one final one before start chasing the dream..
@Subhashdec2 said:19c2*29c2/50c5??
bhai i think it should be 30c2 in place of 29c2 as numbers are from x1 to x50??
shouldnt it bne?
@abhishek.2011 said:bhai i think it should be 30c2 in place of 29c2 as numbers are from x1 to x50??shouldnt it bne?
kar diya bhai edit
@albiesriram said:19c2*30C2 /50c5 is there in the options
yup didnt see i think m too sleepy good night friends u guys carry on :)
@abhishek.2011 said:yup didnt see i think m too sleepy good night friends u guys carry n
i am sleepy as well, good night guys . yes option B is the OA
An article was marked at a certain percentage above the cost price.It is then sold for a profit offering a discount.The sum of the profit and the discount equals the sum of its cost price and the differnce of twice its marked price and thrice its cost price.find the mark up percentage.
@nole said:An article was marked at a certain percentage above the cost price.It is then sold for a profit offering a discount.The sum of the profit and the discount equals the sum of its cost price and the differnce of twice its marked price and thrice its cost price.find the mark up percentage.
CP=CP
MP=xCP
SP=yxCP
Profit=(xy-1)CP
discount=xCP(1-y)
CP(xy-1+x-xy)=CP+2xCP-3CP
x-1=1+2x-3
x=1
since it was sold on profit hence we cannot consider this case
x-1=1-2x+3
3x=5
x=5/3
5/3-1=2/3 =67.67%
MP=xCP
SP=yxCP
Profit=(xy-1)CP
discount=xCP(1-y)
CP(xy-1+x-xy)=CP+2xCP-3CP
x-1=1+2x-3
x=1
since it was sold on profit hence we cannot consider this case
x-1=1-2x+3
3x=5
x=5/3
5/3-1=2/3 =67.67%
@Subhashdec2 said:-11-5=-4210/-5=-2bhai,, why we r taking q = 2 ; in dat case it cumes -2/5.....i mean for min val q = 2correct me if m wrong...
Consider all three-digit numbers that are greater than the sum of the squares of their digits by exactly 543. What are the last three digits of the sum of these numbers?
@jain4444 said:Consider all three-digit numbers that are greater than the sum of the squares of their digits by exactly 543. What are the last three digits of the sum of these numbers?
373 ?
677 and 696
@jain4444 said:what about 637 ? any general approach ?
oh yes... missed it..
i did it like...
100a +10b +c - a^2 - b^2 - c^2 = 543
100a +10b +c -543 = a^2 +b^2 +c^2
=> 100(a-5) +10 (b-4) +c-3 = a^2+ b^2 +c^2
if a=7.. then LHS will be 200 + which will be improbable if we analyse b and c..
so a =6...
then did some hit and trial.. for some values of b and c.. 
@jain4444 said:Consider all three-digit numbers that are greater than the sum of the squares of their digits by exactly 543. What are the last three digits of the sum of these numbers?
543+a^2 +b^2+ c^2 =100a+10b+c
A can be only 6
so take a=6
543+36+b^2+c^2 =600+10b+c
b^2+C^2-10b-c=21
rest by hit and trial .
four no. are possible -> 616+637+677+696=2626
Ans 626
A can be only 6
so take a=6
543+36+b^2+c^2 =600+10b+c
b^2+C^2-10b-c=21
rest by hit and trial .
four no. are possible -> 616+637+677+696=2626
Ans 626
@Faruq said:543+a^2 +b^2+ c^2 =100a+10b+cA can be only 6so take a=6543+36+b^2+c^2 =600+10b+cb^2+C^2-10b-c=21rest by hit and trial .four no. are possible -> 616+637+677+696=2626Ans 626
how can we asuume a has to be 6..pls explain ?
@rnishant231 said:how can we asuume a has to be 6..pls explain ?
Put a=7
Check with b,c=0,0 and b,c=9,9
In either case, the RHS > LHS
@Faruq, could you please comment?