1. E
2. D
3. D
4. A
5. D
wat r the OAs?
Hey srikanth267....Alchemist-mba has xplained the ans. plz refer that post
Hey srikanth267....Alchemist-mba has xplained the ans. plz refer that post
A professional gambler has won 40% of his 25 poker games for the week so far. If all of a sudden, his luck changes and he begins winning 80% of the time, how many more games must be play to end up winning 60% of all his games for the week?
1) 15
2) 20
3) 25
4) 30
5) 35
A professional gambler has won 40% of his 25 poker games for the week so far. If all of a sudden, his luck changes and he begins winning 80% of the time, how many more games must be play to end up winning 60% of all his games for the week?
1) 15
2) 20
3) 25
4) 30
5) 35
My Pick 'C'..
games played = 25.. won 40% = 10.
let 'x' be extra number of games need to be played to reach 60%.
Total number of games won from (25+x) would be (25+x) *(60/100).
We already know.. for the first 25 he won 10. for the next 'x', since the winning % is 80.. number of games won would be x * (80/100)..
=> (25+x) * (6/10) = 10 + x *(8/10)
=> solving, we get x = 25..
Hope it helps..
krajkumar6,
If you were to follow this approach.. I guess you should include (in you minus calcuation) the following scenario too -- when two dwarfs (elves) sit together and one dwarf (elv) sits alone... for eg.. DDEEED.
you are very correct.
my mistake...
ORIGINAL QUESTION
What is the reminder when 9^1 + 9^2 + 9^3 + ...... + 9^9 is divided by 6?
(1) 0
(2) 3
(3) 4
(4) 2
(5) None of these
Okay, this is an age old problem from this same forum from sometime around 2007..
No OA was posted and the answer was taken to be E: None of these..
According to me, that is not the answer; rather, the answer should be B: 3. I am giving my explanation.. and I am posting this to confirm whether my methodology is right.
9, 9^2,...9^9.. each of them divided by 6 will give the remainder 3.
adding the remainders, 27/6 will give us the remainder 3.
My reasoning why the first quoted method is wrong:
Because remainder or 9/6 is not the same as that of 3/2.
Edit: Ok I read some newer posts and this issue had been addressed earlier. Sorry!
What is the reminder when 9^1 + 9^2 + 9^3 + ...... + 9^9 is divided by 6?
(1) 0
(2) 3
(3) 4
(4) 2
(5) None of these
Ok, the answer to this must be (5)None of these. Here is what I did:
Numerator = 9^1 + 9^2 + 9^3 + ...... + 9^9 = 9 (1+ 9^1 + 9^2 + 9^3 + ...... + 9^8 )
= 3x3 (1+ 9^1 + 9^2 + 9^3 + ...... + 9^8]
Denominator = 6 = 2x3
Cancel 3 from both num. and den...so you are left with 2 in the denominator. If you divide any integer by 2, the remainder will either be always zero or 1. Hence, options 2,3 and 4 are ruled out.
Remainder will be zero if the numerator is even number.
Lets find out-
Numerator after cancelling 3 = 3 (1+ 9^1 + 9^2 + 9^3 + ...... + 9^8]
9 raised to any integer will always be odd.
Hence, the expression inside the bracket evaluates to (1 + even number) = odd number.
Inside the bracket, I wrote even number because when 8 odd numbers are added, it will give even number.
Numerator = 3 x (some odd number) = odd number. Hence, if you divide this number by 2, you will get a remainder of 1. Hence, none of the options given.
Hope this makes sense. If neone has any other better solution for dis, pls post!
Cheers!
Satish
ans: rem=3
approach -
rem when 9 div 6 = 3
rem when 9^2 div 6 = rem when 3^3 div by 6 =3
.... till 9^9..
therefore rem=rem when 3*9 div 6 =3
TanBhai, it's E)None of these. Please follow mailsatk's post above:
http://www.pagalguy.com/654153-post126.html
Okay, this is an age old problem from this same forum from sometime around 2007..
No OA was posted and the answer was taken to be E: None of these..
According to me, that is not the answer; rather, the answer should be B: 3. I am giving my explanation.. and I am posting this to confirm whether my methodology is right.
9, 9^2,...9^9.. each of them divided by 6 will give the remainder 3.
adding the remainders, 27/6 will give us the remainder 3.
My reasoning why the first quoted method is wrong:
Because remainder or 9/6 is not the same as that of 3/2.
Edit: Ok I read some newer posts and this issue had been addressed earlier. Sorry!
can someone please explain this one.............
if X^4 + y^4 =100 then greatest posiible value of x is between
1) 0 and 3
2)3 and 6
3) 6 and 9
4) 9 and 12
5) 12 and 15
as per my calculation it comes to be 0 and 3but that is not the OA
Q.1) If A is a positive integar and if the unit's digit of A square is 9 then units digit of square of (A+1) is 4, what is the units digit of square of(A+2)?
Ans. 1
3
5
7
9
Q.2) If (x) is the greatest integar less than or equal to x, then what is the value of (-1.6) + (3.4) + (2.7)?
Ans. 3
5
1
7
6
Q.3) If F jeans regularly sell for $15 a pair and P jeans for $ 18. during sale these regular unit prices are discounted at different rates so that a total of $ 9 are saved by purchasing 5 jeans ( 3F & 2P). If the sum of two discounts is 22 % then the rate of discount on P is ?
Ans. 10%
11%
12%
13%
20%
Let total no of games equal to n ,
then .4(25) + .8(n-25) = .6n
n=50
therefore he must play 25 more games( n-25)
answer is C
Q.1) If A is a positive integar and if the unit's digit of A square is 9 then units digit of square of (A+1) is 4, what is the units digit of square of(A+2)?
Ans. 1
3
5
7
9
Q.2) If (x) is the greatest integar less than or equal to x, then what is the value of (-1.6) + (3.4) + (2.7)?
Ans. 3
5
1
7
6
Q.3) If F jeans regularly sell for $15 a pair and P jeans for $ 18. during sale these regular unit prices are discounted at different rates so that a total of $ 9 are saved by purchasing 5 jeans ( 3F & 2P). If the sum of two discounts is 22 % then the rate of discount on P is ?
Ans. 10%
11%
12%
13%
20%
the answers are
Q 1) 1
Q 2)3
Q 3) 20%
what are OAs
The OA is (C). Thanks a lot for the explanation.
My Pick 'C'..
games played = 25.. won 40% = 10.
let 'x' be extra number of games need to be played to reach 60%.
Total number of games won from (25+x) would be (25+x) *(60/100).
We already know.. for the first 25 he won 10. for the next 'x', since the winning % is 80.. number of games won would be x * (80/100)..
=> (25+x) * (6/10) = 10 + x *(8/10)
=> solving, we get x = 25..
Hope it helps..
Q.1) If A is a positive integar and if the unit's digit of A square is 9 then units digit of square of (A+1) is 4, what is the units digit of square of(A+2)?
Ans. 1
3
5
7
9
Q.2) If (x) is the greatest integar less than or equal to x, then what is the value of (-1.6) + (3.4) + (2.7)?
Ans. 3
5
1
7
6
Q.3) If F jeans regularly sell for $15 a pair and P jeans for $ 18. during sale these regular unit prices are discounted at different rates so that a total of $ 9 are saved by purchasing 5 jeans ( 3F & 2P). If the sum of two discounts is 22 % then the rate of discount on P is ?
Ans. 10%
11%
12%
13%
20%
My Picks:
1. unit of A^2 is 9 => unit place can be either '3' or '7'
unit of (A+1)^2 is 4 => unit place must be '7'.
Hence unit of (A+2)^2 would be (7+2)^2 => '1'.
2. (-1.6) + (3.4) + (2.7) = -2 + 3 + 2 = 3
3. Let reduction on F be 'f' and reduction on P be 'p'.
current price of 'F' => 15 (1-f/100)
current price of 'P' => 18 (1-p/100)
Given:
a total of $ 9 are saved by purchasing 5 jeans ( 3F & 2P).
=> Original Price (3F & 2P) = 81
Discounted Price = 81 - f *(45/100) - p*(36/100)
=> f *(45/100) - p*(36/100) = 9
=> 5f + 4p = 100
Given: sum of two discounts is 22 => f + p = 22
solving these two equations => f = 12, p = 10.
What are the OA's?
can someone please explain this one.............
if X^4 + y^4 =100 then greatest posiible value of x is between
1) 0 and 3
2)3 and 6
3) 6 and 9
4) 9 and 12
5) 12 and 15
as per my calculation it comes to be 0 and 3but that is not the OA
X can have a maximum value when y=0 => x^4 = 100,
when x = 3 => x^4 = 81.. and x^4 = 256.. it should be 3 2'.
Hope this helps.
You are absolutely right!!!
i still cudnt understand the last problem ...
My Picks:
1. unit of A^2 is 9 => unit place can be either '3' or '7'
unit of (A+1)^2 is 4 => unit place must be '7'.
Hence unit of (A+2)^2 would be (7+2)^2 => '1'.
2. (-1.6) + (3.4) + (2.7) = -2 + 3 + 2 = 3
3. Let reduction on F be 'f' and reduction on P be 'p'.
current price of 'F' => 15 (1-f/100)
current price of 'P' => 18 (1-p/100)
Given:
a total of $ 9 are saved by purchasing 5 jeans ( 3F & 2P).
=> Original Price (3F & 2P) = 81
Discounted Price = 81 - f *(45/100) - p*(36/100)
=> f *(45/100) - p*(36/100) = 9
=> 5f + 4p = 100
Given: sum of two discounts is 22 => f + p = 22
solving these two equations => f = 12, p = 10.
What are the OA's?
You are absolutely right!!!
i still cudnt understand the last problem ...
Hope this
helps
Total actual amount for 3F and 2P is 81$
if u subtract this from the discounted price u get 9$
Let the discount for F be f% and P be p%
81- 3*5*(1-f/100) - 2*18*(1-p/100) = 9
so this boils down to 5f+4p = 100----------1
they say that the total discount is 22%
f+p=22--------2
solve 1 and 2
p=10
f=12
I found some tricks for solving"hard' questions in NOVA..like to know whether these tricks really work on D day..?
3,k,2,8,m,3
The AM of the list of numbers above is 4. If k & m are integars and K is not equal to m, what is the median of the list?
Ans. 2, 2.5, 3, 3.5, 4
I dont knw whether such tricks work or not but iam surely interested to know these tricks!!!
ganesh99 SaysI found some tricks for solving"hard' questions in NOVA..like to know whether these tricks really work on D day..?
3,k,2,8,m,3
The AM of the list of numbers above is 4. If k & m are integars and K is not equal to m, what is the median of the list?
Ans. 2, 2.5, 3, 3.5, 4
A.M = (3+k+2+8+m+3) / 6 = 4 => m+k = 8.... (m,k) = (1,7)(2,6)(3,5) or (7,1)(6,2)(5,3) --> in all cases median is '3'.
GMAT problem solving questions are designed more to test your understanding of underlying mathematical concepts than to test your ability to actually carry out quantitative procedures accurately.The Problem solving tips are Read the questions carefully,Use your scrap paper for every question,Do not get bogged down with complicated or lengthy calculations,The "guesstimating" technique is extremely effective on this exam,Learn how to work backwards,Convert quantities freely and Use process of elimination as a last resort.
------------------------------
gillberk
LINK BUILDING
1.On a certain scale of intensity, each increment of 10 in magnitude represents a tenfold increase in intensity. On this scale an intensity corresponding to a magnitude of 165 is how many times an intensity corresponding to a magnitude of 125?
2.
Harry started a 6-mile hike with a full 10-cup canteen of water and finished the hike in 2 hours with 1 cup of water remaining in the canteen. If the canteen leaked at the rate of 1 cup per hour and Harry drank 3 cups of water during the last mile, how many cups did he drink per mile during the first 5 miles of the hike?