# QUESTIONS ON AVERAGE (PART-1)

## QUESTIONS ON AVERAGE (PART-1)

#### QUERY 1

**The average of 7 consecutive integers is 7. Find the average of squares of these integers.**

A) 35

B) 45

C) 54

D) 53

**Amit jha
**Let the smallest number is x then the integers will be x, x+1, x+2, x+3, x+4, x+5, x+6 So their sum = 7x+21

And their average would be (7x+21)/7 = x+3

But average of the numbers is 7 (given)

So x+3 = 7

=> x = 4

Therefore the numbers are 4, 5, 6, 7, 8, 9, 10

Now Sum of the square of “n” natural numbers = n(n + 1)(2n + 1)/6

Hence the sum of squares of 1 to 10 = 10 x 11 x 21/6 = 385

And the sum of squares of 1 to 3 = 3 x 4 x 7/6 = 14

Thus the sum of squares of 4 to 10 = 385 – 14 = 371

So the average = 371/7 = 53 (option ‘D’)

#### QUERY 2

**The average of 5 consecutive integers starting with m is n. What is the average of 6 consecutive integers starting with (m+2)**

A) n+3

B) (2n+9)/2

C) (2n+5)/2

D) n+2

**MAHA GUPTA**

Case-I: Average = [m + (m+1) + (m+2) + (m+3) + (m+4)]/5 = (5m+10)/5 = m+2

But it’s given that m+2 = n

So case-II: Average = [n + (n+1) + (n+2) +(n+3) + (n+4) + (n+5)]/6 = (6n + 15)/6 = (2n+5)/2 (option ‘C’)

#### QUERY 3

**The average of 8 numbers is 20. The average of first two numbers is 31/2 and that of the next three is 64/3. If the sixth number be less than the seventh and eighth numbers by 4 and 7 respectively, then the eighth number is?**

A) 25

B) 27

C) 35

D) 17

**MAHA GUPTA**

If the total of last 3 numbers is known, the problem is solved.

Now, total of last 3 numbers = Total of all – (total of first 2 + total of next three)

= 20*8 – {(31/2)*2 + (64/3)*3}

= 160 – (31 + 64)

= 65

Now it’s understandable that if 4+7 = 11 is removed from 65 and divided by 3, the result will be the sixth number. So the eighth number will be 7 more of that.

So the eighth number = (65 – 11)/3 + 7= 54/3 + 7 = 18 + 7 = 25 (option ‘A’)

#### QUERY 4

**A man spends an average of Rs. 1,694.70 per month for the first 7 months and Rs.1,810.50 per month for the next 5 months. His monthly salary if he saves Rs. 3,084.60 during the whole year is?**

A) Rs 1000

B) Rs 2000

C) Rs 2400

D) Rs 3000

**JAYANTCHARAN CHARAN**

Average expenditure is between 1694.70 and 1,810.50 (I hope that is easy to make out!) . Also average saving per month is about 3084/12 = 257

Now the average salary should be around 250 + something between 1700 and 1800, so it is around 2000 (not going for exact answer as it is not required, though you can do it I guess, hope it is clear). So option ‘B’ is the correct answer

#### QUERY 5

**The total expenses of a boarding house are partly fixed and partly varying linearly with the number of boarders. The average expense per boarder is Rs 700 when there are 25 boarders and Rs 600 when there are 50 boarders. What is the average expense per boarder when there are 100 boarders?**

A) 500

B) 550

C) 600

D) 650

**MRITYUNJAY MAURYA**

Let x be the fixed part and y be the variable part of the total average expense per boarder.

Now in first case of 25 boarders the average expense per boarder = x+(y/25) = 700 …………………………………….1

In second case of 50 boarders the average expense per boarder = x+(y/50) = 600 ………………………………………2

Solving equations 1 and 2

x=500, y=5000

Now for 100 boarder we will have the average expense per boarder = x+(y/100) = 500+(5000/100) = 550 (option ‘A’)

PS: Remember …in the question it’s given that one part of the average expense is linearly varying and from the two cases given in the question we can see that that part is linearly decreasing. That’s why I have used (y/25), (y/50) and (y/100)

#### QUERY 6

**A man’s pension on retirement from service is equal to the half the average salary during last 3 years of his service. His salary from 1.1.83 is Rs 380 a month with increment of Rs 40 due on 1.10.83, 1.10.84 and 1.10.85. If he retires on 1.1.86, what pension does he draw?**

A) 205

B) 215

C) 225

D) 230

**MAHA GUPTA**

As the man retires on 1.1.86 and we have to know his last three years’ salary; means we need to know his salary for 1.1.83 to 31.12.83, 1.1.84 to 31.12.84 and 1.1.85 to 31.12.85

His salary from 1.1.83 to 30.9.83 = 380*9 = 3420

His increment on 1.10.83, 1.1084 and 1.1.85 = Rs 40

Therefore his salary from 1.10.83 to 30.9.84 = 420*12 = 5040 —–his monthly salary is now 380+40 i.e. 420 a month

And his salary from 1.10.84 to 30.9.85 = 460*12 = 5520 —–his monthly salary is now 420+40 i.e. 460 a month

And his salary from 1.10.85 to 31.12.85 = 500*3 = 1500 —–his monthly salary is now 460+40 i.e. 500 a month

Thus his salary for three years = 3420 + 5040 + 5520 + 1500 = 15480

Hence his monthly average salary = 15480/36 = 430

But his pension is half of average salary

So it is 430/2 = 215 (option ‘B’)\

#### QUERY 7

**The mean weight of 150 students in a certain class is 60 kg. The mean weight of the boys from the class is 70 kg, while that of girls is 55 kg. What is the number of girls in the class?**

A) 105

B) 100

C) 95

D) 60

**MAHA GUPTA**

Let the number of girls in the class = x

Thus, number of boys = 150 – x

Now according to the question

55x + 70*(150 – x) = 150*60

=> 55x + 10500 – 70x = 9000

=> -15x = -1500

=> x = 100 (option ‘B’)

ALLEGATION METHOD

**BOYS GIRLS**

70 | 55 | |||

60 | ||||

60 – 55 = 5 | 60 – 70 = 10 |

So, BOYS : GIRLS = 5 : 10 = 1 : 2

Therefore, number of girls = ^{2}⁄_{3} × 150 = 100 (option ‘B’)

TRICK

Check with answer options. Let’s take 100 first as calculations with 100 are easier.

Number of boys accordingly = 150 – 100 = 50

70*50 + 55*100 = 60*150

3500 + 5500 = 9000

We see LHS = RHS, therefore, 100 is the correct answer option.

#### QUERY 8

**In a family of 8 adults and some minors, the average consumption of rice per head per month is 10.8 kg; while the average consumption for adults is 15 kg per head and for minors it is 6 kg per head. The number of minors in the family is?**

A) 8

B) 6

C) 7

D) 9

**MAHA GUPTA**

Let the number of minors in the family = x

Thus, total members of the family = 8 + x

Now according to the question

6x + 15*8 = 10.8*(8 + x)

=> 6x + 120 = 86.4 + 10.8x

=> 4.8x = 33.6

=> x = 7 (option ‘C’)

ALLEGATION METHOD

**ADULTS MINORS**

15 | 6 | |||

10.8 | ||||

10.8 – 6 = 4.8 | 10.8 – 15 = 4.2 |

So, ADULTS : MINORS = 4.8 : 4.2 = 8 : 7

Number of adults = 8 ratios = 8

Hence, 1 ratio = 1

Therefore, number of minors = 7*1 = 7 (option ‘C’)

TRICK

Check with answer options. Let’s take 8 first.

Total members of the family = 8 + 8 = 16

15*8 + 6*8 = 10.8*16

120 + 48 = 172.8

168 ≠ 172.8

We see LHS ≠ RHS, therefore, 8 is an incorrect answer option. Likewise we can check others.

#### QUERY 9

**The average of the test scores of a class of ‘m’ students is 70 and that of ‘n’ students is 91. When the scores of both the classes are combined, the average is 80. What is ^{n}⁄_{m }?**

A) ^{10}⁄_{13}

B) ^{10}⁄_{11}

C) ^{11}⁄_{10}

D) ^{13}⁄_{10}

**MAHA GUPTA
**According to the question

70m + 91n = 80*(m+ n)

=> 70m + 91n = 80m + 80n

=> 10m = 11n

=>

^{n}⁄

_{m}=

^{10}⁄

_{11}(option ‘B’)

ALLEGATION METHOD

**CLASS M CLASS N**

70 | 91 | |||

80 | ||||

80 – 91 = 11 | 80 – 70 = 10 |

So, m : n = 11 : 10

Therefore, ^{n}⁄_{m} = ^{10}⁄_{11} (option ‘B’)

#### QUERY 10

**5 members of a team are weighed consecutively and their average weight calculated after each member is weighed. If the average weight increases by 1 kg each time. How much heavier is the last player than the first?**

A) 5 kg

B) 9 kg

C) 8 kg

D) 11 kg

**MAHA GUPTA
**In such a question TO FIND ANY NUMBER’s weight just add to it a number which is less by 1 to that. Suppose we have to find the weight of a guy who is 17th, just add 16 to that. So his weight is 17 + 16 = 33

SOLUTION TO THE ABOVE QUESTION:

To get the desired answer we must have the weights of the last and the first member

Now according to the formula given above weight of the fifth member = 5 + 4 = 9 kg

And weight of the first member = 1 + 0 = 1 kg

Hence, difference of their weights = 9 – 1 = 8 kg (option ‘C’)