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  • IBPS Clerk CWE Exam Quantitative Aptitude 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 Quantitative Aptitude Averages for IBPS Clerk CWE Exam student. This free online Quantitative Aptitude study material for IBPS Clerk CWE Exam will help students in learning and doing practice on Averages topic of IBPS Clerk CWE Exam Quantitative Aptitude. The study material on Averages, help IBPS Clerk CWE Exam Quantitative Aptitude students to learn every aspect of Averages and prepare themselves for exams by doing online test exercise for Averages, as their study progresses in class. Kidsfront provide unique pattern of learning Quantitative Aptitude with free online comprehensive study material and loads of IBPS Clerk CWE Exam Quantitative Aptitude Averages exercise prepared by the highly professionals team. Students can understand Averages concept easily and consolidate their learning by doing practice test on Averages regularly till they excel in Quantitative Aptitude Averages.


Averages
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.
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