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21 litres

Let x be the total vol. B is 5x/12 if we take out 9 litres. then amount of B taken out is 5*9/12=15/4 and we are adding 9 litres of B. but the final volumes remains the same i.e x. And in final volume B is 9*x/16. so the equation will be 5x/12 - 15/4 + 9 = 9*x/16 this gives x= 36. And initial amount...

by aa2kash

Mon Sep 21, 2009 6:45 am
Forum: Problem Solving
Topic: Mixture Question
Replies: 2
Views: 1004

AS per my understanding; the worm ate 4 books completly and the front cover of the last book + back cover of the first book. each book is 1.5 inch thick. 4 books = 6 inch front cover= .25 inch last cover for the first book= .25 inch I am getting a total of 6.5 inches. which is none of the options. e...

by aa2kash

Mon Sep 21, 2009 4:58 am
Forum: Problem Solving
Topic: Help with math problem!
Replies: 6
Views: 1634

@Viju
Its mention in the question that they start simultaneously. It means they when they are meeting they have taken the same time to travel their respective distances.So, we can equate the time taken by both of them.
Hope it clears your query now.

by aa2kash

Sun Sep 20, 2009 1:22 pm
Forum: Problem Solving
Topic: Speed, Time prob #1
Replies: 7
Views: 2352

30 min

Let x be the normal speed. and t is the normal time
then equation 1 can be
xt=15

if he is 1.5times slower than normal speed.then new speed is 2x/3. and as he takes 15 min extra 1.e 1/4 hrs

equation 2 will be
2x/3=15/(t+1/4)
solving these 2 u get
x=30 and t=1/2 i.e 30 mins

by aa2kash

Sun Sep 20, 2009 7:42 am
Forum: Problem Solving
Topic: SDT
Replies: 1
Views: 1245
by aa2kash

Sat Sep 19, 2009 9:24 am
Forum: Problem Solving
Topic: Power Prep Exponent Questions
Replies: 10
Views: 1785

Answer is B

draw a line from the centre of the circle to the point D. this is equal to the radius of the circle.

Now u have a right angle traingle with the sides
6, r and r-2
solve it and get r=10.
area of semicircle is pi*r*r/2 = 50pi

by aa2kash

Sat Sep 19, 2009 9:03 am
Forum: Problem Solving
Topic: Tough Geom2
Replies: 4
Views: 2227

ok. I misread the last line.we have to tell the number of stamps after kay gave 10 stemps to alberto.
as we have solved x=30.
Kay has 5x-10= 140 stamps
Alberto has 3x+10= 100 stamps

difference is 40 stamps.
Answer is c

by aa2kash

Sat Sep 19, 2009 8:54 am
Forum: Problem Solving
Topic: GMAT Prep Questions
Replies: 4
Views: 1743

thanks Ali for the different and timesaving approach.

I did it with the basic formula, time=dis/speed
since they are meeting after sometime say T. thefore i compared the timetaken by both peter and paul.
70+x is covered by peter and 70-x is covered by paul.

(70+x)/52=(70-x)/39
this gives x=10.

by aa2kash

Sat Sep 19, 2009 3:48 am
Forum: Problem Solving
Topic: Speed, Time prob #1
Replies: 7
Views: 2352

number one: As the stamps are in the ratio 5:3 say kaye has 5x and alberto has 3x. then after kaye gave 10 stemps to alberto. kaye= 5x-10 alberto=3x+10 and the new ratio is 7:5 solve this: 5x-10/3x+10=7/5 u will get x=30. the difference in the stemps is 5(30)-3(30)=60 the Answer is D 60. number two ...

by aa2kash

Sat Sep 19, 2009 1:35 am
Forum: Problem Solving
Topic: GMAT Prep Questions
Replies: 4
Views: 1743

Hi Viju
For better understanding of this try to analyse it by putting different values of n. for eg n=5
n!+1 --> 5! +1 =121 which is not divisible by 2,3,4,5 i.e. any number less than or equal to 5 . it is divisible by 11 .

Its a general rule therefore just remember this.

by aa2kash

Fri Sep 18, 2009 2:30 pm
Forum: Problem Solving
Topic: Products of consecutive intervals
Replies: 4
Views: 1170

There is a simple rule that states that n!+1 cannot be divisible by any number less than or equal to n. modulo arithmetic: n! mod x (where x <= n) = 0 n! + 1 mod x = 1, therefore not divisible by x. In this particular question the function can be written as: h(100) + 1 = (2^50)*50! + 1 therefore it'...

by aa2kash

Fri Sep 18, 2009 3:51 am
Forum: Problem Solving
Topic: Products of consecutive intervals
Replies: 4
Views: 1170

Answer is B as explained above.

by aa2kash

Fri Sep 18, 2009 2:43 am
Forum: Problem Solving
Topic: PS - Question 55.3.12
Replies: 2
Views: 1049