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28 Feb 2016, 09:40
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
75% (01:23) correct
25%
(01:44)
wrong
based on 454
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David and Stacey are riding bicycles on a flat road at a constant rate. If Stacey is now three miles ahead of David, in how many minutes will Stacey be just two miles ahead of David?
(1) Stacey is traveling at rate of 10 mph and David is traveling at a rate of 12 mph.
(2) 45 minutes ago Stacey was 4.5 miles ahead of David.
(1) Stacey is traveling at rate of 10 mph and David is traveling at a rate of 12 mph.
(2) 45 minutes ago Stacey was 4.5 miles ahead of David.
02 Mar 2016, 10:55
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IMO D
In option A the difference in speed is 2(i.e 12-10= 2 mph). So every hour david is gaining 2 mile.
Since we have to maintain a distance of two miles between David and Stacey(presently three), the 1 mile will be covered by David in half n hour(as david is gaining 2 mile /hr,so to gain 1 mile it will take David 1/2 hr and the dist will be 2mile b/w david and stacey).==> Suff
In option B
45 mins ago the dist was 4.5 mile and now dist is 3 mile. Therefore david covered 1.5(4.5-3) mile in 45 min so speed of david catching stacey is 2mph.
By same logic as for opton A==> Suff
SO D
In option A the difference in speed is 2(i.e 12-10= 2 mph). So every hour david is gaining 2 mile.
Since we have to maintain a distance of two miles between David and Stacey(presently three), the 1 mile will be covered by David in half n hour(as david is gaining 2 mile /hr,so to gain 1 mile it will take David 1/2 hr and the dist will be 2mile b/w david and stacey).==> Suff
In option B
45 mins ago the dist was 4.5 mile and now dist is 3 mile. Therefore david covered 1.5(4.5-3) mile in 45 min so speed of david catching stacey is 2mph.
By same logic as for opton A==> Suff
SO D
02 Mar 2016, 23:11
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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.
David and Stacey are riding bicycles on a flat road at a constant rate. If Stacey is now three miles ahead of David, in how many minutes will Stacey be just two miles ahead of David?
(1) Stacey is traveling at rate of 10 mph and David is traveling at a rate of 12 mph.
(2) 45 minutes ago Stacey was 4.5 miles ahead of David.
In the original condition, there are 2 variables(you need to figure out the speed of David and Stacey as you can figure out the time when they are 1 mile apart by knowing their speed). In order to match with the number of equations, you need 2 equations. For 1) 1 equation, for 2) 1 equation, which is likely to make C the answer.
When 1) & 2), for 1), they are 2 miles apart at 1 hour. So, to be 1 mile apart, 30 minutes should pass, which is sufficient.
For 2), 45 minutes:1.5 miles=x;1 mile, x=30 minutes is derived, which is unique and sufficient.
Since 1) = 2), the answer is D.
For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
David and Stacey are riding bicycles on a flat road at a constant rate. If Stacey is now three miles ahead of David, in how many minutes will Stacey be just two miles ahead of David?
(1) Stacey is traveling at rate of 10 mph and David is traveling at a rate of 12 mph.
(2) 45 minutes ago Stacey was 4.5 miles ahead of David.
In the original condition, there are 2 variables(you need to figure out the speed of David and Stacey as you can figure out the time when they are 1 mile apart by knowing their speed). In order to match with the number of equations, you need 2 equations. For 1) 1 equation, for 2) 1 equation, which is likely to make C the answer.
When 1) & 2), for 1), they are 2 miles apart at 1 hour. So, to be 1 mile apart, 30 minutes should pass, which is sufficient.
For 2), 45 minutes:1.5 miles=x;1 mile, x=30 minutes is derived, which is unique and sufficient.
Since 1) = 2), the answer is D.
For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
