Advanced Quant Flashcards
Diagonal of a cube
side * sqrt (3)
Diagonal of rectangular solid
Diagonal ^ 2 = L ^2 + W ^ 2 + H ^ 2
An area of a rectangle for a fixed parameter is maximized if that rectangle is square
Addinganumbertotheset such thatthe number isveryclose to themean generally reducestheSD.
SPEED is distance over time. over time. like rate. it’s over time. over time. over time. over time. over time.
If we say the varying distances are the same and the body travels with two varying speeds v1 and v2 over two equal distances, how can we compute the average speed quickly?
2 (v1 * v2 )/ (v1 + v2)
Successive percentage change (e.g., A’s salary is increased by 10% and then decreased by 10%. The change in salary is): x%->y%
(x+ y + (xy) /100) or (x - y - (xy) /100) thus (x - y - (xy)/100)>0 val2 positive.
1. {10 – 10 – (10 x 10)/100} = -1%
Viete’s Theorem
ax^2 + bx + c = 0 —> r1 + r2 = -b/a and r1 * r2 = c/a
Machine A is 50 percent faster than Machine B:
- RATE * TIME = WORK
- If machine B is 50 percent faster than machine A we can write the rate as: R_B = 3/2 R_A
AVERAGE is always…
Between Min and Max if Min equals to our average then all the same and if max equals to our average then all the same.
Evenly Spaced Sets:
{1,2,3,4}, {2,4,6,8}, {0,5,10,15}.
In any evenly spaced sets median and mean are equal! You can again use the backend technique (lowest + highest) / 2 to get the mean.
Catch up time when both objects moving the same direction:
catch up time when both r moving in same direction=?
9. CATCH UP IS ALL ABOUT DISTANCE and SPEED difference!
10. Example: Car A is 20 miles behind car B, which is traveling in the same direction along the same route as Car A.Car A is traveling at a constant speed of 58 miles per hour and Car Bis traveling at a constant speed of 50miles per hour. How many hours will it take for Car A to overtake and drive 8 miles ahead of Car B?
Ask yourself: which object is moving faster.
catch up time when both r moving in same direction = (distance between the two ) / (Speed of Car A - Speed of Car B)
Relative speed of car A is 58-50=8 miles per hour, to catch up 20 miles and drive 8 miles ahead so to drive 28 miles it’ll need 28/8=3.5 hours.
If we wanted to just to catch up but not pass we would get: 20/8
catch up and catch up pass similar to meeting.
f the difference is not in time but rather distance DS = DJ + 120ft (80 ft behind and needs to get 40ft ahead)
Time to catch up and pass plug and chug: ** Time = (delta x)/ (delta v) ** —> JOON is 40ft behind needs to get 80 ahead of Sarah. DJ / DS + 120
Example: We are given that car A is 20 miles behind car B and we need to determine the time when car A is 8 miles ahead of car B. Thus, we can say that the change in distance is 20 + 8 = 28 miles. We are also given that car A travels at a constant speed of 58 mph and car B travels a constant speed of 50 miles per hour. Thus, we can say that the change in rate is 58 – 50 = 8 mph. time = 28/8 = 7/2 = 3.5 hours
Inscribing a square inside another square
The square is its smallest when it touches the midpoints of the side. In other words when I inscribe a square inside another square, the minimum area that the inscribed square can have is half the circumscribed square. Same with the shaded area.
The shaded area can be at most 1/2 of the larger square (Manhattan advanced quant), which occurs when the smallest possible square is inscribed in the larger square. This gives us a great cutoff!
