CALENDERS 1. 100 years contains '5' odd days. 200 years contains '3' odd days. 300 years contains '1' odd days. 400 years contains '0' odd days. 2. Sunday -------> '0' odd day. Monday -------> '1' odd day. . . . . 3. One leap year contains '2' odd days. 4. The years which are mul of '4' are called leap years. 5. Leap year -------> 366 days (Feb --> 29 days). Ordinary year -------> 365 days. 6. Leap year ------> '52' weeks + '2' odd days. Ordinary year ------> '52' weeks + '1' odd day. BANKERS DISCOUNT B.D S.I T.D B.G
-----------> -----------> -----------> ----------->
Bankers Discount Simple Interest True Discount Bankers Gain
1. On bill for unexpired time, B.D = S.I 2. B.G = B.D - T.D 3. B.G = S.I on T.D 4. T.D = √(P.W) * (B.G) 5. B.G = (T.D)2/(P.W) 6. B.D = (A * R * T)/100 7. T.D = (A * R * T)/[100 + (R * T)] 8. A = (B.D * T.D)/(B.D - T.D) 9. T.D = (B.G * 100)/(R * T) 10. Sum due = (B.D * T.D)/(B.D - T.D) = (B.D * T.D)/B.G Sum due = Amount 11. T.D/B.G = Sum/B.D 12. B.D - T.D = A * {(R + T)2/[100(100 + (R * T))]}
TRUE DISCOUNTS T.D P.W S.I A R T
-----------> -----------> -----------> -----------> -----------> ----------->
True Discount Present Worth Simple Interest Amount Rate Time
1. A = P.W + T.D 2. P.W = (100 * amount)/[100 + (R * T)] 3. T.D = (P.W * R * T)/100 4. T.D = (A * R * T)/[100(R + T)] 5. S.I on T.D = S.I - T.D 6. Sum = (S.I * T.D)/(S.I - T.D) 7. When the sum is put at C.I, P.W = A/[1 + (R/100)]T 8. T.D = S.I on P.W 9. P.W = (100 * T.D)/(R * T) 10. T = (100 * T.D)/(P.W * R) 11. When the interest is at C.I, T.D = P.W[1+ (r/100)]t - P.W CLOCKS x -----> first given time. 1. For coinciding the hands , (5x) * (12/11) 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) 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
PIPES AND CISTERNS 1. t(A + B) = (tA * tB)/(tA + tB) 2. tA = (tB * t(A + B))/(tB - t(A + B)) 3. Time for filling, (Filling pipe is bigger in size.) F = (e * f)/(e - f) 4. Time for emptying, (emptying pipe is bigger in size.) E = (f * e)/(f - e) 5. T(A + B + C)=L/[(L/tA) + (L/tB) + (L/tC)] 6. Pipes 'A' & 'B' can fill a tank in f1hrs & f2hrs respectively. Another pipe 'C' can empty the full tank in 'e'hrs. If the three pipes are opened simultaneously then the tank is filled in , F = L/[(L/f1) + (L/f2) - (L/e)] 7. Two taps 'A' & 'B' can fill a tank in 't1' & 't2' hrs respectively. Another pipe 'C' can empty the full tank in 'e'hrs. If the tank is full & all the three pipes are opened simultaneously . Then the tank will be emptied in, E = L/[(L/e) - (L/f1) - (L/f2)] 8. A filling tap can fill a tank in 'f'hrs. But it takes 'e'hrs longer due to a leak at the bottom. The leak will empty the full tank in , E = [t(f * e) * tf]/[t(f + e) - tf] 9. Capacity of the tank is , F = (f * e)/(e - f) 10. tc = [t(A + B) * t(A + B + C)]/[t(A + B) - t(A + B + C)] 11. T = (xyz)/[(xz) + (yz) - (xy)] 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' = √ab DECIMAL FRACTIONS 1. [(a2-b2)/(a+b)] = [ a-b ] 2. [(a2-b2)/(a-b)] = [ a+b ] 3. [(a3+b3)/(a2-(a*b)+b2)] = [ a+b ] 4. [(a3-b3)/(a2-(a*b)+b2)] = [ a-b ] 5. [(a+b)2+(a-b)2 /(a2+b2)] = 2 6. [(a2+b2-(2*a*b) )/(a-b)] = [ a-b ] 7. [(a2+b2+(2*a*b) )/(a+b)] = [ a+b ]
MENSURATIONS 1. Square:-figure (1) Area = a2Sq units. (or) P2/16 (2) Perimeter = P =4a (or) a = P/4 (3) Diagonal (or) length of the rod that can be placed= a = P/4 2. Rectangle:-figure (1)Area = l*b (2)Perimeter = P = 2(l+b) (3)Diagonal = d = √l2+b2 b = √d2-l2 l = √d2-b2 3. If area of plot is given as 'z'm2 and the ratio of l:b is given as x:y, then l = x * [√z/(x*y)] b = y * [√z/(x*y)] 4. length required=(length * breadth of a room)/width of the carpet 5. No: of stones = (length * breadth of a room)/(length * breadth of a stone) 6. A2/A1= (a2/a2)2 = (d2/d1)2 7. P2/P1=√A2/A1 8. Circle:-figure (1)Area = Pr2 (or) P(d2/4) (2)d = 2r (3)Perimeter (or) Circumference =2Pr = Pd P = 22/7 (or) 3.14 (4)A = c2/(4 P) (or) c = √4 PA = 2√PA 9. % dec in Area = fs [(r12 – r22)/r12] * 100 10. Distance travelled in 'N' revolutions is, D = N * Pd (or) N = D/(Pd) 11. Area left ungrazed = a2(1 - P/4) 12. Area of the road (1)Road out of the garden:-figure Area of the road =2w[l+b+2w] = [(l+2w)(b+2w)]-(l*b) (2)Road inside the garden:-figure Area of the road = 2w[(l+b)-2w] (3)Two parallel roads:-figure Area of the road = w[(l+b)-w]
13. Triangles:(1)Right angled triangles:-figure Area = (1/2)*b*h d = √b2+h2 (2)Equilateral triangles:-figure Area= (√3/4)a2 Perimeter = P = 3a Height= (√3/2)a (3)Scalene triangle:-figure Perimeter = P = 2s=a+b+c s= (a+b+c)/2 Area = √s(s-a)(s-b)(s-c) (4)Isosceles triangle:-figure Perimeter = P = 2a+b Area = b/4(√4a2-b2) 14.Volumes:(a)Cube:-figure (1)Lateral surface area =4a2 (2)Total surface area = 6a2 (3)Volume of a solid = Base area * Height = a2 * a = a3 (4)Diagonal (or) Longest pole = d = √3a (b)Cuboids:-figure (1)Lateral surface area = AL = 2h[l+b] (2)Total surface area = AT = 2[lb+lh+bh] (3)Volume = V = lbh (4)Diagonal = d = √l2+b2+h2 (5)No:of boxes = (lbh)/l1b1h1 = (Volume of big box)/(Volume of small box) 15. a3 = v13+v23+v33 16. a1/a2 = (v1/v2)1/3 17.No:of boxes(if areas are given) = a3/a13 = (a/a1)3 18.Cylinder:-figure (1)Lateral surface area = AL = 2Prh (2)Total surface area = AT = 2Pr(h+r) (3) AT/AL = (h+r)/h (4)Volume = v = Pr2h (5)Area of each flat surface i.e of ends =Pr 2 19.Cone:-figure (1)Slant height = l = √h2+r2 (2)Volume of the cone = 1/3(Pr2h) (3)Curved surface area of cone = Prl (4)Total surface area = Pr(l+r) (5) v1/v2=(r1/r2)2 * h1/h2
20. H-h = (4/3) * rs3/rd2 21.Area of circle inscribed in an equilateral triangle is r2. It's height is, h = 3r 22.Sector:-figure (1)l= (q/360)*2P r (2)A = (q/360)*Pr2 (3)Circumference, c =l+2r 23.Four circular cardboard pieces, each of radius 'r'cm are placed in such a way that each piece touches two other pieces. The area of the space enclosed by four pieces is, (2r)2 [1-P/4]cm2 24. Rhombus:-figure (1) 4a2 = d12 + d22 (2)Area = (1/2)d1d2 (3)Perimeter = P = 4a 25. Parallelogram:-figure (1)Area of Dle ABC = 1/2(bh) (2)Area of Dle ACD = 1/2(b/h) (3)Area of parallelogram = bh 26. Trapezium:-figure Area of Trapezium=Area of (DleABC + DleACD) = 1/2(ah) + 1/2(bh) = [(1/2)h][a + b] 27. Sphere:-figure (1)Surface area = 4Pr2 (2)Volume = 4/3(Pr3) (3)A1/A2 = (r1/r2)2 (4)v1/v2 = (r1/r2)3 (5)v1/v2 = (A1/A2)3/2 (6)A1/A2 = (v1/v2)2/3 28.Area of four walls = 2 * (length + breadth) * height AGES No separate formulas, But problems are done by logical method. Each part = Total Age/Sum of ratio's of Age's SIMPLIFICATIONS In this chapter, we must simplify the problems in the following order only. V ---> - (Veruculum), B ---> () (Bracket), O ---> of (of), D ---> % (division), M ---> * (Multiplication), A ---> + (Addition), S ---> - (Subtraction)
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' articles 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 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. D ---> 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.D=[T(xy)]/x+y=[T(b2-c2)]/2b 7.T=(D*2b)/b2-c2 8.T=(D/d)+(D/u)=[D/(b+c)]+[D/(b-c)]
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 quantity of pure milk. y -----> Quantity drawn out each time. n -----> No. of operations. 10.No.of rabbits (4 legs) = [ No. of legs given - (No. of heads given * 2) ]/2 No. of pigeons = [ No. of heads given - No. of rabbits ] 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 liters ]
TIME AND WORK 1.tA+B = (tA * tB)/tA 2.tB = (tA * tA+B)/tA - (tA+B) 3.tA+B+C =[ L/(L/tA) + (L/tB) + (L/tC) ] L ---> L.C.M of tA,tB,tC. 4.tC =[ L/(L/tA+B+C) - (L/tB) - tB) ] 5.If A+B, B+C, A+C are given then A+B+C=? (i)tA+B+C = 2L/[ (L/tA+B) + (L/tB+C) + (L/tC+A) ] (ii)tC = 2L/[ (L/tB+C) + (L/tC+A) + (L/tA+B) ] (iii)tB = 2L/[ (L/tA+B) + (L/tB+C) + (L/tA+C) ] 6.S1d1 = S2d2 7.wA+B = [ (wA * wB)/(wA+wB) ] 8.Working alternatively, 2 * tA+B = 2 * [ (tA.tB)/(tA+tB) ] DISCOUNTS 1. Gain = x-d-(x*d/100) x -----> Extra percentage added to C.P to fix M.P d -----> Discount offered on M.P g -----> Gain% obtained 2.Discount = M.P-S.P 3.d% = [ (M.P-S.P)/100] * 100 4.Discount = M.P * (d%/100) 5.Successive Discounts, [ D = (d1+d2)-(d1.d2)/100 6. (C.P/M.P) = (100-d)/(100+g) 7.M.P=(S.P2-S.P1)/(d2-d1) * 100 8.S.P=M.P * (100-d)/100 9.S.P = M.P * [ (100-d1)/100 ] * [ (100-d2)/100 ] 10.Difference of discounts = [M.P * [ d1.d2/(100*100) ] 11.[ (100-d1)/(100-d2) ] = [ (100+g1)/(100+g2) ] 12.Number of shirts = [ Total Discount/Discount on each shirt ] 13.g% = [ (S.P-C.P)/C.P * 100 ] = [ (gain/C.P) * 100 ] 14.C.P = (g/g%) * 100 15.S.P = (g/g%) * (100+g) 16.C.P = [S.P/(100+g)] * 100 17.M.P = [C.P/(100-d)] * 100 18.G = [ (G1+G2)+(G1.G2)/100] 19.[(100-d)(100+g )* M.P ] = [S.P * (100)2] (S.P/M.P) = [ (100-d) * (100+g) ]/(100)2
COMPOUND INTEREST 1.A = P [ 1 + [ R/(100*n)n*t] ] P -----> Principle R -----> Rate % per annum n -----> No. of conversions 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 annually 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 of interest is R1% R2% R3% for Ist year, IInd year, IIIrd 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 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 =[ √A23/A6 ] =[ √A13/A3 ] =[ √A33/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. 1/T 15.R = [ (A/P) - 1 ] * 100 16.P = [ A2 * [100/(100+R)]2 ] 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)
CHAIN RULE 1. M1D1T1S1W2 A2F2 = M2D2T2 S2W1A1F1 M -----> Men/labor 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 ---> Days L ---> Length B ---> Broad (or) Breadth H ---> Deep 3.Additional Men = M2-M1 AVERAGES 1.Average = [ Total of observations/No. of observations ] 2. When a person joins a group (i) in case of increasing average Age (or) 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 person 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 nobody 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 y 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) ]
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 = √(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/Rd.P) 16.R = (100.Id/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] + .......] = D a ---> Annual installment. D ---> Amount due 21.A = P * [ (100+R1+R2+R3)/100] PERCENTAGES 1. 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 IInd 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. (i) If the sides of the triangle, rectangle, square, circle, rhombus etc is increased by x%. Its area is increased by 2x+(x2/100) (ii)If decreased x%. Its area is decreased by, -2x+(x2/100)
4.If 'A' is x% of 'C' and 'B' is y% of 'C' then 'A' is (x/y) * 100% of 'B'. 5. % 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] 6. 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) 7. 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% 8.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 9. (i)If the price of commodity increases by r% then reduction in consumption so as not to increase the expenditure is (r/100+r) * 100 (ii)If the price of commodity decreases by r% then,(r/100-r)*100 10.If the population of town (or) length of a tree is 'p' and its annual increase is r% then, (i)population (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] 11.If the population of town (or) value of a machine is 'p' and annual decrease is r% then, (i)population (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] 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 numerator of 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)]
