QUANT Flashcards
25%=
50%=
75%=
100%=
125%=
150%=
175%=
200%=
25%=25/100=1/4=0.25
50%=50/100=1/2=0.5
75%=75/100=3/4=0.75
100%= 1
125%=125/100=5/4=1.25
150%=150/100=3/2=1.5
175%=175/100=7/4=1.75
200%=200/100=2
2x2-18=
2(x-3)(x+3)
196n2-144=
4(7n+6)(7n-6)
294r2-150=
6(7r+5)(7r-5)
150k2-216=
6(5k+6)(5k-6)
16m6-n6=
(4m3-n3)(4m3+n3)
4x4m-36y4m=
m(4x4-36y4) = m(2x2-6y2)(2x2+6y2)
or
4m(x4-9y4) = 4m(x2-3y2)(x2+3y2)
- in general, if you see a common factor between the 2 elements –> take out the common factor and then do the factorization
x2 −7x−18=
(x − 9)(x + 2)
- find the factor pair of the costant that when added gives you the coefficient of x
- if the costant is positive –> the 2 numbers int the factored form must have EQUAL SIGN (both positive: if the x coeff. is + or both negative: if the x coeff. is -)
- if the constant is negative (like in this e.g.) –> in the factored form one number is positive and one number is negative. (if the x coeff. is negative –> the > number is - ; if the x coeff. is positive –> the > number is +)
- if x2 has a coefficient –> look for common factors tu pull out before doing the factorization
- avoid fractional coefficients at all costs
- if it’s an EQUATION (and not an expression): put all numbers on one side, factorize, put each expression = 0 and find the values of x (if x2–> 2 values)
7x2−31x−20=
(7x + 4)(x − 5)
no rules (provale tutte)
7x2 −45x−28=
(7x + 4)(x − 7)
2b2+17b+21=
(2b + 3)(b + 7)
A doctor prescribed 18 cm3 of a certain drug to a patient whose body weight was 120 punds. If the typical dosage is 2 cm3 per 15 pounds of body weight, by what percent was the prescribed dosage greater than the typical dosage?
(A) 8%
(B) 9%
(C) 11%
(D) 12.5%
(E) 14.8%
CLUE: BY WHAT PERCENT WAS X GREATER THAN Y?
%DIFFERENCE= DIFFERENCE/BASIS FOR COMPARISON
- the denominator is always what comes after THAN
1) 18 cm3/120p 2cm3/15p
make thetwo fractions comparable: 18cm3/120p = 16cm3/120p
Typical dosage for a 120 pounds man: 16cm3
2) % difference?
[(18-16)/16]*100 = 12.5%
1) Draw a new line segment BD, splitting the quadrilateral into two right triangles
2) In triangle BCD, either recognize the common right triangle: 3–4–5 or use the Pythagorean theorem to find the length of BD (which is the hypothenuse):
32 + 42 = BD2
9 + 16 = BD2
25 = BD2
5 = BD
3) In triangle ABD, you must use the Pythagorean theorem to find the length of AB (BE CAREFUL! In this case BD isn’t the hypothenuse: because it’s one of the perpendicular legs of the triangle):
12 + AB2 = 52
1 + AB2 = 25
AB2 = 24
AB = sqr24
AB = sqr(4*6)
AB = 2*sqr6
This time, the unknown side was not the hypotenuse, but one of the perpendicular legs of the triangle. Missing this detail might lead to wrong answer (A): 12 + 52 = 26, so if AB were the hypotenuse it would be sqr26. This is not correct!
NOTE: i cateti sono quelli che fanno l’angolo retto
Enclosed by a rectangular fence is code for perimeter, which is 2(w + L) for a rectangle.
However, only three-fourths of the total lawn area is to be enclosed by a fence, and the problem discusses two different ways the enclosed area is created: full width and reduced length, or full length and reduced width.
1) Full width and reduced length implies w = 30 and L=34×40=30. The perimeter of this rectangle is 2(w + L) = 2(30 + 30) = 2(60) = 120 feet.
2) Full length and reduced width implies L = 40 and w=34×30=22.5. The perimeter of this rectangle is 2(w + L) = 2(22.5 + 40) = 2(62.5) = 125 feet.
* The difference between the two perimeters is 125 – 120 = 5 feet. Enclosing the shaded rectangles in the sketches would require 5 fewer feet of fence in the first case than in the second.
