The last problem on the no calculator problem of the PSAT requires that you solve 2 equations and 2 variables and give a free answer. The problem works out to H + S = 60 and 54H + 35S = 3360. It requires a significant amount of calculations. I guess some of the problems in the calculator section require more calculations. I guess they are doing this because calculations are required in college classes. I never used a calculator when taking old SAT practice tests, but now it may be needed.
I guess you do need to write out some stuff for that problem.
I have not seen the problem. But for fun, from the equations you wrote, can I guess that it was something like:
You have a total of 60 thingies, some h-type and some s-type. The h-type thingies cost 54 cents each and the s-type thingies cost 35 cents each. If the total cost of the collection is $33.60, how many h-type thingies were there?
Was I even close?
And are you sure those were the numbers? Those are some ugly calculations…
Some calculations needed, but I don’t think it’s that much. Multiply the first equation by 35, take their difference. However one has to be careful to avoid silly arithmetic errors (I tend to make these quite a bit).
Take their difference…and then? Yuck.
@pckeller 35H + 35S = 60*35 = 2100
==> 19H = 3360 - 2100 = 1260
H = 1260/19
S = 60 - (1260/19) = (1140 - 1260)/19 = -120/19, solved.
Some slightly annoying calculations IMO and lots of places to trip up, but nothing too awful.
I think I was slightly off on the numbers, as it came out evenly. S=36. I wrote it out from memory.
@sattut yeah, the solution to this system is (H,S) = (1260/19, -120/19).
But oftentimes systems of equations that are known to have integer solutions will usually have faster ways than doing all of the calculations.
Yeah, I kind of figured the numbers had to be easier…
If you want, you can puzzle this out without equations. These two-category problems all fit the same template.
Start by making all 60 the smaller type. 36x60=2160. That’s too small by 1200. So you need to replace some of the smaller 36-types with bigger 54-types. Each replacement gains you 54-36=18. So you need to make 1200/18 replacements.
It’s the same arithmetic you would do if you were using algebra but without the algebra.
Sorry, the numbers were H+S = 60, 56H + 35S = 2604. So 35H + 35S = 2100, 21H = 504. H = 24, so S = 36. I didn’t find the calculations difficult, but many students do without a calculator. It seemed like worse calculations than were on the old SAT where calculators were always permitted.
You should reduce 35s+56h=2604 by dividing by 7 as the first step. Do not multiply by 35 – it’s too big.
5s+8h=372
Then multiply (s+h=60) by 8
8s+8h=480
5s+8h=372 subtract equations
3s=108 divide by 3
s=36
It’s always better to divide before you multiply when working without a calculator.
You can tell the equation will be reducible because 35 and 56 are both multiples of 7.