Time and Distance

1. Speed =     Distance/ Time      ,  Time=      Distance/ 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/a:1/b  or b:a.

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.

                              

PIPES AND CISTERNS

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)

Time and work

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.

Profit and Loss:

1.gain=s.p-c.p

2.loss= -(s.p-c.p)

3.gain= -loss

4.(gain||loss)%=[(gain||loss)x100]/C.P

5.S.P=[(100+gain%)xC.P]/100

6.if an article is sold at a gain of  p% ten S.P=(100+p)% of C.P 

7.if an article is sold at a loss of  p% ten S.P=(100-p)% of C.P 

8.when a person sells two similaritems,ane at a  gain of p%,and the other at a loss of p%,then seller always incurs a loss given by:

Loss%=(p/10)^2

9.if a trader professes to sell his goods at coast price,but uses false weights,then

Gain%=[{(false gain i.e Error)x100}/(true value-Error)]%

Partership:

1.if A and B invest Rs.p and Rs.q for same time t in a business,

then end of the time t:-

(A’s share of profit):(B’s share of profit)= p:q

2.If A invests Rs.p for m time and B invests Rs.q for n time   

then :-

 (A’s share of profit):(B’s share of profit)= pxm:qxn

Ratio and Proportion

1. if  a:b=c:d then we write a:b::c:d and say a,b,c,d are in proportion.

    Here  ’a’  and  ‘d’ are called extrems while ‘b’ and ‘c’ are called means.

     2. Product of means(bxc)=product of extreams(axd)

     3.  If  a:b=c:d then ‘d’ is called fourth proportional .

     4.  If a:b=b:c then ‘c’ is called third proportional

     5.  (ab)^(1/2) is mean proportional between a and b.

     6.   If a:b=c:d

then componendo rule:- (a+b)/b=(c+d)/d

   dividendo rule :-              a/(a-b)=c/(c-d)

componendo and dividend rule:- (a+b)/(a-b)=(c+d)/(c-d)

Chain Rule

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

Sol:

No. of toys                    total  cost

  15                                  Rs.234

  35                                    A

A=[15 toys then Rs234, 35(more)  toys  A( more)  cost  so  min(15) will be denominater]

   =(234x35)/15

Hence cost of 35toys is Rs.546/-

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

Sol:

No. of men               working hrs

36                                    25hrs

15                                   A

A=[36 men then 25hrs ,15(less)men A(more) hrs so 15(min) will be denominater]

  =(36x25)/15=60hrs

Hence 15 men can do  a work in 60hrs/-.

3. If the wages of 6men for 15days be Rs.2100,then find the wages of 9 men for 12 days.

Sol:

Men       days                 wages

6             15                       2100

9             12                         A

A=[6men  then wages=Rs2100,9 men then more(A) wages so min (6)be denominater]&&[15days

      then wages=Rs2100,12days then less(A) wages so max(15) will be denominater ]

=[9x12x2100]/(6x15)

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

Sol:

No. of engines                 hrs in a day             consumption ratio              tonnes

9                                          8                                      1/3                                        24

8                                         13                                     1/4                                        A

                                  

A=[9engn=24tonnes then 8engn=less tonnes,so more (9) will be denominater]&&

    [8hrs=24tonnes then 13hrs=more(A)tones so min(8) will be denominater]&&

    [1/3=24 then ¼=less(A)tones so more(1/3) will be denominater]

=[{8x13x(1/4)}x24]/{9x8x(1/3)}=26tonnes