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  • UIIC ASSISTANT RECRUITMENT Numerical Ability Study Material

Digitization help student to explore and study their academic courses online, as this gives them flexibility and scheduling their learning at their convenience. Kidsfront has prepared unique course material of Numerical Ability Average for UIIC ASSISTANT RECRUITMENT student. This free online Numerical Ability study material for UIIC ASSISTANT RECRUITMENT will help students in learning and doing practice on Average topic of UIIC ASSISTANT RECRUITMENT Numerical Ability. The study material on Average, help UIIC ASSISTANT RECRUITMENT Numerical Ability students to learn every aspect of Average and prepare themselves for exams by doing online test exercise for Average, as their study progresses in class. Kidsfront provide unique pattern of learning Numerical Ability with free online comprehensive study material and loads of UIIC ASSISTANT RECRUITMENT Numerical Ability Average exercise prepared by the highly professionals team. Students can understand Average concept easily and consolidate their learning by doing practice test on Average regularly till they excel in Numerical Ability Average.


Average
The average of 12, 22, 32, 42, 52, 62, 72, 82, 92, 102 is

a) 46.5
b) 50.5
c) 38.5
d) 42.5



Answer
Solution
Correct Answer Is : 38.5
Solution Is : Sum of square of n natural numbers = 1/6n ( n+1 ) ( 2n + 1 ) Average of square of n natural numbers 1/6n (n+1) (2n+1)/n =1/6 (n+1) (2n+1) putting n = 10, Required sum = 1/6 ( 10 +1 ) ( 2 * 10 + 1 ) 1/6 * 11 * 21 = 38.5
There are three baskets of fruits. First basket has twice the number of fruits in the 2nd basket. Third basket has 3/4 th of the fruits in the first. The average of the fruits in all baskets in 30. What is the number of fruits in the first basket ?

a) 20
b) 30
c) 35
d) 40



Answer
Solution
Correct Answer Is : 20
Solution Is : Suppose number of fruits in 2nd basket be x. Number of fruits in 1st basket = 2x and number of fruits in 3rd basket = 2x * 3/4 = 3x/2 then, x + 2x + 3x/2 = 30 * 3 2x + 4x + 3x = 90 9x = 90 x = 90/9 = 10
Geetha weighs 11.235 kg. Her sister weighs 1.4 times her weight. Find their combined weight?

a) 15.729 kg
b) 25.964 kg
c) 26.964 kg
d) 26.964 kg



Answer
Solution
Correct Answer Is : 26.964 kg
Solution Is : Geetha`s weight = 11.235 kg Her sister`s weight = 1.4 (11.235) = 15.729 Combined weight = 15.729 + 11.235 = 26.964
The average age of 27 students of a class is 22. If the age of the teacher is also added to their ages, then the average increases by one. Find the age of the teacher.

a) 52
b) 58
c) 50
d) 62



Answer
Solution
Correct Answer Is : 50
Solution Is : The average age of 27 students of a class is 22. Total age of students = 27 * 22 = 594 Let age of teacher be x If age of teacher is added , average increases by 1 , -----------> (594 + x) / 28 = 22 + 1 594 + x = 23 *28 x = 644 - 594 = 50
The runs scored by two batsmen over 7 matches are given below. Which batsman`s average was better?

a) Batsman 1 : 39.3
b) Batsman 1 : 45.9
c) Batsman 1 : 43.2
d) Batsman 2 : 45.9



Answer
Solution
Correct Answer Is : Batsman 2 : 45.9
Solution Is : Average of batsman 1 = (42 + 51 + 09 + 78 + 63 + 20 + 12 ) / 7 = 275 / 7 =39.2 Average of batsman 2 = (30 + 22 + 91 + 76 + 84 + 11 + 07 ) / 7 = 321 / 7 = 45.85 = 45.9 Average of batsman 2 is greater .
The average of the first 9 positive even number is

a) 12
b) 9
c) 10
d) 10.5



Answer
Solution
Correct Answer Is : 10
Solution Is : The average of the first n positive even number = n + 1 So, the required average =x+1=9+1=10
The average of 1, 3, 5, 7, 11, ______ 25 terms is

a) 625
b) 25
c) 125
d) 50



Answer
Solution
Correct Answer Is : 25
Solution Is : Here a = 1, d = 2, n = 25 Tn = a + ( n-1 )d = 1 + ( 25 - 1 ) * 2 = 1 + 24 * 2 = 49 Since, the nth term is 49, So l = 49 Sn = n/2 [ a + l ] = 25/2 [1+49] = 25/2 * 50 = 25 * 25 = 625 Average => 625/25 = 25.
A batsman makes a score of 87 runs in the 17th inning and thus increased his average by 3. Find his average after 17 th inning.

a) 84
b) 87
c) 90
d) 39



Answer
Solution
Correct Answer Is : 39
Solution Is : A batsman makes a score of 87 runs in 17th inning and his average increases by 3. Score of batsman before 17th inning = x Score of batsman after 17th inning = x + 87 = x + 87/17 - x/16 = 3 16 ( x+87) - 17x = 3*16*17 16x + 87 * 16 - 17x = 3 * 16 * 17, 87 * 16 - 3 * 16 * 17 = x 16 ( 87 - 51 ) = x, x = 16 * 36 = 576 Now, average after 17th inning =663/17 = 39.
The average of 12, 22, 32, 42, 52, 62, 72, 82, 92, 102 is

a) 46.5
b) 50.5
c) 38.5
d) 42.5



Answer
Solution
Correct Answer Is : 38.5
Solution Is : Sum of square of n natural numbers = 1/6n ( n+1 ) ( 2n + 1 ) Average of square of n natural numbers 1/6n (n+1) (2n+1)/n =1/6 (n+1) (2n+1) putting n = 10, Required sum = 1/6 ( 10 +1 ) ( 2 * 10 + 1 ) 1/6 * 11 * 21 = 38.5
There are three baskets of fruits. First basket has twice the number of fruits in the 2nd basket. Third basket has 3/4 th of the fruits in the first. The average of the fruits in all baskets in 30. What is the number of fruits in the first basket ?

a) 20
b) 30
c) 35
d) 40



Answer
Solution
Correct Answer Is : 20
Solution Is : Suppose number of fruits in 2nd basket be x. Number of fruits in 1st basket = 2x and number of fruits in 3rd basket = 2x * 3/4 = 3x/2 then, x + 2x + 3x/2 = 30 * 3 2x + 4x + 3x = 90 9x = 90 x = 90/9 = 10