The average score in a bank examination of 13 students of a class is 50. If the scores of the top five students are not considered, the average score of the remaining students falls by 5. The pass mark was 35 and the maximum mark was 100. It is also known that none of the students failed. If each of the top five scorers had distinct integral scores, the maximum possible score of the topper is. A) 85 B) 90 C) 95 D) 100

The average score in a bank examination of 13 students of a class is 50. If the scores of the top five students are not considered, the average score of the remaining students falls by 5. The pass mark was 35 and the maximum mark was 100. It is also known that none of the students failed. If each of the top five scorers had distinct integral scores, the maximum possible score of the topper is. A) 85 B) 90 C) 95 D) 100
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1 Answer

Answer :

D

Average = Sum of observations/Number of observations

Given, average score in a bank examination of 13 students of a class is 50.

Sum of total scores = 13 × 50= 650

Given, if the scores of the top five students are not considered, the average score of the remaining students

falls by 5.

Sum of scores of remaining 8 students = 8× 45= 360

Sum of scores of the top 5 students = 650-360= 290

Let the scores of the top 5 students be a, b, c, d and e.

Let ‘e’ be the maximum possible score of the topper.

Also, each of the top five scorers had distinct integral scores.

Thus, for ‘e’ to be the maximum possible score, the collective score of (a + b + c + d. should be least possible.

Since average given was 45, so minimum score for highest scorers will be 46 atleast.

Thus,

a = 46, b = 47, c = 48, d = 49

⇒a + b + c + d + e = 290

⇒46+ 47 + 48+ 49 +e = 290

⇒e = 100


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Description : There are two groups of a class consisting of 72 and 88 students respectively. If the average weight of 1st group is 80kg and the weight of 2nd group is 70kg. Find the average weight of whole class? A) 64kg B) 74.5kg C) 84.5kg D) 54kg

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Description : In an examination, 900 students appeared. Out of these students; 56 % got first division, 27 % got second division and the remaining just passed. Assuming that no student failed; find the number of students who just passed. a) 225 b) 153 c) 245 d) 148 e) 298

Last Answer : Answer: B The number of students with first division = 56 % of 900  = 56/100 × 900 = 50400/100  = 504 And, the number of students with second division = 27 % of 900  = 27/100 × 900  =24300/100  = 243 Therefore, the number of students who just passed = 900 – (504 + 243)  = 900- 747  = 153

Students of two colleges appeared for Talent test carrying 250marks as maximum. The average of their marks for college1 & college 2 are 160 & 180respectively. If the number of students of college 1 is the half of the number of students of college 2, then what is the average marks of all students of both the college? A) 170 B) 110 C) 173.33 D) 177.33

Description : Students of two colleges appeared for Talent test carrying 250marks as maximum. The average of their marks for college1 & college 2 are 160 & 180respectively. If the number of students of college 1 is the half of the ... marks of all students of both the college? A) 170 B) 110 C) 173.33 D) 177.33

Last Answer : C) let the number of students of college 2 be '2N' then the number of students of college1 is 'N' the average marks for college1 is 160 the average marks for college 2 is 180 total marks of ... = 360N average marks of all students of both the colleges = (160N + 360N)/N+2N = 173.33marks

Students of two university appeared for a common test of maximum 60 marks. The average of their marks for university 1 & university 2 are 39 & 42 respectively. If the no of students of university 1 is twice the no of students of university 2. Then what is the average marks of all students of both the university? A) 40 B) 42 C) 26 D) 36

Description : Students of two university appeared for a common test of maximum 60 marks. The average of their marks for university 1 & university 2 are 39 & 42 respectively. If the no of students of university 1 is twice the no ... the average marks of all students of both the university? A) 40 B) 42 C) 26 D) 36

Last Answer : Answer: A) Let number of students of university2 be N and the no of students of university 1 be 2N the average of university1 and university 2 is 39 and 42 total marks of university 1 students and university 2 ... 78N and N*42 =42N average of both university,  =(78N+42N) /(N+2N) = 40 marks

Find the average of first 85 natural numbers? A) 43 B) 41 C) 45 D) 47

Description : Find the average of first 85 natural numbers? A) 43 B) 41 C) 45 D) 47

Last Answer : A) Sum of 1st n natural numbers = n(n+1)/2 So, sum of 1st 85 natural numbers = 85(85+1)/2 =85(86)/2 =7310/2 =3655 Required average = 3655/85 =43.

A candidate scoring 50% in an examination fails by 60 marks , while another candidate scores 75 % mark, gets 40 marks more than the minimum pass marks . Find the minimum pass mark. a) 125 b) 220 c) 140 d) 260 e) 298

Description :  A candidate scoring 50% in an examination fails by 60 marks , while another candidate scores 75 % mark, gets 40 marks more than the minimum pass marks . Find the minimum pass mark. a) 125 b) 220 c) 140 d) 260 e) 298

Last Answer : Answer: B  Let x be the maximum marks,  Then (50% of x)+60 = (75% of x)-40  x/2 +60 = 3x/4 -20  60+20 = 3x/4 – x/ 2 X=320  Hence maximum marks = 320  Minimum pass marks = 320/2 + 60 = 220

The marks obtained by 15 students in an examination are given below : 40, 20, 24, 19,20, 35, 12, 48, 29, 40, 45, 48, 42, 23, 35. Find the average mark

Description : The marks obtained by 15 students in an examination are given below : 40, 20, 24, 19,20, 35, 12, 48, 29, 40, 45, 48, 42, 23, 35. Find the average mark

Last Answer : The marks obtained by 15 students in an examination are given below : 40, 20, 24, 19,20, 35, 12, ... , 23, 35. Find the average mark of the students.

Which of the following statements accurately describes test-retest reliability? a. Measure of consistency of test scores over time b. Measure of consistency of scores obtained from two equivalent halves of same test c. Measure of consistency with which a test measures a single construct or concept d. Measure of degree of agreement between two or more scorers, judges, or raters

Description : Which of the following statements accurately describes test-retest reliability? a. Measure of consistency of test scores over time b. Measure of consistency of scores obtained from two equivalent halves of ... concept d. Measure of degree of agreement between two or more scorers, judges, or raters

Last Answer : a. Measure of consistency of test scores over time

Of the 5 numbers, the first is twice the second, the second is one – third of the third, the third is 5 times of the fourth, and the fourth is three – seventh of fifth. The average of the numbers is 35. The largest of these number is. A) 30.63 B) 65.625 C) 43.75 D) 75.356

Description : Of the 5 numbers, the first is twice the second, the second is one – third of the third, the third is 5 times of the fourth, and the fourth is three – seventh of fifth. The average of the numbers is 35. The largest of these number is. A) 30.63 B) 65.625 C) 43.75 D) 75.356

Last Answer : B) Let the fifth number be x'. 4th number = 3/7 x 3rd number = 5 (3/7 x) = 15/7x 2nd number = 1/3 (15/7 x) = 15 /21 x 1st number = 2 (15/21 x) = 30/21 x X + 3/7 x + 15 /7 ... 120x = 3675 X = 30.625 So the numbers are 30.62, 13.125, 65.625, 21.875, 43.75. Therefore largest number is 65.625.

A ship sails out to a mark at the rate of 10 km per hour and sails back at th rate of 15 km per hour. What is its average rate of sailing? (a) 10 km. p.h. (b) 12 km. p.h. (c) 15 km. p.h. (d) 11 km. p.h.

Description : A ship sails out to a mark at the rate of 10 km per hour and sails back at th rate of 15 km per hour. What is its average rate of sailing? (a) 10 km. p.h. (b) 12 km. p.h. (c) 15 km. p.h. (d) 11 km. p.h.

Last Answer : 12 km. p.h.

When the average age of a father, mother and their son was 90 years, the son got married and a child was born just 6 year after the marriage when child turned 14 years the average age of the family is 80yrs. Find the age of daughter - in - law at present? A) 55 B) 56 C) 57 D) 58

Description : When the average age of a father, mother and their son was 90 years, the son got married and a child was born just 6 year after the marriage when child turned 14 years the average age of the family is 80yrs. Find the age of daughter - in - law at present? A) 55 B) 56 C) 57 D) 58

Last Answer : B) total age of father, mother and son at the time of son's marriage = 90*3 =270 present age of family father, mother, son, daughter-in-law, child = (father, mother, son age at the time of marriage ... 60)+ daughter-in-law present age + 14 = 80*5 =400 daughter-in-law present age= 400-344 = 56yrs

Rohit married 21 years ago at the age of 90 years. His wife's present age 99years. If 24 years later the average age of rohit, his wife and their son was 90 years, then what is son's present age? A) 15yr B) 14yr C) 13yr D) 12yr

Description : Rohit married 21 years ago at the age of 90 years. His wife's present age 99years. If 24 years later the average age of rohit, his wife and their son was 90 years, then what is son's present age? A) 15yr B) 14yr C) 13yr D) 12yr

Last Answer : Answer: D) Rohit present age = 111years total age of family after 24 yrs = 90*3= 270 present age of rohit & his wife = 111+99 = 210yrs present age of rohit, his wife & his son= 270- 24*3 =198 present age of son= 210-198=12yrs

The average of 35 results is 28. The average of first 17 of them is 24 and that of last 17 is 21. Find the 18th result? A) 125 B) 215 C) 512 D) 521

Description : The average of 35 results is 28. The average of first 17 of them is 24 and that of last 17 is 21. Find the 18th result? A) 125 B) 215 C) 512 D) 521

Last Answer : B) Clearly 18th result = (sum of 35 results)- (sum of 34 results) =(35 * 28) – { (17 *24) +(17 *21)} = 980 – (408 + 357 =980 – 765 =215

The average weather(rainfall) for the first 6 days out of 8 days recorded to be 9 cm. The rainfall on last 2 days was in the ratio 2:3. the average of 8 days was 14.2 cm. What was the rainfall on the last day? A) 7.8 B) 34.4 C) 35.76 D) 8.9

Description : The average weather(rainfall) for the first 6 days out of 8 days recorded to be 9 cm. The rainfall on last 2 days was in the ratio 2:3. the average of 8 days was 14.2 cm. What was the rainfall on the last day? A) 7.8 B) 34.4 C) 35.76 D) 8.9

Last Answer : C) according to question (6×9+3x+2x)/8=14.2 54+5x=113.6 X=11.92 Last day rainfall=3×11.92=35.76cm

An industry employed 1200 men and 800 women and the average salary was Rs 51 per day. If a woman got Rs 10 less than a man , then what are their daily salary?(mens and women respectively) A) men Rs 25 and women Rs 35 B) men Rs 45 and women Rs 55 C) men Rs 35 and women Rs 25 D) men Rs 55 and women Rs 45

Description : An industry employed 1200 men and 800 women and the average salary was Rs 51 per day. If a woman got Rs 10 less than a man , then what are their daily salary?(mens and women respectively) A) men Rs 25 and women Rs ... Rs 45 and women Rs 55 C) men Rs 35 and women Rs 25 D) men Rs 55 and women Rs 45

Last Answer : D) Let the daily salary of man be Rs’x’ Then the daily salary of a women = Rs (x-100 Now 1200x + 800(x-10) = 51 *(1200+800) 1200x+800x -8000 = 51*2000 2000x = 102000+8000 =110000 X=55 Therefore mens daily salary Rs 55 and womens daily salary Rs 45

There are 50 compartments in a Chennai express carrying an average of 70 passengers per compartments. At least 24 passengers were sitting in each compartment, not any compartment has equal number of passengers, and any compartment does not exceed the number of average passengers expect 50th compartment. Find how many passengers can be accommodated in 50th compartment? A. 748 B. 705 C.739 D.cannot be determined

Description :  There are 50 compartments in a Chennai express carrying an average of 70 passengers per compartments. At least 24 passengers were sitting in each compartment, not any compartment has equal number of passengers, ... can be accommodated in 50th compartment? A. 748 B. 705 C.739 D.cannot be determined 

Last Answer : C) Total number of passengers in Chennai express = 50*70 = 3500 Total number of candidates from 1 to 49 compartments = 24+25+....+70 = (70*71)/2 +(23*24)/2 = 2761 number of passengers in 50th compartment = 3500 - 2761= 739