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

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
by

1 Answer

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
by

Related questions

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

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.

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 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

Description : 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

Last 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

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.

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 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).

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

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 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

Description : 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 ... the cricketer average score in all the matches together? A) 336 B) 146 C) 168 D) 189

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 .

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 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.

Last Answer : Sum of 5 numbers = 9 x 5 = 45 Sum of last three numbers = 15 The average of 1st two numbers = 45-15/2 = 30/2 = 15.

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.

Ronit spends his money from his saving in different way that is he spends 25% on travel,15% on food,20% on wages and 17% on shopping a clothes and after that all expenditure he saved 6900.Find the how much he spent on clothes. A) 4400 B) 5100 C) 4000 D) 6000

Description : Ronit spends his money from his saving in different way that is he spends 25% on travel,15% on food,20% on wages and 17% on shopping a clothes and after that all expenditure he saved 6900.Find the how much he spent on clothes. A) 4400 B) 5100 C) 4000 D) 6000 

Last Answer :  B) Let the total income of Ronit x then total expenditure from income X x (25%+15%+20%+17%)=X*77% X x 77% =Total savings= X x 23%  X = 6900 x 100 / 23 = 30000 expenditure on clothes = 17% so,30000 x 17 / 100 = 5100

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.

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

Description : 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

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 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

Find the average of even numbers above 1000 & upto 1 lakh? A) 50500 B) 50000 C) 49500 D) 51500

Description : Find the average of even numbers above 1000 & upto 1 lakh? A) 50500 B) 50000 C) 49500 D) 51500

Last Answer : Answer: A) No of even numbers upto 1lkah = 50000 No of even numbers upto 1000 = 500 No of even numbers above 1000& upto 1 lakh= 49500 sum of first N even numbers = N*N sum ... - 500*500 = 2499750000 average of even numbers above 1000& upto 1 lakh = 2499750000/ 49500 = 50500

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 monthly income of certain fashion designers is F and that other workers is N. The number of fashion designer workers is 33 times that of other workers. The average monthly income (in rs) of all the worker is. A) F+N/2 B) F+33N/12 C) 33F+N/22 D) (33F+N)/34

Description : The average monthly income of certain fashion designers is F and that other workers is N. The number of fashion designer workers is 33 times that of other workers. The average monthly income (in rs) of all the worker is. A) F+N/2 B) F+33N/12 C) 33F+N/22 D) (33F+N)/34

Last Answer : D) Let the number of other workers be ‘x’. The number of fashion design workers = 33x Total number of workers = 34x Average monthly income = (F * 33x)+(N*x)/34x =x(33F + N)/34x =(33F + N)/34.

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

If the arithmetic mean of 150 numbers is calculated 70. If each number is increased by 10, then mean of new number is. A) 70 B) 80 C) 90 D) 60

Description : If the arithmetic mean of 150 numbers is calculated 70. If each number is increased by 10, then mean of new number is. A) 70 B) 80 C) 90 D) 60

Last Answer : B) Arithmetic mean of numbers = 70 Sum of 150 numbers = (70 * 150) =10500 Total increase = (150 * 10) = 1500 Increased sum = 10500 + 1500 = 12000 Increased average = 12000/150 =80

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

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

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

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

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.

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

Description : 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 ... what is the average mark obtained by mahi, nitesh and ritesh? A) 247 B) 248 C) 264 D) 284

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.

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

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

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.

The average age of a morning class is 108, if the average age of 72 ladies in the class is as same as the total average and the number of gents in the class is 2 more than the ladies in the class, what is the average age of gents in the class? A) 152 B) 108 C) 156 D) 154

Description : The average age of a morning class is 108, if the average age of 72 ladies in the class is as same as the total average and the number of gents in the class is 2 more than the ladies in the class, what is the average age of gents in the class? A) 152 B) 108 C) 156 D) 154 

Last Answer : B) According to question, (72+74)X=72×108+74X 146x-74x=7776 72x=7776 X=7776/72= 108years

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

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.

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

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.

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

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

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 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.

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

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

The average age of students of a college is 31.6 yrs. The average age of boys in the class is 32.8 yrs and that of girls is 30.8. The ratio of number of boys to the number of girls in the class. A) 2:3 B) 3:1 C) 1:3 D) 3:2

Description : The average age of students of a college is 31.6 yrs. The average age of boys in the class is 32.8 yrs and that of girls is 30.8. The ratio of number of boys to the number of girls in the class. A) 2:3 B) 3:1 C) 1:3 D) 3:2

Last Answer : A) Let the ratio be D:1.then, (D * 32.8) + ( 1* 30.8)= (D+1) * 31.6 32.8 D +30.8 = 31.6 D + 31.6 32.8 D – 31.6D = 31.6 – 30.8 1.2D = 0.8 D = 0.8/1/2 = 0.4/0.6 D = 2/3 Required ratio = 2/3 :1 = 2:3.

The average of the 3 digit number, which remain the same when the digits interchange their positions is. A) 444 B) 555 C) 666 D) 777

Description : The average of the 3 digit number, which remain the same when the digits interchange their positions is. A) 444 B) 555 C) 666 D) 777 

Last Answer : B) Average = (111+222+333+444+555+666+ 777+888+999)/9 =(111+999)+(222+888)+ (333+777)+(444+666)+ 555/9 =(4*1110)+555/9 =4440 + 555/9 =4995/9 =555.

The average of marks obtained by 60 students in a computer examination is 18. If the average marks of passed students is 20and that of the failed students is 8, what is the number of students who passed the examination? A) 100 B) 75 C) 50 D) 85

Description : The average of marks obtained by 60 students in a computer examination is 18. If the average marks of passed students is 20and that of the failed students is 8, what is the number of students who passed the examination? A) 100 B) 75 C) 50 D) 85 

Last Answer : C Let the number of passed students be x. Then total marks = 60 × 18= 20x + (60– x) × 8 1080= 20x + 480– 8x 12x = 600 ∴ x = 50 ∴ number of passed students = 50

The average salary per head of all the employees in a company is Rs. 180. The average salary of 24 employees is Rs. 1040and the average salary per head of the rest is Rs. 160. Find the total number of employees in the workshop. A) 2841 B) 1056 C) 1195 D) 2145

Description : The average salary per head of all the employees in a company is Rs. 180. The average salary of 24 employees is Rs. 1040and the average salary per head of the rest is Rs. 160. Find the total number of employees in the workshop. A) 2841 B) 1056 C) 1195 D) 2145

Last Answer :  B) Average = Sum of observations/Number of observations Let the number of employees be x. Given, the average salary per head of all the employees in a company is Rs. 180 So, the sum of salary of all the employees in ... (x - 24) = 180x ⇒ 24960+ 160x - 3840 = 180x ⇒ 20x = 2112 ⇒ x = 1056

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

Description : 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 ... is the average age of the remaining students in the class? A) 102 B) 118 C) 112 D) 108

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

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

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