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We have two statements here. 1. Bob is as old as John will be when Bob is twice as old as John was when Bob was half as old as the sum of their current ages. Here the statement talks about 4 ages of Bob and John. Their current age and ages at 3 different times of their life. From the First line Bob ...
- by Alara533
Wed Mar 18, 2009 3:31 pm- Forum: Problem Solving
- Topic: their current ages
- Replies: 3
- Views: 3299
I think the answer is E. 1. Says Joan's estimated for the distance was within 5 miles of the actual distance. To cover this 5 miles difference in half an hour the average speed should be greater than or equal to 10 miles/hr. But we don't know the speed, hence insufficient. 2. Says Joan's estimate fo...
- by Alara533
Thu Mar 12, 2009 4:08 pm- Forum: Data Sufficiency
- Topic: Estimates questions
- Replies: 1
- Views: 1683
Hi...
Here you have assumed n to be positive and the graph (y-x > n) is drawn as if y is always greater than x. It can also happen that x is greater than y and n is a -ve value. In that case the feasible regions of the two graphs will have common points and that is happening here.
- by Alara533
Thu Mar 12, 2009 3:20 pm- Forum: Data Sufficiency
- Topic: Graphical solution to OG DS #109
- Replies: 3
- Views: 1581
In 0.097531097531, the pattern repeats after 6 digits. So we have 6th digit as 1, 12th digit again as 1, 18th digit as 1..etc etc..and 42th digit as 1.
Then 43rd digit will be 0 and 44th digit will be 9.
- by Alara533
Thu Mar 12, 2009 3:03 pm- Forum: Problem Solving
- Topic: Repeating decimal
- Replies: 3
- Views: 2217
Is the answer 8? Approach- For 130/x to be an integer, x must be a factor of 130 Find prime factors of 130 -> 2,5,13 --> 3 factors The multiples of these factors will also be factors of 130, ie 5 x 2 = 10 5 x 13 = 65 2 x 13 = 26 5 x 2 x 13 = 130 and finally 1 is also a factor of 130...so total 8 pos...
- by Alara533
Thu Mar 12, 2009 2:53 pm- Forum: Problem Solving
- Topic: Arithmetic
- Replies: 6
- Views: 1901
The correct idiom is 'need for X'
or 'need something'.
We can use need in the sentence...I need that. Here that just replaces something. It's used as a pronoun.
In the given question that is used to connect the sentence and not as a pronoun. Hence C is correct.
- by Alara533
Thu Mar 12, 2009 2:42 pm- Forum: Sentence Correction
- Topic: As litigation grows more complex.
- Replies: 9
- Views: 6118
f is the product of first 30 numbers. This product will have 7 zeros (Zero comes in the product when any number is multiplied by 10,20 and 30 and any even number is multiplied by 5, 15 and 25) so we have 5 x 2 = 10 then the number 10 15 x 6 = 90 then the number 20 25 x 4 = 100 then the number 30 So ...
- by Alara533
Thu Mar 12, 2009 2:31 pm- Forum: Data Sufficiency
- Topic: Integer question
- Replies: 5
- Views: 2227
Could you please provide the source of the question, because I find all the answers wrong. I found 26 digits which are not divisible by 5, 3 or 2. I even listed them down to make sure the answer is correct. My approach - There are total of 101 numbers There are 51 multiples of 2 between 200 and 300 ...
- by Alara533
Tue Feb 24, 2009 3:13 pm- Forum: Problem Solving
- Topic: 200 and 300 (both inclusive)
- Replies: 16
- Views: 24584
example: (EE) {EDFATD} The 6 alphabets in the curly braces, I consider (EE) as 1.So, the number comes to 7(Which can hhave 7! combinations). Also out of the 7,D comes twice.....so total number of times a number formed witj E along with one or more will be ==7!/2! Suppose we substitute EE with X, th...
- by Alara533
Mon Feb 09, 2009 11:53 am- Forum: Problem Solving
- Topic: tough arrangement (experts pls help)
- Replies: 9
- Views: 1588
If all the E's are taken together as one, then we have only 6 alphabets ( 2Ds,F,A,T and one grp of Es). The no. of possible arrangements would be 6!/2! so no. of arrangements in which 3Es aren't together = (8!/3!*2!) - (6!/2!) = 3000 Where am I going wrong??? There are cases where 2 'E's are togeth...
- by Alara533
Mon Feb 09, 2009 11:20 am- Forum: Problem Solving
- Topic: tough arrangement (experts pls help)
- Replies: 9
- Views: 1588
B)Getting occurences when E is together: 7!/2! for any pattern like EE(------), we will have 7! ways.dividing by 2 as D comes 2 times.This will cover both the scenarios when 2 or more E come together. Subtracting B from A,we will get 840. :) Suhail, could you please explain how you got 7!/2!. If yo...
- by Alara533
Mon Feb 09, 2009 11:10 am- Forum: Problem Solving
- Topic: tough arrangement (experts pls help)
- Replies: 9
- Views: 1588
- by Alara533
Mon Feb 09, 2009 12:37 am- Forum: Sentence Correction
- Topic: tough arrangement (experts pls explain)
- Replies: 2
- Views: 1205
Since all the corresponding angles are same in both the triangles, they are similar. For similar triangles, the length of the sides will be in same ratio. Check the attachment. Suppose the ratio is a value 'k'. Now we have the area of small triangle = (1/2) * s * h and area of big triangle = (1/2) *...
- by Alara533
Sun Feb 08, 2009 11:53 pm- Forum: Data Sufficiency
- Topic: tough geom ds
- Replies: 4
- Views: 1423
Suppose the cost for 100 ounce of candy bar was X.
Which means cost for 1 ounce = X/100
Since the weight was reduced by 20% and priced remained same, now 80 ounce of candy bar costs X. i.e cost of 1 ounce is X/80.
Now % increase is given [(X/80) - (X/100)] / (X/100) = 0.25 = 25%
- by Alara533
Sun Feb 08, 2009 11:32 pm- Forum: Problem Solving
- Topic: % question
- Replies: 2
- Views: 1242
114. How many numbers between 1 and 100 are ... I have a doubt here, since its mentioned between , should we include 100? In that case, the number of multiples of 2 will be 49 and not 50!! you are correct. by default, "between" is exclusive of the stated endpoints of the interval. if the ...
- by Alara533
Sun Feb 08, 2009 3:55 pm- Forum: Problem Solving
- Topic: numbers between 1-100
- Replies: 9
- Views: 72990