1000 PS- section 2- #12

Curious where are you getting these questions?

1) because the distance remains the same you can say rt=(r+5)(t-1)
so... rt=rt-r+5t-5, subtract rt, move r, and you have your R(rate)=5t-5. Put that into the D=RT equation: 450=(5t-5)t, solve for t and get 5t^2-5t-450=0. Divide by 5, and get t^2-t-90=0 which is the same as (t-10)(t+9)=0, Time cannot be negative so T must be equal to 10 solve in original equation R10=450 so R=45.

2) I'm not too sure about this one. Because you could say that c=1, so 1x2x3=6 which is a multiple of 3 but 6/24 does not equal an integer. 1 is odd so statement B doesn't help. Are you sure that you wrote that correctly

1. C

24. If b is the product of three consecutive positive integers c, c + 1, and c + 2, is b a multiple of 24 ?

(1) b is a multiple of 3,
(2) c is odd.

2. try c=3 and c=7 diff results so no C
1. b is multiple of 3
product of three cons. positive integer is always a multiple of 3.
i wonder how this helps. NO B

ans shud be E

Hi.

for question 12 I would use backsolving.

you know that distance = time x speed and distance is given 450.
and that for s+5 t ould be t-1.

plug in the answers for speed and determine the time.

for b) 450 = 40xT, so T=11+1/4, if s were 45 T= 10, not good 11+1/4-10 > 1

for d) 450=50xT, so T= 9, if s were 55 T=8+2/11, not good either 9-8+2/11 < 1

so try for c)
450=45xT, so T=10, if s were 50 T=9, good, 10-9 = 1, C is the answer.

this method can produce very quick results.

for the data sufficency question,
unless you forgot to mention anything i would agree with devesh that the best answer is E.