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GRE Problem Solving Select Many Sample Questions 1 | GRE Problem Solving Select Many Sample Questions 2 | GRE Problem Solving Select Many Sample Questions 3 | GRE Problem Solving Select Many Sample Questions 4 | GRE Problem Solving Select Many Sample Questions 5 | GRE Problem Solving Select Many Sample Questions 6 | GRE Problem Solving Select Many Sample Questions 7 | GRE Problem Solving Select Many Sample Questions 8 | GRE Problem Solving Select Many Sample Questions 9 | GRE Problem Solving Select Many Sample Questions 10
1. Question:Ritu goes to college by bus at 16 km/hr speed. She returns at 9 km/hr by a bicycle. Which of the following is true? Indicate all such choices.
A. Her average speed for the journey cannot be determined
B. Her average speed for the journey is (16+9)/2 km/hr
C. Her average speed would remain the same if she went to college by a bicycle and returned by a bus at the same speeds respectively.
D. Her average speed does not depend on the distance travelled
E. Her average speed is 11.52 km/hr
Correct Answer: C and E
Explanation:Let the distance covered be x km and t1 and t2 be the time taken for the onward and return journeys
Speed = distance/time
16 = x/t1 and 9 = x/t2
x = 16*t1 and x = 9*t2
Average speed = total distance/total time
= (x+x)/(t1+t2)
= 2x/(x/16+x/9)
= 2x/ (9x+16x)/(9*16)
= 2/25*9*16
= 11.52 km/hr
Options A and B are false and E is true.
Option C is true.
Option D is false.
2. Question:Which of the following is equal to C(6,4)? Indicate all such choices.
A. 6!/4!
B. (6+4)!
C. 4!/(6!2!)
D. 6!(4!2!)
E. 15
Correct Answer: D and E
Explanation:C(6,4) = 6!/[4!(4-2)!]
= 6!/(4!2!)
= 6*5/2 = 3*5 = 15
Options D and E are true.
3. Question:A and B invested a total of $7514 at 10% rate of compound interest. A received the same amount in 7 years as B did in 9 years.
Which of the following is true? Indicate all correct statements.
A. A invested $4114
B. B invested $4114
C. A invested $2057
D. B invested $3400
E. A invested $3400
Correct Answer: A and D
Explanation:Let the share of A be $x
B's share will be (7514-x)
The amount after t years at r% rate of interest for a sum P is given by
A=P(1+r/100)^t
According to the conditions,
x(1+10/100)^7 = (7514-x)(1+10/100)^9
x=(7514-x)(1+10/100)^(9-7)
x=(7514-x)(11/10)^2
x=(7514-x)(121/100)
x=7514*121/100-121x/100
x+121x/100 = 7514*121/100
221x/100=7514*121/100
x=7514*121/221
x= 4114
A invested $4114
B invested $(7514-4114) = $3400
Options A and D are true.
[100^t=100*100*....t times]
GRE Problem Solving Select Many Sample Questions 1 | GRE Problem Solving Select Many Sample Questions 2 | GRE Problem Solving Select Many Sample Questions 3 | GRE Problem Solving Select Many Sample Questions 4 | GRE Problem Solving Select Many Sample Questions 5 | GRE Problem Solving Select Many Sample Questions 6 | GRE Problem Solving Select Many Sample Questions 7 | GRE Problem Solving Select Many Sample Questions 8 | GRE Problem Solving Select Many Sample Questions 9 | GRE Problem Solving Select Many Sample Questions 10
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