KEY 4 WIN: AVERAGE SHORTCUTS FOR QUANTITATIVE APTITUDE

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### AVERAGE SHORTCUTS FOR QUANTITATIVE APTITUDE

AVERAGE is also called mean. suppose a,b,d,e(in ascending order) are the given nos.and c is the average of given nos=a,b,d,e.
the average value c is such a value ,so that sum of the individual distances of nos.from avg. value c lying below the avg. value. c and sum of the individual distances of nos.from avg. value c lying above c are equal.e.g. take nos.1,2,3,4,5
.their avg.is 3. distance of 1 from 3=3-1=2(a)
distance of 2 from 3=1(b)
distance of 3 from 3=0
distance of 4 from 3=1(c)
distance of 5 from 3=2(d)
here a+b=c+d

normally average is also called arithmetic mean.

avg.=total of items/number of items
or,
formula for average 'c'=(a+b+d+e)/4

there is one more way of calculating average.take one example.
numbers given are.21,25,15,20,26,31,18
we dont know the avg.. let us take 20 as an avg. we could have taken any no.like 21 ,22,10,40 ...anything. now take the summation of distances between the assumed avg.20 and given numbers.for the nos. below 20 take the differences with negative sign and for the nos. above 20 take the differences with positive sign.now add them and divide the summation with the total no. of numbers given.if the resulting no.is negative ,substract it from the asumed avg. to get the original avg. and if the resulting no. is positive add to the assumed avg. to get the original avg.

in the given problem,
sum of the differences=1+5+(-5)+0+6+11+(-1)=16
16/7=2.28

now add this to assumed avg. 20=20+2.28=22.28(original avg.)

Now let us solve some of the questions to clarify the concept.
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Q1. the avg. age of 30 boys of a class is 14 years.when the age of the class teacher is included the average becomes 15 years. find the age of clas teacher?
solution:summation of the ages of 30 boys=avg. age of boys x no.of boys
= 14x30=420 yrs-(i).
summation of ages of 30 boys & 1 teacher(total 31 items)=15x31=465 yrs(ii)
now difference of ii and i will give the age of teacher=465-420=45 yrs.
there is one more way of doing this problem.
when avg. is calculated the total value of items is equally distributed among the items whose values has been totalled.in the above problem when avg. age was calculated 14 yrs was distributed equally among 30 boys. when the age of teacher is added the avg. age increases by 1 yr. in other way we can say,each boy was given 1 yr. so total of 30 yrs from the teacher's age was given to the boys. then we have to keep 15 yrs for the teacher(as he also joins the group) so as to maintain the new avg. age of 15 yrs.
so,our answer is 45 yrs(30+15). out of which 30 yrs has been given to 30 boys and remaining 15 yrs is kept to maintain avg. of 15 yrs for the group of 31 people.
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Q2.the avg. age of 4 men is increased by 3 kg when one of them who weighs 120 kg is replaced by another man. what is the weight of the new man?
solution: let the avg. age of four men before exit of man with 120kg weight be 'A'.when he exits and new person enters,avg.age becomes 'A+3'. it means the newcomer brings with him weight which is more than 120 kg. the extra amount brought by him must be 12 yrs, as he has to give 3 yrs each to each of four members including himself(4x3=12).so his age must be 120+12 yrs=132 yrs.
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Q3.one third of a certain journey is covered at rate of 25km/hr, one fourth at the rate of 30 km/hr and the rest at 50 km/hr. find the average speed for the whole journey?
solution:let us take the total distance be 120km(as it is divisible by both 3 and 4)
so,distance travelled @25km/hr=1/3 x 120=40 km. time taken to travel this distance=40/25=1.6 hr
distance travelled @30km/hr=1/4 x 120=30 km. . time taken to travel this distance=30/30=1 hr
distance travelled @50km/hr=120-40-30=50 km. . time taken to travel this distance=50/50=1hr

so, total time taken=1.6+1+1=3.6 hours.
total distance travelled=120 km
avg. speed for the journey=120/3.6=33.33 km/hr
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Q4.a cricketer has completed 10 innings and his average is 21.5 runs.how many runs he must make in his next inning so as to raise his avg.to 24?
solution:keeping in mind the solutions given in Q1 and Q2, from the runs scored in the 11th inning 2.5 runs is to be given to each of the 10 innings so as to raise their avg. score to 24 runs and in addition to that 11th inning has to keep 24 runs to maintain the avg. score per inning(for 11 innings) at 24 runs.
so,total runs to be scored in the 11th inning is =2.5x10+24=49 runs.
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Q5.the avg. of 11 results is 30,that of the first five is 25 andthat of the last five is 28. find the value of sixth number?
solution:total of 11 results=11x30=330 - 'a'
total of the first five results=5x25=125
total of the last five results=5x28=140
total of the first five results & total of the last five results=125+140=265-'b'
in this operation,our sixth result has been left as
1 2 3 4 5 6 7 8 9 10 11
1st five last five

so difference of 'a' and 'b' will give the value of 6th result =330-265=65
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Q6the average of first 61 natural no.is?
solution:first of take the total of first 61 natural nos.={n(n+1)}/2

n=61 so, 61(61+1)/2=1891
now for avg. of 61 nos.= 1891/61=31