+111 At a competition you need to finish 10 questions.
For every correct question you get 10 points and for every incorrect question you lose 3 points.
To get an award you need 51 points min. How many (minimal) correct awnsers do you need to get an award.
(I think that it is 7 but they say that it is not correct)
AWNSER THIS FAST!
+129899 Let x be the number of questions answered correctly.....then, (10-x ) are the number answered incorrectly......and the questions answered correctly times their point value (10) plus the questions answered incorrectly times their point value (-3) must be greater or equal to 51
10x + (-3)(10-x) ≥ 51 simplify
10x - 30 + 3x ≥ 51 combine like terms
13x - 30 ≥ 51 add 30 to both sides
13x ≥ 81 divide both sides by 13
x ≥ about 6.2 → 7
Mmmm.....I get the same answer you did.....if we only answered 6 qusetions correctly that would be 6(10)- 3(4) = 48 points.......and that's not enough !!
Maybe someome else sees something I'm missing???
+129899 Let x be the number of questions answered correctly.....then, (10-x ) are the number answered incorrectly......and the questions answered correctly times their point value (10) plus the questions answered incorrectly times their point value (-3) must be greater or equal to 51
10x + (-3)(10-x) ≥ 51 simplify
10x - 30 + 3x ≥ 51 combine like terms
13x - 30 ≥ 51 add 30 to both sides
13x ≥ 81 divide both sides by 13
x ≥ about 6.2 → 7
Mmmm.....I get the same answer you did.....if we only answered 6 qusetions correctly that would be 6(10)- 3(4) = 48 points.......and that's not enough !!
Maybe someome else sees something I'm missing???
+111 Thanks. I got the awnser like this:
$${\mathtt{x}} = {\mathtt{10}}$$
$${\mathtt{y}} = {\mathtt{3}}$$
$${f}{\left({\mathtt{a}}\right)} = {\mathtt{xa}}{\mathtt{\,\small\textbf+\,}}{y}{\left({\mathtt{10}}{\mathtt{\,-\,}}{\mathtt{a}}\right)}$$
$${f}{\left({\mathtt{6}}\right)} = {\mathtt{48}}$$
$${f}{\left({\mathtt{7}}\right)} = {\mathtt{61}}$$
