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

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
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1 Answer

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
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Related questions

A cricket player had a certain average of runs for his 43 innings. In his 44th innings, he is bowled out for no score on his part. This brings down his average by 4runs. His new average of runs is ? A) 172 B) 182 C) 166 D) 162

Description : A cricket player had a certain average of runs for his 43 innings. In his 44th innings, he is bowled out for no score on his part. This brings down his average by 4runs. His new average of runs is ? A) 172 B) 182 C) 166 D) 162

Last Answer : A) Let the cricket player’s average of runs for his 43 innings be X runs. Total number of runs in 43innings=43x According to the question, [43x + 0]/ 44= x-4 43x = 44x – 176 x = 176 new average of runs = 176-4 =>172

The average weight of M,N and O is 168 kg. If P joins the group, the average weight of the group becomes 160 kg. If another man Q who weighs 6kg more than P replaces M, then the average of N,O,P and Qbecomes158 kg. What is the weight of M? A) 175 B) 140 C) 145 D) 150

Description : The average weight of M,N and O is 168 kg. If P joins the group, the average weight of the group becomes 160 kg. If another man Q who weighs 6kg more than P replaces M, then the average of N,O,P and Qbecomes158 kg. What is the weight of M? A) 175 B) 140 C) 145 D) 150 

Last Answer : D) M+N+O =168 * 3 =504 M+N+O+P=160*4=640……(1) P=136 kg and Q =142 kg N+O+P+Q =158*4 =632 N+O+P =632-142 =490…..(2) from 1 & 2, M=640-490 =150kg

A person’s age is 150% of what it was 15 years ago, but 76% of what it will be after 15years. What is his present age? A) 39.7 B) 45.8 C) 42.8 D) 37.8

Description : A person’s age is 150% of what it was 15 years ago, but 76% of what it will be after 15years. What is his present age? A) 39.7 B) 45.8 C) 42.8 D) 37.8

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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 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

Last Answer : A) Let the highest score be ‘x’ Then lowest score = x-64 Then (25 *20)- 18 * 24 = 432 X+X-64=68 X=66

Average age of A, B, C is 90yrs, however when D joins them, then the average comes down to 80 yrs. Now a new person E, whose age is six-fifth of the age of D, replaces A and the new average is 78. What is age of A? A) 64 B) 68 C) 62 D) 66

Description : Average age of A, B, C is 90yrs, however when D joins them, then the average comes down to 80 yrs. Now a new person E, whose age is six-fifth of the age of D, replaces A and the new average is 78. What is age of A? A) 64 B) 68 C) 62 D) 66

Last Answer : B) according to the question A+B+C/3=90 => A+B+C =270…………..(1) A+B+C +D/4=80 => A+B+C +D=320………(2) From the equation 1 and 2 D=50 E =6/5×50=60 E+B+C +D/4=78 => E+B+C +D =312 B+C +D =312-60=252……………..(3) From 2and 3 The age of A=320-252=68

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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Last Answer : C) Total age of 160 boys = 160* 58= 9280 total age of 30 boys = 30 * 42= 1260 total age of next 50boys = 50 * 36= 1800 average of the remaining boys = [(9280-{1260+1800})/[160 - (30 + 50)] =>9280-3060/80 =>6220/80 =77.75yrs

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

Description : 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 ... integral scores, the maximum possible score of the topper is. A) 85 B) 90 C) 95 D) 100

Last 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 ... b + c + d + e = 290 ⇒46+ 47 + 48+ 49 +e = 290 ⇒e = 100

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 students maks were wrongly entered as 72 instead of 52. Due to that the average marks for the class got increased by half. The number of students in the class is. A) 50 B) 56 C) 46 D) 40

Description : A students maks were wrongly entered as 72 instead of 52. Due to that the average marks for the class got increased by half. The number of students in the class is. A) 50 B) 56 C) 46 D) 40

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The average marks scored by two Class A1 and A2 students are 120 and 130 respectively. If 8 students are moved from Class A2 to Class A1 and the average marks of the two Class get interchanged. Find the total number of students in two Class put together, if the average marks scored by the 8 students who moved is 150 A) 25 B) 30 C) 35 D) 40

Description : The average marks scored by two Class A1 and A2 students are 120 and 130 respectively. If 8 students are moved from Class A2 to Class A1 and the average marks of the two Class get interchanged. Find the total number ... average marks scored by the 8 students who moved is 150 A) 25 B) 30 C) 35 D) 40

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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

Description :  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 ... marks of the remaining students of the class A) 67.54 B) 68.26 C) 66.91 D) 69.06 

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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 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

Description : 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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Saravanan has scored an total of 90% in an examination with five subjects in the ratio 18:14:15:21:22. If 85% is the marks required to get an ‘A’ grade in each subject, then find in how many subjects did he get an ‘A’ grade. Given that the maximum marks in each subject is 120. A) 2 B) 3 C) 4 D) 5

Description : Saravanan has scored an total of 90% in an examination with five subjects in the ratio 18:14:15:21:22. If 85% is the marks required to get an ‘A’ grade in each subject, then find in how many subjects did he get an ‘A’ grade. Given that the maximum marks in each subject is 120. A) 2 B) 3 C) 4 D) 5 

Last Answer : Answer: B) Maximum total marks = 5 120 = 600 Marks scored by Saravanan = 90/100*600=540 Let the marks scored by him in the 5 subjects be 18x,14x,15x,21x and 22x respectively. 18x + 14x + 15x + 21x + ... A' grade i.e. 0.85 120 = 102 Therefore, he has got an A' grade in 3 subjects.

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

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

Description : 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 

Last 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

The average scores of a batch of students in a test is 84. The 36% of students's average score is 57 and 42% of students's score is 84. Then find the average score of remaining 22% of students approximately, A) 128 B) 131 C) 119 D) 106

Description : The average scores of a batch of students in a test is 84. The 36% of students's average score is 57 and 42% of students's score is 84. Then find the average score of remaining 22% of students approximately, A) 128 B) 131 C) 119 D) 106 

Last Answer : Answer: A)  let the number of students be 100, then, 100*84 = 36*57 + 42*84 + 22*x 22x = 8400-2052-3528 x = 128.18

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

Description : 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

Last Answer : C) Let the requird mean score be ‘x’ Then (30*70)+(15*41)+55x = 63*100 2100+615+55x=6300 2715+55x=6300 55x=3585 X=65.181

After replacing an old person by a new person, it was found that the average age of five persons of a assembly is the same as it was 6 years ago. What is the difference between the ages of the replaced and the new persons? A) 24 B) 36 C) 30 D) 15

Description : After replacing an old person by a new person, it was found that the average age of five persons of a assembly is the same as it was 6 years ago. What is the difference between the ages of the replaced and the new persons? A) 24 B) 36 C) 30 D) 15 

Last Answer : C) Let the ages of the five persons at present be a, b, c, d & e years. And the age of the new persons be f years. So the new average of five members' age = (a + b + c + d + f)/5 - ... - e + f) = 5x Solving e - f = 30 years. Thus the difference of ages between replaced and new person = 30years.

The average age of the men in a team is 76 years and that of the women is 75 years. The average age for the whole team is. A) 75 yrs B) 75.5 yrs C) Cannot be computed with the given information D) None of the above

Description : The average age of the men in a team is 76 years and that of the women is 75 years. The average age for the whole team is. A) 75 yrs B) 75.5 yrs C) Cannot be computed with the given information D) None of the above

Last Answer : C) Clearly to find the average, we ought to number of men, women or total persons in the class, neither of which has been given. So the data provided is inadequate.

The average monthly salary of 24 labours and 6 supervisor in a factory was Rs. 1200. When one of the supervisor whose salary was Rs. 1440, was replaced with a new supervisor, then the average salary of the team went down to 1160. What is the salary of the new supervisor? A) 170 B) 420 C) 390 D) 240

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Last Answer : D) The total salary amount = 30 1200=36000 The salary of the exiting supervisor = 1440 Therefore, the salary of 24 labours and the remaining 5 supervisors: =36000-1440=34560 When a new ... 34800=1200 less than that of the old supervisor who left the company, which is equal to 1440−1200=240

A professional basketball player averaged 1919.35 points scored each season. About how many points did he score during his twenty year career?

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How many students studying Dentistry know only either English or Hindi? (1) 166 (2) 162 (3) 308 (4) 198 (5) 248

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Last Answer : 4) 198

According to which Article of Constitution of India, the Chief Minister is appointed by the Governor of A State? (1) Article 163 (2) Article 164 (3) Article 165 (4) Article 166

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Last Answer : (2) Article 164 Explanation: Under Article 164 of Constitution of India, the chief minister is appointed by the governor of a state.

In an exam the average marks obtained by a candidate is 82 per paper. If he had obtained 32 marks more in science paper & 28more marks in social paper, then his average per paper is increased by 15 marks. Then how many papers when there in examination? A) 10 B) 6 C) 4 D) 8

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Last Answer : C)  let the number of paper be x total mark earned him = 82x then,, 82x+32+28= 97x 15x=60 x= 4 =number of subjects

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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What is the average score in an exam taken by 500 students where the minimum score is 200? a. Half the students scored above 700. b. Half the students scored below 700. c. The maximum score in the exam was 850, scored by exactly 42 students.

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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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Mr. Sebastine, a famous author, recently got his new novel released. To his utter dismay, he found that for the 1974 pages on an average there were 3 mistakes in every page. While, in the first 1024 pages there were only 2122 mistakes, they seemed to increase for latter pages. Find the average number of mistakes per page for the remaining pages. A) 5 B) 2 C) 3 D) 4

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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

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

Last Answer :  B)  Total weight of (72 + 88) = (72*80) + (88 * 70)kg = (5760 + 6160)kg  = 11,920kg  Average weight of the whole class = 11920 / 160  = 74.5 kg

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

In an competitive examination, the average was found to be 72 marks. After detecting the errors the marks of 94 candidates had to be charged from 90 to 66 each, and the average is reduced to 64 marks. Find the total number of candidates who took the exam. A) 282 B) 382 C) 828 D) 200

Description : In an competitive examination, the average was found to be 72 marks. After detecting the errors the marks of 94 candidates had to be charged from 90 to 66 each, and the average is reduced to 64 marks. Find the total number of candidates who took the exam. A) 282 B) 382 C) 828 D) 200 

Last Answer :  A)  let the number of candidates be A total marks of candidates = 72A after detecting error the change of marks for one candidate = 90 - 66 = 24 marks change of marks for 94 candidates = 94*24 = ... after detecting error of N candidates = 64A then, 72A- 2256= 64A , 8A=2256 A=282

1/2of a certain travel is covered at the rate of 30km/hr, one-third at the rate of 40 km/hr and the rest at 35km/hr. Find the average speed for the whole travel. A) 33 1/3 B) 33 3/5 C) 34 3/7 D) None of these

Description : 1/2of a certain travel is covered at the rate of 30km/hr, one-third at the rate of 40 km/hr and the rest at 35km/hr. Find the average speed for the whole travel. A) 33 1/3 B) 33 3/5 C) 34 3/7 D) None of these

Last Answer : B) let the total travel be X km.  Then X/2 km at the speed of 30 km/hr and X/3 km at 40 km/hr and the rest distance( X - X/2 - X/3) =1/6 X at the speed of 35km/hr. Total time taken during the travel ... 2*30 hrs+ X/3*40hrs+ X/6*35hrs = 5X/168 hrs Average speed =X/(5x/168) = 168/5 =33 3/5km hr

The average of 12 numbers is 7.9. The average of 5 of them is 6.8, while the average of other five is 7.7. what is the average of the remaining 2 numbers? A) 10.25 B) 11.15 C) 12.65 D) 13.25

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In a famous restaurant rooms were numbered from 401 to 430, each room gives an earning of Rs. 4675/ day for the first 15 days a month and for the later half, Rs. 5370/ day per room. Find the average income room per day over the month of 30 days? A) 5700.5 B) 5872.7 C) 5900.5 D) 5022.5

Description : In a famous restaurant rooms were numbered from 401 to 430, each room gives an earning of Rs. 4675/ day for the first 15 days a month and for the later half, Rs. 5370/ day per room. Find the average income room per day over the month of 30 days? A) 5700.5 B) 5872.7 C) 5900.5 D) 5022.5  

Last Answer : Answer: D) Total number of rooms = 30 earning in 1st 15 days for 30 rooms  = 15 * 4675* 30 = Rs. 2103750 earning in 2nd 15 days for 30 rooms  = 15 * 5370 * 30 = Rs. 2416500 Average income per ... day over the month of 30 days  = [ 2103750 + 2416500] / [ 30 * 30]  = Rs. 5022.5

How to Throw a Javelin

Description : How to Throw a Javelin

Last Answer : How to Throw a Javelin Javelin throwing may seem simple at first, but much technique is involved for you to make a successful throw. Without the right form, the javelin won't travel far, possibly ... 's just a matter of time before you find yourself on the podium, being awarded a gold medal.

The average age of Mr and Mrs Rahim at the time of their marriage in 1981 was 56 yrs. On the occasion of their anniversary in 1986, they observed that the average age of their family had come down by 15 yrs compared to their age at the time of their marriage. This was due to the fact that their daughter varshini was born. What was the age of varshini in 1990? A) 6 B) 5 C) 4 D) 1

Description : The average age of Mr and Mrs Rahim at the time of their marriage in 1981 was 56 yrs. On the occasion of their anniversary in 1986, they observed that the average age of their family had come down by 15 yrs compared to ... varshini was born. What was the age of varshini in 1990? A) 6 B) 5 C) 4 D) 1

Last Answer : B) Sum of the ages of Mr and Mrs Rahim=56×2=112years Sum of the ages in 1986=41×3=123 years Sum of the ages of Mr and Mrs Rahim =112+10=122 Daughters age in 1986=123-122=1 years  Daughters age in 1990 =1+4=5 years.

Gowshik is going to market from his home by bike at a speed of 20kmph. While he comes back to his home with a speed of X kmph, what should be the value of x so that his average speed is 24kmph ? A) 24 B) 34 C) 30 D) 32

Description : Gowshik is going to market from his home by bike at a speed of 20kmph. While he comes back to his home with a speed of X kmph, what should be the value of x so that his average speed is 24kmph ? A) 24 B) 34 C) 30 D) 32

Last Answer : C) Let speed be X Average speed =2xy/(x+y) =>2×X×20/(x+20)=24 40x=24x+480 16x=480 X=30 kmph.

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

Monica's average expenses for 4 days is Rs 6.0. She spent Rs 7.70 on first day, Rs 6.30 on second day. If she spent 10 on third day , How much did she spend on the 4th day? (a) Rs 2 (b) Rs 3 (c) Rs 4 (d) Rs 0

Description : Monica's average expenses for 4 days is Rs 6.0. She spent Rs 7.70 on first day, Rs 6.30 on second day. If she spent 10 on third day , How much did she spend on the 4th day? (a) Rs 2 (b) Rs 3 (c) Rs 4 (d) Rs 0 

Last Answer : Rs 0 Hint : Required Amount = 24 - ( 7.70 + 6.30 + 10)

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

The average of 3,8,7 and a is 6 and the average of 19,2,7, a and b is 11. What is the value of b? A) 11 B) 21 C) 31 D) 41

Description : The average of 3,8,7 and a is 6 and the average of 19,2,7, a and b is 11. What is the value of b? A) 11 B) 21 C) 31 D) 41 

Last Answer : B) We have (3+8+7+a+6/5)=6 24 =a = 30 A=6 Also (19+2+7+a+b/5)=11 19+2+7+6+b=55 34+b=55 b=21

The average of 7 consecutive odd numbers is 41. Find the largest of these numbers A) 40 B) 57 C) 47 D) 43

Description : The average of 7 consecutive odd numbers is 41. Find the largest of these numbers A) 40 B) 57 C) 47 D) 43

Last Answer : C) Let the numbers be x , x+2,x+4,x+6, x+8, x+10 and x+12 Then x+x+2+x+4+x+6+x+8+x+10+x+12/7 7x+42=287 7x=245 X=35 Largest number = x+12 = 35+12 =47

The average state of the city in the first 4 days of a month was 116 degrees. The average for the second, third, fourth and fifth days was 120 degrees. If the state of the first and fifth days were in the ratio 7:8 then what is the state on the fifth day. A) 112degrees B) 128degrees C) 132degrees D) 130degrees

Description : The average state of the city in the first 4 days of a month was 116 degrees. The average for the second, third, fourth and fifth days was 120 degrees. If the state of the first and fifth days ... :8 then what is the state on the fifth day. A) 112degrees B) 128degrees C) 132degrees D) 130degrees

Last Answer :  B)  Sum of states on 1st, 2nd, 3rd, 4th days = (116 *4) = 464 =======>A Sum of states on 2nd, 3rd, 4th and 5th days = (120 * 4)= 480 ======>b From a and b Subtracting each other ... be 7x and 8x degrees respectively Then 8x - 7x = 16 x = 16 therefore state on the 5th day = 8x = 128 degrees.

20 years ago, the average age of a family of 4 member was 48 years. 2 children having been born (with the age difference of4 yrs) the present average age of the family is the same. The present age of the youngest child is. A) 2yrs B) 4 yrs C) 6yrs D) 8yrs

Description : 20 years ago, the average age of a family of 4 member was 48 years. 2 children having been born (with the age difference of4 yrs) the present average age of the family is the same. The present age of the youngest child is. A) 2yrs B) 4 yrs C) 6yrs D) 8yrs

Last Answer : C) Total age of 4 members, 20 yrs ago = (48 * 4)=192 yrs Total age of 4 members now = 192 + 20 * 4 = 272 yrs Total age of 6 members now = (48*6)=288yrs Sum of the age of children = 288 - 272 = 16 ... the age of the elder child 4 +x So x + 4 +x = 16 2x=12 X=6 Age of younger child = 6years

The average age of Krishnan family of 10 members was 34 yrs, 3 yrs ago. A baby having been born, the average age of the family is the same today. The present age of the baby. A) 1yr B) 3yrs C) 4yrs D) 2yrs

Description : The average age of Krishnan family of 10 members was 34 yrs, 3 yrs ago. A baby having been born, the average age of the family is the same today. The present age of the baby. A) 1yr B) 3yrs C) 4yrs D) 2yrs

Last Answer : C) Total age of 10 members , 3 yrs ago = (34*10)yrs = 340 yrs Total age of 10 members now = (340 +3*10) =370yrs Total age of 11 members now = (34*11)yrs =374yrs Therefore age of the baby= 374 – 370 =4yrs.