Josh rode his bike from his house to his grandma’s at 10 mph, he then rode home later on at 15 mph, it took a total of 30 minutes. How far does he live from his grandma?
I am VERY bad at these kinds of questions (involving time/speed) please be thorough as possible
d = r*t, so t = d/r
t = t1 + t2 = d/r1 + d/r2 = d/(10 mph) + d/(15 mph) = 1/2 hr
d * (1/10 + 1/15) = 1/2
d = 1/2 / (1/10 + 1/15) = 3 miles
The key here is to understand that 30 minutes isn’t portional with what was given. First convert the minutes into hours.
Also, If the Yoga club has 42 members and the Sports club has 25 members and there are 18 students that belong to both clubs, how many people total will be at a joint meeting of both clubs, if all members attended? Steps as well please
That would be 49 members. 42 + 25 = total amount of members, but 18 is being counted twice, so you have to subtract it.
Also,
The circumference of a circle is C and the perimeter of a square is P such that P = C/Pi. If the Area of the circle is A, aand the area of the square is A(the letters are notations)s then As=
A) Ac
B) Ac/pi
C) Ac/2pi
D) Ac/3pi
E) Ac/4pi
explain how please?
C = 2Pir, Ac = Pir^2, so:
Ac = Pi(C / 2Pi)^2 = Pi * C^2 / 4Pi^2 = C^2 / 4Pi
P = C/Pi = 4s, so s = C / 4Pi
As = s^2 = (C / 4Pi)^2 = C^2 / 16Pi^2 = Ac / 4Pi
(E)
@avneety I feel like you mistyped the question - “A” should be replaced with “Ac.” Because the quantity A*C (as I first interpreted it) has dimensions (length)^3, which cannot possibly be an area.
A good way to solve this is to break it down bit by bit, after skimming the question once:
Then C = d*pi, where d is the diameter.
So P = d, and the side length of the square is d/4.
[quote]
If the Area of the circle is Ac
/quote
The radius is d/2, so A_c = pi(d/2)^2 = pId^2 / 4
A_s = d^2 / 16.
However the choices are in terms of Ac. We know Ac = pi*d^2 / 4. Dividing by 4pi, we have (Ac)/4pi = d^2 / 16 = As. So (E) is the answer.
If you’re not as confident with algebra and manipulating expressions, you can let C = 4pi and determine P, Ac, As and plug in the answer choices. You will still obtain the correct answer. Even for the students that are very comfortable with algebra, I sometimes recommend doing this simply to check your answer (if you have time).
Compare to this, posted just recently:
http://talk.collegeconfidential.com/sat-preparation/1850343-help-with-math.html#latest
If your question has the answer choices, than the trial and error method would probably be the easiest to solve this problem. (Credit goes to @pckeller and @DrSteve.)
When you get to the answer choice 3 miles, then travel time one way is 3mi/10mph=.3hr; time coming back is 3mi/15mph=.2hr; together =.5hr, or 30min. It works.
If it’s a grid-in question, there are other solutions posted in that thread (@goldenbear2020’s is similar to one of them); pick one that is easy to follow and recreate.
It pays to read other people’s threads! 
@avneety , you seem to struggle a lot with basic verbal problems. If it’s OK, I would like to recommend a resource for you: http://www.barnesandnoble.com/w/how-to-solve-word-problems-in-algebra-2nd-edition-mildred-johnson/1122989540?ean=9780071343077
I bought it the last time I taught Algebra I, so I would have more problems that worked out nicely than were in our textbook. I think they do a good job with the explanations.