@ChirpiBird said:@jashholmesIn a city 60 % of the population owns a car, 75 % owns a rickshaw, 80 % owns a scooter, and 95% owns a bicycle, what person at least owns all four of them???and 20%???
@nole said:@Estallar12 I didn't get your explanation.can you please explain again. I tried it solving by finding the total sum and then trying to divide by 2 ( i couldn't do that coz finding total sum was difficult) . A must be a big number,how can 2^8 be the highest power. ? I mean that means 256(2^8) * k = A where K is not divisible by 2. isn't it ?
@sbharadwaj said:Nice soln..Bhai I have a doubt.. We got A = 10!( Sum of something say X).. Now there's a possibility that X may add up to a 2's multiple..in which case, answer may differ..! Plz correct me if I'm wrong..
@akashgupta1987 said:@ScareCrow28 yar change in avg se nikalna hai na? zara approach batao...is it 6/n=1/4??
@jashholmes said:A cricketer had played a certain number of innings during the year. He finds that if he had not played the last game when he made 20 runs, or if he had played an extra innings and made 56 runs, he would have increased his average by 1. What was his average?
@scrabbler said:Orally:The difference of two innings (ek kum se ek jyada) gives us 56 + 20 = 76 runs, keeping the average same. So that average must be 38. This is 1 more than the actual average, so actual must be 37.regardsscrabbler
@DeAdLy said:good @scrabbler , i am seeing u are solving most of the questions by simple logic or orally...keep it up. u seem to be a veteran of CAT
Couldn't do@ScareCrow28 said:@scrabbler sir!
Sir nahin!@ScareCrow28 said:Wo Factorial wala kaise hoga? Couldn't do
@scrabbler said:Orally:The difference of two innings (ek kum se ek jyada) gives us 56 + 20 = 76 runs, keeping the average same. So that average must be 38. This is 1 more than the actual average, so actual must be 37.regardsscrabbler
@scrabbler said:Sir nahin!Yeh waala?If A= 10 ! + 12! + 14!........ 100! . Then find the highest power of 2 in A. How to solve this ?10! mein there are 5+2+1 = 8 2s. All the later terms have more than 8 2s so can be treated as 10! * E (When E stands for some Even number)So the given expression is 10!* (1 + E + E +E...+E) = 10! * (an odd number). Hence it will have 8 2s only I feel...regardsscrabbler
I was writing it as 10! ( 1 + 11 + 11*12 +...) There was a sum in elitmus exam .......... 1024*3672 + 4152*5623 + 6413*5241 .......... this multiplication is in octal form (i.e base 8) ......... The question was ....... How many digit 6 are there in the above multiplication !! ........ Please help me about how to solve this problem !!
@iLoveTorres said:bro can you generalize this if i change the question as he scores 20 and 56 thereby increasing his average by 1.. in that case what is to be done?
@DeAdLy said:good @scrabbler , i am seeing u are solving most of the questions by simple logic or orally...
@nole said:@scrabbler Though it won't have any effect on answer but i think 10! * (1 + E+E+E....) there all numbers won't be E right ? they should be different even numbers. right ?
@scrabbler said:My basic mantra is laziness (am a Goan after all!) So I try to write as little as possible. It is a good habit, especially in practice....when trying to solve orally I am forced to think clearly and sometimes see stuff I would otherwise miss...regardsscrabbler
@scrabbler said:No then I need to know the initial average....See if he has played n innings averaging a, then he's scoring 20 + 56 = 76 , in 2 innings....so out of that 2a will go just to maintain the average. The rest will be distributed across the (n+2) innings....so if average increase by 1, then these extra runs equal (n+2)*1 and so 76 must be 2a + (n+2). We need either one of n and a to find the other.In the given question, n was not relevant as the average was the same in both cases - 1 innings ago and one innings later. Hence the extra runs scored in those two innings was exactly 2a.regardsscrabbler
@ScareCrow28 said:@scrabbler Sir If it had been A = 10! + 11! + 12! + ..... Then could we have come to an answer??
