devil2043

fiirst i have calculated the numbers that are divisible by 2, in that case i have included 0 for unit place sine when we use 0 on unit place ,the number will be divisible by both 2 and 5 .so while calulating numbers divisible by 5 i just ignored te case when 0 will be at unit place .since that case has already been included in previous case (divisible by 2) just draw a ven diagram .there are few numbers divisible by 2 ,5 and both..there are actually 500 numbers that are divisible by both 2 and 5 .you are just considering numbers again.thats why your answer come out to be 2000 instead of 1500. even from the given problem when we consider the 23450 its divisible by both 2 and 5 .you are just considering the number in both cases. or lets say how many numbers from 0 to 20 are divisible by both 2 or 5 or both number divisible by 2 =2,4,6,8,10,12,14,16,18,20 divisible by 5= 5,10,15,20 10 n 20 come in both cases u just cant count them twice i hope u understand it now

possible sides comes out to be 6.5,6.5,13.5 n 11,11,2 here 6.5+6.5=1311 so such triangle is possible thus perimeter =2+11+11=24 thus ans is b

answer is b

ques 2 for a number to be divisible by 2 last digit must be even or 0 but if the last digit is zero it ll also be divisible by 5 case 1 divisible by 2 4 5 5 5 2 =4*5*5*5*2= 1000 _ _ _ _ _ i.e for unit place we have 2 choices either 2 or 4 (remember dont include 0 here ,as it will also contain numbers divisible by 5) ,similarly for 10's place 5 and so on case 2 no divisible by 5 for this unit place have to be 0 4 5 5 5 1 =4*5*5*5*1=500 here 4 unit place we have only 1 choice _ _ _ _ _ thus total such numbers will be 1000+500=1500

i hope i am not late GRE - Big Book.pdf - 4shared.com - document sharing - download :pirate: