I am stuck with the below caselet of averages:
In the island of Hoola Boola Moola , inhabitants have a strang process of calculating their average incomes and expenditures. According to an old legend prevalent on that island. the average monthly income had to be calculated on the basis of 14 months in a calendar year while the average monthly expenditure was to be calculated on the basis of 9 months per year. This would lead to people having an underestimation of their saving since there would be an underestimation of the income and overestimation of their expenditure per month.
Qstn) Mr. Boogie Woogle comes back from the USSR and convinces his community comprising 273 families to start calculating the average incomeon on the basis of 12 months per calendar year.Now if it is known that the average estimated income in his community is(according to the old system) 87 Moolahs per month(Moolah - official currency of Hoola Boola Moola). Then what will be the change in the average estimated savings for the island of Hoola Boola Moola. (Assume that there is no other change).
a) 251.60 Moolahs b) 282.75 Moolahs c) 312.75 Moolahs d) Cannot be determined
My dbt here is that we do not know the expenditure here. Assuming that is constant:
E = expenditure per month
Savings/mnth= 87-E (Before cuming of Mr Boogie )
Savings/mnth(new) = 87*14/12-E
change in savings = 87*14/12 - 87
=87*1/6
considering the whole community size as only 273 .. the ans shud be 87*1/6*273 = 3958.5(not in any option)
I am unable to get any other way to solve it. help
Hi
I also have a dbt on this qstn. Can sum1 help me solve the whole caselet? Qstn 40-48 of Arun sharma LOD3 for Averages.
Hi
I also have a dbt on this qstn. Can sum1 help me solve the whole caselet? Qstn 40-48 of Arun sharma LOD3 for Averages.
Dude .. If you don't have time to type the question how do you expect others to take time out for you and hunt for the book for solving your question
a humble request to post the complete question
Hi can any1 help me wid d above qstn?
In the island of Hoola Boola Moola , inhabitants have a strang process of calculating their average incomes and expenditures. According to an old legend prevalent on that island. the average monthly income had to be calculated on the basis of 14 months in a calendar year while the average monthly expenditure was to be calculated on the basis of 9 months per year. This would lead to people having an underestimation of their saving since there would be an underestimation of the income and overestimation of their expenditure per month.
Qstn) Mr. Boogie Woogle comes back from the USSR and convinces his community comprising 273 families to start calculating the average incomeon on the basis of 12 months per calendar year.Now if it is known that the average estimated income in his community is(according to the old system) 87 Moolahs per month(Moolah - official currency of Hoola Boola Moola). Then what will be the change in the average estimated savings for the island of Hoola Boola Moola. (Assume that there is no other change).
a) 251.60 Moolahs b) 282.75 Moolahs c) 312.75 Moolahs d) Cannot be determined
Hi pls help me wid the mentioned qstn?
.
Hey Thanks pal, me too was getting the same answer so just wanted to confirm with you puys. The answer in the book is given as 4373, and i think they have included all the digits to get that answer, ie including double and single digit numbers, so there are two double digit numbers, viz 12,21 and no single digit number divisible by 3, so the final answer in this case is 4371+2 = 4373. So now am convinced that its a typo, THANK YOU very much for the help!!
I am getting 4371. Not sure why it is not in options.
First we need to find out how many numbers we can form.
It is like forming numbers in ternary system.
So, we can take numbers till 8-digit numbers => 3^8 numbers from 0 to 3^8 - 1.
And then, for 9 digit numbers, we can take only 1 as first digit and last digits can be filled in 3^8 ways.
So, we will have total 2*3^8 numbers.
Out of these, (1/3)rd will be divisible by 3.
So, numbers divisible by 3 = 2*3^7 = 4374
But we need to remove 1 and 2 digit numbers.
So, we need to remove 0, 12 and 21. ... (We have taken numbers from 0)
So, total cases = 4374 - 3 = 4371
Hey Thanks pal, me too was getting the same answer so just wanted to confirm with you puys. The answer in the book is given as 4373, and i think they have included all the digits to get that answer, ie including double and single digit numbers, so there are two double digit numbers, viz 12,21 and no single digit number divisible by 3, so the final answer in this case is 4371+2 = 4373. So now am convinced that its a typo, THANK YOU very much for the help!!
please help me with the following
consider a circle with unit radius. there are 7 adjacent sector, s1,s2....s7...in the circle such that their total area is 1/16 of the area of the circle...
further, area of jth sector is twice that of the (j-i)th sector...for j=2....7.
what angle , in radians, subtended by the arcs of s1 at the centr of the circle
?
thanks in advance...
So the sum of s1,s2...s7= (pie)/16
Let area of s1=x ; also the angle subtended by s1 at center be y.(x=y*r/2=y/2 radians)
s2 = 2x
s3 = 4x ....so on till s2 = 2^6x.
sum of areas of s1 till s7 = x + 2x + 4x + ....... + 2^6x.
Hence x*(2^7-1) = (pie)/16
or x = (pie)/(127*16)
or y/2 = (pie)/(127*16)
hence y= (pie)/(127 * 8 = (pie)/1016 radians.
...50/5=10, 10/5=2 so10+2 =12
hi puys
pls help in solving below ques of progression
1/1*5 + 1/5*9 + 1/9*13 + ...........+ 1/221*225
a. 28/221
b. 56/221
c. 56/225
d. none of these
this is quest no. 31 on page 72...
i have few more question similar to this (32,33 and 35)..so please letme knw the basic approach to solve these kind of questions
hi puys
pls help in solving below ques of progression
1/1*5 + 1/5*9 + 1/9*13 + ...........+ 1/221*225
a. 28/221
b. 56/221
c. 56/225
d. none of these
My take is 56/225.
1/(1*5) + 1/(5*9) + .... + 1/(221*225)
=1/4 *
........ This is because (1/1 - 1/5) = 4/(1*5) and same approach for other terms.
So, all therms in bracket, except first one and last one, will cancel out.
So, sum = 1/4 *
= 56/225
ithinks its correct...because 4c2 for women..4c3 for men then ....2 ways for selecting p or vp then 2c1 between selected men and 3 ways for selecting left out position
6*4*2*2*3= 288
similarly 32
and how abt this one
1/1 + 1/3 + 1/6 + 1/10 + 1/15.....
a.2
b.2.25
c.3
d.4
and how abt this one
1/1 + 1/3 + 1/6 + 1/10 + 1/15.....
a.2
b.2.25
c.3
d.4
It should be 2.
S = 1/1 + 1/3 + 1/6 + 1/10 + 1/15 + .....
= 1/1 + 1/(1+2) + 1/(1+2+3) + ...
nth term = 2/(n(n+1)) = 2
=> S = 2
= 2
= 2
It should be 2.
S = 1/1 + 1/3 + 1/6 + 1/10 + 1/15 + .....
= 1/1 + 1/(1+2) + 1/(1+2+3) + ...
nth term = 2/(n(n+1)) = 2
=> S = 2
= 2
= 2
thanks a lot!
i could not have imagined it's solution..!!
s=1/1+1/3+1/6+1/10+1/15
= 1+1/3(1+1/2)+1/5(1/2+1/3)+1/7(1/3+1/4).........
= 1+1/3(3/2)+1/5(5/6)+1/7(7/12).........
= 1+1/2+1/6+1/12.........
= 1+1=2
s=1/1+1/3+1/6+1/10+1/15
= 1+1/3(1+1/2)+1/5(1/2+1/3)+1/7(1/3+1/4).........
= 1+1/3(3/2)+1/5(5/6)+1/7(7/12).........
= 1+1/2+1/6+1/12.........
= 1+1=2
pls check for the solution of below question...it's on page 105 chapter 3,averages...question number 21
the avg of 71 results is 48. if the avg of the first 59 results is 46 and that of the last 11 is 52. find the 60th result
a)132 b)122 c)134 d)128 e)136
my answer is 72
also check for question no 18 on the same page...
there were 42 students in a hostel...due to the admission...........
my answer is 632.94 so it should be (e) but answer is (a)