Question 1

P, Q, R, S, T and U are six consecutive odd numbers and their average is 52. What is the product of P and U ?

Accepted Answer

Let x, x+2, x+4, x+6, x+8 and x+10 are six consecutive odd numbers. Given that their average is 52 Then, x + x+2 + x+4 + x+6 + x+8 + x+10 = 52×6 6x + 30 = 312 x = 47 So Product = 47 × 57 = 2679

Question 2

In a class of 78 students 41 are taking French, 22 are taking German. Of the students taking French or German, 9 are taking both courses. How many students are not enrolled in either course?

Accepted Answer

You could solve this by drawing a Venn diagram. A simpler way is to realize that you can subtract the number of students taking both languages from the numbers taking French to find the number taking only French. Likewise find those taking only German. Then we have:Total = only French + only German + both + neither 78 = (41-9) + (22-9) + 9 + neither. Not enrolled students = 24

Question 3

The average of 9 results is 50. The average of the first four results is 52 and the average of the last four results is 49. what is the fifth result?

Accepted Answer

Total of first four results = 52 x 4 = 208 Total of Last four results = 49 x 4 = 196 Hence the fifth result = 450 - (208 + 196) = 46

Question 4

The average price of three items of furniture is Rs. 15000. If their prices are in the ratio 3:5:7, the price of the cheapest item is :

Accepted Answer

Let their prices be 3x, 5x and 7x. Then, 3x + 5x + 7x = (15000 * 3) or x = 3000. Cost of cheapest item = 3x = Rs. 9000.

Question 5

In a mens hostel, there were 95 students. To accommodate 30 more students the average is decreased by rupees 5. But total expenditure increased by Rs.500. Find the total expenditure of the hostel now ?

Accepted Answer

Let the expenditure on the 95 students be 'x' Then, 95x + 500 = 120(x – 5) 95x + 500 = 120x - 600 120x - 95x = 1100 25x = 1100 => x = 44 Therefore, total expenditure of the hostel becomes 95 x 44 + 500 = 4680.

Question 6

David obtained 76, 65, 82, 67 and 85 marks (out of 100) in English, Mathematics,Physics, Chemistry and Biology What are his average marks ?

Accepted Answer

Average = (76 + 65 + 82 + 67 + 85 )/ 5 = 375/5 = 75.

Question 7

The mean of 100 observations is 40. It is found that an observation 84 was misread as 48. Then the correct mean is ?

Accepted Answer

Given, mean of 100 observations is 40 => Total of 100 observations = 40 x 100 = 4000 84 is misread as 48 => Difference = 84 - 48 = 36 => Now, new total of 100 observations = 4000 + 36 = 4036 Correct Mean = 4036/100 = 40.36

Question 8

Without any stoppage, a person travels a certain distance at an average speed of 42 km/h, and with stoppages he covers the same distance at an average speed of 28 km/h. How many minutes per hour does he stop?

Accepted Answer

Let the total distance to be covered is 84 kms. Time taken to cover the distance without stoppage = 84/42 hrs = 2 hrs Time taken to cover the distance with stoppage = 84/28 = 3 hrs. Thus, he takes 60 minutes to cover the same distance with stoppage. Therefore, in 1 hour he stops for 20 minutes.

Question 9

The average age of 7 boys is increased by 1 year when one of tem whose age is replaced by a girl. What is the age of the girl ?

Accepted Answer

Let the avg age of 7 boys be 'p' years Let the age of the girl be 'q' years From given data, The age of 7 boys = 7p years Now the new average = (p + 1) when 22 yrs is replaced by q Now the age of all 7 will become = 7(p + 1) yrs Hence, 7p - 22 + q = 7(p + 1) yrs 7p - 22 + q = 7p + 7 q = 22 + 7 = 29 Therefore, the age of girl = q = 29 years.

Question 10

In Arun's opinion, his weight is greater than 65 kg but leas than 72 kg. His brother does not agree with Arun and he thinks that Arun's weight is greater than 60 kg but less than 70 kg. His mother's view is that his weight cannot be greater than 68 …

Accepted Answer

Let Arun's weight be X kg. According to Arun, 65 < X < 72. According to Arun's brother, 60 < X < 70. According to Arun's mother, X < 68. The values satisfying all the above conditions are 66 and 67. Required average = (66 + 67) / 2 = 66.5 kg

Question 11

The average age of 15 students of a class is 15 years. Out of these, the average age of 5 students is 14 years and that of the other 9 students is 16 years, The age of the 15th student is :

Accepted Answer

Age of the 15th student = [15 * 15 - (14 * 5 + 16 * 9)] = (225-214) = 11 years.

Question 12

The ratio of speeds of a car, a train and a bus is 5:9:4. The average speed of the car, the bus and the train is 72 km/hr together. What is the average speed of the car and the train together ?

Accepted Answer

Let the speeds of the car, train and bus be 5x, 9x and 4x km/hr respectively. Average speed = 5x + 9x + 4x/3 = 18x /3 = 6x km/hr. Also, 6x = 72 => x = 12 km/hr Therefore, the average speed of the car and train together is = 5 x + 9 x 2 = 7x = 7 x 12 = 84 km/hr.

Question 13

Out of three numbers, the first is twice the second and is half of the third. If the average of the three numbers is 56, what will be the sum of the highest number and the lowest number?

Accepted Answer

Let the three numbers be x, y, z. From the gien data, x = 2y ....(1) x = z/2 => z = 2(2y) = 4y .....(From 1) ...........(2) Given average of three numbers = 56 Then, x + y + z 3 = 56 2 y + y + 4 y 3 = 56 ( From 1 & 2 ) 7 y = 56 x 3 y = 24 Now, x = 2y => x = 2 x 24 = 48 z = 4y = 4 x 24 = 96 Now, the highest number is z = 96 & smallest number is y = 24 Hence, required sum of highest number and smallest number = z + y = 96 + 24 = 120.

Question 14

The average age of a group of persons going for picnic is 16 years. Twenty new persons with an average age of 15 years join the group on the spot due to which their average age becomes 15.5 years. The number of persons initially going for picnic is :

Accepted Answer

Let the initial number of persons be x. Then, 16x + 20 * 15 = 15.5 (x + 20) <=> 0.5x = 10 <=> x = 20.

Question 15

A motorist travels to a place 150 km away at an average speed of 50 km/hr and returns at 30 km/hr. His average speed for the whole journey in km/hr is :

Accepted Answer

Average speed = (2xy) /(x + y) km/hr = (2 * 50 * 30) / (50 + 30) km/hr. 37.5 km/hr.

Question 16

After replacing an old member by a new member, it was found that the average age of five members of a club is the same as it was 3 years ago. The difference between the ages of the replaced and the new members?

Accepted Answer

Let avg ages of 5 members at present is G And age of new member is m and an older is n Required = (5G + m - n)/5 = G - 3 m - n = 15 years

Question 17

The average monthly salary of 20 employees in an organisation is Rs. 1500. If the manager's salary is added, then the average salary increases by Rs. 100. What is the manager's monthly salary ?

Accepted Answer

Manager's monthly salary Rs. (1600 * 21 - 1500 * 20) = Rs. 3600.

Question 18

In a mixture of three varities of oils , the ratio of their weight is 4 : 5 : 8. If 5 kg of oils of the first variety, 10 kg of the second variety and some quantity of oils of the third variety are added to the mixture, the ratio of the weights of t…

Accepted Answer

Let the ratio of initial quantity of oils be 'x' => 4x, 5x & 8x. Let k be the quantity of third variety of oil in the final mixture. Let the ratio of initial quantity of oils be 'y' From given details, 4x + 5 = 5y ..... (1) 5x + 10 = 7y .....(2) 8x + k = 9y ......(3) By solving (1) & (2), we get x = 5 & y = 5 From (3) => k = 5 Therefore, quantity of third variety of oil was 9y = 9(5) = 45kg.

Question 19

What will be the cost of building a fence around a square plot with area equal to 289 sq ft, if the price per foot of building the fence is Rs. 58 ?

Accepted Answer

Let the side of the square plot be 'a' ft. Given area of the plot (a x a) = 289 => a = 17 Length of the fence = Perimeter of the plot = 4a = 68 ft. Cost of building the fence = 68 x 58 = Rs. 3944.

Question 20

It costs Rs. p each to send the first thousand messsages and Rs. q to send each subsequent one . If r is greater than 1,000, how many Rupees will it cost to send r messages ?

Accepted Answer

We need to find the total cost to send r messsages, r > 1000. The first 1000 messsages will cost Rs.p each (Or) The total cost of first 1000 messsages = Rs.1000p The remaining (r - 1000) messsages will cost Rs.q each (Or) The cost of the (r - 1000) = Rs.(r - 1000)y Therefore, total cost = 1000p + rq - 1000q = 1000(p - q) + qr

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