here is a numerical question i got not idea. i can only guess the answer is 565 as it’s asking the least possible number of exams. but i don’t know how to work it out. please discuss.
Hmm, this is a tough one. I think I’d probably skip it if I were on a time scale
Okay lets give it a go:
We know that EVERYONE (138 people) has failed two tests during their time.
We know, THIS YEAR. 31 people have completed these tests
Each person in grade II must have done grade one first. so that’s (35+7=42…*2= 84)
Again, each person in GIII must have completed 1+2 first. So that’s (55+10=65…*3=195)
That comes to 586 (276+31+84+195) over here… That was a rough attempt, maybe someone can put some light on what I missed - or whether you can answer it!
I, like you would guess the same and assume I rounded something incorrectly (on a time scale)
I think your solution is almost correct except for the fact that in steps Grade I, Grade II, Grade III you should count only those students who completed their last exams successfully:
in grade I, only 27 people passed the test, not 31
in grade II, 35 not 42
in grade III, 55 not 65
so 586 (your answer) - 4 - 7 - 10 = 565
You can also calculate it in this (longer) way ((273 + 42) + (353 + 72 + 35 + 7) + (553+102 + 552 + 102) = 565
I hope this is detailed enough and doesn’t need explanation
In other words, the solution is:
“We know that EVERYONE (138 people) has failed two tests during their time.
= 276” [I’m so lazy=)]
“We know, THIS YEAR. 27 people have PASSED these tests”
"Each person in grade II must have done grade one first. so that’s " 35 + (35+7) = 77
“Again, each person in GIII must have completed 1+2 first. So that’s” 55 + (55+10)*2 = 185
and finally, 276 + 27 + 77 + 185 = 565
Ah good spot, I didn’t consider taking into account that one of the 2 fails would be this year.
That’s a painful one. Good to get the practise in.