model 1 basic average questions Section-Wise Topic Notes With Detailed Explanation And Example Questions
MOST IMPORTANT quantitative aptitude - 9 EXERCISES
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Basic Average Questions Practice Test, Shortcuts, Tricks, PDF »
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New Average Aptitude MCQ Average of Consecutive Numbers »
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Top 99+ Average Aptitude: Twice, One Third of Numbers MCQ »
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New +99 Average Aptitude MCQ Find nth Average of Numbers »
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99+ Average Aptitude Find New Average from Error MCQ Test »
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Top 99+ Aptitude MCQ Test on Average With Excluded Number »
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Top 99+ Aptitude Average on Ages & Weights MCQ Test Quiz »
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Top Aptitude Average MCQ Find Monthly Income for Average »
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Top 149+ Aptitude Average on Cricket & Exam MCQ Test Quiz »
The following question based on average topic of quantitative aptitude
(a) $a+(2^(n-1)-1)/n$
(b) $a+2(2^n-1)/n$
(c) $a+2^(n-1)/n$
(d) $a+(2^n-1)/n$
The correct answers to the above question in:
Answer: (b)
Sum of new numbers
= na + (2 + 4 + 8 + 16 ..... to n terms)
Now, S = 2 + 4 + 8 + 16 + ..... to n terms
Here, a = first term = 2
r = common ratio = $4/2$ = 2
It is a geometric series.
∴ S =${a({r^n} - 1)}/{r-1}$= ${2({2^n} - 1)}/{2-1}$
= 2 ($2^n$ –1)
∴ Required average
= ${na +2({2^n}-1)}/n$
a +${2({2^n}-1)}/n$
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Read more simple average Based Quantitative Aptitude Questions and Answers
Question : 1
a, b, c, d, e, f, g are consecutive even numbers. j, k, l, m, n are consecutive odd numbers. The average of all the numbers is
a) $3(\text"a+n"/2)$
b) $(\text"l+d"/2)$
c) $\text"a+b+m+n"/4$
d) $\text"j+c+n+g"/4$
Answer »Answer: (b)
According to the question consecutive even numbers = a, b, c, d, e, f, g
Consecutive odd numbers = j, k, l, m, n
Consecutive even number = 2, 4, 6, 8, 10, 12, 14
$\text"2+4+6+8+10+12+14"/7=56/7=8$ middle term
Consecutive odd numbers 1, 3, 5, 7, 9
$\text"1+3+5+7+9"/5=25/5$ middle term
Same as in the above situation.
Average of a, b, c, d, e, f, g = d
Average of j, k, l, m, n, = l
∴ Required average = ${d+l}/2$
Question : 2
The average of nine consecutive numbers is n. If the next two numbers are also included the new average w
a) increase by 2
b) remain the same
c) increase by 1.5
d) increase by 1
Answer »Answer: (d)
Given,
$\text"(a + a + 1 + a + 2 + a + 3 + a + 4 + a + 5 + a + 6 + a + 7 + a + 8)"/9 = n$
$=> \text"(9a + 36)"/9 = n$
=> a + 4 = n -------------------- ( 1 )
If the next 2 numbers are included , let new average = k
$\text"(a + a + 1 + a + 2 + a + 3 + a + 4 + a + 5 + a + 6 + a + 7 + a + 8 + a + 9 + a + 10)"/11 = k$
$\text"(11a + 55)"/11 = k$
a + 5 = k -------------------- ( 2 )
Subtracting ( 1 ) from ( 2 ) , we get :
a + 5 - a - 4 = k - n
=> k - n = 1
=> k = n + 1
Therefore the new average is n + 1
Question : 3
The average of 5 consecutive integers starting with ‘m’ is n. What is the average of 6 consecutive integers starting with (m + 2) ?
a) $\text"2n+5"/2$
b) $n+2$
c) $n+3$
d) $\text"2n+9"/2$
Answer »Answer: (a)
According to the question let m = 1
∴ 5 consecutive integers are = 1,2,3,4,5
$\text"1+2+3+4+5"/5=n$
$n = 15/5=3$
6 consecutive integers starting with (m + 2) are = 3,4,5,6,7,8
∴$\text"3+4+5+6+7+8"/6=33/6=11/2$
Now check from option to put n = 3
While in Option :(a) $\text"2n+5"/2 = 11/2 $
Question : 4
In a class, average height of all students is ‘a’ cms. Among them, average height of 10 students is ‘b’ cms and the average height of the remaining students is ‘c’ cms. Find the number of students in the class. (Here a > c and b > c )
a) $(a(b-c))/(a-c)$
b) $(b-c)/(a-c)$
c) $(b-c)/\text"10(a-c)"$
d) $\text"10(b-c)"/(a-c)$
Answer »Answer: (d)
Let the total number of students in the class be n.
According to the question,
an = 10 × b + (n – 10) c
⇒ an = 10b + nc – 10c
⇒ an – cn = 10b – 10c
⇒ n (a – c) = 10 (b – c)
$⇒ n =\text"10 (b - c )"/ ( a - c )$
Question : 5
The average of $x$ numbers is $y^2$ and the average of $y$ numbers is $x^2$. So the average of all the numbers taken together is
a) $(x^3+y^3)/\text"x+y"$
b) $xy$
c) $(x^2+y^2)/\text"x+y"$
d) $xy^2+yx^2$
Answer »Answer: (b)
According to the question,
Average of x number is y2
∴ Sum of x number is = xy2
Average of y number is = x2
∴ Sum of y number is = yx2
Average of all number is
= ${xy^2+yx^2}/{x+y}$
= ${xy(x+y)}/{x+y}$
= xy
Question : 6
If average of 20 observations x1,x2, ....., x20 is y, then the average of x1 – 101, x2 – 101, x3 –101, ....., x20 –101 is
a) y-20
b) y-101
c) 20y
d) 101y
Answer »Answer: (b)
${X_1 + X_2 + X_3 + X_4 + ..... + X_20}/20 = Y$
$X_1 + X_2 + X_3 + X_4 + ..... + X_20 = 20Y$
$= {X_1 - 101 + X_2 - 101 + X_3 - 101 + X_4 - 101 + ..... + X_20 - 101}/20$
$= (X_1 + X_2 + X_3 + X_4 + ..... + X_20) - 20 x 101/20$
$\text"20Y - 20 x 101"/20$
= Y - 101
GET average PRACTICE TEST EXERCISES
model 1 basic average questions
model 2 average of consecutive numbers
model 3 twice, thrice, one third etc. of numbers
model 4 find nth average from 1st & last number
model 5 find new average from error
model 6 find average of excluded number
model 7 average on ages/weight
model 8 find monthly income
model 9 average on cricket/exam
average Shortcuts and Techniques with Examples
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model 1 basic average questions
Defination & Shortcuts … -
model 2 average of consecutive numbers
Defination & Shortcuts … -
model 3 twice, thrice, one third etc. of numbers
Defination & Shortcuts … -
model 4 find nth average from 1st & last number
Defination & Shortcuts … -
model 5 find new average from error
Defination & Shortcuts … -
model 6 find average of excluded number
Defination & Shortcuts … -
model 7 average on ages/weight
Defination & Shortcuts … -
model 8 find monthly income
Defination & Shortcuts … -
model 9 average on cricket/exam
Defination & Shortcuts …
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