rajiv lend out rs 9 to anni on condition that the amount is payable in 10 months by 10 equal installments of re. 1 each payable at the start of every month.what is the rate of interest per annum if the first instalment has to be paid one month from the date the loan is availed?
rajiv lend out rs 9 to anni on condition that the amount is payable in 10 months by 10 equal installments of re. 1 each payable at the start of every month.what is the rate of interest per annum if the first instalment has to be paid one month from the date the loan is availed?please suggest.
General
lets say initial value is x and each instalment is y,rate of interest be r
say after 1 year u pay y ==> its inital value at the time of taking loan =y/r
after 2 years u pay y ==> its value at the time of taking loan=y/r^2
similarly...for others
here let monthly interest be r then annual interest will be 12r ,which would be the required answer.
rajiv lend out rs 9 to anni on condition that the amount is payable in 10 months by 10 equal installments of re. 1 each payable at the start of every month.what is the rate of interest per annum if the first instalment has to be paid one month from the date the loan is availed?please suggest.
I think it will a simple interest question (else calculations will be complex)
Consider all possible seven-digit binary numbers having four 1s and three 0s. What is the sum of all such numbers? ???
All the numbers will be of form 1abcdef, where three of a, b, c, d, e, f are 1 and other three are 0.
So, C(6, 3) = 20 such numbers
Now, for every number 'abcdef', there will be a number '111111 - abcdef', which satisfies the required condition. So, sum will be 111111*10 (only for last 6 digits)
So, sum of all numbers = (2^6)*20 + 63*10 = 1280 + 630 = 1910
rajiv lend out rs 9 to anni on condition that the amount is payable in 10 months by 10 equal installments of re. 1 each payable at the start of every month.what is the rate of interest per annum if the first instalment has to be paid one month from the date the loan is availed?please suggest.
As case is for periodic installment We have to consider periodic interest on each installment given to rajiv by anni Anni gave rs 1 for 9 months, another rs 1 for 8 months, another for 7 ...for 1 month and at last only rs 1 Total money received by rajiv=1*r*9/(12*100) + 1*r*8/(12*100) +...1*r*1/(12*100) + 10 = 10+(9*10/2)r/(100*12)=10+45r/1200---2 Anni should give 9+interest on rs 9 for 10 months(9*r*10/1200)---2 Equating both. 9+90r/1200=10+45r/1200 45r/1200=1 r=1200/45 pa
A similar question:-Find the sum of all ten - digit binary strings having five zeros and five ones.Note:- Strings can have a leading zero
1111100000 1 in tenth+1 in 9th+.....+1 at last Coz' 0 can appear at the beginning, for all cases, possible ways= 9!/4!*5! 9!/(4!*5!)*2^9+9!/(4!*5!)*2^8+...+9!/(4!*5!)*2^0
10 boxes each has n balls, you have a spring balance which can measure accurate weights . Each ball is 100gm.Only one box contains lighter balls of 90gm each.You can take out any number of balls from any number of boxes. m=number of minimum weighings find m ,when n= 1)9 2)8 3)1 4)2 5)3ďťż
10 boxes each has n balls, you have a spring balance which can measure accurate weights . Each ball is 100gm.Only one box contains lighter balls of 90gm each.You can take out any number of balls from any number of boxes.m=number of minimum weighings find m ,when n=1)92)83)14)25)3ďťż
10 boxes each has n balls, you have a spring balance which can measure accurate weights . Each ball is 100gm.Only one box contains lighter balls of 90gm each.You can take out any number of balls from any number of boxes.m=number of minimum weighings find m ,when n=1)92)83)14)25)3
n = 1.
take out p balls from from box no. p
and say kth box has 90 gm balls
So the final weight should be = 100*(1+2+3+4+5....10) - k*10 = 10*(55-k)
hence based on the final weight observed we can find the value of 'k'.
10 boxes each has n balls, you have a spring balance which can measure accurate weights . Each ball is 100gm.Only one box contains lighter balls of 90gm each.You can take out any number of balls from any number of boxes.m=number of minimum weighings find m ,when n=1)92)83)14)25)3
For n=9 only 1 For n=8 only 2 For rest(n=1) only 4 For n=2,3 only 3 Edit ???
n = 1.take out p balls from from box no. pand say kth box has 90 gm ballsSo the final weight should be = 100*(1+2+3+4+5....10) - k*10 = 10*(55-k) hence based on the final weight observed we can find the value of 'k'.ATDH.
sir can u plz explain a bit more.....here we have to find m........for different values of n(when,n= 9,n=8,n=1,n=2,n=3).....
sir can u plz explain a bit more.....here we have to find m........for different values of n(when,n= 9,n=8,n=1,n=2,n=3)..... to find m........for different values of n(when,n= 9,n=8,n=1,n=2,n=3)..... discuss ur approach plz......as I dont hav d OA for dis
3) 1
When n = 9 the number of balls in each box = 0,1,2,3,4,5,6,7,8,9 So 1 weighing of all the boxes together will give the answer. Ideal sum = 100 * 36 = 3600 If box 1 has 90g balls, then weight = 3600 If box 2 has 90g balls, then weight = 3590 If box 3 has 90g balls, then weight = 3580 and so on..
sir can u plz explain a bit more.....here we have to find m........for different values of n(when,n= 9,n=8,n=1,n=2,n=3)..... to find m........for different values of n(when,n= 9,n=8,n=1,n=2,n=3)..... discuss ur approach plz......as I dont hav d OA for dis
10 boxes are there each has n balls
if n=9
Make two set A(first 9 boxes),B(last box)
From set A take 1 ball from first box 2 balls from 2nd box ... 9 balls form 9th box
If first box contains 90grms balls then total weight(TW)=90+(2+3+4..9)*100
And hence from the value of TW we can find then defective box in set A, and hence for total if set A doesn't contains defective then B that's 10th ball will be defective
So weighting required will be 1
If n=8
Make two sets A(first 8 boxes),B(last 2 boxes)
Apply same procedure for A to find defective box. Take out 1 balls from 1, 2 from 2nd and so on If A don't have defective box, then B will have defective box(worst case)
Again a weighting have to be done to identify the defective among two boxes in B
So weighting required will be 2 edit
If n=1
Make 2 sets A(first 5) B(next5) Find total weight of A if !=5*100 then B will have defective box else A In any case problem will reduce to 5 boxes 1 weighting will reduce prblem to 5 boxes 2 weighting will reduce to either 2 or 3 Take worst 3 now 2 weightings will be required total of 4 weightings
If n=2 or If n=3Edit
Make 2 sets A(first 5) B(next5) Find total weight of A if !=5*100 then B will have defective box else A In any case problem will reduce to 5 boxes 1 weighting will reduce prblem to 5 boxes 2 weighting will reduce to either 2 or 3 Take worst 3 from first take 1 ball, 2nd 2 balls and from 3rd three balls and measure the weight. For n=2 take out 0 balls 1st, 1 from 2nd 3 from 3rd and so on. Now here only 1 weighting will be required So total 3 weightings
3) 1When n = 9the number of balls in each box = 0,1,2,3,4,5,6,7,8,9So 1 weighing of all the boxes together will give the answer.Ideal sum = 100 * 36 = 3600If box 1 has 90g balls, then weight = 3600If box 2 has 90g balls, then weight = 3590If box 3 has 90g balls, then weight = 3580and so on..This will be true for any n
Follow your approach if n=1, that's each box will have only 1 ball