GRE Questions

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Using Pythagoras theorem we can see that 12² + 15² is not equal to 18²
Neither is 10² + 24² = 25²

Time from 12 p.m. on Tuesday to 2 p.m. on the following Tuesday = 7 days 2 hours.
= 170 hours.
The watch gains = (2 + 4 x 4/5) min
= 34/5 min. in 170 hrs.
Now, 34/5 min are gained in 170 hrs.
Then, 2 min are gained in (170 x 5/34 x 2) hrs.
Watch is correct after 2 days 2 hrs after 12 p.m. on Tuesday, i.e., it will be correct at 2 p.m. on Thursday.

Despite, Its meaning is "without effected by ", without effecting by their poverty, Kavitha still dresses more appropriately than most of her friends.

Parallel construction error in the original statement. A similar form is required on each side of the comparison. " Diligence - brightness ".

Computer is the correct word.

Let us assume A is the faster and B is slower, A takes 10s to finish a lap of 50 m, B takes 17s(approx) to finish a lap of 50 m.
So at 11th sec A starts his 2nd lap by the time B will be in between so that is the first time they will meet each other, at 21st sec A starts his 3rd lap by the time B will be coming back to finish his 2nd lap. So that is the second time they meet in opposite. Then at 31st sec A will start his 4th lap by the time B would not even finish 2nd lap as he take 34(2*17) sec to finish 2 laps. So at 35th sec B will start his 3rd lap by the time A will be coming back to finish his 4th lap in the half way so they will meet 3rd time. At 41st sec A will start its fifth lap by the time B will be going to finish 3rd lap in same direction. At 50th second both will complete A's 5th lap and B's 3rd lap. So they will start next lap(A's 6th and B's 4th) from the same point in same direction. And A will finish the race after completing 300m(6*50) total distance and then they wont meet opposite to each other.

So totally they met 3 times in opposite direction.

circumference of A = 2πr = 8 => so r = 4π
volume = π×4π2×10 = 160π
circumference of B = 2πr   = 10 => so r = 5π
volume = π×5π2×8 =  200π
so ratio of capacities = 160/200 = 0.8
so capacity of A will be 80% of the capacity of B.

Let S be the sample space.
Here n(S)= 23 = 8
Let E be the event of getting atmost two heads. Then,
n(E) = {(H,T,T), (T,H,T), (T,T,H), (H,H,T), (T,H,H), (H,T,H)}
Required probability = n(E)/n(S) = 6/8 = 3/4.