- The average of n terms equals $$\frac{x_1+x_2+. . .+x_n}{n}$$
- The weighted average of n terms equals $$\frac{w_1*x_1+ w_2*x_2+. . .+ w_n*x_n}{ w_1+ w_2+ . . . w_n}$$
Averages
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Question 1
Onion is sold for 5 consecutive months at the rate of Rs 10, 20, 25, 25, and 50 per kg, respectively. A family spends a fixed amount of money on onion for each of the first three months, and then spends half that amount on onion for each of the next two months. The average expense for onion, in rupees per kg, for the family over these 5 months is closest to
Solution
Let us assume the family spends Rs. 100 each month for the first 3 months and then spends Rs. 50 in each of the next two months.
Then amount of onions bought = 10, 5, 4, 2, 1, for months 1-5 respectively.
Total amount bought = 22kg.
Total amount spent = 100+100+100+50+50 = 400.
Average expense = $$\frac{400}{22}=Rs.18.18\approx\ 18$$
Question 2
In a football tournament, a player has played a certain number of matches and 10 more matches are to be played. If he scores a total of one goal over the next 10 matches, his overall average will be 0.15 goals per match. On the other hand, if he scores a total of two goals over the next 10 matches, his overall average will be 0.2 goals per match. The number of matches he has played is
Solution
Correct Answer: 10
Let Total matches played be n and in initial n-10 matches his goals be x
so we get $$\frac{\left(x+1\right)}{n}=0.15$$
we get x+1 =0.15n (1)
From condition (2) we get :
$$\frac{\left(x+2\right)}{n}=0.2$$
we get x+2 = 0.2n (2)
Subtracting (1) and (2)
we get 1 =0.05n
n =20
So initially he played n-10 =10 matches
Question 3
The arithmetic mean of scores of 25 students in an examination is 50. Five of these students top the examination with the same score. If the scores of the other students are distinct integers with the lowest being 30, then the maximum possible score of the toppers is
Solution
Correct Answer: 92
Let sum of marks of students be x
Now therefore x = 25*50 =1250
Now to maximize the marks of the toppers
We will minimize the marks of 20 students
so their scores will be (30,31,32.....49 )
let score of toppers be y
so we get 5y +$$\frac{20}{2}\left(79\right)$$=1250
we get 5y +790=1250
5y=460
y=92
So scores of toppers = 92
Question 4
The average weight of students in a class increases by 600 gm when some new students join the class. If the average weight of the new students is 3 kg more than the average weight of the original students, then the ratio of the number of original students to the number of new students is
Solution
Let the original number of students be 'n' whose average weight is 'x'
Let the number of students added be 'm' and the average weight will be x + 3
We need to find the value of n : m
It is given, average weight of students in a class increased by 0.6 after new students are added.
Therefore,
$$\ \frac{\ nx+m\left(x+3\right)}{n+m}=x+0.6$$
$$\ \ nx+mx+3m=mx+nx+0.6n+0.6m$$
$$2.4m=0.6n$$
$$4m=n$$
$$\frac{n}{m}=\frac{4}{1}$$
The answer is option C.
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