## RATIO AND PROPORTION SHORTCUTS FOR QUANTITATIVE APTITUDE

RATIO AND PROPORTION SHORTCUTS FOR QUANTITATIVE APTITUDE
o   RATIO- t he ratio of two quantities of the same kind is the fraction that one quantity is of the other, in other words to say, how many times a given number is in comparison to another number. A ratio between two nos.x A and B is denoted by A/B
o   Some of the points to be remembered :
1.   The two quantities must be of the same kind.
2.   The units of the two quantities must be the same.
3.   The ratio has no measurement.
4.   The ratio remains unaltered even if both the antecedent(A) and the consequent(B)are multiplied or divided by the same no.
o   If two different ratios ( say A /B and C/D) are expressed in different units, then if we are required to combine these two ratios we will follow the following rule=
A xC / B xD
The required ratio is AC / BD
o   The duplicate ratio of A/B is A2/B2 the triplicate ratio of A/B is A3/B3
o   The subduplicate ratio of A/B is sq.root of A/ sq.root of B
o   The subtriplicate ratio of A/B is cube root of A/ cube root of B
o   To determine which of the given two ratio  A/B and C/D is greater or smaller ,we compare A xD and B xC provided B>0 and D>0;
if AxC> B xD  then A/B > C/D and vice versa,but if A xC= B xD  then A/B = C/D
o   Properties of ratios.
1.   Inverse ratios of two equal ratios are equal, if A/B=C/D then B/A = D/C.
2.   The ratios of antecedents and consequents of two equal ratios are equal if A/B=C/D then A/C=B/D
3.   If A/B=C/D THEN A+B/B=C+D/D
4.   If A/B=C/D THEN A-B/B=C-D/D
5.   If A/B=C/D THEN A+B/A-B=C+D/C-D
6.   If A/B=C/D=E/F.....so on then each of the ratio( A/B, C/D.....etc) is equal to
sum of th numerators/sum of the denominators=A+C+E...../B+D+F......=k

PROPORTION
o   Two ratios of two terms is equal to the ratio of two other terms, then these four terms are said to be in proportion i.e. if A/B=C/D then A,B,C and D are in proportion.
A,B,C and D are called first, second,third and fourth proportionals respectively.
A and D are called Extremes and B and C are called the Means
and it follows that A xD=B xC
o   Continued proportion: when A/B=B/C then A, B and C are said to be in continued proportion and B is called the geometric mean of A and C so it follows,
A xC=B2 ,OR square root of (A xC)=B
o   Direct proportion: if two quantities A and B are related and an increase in A decreases B and vice-versa then A and B are said to be in direct proportion.Here A is directly proportional to B is written as AµB.when a is removed equation comes to be
A = kB,where k is constant.
o   Inverse proportion: if two quantities A and B are related and an increase in A increases B and vice-versa then A and B are said to be in inverse proportion. Here A is inversely proportional to B is written as Aµ1/B or, A=k/B,where k is constant.
o   Propotional division:
It simply means a method by which a quantity may be divided into parts which bear a given ratio to one another .The parts are called propotional parts.
e.g.divide quantity "y" in the ratio  a:b:c then
first part= a/(a+b+c)=y           second part=b/(a+b+c)=y      third part=c/(a+b+c)=y

Now let us work out some questions to understand the underlying concept.
Q1. Find the three numbers in the ratio of 1:2:3 so that the sum of their squares is equal to 504?
Ans:let 1st no. be 1x,2nd no. be 2x and 3rd no. be 3x
their squares-  x2 , (2x)2 and (3x)2
as per the question, x2 + (2x)2+(3x)2 = 504
x2+4x2+9x2=504
14x2=504
x2=504/14=36
so, x=6
So the three no. are 1x=6,2x=12 and 3x=18

Q2.A,B,C and D are four quantities of the same kind such that A:B=3:4,B:C=8:9 and C:D= 15:16xfind ratio a)A:D b)A:B:C:D
ans: a)A/D=A/B x B/C x C:D=3/4 x 8/9 x 15/16=5/8
so A:D=5:8
A :  B    =  3 : 4
B :  C    = 8 : 9
C :  D   = 15  :16
in A:B:C:D value of A will be given by product of ABC .
value of B will be given by product of BBC
value of C will be given by product of BCC
value of D will be given by product of BCD
so A:B:C:D is 3x8x15:4x8x15:4x9x15:4x9x16
or,30:40:45:48

Q3.if a carton containing a dozen mirrors is dropped, which of the following cannot be the ratio of broken mirrors to unbroken mirrors?
options:a)2:1 b)3:1 c)3:2 d) 1:1 e)7:5
There are 12 mirrors in the cartonx in the given options antecedents tell the broken mirrors and consequents  tell the unbroken mirrorsx so, the sum of antecedent and consequent in each ratio should divide the noxof mirrors perfectlyxout of the given options option 'c' which totals 5 cannot divide 12, cannot be the ratio of broken mirrors to unbroken mirrorsx

Q4.find the fourth proportional to the numbers 6,8 and 15?
ans: let K be the fourht proportional, then  6/8=15/K
solving it we get K=(8x15)/6= 20

Q5. find the mean mean proportion between 3 and 75?
ans. this is related to continued proportion.let x be the mean proportionalx then we have
x2=3x75 or x=15

Q6.divide Rs 1350 into three shares proportional to the numbers 2, 3 and 4?
ans: 1st share= Rs 1350x(2/2+3+4)=Rs 300
2nd share = Rs 1350x(3/2+3+4)=Rs 450
3rd share= Rs 1350x(4/2+3+4)=Rs 600

Q7. a certain sum of money is divided among A,B and C such that for each rupee A has ,B has 65 paise and C has 40 paisex if C's share is Rs 8, find the sum of money?
ans: here A:B:C = 100:65:40 = 20:13:8
now 20+13+8=41
as 8/14 of the whole sum=Rs 8
so, the whole sum=Rs 8x41/8=Rs 41

Q8.in 40 litres mixture of milk and water the ratio of milk and water is 3:1. how much water should be added in the mixture so that the ratio of milk to water becomes 2:1.?
ans:here only amount of water is changing. the amount of milk remains same in both the mixtures. so, amount of milk before addition of water =(3/4)X40=30 ltrs. so amount of water is 10 ltrs.
After addition of water the ratio changes to 2:1.here the mixture has two ltrs of milk for every 1 ltr of water. since amount of milk is 30 ltrs the amount of water has to be 15 ltr so that the ratio is 2:1. so the amount of water to be added is 15-10=5 ltrs.

Q9. three quantities A, B and C are such that AB=kC ,where k is constant. when A is kept constant, B varies directly as C: when B is kept constant, A varies directly C and when C is kept constant, A varies inversely as B.
initially A was at 5 and A:B:C was 1:3:5. find the value of A when B equals 9 at constant B?
solution: initial values are A=5,B=15 and C=25.
thus we have 5x15=kx25         hence, k=3
thus the equation becomes AB=3C.
for the problem C is kept constant at 25. then,
Ax9=3x25 A=75/9=8.33