| Updated On September 6th, 2019 at 11:38 am
Quantitative Aptitude for Bank Exams 2019
Quantitative Aptitude is hard in most cases especially in exams like Banks and Insurance. Many Banks exam has a two-tier examination pattern i.e., Prelims and Mains. Most of them have changed their exam patterns and set a sectional timing of 20 minutes on each section. Quantitative aptitude is important for every Bank Exams because proper strategy and enough practice can help you score full marks in this section. The level of changes may not be assured in the language section and you may be stuck while solving reasoning questions but quants is a scoring subject and assure full marks if the calculation is correct. So to help you ace the quants and to save your precious time during exam hours Adda247 providing some quant tricks to help aspirants.
Today we learn some more concepts of average
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Average of the series in AP.
Average of AP or the consecutive number is (First term + Last term)/2
Example–101, 104, 107, 110, 113
Average is= 101+113=214/2=107
Now since the example has 5 terms (ODD terms) the average is the middle number
And here notice that if we take 104 and 110 then also average is 107
So, we can say an average of consecutive series having odd terms, the average is the middle term
And when the average of the number having even terms then the average is calculated by the help of an example
Example- 101, 104, 107, 110, 113, 116
Average is= 101+116= 108.5 which lies somewhere between the middle terms (107, 110)
Example based on this concept-
Example- The average of 5 consecutive integers starting with m is n. What is the average of 6 consecutive integers starting with (m+2).
Solution– The numbers are m, m+1, m+2, m+3, m+4
Average is given=n
We know the average of a consecutive odd number is middle term= m+2
Now the average of 6 consecutive numbers stating with m+2
m+2, m+3, m+4, m+5, m+6, m+7
Here the number of terms is even so the average lies between the middle term or average is (last term+ first term)/2.
(m+4+m+5)/2= (2m+9)/2=m+4.5 or 2n+5/2 is answer
Example2– Find the average of 5 consecutive even numbers.
Statement1: The difference of fifth number and the first number is 8.
Statement2: The sum of first two numbers is 6 more than the fifth number.
Solution– let the even number is x, x+2, x+4, x+6, x+8
we can’t find solution by stmt 1
average of 5 number is middle number i.e.= 16
so only 2 is required
Example3- Smallest of a five-digit consecutive odd number is 3 more than second largest number of five consecutive even number. Find the average of consecutive even number is how much less than the average of an odd number.
Solution-let the odd number is- x, x+2, x+4, x+6, x+8
Average of odd number= x+4
smallest of five-digit consecutive odd number is 3 more than the second-largest even number
so the even number is
x-9, x-7, x-5, x-3, x-1
Now the average is
Average of even number= x-5
Average of an odd number is 9 more than the average of even number
Example4– The sum of five consecutive even number of Set A is 220.
What is the different Set B of five consecutive positive integers whose 2nd lowest number is 37 less than double of the lowest of Set A.
Solution-First find the average= 220/5=44
we know the average of 5 consecutive even number is the middle term
middle term= 44
so the five numbers of SetA are 40, 42, 44, 46, 48
for set B
given that 2nd lowest number of SetB= 2*40-37=43
i.e Set B is 42, 43, 44, 45, 46
Example5– The average of the group of 8 friends is 52 kg. What will be the average weight when a person of weight 76 kg joins the group?
Solution- If you know the concept of deviance you can solve it in seconds
here average is 52 kg means all 8 person is 52 kg
when a new man joins its weight is (76-52) kg more than the group=24 kg more
this extra 24 kg is divided into 9 persons now
Average = 52+ 24/9
You can post your solutions and methods in the comment section.
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