Booster PT- QR

TRICK QUES: quantity A: √49 quantity B: the val of x in x^2= 49

CANNOT BE DETERMINED!!!!!(?) for the DAT's sake, √49= +7 ONLY (neg roots arent true roots i suppose) and x^=49 will yield both +/- 7, which makes it difficult to determine if A is greater (if B is -7) or if theyre equal (if B is +7) so results are inconclusive.

a train is traveling @ 120km/hr, how fast is it traveling in ft/sec (1in=2.54cm)

120km/hr(hr/3600sec)(1x10^3m/km)(cm/1x10^-2m)(in/2.54cm)(ft/12in) km, hr, m, cm, and in cancel out, leaving ft/sec so u have 120/3600(1x10^3/1x10^-2)(1/3ish)(1/12) (using calc it gives 109, which ~110)

.5% of 12% of 150

150x.12x.005= .09 (use calc)

HIGH YIELD: a dice is rolled twice. quantity A: prob of rolling 2 odd #s quantity B: prob of rolling odd and even #

A. rolling odd AND odd: (1, 3, 5) = 3/6 x 3/6 = 9/36 = 1/4 B1. can either roll odd AND even: (2, 4, 6) = 3/6 x 3/6= 1/4 B2. or roll even AND odd: also 1/4 total for B: 1/4 +1/4 (taking both possibilities into acc) = 2/4 = 1/2 1/4 from A < 1/2 from B, so B is greater

V(avg)=

D(tot)/t(tot)

HY: if 16 tennis players are in a tournament, and each player must play every other player once, how many matches must be played?

SIMILAR to the "how many lines can be drawn to 8 different points" ques, bc no repeats there as well u do 16-1= 15, and then add the consecutive descending numbers. so 15+14+13+12+11+10+9+8+7+6+5+4+3+2+1= 120

HIGH YIELD QUESTION: currently tommy is 5x as old as bianca. in 8yrs, tommy will only be 3x as old as bianca. how old is tommy rn?

T= 5B T+8=3(B+8) [THINK: T IS 3X AS OLD AS B AFTER THE 8YRS PASS) so u substit T for 5B, and distrib the 3, which would yield: 5B+8=3B+24 24-8= 16 5B-3B= 2B so 2B= 16 B=8 then PLUG BACK IN to the T=5B equat Tommy is 40yrs old

the vol n surf area of a cube are equal. what is the edge length of the cube

V(cube)= s^3 SA(cube)= 6s^2 bc theyre equal, u set them equal to each other, so s^3=6s^2 divide s^2 on both sides to yield s=6

HY: a basketball player who only scores 40% of the time, shoots the ball three times. quantity A: 30% quantity B: the prob that he scores exactly 2 out of 3 shots

if they score 40% of the time, then they miss 60% of the time. prob of scoring 2 out of 3 shots: P(score/score/miss)= .4 x .4 x .6= .096 P(score/miss/score)= .4 x .6 x .4= .096 P(miss/score/score)= .6 x .4 x .4= .096 tot P= .096 x 3= .288 aka 28.8% which is LESS than 30% quantity A > quantity B

HY: max sells stickers to make extra cash. he spends $50 to purchase 200 stickers. when the stickers arrive, he notices that 20% of them are damaged and unable to be sold. if max wants to make a 60% profit, what should he charge per sticker?

50/200= .25 per sticker 50x160% (***bc PROFIT***)= 50x1.6= 80 wants to make 200x.2= 40 damaged, but also 160 not damaged 160x= 80 80/160= .50 he must charge 50 cents each to gain a 60% profit

HIGH YIELD: if f(x)= -x^3 + 2x^2 - x, and g(x)= -x^2 + 6, what's f(g(3))?

P.I. 3 into g(x) = -(3)^2 is -9, +6= -3 then P.I. -3 for f(x) -(-3)^3 + 2(-3)^2 - (-3) -(-27) + 2(9) + 3 = 27+18+3 =48 (trick: expon ONLY goes to what's in the parentheses, thEN u add the sign on the outside)

HIGH YIELD: f(x)= √(x^3+9) g(x)= (2f(x))^2 +4 what is g(3)

P.I. f(x) where it pops up in g(f(x)) form g(x)= (2(√x^3+9))^2 +4 THEN P.I. the 3 in g(3) = (2(√(3)^2 +9))^2 + 4 = (2√36)^2 +4 = (2x6)^2 +4 = 12^2 +4 = 148

what is 15% more than 10% less than 40% of $5,000?

calc: 5000(.4)= 2000(.1)= 200 LESS THAN= 2000-200= 1800 1800(.15)= 270 MORE THAN= 1800+270= 2070

an equil triangle has sides of length 12. what's the area

can either: (quick) use the EQUIL TRIANGLE AREA FORM: (s^2√3)/4 which would give u (12^2√3)/4= (144√3)/4= 36√3 or: (slow) bisect the triangle into 2 separate 30:60:90 triangles with bases of 6 and hyp of 12. the RULE of 30-60-90 triangles is that the sides= 1:2:√3. so side 6 is 1, hyp is 2, and leg that u bisected is √3. BE CAREFUL bc √3 is ACTUALLY 1√3, so leg is actually 6√3. then P.I. to regular triangle area form, 1/2(b x h) 1/2(12)(6√3) = 1/2 x 72√3 = half of 72 is 36 = 36√3

HIGH YIELD: C and D are positive integers. is C odd? 1) C+D is even 2) C^2 x D is odd

choice 1 can be true (C+D=even), but it's inconclusive bc in both cases where C+D=even, C itself could be odd or even. (1=3=4, 2+4=6) so statemt 1 alone is INSUFFICIENT. for choice 2, for a product to be odd, each factor needs to be odd as well. so C^2 and D are odd. note that C^2 is a prod of CxC, both of which are also odd, which makes C as a whole, DEFINITIVELY, an odd #. only statement 2 is sufficient to answer this question. (basically, for statement 1 to be even, C could be EITHER even or odd, which doesnt answer the question. however, for statement 2 to be odd, C can ONLY be odd, which answers the question.)

***HIGH YIELD (kinda long tho): chris can shovel a parking lot in 6hrs, alex can shovel the same parking lot in 5hrs. if they both work tg, how many hrs will it take them to shovel the parking lot if alex only works for 2hrs?

find combined work rate 1/6 + 1/5= 5/30 + 6/30= 11/30th of the work tg in 1hr now we mult by 2 bc theyre working tg for 2hrs 11/30 x 2= 11/15 x 1= 11/15th of the parking lot gets done in 2hrs, so then chris has 4/15ths of the lot to finish by himself. we know that he can finish 1 lot/6hrs, and he can do 4/15ths of the job in x amount of hrs. set up a proportion: 1/6 = (4/15)/x = 6 x 4/15= 1x = 6 and 15 both go into 3 = 2 x 4/5 = 8/5 = 1.6hrs so it'll be 2hrs w/ alex and chris both working, and 1.6 with chris finishing the lot by himself= 3.6hrs = 3 and 6/10hrs = 6/10 of an hr x 60min (to convert hrs to min) = 360/10= 36min so it'll take chris and alex 3hrs and 36min to complete the job if alex only works 2hrs.

HY: given the following expression, what is 16^x/2^y? (y+10)/4 = x

so solve for y in terms of x y= 4x-10 P.I. for y in the expression 16^x/2^4x-10 substit 16^x into 2^4x so it's 2^4x/2^4x-10 = 2^4x-(4x-10) = 2^4x+4x+10 = 2^10

HIGH YIELD: sarah bought a sandwich in nickels, dimes, and quarters. she used more nickels than dimes, and more dimes than quarters. what's the minimum amount of coins she could have used if the order was $3.45?

let x= quarters (bc used the least) x+1= dimes and x+2= nickels (bc used the most) then multiply each by the amount per coin, and set equal to tot amount to solve .25x + .10(x+1) + .05(x+2) = 3.45 .25x+10x+.10+.05x+.10=3.45 3.45-.20= 3.25 .25x+.10x+.05x= .40x so .40x=3.25 quick: round 3.25 to 3.20/.40 = 32/4= 8 now P.I. to equat from before to see if u need to add any more coins to equal $3.45 .25(8)+.10(9)+.05(10) 2+.9+.5= $3.40 we need $3.45, so we only need 1 more nickel = 8 quarters, 9 dimes, and 11 nickels= 28 coins.

HIGH YIELD: 5 fair coins are tossed. what is the prob of the toss yielding @ least 4 heads

ok so total # of combin that can be made= outcome^# of items/chances= 2^5= 32 tot combos in the case of 4 heads and 1 tail: n!/(n-k)!= 5!/(5-1)!, bc HHHHT, HHHTH, HHTHH, HTHHH, and THHHH. so 5 permutations that would satisfy the ques, and 1 will be picked, so that's why it's 5!/(5-1)!= 5!/4!= 5 bc said @ LEAST 4, must take into account the chance that all 5 are heads: HHHHH, aka only 1 chance that this will occur. in all, there are 6 different chances out of the 32 tot combos that meet the requirements. 6/32 simplify into 3/16.

HIGH YIELD: 42% of 83 is 17% of what?

on calc: .42x83= 34.86/.17= 205

HY: after flipping a coin 4 times, what is the prob of not getting all tails?

per coin, there's a 1/2 chance of getting tails. the trick here is that u need to keep in mind: 1 - P(all tails) = P(not all tails) so find the off-chance FIRST, then subtract by 1 to get what theyre asking u coin 1= 1/2 T, 1/2 T coin 2= 1/2 T, 1/2 T (bc these ARENT separate flips of the coins, u mult all of them as 1 flip of BOTH coins) 1/2 x 1/2 x 1/2 x 1/2= 1/4 x 1/4= 1/16 for P(all T) now we do 1- P(all T) to get P(NOT all T) 16/16 - 1/16 = 15/16

test2, #23: (look to see if in rom's book bc long too, or if rom's way is better bc sheesh) what is the vertex form of the parabola expressed by the equat: 3y - 12 = -3x^2 + 6x

put it in standard y= ax^2 + bx + c 3y= -3x^2 + 6x + 12 y= -x^2 + 2x + 4 then put in VERTEX FORM y= a(x-h)^2 + k where h is x-coor, and. k is y-coor only focus on "a" and "b," aka -x^2 + 2x, making a= -1 and b= 2 so bc vertex from has x-h, and we have a + sign in between -x and (+) 2x, we have to get a neg sign in b/w them. to fit the x(-)h format so we factor out the -1 y= -(x^2 -2x +___) + 4 - ____ this makes "a" now = 1 and "b" now = -2 b is assoc w/ x, so we take b, divide it by 2, and square it = (b/2)^2 bc b= -2 in the vertex form, we P.I. -2 = (-2/2)^2 = (-1)^2 = 1 goes in the gaps y= -(x^2 -2x + 1) +4 - 1 *NOTE*: bc of the neg sign in the front negates the 1 in the parenth, we need to invert the sign on the outside 1, so it can cancel out the -1 after distrib the - outside the parenth. so it's actually y= -(x^2 -2x +1) +4 +1 now convert into a squared binom what adds to -2 and mult to +1, and square it. then simplify/add on what's outside the parenth y= -(x-1)^2 + 5

quantity A: the numerical val for the internal angle of a reg 14-gon quantity B: the numerical val for the are of a circle w/ circum of 16pi

quantity A: form to find sum of internal angles: 180(s-2) so 14-2= 12x180= 2160/14 angles= 154.3 for 1 internal angle quantity B: C= 16pi, so diam=16, radius must= 8. A= pi x r^2= 64 x pi= 201 so quanB is greater

a bowl contains 8 white marbles, 6 red, and 2 black. 2 marbles are drawn, w/out replacement. quantity A: the prob that both marbles drawn are white quantity B: the prob that only 1 of the 2 drawn is black

quantity A: 8white/16tot x 7white/15tot = 1/2 x 7/15= 7/30 quantity B: u must take into acc BOTH scenarios, where u first draw a black marble first and a red or white marble 2nd, OR a red/white marble 1st, and a black marble 2nd. P (black and non-black)= 2/16x14/15= 28/240 simplified is 7/60 P (non-black and black)= 14/16x2/15= 28/240 simplified is 7/60 then add the two, so 7/60+7/60= 14/60= 7/30 quantity A= quantity B

what is the value of x? log5 (7-x) - log5 (3) = log5 (x+5/2)

quotient rule: loga(x) - loga(y)= loga(x/y) equality rule: loga(x)= loga(y), then x=y quotient rule 1st: log5 (7-x) - log5 (3) =log5 (7-x/3) set = to orig prod log5 (7-x/3) = log5 (x+5/2) now use equality rule: (7-x/3) = (x+5/2) cross mult and simplify 2(7-x) = 3(x+5) 14-2x=3x+15 5x= -1 x= -1/5

HIGH YIELD: evaluate 5log₂(32^3)

recall the POWER LOG RULE: log(x^y)=y*log(x) and logs are the opp of powers, so log₂x= 2^y=x so 5 x log₂(32^3)= 5 x 3 (bc opp power) x log₂32 log₂32= 2 to WHAT POWER will = 32, so that =5 so 5 x 3 x 5 = 25 x 3= 75.

***HY: the grades on a chem exam are normally distributed w/ a mean of 82% and a SD of 6. what is the prob that a student has scored 88% or higher on the exam?

remember the 68%:95%:99.7% rule with SD for 1SD:2SD:and 3SD, respectively know that bc given SD is 6, +1SD is 82+6= 88, and 82-6= 76. w/in 1SD of the mean is 68% (34% above and below) to find % data that is outside that 68% (higher than the 88 that's w/in 1SD), u subtract 100%-68%= 32% we have to think 32% above AND below. we only need above, so it's 32/2= 16% of the grades fall ABOVE 88% (as well as 16% falling below)

HY: max can pack 3 lunched in 10min. jeff can pack 5 lunched in 20min quantity A: # of mins it takes max to pack 12 lunches quantity B: # of mins it takes jeff to pack 11 lunches

use dimensional analysis here. max: (12 lunches)(10mins/3lunches)= 40min jeff: (11 lunches)(20min/5lunches)= 44mins it takes jeff more time to pack 11 lunches than it takes max to pack 12 lunches. OR: set up a proportion: 3lunches/10mins = 12lunches/x, x(max)= 40min 5lunch/20min = 11lunch/x, x(jeff)= 44min either way, quantity B > quantity A

HIGH YIELD: a committee of 4ppl is to be selected from 3 men n 6 women. if the selection is made randomly, what is the prob that there are 2 men and 2 women chosen to be on the committee?

(***omit tho if not in romano's book bc LONGGGGG ASF***) find tot # of outcomes, nCk= n!/(n-k)!k! (think: No! No calvin Klein! oK!) or: tCa= t!/a!(t-a)! (think: combo= T!/A! T-A!, like "tada!") where t is total and a is what's being asked. bc there are 3 men and 6 women= 9ppl tot and 2 men and 2 women chosen= 4ppl that ur asking for = 9!/4! (9-4)! 9!/ 4! x 5! 9x8x7x6/4x3x2x1 (the 3x2 cancel w/ 6) 9x8x7/4 =126 committees that can be created, but now we have to narrow it down to committees w/ 2 men and 2 women so that's 2 men out of 3, and 2 women out of 6. tCa with 3 tot men and 2 asking for = 3!/2!(3-2)! =3!/2! =3 ways to chose 2 men tCa, 6 tot women w/ 2 asking for 6!/2!(6-3)! =6!/2!4! = 6x5/2x1 =15 ways to chose 2 women final prob= mult the ways to chose 2 men AND 2 women. then divide that by tot # committees. (like part/whole) 15x3 men AND women/126 tot committees =45/126 (divisible by 9) =5/14

HIGH YIELD: if 5a-10b=61, and 12a-3b=131, what's a=b?

(12a-3b=131) - (5a-10b=61) = 7a+7b=70 simplify: 7a/7 + 7b/7= 70/7 so a+b= 10

HIGH YIELD: if a 4L soln is 80% conc lime juice and 20% water, how much 30% conc lime juice must be added so that the conc of the resulting soln is 50% conc lime juice?

(DOUBLE CHECK IF ROMANO'S WAY IF QUICKER) 4L x .8= 3.2 lime juice lime juice/total vol= 50% or 1/2 so set up like a proportion lime juice/tot vol= 1/2 plug in 3.2 + .3x (for lime juice added) to take into acc all the lime juice. set that over 4L + (x)L (to take into acc both vol we started w/ and unknown vol of conc lime juice) for tot/resulting vol and everything must be equal to the 50%, or half soln lime juice 3.2L x .3(x) / 4L + (x)L = 1/2 2(3.2L + .3x L) = 4L + x L 6.4L + .6x L = 4L + x L 6.4-4= 2.4L 1x - .6x= .4x L 2.4= .4x x= 6L

if x and y are odd integers, which of the following must be even? x, y>1

(x-1)(y+1) let x= 3, and y=5 integer= non-decimal whole # (3-1)(5+1)= 2x6= 12, satisfies the ques bc even #

solve the following equat

-7x/2x+6 = 4x/6x+18 + 3x-2/x+3

5 students arrive at a math class. they learn that only 4 desks are in the classroom. quantity A: different ARRANGEMENTS of students that can sit in the 4 desks quantity B: different GROUPS of students that can sit on 4 desks

A talks about the students and their ARRANGEMENTS, which means ORDER MATTERS, and a permutation will be used. large#Psmall#= l!/(l-s)! 5P4= 5!/(5-4)!= 5!/1!= 120 B talks about the GROUPS, and order doesnt matter bc they dont have to be arranged in a certain way, so a combination will be used. large#Csmall#= l!/s!(l-s)! 5C4= 5!/4!(5-4)!= 5!/4!(1)!= 5x4!/4!= 5 A>B

HY: a truck driver begins traveling @ 40mi/hr. after driving @ a constant speed for 1hr, the driver lowers his speed by 10mi/hr every 15min until he comes to a complete stop. quantity A: # of mi the truck traveled during the 1st hr quantity B: twice the # of mi the truck traveled after the first hr until it stopped

A: d=vt d= (40mi/hr)(1hr)= 40mi B: after the first hr, the speed decreases to 30mph. 15min later, it's at 20mph. 15min after that, it's at 10mph, and will stop after another 15min. so 15min (1hr/60min)= 1/4hr dist traveled during the first 15min: 30mph (1/4hr)= 7.5mi 20mph (1/4hr)= 5mi 10mph (1/4hr)= 2.5mi tot dist traveled: 7.5 + 5 + 2.5= 15mi B. 15 x 2= 30mi so A>B

***HIGH YIELD: an airplane going from LA to chicago travels @ 600mi/hr. however, the same airplane travels @ 400mi/hr on the way back due to turbulence. quantity A: the avg speed of the airplane for the round trip quantity B: 500mi/hr

D= vt, so V(avg)= D(tot)/t(tot) to find D(tot), we know the same dist was traveled from LA to chic as it is from chic to LA, so distance= 2D now we find time: t(LA to chic)= D/600 t(chic to LA)= D/400 so t(tot)= (LA to chic) + (chic to LA) t(tot)= D/600 + D/400 but need to find common denom. t(tot)= 2D/1200 + 3D/1200= 5D/1200 now that we have D(tot) and t(tot), we P.I. 2D/(5D/1200) 2D x 1200/5D= 2400D/5D, the D's cancel out, calc for 2400/5= 480mph

HY: solve -7x/2(x+3) = 4x/6(x+3) + 3x-2/x+3

LCD is 6(x+3), bc 2(x+3), 6(x+3), and x+3 all go into it evenly. next, MULT all frac by the common denom to cancel out the (x+3)'s: -7x/2(x+3) x 6(x+3) = -7x/2 x 6= -7x (6/2) 4x/6(x+3) x 6(x+3) = 4x 3x-2/x+3 x 6(x+3) = (3x-2)6 now it becomes: -7x(3) = 4x + 18x - 12 -21x= 22x - 12 -21x-22x= -43x -43x = -12 12/43 = x

HY: the avg age of 30 randomly selected dentists is 42yrs. one of those dentists is replaced w/ a new graduate, who is 26yrs old. quantity A: the avg age of the 30 dentists, including the new grad quantity B: the avg age of the orig 30 dentists

TRICK: there's not enough info to answer the ques!!! the dentist that was replaced w/ the new 26yr old graduate could have been younger, older, or the same age as them. we cant tell how the avg will be affected bc we know too little about the replaced dentist. if the replaced dentist was younger than the 26yr old grad, then the avg would have gone up. if the replaced dentist was older, than the avg would have gone down. there would've been no change if the replaced dentist was the same age as the grad.

HIGH YIELD: if 2x-4y=20 and 4x+y=22, what is x+y?

bc in the 2nd equat already has y by itself, we can isolate that and P.I./substit that for y in the 1st equat. 4x+y=22 y=22-4x P.I. to 1st equat, so everything is in terms of x 2x - 4(22-4x)=20 2x-88+16x= 20 combine like terms 20+88=108 2x+16x=18x 18x=108 x=6 then u P.I. x into either orig equat 4(6)+y=22 24+y=22 y=24-22= -2 x+y= 6+(-2)= 4 (basically: if there is a variable that is by itself in one of the equat, isolate it, P.I. to the other equat to get everything in terms of the other variable, solve for that variable.. then P.I. to either one of the orig equat to get the other variable that u isolated earlier. then find x+y.)

HY: 7ppl line up in front of a store. if the front 3 in the line must stay tg and must remain at the front of the line, how many ways can ppl in the line be arranged?

bc the first three ppl MUST stay tg and cant rearrange themselves with the other 4 ppl, the first 3 spots should be taken up by: 3 x 2 x 1= 6 and the same rule applies for the last 4 ppl, they can only rearrange with themselves for the last 4 spots. so it's: 4 x 3 x 2 x 1= 24 now bc it's 7 spots tot, 3 spots AND 4 spots= 6 x 24= 144

HIGH YIELD: a sock drawer contains 7 pairs red, 4 blue, and 2 yellow. what's the probability that 2 socks selected @ random w/out replacement, are both red?

bc the socks are in PAIRS, u multiply the amounts by 2. 7x2= 14 red 4x2= 8 blue 2x2= 4 yellow so actually 26 socks total 14/26 x 13/25, simplifies into 7/13 x 13/25= 7/25 chance

HY: jay reaches into a bag containing the letters of the word "NOTEBOOKS" and randomly pulls out 2 letters. what is the prob that 1 letter is a vowel and the other is a consonant in any order

bc they dont state, assume NO REPLACEMENT!!! P (vowel, consonant)= 4/9 x 5/8 = cross-reduce the 4 and 8 to yield 1/9 x 5/2 = 5/18 P (consonant, vowel)= 5/9 x 4/8 = reduce 4/8 to 1/2= 5/9 x 1/2 =5/18 P(tot)= 5/18 + 5/18= 10/18 = 5/9

how many 4 digit #s can be made from the following digits if no repeats? 0,3,5,7

bc they need a 4 digit #, so 0 CANNOT be included in the choices for the first digit (like 0357= just 357). so there are only 3 REAL choices for the first digit. so u would do 1st digit x 2nd x 3rd x 4th possible choices = 3x3x2x1= 18 choices

HIGH YIELD: √0.000016/√0.0004

break it down into EVEN exponents (so they'll break down into halves when sqrt): √16x10^-6= 4x10^-3 √4x10^-4= 2x10^-2 -3-(-2)= -3+2= -1 4/2= 2 so the answer is 2x10^-1 aka 0.2

HIGH YIELD: what is the vertex of the parabola given by the equation: y + 7 = x^2 - 2x = 8

convert the equat into ax^2+bx+c=0 y+7= x^2-2x-8 y= x^2-2x-15 x^2-2x-15=0 now factor and solve (x-5)(x+3)=0 x=5 and -3 the HALFWAY POINT in b/w these #s, aka the VERTEX, is @ x= 1 P.I. 1 for x in the "y=" equat from earlier y= x^2-2x-15 y= (1)^2-2(1)-15 y= -16 so the vertex is at (1, -16)

***HY: the points (x,-4) and (9,2) have a distance of 6 between them. find x.

distance= (X1-X2)^2 + (Y1-Y2)^2 (6)^2= (x-9)^2 + (-4-2)^2 36= (x-9)^2 + (-6)^2 36= (x-9)^2 + 36 √0= √(x-9)^2 0= x-9 x= 9

how many distinct lines can be drawn connecting 8 points on a plane

draw a diagram of 8 dots from point A, we can draw 7 lines to every other point u dont want repeating lines (B->A = A->B), so every line will have 1 less line than the one before so A= 7 B= 6 C= 5 D= 4 E=2 F= 1 and G= all repeats, so dont count 7+6+5+4+3+2+1= 28 lines (rule of thumb: take # of points asked, subtract 1, and then add the rest of the desc consecutive #s down to 1)

what is the price of concert ticket A if the price of concert tickets A and B tg is $400 1) the price of ticket B is $250 2) the price of 2 ticket A's and 4 ticket B's costs $1300

each statement alone is sufficient. for statement 1, if B is $250, then A is $400-$250= $150 bc statement 1 is sufficient, we know that it's either sufficient alone or B is also sufficient alone. cancel out anything that says A is NOT sufficient or insufficient by ITSELF. for statement 2, we'll use the equat/logic: A+B=400 (from the given), and 2A+4B=$1300 (from stmnt 2) isolate a variable. bc the ques is asking about ticket A, we wanna keep everything in terms of A. so B= 400-A, substitute that into stmnt 2, and solve for A. 2A+4(400-A)=1300 2A+1600-4A=1300 1300-1600= -300 2A-4A= -2A -300/-2 = $150=A, which matches stmnt 1. in all, both statements ALONE will solve the problem. (thinking in hindsight: the ques is asking for ticket A, and gives the tot of A+B. choice 1 gives u the price of ticket B, aka the other missing variable. it is sufficient. choice 2 gives u multiples of A and B, and a total of those multiples. u also have the prices of A and B -in terms of each- from the given. all u have to do is plug in, and ur done. no need to solve, just think thru it.)

HIGH YIELD: out of a 50 member class, 36 play basketball, 18 play baseball, and 8 play both. how many play neither?

find out who plays each sport individually. subtract both from each. 36-8= 28 basketball only 18-8=10 baseball only so overall, 10 bsebll + 28 bsktbll + 8 both = 46 ppl that play @ LEAST 1 sport OR (quicker) u can add both givens per sport and just subtract the OVERLAP (aka both), so players arent counted twice. 36+18-6= 46 ppl play @ least 1 sport then subtract the tot # ppl by those who play @ least 1 sport to get those who play neither 50-46= 4 dont play a sport

HY: (f(3)-2)^2 f(x)= √x^2+9x

first P.I. 3 for x and solve f(3)= √(3)^2+9(3) = √9+27 = √36 = 6 then subtract 2 6-2= 4 and square it 4^2= 16

which of the following equat forms a perp line to 16=3y+5x

first get it into y=mx+b form so y= -5x/3 +16/3 perp= neg recip so find equat w/ simplified m val of 3/5, and that's the answer

HIGH YIELD: solve the following ineq: 4| 14-2(-x+3) | <48

first isolate abs val by dividing both sides by 4 =| 14-2(-x+3) | <12 then dist the -2 to get rid of parentheses =| 14+2x-6 | <12 combine like terms =| 8+2x | <12 to get rid of the abs val, we write it as a COMBINED ineq = -12< 8+2x <12 then isolate the x by subtracting 8 from both sides = -20<2x<4 divide everything by 2 to get x by itself = -10<x<2

***GO OVER (test2 #9) HIGH YIELD: a scientist has 5oz soln (10% alc) in a beaker. if the scientist adds more water to the soln to dilute it down to 5% alc, what's the tot amount of the final soln in oz?

first we need to find how much soln the scientist added to dilute it from 10% to 5% alc. alc/tot vol = X oz/5oz = 10% X oz/5oz= .10 X= .5oz (amount of alc the orig soln had) now we need to figure out how much water was added to create the new diluted soln. (our soln has .5oz of alc, and only water was added to the soln, to the amount of alc stays the same) 0.5oz/Y oz = 5% Y= 10oz OR this way: 5oz 10% alc X oz water 0% alc (the diluter) ------------------------- 5+X 5% alc final soln ironically, 5% is right in b/w 0% and 10%. this means that u had to add the same amount as the starting soln bc the conc of the final soln is halfway b/w (2 halves= a whole). this means that 5oz of water was added to orig 5oz.= 10oz is the vol of the new soln.

the dimensions of sam's backyard are 60ft by 32ft. if sam's pool is a circle with A= 314sqft, how many pools of the same size can he fit in his backyard

first, find the radius of the pool by doing the area form in reverse: 314sqft= (3.14)r^2 314/3.14= r^2 100= r^2 r= 10ft bc the r= 10ft, then diam= 20ft bc the backyard is 60ft by 32ft, we know that 3 circles of 20ft in diam can fit the span of the backyard. so 3 pools can fit in his backyard

HIGH YIELD: which of the following equat is parallel to 2y-6= -8x+4 and passes thru the point (1,3) A. y= -4x+7 B. y= -4x+5 C. y= 4x+6 D. 1/4x +5 E. 1/4x-3

first, put into y=mx+b form 2y(/2) -6(+6) = -8x+4(+6) / 2 y= -4x+5 (narrowd down to y= -4x+7 and y= -4x+5, bc same slope) now P.I. (1) for the x val to see if it'll equal (3) in the y val y= -4(x)+5 y= -4(1)+5 -4+5= 1 1 doesnt equal 3. doesnt work. so now we try the other choice w/ a slope of -4 y= -4(x)+7 y= -4(1)+7 -4+7= 3 3=3, so A is the answer. (basically put into y=mx+b form, then P.I. x coor to see if it equals y coor)

HIGH YIELD: solve: |4(2+x)| < 24

first, solve what's in the abs val sign. |8+4x| < 24 bc it's absolute val, u write it as a COMBINED INEQ, so it negates/cancels out the abs val. -24 < 8+4x < 24 simplify by subtracting 8 from both sides of the ineq -32 < 4x < 16 then simplify fully by dividing all sides by 4 to yield the answer: -8<x<4

HIGH YIELD: the mean # of classes taken per semester is 4, with a SD of 1. assuming normal dist, what is the percentage of students that are taking 6 or LESS classes per semester

rule of thumb: 68:95:99.7 rule 68% falls w/in +/- 1SD of the mean 95% falls w/in +/- 2SD of the mean 99.7% falls w/in +/- 3SD of the mean that being said: if students are taking 6 classes OR LESS, when the mean is 4. so that means that u wanna shoot fro 3SD fo the mean, aka 99.7% also bc 95% falls w/in 2SD, then 5% falls outside that range. bc u take BOTH pos n neg ranges, u divide 5% by 2= 2.5% students take more than 6 classes, and 97.5% take LESS than 6 classes per semester.

HY: x - y = 5 x + 3 = -y quantity A: x^2 - y^2 quantity B: x^2 + 3y

set equat 1= equat 2 to find out what x and y are. x-y=5 + x+y=-3 ---------- 2x=2 so x= 1 we can choose an equat to P.I. x for. (1)-y=5 y= -4 then P.I. to quantities A and B A: (1)^2 - (-4)^2= -15 B: (1)^2 + 3(-4)= -11 B>A

HY: an investment of $1,800 decreases by a third. its new value is 20% of the value of a 2nd investment. what's the value of the 2nd investment?

so the investment was decreased by a third= 1800/3= 600 new value= 1800-600= $1200 bc the new val was 20% OF the 2nd investment, set up a proportion: is/of= %/100, where "of" is the 2nd investment, and "is" is the investment after it was decreased by a third. 1200/x = 20/100 (cancel out 0's) 120/x = 2/10 120x10= 1200/2= $6000 for the 2nd investment (basically, find the amount of the investment after it was decreased, and then P.I. to a proportion if they give u a "x% OF" question)

HY: alfred buys a new book from a local bookstore and receives a 30% discount off its orig price. when paying for the book, he pays an addit 8% in sales tax. if alfred paid x dollars to the cashier, which of the following represents the orig price of the book?

so think of this in terms of him PAYING 70% of the orig price, and 108% of the DISCOUNTED PRICE. (bc 1-.30 discount= .70 paid, and 1+.08tax after discount= 1.08) X paid= (.70discount)(y orig price) x (1.08tax) so y orig price= X/(1.08)(.7)

HY: if x=3y, what amount of x equals 6/5 of y?

solve what they gave u 1st to get everything in terms of y: if x= 3y, then y= x/3 now bc x= 6/5y, we can P.I. from earlier. 6/5(x/3)= 6/15x (div by 3) = 2/5x

new machine can prod X dolls in Y min, older machine can prod Z dolls in A hrs. if both machines work tg, how many dolls can be prod in H hrs?

start w/ old machine bc already in hrs old mach can prod z/a dolls/hr new mach can prod x/y dolls/min, but 60min= 1hr, so actually= 60x/y dolls/hr add them for the combined work done= (60x/y + z/a) finally, multiply combined term by h to find out how many dolls they make in h hrs final answer: h(60x/y + z/a)

HIGH YIELD QUES: what's the area of a triangle enclosed by the lines 1) y=8 2) y=3x/2 + 5 3) y= -x

start w/ the pnts where the 3 lines intersect. start w/ most definitive val given, and set it equal to all other val of y. in this case, it's y=8. set it equal to equat 2, which will give u 8=3/2x + 5, which gives an answer of x=2. so the first coor is (8,2). do the same for equat 3, so 8= -x, so x= -8. 2nd coor is (-8,8). lastly, bc we've set equat 1= equat 2, equat 1= equat 3, now we have to set equat 2= equat 3 to get the 3rd coor. -x=3/2x + 5 -5= 5/2x x= -2 3rd coor is (-2,2). plot graph, see that base= 10, height= 6 area= 1/2(10)(6)= 30

see if rom has a similar ques!!! -6 < x < 0 quantity A: 10x^2 + 5 quantity B: -(10x^3 - 1)

the given says that x can = -1 to -5 use the 2 extremes: set x= -1 and -5 if x= -1, A>B. but if x= -5, B>A the results are inconclusive. maybe too wide of a range?

a license plate can be made of any 4 letters (excluding M, N, O, P, Q, and R) followed by any 2 digits (excluding 5). what's the prob for a license plate to start w/ a B and end with an odd digit?

there are 26 letters in the alphabet, but minus the 6 that are listed, there are only 20 possible combos for the letters, excluding the first choice bc that HAS to be B. so it's B and ___ and ___ and ___. bc it's AND, it has to be mult. so 1x20x20x20 for the letters. for the numbers, we exclude 5, and have 2 places: ___ and an odd digit. bc these are digits, u have 9 of them (0, 1, 2, 3, 4, 6, 7, 8, and 9). the last digit HAS to be odd, so u only have 4 digits to choose from (1, 3, 7, and 9). that leaves 9x4 for the digits. in all, 1x20x20x20x9x4 calc= 288,000= # of events now we need to figure out the tot possibilities of license plates (***W/OUT specification that first letter has to be B or last # has to be odd***). first 4 letters can be any except for M, N, O, P, Q, and R, so each has 20 possibilities. looking at the digits, the 2 digits can be any except #5, so each has 9 possibilities. so it's 20x20x20x20x9x9= 12,960,000 so 288,000/12,960,000= 0.22

HY: how many 5-digit #s contain 5 different digits? assume that every 5-digit # is pos

there is a max of 10 different #s to choose from bc it can be anything from (0-9) HOWEVER, there are only 9 possible digits for the first # bc 0 cant be the first #!!! so instead of 10x9x8x7x6, it's 9x9x8x7x6 for the 5 digits/spots =27,216

HIGH YIELD: station X and station Y are 480mi apart. a train @ a certian time from station X headed toward station Y @ 60mph. 2hrs later, a train departs from station Y headed towards station X @ 48mph. how far from station X do the 2 trains meet?

this is a DISTANCE AND SPEED problem. let time traveled by train X from station X= t, and time traveled by train Y= t-2 (t is in hrs, and -2 bc it was 2hrs LATER). by the time the trains meet, they must have covered the entire 480mi distance. set up equat: 60(t) + 48(t-2) = 480 60t + 48t - 96 = 480 108t= 576= a lil over 5ish hrs (rounded down) then use d=vt to calc tot distance from station X 5ish hrs (60mi/hr)= 300ish mi (but more bc we rounded down earlier) actual answer= 320mi (or use calc for 576/108 and mult by 60)

for the month of june, kevin sold an avg of 18 popsicles on sunny days and avg of 6 popsicles on rainy days. if it only rained 1 in every 5 days in june, how many popsicles did he sell on avg/day?

this is a MULTIPLE AVG ques., so we have to find the tot # of popsicles sold and divide that by # of days in june (30) bc it rained 1 in every 5 days, then: 30days in june/5= 6 rainy days to find sunny days: 30days-6 rainy= 24 sunny he sold avg 16 on sunny and 6 on rainy days. so: 24sun(18pop)+6rain(6pop)= tot pop sold 432+36= 468 popsicles sold in june, now we need how many per DAY 468/30, round down to 450/30= 15, but more bc we rounded down actual answer= 16 (or use calc)

HIGH YIELD: the prod that joe wins his 1st boxing match is .47, and prob that he wins the 2nd boxing match is .85, so what's the prob that he wins @ LEAST 1 of the 2 matches?

this is a PROB W/ CONDITIONS ques. it's easier to calc the prob of events that ARENT happening, and subtract that from 1. prob that he loses 1st match is .53, and that he loses 2nd match is .15 prob that he loses both (lose AND lose)= MULT them tg .53x.15= .0795 take this and subtract it by 1 to get prob that joe wins @ least 1 match 1-.0795= .9205, or .92

HIGH YIELD: @ the end of 2014, john deposited $1,000 into his bank account. if he didnt deposit any more money into his acc, and the acc earns 4.2% interest/yr, how much money is in the acc by the end of 2019, to the nearest cent?

this is a REPEATED PERCENTAGE CHANGE aka INTEREST problem. if the acc earns 4.2% interest/yr, then the amount in the acc has to be MULTIPLIED by 1.042 every yr from 2014 onward. final amount= Principal (original) x (1+r)^n (think: "FAPx1R^N") or final= O1rn (think: "final o1rion," like orion's belt) where final amount= money in acc when allotted time is over, original= beginning sum, r is the rate in DEC, and n= number of increases 1000x(1+.042)^5 on calc: 1.042x1.042x1.042x1.042x1.042 = 1.228, x1000 = $1228

HY: there are 30 students in a classroom, 12 boys and 18 girls. the avg weight of the 30 students is 165lbs. the avg weight of the boys is 180lbs. quantity A: avg weight of the girls quantity B: 155lbs

to find avg weight of girls in class: 1) set up an equat for avg weight of tot students: x/30= 165 x= 4,950 2) set of another equat for avg weight of tot boys: x/12= 180 x= 2160 3) subtract tot from weight of boys, then set up another equat (set x= to tot - boys) and divide by 18 to find weight for girls 4950- 2160= 2790 (tot weight of girls) so 2790 tot weight girls/18 # of girls= x x= 155 avg weight girls quantity A= quantity B

***HIGH YIELD: u roll 2 dice. if roll a total of 5 or 10 on ur first roll, u win. if u roll a total of 2 or 12, u lose. all other #s are worth 1pt and the game goes on. based on these rules, what's the proc of winning/losing on the first roll?

total combos that the dice can give= 6x6=36 win: 4 combos to roll sum 5: 1,4 4,1 2,3 3,2 3 combos to roll sum 10: 4,6 6,4 5,5 tot combos to win: 7 lose: 1 combo to roll sum 2: 1,1 1 combo to roll sum 12: 6,6 tot combos to lose: 2 bc we can win OR lose, we add them. then divide by total (part/whole) (7+2)/36= 9/36= 1/4 (basically, u see how many combos to win, add it to how many combos to lose, and divide all by tot)

see if rom has a similar ques! a circle and an isosceles right triangle have the same area. what's the ratio of the radius of the circle to the hypotenuse of the isosceles right triangle?

triangle is in a 45:45:90 angle ratio, and therefore, is in a 1:1:1√2 side ratio if the hypotenuse is h= 1/√2, then both legs are h/√2. bc the area of a triangle is 1/2bh, then area= 1/2(h/√2)(h√2)= (h√2)^2 /2 = h^2/2 /2= h^2/2 x 1/2= h^2 /4 now we can equate the area of a right triangle to the area of a circle, to get r:h, or r/h (pi)r^2 = h^2/4 4(pi)r^2 = h^2 (move 4(pi) to right side and h^2 over to left side to get r/h) r^2/h^2 = 1/4(pi) r/h= 1/2√pi

high yield on booster exams: x<0, y =/= 0 quantity A: x^4 + 3x^3/2x^5 + 6x^4 quantity B: x^3y + 2x^2y/3x^4y + 6x^3y is A greater, B greater, they're equal, or cant be determined?

try and see if u can SIMPLIFY, to make it easier to determine which is greater/if equal quantity A: x^3(X+3)/2x^4(x+3) = 1/2x quantity B: x^y(x+2)/3x^2y(x+2) = 1/3x bc x is neg bc it's less than 0, just P.I. -1 to test this. -1/2 < -1/3 -1/3 is greater than -1/2, bc it's a smaller neg val, so quantity B is greater.

HIGH YIELD: if u pick 3 letters from the word ANTEATER w/ replacement, what's the prob that all 3 letters are vowels

vowels: AEAE consonants: NTTR P(vowel)= 4vowels/8 letters= 1/2 since there IS replacement, the prob does NOT change for the 2nd pick, and so on. so the prob of picking 3 vowels for 3 letters is seen as: 1/2 x 1/2 x 1/2= 1/8

LONG ASF a bag that contains walnuts, pistachios, and cashews weighs 3.5lbs. 1/4 of the bag is walnuts, and 3/8 of the bag is pistachios. the tot cost of the bag is $30. if the walnuts cost $10/lb, and the pistachios cost $12/lb, what's the price/lb for cashews.

we need to find how much of the bag is taken up by cashews: walnuts take up 1/4 of the bag, and pistachios take up 3/8 of the bag. common denom= 8 so 2/8 walnuts + 3/8 pistachios = 5/8 so cashews MUST take up the remaining 3/8 of space to equal 8/8, or 100% of the bag now we figure out how much eat type of nut sells for per lb: we know that walnuts take up 1/4 of the bag, so 1/4(3.5lbs)= .875lbs walnuts x $10/lb= $8.75 per lb of walnuts we also know that 3/8 of the bag is pistachios, so 3/8(3.5)= 1.31lbs (rounded down a lil) pistachios x $12/lb= $15.72 $8.75+$15.72= $24.47 $30- $24.47= $5.53 for cashews now we need to find how much PER LB so .875+1.31+x(cashews) = 3.5 2.19+x= 3.5 x= 1.31lbs cashews (also we can just assume that cashews= pistachios bc it also takes up 3/8s of the bag) $5.53 (rounded up)/1.31lb cashews =$4.22 (around that bc we rounded a couple times) actual answer: $4.19

HY: the avg age of 3 students in a high school english class. is 15yrs. if the teacher is 39yrs old, what's the avg age of all 4ppl combined?

work backwards w/ the avg of the students. if the avg is 15, and there are 3 students, then 15x3=45yrs for the tot of their ages. the teacher is 39yrs old. add that to the tot age of the students, 39+45= 84, and divide that by 4 bc u want the avg. 84/4= 21 of all of them combined

HIGH YIELD: d (distance)=

√(X2-X1)^2 + (Y2-Y1)^2 (pretty much sqrt of delta X^2 + delta Y^2) (can be used when X person lives W blocks west= -W, V blocks north= +V. Y person lives U blocks south= -U, Z blocks east= +Z) (think of it in terms of the x and y axis, north is pos, south is neg, east is pos, west is neg.)