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

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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1 Answer

Answer :

B)

Sum of the remaining 2 numbers = [(12 * 7.9) – (586.8) – (5*7.7)]

 = 94.8 – 34 – 38.5

 = 22.3

 Required average = 22.3 / 2

 = 11.15
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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

Description : 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 ... Rs. 30, Rs.20 B) Rs. 8, Rs. 30 C) Rs. 10, Rs. 16 D) Rs. 12, Rs. 20

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

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.

There are five numbers, the second number is 25% more than the first or third number, the 4th number is 5/4 of the third number and the fifth number is 3/2of the 3rd number. What is the average of 5 numbers if the first number is 10? A) 12.5 B) 12 C) 13 D) 10

Description : There are five numbers, the second number is 25% more than the first or third number, the 4th number is 5/4 of the third number and the fifth number is 3/2of the 3rd number. What is the average of 5 numbers if the first number is 10? A) 12.5 B) 12 C) 13 D) 10

Last Answer : B) Let us consider 5 numbers are A,B,C,D and E B=125/100×A B=125/100×C 125/100×A=125/100×C A/C=1/1 D=5/4×c E=3/2*C First number is 10, A=10=x B=25% ×10+10=12.5 C=10 D=5/4×10=12.5 E=3/2×C=15 Required average=10+12.5+10+12.5+15/5=12

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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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

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).

If the average of 4 observations x, x+1, x+2, x+3 is 12 then the average of last 2 observations is A) 10 B) 11 C) 12 D) 13If the average of 4 observations x, x+1, x+2, x+3 is 12 then the average of last 2 observations is A) 10 B) 11 C) 12 D) 13

Description : If the average of 4 observations x, x+1, x+2, x+3 is 12 then the average of last 2 observations is A) 10 B) 11 C) 12 D) 13

Last Answer :  D) We have (x+x+1+x+2x+3/40=12 4x+6=48 4x=42 x=10.5 so the numbers are 10.5,11.5, 12.5, 13.5 required average = 12.5+13.5/2 =26/2 =13.

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

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.

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

Description : 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

Last Answer : C) According to the question, 63=3/7×49+4/7×y 63=147/7+4y/7 441=147+4y 4y=294 Y=73.5

11 girls went to a canteen. 10 of them spent 10 each and the 11th girl spent 25 more than the average expenditure of all. Find the total money spent by them? A) 172 .5 B) 178 .5 C) 137.5 D) 165.5

Description : 11 girls went to a canteen. 10 of them spent 10 each and the 11th girl spent 25 more than the average expenditure of all. Find the total money spent by them? A) 172 .5 B) 178 .5 C) 137.5 D) 165.5

Last Answer : C)  Let the average money spent by the 11girls = x  Money spent by the 11th girl = (x + 25)  Money spent by the other 10 girls = (10*10) = 100  Total money spent by the 11 girls = (100 + x + 25) =(x + ... = (x + 125)/ 11 =>10x=125  x =12.5 Total money spent by the 11 girls = 11*12.5 = 137.5.

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

Description : 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 

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 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

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

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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.

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

Description : 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 ... the average number of mistakes per page for the remaining pages. A) 5 B) 2 C) 3 D) 4

Last Answer : D According to the question X is the remaining pages average. 2122+(1974-1024)×x/1974=3 950x=5922-2122=3800 X=3800/950=4

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

There were 32 boys in a hostel. If the number of boys be increased by 15, then the expenditure on food increases by Rs. 43 per day while the average of expenditure of boys is reduced by Rs. 2. What was the initial expenditure on food per day? A) Rs.108.2 B) Rs.125.6 C) Rs.135.8 D) Rs.144.5

Description : There were 32 boys in a hostel. If the number of boys be increased by 15, then the expenditure on food increases by Rs. 43 per day while the average of expenditure of boys is reduced by Rs. 2. What was the initial expenditure on food per day? A) Rs.108.2 B) Rs.125.6 C) Rs.135.8 D) Rs.144.5

Last Answer : Answer: A)  Let initial expense of 32 boys = Rs. x average expense of 32 boys = x/32 average expense of 47 boys = (x + 43)/ 47 then, (x/32) - ((x+43)/47) = 2  x= Rs.108.28

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

Description : 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

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

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 

Last Answer : D) average of entire class = 69marks average of 28 % of the students = 66 marks average of 10% of the students = 77 marks then % of remaining students = (100- 10 - 28) = 62% let the average of 62% of the ... (62*x)+770+1848= 6900 62 * x = 6900-770-1848 62*x = 4282 x = 69.06

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

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 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

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

Description : 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 

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

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50'. What number is missing? -Riddles

Description : 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50'. What number is missing? -Riddles

Last Answer : 22

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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Last Answer : C) 150 students average 150×40/100=60 years According to the question, 150×60=50×32+50×36+50×X 50×X=9000-1600-1800 50x=5600 X=112

When a girl weighing 90 kgs left a class, the average weight of the remaining 119 students increased by 400 g. What is the average weight of the remaining 119 students? A) 124 B) 138 C) 145 D) 116

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Last Answer :  B Let the average weight of the 119 students be X. Therefore, the total weight of the 119 of them will be 119X. The questions states that when the weight of this student who left is added, the total ... , the average weight decreases by 0.4 kgs. (119X+90)/120=X−0.4 ⇒119X+90=120X−48  ⇒X=138

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

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If the average of two numbers are 138& product is 2241. Then find the difference of both numbers? A) 225.40 B) 259.25 C) 267.81 D) 294.57

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Last Answer : B) let the two number be a & b sum of two numbers, a+b = 138*2 =276---1 product of two numbers, a*b=2241 (a+b)2 = a2 + 2ab + b2 ------2 (a-b)2 = a2 -2ab +b2 ------3 solving 2 & 3, 76176 - (a-b)2 = 4*2241 a-b = 259.25

The average of 5 numbers is 9 and the average of the last three numbers is 5.Find the average of the first two numbers.

Description : The average of 5 numbers is 9 and the average of the last three numbers is 5.Find the average of the first two numbers.

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The term total milk solids describes the remaining ____ percent of milk. a. 10–11 c. 5–7 b. 12–13 d. 8–9

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Last Answer : b. 12–13

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

Description :  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 

Last Answer : Answer: D) let the initial strength of the batch be X total height of the batch initially = 344X total height of 16 new students = 320 * 16 = 5120 cm Then, the average height of X+ 16students = 332  332 = [5120 + 344 X]/ [X + 16]  X= 16 students

What is the output of the following? i = 5 while True: if i%0O9 == 0: break print(i) i += 1 a) 5 6 7 8 b) 5 6 7 8 9 c) 5 6 7 8 9 10 11 12 13 14 15 …. d) error

Description : What is the output of the following? i = 5 while True: if i%0O9 == 0: break print(i) i += 1 a) 5 6 7 8 b) 5 6 7 8 9 c) 5 6 7 8 9 10 11 12 13 14 15 …. d) error

Last Answer : d) error

The average marks obtained by a student in Tamil, english, maths, science, and social together is 65% above the average mark obtained in maths and science together. How many more marks exceeded by the average of maths and science together than the sum of the tamil and english together? A) 50 B) 65 C) Data insufficient D) None of these

Description : The average marks obtained by a student in Tamil, english, maths, science, and social together is 65% above the average mark obtained in maths and science together. How many more marks exceeded by the average of ... of the tamil and english together? A) 50 B) 65 C) Data insufficient D) None of these

Last Answer : C) According to the question  There is no other information about the marks in any one of the subject. The data gives is insufficient to answer the question.

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

The average monthly expenditure of Mr. Abi family for the first 4 months is Rs.4210, next 4 month expenditure Rs.4450 for the last 4 months Rs. 4360 . If his family saves Rs. 6800 for 12 months, find the average monthly income of the family for the 12 months? A) Rs. 4707.25 B) Rs.4564.75 C) Rs. 4906.66 D) Rs. 4806.50

Description : The average monthly expenditure of Mr. Abi family for the first 4 months is Rs.4210, next 4 month expenditure Rs.4450 for the last 4 months Rs. 4360 . If his family saves Rs. 6800 for 12 months, find the average monthly ... the 12 months? A) Rs. 4707.25 B) Rs.4564.75 C) Rs. 4906.66 D) Rs. 4806.50

Last Answer : C) Mr Abi family first 4 month expenditure=4210*4=16840 Mr Abi family next 4 month expenditure=4450*4=Rs17800 Mr Abi family last 4 month expenditure=4360*4=Rs.17440 Mr Abi family total ... 17440+6800=Rs.58880. Mr Abi family average expenditure in 12 months =58880/12=Rs.4906.66

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

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

Description : 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

Last Answer : B) siva:sasi=x:3x siva+sasi/2=somu + ramu/2 X+3X=3/2×X+25 5X=50 X=10 Siva’s age is 10 then somu's age = 3/2×10=15 years

There were 126 people in a Inn. Due to the entry of 39 new people, the expenses of the inn increase by Rs. 93 per day while the average expenditure per head diminished by Rs. 9. What was the original expenditure of the inn? A) Rs. 6343 B) Rs.5098 C) Rs.6215 D) Rs. 5476

Description : There were 126 people in a Inn. Due to the entry of 39 new people, the expenses of the inn increase by Rs. 93 per day while the average expenditure per head diminished by Rs. 9. What was the original expenditure of the inn? A) Rs. 6343 B) Rs.5098 C) Rs.6215 D) Rs. 5476

Last Answer : Answer: B)  Let per day average expense = x 126x + 93 = 165(x-9) 126x+93=165x-1485 1578=39x X=40.46 original expenditure = 126*40.46 = Rs 5098

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

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

Last Answer :  D)  let the number of students in Class A1 be x and class A2 be y. Total marks scored by the students will be 120x and 130y, the average gets interchanged after moving student from y Thus we get, 130y- ... (x+8) 120x+1200=130x+1040 10x=160 X=16 Thus the total number of students =24+16=40

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 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.

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

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 age of 11 daughters of a family is 12yrs. Average age of the daughters together with their parents is 27yrs. If mother is 14yrs younger than father. Then find the father's age A) 127 yrs B) 116.5 yrs C) 168yrs D) 170yrs

Description : The average age of 11 daughters of a family is 12yrs. Average age of the daughters together with their parents is 27yrs. If mother is 14yrs younger than father. Then find the father's age A) 127 yrs B) 116.5 yrs C) 168yrs D) 170yrs

Last Answer : Answer: B) Total age of 11 daughters of a family = 11*12 = 132 father age = 14 +mother age total age of the family (11 daughters + father + mother) is, 132+mother +14+ mother = 27 * 13 mother = 102.5 father age = 14+102.5 =116.5years

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.

I have some matchsticks set up like this: I + XI = X. If you do not know Roman numerals, here is a quick starter: I=1 II=2 III=3 IV=4 V=5 VI=6 VII=7 VIII=8 IX=9 X=10 XI=11 XII=12 XIII=13 XIV=14 XV=15...... Of course, I + XI = X is wrong. You are told to move the minimum number of matchsticks to make the question correct. So, how many do you have to move? -Riddles

Description : I have some matchsticks set up like this: I + XI = X. If you do not know Roman numerals, here is a quick starter: I=1 II=2 III=3 IV=4 V=5 VI=6 VII=7 VIII=8 IX=9 X=10 XI=11 ... are told to move the minimum number of matchsticks to make the question correct. So, how many do you have to move? -Riddles

Last Answer : Most people would get to one move at the least, but the real answer is zero moves. If you rotate the whole question around 180 degrees, it will display this: X = IX + I

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

Description : 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

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

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.