BOATS AND STREAMS
b ---> Boat speed/Man speed in water.
c ---> Current Speed/Speed of the River.
d ---> Down stream speed.
u ---> Up stream speed.
Dis ---> Total distance travelled.
T ---> Total time.
1. d=b+c |
2. u=b-c |
3. b=(d+u)/2 |
4. c=(d-u)/2 |
5. Average Speed=(2xy)/x+y i.e |
(b2-c2)/b |
6. Dis=[T(xy)]/x+y=[T(b2-c2)]/2b |
7. T=(Dis*2b)/b2-c2 |
8. T=(Dis/d)+(Dis/u)=[Dis/(b+c)]+[D/(b-c)] |
STOCK AND SHARES
Formulae on Stock & Shares:-
1.No: of stock =
Total Stock/100
Purchase Cost / (Mk.Vl + BR)
Sale Realisation / (Mk.Vl - BR)
Annual Income/Rate%
Formulae on Debentures:-
2.No: of Share =
Investment (or) Purchase Cost/[MK.VL(1 + B%)]
Sale Realisation/[Mk.Vl(1 - B%)]
(Annual Income * 100)/(Divident% * Face value)
CLOCKS
1.For coinciding the hands , (5x) * (12/11)
x -----> first given time.
2.Right angles at each other , (5x ± 15) * (12/11)
3.Opposite Direction , (5x - 30) * (12/11)
4.For finding time when it is 't'min space apart , (5x ±t) * (12/11)
5.For finding the angle between the hands of a clock is ,
30 * [HRS - (MIN/5)] + (MIN/2) or Angle = 30H-11(M)/2
PERCENTAGES
- If x% is deducted on tax and y% of the remaining is spent on
education and still there is a balance, the formula is :-
Balance * [ 100/(100-x) ] * [ 100/(100-y) ] * [ 100/(100-z) ]
2. The population of a town is 'P'. It increased by x% during Ist year,
increased by y% during IIst and again increased by z% during Ist.
The population after 3 years will be,
P * [ (100+x)/100 ] * [ (100+y)/y ] * [ (100+z)/100 ]
3. % of effect
(i) Inc of x% Dec of x% x-y-[(x*y)/100]
(ii) Inc of x% Inc of y% (x+y)+[(x*y)/100]
(iii) Dec of x% Inc of y% [-x2/100]
(iv) Dec of x% Dec of y% (-x-y)+[(x*y)/100]
(v) Inc of x% Dec of x% [-x2/100]
(vi) Inc of x% Inc of x% 2*x+[x2/100]
4. (i) If the sides of the traingle,rectangle,square,circle,rhombus etc
is increased by x%.Its area is increased by ----------2x+(x2/100)
(ii)If decreased x%.Its ares is decreased by, ------2x+(x2/100)
5. In an examination x% failed in Hindi and y% failed in Science, if z%
of the candidates failed in both of the subjects. The percentage of
students who passed in both of the subjects is, ------100-(x+y-z)
6. If A's income is r% more than B's income, the B's income is less
than A's income by
(r/100+r) * 100%
7.If A's income is r% less than B's income, then B's income is more than A's
income by
(r/100-r) * 100
8. (i)If the price of comodity increases by r% then reduction in consumption
so as not to increase the expenditure is --------------(r/100+r) * 100
(ii)If the price of comodity decreases by r% then, -------(r/100-r)*100
9.If the population of town (or) length of a tree is 'p' and its annual increase
is r% then,
(i)populaton (or) length of a tree after 'n' years is, ----p[1+(r/100)]
(ii)population (or) length of a tree 'n' years ago is, ---- p/[1+(r/100)n]
10.If the population of town (or) value of a machine is 'p' and annual decrease
is r% then,
(i)populaton (or) value of machine after 'n' years is, -------p[1-(r/100)n]
(ii)population (or) value of a machine 'n' years ago is, -----p/[1-(r/100)n]
11.If 'A' is x% of 'C' and 'B' is y% of 'C' then 'A' is (x/y) * 100% of 'B'.
12. If two values are respectively x% and y% more than a third value, then
the first is
[(100+x) / (100+y)] * 100% of second
13.Total no.of votes = (Difference in votes/Difference in %) * 100
14.Maximum marks = [(pass marks/pass %) * 100]
15.Total marks = (Difference in marks / Difference in %)*100
16. (i)Reduced rate = [(Amount/Quantity more) * (Reduction % /100)]
(ii)Original rate (or)
previous rate = [(Amount/Quantity more) * (Reduction % /100-reduction%)]
17. (i)Increased rate = [(Amount/Quantity less) * (increase % /100)]
(ii)Original rate (or) previous rate = [(Amount/Quantity less) * (Increase % /100-increase%)]
18. If the numeratorof fraction is increased by x% and its denominator is diminished by y% ,the value of the fraction is A/B.Then the original fraction is, --------
----(A/B) * [(100-y) / (100+x)]
SIMPLE INTEREST
1. S.I = PTR/100
P -----> Principal
T -----> Time (in yrs)
R -----> Rate % per anum
2. Amount = P+S.I
3.TO find the rate of interest per annum when a sum double/triple
etc itself in x years.Then, [R * T = 100 * (n-1)]
4. [(R1*T1)/R2*T2) = (N1-1)/(N2-1)] | 5. [(A/S.I = (100/R*T)+1] | |
6. [R(or)T = Ö(100*S.I)/P] | 7. [(R1-R2) = (More interest * 100/(P*t))] | |
8. A=[(P+S.I) = P(1+(T.R/100))] | 9. [P=(A1*T2-A2*T1)/T2-T1] | |
A ---> Amount T ---> Time | ||
10. R=[(A2-A1)/(A1*T2-A2*T1)] * 100 | 11. [I = ATR/(100+TR)] | |
12.If I1= I2, [(P1/P2) = (T2.R2)/T1.R1] | ||
13. [P = (100/Id)/(Rd.T)] | 14. [T = (100.Id/Pd.R)] | |
15. [T = (100.Id/P R.d)] | 16. [R = (100.Is/Td.P)] | |
17. [Gain = P.Rd.T/100] | 18. [R = (100.ITotal)/(P1.T1+P2.T2+P3.T3)........] | |
19. P=(100.ITotal/(R1.T1+R2.T2+R3.T3+......)] | 20. a[ [100/100] + [(100+R)/100] + (100+2R)/100] + .......] = 0 | |
a ---> Annual instalment. D ---> Amount due | ||
21. A = P * [ (100+R1+R2+R3)/100] | ||
CHAIN RULE
1. M1D1T1S1W2 A2F2 = M2D2T2 S2W1A1F1
M -> Men/labour
D --> Days
T --> Time (in hrs)
S -> Speed
W --> part of work done/wages
A ---> Amount earned
F -----> Food consumed/Milk used/coal required for
Machines/Diesel required for pumps.
2. D1W1 = D2W2
i.e., D1(L2B2H2) = D2(L1B1H1)
D---> Day
L--> Length
B---> Broad (or) Breadth
H---> Deep
3.Additional Men = M2-M1
NUMBER SERIES
1.The difference between the no: and the no: obtained by interchanging
the digits is 'x'.The difference between digits is , -------------diff = x/9
2.The sum of the no: and the no: obtained by interchanging the digits
is 'y'.The sum of the digits is ,
sum = y/11
3.The sum of two numbers is 'x' and their difference is 'y'.The product
of the no: is , ----
[(x + y)2 - (x - y)2]/4
4. Dividend = (Divisor * Quotient) + Remainder
PROFIT AND LOSS
1. Profit = S.P - C.P | 2. Loss = C.P - S.P |
3. Gain% = (Gain/C.P)*100 | 4. Loss% = (Loss/C.P)*100 |
5. S.P = [(100+Gain%)/100]*C.P | 6. C.P=S.P*[100/(100+Gain%)] |
7. S.P= [(100-Loss%)/100]*C.P | 8. C.P= S.P*[100/(100-Loss%)] |
9.By selling an article for Rs/ '-S'1 , a man looses 'L%'.In order to gain 'G%' he uses the following formula, S1/(100-L%)=S2(100-G%)
10. If C.P of 'x' articies is equal to the S.P of 'y' articles,the profit% is: [(x-y)/y]*100
11. Gain%=[Error/(truevalue-error)]*100
12. C.P = S.P/(1-losspart)
13. C.P=S.P*[100/(100+g1)]*[100/(100+g2)]*[100/(100+g3]
14. S.P=C.P*[(100+g1)/100]*[(100+g2)/100]*[(100+g3)/100]
15. C.P = [(S.P1-S.P2)/x2-x1]*100
x1 ---------> gain1 (or) loss1 x2 ---------> gain2 (or) loss2
16. S.P=C.P + [(C.P*g)/100]
17. Overall gain or loss = (x1*g1)-(x2*L1)+(x3*g3)
Where x1,x2,x3 ----------> Parts of items sold
AGES
No seperate formulas,But problems are done by logical method.
Each part = Total Age/Sum of ratio's of Age's
RATIO AND PROPORTION
1. If a:b = c:d , then Product of Means=Product of Extremes i.e 2ndterm*3rdterm=1stterm*4thterm
2. Each part = Total Amount/Total of Ratios
3. If a:b = x:y & b:c = p:q ,then a:b:c = xp:yp:yq
4.Third proportion to 'x' & 'y' = y2/x
5.The mean proportion between 'a' & 'b' = sqrt(ab)
PARTNERSHIP
1.Part of A/Part of B =
[(Amount invested by A*No.of months invested)/(Amount invested by B*No.of months invested)]
2.Each part = (Total profit/Total of Ratios)
COMPOUND INTEREST
1. A = P [ 1 + [ R/(100*n) n*t] ]
P -> Principle
R --> Rate % per annum
n ---> No.of convertions per year
T--> Time in years
2.C.I=A-P i.e., [ P [ 1+(R/(100*n)n*t] - 1 ]
3.When interest is calculated anually n=1,
A = P[1+(R/100)t]
4.When time is in fraction, t = x * (1/y) year:
A = P[1+(R/100)x] + [1 + (1/y)*R/100 ]
5.When rate od interest is R1% R2% R3% for 1st year,2ndyear,
3rd year respectively then amount,
A = P [ 1 + (R1100) * [1+(R2/100)] [1+(R3/100)]
6.When difference between C.I and S.I on certain sum at rate% on Rs.x,
[ C.I - S.I = sum * (r/100)2 ]
i.e., [ D = P * (r/100)2 ]
Note: Applicable only for two years.
7.D =[ (P*R2)(300+R)/1003 ]
Note:Applicable only for 3 years.
8.[ C.I/(200+R) = S.I/200 ]
Note: Applicable only for 2 years.
9.R = [ (2*difference of C.I and S.I)/S.I ] * 100
10.R% amounts after 2 successive years we given:-
R = [ (An+1-An)/An ] * 100
An+1 -----> Amount after (n+1) years. An -----> Amount after (n+1) years.
11.P =[ A32/A6 ] = [ A22/A4 ] = [ A12/A2 ] = [A42/A8 ]
Note: Double the years.
12.P =[ A23/A32 ] = [ A34/A43 ] = [ A45/A54 ]
Note: Consecutive years.
P =[ ÖA23/A6 ] = [ ÖA13/A3 ] = [ ÖA33/A9 ]
13.R =[( A6/A3)1/3 - 1 ] = [ ( A4/A2 )1/2 - 1 ] = [ ( A5/A2 )1/3 - 1 ]
R =[( A7/A2)1/3 - 1 ] = [ ( A10/A2 )1/8 - 1 ] = [ ( A10/A7 )1/3 - 1 ]
14.Installment problems:a [ 100/(100+R) + 100/(100+R)2 + 100/(100+r)3 + ....... ] = B
a -----> Annual installment B -----> Borrowed amount.
15.R = [ (A/P)1/T - 1 ] * 100
16.P = [ A2 * [100/(100+R)]2 ]
AVERAGES
1. Average = [ Total of observations/No.of observations ]
2.(i)When a person joins a group in case of increasing average Age weight of new comer =
[ (Previous Age + No.of persons) * Increase in Avg ]
(ii)In case of decreasing Average, Age (or) weight of new comer =
[ (Previous Age - No.of persons) * Decrease in Avg ]
3.When a persom leaves a group and another person joins the group in
the place of person left, then
(i)In case of increasing average, Age (or) weight of new comer =
[ (Age of person left + No.of persons) * Increase in Avg ]
(ii)In case of decreasing Average, Age (or) weight of new comer =
[ (Age of person left - No.of persons) * Decrease in Avg ]
4.When a person leaves the group but no body joins this group, then
(i)In the case of increasing Average, Age (or) weight of man left =
[ (Previous Age - No.of present persons) * Increase in Avg ]
(ii)In case of decreasing Average, Age (or) weight of new comer =
[ (Previous Age + No.of present persons) * Decrease in Avg ]
5.If a person travels a distance at a speed of x Km/hr returns to the
original place of y Km/hr then average speed is -----------
[ 2.x.y/(x+y) ]
6.If half of the journey is travelled at speed of x km/hr and the next
half at a speed of x km/hr. Then average speed during the whole
journey is ------------[ 2.x.y/(x+y) ]
7.If a person travels 3 equal distances at a speed of x Km/hr,
y Km/hr,z km/hr.Then average speed during whole journey is =-----
[ 3.x.y/(x.y+y.x+z.x) ]
8.A½ [ 3.x.y/(2x*y) ]
9.A½ [ 3*L/[ (L/S1)+(L/S2)+(L/S3) ]
10.A½ 4L/[ (L.S1)+(L/S2)+L/S3)+(L/S4) ]
11.A½ 1/[(x/100) * (1/S1)]+[y/100)*(1/S2)]+[(z/100)*(1/S3)]
L.C.M AND H.C.F
1.H.C.F of fractions = [H.C.F of Numerators/L.C.M of Denominators]
2. (i)which will be divided - L.C.M (ii)Which divides - H.C.F
3.The greatest number which can divide x, y and z leaving the same remainder 'A' in each case is X-A = ?, Y-A = ?, Z-A = ? and Find the
H.C.F of obtained numbers.
4.The greatest number by which if x and y are divided.The remainder
will be A&B respectives is, x.A = ? , y-B = ? Find the H.C.F of
obtained numbers.
5.L.C.M of fractions = [L.C.M of Numerators/H.C.F of Denominators]
6. [ H.C.F * L.C.M = n1 * n2 ]
7. The least number which when divided by x,y and z leaves the remainder A,B and C respectively is, x-A = ? , y-B = ? , z-C = ?. Here, there will be equal difference between them i.e., D.
Required no = [ L.C.M of x,y and z ] - D
8.The smallest number which when diminished by A, is divisible by p,q,r,s is,
Smallest no = [ (L.C.M of p,q,r,s) + A ]
ALLIGATION AND MIXTURES
1. C.P = [ S.P/(100+g) * 100 ]
2.Mean rate of interest, R = [ (100*I)/P*T) ]
3.Final % of Alcohol = [ (Qi/Pi)/(Qi+Qw added) ]
Pi -----> Initial percentage
Qw -----> Quantity of water added
4.Final % of alcohol = [ (Qi*Pi)/(Qi-Qw evoparated) ]
Qw -----> Quantity of water evoparated.
5.Quantity of water to be added = [Qmix*[(P2-P1)/(100-P2)]]
P1 and P2 are percentages of water.
6.Other than water = [ Qmix * (P1-P2)/P2 ]
P1 and P2 are the % of constituent other than water
(i.e., salt,alcohol etc)
7.Ratio of water to milk =g/100
8. Percentage of water =[ (100*g)/(100+g) ]
9.[ 1- (y/x) ]n * x
x -----> Capacity of container (or) Initial quatity of pure milk
y -----> Quantity drawn out each time.
n -----> No.of operations.
10.No.of rabits (4 legs) =[No.of legs given-(No.of heads given*2)]/2
No.of pigeons =[ No.of heads given - No.of rabits ]
11. The mixture drawn out and replaced with water, so that the mixture
may be half water and milk is = [(1/2)*(difference in parts/greater part)]
12.One gallon = [ 100 litres ]
Blood Relations
1. Relations of Paternal side:
Father's father → Grandfather
Father's mother → Grandmother
Father's brother → Uncle
Father's sister → Aunt
Children of uncle → Cousin
Wife of uncle → Aunt
Children of aunt → Cousin
Husband of aunt → Uncle
2. Relations of Maternal side:
Mother's father → Maternal grandfather
Mother's mother → Maternal grandmother
Mother's brother Maternal uncle
Mother's sister → Aunt
Children of maternal uncle → Cousin
Wife of maternal uncle → Maternal aunt
Relations from one generation to next:
Calculating the day of the week
Examples
Examples
Now for an example of the complete algorithm, let's use April 24, 1982.
- Look up the 1900s in the centuries table: 0
2. Note the last two digits of the year: 82
3. Divide the 82 by 4: 82/4 = 20.5 and drop the fractional part: 20
4. Look up April in the months table: 6
5. Add all numbers from steps 1–4 to the day of the month (in this case, 24): 0+82+20+6+24=132.
6. Divide the sum from step 5 by 7 and find the remainder: 132/7=18 remainder 6
Find the remainder in the days table: 6=Saturday.
1. Look up the 2000s in the centuries table: 6 |
2. Note the last two digits of the year: 00 |
3. Divide the 00 by 4: 0/4 = 0 and drop the fractional part: 0 |
4. Look up January in the months table: 6 (leap) |
5. Add all numbers from steps 1–4 to the day of the month (in this case, 1): 6+00+0+6+1=13. |
6. Divide the sum from step 5 by 7 and find the remainder: 13/7=1 remainder 6 |
7. Find the remainder in the days table: 6=Saturday. |
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