 How to Solve Average Questions? Important Tips & Tricks & Formulas ## How to Solve Average Questions? Tips & Tricks and Formulas

Average Question with Solution Free PDF for SSC, Railway & Banking Exam. How to Solve Average Questions Problems? Short Tricks to Solve Average Questions. As we all know in many competitive exams like SSC, Railways, Banking, FCI, CWC, Insurance Exams, and other state exams, Average Questions are asked repeatedly. Average Questions are a very important topic of the Quantitative Aptitude section for every exam, so every student must know How to Solve Average Questions problems? Tips & Tricks & Formulas.

To make the Average chapter easy for you all, we are providing you all some Important Short Tricks to How to Solve Average Questions which will surely make the chapter easy for you all. Use these below given short Tips & Tricks to solve Average questions within minimum time.

## How to Solve Average Questions Problems? Tips & Tricks

These shortcuts (How to Solve Average Questions problems?) will be very helpful for your upcoming All SSC, Railways, and Banking Exams.

The average or mean or arithmetic mean of a number of quantities of the same kind is equal to their sum divided by the number of those quantities. Arithmetic average is used for all averages like:-

• Average income, average profit, average age, average marks etc.
• It is defined as the sum total of all volumes of items divided by the total number of items. To calculate the sum of observations, they should be in the same unit.

Example 1 : A man purchased 5 toys at the rate of `200 each, 6 toys at the rate of `250 each and 9 toys at the rate of `300 each. Calculate the average cost of one toy.

Solution :

Price of 5 toys = 200 × 5 = 1000

Price of 6 toys = 250 × 6 = 1500

Price of 9 toys = 300 × 9 = 2700 Example 2 : In three numbers, the first is twice the second and thrice the third. If the average of these three numbers is 44, then the first number is :

Solution : Let the three numbers be x, y and z

Therefore, x = 2y = 3z, Example 3 : The average of five consecutive odd numbers is 61. What is the difference between the highest and lowest numbers?

Solution : Let the numbers be x, x + 2, x + 4, x + 6 and x + 8. or 5x + 20 = 305 or x = 57.

So, required difference = (57 + 8) – 57 = 8.

How to Solve Average Questions? Average of a group consisting two different groups when their averages are known :

1. Let Group A contains m quantities and their average is a and Group B contains n quantities and their average is b, then average of group C containing a + b quantities Example  : There are 30 student in a class. The average age of the first 10 students is 12.5 years. The average age of the next 20 students is 13.1 years. The average age of the whole class is :-

Solution :

Total age of 10 students =12.5 × 10 = 125 years

Total age of 20 students = 13.1 × 20 = 262 years

Average age of 30 students = (125+262) / 30 = 12.9 Years

1. It average of m quantities is a and the average of n quantities out of them is b then the average of remaining group rest of quantities is Example : Average salary of all the 50 employees increasing 5 officers of a company is `850. If the average salary of the officers is `2500. Find the average salary of the remaining staff of the company.

Solution : Here, m = 50, n = 5, a = 850, b = 2500 How to Solve Average Questions Problems?  – Weighted Average

If we have two or more groups of members whose individual averages are known, then the combined average of all the members of all the groups is known as weighted average. Thus if there are k groups having member of number n1, n2, n3,…….nk with averages A1, A2, A3……..Ak respectively then weighted average. Example : The average monthly expenditure of a family was `2200 during the first 3 months; `2250 during the next 4 months and `3120 during the last 5 months of a year. If the total saving during the year were `1260, then the average monthly income was

Solution : Total annual income

= 3 × 2200 + 4 × 2250 + 5 × 3120 + 1260

= 6600 + 9000 + 15600 + 1260 = 32460

So Average monthly income = 32460/12 = 2705 If, in a group, one or more new quantities are added or excluded, then the new quantity or sum of added or excluded quantities = [Change in no. of quantities × original average] ± [change in average × final no. of quantities] Take +ve sign if quantities added and take –ve sign if quantities removed.

Example : The average weight of 29 students in a class is 48 kg. If the weight of the teacher is included, the average weight rises by 500 g. Find the weight of the teacher.

Solution : Here, weight of the teacher is added and final average of the group increases.

Change in average is (+)ve, using the formula Sum of the quantities added weight of teacher = (1 × 48) + (0.5 × 30) = 63 kg.

weight of teacher is 63 kg.

Geometric Mean Or Geometric Average Example: The production of a company for three successive years has increased by 10%, 20% and 40% respectively what is the average increase of production.

Solution :
G.M. = (10 × 20 × 40)1\3 = 20%

### How to Solve Number Series Problems? Tips and Tricks – Click Here

How to solve average questions?

Average of numbers= {sum of numbers/total No. of numbers}

Or you can use the below formula

Average= (sum of 1st term+sum of last term)/2

Suppose you are given a question as given below:

Find the average of 5,7,9,11,13?

Solution-

In this type of questions, first look for the difference of numbers. Since the difference is constant throughout i.e. 2. And given numbers are odd in numbers i.e. total 5 types of numbers.

Then in this type of questions, the answer is always the middle term i.e. 9 (in above question).

But if the numbers were given even in numbers means there are six numbers like 5, 7, 9 , 11, 13,15. Then the answer is always sum of middle terms or sum of opposite terms divided by 2 i.e. (5+15)/2 or (7+13)/2 or (9+11)/2. The answer is always same. In the above case the average will be 10.

Example :

The average mark of 70 students in a class is 80. Out of these 70 students, if the average mark of 40 students is 75, what is the average mark of the remaining 30 students?

Average of 70 students = 80.

Hence total marks of all the 70 students = 70×80 = 5600

Out of these 70, average of 40 students = 75

Hence total marks of these 40 students = 40×75 = 3000.

So the total marks of the remaining 30 students = 5600-3000 = 2600

Hence the average of the remaining 30 students = 2600/30 = 86.66

Now there will be a serious difference in the time taken if the numbers given here are not multiples of 10.

### Average Practice Question with Solution

1. 12 men and 16 women together can complete a work in 26 days while 10 women and 18 children together can complete the work in 20 days. Also, 8 men and 16 children can complete the work in 25 days. 1 man, 1 woman and 1 child start the work and from 2nd day onwards every day 1 new man, 1 new woman and 1 new child join the work. On which day the work shall be completed?
A. 20th day
B. 21st day
C. 22nd day
D. None of the above
E. Cannot be determined

B. 21st day

LCM of 26, 20 and 25 = 1300
Let the total work be 1300 unit
Let ‘m’ be the units of work done by 1 man in 1 day
‘w’ be the units of work done by 1 woman in 1 day
and ‘c’ be the units of work done by 1 child in 1 day
26(12m + 16w) = 1300 ………(i)
20(10w + 18c) = 1300 ………(ii)
25(8m + 16c) = 1300 ……….(iii)
On solving equation (i), (ii) and (iii);
We get, m = 1.5 units, w = 2 units and c = 2.5 units
Work done on day 1 = 1.5+2+2.5 = 6*1 units
Work done on day 2 = (6+6*1) = 6*2 units

Work done on day 3 = 6*3 units
Let the work is completed on nth day
6*1 + 6*2 + …………. + 6n ≥ 1300
6n(n+1)/2 ≥ 1300
3n(n+1) ≥ 1300
The smallest integer value of n satisfying the above equation will be the required number of days
For n = 20
3n(n+1) = 1260
For n = 21
3n(n+1) = 1386 ≥ 1300
Therefore, the work shall be finished on 21st day.

2.The ages of A, B, C, D and E are 35, 33, 30, 27 and 25 years respectively. A, B, C, D and E bought furniture carrying MRP in the ratio of 2:3:4:5:6 respectively. A bought furniture carrying Rs. 1500 MRP. At the time of billing each of them received a surprise discount equal to their age but not more than 30%. Had they known the discount scheme earlier and exchanged their purchase among themselves to maximize combined discount, how much combined saving could have been done apart from the discount given?
A. Rs. 105
B. Rs. 135
C. Rs. 165
D. Rs. 195
E. Rs. 215

D. Rs. 195

MRP of A’s purchase = 1500 (Discount: 30%)
Discount amount = 1500*30% = Rs. 450
MRP of B’s purchase = 2250 (Discount: 30%)
Discount amount = 2250*30% = Rs. 675
MRP of C’s purchase = 3000 (Discount: 30%)
Discount amount = 3000*30% = Rs. 900
MRP of D’s purchase = 3750 (Discount: 27%)
Discount amount = 3750*27% = Rs. 1012.5
MRP of E’s purchase = 4500 (Discount: 25%)
Discount amount = 4500*25% = Rs. 1125
Total discount = 450+675+900+1012.5+1125 = 4162.5

Maximum combined discount is possible when C,D and E receive 30% discount, A receives 25% discount and B receives 27% discount.
Maximum combined discount = 4500*30% + 3750*30% + 3000*30% + 2250*27% + 1500*25% = 4357.5
Therefore, the apart from the given discount, Rs. (4357.5-4162.5) = Rs. 195 could have been saveD.

3.A class comprises of 6 boys and some girls. If there are 281 ways of forming a group of 5 students such that group constitutes at most 2 girls, what is the probability of forming a group of 5 students with at least 2 girls?
A. 81/154
B. 73/154
C. 127/154
D. 27/154
E. None of the above

C. 127/154

Let the number of girls be n
A/Q
6C5 + 6C4 * nC1 +6C3 * nC2 = 281
6 + 15n + 10n(n-1) = 281
=> 10n2+ 5n – 275 = 0
=> 2n2+ n – 55 = 0
=> 2n2+ 11n – 10n – 55 = 0
=> n(2n+11) – 5(2n+11) = 0
=> (n-5) (2n+11)= 0
Here n is a positive integer so, n = 5
Probability of forming a group of 5 students with at most 1 girl =(6C5 + 6C4 * 5C1 )/11C5 = 81/462= 27/154
Required probability = 1 – 27/154 = 127/154

4.A and B are standing on points P and Q respectively which are 200 km apart. A starts moving to and from between P and Q at a speed of 20 km/h while B starts moving to and from between Q and P at a speed of 30 km/h. What will be the difference between the total distance travelled by A and B when they meet for the 10th time?
A. 380 km
B. 570 km
C. 650 km
D. 760 km
E. 820 km

D. 760 km 5.The ratio of Ekta’s and Reema’s income last year was 10:3. The ratio of Ekta’s this year income and last year income is 6:5 and the ratio of Reema’s this year income and last year income is 2:3. If the sum of Ekta’s and Reema’s present incomes is Rs. 5,124, what was Reema’s income last year?
A. Rs. 1464
B. Rs. 1830
C. Rs. 1372
D. Rs. 1654
E. Rs. 1098

E. Rs. 1098

Let Ekta and Reema’s current income be 6x and 2y respectively.
Hence, their income last year was 5x and 3y respectively.
Since the ratio of their income last year was 10 : 3;
5x : 3y = 10 : 3
=> x = 2y

The sum of their present incomes is Rs.5,124.
=> 6x + 2y = 5124
=> 7x = 5124
=> x = 732 and y = 366
=> Reema’s income last year = 3y = Rs. 1,098

6.Two trains while moving in opposite direction on parallel tracks, take 10 seconds to cross each other. However, if they travel in the same direction, the longer train, which is faster than shorter train, crosses the shorter train in 30 seconds. If the length of the longer train is decreased by 75%, it takes 12 seconds less to cross the shorter train traveling in the same direction. Find the time taken by the longer train to cross a tunnel twice its length, if the difference between the lengths of the trains is 50 m?
A. 20 seconds
B. 22 seconds
C. 24 seconds
D. 32 seconds
E. 36 seconds

C. 24 seconds

Let the length of shorter and longer trains be ‘x’ and (x+50) respectively
And their speed be v and u m/s, respectively
In first case: If trains travel in opposite direction
(x+(x+50))/10 = u+v => (2x+50)/10 = u+v ……………..(i)
In second case: If trains travel in same direction
(2x+50)/30 = u-v ……………………(ii)
In third case: If trains travel in same direction with decreased length
[x+{(x+50)/4}]/18 = u-v …………………(iii)
On solving (i) and (ii)
u = (2x+50)*2/10*3 = 2(2x+50)/30 …………………. (iv)
v = (2x+50)/30 ……………………..(v)
on solving (iii),(iv) and (v)

x = 350 m, u = 50 m/s and v = 25 m/s
Therefore, the length of longer train = 350+50 = 400 m
Length of tunnel = 2*400 = 800 m
Total distance = (400+800) = 1200 m
Time taken to cross tunnel = 1200/50 = 24 seconds

7. Average age of a family of four people, 5 years ago, was 40 years. 5 years from now, the ratio of oldest member’s age to the youngest member’s age will be 18:7 and the ratio of present ages of other two member is 7:3. If the youngest member is 40 years younger than second oldest member, what is the present age of oldest member?
A. 57 years
B. 62 years
C. 67 years
D. 72 years
E. 77 years

C. 67 years

Let the current age be ‘a’ of oldest member, ‘b’ of the second oldest member, ‘d’ of youngest member and ‘c’ of the fourth member of the family.
a-5 + b-5 + c-5 + d-5 = 40*4
a + b + c + d = 180 ……….(i)
(a+5)/(d+5) = 18/7
=> 7a-18d = 55 ………(ii)
d+40 = b………(iii)
b/c = 7/3………(iv)
Solving the above four equations we get,
a = 67 years

8. A watch merchant bought 140 watches at the rate of Rs. 900 per dozen watches. He marked the watches at Rs. 120 per piecE. He sold 75% of the watches at 10% discount on marked price and the remaining watches were sold at 20% profit on cost price. Had he not given any discount, by what percentage his profit would have increased?
A. 12.4%
B. 13.3%
C. 14.6%
D. 15.9%
E. 16.4%

D. 15.9%

Unit cost price = 900/12 = Rs. 75
Total cost price = 140*75 = Rs. 10500
Total selling price = 140*0.75*120*0.9 + 140*0.25*75*1.2 = 11340 + 3150 = 14490
Profit = 14490 – 10500 = Rs. 3990
If there will be no discount, total selling price = 120*140 = 16800
Therefore, required percentage = 100*(16800-14490)/14490 = 15.9%

9.’Bhoomi’ is a rectangular shaped construction sitE. The ratio of the square of the perimeter of ‘Bhoomi’ and the sum of the squares of the diagonals of ‘Bhoomi’ is 98:25. What is the ratio of the sum of the adjacent sides of ‘Bhoomi’ and difference between adjacent sides of ‘Bhoomi’?
A. 2
B. 3
C. 5
D. 7
E. 9

D. 7

Let the length and the breadth of the rectangle be l and b respectively
Then, (2(l + b))2/2(l2+ b2) = 98/25
(l2+ b2+2lb)/(l2+b2) = 49/25
50*l*b = 24(l2+b2)
12l2-25*l*b + 12b2= 0
(l/b) = (3/4) or (4/3)
But l cannot be less than b
=> (l/b) = (4/3)
Required ratio = (l + b)/(l – b) = ((4b/3 + b))/((4b/3 – b)) = 7

10.What is the remainder when 1! + 2! + 3! + 4! + … + 77! is divided by 15?
A. 3
B. 5
C. 7
D. 9
E. 11

A. 3

n! = n × (n − 1) × (n − 2) × … × 3 × 2 × 1
Thus, every term from 15! to 77! will have a ‘15’ in it.
Thus, every term from 15! to 77! is divisible by 15.
Since 15 = 5 × 3, any x! that has a term (5 × 3) in it will also be divisible by 15.
5! = 5 x 4 × 3 × 2 × 1
Since every subsequent term from 5! to 14! will have a 5! in it, each term from 5! to 14! is divisible by 15.
Therefore, required remainder = remainder of [ (1! + 2! + 3! + 4!)/15 ] = remainder of [ (1+2+6+24)/15 ] = 3

Hope you all like this “How to Solve Average Questions problems? Tips & Tricks & Formulas” article and this will also help in upcoming  all competitive exams like – Banking, SSC, Railway and other exams. If you want to practice more Average Question based on latest pattern, you can download 350+ Average Question with Solution Free PDF from below link – 