1) sp = cp(1+pf)
2) sp = cp(1 - lf)
3) when error in weighing, x gm is sold as y gm(x < y) then profit % is ((y-x)/x)*100
4) when two items are sold at same price SP. sold item 1 at profit of x% and item 2 at loss of y%
(2(1+xf)(1-yf) /( 2+xf-yf) -1)*100
when xf and yf are same a^2/100
5) sp = mp(1-df)
6) Two discount of a% and b% on an item is equal to single discount of
df = 1-(1-af)(1-bf)
7) If tax is t% then Pnew = Porg (1+tf)
8)Revenue = Price * Volume
cp = cost price
sp= selling price
pf = profit fraction
lf = loss fraction
Formula 1
If sp is selling price, cp cost price and pf profit fraction then
sp = cp(1+pf)
Proof
p% = sp-cp/cp *100
pf = sp-cp/cp
cp *pf +cp = sp
sp=cp(1+pf)
Formula 2
sp = cp(1 - lf)
proof similar to above
Formula 3
when error in weighing, x gm is sold as y gm(x < y) then profit % is
((y-x)/x)*100
Assume t to be price/gm
cost price for x gm = xt
selling price is yt
so (yt - xt/xt)*100 = ((y -x) /x)* 100
Formula 3
when two items are sold at same price SP. sold item 1 at profit of x% and item 2 at loss of y%
cp1
\ ---------x% profit
SP
/ -------- y% loss
cp2
consider cost price of item 1 as cp1 and of item 2 as cp2
Assuming profit
overall gain % = {(total selling price of two item) - (total cost price of two item)} / (total cost price of two item) * 100
i.e ( (sp+sp) - (cp1+cp2) ) / (cp1+cp2) * 100 or
(2sp /(cp1+cp2) - 1) *100 --------- (1)
we can calculate cost price in terms of selling price for each item since we have been given profit/loss%
sp = cp1(1 + xf) => cp1=sp/(1+xf) -------------------(2)
sp = cp2(1 - yf) => cp2=sp(1+yf) --------------------(3)
putting in equation 1 we get
( 2sp / ( (sp/1+xf ) + (sp/1-yf)) -1 )* 100
(2(1+xf)(1-yf) /( 1+xf+1-yf) -1) *100
(2(1+xf)(1-yf) /( 2+xf-yf) -1)*100 -------------- (4)
when xf and yf are same, call it af
((1-af2) - 1 ) *100 = > - (af2)*100 // negative number indicates he is at loss
or
loss in percentage a2/100
Problem 1 A person sells two items at Rs 100 one at profit of 10% and other at loss of 10%.Find overall gain% or loss%
Solution
From our formula above 1% loss. Lets verify it
the item on which he gets profit its cost price CP1 = 100/(1.10) = ~ 90.90
the item on which he gets loss its cost price CP2 = 100/(.90) = ~ 111.11
total cost price = 111.11+90.90 = 202.01
total selling price = 2*100 =200
total loss % = 202.01 - 200 / 202.01 * 100 ~= 1
Problem 2 A person sells two item at Rs 100 one at profit of 20% and other at loss of 15%.Find overall gain or loss percentage
Solution
From equation 4 above
we have (2(1+.20)(1-.15) / (2+.20-.15) ) - 1 * 100 =~ -(.49)%
that is loss of .49%
Formula 4
Discount % = (mp - sp) / mp * 100 or
sp = mp(1-df)
Problem Orange are available at Rs 8 per dozen and discount of 10% on that price. How many oranges can u buy for 2.4 rs
Solution sp = 8(1-.10) = 7.2 per dozen
simple proportion 7.2rs -----12 oranges
2.4rs ----------?
4 oranges
Formula
Two discount of a% and b% on an item is equal to single discount of
Discount of b% on item then
sp =(1-bf)mp
discount of a% on it
(1-af)((1-bf)mp)
consider single discount d then
(1-df)mp = (1-af)(1-bf)mp
(1-df)=(1-af)(1-bf)
df = 1-(1-af)(1-bf)
Problem discount of of 20 and 40% on an item equals discount of
df = 1-(1-.20)(1-.40) = 1-.48 = .52 or 52%
Formula If tax is t% then
Pnew = Porg (1+tf)
Problem A tradesman gives 4% discount on the marked price and gives article free for buying every 15 articles and thus gains 35%. The marked price is above the cost price by
Solution
consider cost price of an item as cp and marked price as mp
4% discount on item mp makes its sp = (1- .04)mp
for 15 items its total selling price would be 15(1-.04)mp and trademan gives an item free on this but he incurs the cost of this extra item
total cost price for this 16 item = 16cp
he incurs a gain of 35%
(15(1-.04)mp - 16cp)/16cp *100 = 35
14.4*mp = (.35+1)16cp
mp = 1.5 cp
Pipe and Cistern
1)we add or subtract the amount filled in 1hr
or we add or subtract the amount only and not time and that too on normalized scale
Direct proportion of time it takes and amount that is filled
)If tank can be filled in xhrs then part filled in 1hr is 1/x
or If in 1hr, 1/x of tank is filled then it take x hr to fill complete tank
If a tank is completely filled in x hrs then
time it takes to fill 1/p of tank is (1/p)*x
Additive property for amount of tank filled
1)if in 1hr 1/x of tank is filled by a pipe then with k similar pipes opened simultaneously
amount filled would be k*(1/x)
2)when two inlet open simultaneously
one can fill in x hrs other in y hrs then time it takes to fill tank is xy/(x+y) because
in 1hr both can fill 1/x + 1/y = (x+y)/xy so can be completely filled in xy/(x+y)
3)inlet fill in x hr and outlet empties in y hr
If x>y that means it takes more time to fill the tank than to empty it
If tank is already 1/p filled then we need to empty this much part
part emptied in 1hr is 1/y-1/x
time to empty 1/p part is
4)If pipe A fills tank t hrs faster than pipe B it means time taken by A is 10 greater than B
A= x+10 hrs
B=x hrs
5)part filled by A and B simultaneously and rest filled with C, if total time taken is xhr then
part filled with A+B in t min + part filled with C in x-t min = 1
|-----t min----|----------30-t min ---------|
A+B C
Train
conversion from kmph to mph multiply by 5/18 1)If two trains start from point A and B and move towards each other
A|-------------------------|B
d
and after they cross each other it takes a hrs for train from A to reach B and b hr for train from B to reach A
A|---------t|-------------|B
d1 (d-d1)
let they meet after time t then assume distance traveled by A is d1 and by B is d-d1
and speed of train from A is x and from B is y km/hr
then
d1= xt and d-d1 = ty
y/x = (d-d1)/d1 ---------------(1)
Now to travel d-d1 by train from A takes a hrs to
d-d1 = xa and d1= yb -------(2)
equating value of d-d1 and d1 in 1 we get
y/x = xa /yb
or y^2/x^2 = a/b ------------(3)
2) If a person sitting in train A moving at x kmph and train B of length l crosses men in t time then
relative speed of train B is l/t
absolute speed of train when moving in opposite direction to A is x - l/t
3) If two trains are moving together either in opposite or same direction then find their relative speed and do the usual calculation
like if trains A is l1 in length and train B is l2 in length then if they are moving in opp direction its same as passing l1+l2 with relative speed
Clock
relation between speed of minute and hr hand and min hand and sec hand
in time of 60 min, hr hand moves 5unit
in time of 60 min, min hand moves 60 units. So,
hr hand is 12 times slower than min hand
in time of 60 min, sec hand moves 3600 units so
sec hand is 60 times faster than min hand
angle traversed in t hrs
min hand traverses 360 degree in 60 min
so in t hr angle traversed is
m=360*t
hr hand will traverse (360/12) degree in 60min
so in t hrs it will traverse
h=30t
second hand traverses 360*60 degree in 60min
so in t hrs
s=21600t
Problem When the hands of clocks coincide?
Solution
For hr hand and min hand to coincide
m-h = 360*n where n is integer >=0
360t - 30t = 360n
11t =12n
so at t = 12n/11 they overlap
For sec and hr hand to coincide
s-h =360m
21600t - 30t = 360m
21570t = 360m
719t =12m
so at t = 12/719 * m hr sec and hr hand coincide
For all to overlap
12n/11= 12m/719
719n = 11m
Problem How many times hr hand and sec hand coincide in 12hr
Solution
we know that they coincide at time t=12n/11 when n integer >=0
so considering the starting time as 12 o'clock for
n=0 or 12hrs at which they coincide
for n=1 t=12/11 or 1hr 5.45min
for n=2 t=24/11 or 2hr 10.90min
for n=10 t= 120/11 = 10 hr 54.54mins
and for n=11 its same as n=0 that is 12 o'clock
so they overlap 11 times in 12 hrs
60
Problem At what time between 4 and 5 oclock hands of clock are at right angle
360t-30t= 90n
330t = 90n
11t=3n
t=3n/11
for 4
for n =15
t= 3*15/11 = 4hr 60/11 min
A watch that gains and loses time
Problem
A watch which gains 5 seconds in 3 minutes was set right at 7 a.m. In the afternoon of the same day, when the watch indicated quarter past 4 o'clock, what is true tim
Solution
for each 3 min it gains 5 sec means
3 min on correct clock is 3min 5sec on this clock
or 1/60*(3+5/60)hr is 3/60 hr on correct clock
or 37/720 hr on this clock is 1/20 on correct clock
time spent from 7 am to 4.15pm is 9hr 15mins or 37/4hr
so 37/4 hr on this clock means time spent on correct clock is (1/20*37/4*720/37) or 9hrs
find the time spent on incorrect clock and then calculate proportionately time on correct clock
Problem A watch gains uniformly, at 8am on sunday it is 5mins behind and the following sunday at 8pm its 5min 48sec ahead .when was it correct
Solution
It will show the correct time when it has gained 5mins by which its lagging
time spent by incorrect clock is 7day 12 hrs or 180hrs
in 180 hrs watch gained 5+(5+48/60) = 54/5
total time gained = the time by which it was lagging + new time it has gained
therefore 5mins are gained in (180*5)/(54/5) = 83hr 20minAllegation and mixture
cp of cheaper cp of dearer
c d
now required price is m
for a unit quantity how much we should add cheaper quantity and dearer quantity to make the price m
lets add cheaper quantity x amount
then dearer quantity has to be added 1-x
now
m*1 = cx +(1-x)d
(d-c)x = d-m
x = (d-m)/(d-c)
and
y= (m-c)/(d-c)
ratio of cheaper/dearer quantity = (d-m)/(m-c)
Problem in what ration water must be mixed so as to gain 20% by selling at cost price?
Solution
cost price should be such that it gives profit % of 20 that is
sp= (1+.2)cp or
cp = 10/12 * sp =5/6 sp
this is the price that he is getting now when he sells at sp he gets profit of 20%
water milk
0 cp1
mean 5/6 * sp
cp1 = sp because he is now selling the mixture at the cost of milk that is what was R rs/lt for milk is not R rs/lt for mixture
so ratio of water to milk = (1-5/6)/5/6 = 1/5
Time and work
A can do a unit of work in n days then working uniformly A's one day work = 1/n -----------------(1)
Almost all work problems will make use of the formula below
if for a unit of work
A --------- t1 days
B ---------- t2 days
we find one days work in such case here it is
A's --------one day work ---------- 1/t1
B's -------- one day work ---------- 1/t2
we add the work of persons not their time
A+B ----------one day work ---------- 1/t1+1/t2 --------------(2)
A+B ----------1/days work ------------ t1t2/t1+t2 days ---------------------(3)
For a unit of work if
n person ------- tdays
m person(m<n) ------- t (n/m) days
(m>n) ------- t(m/n) day
Example It takes 4 men to finish work in 10 day. How many days will be required if only i)2 men work or ii)10 men work
For less men we would need more days
2 men would need 10(4/2) = 20 days
10 men would need 10(4/10) = 4 days
If na1 persons in group A can do a piece of work in ta1 days. It takes na2(na2 > na1) persons from group B to do same piece of work in tb2 days. If na2 persosns from group A and nb2 persons from s group B works together how much time will it take
For a unit of work
na1 --------------- ta1 days
na2(na2 > na1) --------------- ta1(na1/na2) days
nb1--------------tb1 days
nb2(nb2<na1) <="" p="">
If 4 men or 6 women or 10 girls can do a piece of work in 10 days How many days are required for 3 men 8 women and 6 girls
Solution
For a unit of work
4men --------------10days
3men--------------10(4/3)days
6 women ----------------10days
8 women----------------10(6/8)days
10girls-----------------10days
6girls-------------------10(10/6)days
3men+8wome+6girl -------one days work ------- 3/40 + 8/60 + 6/100
for a unit of work
3men+8wome+6girl ------------------ ------- 1/(3/40 + 8/60 + 6/100)
If B is p% more efficient than A then
for unit of work
B--------------xdays
A-------------x(1+pf)days
If B is 100% more efficient than B then
A------------x(1+1) =2x A takes twice the time B takes
If B is 40% more efficient than B then
A ------------x(1+.4)=1.4x
Poblem A is thrice as good a workman as B and therefore is able to finish a job in 60 days less than B. Working together, they can do it in:
Solution
let time taken by A to finish work be x
time taken by B would be 3x
and difference of their time is given as 3x - x = 60
x=30
so B time is 90
one days work of A+B = 1/30+1/90
Nice questions ....Please tell me Which book do you prefer for CAT preparation?
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