. Find the simple interest on Rs. 3000 at 6 1/4% per annum for the period from 4th February 2005 to 18th April 2005.

Time = (24+31+18)days = 73 days = 73/365 years = 1/5 years.
P = Rs.3000
R = 6 ¼ % p.a = (25/4)% p.a
S.I = (3,000*(25/4)*(1/5)*(1/100)) = 37.50 R.s

Remark : The day on which money is deposited is not counted while the day on which money is withdrawn is counted .

3. A sum at simple interests at 13 ½ % per annum amounts to Rs.2502.50 after 4 years find the sum.

Let sum be Rs. x then , 
S.I. = Rs.(x*(27/2) *4*(1/100) ) = Rs.27x/50
amount = (Rs. x+(27x/50)) = Rs.77x/50
       =  77x/50 = 2502.50 => 77x = 2502.50 * 50 =>  x = 1625

4. A sum of Rs. 800 amounts to Rs. 920 in 8 years at simple intere interest rate is increased by 8%, it would amount to bow mucb ?

S.l. = Rs. (920 - 800) = Rs. 120
P = Rs. 800
T = 3 yrs. 
R = ((100 x 120)/(800*3)) % = 5 %.

New rate = (5 + 3)% = 8%.
New S.l. = (800*8*3)/100 = 192 Rs.
New amount = (800+192) = 992 Rs.

5. Adam borrowed some money at the rate of 6% p.a. for the first two years , at the rate of 9% p.a. for the next three years , and at the rate of 14% p.a. for the period beyond five years. 1£ he pays a total interest of Rs. 11, 400 at the end of nine years how much money did he borrow ?

Let the sum borrowed be x. Then,

(x*2*6)/100 + (x*9*3)/100 + (x*14*4)/100 = 11400
=> (3x/25 + 27x/100 + 14x / 25) = 11400       
=> 95x/100 = 11400 
=> x = (11400*100)/95 = 12000.
Hence, sum  borrowed = Rs.12,000.

6. A certain sum of money amounts to Rs. 1008 in 2 years and to Rs.1164 in 3 ½ years. Find the sum and rate of interests.

S.I. for 1 ½ years = Rs.(1164-1008) = Rs.156.
S.l. for 2 years = Rs.(156*(2/3)*2)=Rs.208         
Principal = Rs. (1008 - 208) = Rs. 800.
Now, P = 800, T = 2 and S.l. = 208.
Rate =(100* 208)/(800*2)% = 13%

7. At what rate percent per annum will a sum of money double in 16 years.

Let principal = P. 
Then, S.l. = P and T = 16 yrs.
Rate = (100 x P)/(P*16) % = 6 1/2 % p.a.

The simple interest on a sum of money is 4/9 of the principal .Find the rate percent and time, if both are numerically equal.

Let sum = Rs. x. Then, S.l. = Rs. 4x/9
Let rate = R% and time = R years.
Then, (x*R*R)/100=4x/9 or R2 =400/9 or R = 20/3 = 6 2/3.
Rate = 6 2/3 %   and Time = 6 2/3 years = 6 years 8 months.

9. The simple interest on a certain sum of money for 2 l/2 years at 12% per annum is Rs. 40 less tban the simple interest on the same sum for 3 ½ years at 10% per annum. Find the sum.

Let the sum be Rs. x Then, ((x*10*7)/(100*2)) – ( (x*12*5)/(100*2)) = 40                     
                => (7x/20)-(3x/10) = 40      
                => x = (40 * 20) = 800.
Hence, the sum is Rs. 800.

10. A sum was put at simple interest at a certain rate for 3 years. Had it been put at 2% higher rate, it would have fetched Rs. 360 more. Find the sum.

Let sum = P and original rate = R.
Then, [ (P*(R+2)*3)/100] – [ (P*R*3)/100] = 360.
=> 3PR + 6P - 3PR = 36000 Û 6P=36000 Û P=6000
Hence, sum = Rs. 6000.

11. What annual instalment will discharge a debt of Rs. 1092 due in 3 years at 12% simple interest?

Let each Instalment be Rs. x
Then, ( x + ((x*12*1)/100) ) + ( x + ((x*12*2)/100) ) + x = 1092
=> ((28x/25) + (31x/25) + x) = 1092   
=> (28x + 31x + 25x) = (1092*25)
=> x = (1092*25)/84 = Rs.325.  
Each instalment = Rs. 325.

12. A sum of Rs. 1550 is lent out into two parts, one at 8% and another one at 6%. If the total annual income is Rs. 106, find the money lent at each rate.

Let the sum lent at 8% be Rs. x and that at 6% be Rs. (1550 - x).
((x*8*1)/100) + ((1550-x)*6*1)/100 = 106
=> 8x + 9300 – 6x = 10600 
=> 2x = 1300  
=> x = 650.
Money lent at 8% = Rs. 650. 
Money lent at 6% = Rs. (1550 - 650) = Rs. 900.

Simple Interest

1. RATIO: The ratio of two quantities a and b in the same units, is the fraction a/b and we write it as a:b.
In the ratio a:b, we call a as the first term or antecedent and b, the second  term or consequent.

Ex. The ratio 5: 9 represents 5/9 with antecedent = 5, consequent = 9.
Rule: The multiplication or division of each term of a ratio by the same non-zero number does not affect the ratio.
Ex.  4: 5 = 8: 10 = 12: 15 etc. Also, 4: 6 = 2: 3.

2. PROPORTION: The equality of two ratios is called proportion.
If a: b = c: d, we write, a: b:: c : d and we say that a, b, c, d are in proportion . Here a and d are called extremes, while b and c are called mean terms.
 Product of means = Product of extremes.
Thus, a: b:: c : d <=> (b x c) = (a x d).

3. (i) Fourth Proportional: If a : b = c: d, then d is called the fourth proportional
            to a, b, c.
   (ii) Third Proportional: If a: b = b: c, then c is called the third proportional to
            a and b.
   (iii) Mean Proportional: Mean proportional between a and b is square root of ab

4. (i) COMPARISON OF RATIOS:
        We say that (a: b) > (c: d) <=>  (a/b)>(c /d).
      (ii) COMPOUNDED RATIO:
            The compounded ratio of the ratios (a: b), (c: d), (e : f) is (ace:bdf)

5.   (i) Duplicate ratio of (a : b) is (a2 : b2).
      (ii) Sub-duplicate ratio of (a : b) is (√a : √b).
     (iii)Triplicate ratio of (a : b) is (a3 : b3).
     (iv) Sub-triplicate ratio of (a : b) is (a ⅓ : b ⅓ ).
     (v) If (a/b)=(c/d), then  ((a+b)/(a-b))=((c+d)/(c-d))    (Componendo and dividendo)

6. VARIATION:
(i) We say that x is directly proportional to y, if x = ky  for some constant k and
     we write, x µ y.
(ii) We say that x is inversely proportional to y, if xy = k for some constant k and
       we write, x∞(1/y)

1. If a : b = 5 : 9 and b : c = 4: 7, find a : b : c.

a:b=5:9 and b:c=4:7= (4X9/4): (7x9/4) = 9:63/4
         a:b:c = 5:9:63/4 =20:36:63.

2. Find: (i) the fourth proportional to 4, 9, 12; (ii) the third proportional to 16 and 36; iii) the mean proportional between 0.08 and 0.18.

i) Let the fourth proportional to 4, 9, 12 be x.

       Then, 4 : 9 : : 12 : x  ó4 x x=9x12 óX=(9 x 12)/14=27;
   Fourth proportional to 4, 9, 12 is 27.

 (ii) Let the third proportional to 16 and 36 be x.

      Then, 16 : 36 : : 36 : x  ó16 x x = 36 x 36 ó x=(36 x 36)/16 =81
      Third proportional to 16 and 36 is 81.
                                           

(iii) Mean proportional between 0.08 and 0.18
      Ö0.08 x 0.18 =Ö8/100 x 18/100= Ö144/(100 x 100)=12/100=0.12

3. If x : y = 3 : 4, find (4x + 5y) : (5x - 2y).

X/Y=3/4 ó (4x+5y)/(5x+2y)= (4( x/y)+5)/(5 (x/y)-2) =(4(3/4)+5)/(5(3/4)-2)
              
              =(3+5)/(7/4)=32/7

4. Divide Rs. 672 in the ratio 5 : 3.

Sum of ratio terms = (5 + 3) = 8.
      First part = Rs. (672 x (5/8)) = Rs. 420; Second part = Rs. (672 x (3/8)) = Rs. 252.

5. Divide Rs. 1162 among A, B, C in the ratio 35 : 28 : 20.

Sum of ratio terms = (35 + 28 + 20) = 83.
       A's share = Rs. (1162 x (35/83))= Rs. 490; B's share = Rs. (1162 x(28/83))= Rs. 392;
       C's share = Rs. (1162 x (20/83))= Rs. 280.

6. A bag contains 50 p, 25 P and 10 p coins in the ratio 5: 9: 4, amounting to Rs. 206. Find the number of coins of each type.

Let the number of 50 p, 25 P and 10 p coins be 5x, 9x and 4x respectively.
       (5x/2)+( 9x/ 4)+(4x/10)=206ó 50x + 45x + 8x = 4120ó1O3x = 4120 óx=40.
                                             
      Number of 50 p coins = (5 x 40) = 200; Number of 25 p coins = (9 x 40) = 360;
      Number of 10 p coins = (4 x 40) = 160.

7. A mixture contains alcohol and water in the ratio 4 : 3. If 5 litres of water is added to the mixture, the ratio becomes 4: 5. Find the quantity of alcohol in the given mixture.

Let the quantity of alcohol and water be 4x litres and 3x litres respectively
    4x/(3x+5)=4/5 ó20x=4(3x+5)ó8x=20 óx=2.5
       Quantity of alcohol = (4 x 2.5) litres = 10 litres.

Ratio and Propotion

1. Direct Proportion: Two quantities are said to be directly proportional, if on the increase (or decrease) of the one, the other increases (or decreases) to the same
 Ex. 1. Cost is directly proportional to the number of articles.
               (More Articles, More Cost)

Ex. 2. Work done is directly proportional to the number of men working on it
               (More Men, More Work)
   2. Indirect Proportion: Two quantities are said to be indirectly proportional,if on the increase of the one, the other decreases to the same extent and vice-versa.
    Ex. 1. The time taken by a car in covering a certain distance is inversely proportional to the speed of the car.
              (More speed, Less is the time taken to cover a distance)

Ex. 2. Time taken to finish a work is inversely proportional to the num of persons working at it.
(More persons, Less is the time taken to finish a job)

Remark: In solving questions by chain rule, we compare every item with the term to be found out.

1. If 15 toys cost Rs, 234, what do 35 toys cost?

Let the required cost be Rs. x. Then,
         More toys, More cost     =>      (Direct Proportion)
     15 : 35 : : 234 : x => (15 * x) = (35 * 234) => x = (35 X 234)/15 = 546
               Hence, the cost of 35 toys is Rs. 546.

2. If 36 men can do a piece of work in 25 hours, in how many hours will 15 men do it ?

Let the required number of hours be x. Then,
        Less men, More hours  (Indirect Proportion)
                                                                                
       15 : 36 : : 25 : x  ó(15 x x) = (36 x 25)  ó(36 x 25)/15 = 60
       
       Hence, 15 men can do it in 60 hours.

3. If the wages of 6 men for 15 days be Rs.2100, then find the wages of for 12 days.

Let the required wages be Rs. x.
               More men, More wages       (Direct Proportion)
               Less days, Less wages      (Direct Proportion)

                Men  6: 9           : :2100:x
                Days 15:12
Therefore (6 x 15 x x)=(9 x 12 x 2100) =>  x=(9 x 12 x 2100)/(6 x 15)=2520

          Hence the required wages are Rs. 2520.

4. If 20 men can build a wall 66 metres long in 6 days, what length of a similar can be built by 86 men in 8 days?

Let the required length be x metres

More men, More length built     (Direct Proportion)                                          

Less days, Less length built      (Direct Proportion)


Men 20: 35
Days 6: 3  : : 56 : x

Therefore (20 x 6 x x)=(35 x 3 x 56)óx=(35 x 3 x 56)/120=49

            Hence, the required length is 49 m.

5. If 15 men, working 9 hours a day, can reap a field in 16 days, in how many days will 18 men reap the field, working 8 hours a day?

Let the required number of days be x.
More men, Less days   (indirect proportion)
Less hours per day, More days (indirect proportion)


        Men 18 : 15
       Hours per day 8: 9   } : :16 : x
        (18 x 8 x x)=(15 x 9 x 16)ó x=(44 x 15)144 = 15
    
        Hence, required number of days = 15.

6. If 9 engines consume 24 metric tonnes of coal, when each is working 8 hours day, bow much coal will be required for 8 engines, each running 13hours a day, it being given that 3 engines of former type consume as much as 4 engines of latter type? Sol.

Let 3 engines of former type consume 1 unit in 1 hour.
        Then, 4 engines of latter type consume 1 unit in 1 hour.
Therefore 1 engine of former type consumes(1/3) unit in 1 hour.

1 engine of latter type consumes(1/4) unit in 1 hour.

Let the required consumption of coal be x units.
Less engines, Less coal consumed  (direct proportion)
More working hours, More coal consumed (direct proportion)
Less rate of consumption, Less coal consumed(direct prportion)

 Number of engines 9: 8
Working hours       8 : 13 } :: 24 : x
Rate of consumption (1/3):(1/4)
[ 9 x 8 x (1/3) x  x) = (8 x 13 x (1/4) x 24 ) => 24x = 624 ó x = 26.
Hence, the required consumption of coal = 26 metric tonnes.

7. A contract is to be completsd in 46 days sad 117 men were said to work 8 hours a day. After 33 days, (4/7) of the work is completed. How many additional men may be employed so that the work may be completed in time, each man now working 9 hours a day?

Sol.Remaining work = (1-(4/7) =(3/7)
                Remaining period = (46 - 33) days = 13days
       Let the total men working at it be x.
               Less work, Less men                        (Direct Proportion)
               Less days, More men                        (Indirect Proportion)
               More Hours per Day, Less men        (Indirect Proportion)

Work  (4/7): (3/7)
Days  13:33         } : : 117: x
Hrs/day 9 : 8
Therefore (4/7) x 13 x 9 x x =(3/7) x 33 x 8 x 117 or x=(3 x 33 x 8 x 117)/(4 x 13 x 9)=198

       Additional men to be employed = (198 - 117) = 81.

8. A garrison of 3300 men had provisions for 32 days, when given at the rate of 860 gns per head. At the end of 7 days, a reinforcement arrives and it was for that the provisions wi1l last 17 days more, when given at the rate of 826 gms per head, What is the strength of the reinforcement?

The problem becomes:
3300 men taking 850 gms per head have provisions for (32 - 7) or 25 days,
How many men taking 825 gms each have provisions for 17 days?
            Less ration per head, more men          (Indirect Proportion)
            Less days, More men                        (Indirect Proportion)
              
 Ration 825 : 850
 Days 17: 25  } : : 3300 : x
(825 x 17 x x) = 850 x 25 x 3300 or x = (850 x 25 x 3300)/(825 x 17)=5000

Strength of reinforcement = (5500 - 3300) = 1700.

Chain Rule

Factorial Notation: Let n be a positive integer.  Then, factorial n, denoted by n!  is defined as:
                           
                        n! = n(n-1)(n-2)........3.2.1.

Examples: (i) 5! = (5x 4 x 3 x 2 x 1) = 120; (ii) 4! = (4x3x2x1) = 24 etc.
We define, 0! = 1.

Permutations:  The different arrangements of a given number of things by taking some or all at a time, are called permutations.

Ex. 1.All permutations (or arrangements) made with the letters a, b, c by taking  two at a time are: (ab, ba, ac, bc, cb).

Ex. 2.All permutations made with the letters a,b,c, taking all at a time are:
(abc, acb, bca, cab, cba).

Number of Permutations: Number of all permutations of n things, taken r  at a time, given by:

                        nPr  = n(n-1)(n-2).....(n-r+1) = n!/(n-r)!

Examples: (i) 6p2= (6x5) = 30. (ii) 7p3 = (7x6x5) = 210.

Cor. Number of all permutations of n things, taken all at a time = n!

An Important Result: If there are n objects of  which p1 are alike of  one kind; p2 are alike of another kind; p3 are alike of third kind and  so on and pr are alike of rth kind, such that (p1+p2+.......pr) = n.

Then, number of permutations of these n objects is:

                        n!  /  (p1!).p2!)......(pr!)

Combinations: Each of the different groups or selections which can be formed by taking some or all of a number of objects, is called a combination.
Ex. 1. Suppose we want to select two out of three boys A, B, C.  Then, possible selections are AB, BC and CA.
Note that AB and BA represent the same selection.


Ex. 2. All the combinations formed by a, b, c, taking two at a time are ab, bc, ca.

Ex. 3. The only combination that can be formed of three letters a, b, c taken all at a time is abc.

Ex. 4. Various groups of 2 out of four presons A, B, C, D are:

                        AB, AC, AD, BC, BD, CD.

Ex. 5. Note that ab and ba are two different permutations but they represent  the same combination.

Number of Combinations: The number of all combination of n things,
taken r at a time is:

            nCr = n! / (r!)(n-r)! = n(n-1)(n-2).....to r factors / r!

Note that: ncr  = 1 and nc0 = 1.

An Important Result:  ncr = nc(n-r).

Example:  (i)  11c4= (11x10x9x8)/(4x3x2x1) = 330.

                 (ii) 16c13= 16c(16-13) = 16x15x14/3! = 16x15x14/3x2x1 = 560.

1. Evaluate: 30!/28!

We have, 30!/28! = 30x29x(28!)/28! = (30x29) = 870.

2. Find the value of (i) 60p3 (ii) 4p4

 (i) 60p3 = 60!/(60-3)! = 60!/57! = 60x59x58x(57!)/57! = (60x59x58) = 205320.

 (ii) 4p4 = 4! = (4x3x2x1) = 24.

3. Find the vale of (i) 10c3 (ii) 100c98 (iii) 50c50

  (i) 10c3 = 10x9x8/3! = 120.

        (ii) 100c98 = 100c(100-98)= 100x99/2! = 4950.

        (iii) 50c50 = 1.      [ncn = 1]

4. How many words can be formed by using all letters of the word “BIHAR”

The word BIHAR contains 5 different letters.

Required number of words = 5p5= 5! = (5x4x3x2x1) = 120.

5. How many words can be formed by using all letters of the word ‘DAUGHTER’ so that the vowels always come together?

Given word contains 8 different letters.  When the vowels AUE are always together, we may suppose them to form an entity, treated as one letter.
Then, the letters to be arranged are DGNTR (AUE).

Then 6 letters to be arranged in 6p6 = 6! = 720 ways.

The vowels in the group (AUE) may be arranged in 3! = 6 ways.

Required number of words = (720x6) = 4320.

6. How many words can be formed from the letters of the word ‘EXTRA’ so that the vowels are never together?

The given word contains 5 different letters.
        Taking the vowels EA together, we treat them as one letter.
        Then, the letters to be arranged are XTR (EA).
        These letters can be arranged in 4! = 24 ways.
        The vowels EA may be arranged amongst themselves in 2! = 2 ways.
        Number of words, each having vowels together = (24x2) = 48 ways.
       Total number of words formed by using all the letters of the given words
                           = 5! = (5x4x3x2x1) = 120.

Number of words, each having vowels never together = (120-48) = 72.

7. How many words can be formed from the letters of the word ‘DIRECTOR’ So that the vowels are always together?

       In the given word, we treat the vowels IEO as one letter.
       Thus, we have DRCTR (IEO).
       This group has 6 letters of which R occurs 2 times and others are different.
       Number of ways of arranging these letters = 6!/2! = 360.
       Now 3 vowels can be arranged among themselves in 3! = 6 ways.
       Required number of ways = (360x6) = 2160.

8. In how many ways can a cricket eleven be chosen out of a batch of 15 players ?

Required number of ways = 15c11 = 15c(15-11) = 11c4

                                                   = 15x14x13x12/4x3x2x1 = 1365.

9. In how many ways, a committee of 5 members can be selected from 6 men and 5 ladies, consisting of 3 men and 2 ladies?

(3 men out 6) and (2 ladies out of 5) are to be chosen.

Required number of ways = (6c3x5c2) = [6x5x4/3x2x1] x [5x4/2x1] = 200.

Permutations & Combinations

1.Experiment :An operation which can produce some well-defined outcome is called an experiment

2.Random experiment: An experiment in which all possible outcome are known and the exact out put cannot be predicted in advance is called an random experiment
Eg of performing random experiment:
(i)rolling an unbiased dice
(ii)tossing a fair coin
(iii)drawing a card from a pack of well shuffled card
(iv)picking up a ball of certain color from a bag containing ball of different colors

Details:
(i)when we throw a coin. Then either a head(h) or a tail (t) appears.
(ii)a dice is a solid cube, having 6 faces ,marked 1,2,3,4,5,6 respectively when we throw a die , the outcome is the number that appear on its top face .
(iii)a pack of cards has 52 cards it has 13 cards of each suit ,namely spades,  clubs ,hearts and diamonds
       Cards of spades  and clubs are black cards
       Cards of hearts  and diamonds are red cards
       There are 4 honors of each suit
       These are aces ,king ,queen and jack
       These are called face cards

3.Sample space :When we perform an experiment ,then the set S of all  possible outcome is called the sample space
eg of sample space:
(i)in tossing a coin ,s={h,t}
(ii)if two coin are tossed ,then s={hh,tt,ht,th}.
(iii)in rolling a die we have,s={1,2,3,4,5,6}.

4.event:Any subset of a sample space.
5.Probability of  occurrence of an event.
let S be the sample space and E be the event .
then,EÍS.
P(E)=n(E)/n(S).

6.Results on Probability:
(i)P(S) = 1  (ii)0


1. In a throw of a coin ,find the probability of getting a head.

Here s={H,T} and E={H}.
P(E)=n(E)/n(S)=1/2


2. Two unbiased  coin are tossed .what is the probability of getting atmost one head?

Here S={HH,HT,TH,TT}
Let Ee=event of getting one head
E={TT,HT,TH}
P(E)=n(E)/n(S)=3/4


3. An unbiased die is tossed .find the probability of getting a multiple of 3

Here S={1,2,3,4,5,6}
Let E be the event of getting the multiple of 3
then ,E={3,6}
P(E)=n(E)/n(S)=2/6=1/3


4. In a simultaneous  throw of pair of dice .find the probability of getting the total more than 7

Here n(S)=(6*6)=36
let E=event of getting a total more than 7
              ={(2,6),(3,5),(3,6),(4,4),(4,5),(4,6),(5,3),(5,4),(5,5),(5,6),(6,2),(6,3),(6,4),(6,5),(6,6)}
P(E)=n(E)/n(S)=15/36=5/12.


5. A bag contains 6 white and 4 black balls .2 balls are drawn at random. find the probability that they are of same colour.

let S be the sample space
Then n(S)=no of ways of drawing 2 balls out of (6+4)=10c2=(10*9)/(2*1)=45
Let E=event of getting both balls  of same colour
Then n(E)=no of ways(2 balls out of six) or(2 balls out of 4)
               =(6c2+4c2)=(6*5)/(2*1)+(4*3)/(2*1)=15+6=21
P(E)=n(E)/n(S)=21/45=7/15


6.Two dice are thrown together .What is the probability that the sum of the number on the two faces is divided by 4 or 6

Clearly n(S)=6*6=36
Let E be the event that the sum of the numbers on the two faces is divided by 4  or 6.Then

E={(1,3),(1,5),(2,2),(2,4),(2,6),(3,1),(3,3),(3,5),(4,2),(4,4),(5,1),(5,3),(6,2),  
        (6,6)}
n(E)=14.
Hence p(e)=n(e)/n(s)=14/36=7/18


7.Two cards are drawn at random from a pack of 52 cards.what is the probability that either both  are black or both are queen?

We have n(s)=52c2=(52*51)/(2*1)=1326.
Let A=event of getting both black cards
      B=event of getting both queens
A Ç B = event of getting queen of black cards
n(A) = 26c2 = (26*25)/(2*1) = 325,
n(B) = 4c2 = (4*3)/(2*1) = 6 and
n(A Ç B) = 2c2 = 1
P(A) = n(A)/n(S)=325/1326;
P(B) = n(B)/n(S)=6/1326 and
P(A Ç B) = n(A Ç B)/n(S)=1/1326
P(A È B) = P(A)+P(B)-P(A Ç B) = (325+6-1/1326) = 330/1326 = 55/221

Probability

The Face or dial 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 one is called the minute hand or long hand..
  i) In 60 minutes, the minute hand gains 55 minutes on the hour hand.
 ii) In every hour, both the hands coincide once.
iii) The hands are in the same straight line when they are coincident or opposite to each other.
 iv) When the two hands are at right angles, they are 15 minute spaces apart.
  v) When the hand's are in opposite directions, they are 30 minute spaces apart.
 vi) Angle traced by hour hand in 12 hrs = 360°.
vii) Angle traced by minute hand in 60 min. = 360°.

Too Fast and Too Slow: If a watch or a clock indicates 8.15, when the correct time , 8 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 15 minutes too slow.

1. Find the angle between the hour hand and the minute hand of a clock when 3.25.

angle  traced by the hour hand in 12 hours = 360°
Angle traced by it in three hours 25 min (ie) 41/12 hrs=(360*41/12*12)° =102*1/2°
angle traced by minute hand in 60 min. = 360°.
Angle traced by it in 25 min. = (360 X 25 )/60= 150°
Required angle = 1500 – 102*1/2°= 47*1/2°

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

At 2 o'clock, the hour hand is at 2 and the minute hand is at 12, i.e. they are 10 min  spaces apart.
To be together, the minute hand must gain 10 minutes over the hour hand.
Now, 55 minutes are gained by it in 60 min.
10 minutes will be  gained in (60 x 10)/55  min. = 120/11 min.
The hands will coincide at 120/11 min. past 2.

3. At what time between 4 and 5 o'clock will the hands of a clockbe 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 min. spaces behind the hour hand:
In this case min. hand will have to gain (20 - 15) = 5 minute spaces. 55 min. spaces are gained by it in 60 min.

 5 min spaces will be gained by it in 60*5/55  min=60/11min.

:. They are at right angles at 60/11min. past 4.
Case II. When the minute hand is 15 min. spaces ahead of the hour hand:
To be in this position, the minute hand will have to gain (20 + 15) = 35 minute spa' 55 min. spaces are gained in 60 min.
35 min spaces are  gained in (60 x 35)/55 min =40/11
          
     :. They are at right angles at 40/11 min. past 4.

4. Find at what time between 8 and 9 o'clock will the hands of a clock being the same straight line but not together.

At 8 o'clock, the hour hand is at 8 and the minute hand is at 12, i.e. the two hands_ are 20 min. spaces apart.
To be in the same straight line but not together they will be 30 minute spaces apart. So, the minute hand will have to gain (30 - 20) = 10 minute spaces over the hour hand.
55 minute spaces are gained. in 60 min.
10 minute spaces will be gained in (60 x 10)/55 min. = 120/11min.
:. The hands will be in the same straight line but not together at 120/11 min.

5. At whattime between 5 and 6 o'clock are the hands of a clock 3minapart?

At 5 o'clock, the minute hand is 25 min. spaces behind the hour hand.
       Case I. Minute hand is 3 min. spaces behind the hour hand.
In this case, the minute hand has to gain' (25 - 3) = 22 minute spaces. 55 min. are gained in 60 min.
22 min. are gaineg in (60*22)/55min. = 24 min.
:. The hands will be 3 min. apart at 24 min. past 5.
     Case II. Minute hand is 3 min. spaces ahead of the hour hand.
In this case, the minute hand has to gain (25 + 3) = 28 minute spaces. 55 min. are gained in 60 min.
 28 min. are gained in  (60 x 28_)/55=346/11
The hands will be 3 min. apart at 346/11 min. past 5.

6. Tbe minute hand of a clock overtakes the hour hand at intervals of 65 minutes of the correct time. How much a day does the clock gain or lose?

In a correct clock, the minute hand gains 55 min. spaces over the hour hand in 60 minutes.
To be together again, the minute hand must gain 60 minutes over the hour hand. 55 min. are gained in 60 min.
60 min are gained in  (60 * 60)/55 min = 720/11 min.
                                  
But, they are together after 65 min.
Gain in 65 min = 720/11 - 65 = 5/11min.

Gain in 24 hours =(5/11 * (60*24)/65)min =440/43
The clock gains 440/43  minutes in 24 hours.

7. A watch which gains uniformly, is 6 min. slow at 8 o'clock in the morning Sunday and it is 6 min. 48 sec. fast at 8 p.m. on following Sunday. When was it correct?

Time from 8 a.m. on Sunday to 8 p.m. on following Sunday = 7 days 12 hours = 180 hours

The watch gains (5 + 29/5) min. or 54/5 min. in 180 hrs.
Now 54/5  min. are gained in 180 hrs.
5 min. are gained in (180 x 5/54 x 5) hrs. = 83 hrs 20 min. = 3 days 11 hrs 20 min.
Watch is correct 3 days 11 hrs 20 min. after 8 a.m. of Sunday.
It will be correct at 20 min. past 7 p.m. on Wednesday.

8. A clock is set right at 6 a.m. The clock loses 16 minutes in 24 hours. What will be the true time when the clock indicates 10 p.m. on 4th day?

Time from 5 a.m. on a day to 10 p.m. on 4th day = 89 hours.
      Now 23 hrs 44 min. of this clock = 24 hours of correct clock.

356/15 hrs of this clock = 24 hours of correct clock.
       89 hrs of this clock = (24 x 31556 x 89) hrs of correct clock.
                              = 90 hrs of correct clock.
       So, the correct time is 11 p.m.

9. A clock is set right at 8 a.m. The clock gains 10 minutes in 24 hours will be the true time when the clock indicates 1 p.m. on the following day?

Time from 8 a.m. on a day  1 p.m. on the following day = 29 hours.
       24 hours 10 min. of this clock = 24 hours of the correct clock.
145 /6  hrs of this clock = 24 hrs of the correct clock

29 hrs of this clock = (24 x  6/145 x 29) hrs of the correct clock
= 28 hrs 48 min. of correct clock
The correct time is 28 hrs 48 min. after 8 a.m.
This is 48 min. past 12.

Problems on Clock

Under this heading we mainly deal with finding the day of the week on a particular given date the  process of finding it lies on obtaining the number of odd days.

1. Odd Days : Number of days more than the complete number of weeks in a given Period.

2. LeapYear: Every year which is divisible by 4 is called a leap year.

3. Thus each one of the years 1992, 1996, 2004, 2008, 2012, etc. is a leap year. Every 4th century is a leap year but no other century is a leap year.thuseach one of 400, 800, 1200,' 1600, 2000, etc. is a leap year.
None of 1900, 2010, 2020, 2100, etc. is a leap year.

4. An year which is not a leap year is called an ordinary year.

   a) An ordinary year has 365 days.   b) A leap  year has 366 days.

   Counting of Odd Days:

      i) 1 ordinary year = 365 days = (52 weeks + 1 day).
         :. An ordinary year has 1 odd day.
     ii) 1 leap year = 366 days = (52 weeks + 2 days).
         :. A leap year has 2 odd days.
    iii) 100 years = 76 ordinary years + 24 leap years
                   = [(76 x 52) weeks + 76 days) + [(24 x 52) weeks + 48 days]
                   = 5200 weeks + 124 days = (5217 weeks + 5 days).
         :. 100 years contain 5 odd days.
         200 years contain 10 and therefore 3 odd days.
         300 years contain 15 and therefore 1 odd day.
         400 years contain (20 + 1) and therefore 0 odd day.
         Similarly, each one of 800, 1200, 1600, 2000, etc. contains 0 odd days.
         Remark: (7n + m) odd days, where m < 7 is equivalent to m odd days.
         Thus, 8 odd days ≡ 1 odd day etc.
.-------------.
| days | Day 
|-------------
| 0     | sun   
| 1     | mon   
| 2     | tues  
| 3     | wed   
| 4     | thur  
| 5     | fri   
| 6     | sat   
'-------------'

1.Wbat was the day of the week on, 16th July, 1776?

16th July, 1776 = (1775 years + Period from 1st Jan., 1776 to 16th July, 1776)
Counting of odd days :
1600 years have 0 odd day. 100 years have 5 odd days.
75 years = (18 leap years + 57 ordinary years)
= [(18 x 2) + (57 x 1)] odd days = 93 odd days
= (13 weeks + 2 days) = 2 odd days.

.. 1775 years have (0 + 5 + 2) odd days = 7 odd days = 0 odd day.
Jan.  Feb.  March   April   May   June   July
31   + 29  +   31    +  30   +  31  +  30  +16 = 198days
= (28 weeks + 2 days) =2days
:. . Total number of odd days = (0 + 2) = 2. Required day was 'Tuesday'.

2. What was the day of the week on 16th August, 1947?

15th August, 1947 = (1946 years + Period from 1st Jan., 1947 to 15th
Counting of odd days:
1600 years have 0 odd day. 300 years have 1 odd day.
47 years = (11 leap years + 36 ordinary years)
= [(11 x 2) + (36 x 1») odd days = 58 odd days = 2 odd days.
Jan. Feb. March April May June July Aug.
31 + 28 + 31 + 30 + 31 + 30 + 31 + 15
= 227 days = (32 weeks + 3 days) = 3,
Total number of odd days = (0 + 1 + 2 + 3) odd days = 6 odd days.
Hence, the required day was 'Saturday'.

3. What was the day of the week on 16th April, 2000 ?

16th April, 2000 = (1999 years + Period from 1st Jan., 2000 to 16thA'
Counting of odd days:
1600 years have 0 odd day. 300 years have 1 odd day.
99 years = (24 leap years + 75 ordinary years)
= [(24 x 2) + (75 x 1)] odd days = 123 odd days
= (17 weeks + 4 days) = 4 odd days.
Jan. Feb. March April
31 + 29 + 31 + 16 = 107 days = (15 weeks + 2 days) = 2 odd,
Total number of odd days = (0 + 1 + 4 + 2) odd days = 7 odd days = 0 odd day.   Hence, the required day was 'Sunday'.

4. On what dates of Jull.2004 did Monday fall?

Let us find the day on 1st July, 2004.
2000 years have 0 odd day. 3 ordinary years have 3 odd days.

Jan. Feb. March April May June July

31 + 29 + 31 + 30 + 31 + 30 + 1
= 183 days = (26 weeks + 1 day) = 1 t .
Total number of odd days = (0 + 3 + 1) odd days = 4 odd days.      '
:. 1st July 2004 was 'Thursday',-,-
Thus, 1st Monday in July 2004 _as on 5th July.
Hence, during July 2004, Monday fell on 5th, 12th, 19th and 26th.

5. Prove that the calendar for the year 2008 will serve for the year 20ll

In order that the calendar for the year 2003 and 2014 be the same, 1st  January of  both the years must be on the same day of the week.
For this, the number of odd days between 31st Dec., 2002 and 31st Dec.,2013 must be the same.
We know that an ordinary year has 1 odd day and a leap year has 2 odd During this period, there are 3 leap years, namely 2004, 2008 and 2012 and 8 ordinary years.
Total number of odd days = (6 + 8) days = 0 odd day.
Hence, the calendar for 2003 will serve for the year 2014.

6. Prove that any date in March of a year is the same day of the week corresponding date in November that year.

We will show that the number of odd days between last day of February and last  day of October is zero.    .
March    April    May    June    July    Aug.    Sept.   Oct.
31     +   30  +  31    + 30   + 31 +   31  +    30   +  31
= 241 days = 35 weeks = 0 odd day. ,Number of odd days during this period = O.
Thus, 1st March of an year will be the same day as 1st November of that year. Hence, the result follows.

Problems On Calendar

1. Partnership:When two or more than two persons run a business jointly, they are
    called partners and the deal is known as partnership.

2. Ratio of Division of Gains:
          i) When investments of all the partners are for the same time, the gain or loss is distributed     a  among the partners in the ratio of their investments.
Suppose A and B invest Rs. x and Rs. y respectively for a year in a business, then at the end of the year:
(A’s share of profit) : (B's share of profit) = x : y.
       ii) When investments are for different time periods, then equivalent capitals are calculated for a    unit of time by taking (capital x number of units of time). Now, gain or loss is divided in the ratio of  these capitals.

 Suppose A invests Rs. x for p months and B invests Rs. y for q months, then
 (A’s share of profit) : (B's share of profit) = xp : yq.

3. Working and Sleeping Partners: A partner who manages the business is known . as a working partner and the one who simply invests the money is a sleeping partner.

1. A, B and C started a business by investing Rs. 1,20,000, Rs. 1,35,000 and ,Rs.1,50,000 respectively. Find the share of each, out of an annual profit of Rs. 56,700.

Ratio of shares of A, Band C = Ratio of their investments
= 120000 : 135000 : 150000 = 8 : 9 : 10.
A’s share = Rs. (56700 x (8/27))= Rs. 16800.
B's share = Rs. ( 56700 x (9/27)) = Rs. 18900.
C's share = Rs. ( 56700 x (10/27))=Rs. 21000.

2. Alfred started a business investing Rs. 45,000. After 3 months, Peter joined him with a capital of Rs. 60,000. After another 6 months, Ronald joined them with a capital of Rs. 90,000. At the end of the year, they made a profit of Rs. 16,500. Find the lire of each.

Clearly, Alfred invested his capital for 12 months, Peter for 9 months and Ronald
         for 3 months.
         So, ratio of their capitals = (45000 x 12) : (60000 x 9) : (90000 x 3)
                                             = 540000 : 540000 : 270000 = 2 : 2 : 1.
         Alfred's share = Rs. (16500 x (2/5)) = Rs. 6600
         Peter's share = Rs. (16500 x (2/5)) = Rs. 6600
         Ronald's share = Rs. (16500 x (1/5)) = Rs. 3300.

3. A, Band C start a business each investing Rs. 20,000. After 5 months A withdrew Rs.6000 B withdrew Rs. 4000 and C invests Rs. 6000 more. At the end of the year, a total profit of Rs. 69,900 was recorded. Find the share of each.

Ratio of the capitals of A, Band C
                     = 20000 x 5 + 15000 x 7 : 20000 x 5 + 16000 x 7 : 20000 x 5 + 26000 x 7
                      = 205000:212000 : 282000 = 205 : 212 : 282.


        A’s share = Rs. 69900 x (205/699) = Rs. 20500                        I
        B's share = Rs. 69900 x (212/699) = Rs. 21200;
        C's share = Rs. 69900 x (282/699) = Rs. 28200.

4. A, Band C enter into partnership. A invests 3 times as much as B and B invests two-third of what C invests. At the end of the year, the profit earned is Rs. 6600. What is the share of B ?

Let C's capital = Rs. x. Then, B's capital = Rs. (2/3)x
        A’s capital = Rs. (3 x (2/3).x) = Rs. 2x.
    Ratio of their capitals = 2x : (2/3)x :x = 6 : 2 : 3.
    Hence, B's share = Rs. ( 6600 x (2/11))= Rs. 1200.

5. Four milkmen rented a pasture. A grazed 24 cows for 3 months; B 10 for 5 months; C 35 cows for 4 months and D 21 cows for 3 months. If A's share of rent is Rs. 720, find the total rent of the field.

Ratio of shares of A, B, C, D = (24 x 3) : (10 x 5) : (35 x 4) : (21 x 3)   = 72 : 50 : 140 : 63.                                                                                     
            Let total rent be Rs. x. Then, A’s share = Rs. (72x)/325
                     (72x)/325=720  x=(720 x 325)/72 = 3250                  
            Hence, total rent of the field is Rs. 3250.

6. A invested Rs. 76,000 in a business. After few months, B joined him Rs. 57,000. At the end of the year, the total profit was divided between them in ratio 2 : 1. After bow many months did B join?

Suppose B joined after x months. Then, B's money was invested for (12 - x)
                        (76000 x 12)/(57000 x (12-x) =2/1 
      => 912000=114000(12-x)                                               
 114 (12 - x) = 912 => 12-x=8 => x=4
 Hence, B joined after 4 months.

7. A, Band C enter into a partnership by investing in the ratio of 3 : 2: 4. After 1 year, B invests another Rs. 2,70,000 and C, at the end of 2 years, also invests Rs.2,70,000. At the end of three years, profits are shared in the ratio of 3 : 4 : 5. Find initial investment of each.

Let the initial investments of A, Band C be Rs. 3x, Rs. 2x and Rs. 4x respectively. Then,
(3x x 36) : [(2x x 12) + (2x + 270000) x 24] : [(4x x 24) + (4x +270000) x 12]=3:4:5
                     
1O8x : (72x + 6480000) : (144x + 3240000) = 3 : 4 : 5
          108x /(72x+6480000)=3/4  => 432x = 216x + 19440000
                                   => 216x = 19440000         
                                         x = 90000
            
Hence, A’s initial investment = 3x = Rs. 2,70,000;
   B's initial investment = 2x = Rs. 1,80,000;
   C's initial investment = 4x = Rs. 3,60,000.

Problems on Patnership

PROFIT OR GAIN:IF SP IS GREATER THAN CP,THE SELLING PRICE IS SAID TO HAVE PROFIT OR GAIN.
LOSS: IF SPIS LESS THAN CP,THE SELLER IS SAID TO INCURED A LOSS.
FORMULA :
1.GAIN=(SP)-(CP).          
2.LOSS=(CP)-(SP).
3.LOSS OR GAIN IS ALWAYS RECKONED ON CP
4. GAIN %={GAIN*100}/CP.                                                                             
5.LOSS%={LOSS*100}/CP.
6.SP={(100+GAIN%) /100}*CP.
7.SP={(100-LOSS%)/100}*CP.
8.{100/(100+GAIN%)} *SP
9.CP=100/(100-LOSS%)}*SP
10.IF THE ARTICLE IS SOLD AT A GAIN OF SAY 35%, THEN SP =135% OF CP
11.IF A ARTICLE IS SOLD AT A LOSS OF SAY 35%. THEN SP=65% OF CP.
12.WHEN A PERSON SELLS TWO ITEMS,ONE AT A GAIN OF X% AND OTHER AT A LOSS OF X%.THEN THE SELLER ALWAYS INCURES A LOSS GIVEN:
{LOSS%=(COMON LOSS AND GAIN ) 2}/10.=(X/10) 2
13.IF THE TRADER PROFESSES TO SELL HIS GOODS AT CP BUT USES FALSE WEIGHTS,THEN
GAIN=[ERROR/(TRUE VALUE)-(ERROR)*100]%

1 A man buys an article for rs.27.50 and sells it for rs.28.50. find his gain %.

cp=rs27.50, sp=rs 28.50
gain=rs(28.50 –27.50)=rs1.10
so gain%={(1.10/27.50)*100}=4%

2. If the a radio is sold for rs 490 and sold for rs 465.50.find loss%.

cp=rs490,sp= 465.50.
loss=rs(490-465.50)=rs 24.50.
loss%=[(24.50/490)*100]%=5%

3.find S.P when (i)CP=56.25,gain=20%. (ii)CP=rs 80.40,loss=5%

sol:
(i)SP =20% of rs 56.25 ,=rs{(120/100)*56.25}=rs67.50.
(ii)CP=rs 80.40,loss=5%
 sol: sp=85% of rs 80.40
=rs {(85/100)*80.40}=rs 68.34.

4 find cp when: (i) sp =rs 40.60 : gain=16% (ii) sp=rs51.70:loss=12%

(i)  cp=rs{(100/116)*40.60}=rs 35.
(ii) cp=rs{(100/88)*51.87}=rs58.75.

5 A person incures loss for by selling a watch for rs1140.at what price should the watch be sold to earn a 5% profit ?

let the new sp be rsx.then
  (100-loss%) : (1st  sp) = (100+gain%) (2ndsp)
   =>  {(100-5)/1400} = {(100+5)/x}
   => x = {(105*1140)/95} = 1260.

6 A book was sold for rs 27.50 with a profit of 10%. if it were sold for rs25.75, then what would be % of profit or loss?

sol. SP=rs 27.50: profit =10%.
sol. CP=rs {(100/110)*27.50}=rs 25.
When sp =Rs25.75 ,profit =Rs(25.75-25)=Rs 0.75                       
Profit% = {(0.75/25)*100}% = 25/6% =3%

7 .If the cost price is 96% of sp then whqt is the profit %

sp=Rs100 : then cp=Rs 96:profit =Rs 4.
Profit={(4/96)*100}%=4.17%

Ex.8. The cp of 21 articles is equal to sp of 18 articles.find gain or loss %

CP of each article be Rs 1
CP of 18 articles =Rs18 ,sp of 18 articles =Rs 21.
Gain%=[(3/18)*100]%=50/3%

9 By selling 33 metres of cloth , one gains the selling price of 11 metres . Find the gain percent .

(SP of 33m)-(CP of 33m)=Gain=SP of 11m
SP of 22m = CP of 33m
Let CP of each metre be Re.1 , Then, CP of 22m= Rs.22,SP of 22m=Rs.33.
Gain%=[(11/22)*100]%=50%

10 A vendor bought bananas at 6 for Rs.10 and sold them at Rs.4 for Rs.6 .Find his gain or loss percent .

Suppose , number of bananas bought = LCM of 6 and 4=12
CP=Rs.[(10/6)*12]=Rs.20  ; SP= Rs[(6/4)*12]=Rs.18
Loss%=[(2/20)*100]%=10%

11. A man brought toffees at for a rupee. How many for a rupee must he sell to gain 50%?

C.P of 3 toffees=Re 1; S.P of 3 toffees =150% of Re.1=3/2.
    For Rs.3/2, toffees sold =3, for Re.1, toffees sold = [3*(2/3)] = 2.

12.A grocer purchased 80 kg of sugar at Rs.13.50 per kg and mixed it with 120kg sugar at Rs.16per kg. At what rate should he sell the mixer to gain 16%?

C.P of 200 kg of mixture = Rs. (80 * 13.50+120*16) = Rs.3000.
       S.P =116% 0f Rs.3000 =Rs.[(116/200) *3000]=Rs.3480.
\ Rate of S.P of the mixture =Rs.[3480/200] per kg =Rs.17.40 per kg.

13. Pure ghee cost Rs.100 per kg. After adulterating it with vegetable oil costing Rs.50 per kg, A shopkeeper sells the mixture at the rate of Rs.96 per kg, thereby making a profit of 20%.In What ratio does he mix the two?

Mean cost price =Rs. [ (100/120)*96 ]=Rs.80 per kg.

By the rate of allegation :
:========================================:
:    C.P of 1kg ghee   : C.P of 1kg oil  :
:           100        :     50          :
:......................:.................:
:    Mean price     :
:     80      :
:========================================:  
Required ratio = 30 : 20 = 3 : 2

14. A dishonest dealer professes to sell his goods at cost price but uses a weight of 960 gms for a kg weight . Find his gain percent.

Gain% =[ ( Error/(error value)-(error) )* 100 ]% =  [(40/960)*100] % = 4 1/6 %

15. If the manufacturer gains 10%,the wholesale dealer 15% and the retailer 25%,then find the cost of production of a ,the retail price of which is Rs.1265?

  Let the cost of production of the table be Rs x
 The ,125% of 115% of 110% of x=1265
=>  125/100*115/100*110/100*x=1265=>253/160*x=>1265=>x=(1265*160/253)=Rs.800

16 . Monika purchesed a pressure cooker at 9/10th of its selling price and sold it at 8% more than its S.P .find her gain percent.

Let the s.p be Rs. X .then C.P = Rs.9x/10,Receipt=108% of rs.x=Rs 27x/25
Gain=Rs (27x/25*9x/10)=Rs(108x-90x/100)=Rs18x/100
Gain%=(18x/100*10/9x*100)%=20%

17 An article is sold at certain price. By selling it at 2/3 of its price one losses 10%,find the gain at original price ?

let the original s.p be Rs x. then now S.P=Rs2x/3,loss=10%
now C.P=Rs20x/27*27/20x*100)%=35%

18. A tradesman sold an article at a loss of 20%.if the selling price has been increased by Rs100,ther would have been a gain of 5%.what was the cost price of the article?

Let C.P be Rs x. then (105% of x)-(80 % of x)=100 or 25% of x=100
=> x/4=100 or x=400
=>  so,C.P =Rs 400

19. A man sells an article at a profit of 25%if he had bought it 20% less and sold it for Rs 10.50 less,he would have gained 30%find the cost price of the article.

 Let the C.P be Rs x
      1st S.P=125% of x =125x/100=5x/4;2nd S.P=80% of x=80x/100=4x/5
      2nd S.P=130% of 4x/5=(130/100*4x/5)=26x/25
      => 5x/4-26x/25=10.50óx=(10.50*100)/21=50
   hence C.P=Rs.50

20.The price of the jewel,passing through three hands,rises on the whole by65%.if the first and the second sellers 20%and25% profit respectively find the percentage profit earned by the third seller.

Let the orginal price of the jewel be Rs p and let the profit earned  by the thrid seller be x%
Then,(100+x)% of 125% OF 120% OF P=165% OF P
=> ((100+X)/100*125/100*120/100*P)=(165/100*P)
=>  (100+X)=(165*100*100)/(125*120)=110=>X=10%

Profit And Loss

1. a km/hr= (a* 5/18) m/s.    

2. a m / s = (a*18/5) km/hr.

3 Time taken by a train of length 1 metres to pass a pole or a standing man or a signal  post is equal to the time taken by the train to cover 1 metres.

4. Time taken by a train of length 1 metres to pass a stationary object of length  b metres is the time taken by the train to cover (1 + b) metres.

5. Suppose two trains or two bodies are moving in the same direction at u m / s and v m/s, where u > v, then their relatives speed = (u - v) m / s.

6. Suppose two trains or two bodies are moving in opposite directions at u m / s and v m/s, then their relative speed is = (u + v) m/s.

7. If two trains of length a metres and b metres are moving in opposite directions at u  m / s and v m/s, then time taken by the trains to cross each other =  (a + b)/(u+v) sec.

8.If two trains of length  a metres and b metres are moving in the same direction
at u m / s and v m / s, then the time taken by the faster  train to cross the slower train = (a+b)/(u-v) sec.

9. If two trains (or bodies) start at the same time from points A and B towards each other and after crossing they take a and b sec in reaching B and A respectively, then
(A's speet) : (B’s speed) = (b1/2: a1/2).

1. A train 100 m long is running at the speed of 30 km / hr. Find the time taken by it to pass a man standing near the railway line.

Speed of the train = (30 x 5/18_) m / sec 
                   = (25/3) m/ sec.
Distance moved in passing the standing man = 100 m.
Required time taken = 100/(25/3) = (100 *(3/25)) sec = 12 sec

2. A train is moving at a speed of 132 km/br. If the length of the train is 110 metres, how long will it take to cross a railway platform 165 metres long?

       Speed of train = 132 *(5/18) m/sec = 110/3  m/sec.
       Distance covered in passing the platform = (110 + 165) m = 275 m.       
       Time taken =275 *(3/110)  sec = 15/2 sec = 7 1/2 sec

3. A man is standing on a railway bridge which is 180 m long. He finds that a train crosses the bridge in 20 seconds but himself in 8 seconds. Find the length of the train and its speed?

Let the length of the train be x metres,
Then, the train covers x metres in 8 seconds and (x + 180) metres in 20 sec
x/8=(x+180)/20 => 20x = 8 (x + 180)     <=>   x = 120.
Length of the train = 120 m.
Speed of the train = (120/8) m / sec = m / sec = (15 *18/5) kmph = 54 km

4. A train 150 m long is running with a speed of 68 kmph. In what time will it pass a man who is running at 8 kmph in the same direction in which the train is going?

Speed of the train relative to man
           = (68 - 8) kmph
           = (60* 5/18) m/sec = (50/3)m/sec

Time taken by the train to cross the man  
= Time taken by It to cover 150 m at 50/3  m / sec = 150 *3/ 50  sec  = 9sec

5. A train 220 m long is running with a speed of 59 kmph.. In what will it pass a man who is running at 7 kmph in the direction opposite to that in which the train is going?

Speed of the train relative to man = (59 + 7) kmph
       = 66 *5/18  m/sec = 55/3 m/sec.
Time taken by the train to cross the man
                = Time taken by it to cover 220 m at (55/3) m / sec 
                = (220 *3/55) sec = 12 sec

6. Two trains 137 metres and 163 metres in length are running towards each other on parallel lines, one at the rate of 42 kmph and another at 48 kmpb. In what time will they be clear of each other from the moment they meet?

Relative speed of the trains = (42 + 48) kmph = 90 kmph
                             = (90*5/18) m / sec = 25 m /sec.  
Time taken by the  trains  to'pass each other
   = Time taken to cover (137 + 163) m at 25 m /sec =(300/25) sec = 12 sec

7. Two trains 100 metres and 120 metres long are running in the same direction with speeds of 72 km/hr,In howmuch time will the first train cross the second?

Relative speed of the trains = (72 - 54) km/hr = 18 km/hr
      = (18 * 5/18) m/sec = 5 m/sec.
Time taken by the trains to cross each other
      = Time taken to cover (100 + 120) m at 5 m /sec = (220/5) sec = 44 sec.

8. A train 100 metres long takes 6 seconds to cross a man walking at 5 kmph in the direction opposite to that of the train. Find the speed of the train.?

Let the speed of the train be x kmph.
Speed of the train relative to man = (x + 5) kmph = (x + 5) *5/18 m/sec.
Therefore 100/((x+5)*5/18)=6 <=> 30 (x + 5) = 1800 <=> x = 55      
Speed of the train is 55 kmph.

9. A train running at 54 kmph takes 20 seconds to pass a platform. Next it takes.12 sec to pass a man walking at 6 kmph in the same direction in which the train is going . Find the length of the train and the length of the platform.

Let the length of train be x metres and length of platform be y metres.
Speed of the train relative to man = (54 - 6) kmph = 48 kmph
      = 48*(5/18) m/sec = 40/3 m/sec.
In passing a man, the train covers its own length with relative speed.
     Length of train = (Relative speed * Time) = ( 40/3)*12 m = 160 m.
Also, speed of the train = 54 *(5/18)m / sec = 15 m / sec.
   (x+y)/15 = 20 <=> x + y = 300 <=> Y = (300 - 160) m = 140 m.

10. A man sitting in a train which is traveling at 50 kmph observes that a goods train, traveling in opposite direction, takes 9 seconds to pass him. If the goods train is 280 m long, find its speed.?

 Relative speed = 280/9 m / sec = ((280/9)*(18/5)) kmph = 112 kmph.
    Speed of goods train = (112 - 50) kmph = 62 kmph.

Problems On Trains

1.In water ,the direction along the stream is called downstream and ,the direction against the stream is called upstream.
2.If the speed of a boat in still water is u km/hr and the speed of the stream is v km/hr,then:
             speed downstream=(u+v)km/hr.
             speed upstream=(u-v)km/hr.
3.If the speed downstream is a km/hr and the speed upstream is b km/hr,then :
speed in still water = 1/2(a+b)km/hr
rate of stream = 1/2(a-b)km/hr

1.A man can row upstream at 7 kmph and downstream at 10kmph.find man’s rate in still water and the rate of current.

Sol. Rate in still water=1/2(10+7)km/hr=8.5 km/hr.
Rate of current=1/2(10-7)km/hr=1.5 km/hr.

2. A man takes 3 hours 45 minutes to row a boat 15 km downstream of a river and 2hours30minutes to cover a distance of 5km upstream. find the speed of the river current in km/hr.

rate downstream=(15/3 ¾)km/hr=(15*4/15)km/hr=4km/hr.
Rate upstream=(5/2 ½)km/hr=(5*2/5)km/hr=2km/hr.
Speed of current=1/2(4-2)km/hr=1km/hr

3. a man can row 18 kmph in still water.it takes him thrice as long to row up as to row down the river.find the rate of stream.

Let man’s rate upstream be x kmph.then ,his rate downstream=3xkmph.
So,2x=18 or x=9.
Rate upstream=9 km/hr,rate downstream=27 km/hr.
Hence,rate of stream=1/2(27-9)km/hr=9 km/hr.

4. there is a road beside a river.two friends started from a place A,moved to a temple situated at another place B and then returned to A again.one of them moves on a cycle at a speed of 12 km/hr,while the other sails on a boat at a speed of 10 km/hr.if the river flows at the speed of 4 km/hr,which of the two friends will return to placeA first?

Clearly the cyclist moves both ways at a speed of 12 km/hr.
The boat sailor moves downstream @ (10+4)i.e.,14 km/hr and upstream @ (10-4)i.e., 6km/hr.
So,average speed of the boat sailor = (2*14*6/14+6)km/hr
                                    = 42/5 km/hr=8.4 km/hr.
since the average speed of the cyclist is greater ,he will return ta A first.

5. A man can row 7 ½ kmph in still water.if in a river running at 1.5 km/hr an hour,it takes him 50 minutes to row to a place and back,how far off is the place?

Speed downstream =(7.5+1.5)km/hr=9 km/hr;
Speed upstream=(7.5-1.5)kmph=6kmph.
Let the required distance be x km.then,
x/9+x/6=50/60.
2x+3x=(5/6*18)
5x=15
x=3.
Hence,the required distance is 3km.

6. In a stream running at 2kmph,a motar boat goes 6km upstream and back again to the starting point in 33 minutes.find the speed of the motarboat in still water.

let the speed of the motarboat in still water be x kmph.then,
6/x+2  +6/x-2=33/60
11x2-240x-44=0
11x2-242x+2x-44=0
(x-22)(11x+2)=0
x=22.

7.A man can row 40km upstream and 55km downstream in 13 hours also, he can row 30km upstream and 44km downstream in 10 hours.find the speed of the man in still water and the speed of the current.

let rate upstream=x km/hr and rate downstream=y km/hr.
Then,40/x +55/y =13…(i) and 30/x +44/y =10
Multiplying (ii) by 4 and (i) by 3 and subtracting ,we get:11/y=1 or y=11.
Substituting y=11 in (i),we get:x=5.
Rate in still water =1/2(11+5)kmph=8kmph.
Rate of current=1/2(11-5)kmph=3kmph

Boats and Streams

1. Speed =  Time, Time = Speed , 
   Distance  =  (Speed *  Time)                          
2. x km / hr =  x  *  5/18                          
3. x  m/sec  = (x * 18/5) km /hr          
4. If the ratio of the speeds of A and B is a:b ,
 then the ratio of the times taken by them to cover the same distance is  1/1 : 1/b
5. Suppose a man covers a certain distance at x km/ hr and 
an equal distance at y km / hr . Then , the average speed during
 the whole journey is   2xy/x + y km/ hr.

1. How many minutes does Aditya take to cover a distance of 400 m, if he runs at a speed of 20 km/hr?

Aditya’s speed = 20 km/hr  = {20 * 5/18} m/sec  =   50/9 m/sec
Time taken to cover 400 m= { 400 * 9/50} sec =72 sec = 1 12/60  min 1 1/5min.

2. A cyclist covers a distnce of 750 m in 2 min 30 sec. What is the speed in km/hr of the cyclist?

Speed = { 750/150 } m/sec  =5 m/sec  = { 5  *  18/5 } km/hr =18km/hr

3. A dog takes 4 leaps for every 5 leaps of a hare but 3 leaps of a dog are equal to 4 leaps of the hare. Compare their speeds?

Let the distance covered in 1 leap of the dog be x 
and that covered in 1 leap of the hare by y.
Then , 3x = 4y => x = 4/3y  =>  4x = 16/3  y.
Ratio(speeds of dog and hare) = Ratio(distances covered by them)
=> 4x : 5y => 16/3 y : 5y  => 16/3  : 5  => 16:15

4.While covering a distance of 24 km, a man noticed that after walking for 1 hour and 40 minutes, the distance covered by him was 5/7 of the remaining distance. What was his speed in metres per second?

 Let the speed be x km/hr.
 Then, distance covered in 1 hr. 40 min. i.e., 1  2/3 hrs  = 5x/3  km
 Remaining distance = { 24 – 5x/3 } km.

          5x/3  =  5/7 {  24 -  5x/3  } 
      =>  5x/3  =  5/7 {  (72-5x)/3  }  =>  7x  = 72 – 5x                                            
                                        => 12x = 72  
                                        =>  x = 6
  Hence speed = 6 km/hr ={ 6 * 5/18 } m/sec  =  5/3  m/sec = 1 2/3

5.Peter can cover a certain distance in 1 hr. 24 min. by covering two-third of the distance at 4 kmph and the rest at 5 kmph. Find the total distance.

Let the total distance be x km . Then,
            (2/3)x / 4  + (1/3)x / 5 = 7/5
         => x/6 + x/15 = 7/5
         => 7x = 42
         => x = 6

6.A man traveled from the village to the post-office at the rate of 25 kmph and walked back at the rate of 4 kmph. If the whole journey took 5 hours 48 minutes, find the distance of the post-office from the village.

 Average speed   = 2xy/(x+y) km/hr  = 2*25*4/(25+4) km/hr  = 200  km/hr
 Distance traveled in 5 hours 48 minutes i.e., 5(4/5)  hrs.  =  (200/29)*(29/5) km  = 40km
 Distance of the post-office from the village = 40 / 2    = 20 km

7.An aeroplane files along the four sides of a square at the speeds of 200,400,600 and 800km/hr.Find the average speed of the plane around the field.

Let each side of the square be x km and let the average speed of the plane around the field by y km per hour then ,
 x/200+x/400+x/600+x/800=4x/yó25x/2500ó4x/yóy=(2400*4/25)=384
hence average speed =384 km/hr

8.Walking at 5/7 of its usual speed, a train is 10 minutes too late. Find its usual time to cover the journey.

New speed =5/6 of the usual speed
New time taken=6/5 of the usual time
So,( 6/5 of the usual time )-( usual time)=10 minutes.
=>1/5 of the usual time=10 minutes.
=>  usual time=10 minutes

9.If a man walks at the rate of 5 kmph, he misses a train by 7 minutes. However, if he walks at the rate of 6 kmph, he reaches the station 5 minutes before the arrival of the train. Find the distance covered by him to reach the station.

Let the required distance be x km
Difference in the time taken at two speeds=1 min =1/2 hr
Hence x/5-x/6=1/5<=>6x-5x=6
=> x=6
Hence, the required distance is 6 km

10. A and B are two stations 390 km apart. A train starts from A at 10 a.m. and travels towards B at 65 kmph. Another train starts from B at 11 a.m. and travels towards A at 35 kmph. At what time do they meet?

Suppose they meet x hours after 10 a.m. Then,
   (Distance moved by first in x hrs) + [Distance moved by second in (x-1) hrs]=390.                                                                                                     
   65x + 35(x-1) = 390  => 100x = 425  => x = 17/4
So, they meet 4 hrs.15 min. after 10 a.m i.e., at 2.15 p.m.

11. A goods train leaves a station at a certain time and at a fixed speed. After ^hours, an express train leaves the same station and moves in the same direction at a uniform speed of 90 kmph. This train catches up the goods train in 4 hours. Find the speed of the goods train.

Let the speed of the goods train be x kmph.
     Distance covered by goods train in 10 hours= Distance covered by express train in 4 hours
     10x = 4 x 90 or x =36.
 So, speed of goods train = 36kmph.

12. A thief is spotted by a policeman from a distance of 100 metres. When the policeman starts the chase, the thief also starts running. If the speed of the thief be 8km/hr and that of the policeman 10 km/hr, how far the thief will have run before he is overtaken?

Relative speed of the policeman = (10-8) km/hr =2 km/hr.
Time taken by police man to cover 100m  (100/1000 x 1/2) hr = 1/20  hr.
In 1/20  hrs, the thief covers a distance of 8  x  1/20  km = 2/5  km  = 400 m

13. I walk a certain distance and ride back taking a total time of 37 minutes. I could walk both ways in 55 minutes. How long would it take me to ride both ways?

Sol. Let the distance be x km. Then, 
        ( Time taken to walk x km) + (time taken to ride x km) =37 min.
        ( Time taken to walk 2x km ) + ( time taken to ride 2x km )= 74 min.
 But, the time taken to walk 2x km = 55 min.
 Time taken to ride 2x km = (74-55)min =19 min.

Time and Distance

1. Inlet: A pipe connected with a tank or a cistern or a reservoir,
   that fills it, is known as an inlet.
   Outlet: A pipe connected with a tank or a cistern or a reservoir,
   emptying it, is known as an 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 full 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 emptied in 1 hour = (1/y)-(1/x)

1:Two pipes A and B can fill a tank in 36 bours and 46 bours respectively. If both the pipes are opened simultaneously, bow mucb time will be taken to fill the tank?

Part filled by A in 1 hour = (1/36);
Part filled by B in 1 hour = (1/45);
Part filled by (A + B) 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 10hours and 12 hours respectively while a third, pipe empties the full tank in 20 hours. If all the three pipes operate simultaneously, in how much time will the tank be filled?

Net part filled In 1 hour =(1/10)+(1/12)-(1/20)=(8/60)=(2/15).
The tank will be full in 15/2 hrs = 7 hrs 30 min.

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

let the reservoir be filled by first pipe in x hours.
Then ,second pipe fill it in (x+10)hrs.
Therefore (1/x)+(1/x+10)=(1/12)
=> (x+10+x)/(x(x+10))=(1/12).
=>  x^2 –14x-120=0  
=> (x-20)(x+6)=0
=> x=20 
so, the second pipe will take (20+10)hrs.
(i.e) 30 hours to fill the reservoir

4: A cistern has two taps which fill it in 12 minutes and 15minutes respectively. There is also a waste pipe in the cistern. When all the 3 are opened , the empty cistern is full in 20 minutes. How long will the waste pipe take to empty the full cistern?

Workdone by the waste pipe in 1min  
 =(1/20)-(1/12)+(1/15) = -1/10                           
therefore the waste pipe will empty the full cistern in 10min

5: An electric pump can fill a tank in 3 hours. Because of a leak in ,the tank it took 3(1/2) hours to fill the tank. If the tank is full, how much time will the leak take to empty it ?

work done by the leak in 1 hour=(1/3)-(1/(7/2))=(1/3)-(2/7)=(1/21).
The leak will empty .the tank in 21 hours.

6. 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 tooki 32 minutes more to fill the cistern.When the cistern is full, in what time will the leak empty it?

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

7: Two pipes A and B can fill a tank in 36 min. and 45 min. respectively. A water pipe C can empty the tank in 30 min. First A and B are opened. after 7 min,C is also opened. In how much time, the tank is full?

Part filled in 7 min. = 7*((1/36)+(1/45))=(7/20).
Remaining part=(1-(7/20))=(13/20).
Net part filled in 1min. when A,B and C are opened=(1/36)+(1/45)-(1/30)=(1/60).
Now,(1/60) part is filled in one minute.
(13/20) part is filled in (60*(13/20))=39 minutes.

8: 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.?

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
Therefore x*((1/24)+(1/32))+(18-x)*(1/24)=1   ó (7x/96) + ((18-x)/24)=1.
=> 7x +4*(18-x)=96.
Hence, be must be closed after 8 min.

Pipes and Cisterns

1. If A can do a piece of work in n days, then A's 1 day's work = (1/n).
                                                                                      
2. If A’s 1 day's work = (1/n),then A can finish the work in n days.

3. A is thrice as good a workman as B, then:
   Ratio of work done by A and B = 3 : 1.
   Ratio of times taken by A and B to finish a work = 1 : 3.

1. Worker A takes 8 hours to do a job. Worker B takes 10 hours to do the same Job.How long should it take both A and B, working together but independently, to do the same job?

A’s 1 hour's work = 1/8
B's 1 hour's work = 1/10

(A + B)'s 1 hour's work = (1/8) +(1/10)=9/40
Both A and B will finish the work in 40/9 days.

2. A and B together can complete a piece of work in 4 days. If A alone can complete the same work in 12 days, in how many days can B alone complete that work?

(A + B)'s 1 day's work = (1/4). 
 A's 1 day's work = (1/12).
 B's 1 day's work =((1/4)-(1/12))=(1/6)
       
 Hence, B alone can complete the work in 6 days.

3. A can do a piece of work in 7 days of 9 hours each and B can do it in 6 days of 7 bours each. How long will they take to do it, working together 8 hours a day?

 A can complete the work in (7 x 9) = 63 hours.
 B can complete the work in (6 x 7) = 42 hours.
 A’s 1 hour's work = (1/63) and B's 1 hour's work =(1/42)
 (A + B)'s 1 hour's work =(1/63)+(1/42)=(5/126)
 Both will finish the work in (126/5) hrs.
 Number of days. of (42/5) hrs each =(126 x 5)/(5 x 42)=3 days

4. A and B can do a piece of work in 18 days; Band C can do it in 24 days A and C can do it in 36 days. In how many days will A, Band C finish it together and separately?

(A + B)'s 1 day's work = (1/18)  
(B + C)'s 1 day's work = (1/24)
(A + C)'s 1 day's work = (1/36)
                                                                                                
 Adding, we get:  2 (A + B + C)'s 1 day's work =(1/18 + 1/24 + 1/36) =9/72 =1/8
 (A +B + C)'s 1 day's work =1/16

 Thus, A, Band C together can finish the work in 16 days.
 Now, A’s 1 day's work = [(A + B + C)'s 1 day's work] - [(B + C)'s 1 day work:
      =(1/16 – 1/24)= 1/48

A alone can finish the work in 48 days.
Similarly, B's 1 day's work =(1/16 – 1/36)=5/144
B alone can finish the work in  144/5=28 4/5 days
And C’s 1 day work =(1/16-1/18)=1/144
Hence C alone can finish the work in 144 days.

5. A is twice as good a workman as B and together they finish a piece in 18 days. In how many days will A alone finish the work?

(A’s 1 day’s work):)(B’s 1 days work) = 2 : 1.                                                     
(A + B)'s 1 day's work = 1/18

Divide 1/18 in the ratio 2 : 1.
A’s 1 day's work =(1/18*2/3)=1/27
Hence, A alone can finish the work in 27 days.

6. A can do a certain job in 12 days. B is 60% more efficient than A. How many days does B alone take to do the same job?

Ratio of times taken by A and B = 160 : 100 = 8 : 5.
Suppose B alone takes x days to do the job.
Then, 8 : 5 :: 12 : x = 8x = 5 x 12 =x = 7 1/2 days.

7. A can do a piece of work in 80 days. He works at it for 10 days B alone finishes the remaining work in 42 days. In how much time will A and B working together, finish the work?

Work done by A in 10 days =(1/80*10)=1/8
Remaining work = (1- 1/8) =7/ 8
Now,7/ 8 work is done by B in 42 days.
 Whole work will be done by B in (42 x 8/7) = 48 days.
A’s 1 day's work = 1/80 and B's 1 day's work = 1/48
                                                
(A+B)'s 1 day's work = (1/80+1/48)=8/240=1/30
Hence, both will finish the work in 30 days.

8. A and B undertake to do a piece of work for Rs. 600. A alone can do it in 6 days while B alone can do it in 8 days. With the help of C, they finish it in 3 days. !find the share of each?

C's 1 day's work = 1/3-(1/6+1/8)=24
A : B : C = Ratio of their 1 day's work = 1/6:1/8:1/24= 4 : 3 : 1.
A’s share = Rs. (600 *4/8) = Rs.300, B's share = Rs. (600 *3/8) = Rs. 225.
C's share = Rs. [600 - (300 + 225») = Rs. 75.

9. A and B working separately can do a piece of work in 9 and 12 days respectively, If they work for a day alternately, A beginning, in how many days, the work will be completed?

 (A + B)'s 2 days' work =(1/9+1/12)=7/36
Work done in 5 pairs of days =(5*7/36)=35/36
   Remaining work =(1-35/36)=1/36

  On 11th day, it is A’s turn. 1/9 work is done by him in 1 day.

 1/36 work is done by him in(9*1/36)=1/4 day
Total time taken = (10 + 1/4) days = 10 1/4days.

10 .45 men can complete a work in 16 days. Six days after they started working, 30 more men joined them. How many days will they now take to complete the remaining work?

(45 x 16) men can complete the work in 1 day.
                                  
1 man's 1 day's work = 1/720
                                  
45 men's 6 days' work =(1/16*6)=3/8

 Remaining work =(1-3/8)=5/8

75 men's 1 day's work = 75/720=5/48

Now,5 work is done by them in 1 day.
           
work is done  by them in (48 x 5)=6 days.

11.2 men and 3 boys can do a piece of work in 10 days while 3 men and 2 boys can do the same work in 8 days. In how many days can 2 men and 1 boy do the work?

Let 1 man’s 1 day’s work = x and 1 boy’s 1 day’s work = y.
Then, 2x+3y = 1 and 3x+2y = 1
Solving,we get:  x = 7  and y = 1
(2 men + 1 boy)’s 1 day’s work  = (2 x   7 + 1 x 1  ) = 16
So, 2 men and 1 boy together can finish the work in 25 = 12  days

Time and Work

1. Concept of Percentage : By a certain percent ,we mean that many hundredths.
 Thus x percent means x hundredths, written as x%.

 To express x% as a fraction : We have , x% = x/100.
 Thus, 20% =20/100 =1/5; 48% =48/100 =12/25, etc.
 To express a/b as a percent : We have, a/b =((a/b)*100)%.
 Thus, ¼ =[(1/4)*100] = 25%; 0.6 =6/10 =3/5 =[(3/5)*100]% =60%.
         
2. If the price of a commodity increases by R%, then the reduction in consumption 
so as not to increase the expenditure is [R/(100+R))*100]%.
If the price of the commodity decreases by R%,then the increase in consumption 
so as to decrease the expenditure is [(R/(100-R)*100]%.

3. Results on Population : Let the population of the town be P now and suppose it
 increases at the rate of R% per annum, then :
 1. Population after nyeras = P [1+(R/100)]^n.
 2. Population n years ago = P /[1+(R/100)]^n.

4. Results on Depreciation :  Let the present value of a machine be P.
 Suppose  it depreciates  at the rate R% per annum. Then,
 1. Value of the machine after n years = P[1-(R/100)]n.
 2. Value of the machine n years ago = P/[1-(R/100)]n.

5. If A is R% more than B, then B is less than A by
   [(R/(100+R))*100]%.
   If  A is R% less than B , then B is more than A by
   [(R/(100-R))*100]%.

1. Express each of the following as a fraction :
(i) 56% (ii) 4% (iii) 0.6% (iv) 0.008%

(i)   56% = 56/100= 14/25.            (ii) 4% =4/100 =1/25.
(iii) 0.6 =6/1000 = 3/500.            (iv) 0.008 = 8/100 = 1/1250.

2. Express each of the following as a Decimal :
(i) 6% (ii)28% (iii) 0.2% (iv) 0.04%

(i)   6% = 6/100 =0.06.                  (ii) 28% = 28/100 =0.28.
(iii) 0.2% =0.2/100 = 0.002.             (iv) 0.04%= 0.04/100 =0.004.

3. Express each of the following as rate percent :
(i) 23/36 (ii) 6 ¾ (iii) 0.004

(i) 23/36 = [(23/36)*100]% = [575/9]% = 63 8/9%.
(ii) 0.004 = [(4/1000)*100]% = 0.4%.
(iii) 6 ¾ =27/4 =[(27/4)*100]% = 675%.

Ex. 4. Evaluate :
(i) 28% of 450+ 45% of 280
(ii) 16 2/3% of 600 gm- 33 1/3% of 180 gm

(i) 28% of 450 + 45%  of 280 =[(28/100)*450 + (45/100)*280] = (126+126) =252.
(ii) 16 2/3% of 600 gm –33 1/3% of 180 gm = [  ((50/3)*(1/100)*600) – ((100/3)*(1/3)*280)]gm = (100-60) gm = 40gm.

5. (i) 2 is what percent of 50 ?
(ii) ½ is what percent of 1/3 ?
(iii)What percent of 8 is 64 ?
(iv)What percent of 2 metric tones is 40 quintals ?
(v)What percent of 6.5 litres is 130 ml?

(i) Required Percentage   = [(2/50)*100]% = 4%.
(ii) Required Percentage = [ (1/2)*(3/1)*100]% = 150%.
(iii)Required Percentage = [(84/7)*100]% = 1200%.
(iv) 1 metric tonne = 10 quintals.
Required percentage = [ (40/(2 * 10)) * 100]% = 200%.
(v) Required Percentage  = [ (130/(6.5 * 1000)) * 100]% = 2%.

6. Find the missing figures : (i) ?% of 25 = 20125 (ii) 9% of ? = 63 (iii) 0.25% of ? = 0.04

(i)  Let x% of 25 = 2.125. Then , (x/100)*25 = 2.125
     X = (2.125 * 4) = 8.5.
(ii) Let 9% of x =6.3. Then , 9*x/100 = 6.3
     X = [(6.3*100)/9] =70.
(iii) Let 0.25% of x = 0.04. Then , 0.25*x/100 = 0.04
      X= [(0.04*100)/0.25] = 16.

7. Which is greatest in 16 ( 2/3) %, 2/5 and 0.17 ?

16 (2/3)% =[ (50/3)* )1/100)] = 1/6 = 0.166, 2/15 = 0.133. Clearly, 0.17 is the greatest.

If the sales tax reduced from 3 1/2 % to 3 1/3%, then what difference does it make to a person who purchases an article with market price of Rs. 8400 ?

Required difference = [3 ½ % of Rs.8400] – [3 1/3 % of Rs.8400]
                    = [(7/20-(10/3)]% of Rs.8400 =1/6 % of Rs.8400
                    = Rs. [(1/6)8(1/100)*8400] =  Rs. 14.

9. An inspector rejects 0.08% of the meters as defective. How many will be examine to project ?

Let the number of meters to be examined be x.
Then, 0.08% of x =2
[(8/100)*(1/100)*x] = 2
x = [(2*100*100)/8] = 2500.

10. Sixty five percent of a number is 21 less than four fifth of that number. What is the number ?

Let the number be x.
Then, 4*x/5 –(65% of x) = 21
4x/5 –65x/100 = 21
5 x = 2100
x = 140.

11. Difference of two numbers is 1660. If 7.5% of the number is 12.5% of the other number , find the number ?

Let the numbers be x and y. Then , 7.5 % of x =12.5% of y
X = 125*y/75 = 5*y/3.
Now, x-y =1660
5*y/3 –y =1660
2*y/3= 1660
y =[ (1660*3)/2] =2490.
One number = 2490, Second number =5*y/3 =4150.

12. In expressing a length 810472 km as nearly as possible with three significant digits , find the percentage error.

Error = (81.5 – 81.472)km = 0.028.
Required percentage = [(0.028/81.472)*100]% = 0.034%.

13. In an election between two candidates, 75% of the voters cast thier thier votes, out of which 2% of the votes were declared invalid. A candidate got 9261 votes which were 75% of the total valid votes. Find the total number of votes enrolled in that election?

Let the number of votes enrolled be x. Then ,
Number of votes cast =75% of x. Valid votes = 98% of (75% of x).
75% of  (98% of (75%of x)) =9261.
[(75/100)*(98/100)*(75/100)*x] =9261.
X = [(9261*100*100*100)/(75*98*75)] =16800.

14. Shobha’s mathematics test had 75 problems i.e.10 arithmetic, 30 algebra and 35 geometry problems. Although she answered 70% of the arithmetic ,40% of the algebra, and 60% of the geometry problems correctly. she did not pass the test because she got less than 60% of the problems right. How many more questions she would have to answer correctly to earn 60% of the passing grade?

Number  of questions attempted correctly=(70% of 10 + 40% of 30 + 60% 0f 35)
                                                                          =7 + 12+21= 45
questions to be answered correctly for 60% grade=60% of 75 = 45

therefore required number of questions= (45-40) = 5.

15. if 50% of (x-y) = 30% of (x+y) then what percent of x is y?

50% of (x-y)=30% of(x+y) 
=>(50/100)(x-y)=(30/100)(x+y)
=> 5(x-y)=3(x+y) 
=> 2x=8y 
=> x=4y

required percentage =((y/x) X 100)% = ((y/4y) X 100) =25%

16. Mr.Jones gave 40% of the money he had to his wife. he also gave 20% of the remaining amount to his 3 sons. half of the amount now left was spent on miscellaneous items and the remaining amount of Rs.12000 was deposited in the bank. how much money did Mr.jones have initially?

Let the initial amount with Mr.jones be Rs.x then,
Money given to wife= Rs.(40/100)x=Rs.2x/5.Balance=Rs(x-(2x/5)=Rs.3x/5.
Money given to 3 sons= Rs(3X((20/200) X (3x/5)) = Rs.9x/5.
Balance = Rs.((3x/5) – (9x/25))=Rs.6x/25.
Amount deposited in bank= Rs(1/2 X 6x/25)=Rs.3x/25.
Therefore 3x/25=12000 ó x= ((12000 x 35)/3)=100000
So Mr.Jones initially had Rs.1,00,000 with him.
Short-cut Method : Let the initial amount with Mr.Jones be Rs.x
Then,
(1/2)[100-(3*20)]% of x=12000
=> (1/2)*(40/100)*(60/100)*x=12000
=> x=((12000*25)/3)=100000

18 A salesman`s commission is 5% on all sales upto Rs.10,000 and 4% on all sales exceeding this.He remits Rs.31,100 to his parent company after deducing his commission . Find the total sales?

Let his total sales be Rs.x.Now(Total sales) – (Commission )=Rs.31,100
x-[(5% of 10000 + 4% of (x-10000)]=31,100
x-[((5/100)*10000 + (4/100)*(x-10000)]=31,100
=>x-500-((x-10000)/25)=31,100
=>x-(x/25)=31200 ó 24x/25=31200óx=[(31200*25)/24)=32,500.
Total sales=Rs.32,500

19 Raman`s salary was decreased by 50% and subsequently increased by 50%.How much percent does he lose?

Let the original salary = Rs.100
New final salary=150% of  (50% of Rs.100)=
Rs.((150/100)*(50/100)*100)=Rs.75.
Decrease = 25%

20 Paulson spends 75% of his income. His income is increased by 20% and he increased his expenditure by 10%.Find the percentage increase in his savings?

Let the original income=Rs.100 . Then , expenditure=Rs.75 and savings =Rs.25
New income =Rs.120 , New expenditure =
Rs.((110/100)*75)=Rs.165/2
New savings = Rs.(120-(165/2)) =  Rs.75/2
Increase in savings = Rs.((75/2)-25)=Rs.25/2
Increase %= ((25/2)*(1/25)*100)% = 50%.

21. The salary of a person was reduced by 10% .By what percent should his reduced salary be raised so as to bring it at par with his original salary ?

Let the original salary be Rs.100.
 New salary = Rs.90.
 Increase on 90 = 10 , 
 Increase on 100 =((10/90)*100)% = (100/9)%

22 When the price fo a product was decreased by 10% , the number sold increased by 30%. What was the effect on the total revenue ?

Let the price of the product be Rs.100 and let original sale be 100 pieces.
Then , Total Revenue = Rs.(100*100)=Rs.10000.
New revenue = Rs.(90*130)=Rs.11700.
Increase in revenue = ((1700/10000)*100)%=17%.

23 . If the numerator of a fraction be increased by 15% and its denominator be diminished by 8% , the value of the fraction is 15/16. Find the original fraction?

Let the original fraction be x/y.
Then,
    (115%of x)/(92% of y)=15/16
 => (115x/92y)=15/16
 => ((15/16)*(92/115))=3/4

24. In the new budget , the price of kerosene oil rose by 25%. By how much percent must a person reduce his consumption so that his expenditure on it does not increase ?

Reduction  in consumption = [((R/(100+R))*100]%
                          = [(25/125)*100]%=20%.

25 The population of a town is 1,76,400 . If it increases at the rate of 5% per annum , what will be its population 2 years hence ? What was it 2 years ago ?

Population after 2 years = 176400*[1+(5/100)]^2
=[176400*(21/20)*(21/40)]
= 194481.
Population 2 years ago = 176400/[1+(5/100)]^2
=[716400*(20/21)*(20/21)]= 160000.

26 The value of a machine depreiates at the rate of 10% per annum. If its present is Rs.1,62,000 what will be its worth after 2 years ? What was the value of the machine 2 years ago ?

Value of the machine after 2 years
=Rs.[162000*(1-(10/100))^2] = Rs.[162000*(9/10)*(9/10)]
=Rs. 131220
Value of the machine 2 years ago
= Rs.[162000/(1-(10/100)^2)]=Rs.[162000*(10/9)*(10/9)]=Rs.200000

27. During one year, the population of townincreased by 5% . If the total population is 9975 at the end of the second year , then what was the population size in the beginning of the first year ?

Population in the beginning of the first year
= 9975/[1+(5/100)]*[1-(5/100)] = [9975*(20/21)*(20/19)]=10000.

28. If A earns 99/3% more than B,how much percent does B earn less then A ?

Required Percentage = [((100/3)*100)/[100+(100/3)]]% =[(100/400)*100]%=25%

29 If A`s salary is 20% less then B`s salary , by how much percent is B`s salary more than A`s ?

Required percentage = [(20*100)/(100-20)]%=25%.

30 .How many kg of pure salt must be added to 30kg of 2% solution of salt and water to increase it to 10% solution ?

Amount of salt in 30kg solution = [(20/100)*30]kg=0.6kg
Let x kg of pure salt be added
Then, 
(0.6+x)/(30+x) = 10/100ó60+100x = 300+10x
=> 90x=240 
=>  x=8/3.

31. Due to reduction of 25/4% in the price of sugar , a man is able to buy 1kg more for Rs.120. Find the original and reduced rate of sugar?

Let the original rate be Rs.x per kg.
Reduced rate = Rs.[(100-(25/4))*(1/100)*x}]=Rs.15x/16per kg
120/(15x/16)-(120/x)=1 ó(128/x)-(120/x)=1
ó x=8.
So, the original rate = Rs.8 per kg
Reduce rate = Rs.[(15/16)*8]per kg =  Rs.7.50 per kg

32 In an examination , 35% of total students failed in Hindi , 45% failed in English and 20% in both . Find the percentage of those who passed in both subjects .

Let A and B be the sets of students who failed in Hindi and English respectively .
Then , n(A) = 35 , n(B)=45 , n(AÇB)=20.
So , n(AÈB)=n(A)+n(B)- n(AÇB)=35+45-20=60.
Percentage failed in Hindi and English or both=60%
Hence , percentage passed = (100-60)%=40%

33. In an examination , 80% of the students passed in English , 85% in Mathematics and 75% in both English and Mathematics. If 40 students failed in both the subjects , find the total number of students.

Let the total number of students be x .
Let A and B represent the sets of students who passed in English and Mathematics respectively .
Then , number of students passed in one or both the subjects
= n(AÈB)=n(A)+n(B)- n(AÇB)=80% of x + 85% of x –75% of x
=[(80/100)x+(85/100)x-(75/100)x]=(90/100)x=(9/10)x
Students who failed in both the subjects = [x-(9x/10)]=x/10.
So, x/10=40 of x=400 .
Hence ,total number of students = 400.

In this section, questions involving a set of numbers are put in the form of a puzzle. You have to analyze the given conditions, assume the unknown numbers and form equations accordingly, which on solving yield the unknown numbers.

1. A number is as much greater than 36 as is less than 86. Find the number?

 
Let the number be x. 
Then, x - 36 = 86 - x  
=> 2x = 86 + 36 = 122 
=> x = 61.
Hence, the required number is 61.

2. Find a Number Such that when 15 is subtracted from 7 times the number, the Result is 10 more than twice the number?

Let the number be x. Then,
   7x - 15 = 2x + 10 
=> 5x = 25 =>x = 5.
Hence, the required number is 5.

3. The sum of a rational number and its reciprocal is 13/6. Find the number?

Let the number be x.
Then, x + (1/x) = 13/6
=> (x2 + 1)/x = 13/6
=> 6x2 – 13x + 6 = 0
=> 6x2 – 9x – 4x + 6 = 0
=> (3x – 2) (2x – 3) = 0
=>  x = 2/3 or x = 3/2
Hence the required number is 2/3 or 3/2.

4. The sum of two numbers is 184. If one-third of the one exceeds one-seventh of the other by 8, find the smaller number?

 
Let the numbers be x and (184 - x). 
Then, (X/3) - ((184 – x)/7) = 8 
  => 7x – 3(184 – x) = 168 
  => 10x = 720 => x = 72.
So, the numbers are 72 and 112. 
Hence, smaller number = 72.

5. The difference of two numbers is 11 and one-fifth of their sum is 9. Find the numbers?

Let the number be x and y. 
Then,
 x – y = 11      ----(i)     
 and 
 1/5 (x + y) = 9  
 => x + y = 45   ----(ii)
Adding (i) and (ii), we get: 2x = 56 or x = 28. 
Putting x = 28 in (i), we get: y = 17.
Hence, the numbers are 28 and 17.

6. If the sum of two numbers is 42 and their product is 437, then find the absolute difference between the numbers?

Let the numbers be x and y. Then, 
x + y = 42 and xy = 437
x - y = sqrt[(x + y)2- 4xy]
       = sqrt[(42)2 - 4 x 437 ]
       = sqrt[1764 – 1748]
       = sqrt[16]
       = 4.
Required difference = 4.

7. The sum of two numbers is 16 and the sum of their squares is 113. Find the numbers?

Let the numbers be x and (15 - x).
Then, x2 + (15 - x)2= 113           
 =>  x2 + 225 + X2 - 30x = 113  
 =>  2x2 - 30x + 112 = 0          
 =>  x2 - 15x + 56 = 0
 =>  x - 7) (x - 8) = 0 
 =>  x = 7  or  x = 8.
   
  So, the numbers are 7 and 8.

8. The average of four consecutive even numbers is 27. Find the largest of these numbers?

Let the four consecutive even numbers be 
x, x + 2, x + 4 and x + 6.
Then, sum of these numbers = (27 x 4) = 108.
So, x + (x + 2) + (x + 4) + (x + 6) = 108  
=> 4x = 96  
=> x = 24.
Largest number = (x + 6) = 30.

9. The sum of the squares of three consecutive odd numbers is 2531.Find the numbers?

Let the numbers be x, x + 2 and x + 4.
Then, X2+ (x + 2)2 + (x + 4)2 = 2531 => 3x2+ 12x - 2511 = 0
=>  X2 + 4x - 837 = 0     
=> (x - 27) (x + 31) = 0
=>  x = 27.
Hence, the required numbers are 27, 29 and 31.

10. Of two numbers, 4 times the smaller one is less then 3 times the 1arger one by 5. If the sum of the numbers is larger than 6 times their difference by 6, find the two numbers?

Let the numbers be x and y, 
such that x > y Then, 3x - 4y = 5 ...(i)  
and  x + y) - 6 (x - y) = 6 =>  -5x + 7y = 6     …(ii)
Solving (i) and (ii), we get: x = 59 and  y = 43.
Hence, the required numbers are 59 and 43.

11. The ratio between a two-digit number and the sum of the digits of that number is 4 : 1.If the digit in the unit's place is 3 more than the digit in the ten’s place, what is the number?

Let the ten's digit be x. 
Then, unit's digit = (x + 3).
Sum of the digits = x + (x + 3) = 2x + 3. 
Number = l0x + (x + 3) = llx + 3.
Then,  11x+3 / 2x + 3  = 4 / 1 
 => 1lx + 3 = 4 (2x + 3)   
 =>  3x = 9     
 =>  x = 3.
Hence, required number = 11x + 3 = 36.

12. A number consists of two digits. The sum of the digits is 9. If 63 is subtracted from the number, its digits are interchanged. Find the number?

 Let the ten's digit be x. 
 Then, unit's digit = (9 - x).
 Number = l0x + (9 - x) = 9x + 9.
 Number obtained by reversing the digits = 10 (9 - x) + x = 90 - 9x.
 therefore, 
     (9x + 9) - 63 = 90 - 9x    
  => 18x = 144
  => x = 8.
  So, ten's digit = 8 and unit's digit = 1.
  Hence, the required number is 81.

13. A fraction becomes 2/3 when 1 is added to both, its numerator and denominator. And ,it becomes 1/2 when 1 is subtracted from both the numerator and denominator. Find the fraction.?

Let the required fraction be x/y. Then,
x+1 / y+1 = 2 / 3  =>  3x – 2y = - 1   …(i) 
and 
x – 1 / y – 1 = 1 / 2 =>  2x – y = 1 …(ii)
Solving (i) and (ii), we get : x = 3 , y = 5
therefore, Required fraction = 3 / 5.

14. 50 is divided into two parts such that the sum of their reciprocals is 1/ 12.Find the two parts?

Let the two parts be x and  (50 - x).
Then, 1 / x + 1 / (50 – x) = 1 / 12
 => (50 – x + x) / x ( 50 – x) = 1 / 12
 => x2 – 50x + 600 = 0 
 => (x – 30) ( x – 20) = 0 
 => x = 30 or x = 20.
So, the parts are 30 and 20.

15. If three numbers are added in pairs, the sums equal 10, 19 and 21. Find the numbers?

Let the numbers be x, y and z. Then,
x+ y = 10      ...(i)      
y + z = 19     ...(ii)      
x + z = 21     ...(iii)
Adding (i) ,(ii) and (iii), 
we get:  2 (x + y + z ) = 50 
 =>  (x + y + z) = 25.
Thus, x= (25 - 19) = 6;  
   y = (25 - 21) = 4;
   z = (25 - 10) = 15.

Hence, the required numbers are 6, 4 and 15.