**Question 1**
If the mean of numbers 28, x, 42, 78 and 104 is 62, then what is the mean of 128, 255, 511, 1023 and x?

- 395
- 275
- 355
- 415
- 365

**Explanatory Answer**

The average (arithmetic mean) of the 5 numbers 28, x, 42, 78 and 104 is 62.

Therefore, the sum of these 5 numbers is 62 * 5 = 310

i.e., 28 + x + 42 + 78 + 104 = 310

Hence, x = 310 - 252 = 58.

The average of 128, 255, 511, 1023 and x == 395

Therefore, the sum of these 5 numbers is 62 * 5 = 310

i.e., 28 + x + 42 + 78 + 104 = 310

Hence, x = 310 - 252 = 58.

The average of 128, 255, 511, 1023 and x == 395

**Question 2**
The arithmetic mean of the 5 consecutive integers starting with 's' is 'a'. What is the arithmetic mean of 9 consecutive integers that start with s + 2?

- 2 + s + a
- 2 + a
- 2s
- 2a + 2
- 4 + a

The correct choice is (E) and the correct answer is

**4 + a**.**Explanatory Answer**

The fastest way to solve such questions is to assume a value for 's'.

Let s be 1.

Therefore, the 5 consecutive integers that start with 1 are 1, 2, 3, 4 and 5.

The average of these 5 numbers is 3.

9 consecutive integers that start with 1 + 2 are 3, 4, 5, 6, 7, 8, 9, 10 and 11

The average of these 9 number is 7.

Now, let us take a look at the answer choices and substitute '1' for 's' and '3' for 'a'.

The only choice that provides us with an answer of '7' is choice (E).

Let s be 1.

Therefore, the 5 consecutive integers that start with 1 are 1, 2, 3, 4 and 5.

The average of these 5 numbers is 3.

9 consecutive integers that start with 1 + 2 are 3, 4, 5, 6, 7, 8, 9, 10 and 11

The average of these 9 number is 7.

Now, let us take a look at the answer choices and substitute '1' for 's' and '3' for 'a'.

The only choice that provides us with an answer of '7' is choice (E).

**Question 3**
The average weight of a group of 30 friends increases by 1 kg when the weight of their football coach was added. If average weight of the group after including the weight of the football coach is 31kgs, what is the weight of their football coach in kgs?

- 31 kgs
- 61 kgs
- 60 kgs
- 62 kgs
- 91 kgs

The correct choice is (B) and the correct answer is

**61 kgs**.**Explanatory Answer**

The new average weight of the group after including the football coach = 31

As the new average is 1kg more than the old average, old average without including the football coach = 30 kgs.

The total weight of the 30 friends without including the football coach = 30 * 30 = 900.

After including the football coach, the number people in the group increases to 31 and the average weight of the group increases by 1kg.

Therefore, the total weight of the group after including the weight of the football coach = 31 * 31 = 961 kgs.

Therefore, the weight of the football coach = 961 - 900 = 61 kgs.

As the new average is 1kg more than the old average, old average without including the football coach = 30 kgs.

The total weight of the 30 friends without including the football coach = 30 * 30 = 900.

After including the football coach, the number people in the group increases to 31 and the average weight of the group increases by 1kg.

Therefore, the total weight of the group after including the weight of the football coach = 31 * 31 = 961 kgs.

Therefore, the weight of the football coach = 961 - 900 = 61 kgs.

**Question 4**
The average wages of a worker during a fortnight comprising 15 consecutive working days was $ 90 per day. During the first 7 days, his average wages was $ 87/day and the average wages during the last 7 days was $ 92 /day. What was his wage on the 8

^{th}day?- $ 83
- $ 92
- $ 90
- $ 97
- $ 104

The correct choice is (D) and the correct answer is

**$97**.**Explanatory Answer**

The total wages earned during the 15 days that the worker worked = 15 * 90 = $ 1350.

The total wages earned during the first 7 days = 7 * 87 = $ 609.

The total wages earned during the last 7 days = 7 * 92 = $ 644.

Total wages earned during the 15 days = wages during first 7 days + wage on 8

=> 1350 = 609 + wage on 8

=> wage on 8

The total wages earned during the first 7 days = 7 * 87 = $ 609.

The total wages earned during the last 7 days = 7 * 92 = $ 644.

Total wages earned during the 15 days = wages during first 7 days + wage on 8

^{th}day + wages during the last 7 days.=> 1350 = 609 + wage on 8

^{th}day + 644=> wage on 8

^{th}day = 1350 - 609 - 644 = $ 97.

**Question 5**
The average of 5 quantities is 6. The average of 3 of them is 8. What is the average of the remaining two numbers?

- 4
- 5
- 3
- 3.5
- 0.5

The correct choice is (C) and the correct answer is

**3****Explanatory Answer**

The average of 5 quantities is 6.

Therefore, the sum of the 5 quantities is 5 * 6 = 30.

The average of three of these 5 quantities is 8.

Therefore, the sum of these three quantities = 3 * 8 = 24

The sum of the remaining two quantities = 30 - 24 = 6.

Average of these two quantities =6/2 = 3.

Therefore, the sum of the 5 quantities is 5 * 6 = 30.

The average of three of these 5 quantities is 8.

Therefore, the sum of these three quantities = 3 * 8 = 24

The sum of the remaining two quantities = 30 - 24 = 6.

Average of these two quantities =6/2 = 3.

**Question 6**
The average age of a group of 10 students was 20. The average age increased by 2 years when two new students joined the group. What is the average age of the two new students who joined the group?

The correct choice is (D) and the correct answer is

**32 years**.**Explanatory Answer**

The average age of a group of 10 students is 20.

Therefore, the sum of the ages of all 10 of them = 10 * 20 = 200

When two new students join the group, the average age increases by 2. New average = 22

Now, there are 12 students.

Therefore, the sum of the ages of all 12 of them = 12 * 22 = 264

Therefore, the sum of the ages of the two new students who joined = 264 - 200 = 64

And the average age of each of the two new students = 64/2 = 32 years.

Therefore, the sum of the ages of all 10 of them = 10 * 20 = 200

When two new students join the group, the average age increases by 2. New average = 22

Now, there are 12 students.

Therefore, the sum of the ages of all 12 of them = 12 * 22 = 264

Therefore, the sum of the ages of the two new students who joined = 264 - 200 = 64

And the average age of each of the two new students = 64/2 = 32 years.

**Question 7**
If m, s are the average and standard deviation of integers a, b, c, and d, is s > 0?

1. m > a

2. a + b + c + d = 0

The correct choice is (A). The correct answer is

1. m > a

2. a + b + c + d = 0

The correct choice is (A). The correct answer is

**Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question.****Explanatory Answer**

s will be zero only in two instances :

(i) when all the elements in the set are the same, or

(ii) the set contains only one element, which in this case is not possible.

So, we need to check whether a, b, c, and d are the same integers.

The average will be equal to a, b, c, and d only when a = b = c = d.

Since m > a, all the elements in the set cannot be the same, and therefore, s > 0.

SUFFICIENT

When a = b = c = d = 0, s = 0

When a = -4, b = 0, c = 0, and d = 4, s > 0

NOT sufficient

The correct answer is A

(i) when all the elements in the set are the same, or

(ii) the set contains only one element, which in this case is not possible.

So, we need to check whether a, b, c, and d are the same integers.

**Statement (1)**: m > aThe average will be equal to a, b, c, and d only when a = b = c = d.

Since m > a, all the elements in the set cannot be the same, and therefore, s > 0.

SUFFICIENT

**Statement (2)**: a + b + c + d = 0When a = b = c = d = 0, s = 0

When a = -4, b = 0, c = 0, and d = 4, s > 0

NOT sufficient

The correct answer is A

**Question 8**
Positive integers from 1 to 45, inclusive are placed in 5 groups of 9 each. What is the highest possible average of the medians of these 5 groups?

- 25
- 31
- 15
- 26
- 23

The correct choice is (B) and the correct answer is

**31**.**Explanatory Answer**

We need to maximize the median in each group in order to maximize the average of all the medians.

The highest possible median is 41 as there should be 4 numbers higher than the median in the group of 9.

So, if we have a group that has a, b, c, d, 41, 42, 43, 44, 45, the median will be 41.

In this set, it is essential not to use any more high values on a, b, c, or d as these do not affect the median.

The median of a group that comprises 1, 2, 3, 4, 41, 42, 43, 44, 45 will be 41.

The next group can be 5, 6, 7, 8, 36, 37, 38, 39, 40. The median will be 36.

To maximize medians in all the 5 groups, the medians of the 5 groups will have to be 21, 26, 31, 36 and 41.

The highest possible average of the medians will be the average of these 5 numbers = 31.

The highest possible median is 41 as there should be 4 numbers higher than the median in the group of 9.

So, if we have a group that has a, b, c, d, 41, 42, 43, 44, 45, the median will be 41.

In this set, it is essential not to use any more high values on a, b, c, or d as these do not affect the median.

The median of a group that comprises 1, 2, 3, 4, 41, 42, 43, 44, 45 will be 41.

The next group can be 5, 6, 7, 8, 36, 37, 38, 39, 40. The median will be 36.

To maximize medians in all the 5 groups, the medians of the 5 groups will have to be 21, 26, 31, 36 and 41.

The highest possible average of the medians will be the average of these 5 numbers = 31.

**Question 9**
If the average of 5 positive integers is 40 and the difference between the largest and the smallest of these 5 numbers is 10, what is the maximum value possible for the largest of these 5 integers?

- 50
- 52
- 49
- 48
- 44

The correct choice is (D) and the correct answer is

**48**.**Explanatory Answer**

The average of 5 positive integers is 40. i.e., the sum of these integers = 5*40 = 200

Let the least of these 5 numbers be x.

Then the largest of these 5 numbers will be x + 10.

If we have to maximize the largest of these numbers, we have to minimize all the other numbers.

That is 4 of these numbers are all at the least value possible = x.

So, x + x + x + x + x + 10 = 200

Or x = 38.

So, the largest of these 5 integers is 48.

Let the least of these 5 numbers be x.

Then the largest of these 5 numbers will be x + 10.

If we have to maximize the largest of these numbers, we have to minimize all the other numbers.

That is 4 of these numbers are all at the least value possible = x.

So, x + x + x + x + x + 10 = 200

Or x = 38.

So, the largest of these 5 integers is 48.

**Question 10**
An analysis of the monthly incentives received by 5 salesmen : The mean and median of the incentives is $7000. The only mode among the observations is $12,000. Incentives paid to each salesman were in full thousands. What is the difference between the highest and the lowest incentive received by the 5 salesmen in the month?

- $4000
- $13,000
- $9000
- $5000
- $11,000

The correct choice is (E) and the correct answer is

**$11,000**.**Explanatory Answer**

The arithmetic mean of the incentives of the 5 salesmen is $7000.

Therefore, the sum of the incentives received by all 5 taken together is 5 * 7000 = $35,000.

The median for these 5 observations is $7000.

Let us say their incentives in ascending order are a, b, c, d and e.

So, c = $7000

and a + b + c + d + e = $35,000

The only mode among the 5 observations is $12,000.

So, the maximum number of observations among the 5 will be in $12,000.

We have deduced that c has got $7000.

Hence, the only possibility is that both d and e got an incentive of $12,000 each.

So, incentives received by c + d + e = 7000 + 12000 + 12000 = $31,000

Therefore, a + b = 35,000 - 31,000 = 4000.

a and b have to be two different values as the only mode is $12,000.

So, a has to be $1000 and b has to be $3000.

The difference between the highest and the lowest incentives is, therefore 12,000 - 1000 = $11,000

Therefore, the sum of the incentives received by all 5 taken together is 5 * 7000 = $35,000.

The median for these 5 observations is $7000.

Let us say their incentives in ascending order are a, b, c, d and e.

So, c = $7000

and a + b + c + d + e = $35,000

The only mode among the 5 observations is $12,000.

So, the maximum number of observations among the 5 will be in $12,000.

We have deduced that c has got $7000.

Hence, the only possibility is that both d and e got an incentive of $12,000 each.

So, incentives received by c + d + e = 7000 + 12000 + 12000 = $31,000

Therefore, a + b = 35,000 - 31,000 = 4000.

a and b have to be two different values as the only mode is $12,000.

So, a has to be $1000 and b has to be $3000.

The difference between the highest and the lowest incentives is, therefore 12,000 - 1000 = $11,000