The Cricketer average score in 40 over is 21, if the man scores 23runs in 8over, 26 runs in 10 over and 18runs in 14 over. What is the average score in remaining over? A) 18 B) 19 C) 16 D) 20

The Cricketer average score in 40 over is 21, if the man scores 23runs in 8over, 26 runs in 10 over and 18runs in 14 over. What is the average score in remaining over? A) 18 B) 19 C) 16 D) 20 
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1 Answer

Answer :

A)

According to the question

40*21=23*8+26*10+18*14+8×X

840=184+260+252+8x

8x=144

 X=144/8=18 runs
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Related questions

The batting average for 20 innings of a cricketer is 25 runs. His highest score exceeds his lowest score by 86 runs. If these two innings are excluded, the average of the remaining 18 innings is 22runs. Find out the highest score of the player. A) 85 B) 90 C) 95 D) 100

Description : The batting average for 20 innings of a cricketer is 25 runs. His highest score exceeds his lowest score by 86 runs. If these two innings are excluded, the average of the remaining 18 innings is 22runs. Find out the highest score of the player. A) 85 B) 90 C) 95 D) 100

Last Answer : C) Total runs scored by the player in 20 innings = 20*25 Total runs scored by the player in 18 innings after excluding two innings = 18 22 Sum of the scores of the excluded innings = 20 25 - 18 22 = 104 Given that ... Now x + 86 + x = 104 => 2x = 18 then x = 9 Highest score = 9+ 86 = 95

A cricketer played 3 matches in tournament. The respective ratio between the scores of first and second matches was 3:7 and that between the scores of second and third matches between was 7 :2. the difference between first and third matches was 84 runs, what was the cricketer average score in all the matches together? A) 336 B) 146 C) 168 D) 189

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Last Answer : A) The ratio between first and second matches equal to 3 :7 The ratio between second and third matches=7:2 The difference between first and third matches=3x-2x=84 runs = 84 Required average=1008/3=336 .

A batsman average for 20 innings is 25 runs. His highest score exceeds his lowest score by 64 runs. If these 2 innings are excluded, the average of the remaining 18 innings is 24 runs. The highest score of the player is. A) 66 B) 72 C) 77 D) 88

Description : A batsman average for 20 innings is 25 runs. His highest score exceeds his lowest score by 64 runs. If these 2 innings are excluded, the average of the remaining 18 innings is 24 runs. The highest score of the player is. A) 66 B) 72 C) 77 D) 88

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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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The average age of A and B is 20 years. If C were to replace A, the average would be 19 and id C were to replace B,the average would be 21 . The ages of A, B and C are (in years) (a) 22, 17,16 (b) 22, 18, 20 (c) 30, 18, 15 (d) 23, 17 ,15

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Last Answer : 22, 18, 20 Hint : A + B = 2 x 20 C + B = 2 x 19 A + C = 2 x 21

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The average marks of a class of 90 students is 126. Out Of them, 4 scores zero, first 60 students scored an average of 116, next 24 scored an average of 118. What is the mark obtained by the remaining student in the class? A) 750 B) 862 C) 774 D) 875

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Last Answer : C) According to the question, 90*126=(4*0)+(60*116)+(24*118)+2y 11340=6960+2832+2y 2y=1548 y=774

The average price of 12 note books is Rs.15 while the average price of 10 of these note books is Rs.13. Of the remaining 2 note books, if the price of one note book is 50% more than the price of the other, what is the price of each of these two note books? A) Rs. 30, Rs.20 B) Rs. 8, Rs. 30 C) Rs. 10, Rs. 16 D) Rs. 12, Rs. 20

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Last Answer : Answer: A) Total price of 12 note books = Rs. 180  Total price of 10 note books= Rs. 130 ⇒ The price of 2 note books = Rs. 50 Let the price of each book be x and y. ⇒ x + y = 50 ----- ... note book is 50% more than the other price 150y/100+y=50  y=20 Substituting y value in (1) we get, x=30

on analyzing the result of an competitive exam the teacher found that the average for the entire the class was 69 marks. If we say that average of 10 % of the students scored 77 marks and average of 28 % of the students scored 66 marks, then calculate average marks of the remaining students of the class A) 67.54 B) 68.26 C) 66.91 D) 69.06

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A group of students of arithmetic mean of the marks in a test was 63. The brightest 30% of them secured a mean score of 70 and the dullest 15% a mean score of 41. The mean score of remaining 55 % is A) 63.675 B) 61.785 C) 65.181 D) 66.67

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In an examination mahi scores 64% of marks, nitesh scores 52% of marks and ritesh scores 48% of marks. The maximum mark of the exam is a three digit number, whose sum is 10 and the middle digit is equal to the sum of the other two digits. The number will decreased by 297, if its digit are reversed. approximately what is the average mark obtained by mahi, nitesh and ritesh? A) 247 B) 248 C) 264 D) 284

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Last Answer : A) Maximum mark consist of a three digit number let's consider Unit digit place is Z,ten's place digit is Y and hundred's place digit is X. According to the question, Y=x+z---------1 X+Y ... Total number of Mahi, Nitesh and Ritesh =164/100 451=739.61 Required average =739.64/3=246.54~247.

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

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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

At the average age of siva and sasi is equal to the average age of somu and ramu. if the ratio of age of siva and sasi 1:3 and somu age is equal to the three by two of siva age. What is the present age of somu if ramu age is 25years? A) 12 B) 15 C) 18 D) 21

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The average height of a batch is 344 cm. 16more students with an average height of 320 cm joined the batch therefore decreasing the average height of the batch by 12 cm. Find the total strength of the batch? A) 12 B) 11 C) 14 D) 16

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In a post graduate examination the marks obtained by a student is 75 per paper. If he had obtained 33 marks more in Evs paper & 27 more marks in science paper, then his average per paper is increased by 3 marks. Then how many papers were there in exam? A) 10 B) 12 C) 14 D) 20

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Last Answer : Answer: D)  Let the number of paper be A. Then total marks earned him = 75A from questions, 75A + 33+ 27= 78A 3A = 60=> A=20 = number of subjects

18 friends went to a restaurant for taking their breakfast. 17 of them spent RS.24 each on their breakfast and the last one spent RS.16 more then the average expenditure of all the 18. What was the total money spent by them? A) 228 B) 448.9 C) 458.6 D) 428.0

Description : 18 friends went to a restaurant for taking their breakfast. 17 of them spent RS.24 each on their breakfast and the last one spent RS.16 more then the average expenditure of all the 18. What was the total money spent by them? A) 228 B) 448.9 C) 458.6 D) 428.0 

Last Answer : B)  Let the average expenditure of all the 18 be Rs ‘x’.  Then (24 * 17) + (x + 16) = 18x 424 + x = 18x x = 24.9 therefore total money spent 18x  = 18 * 24.9  = Rs.448.9

f the temperature falls below 18 degrees, the heating is switched on. When the temperature reaches 21 degrees, the heating is switched off. What is the minimum set of test input values to cover all valid equivalence partitions? A. 15, 19 and 25 degrees B. 17, 18, 20 and 21 degrees C. 18, 20 and 22 degrees D. 16 and 26 degrees

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Last Answer : A. 15, 19 and 25 degrees

A thermometer measures temperature in whole degrees only. If the temperature falls below 18 degrees, the heating is switched off. It is switched on again when the temperature reaches 21 degrees. What are the best values in degrees to cover all equivalence partitions? A. 15,19 and 25. B. 17,18 and19. C. 18, 20 and22. D. 16, 26 and 32.

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Last Answer : A. 15,19 and 25.

The average age of 160 boys in a class is 58 yrs. The average group of 30 boys in the class is 42 yrs and the average of another group of 50 boys in the class is 36 years. What is the average age of the remaining boys? A) 72.58 B) 74.25 C) 77.75 D) 75.68

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The average age of three -seventh of class is 49, what should be the average of remaining four-seventh students so that the average of the entire class is 63? A) 24.5 B) 86.5 C) 73.5 D) 25.5

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The average age of 150 students in a class is 40% of the number of students in the class and the average age of a group of 50 students present in the class is 32yrs and the average age of another 50 students in the class is 36yrs. What is the average age of the remaining students in the class? A) 102 B) 118 C) 112 D) 108

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A group of 4 persons joins in javelin throw competition. The best player scored 42.5 points. If he had scored 46 points, the average score for the team would have been 42. The number of points the team scored. A) 164.5 B) 166.5 C) 169.7 D) 162.5

Description : A group of 4 persons joins in javelin throw competition. The best player scored 42.5 points. If he had scored 46 points, the average score for the team would have been 42. The number of points the team scored. A) 164.5 B) 166.5 C) 169.7 D) 162.5

Last Answer : A) Let the total score be ‘x’. (X+46 – 42.5)/4 = 42 X + 3.5 = 42 * 4 X + 3.5 = 168 X = 168 – 3.5 X = 164.5

The batting average of runs of a cricket player of 30 innings was 96. How many runs must he make in his next innings so as to increase his average of runs by 12? A) 468 b) 668 C) 648 D) 486

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Last Answer : A  Given that the total average runs of a cricket player (30 innings) = 96 After 31 innings = 108 Required number of runs = (108 * 31 – (96 *30) =3348 – 2880 =468

Find the range of the data : 14, 15, 16, 18, 19, 25, 30, 41, 26, 16, 13, 18, 20 and 26.

Description : Find the range of the data : 14, 15, 16, 18, 19, 25, 30, 41, 26, 16, 13, 18, 20 and 26.

Last Answer : Find the range of the data : 14, 15, 16, 18, 19, 25, 30, 41, 26, 16, 13, 18, 20 and 26.

The average age of a board of 8 trustless remains the same as it was 3 years ago, when one of them is replaced by a new member. The new member is younger than the truetee in whose place he has been replaced by (a) 24 years (b) 26 years (C) 47 years (d) 32 years

Description : The average age of a board of 8 trustless remains the same as it was 3 years ago, when one of them is replaced by a new member. The new member is younger than the truetee in whose place he has been replaced by (a) 24 years (b) 26 years (C) 47 years (d) 32 years

Last Answer : 24 years 

4 years ago, the average age of x and y was 22 years. With z joining them, the average becomes 26 years. How ols is z now? A) 62yrs B) 50yrs C) 38 yrs D) 54yrs

Description :  4 years ago, the average age of x and y was 22 years. With z joining them, the average becomes 26 years. How ols is z now? A) 62yrs B) 50yrs C) 38 yrs D) 54yrs

Last Answer : C) age of (x+y) 4 yrs ago= (22 * 2)yrs =44 age of (x+y+z)=(26 *3)yrs =78yrs Therefore z’s age = (78-44)yrs =34 yrs. Current age of z =38 yrs

An Aided school has only three classes which contain 70, 36 and 40students respectively. The pass percentage of these classes are30,25 and 20 respectively. The pass percentage of the aided school is A) 23 % B) 15% C) 26 % D) 30%

Description : An Aided school has only three classes which contain 70, 36 and 40students respectively. The pass percentage of these classes are30,25 and 20 respectively. The pass percentage of the aided school is A) 23 % B) 15% C) 26 % D) 30% 

Last Answer : C) Total number of students passed in class1 = 30/100 *70 = 21 Total number of students passed in class 2= 25/100 *36 = 9 Total number of students passed in class 3= 20/100 *40 = 8 Total number of students passed in the school = 38 Thus, pass percentage of the school = 38 /146*100 = 26 %

The average height of the students in a group was 360cm. When 10 students whose height is 292.8cm are newly admitted. The average height of the group was reduced by 24cm.How many students are present in the group? A) 29 B) 25 C) 28 D) 14

Description : The average height of the students in a group was 360cm. When 10 students whose height is 292.8cm are newly admitted. The average height of the group was reduced by 24cm.How many students are present in the group? A) 29 B) 25 C) 28 D) 14

Last Answer : Answer: C) let the number of students initially in the class be A, then, total height = A * 360------ 1 again the no of students increased = A + 10 then, total height A + 10 student = (A +10) * 336 ... 3 (A+10) *336-2928 =A*360 336A+3360-2928=360A A=18 Number of students in class = 18+10=28

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

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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

Harish buys petrol at rs 21, Rs.24 and Rs.27 per litre for 3 successive years. What approximately is the average cost per litre of petrol if he spends Rs.12000 each year? A) 23.74 B) 54.76 C) 24.66 D) 28.99

Description : Harish buys petrol at rs 21, Rs.24 and Rs.27 per litre for 3 successive years. What approximately is the average cost per litre of petrol if he spends Rs.12000 each year? A) 23.74 B) 54.76 C) 24.66 D) 28.99

Last Answer : A)  Total quantity of petrol consumed in 3 yr = (12000/21 +12000/24+12000/27) =12000(1/2 + 1/24 + 1/27) =12000(72+63+56/1512) =12000(191/1512) =(2292000/1512)litres Total amount spent = rs(3 *12000) = Rs.36000 Average cost = Rs(36000*1512/2292000) =Rs.23.74

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

A man wnt uphill with a speed of 20 km.p.h. and came downhill with a speed of 30 km p.h. The average speed for his journey was (a) 25 km. p.h. (b) 22 1/2 km. p.h. (c) 24 km. p.h. (d) 25 1/2 km. p.h.

Description : A man wnt uphill with a speed of 20 km.p.h. and came downhill with a speed of 30 km p.h. The average speed for his journey was (a) 25 km. p.h. (b) 22 1/2 km. p.h. (c) 24 km. p.h. (d) 25 1/2 km. p.h.

Last Answer : 24 km.p.h.

The average of a positive numbers (non-zero) and its square is 8 times the number. The number is A) 15 B) 17 C) 13 D) 19

Description : The average of a positive numbers (non-zero) and its square is 8 times the number. The number is A) 15 B) 17 C) 13 D) 19

Last Answer : A) Let the number be ‘x’ rThen x+x^2/2 = 8x X+x^2=16x X^2=15x X^2 – 15x = 0 X(x-15)=0 X=0 or x= 15 So the number is 15 (because given [non-zero] positive).

What is the median in the set of data 14 19 26 28 18 21 30 27?

Description : What is the median in the set of data 14 19 26 28 18 21 30 27?

Last Answer : What is the answer ?

Hey I need help What is the median of 14, 19, 26, 28, 18, 21, 30, 27?

Description : Hey I need help What is the median of 14, 19, 26, 28, 18, 21, 30, 27?

Last Answer : I dont know

The six freedoms of the Indian citizens have been enshrined in – a) Art-14-18 b)Art-14-35 c) Art-19 d) Art-21-26

Description : The six freedoms of the Indian citizens have been enshrined in – a) Art-14-18 b)Art-14-35 c) Art-19 d) Art-21-26

Last Answer : c) Art-19

Danielle has taken five math tests so far this year. The tests are out of twenty points, and she has gotten the following scores. 17, 19, 20, 14, 16 What is the mode of her scores?

Description : Danielle has taken five math tests so far this year. The tests are out of twenty points, and she has gotten the following scores. 17, 19, 20, 14, 16 What is the mode of her scores?

Last Answer : Its 17

The mean of 8 article was found to be 15. On rechecking, it was found that two article were wrongly taken as 11 and 9 instead of 16 and 14 respectively. Find the correct mean. A) 17.25 B) 13.65 C) 16.54 D) 16.25

Description : The mean of 8 article was found to be 15. On rechecking, it was found that two article were wrongly taken as 11 and 9 instead of 16 and 14 respectively. Find the correct mean. A) 17.25 B) 13.65 C) 16.54 D) 16.25

Last Answer : D) Calculated mean of 8 articles = 15 Incorrect sum of these 8 articles = (15*8) = 120. Correct sum of these 8 articles = (incorrect sum) - (sum of incorrect articles) + (sum of actual articles) = [120 ... (30)] = 130 Therefore, correct mean = 130/8 = 16.25 Hence, the correct mean is 16.25.

Which Bangladeshi cricketer scores the most runs in ODIs and how many runs ?

Description : Which Bangladeshi cricketer scores the most runs in ODIs and how many runs ?

Last Answer : Bangladesh's Tamim has scored the most runs in ODIs. Score: 154.

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

The average age of 14 mens is increased by 1 year when one of them whose age is 44 years is replaced by a lady. What is the age of the lady? A) 54 B) 58 C) 50 D) 56

Description : The average age of 14 mens is increased by 1 year when one of them whose age is 44 years is replaced by a lady. What is the age of the lady? A) 54 B) 58 C) 50 D) 56 

Last Answer : B) suppose the average age of 14 men is x years and the age of lady is y years. the total age of 14 mens will be = 14x years the average age of 14 mens is increased by 1 years who of them whose age is 44 years is replaced by ... 14x - 44 + y = 14* (x+1) 14x - 44 + y = 14x + 14 y = 14+ 44 y=58

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