nilesh376 wrote:

1. A circular rim 28 inches in diameter rotates the same number of inches/sec as a circular rim 35 inches in diameter. If the smaller rim makes x revolution/sec, how many revolution per min does the larger rim make in terms of x?

a. 48Pi/x

b. 75x

c. 48x

d. 24x

e. x/75

Don't give yourself a headache, just set X=1

So the smaller wheel makes 1 revolution/sec. The rate is 28

Pi inches per second.

Convert inches/sec to inches/minute

28 inches per sec = 28

Pi * 60

To save time you are going to be doing division so let just break 28

Pi*60 into factors \(28

Pi*60 = 7*4*6*5*2*

Pi\)

Then the circumference of the big wheel is 35

Pi.

\(35Pi=5*7*Pi\)

To find its revolutions we take our distance per minute divided by the circumference.

\(\frac{7*4*6*5*2*Pi}{7*5*Pi}\)

You can see that the 7, 5, and

Pi cancel.

Leaving use with the answer \(4*6*2=48 rpm\) choice C

nilesh376 wrote:

2. On a certain road , 10% of the motorist exceed the posted speed limit and receive speeding tickets, but 20% of the motorist who exceeded the posted speed limit do not receive speeding tickets .What percent of the motorists on the road exceed the posted speed limit?

a. 10 1/2%

b. 12 1/2%

c. 15%

d. 22%

e. 30%

A=all drivers

S=speeders

So we want to find \(\frac{(S)}{A}\)

We know that speeders are either caught or they are not caught, so lets break them into 2 groups

T=ticketed speeders

L=(lucky)speeders without tickets

also T+ L =S

The question states \(T= .10A\)

Also stated is \(L = .2S\)

Substitute in \(T+L=S\)

\(T+.2S=S\)

\(T=.8S\) <-- this means the ones who are ticketed represent 80% of speeders

.1A = .8S

We want all the speeders or 1S

So multiple both sides by the same number to yield 1S on the left side

\(\frac{10}{8}\frac{1}{10}A=.8S\frac{10}{8}\)

\(\frac{10}{80}A=S\)

to find percent of A \(\frac{10}{80}=\frac{x}{100}\)

\(x=12.5\)

\(12.5%A=S\)

Choice Bnilesh376 wrote:

3. If d>0 & 0<1-c/d<1 which of the following must be true

I c>0

II c/d<1

III c^2 + d^2 >1

a. I only

b. II only

c. I and II only

d. II and III only

e. I, II and III

Go through them 1 at a time.

I. C>0

must be trueif C<0 then \(\frac{c}{d}\) would be negative 1-(negative) will yield a value greater than 1

II. c/d<1

must be trueif \(c/d>1\) then 1-(number greater than 1)= (negative number), contradicting the original statement

Since I, and II must be true, we are down to answer choices c, or e

III c^2 + d^2 >1

The question states D>0 and we know from I that C>0.

quickest way is substitution, try values for c and d that do not satisfy III but still satisfy the rest

c= 1/4

d= 1/2

\(\frac{1}{4}^2=\frac{1}{16}\)

\(\frac{1}{2}^2=\frac{1}{4}\)

\(\frac{1}{16}+\frac{1}{4} < 1\) <--this opposes III

But when we substitute in the original, we see that it still holds true.

\(0<1-\frac{c}{d}<1\)

\(0<1-\frac{.25}{.50}<1\)

Therefore only I and II must be true. Choice C.

nilesh376 wrote:

4.

1234

1243

.

.

.

4321

The addition problem above shows four of the 24 different integers that can be fonrmed by using each of the digits 1,2,3,4 exactly ince in each integer , What is the sum of these 24 integers

a. 24,000

b. 26,664

c. 40,440

d. 60,000

e. 66,660

I solved this one thinking of permutations

1=A

2=B

3=C

4=D

First pick the thousands digits, this will greatly limit you answer choices

To make the four digit number, I have 4 choices for the thousands digits A,B,C,D

If I pick A then my whole number is going to be 1 thousand something.

How many different choices do I have 6 (3*2*1)

So I thought 6000 + ?

If I pick B then my whole number is going to be 2 thousand something.

How many different choices do I have 6 (3*2*1)

So I thought 6000+12000+?

Repeating this same process I end up with thousands

\(6000+12000+18000+24000=60000\)

Just counting the thousands digits you have 60,000 so answers A,B,C are eliminated

Next do the same process with the hundreds place. If the thousands digit is A(1) then how many complete numbers have B in the hundreds (2*1=2) the same for C and D. A thousand 200*2+300*2+400*2=400+600+800=1800.

So we know just with numbers starting 1,??? the hundreds digits added together will equal 1800.

So our final sum must be greater than 61800 (60,000+1800).

There is no need for any further calculation,

choice E is only answer choice that fits