CAT 2001 | Question: 143

Question

I have a total of Rs. $1,000,$ Item A costs Rs. $110,$ item B costs Rs. $90,$ item C costs Rs. $70,$ item D costs Rs. $40$ and item E costs Rs. $45.$ For every item D that I purchase, I must also buy two of item B. For every item A, I must buy one of item C. For every item E, I must also buy two of item D and one of item B. For every item purchased I earn $1000$ points and for every rupee not spent I earn a penalty of $1500$ points. My objective is to maximise the points I earn. What is the number of items that I must purchase to maximise my points?

• A. $13$
• B. $14$
• C. $15$
• D. $16$

Answer 1

Answer will be 14

there are 3 sets

(1) For every item A, I must buy one of item C

so (A,C) in one set and total costs of (A,C) is 110+70=180

(2) Similarly (D,2B) in another set , total costs is 40+180=220

(3) Similarly (E,2D,B) in another set , total costs is 45+80+90=215

Now, to maximize my point , I  have to maximize my perchase , so small amount left in my hand

Now, if we take set (3) two times we get a perchase of Rs.430

Again if we take set (1) 3 times total cost will be Rs. 540

Total spend  430 + 540 =970 Rs.

That is most close range for total Rs.1000

So, total no of items 4*2 + 2*3 =14

Comments

Can, anyone explain me How can we take (E,2D,B) alone, like if we take this set, won’t B will be affected by 2D, as given by for every item of D i purchase i must also purchase 2items of B, therefore (let) #items of E=x, then #item of D=2x, #item of B=x(due to E)+4x(due to D)????wont it be the case???

(A,C)=(110+70)-->180

(D,2B)=(40+90*2)--->220

(E,2D,5B)--->(45+ 2*40 + 5*90)--->575
we can take combination of any of these to maximize, 1000-(575+220+180)= 25 Rs left, and items =8*1+3*1+2*1---->13 items, it this wrong????