@shattereddream we can use permutation and combination also.how did u do? option d is none of these. i think it is B rite?
@nole said:@shattereddream from 10 to 99 we have 90 2 digits number.We will subtract those numbers that don't follow the rule10,20.....90 9 numbers11,21,31.....91 9 numbers12,13......19 8 numbersso total 2690-26= 64 numbers.
@shattereddream said:Actually i have done it with some other method , yours is also a nice method . thanks
@nole according to your method 34 should be such a value of M
but 3! 4! =144
and 7 != 5040
help please!
@nole said:@shattereddream we can use permutation and combination also.how did u do? Is 64 the correct answer ?
yes its correct , i am also getting 64 with my method..... by permutation and combination could you post that approach
@cynara said:@nole according to your method 34 should be such a value of Mbut 3! 4! =144and 7 != 5040help please!@shattereddream the only values of m i could find was 23, 32. so my solution was D.
Will you share your method ?
@shattereddream I subtracted numbers that don't follow x ! * y ! > x! + y! rule
if a number has 0 or 1 in it,it won't follow that rule.so i counted all those numbers
number ca n be of the form xy
lets consider case with 0 . x can't be 0(coz 2 digit nu ) with y=0, x can have 0,1,2,......9 total 9 values
now case with 1 with x=1 y can have 1,2,........9 values
with y=1 x can have 1,2,.......9 values but 11 gets repeated here so 9-1=8
so total 26
90-26 =64
@cynara for 3! * 4! u are comparing with (3+4)! , compare it with 3! +4!. because it is "sum of factorial of digits .what u are doing is factorial of the sum of the digits.
@nole said:@shattereddream I subtracted numbers that don't follow x ! * y ! > x! + y! ruleif a number has 0 or 1 in it,it won't follow that rule.so i counted all those numbersnumber ca n be of the form xylets consider case with 0 . x can't be 0(coz 2 digit nu ) with y=0, x can have 0,1,2,......9 total 9 valuesnow case with 1 with x=1 y can have 1,2,........9 values with y=1 x can have 1,2,.......9 values but 11 gets repeated here so 9-1=8so total 2690-26 =64
even i have applied this method
@nole said:@cynara for 3! * 4! u are comparing with (3+4)! , compare it with 3! +4!. because it is "sum of factorial of digits .what u are doing is factorial of the sum of the digits.
okay, got it :thumbsup:
1)A man earns x% on the first 5000 rupees of his investment and y% on the rest of his investment . If he earns Rs 1250 from Rs 7000 and Rs 1750 from Rs 9000 invested .Find x
a)20 b)15c)25d)none
"The sum of ten consecutive odd natural numbers is 180 more than the average of the numbers. Find the sum of the largest and the smallest of the ten numbers."
a,a+2,a+4..a+18 so a=11 largest is 29 sum is 40
the odd numbers are in a.p so let the terms be a,a+2 ,a+4 and so on on calculating smallest is 11 and largest is 29 add both 😃
@bs0409 said:Amirchand is selling some articles. Amirchand is offering a discount of 33.33% if one pays by credit card. Amirchand has marked on the article in such a way that after giving a discount, he still manages to get a profit of 25%. Garibchand uses false weighing balance and deceives Amirchand by 20% and he also pays the amount by credit card. If Garibchand gives the same article to another customer at 40% discount on marked price of Amirchand and Garibchand has a profit of Rs 20, then what is the cost price of the article in rupees? (1) 160 (2) 200 (3) 240 (4) 300 (5) none of these
160 hoga kya?
@techsurge said:@naga25french Sir, pls helpCan this question be solved by application of fibonnaci series
I dont think we can use fib series here .
@veertamizhan said:"The sum of ten consecutive odd natural numbers is 180 more than the average of the numbers. Find the sum of the largest and the smallest of the ten numbers."
(2a+1) +(2a+3)+.....(2a+19) = [ (2a+1) +(2a+3)+.....(2a+19)]/ 10 + 180
==> 20a+ 100 = 2a+10+180 ==> a=5
2a+1+2a+19 = 4a+20 =4*5+20 = 40
==> 20a+ 100 = 2a+10+180 ==> a=5
2a+1+2a+19 = 4a+20 =4*5+20 = 40