stonecold wrote:
chetan2u wrote:
Vinayprajapati wrote:
Pam owns an inventory of unopened packages of corn and rice, which she has purchased for $17 and $13 per package, respectively. How many packages of corn does she have ?
1. She has $282 worth of packages
2. She has twice as many packages of corn as of rice
hi,
The Q puts to you a C trap..
the first statement gives you SUM of two variables..
AND the II gives you the ratio of two variables..
and there it is C, BUT more than often the SUM gives you enough INFO to be sufficient..
let the number of packages of corn be C and that of rice be R..
I. 17C + 13R = 282..check for values of C and see if any integer value of R is possible ..
C can take values from 1 to 282/17 = 16..
Only C as 12 and R as 6 is possible..
SUff
II. She has twice as many packages of corn as of riceC = 2R..
various answers possible..
A
Hey Chetan I have a doubt in this Question..
I marked the correct answer as i thought in my head that 13 and 17 are prime so the common factor would repeat after 13*17 hence only one value satisfy.
Now my question is the process of putting the integers as 1,2,3,4...takes a loooooong time..
Is there any other shorter way to tackle this question .
regards
S.C.S.A
Hi
MeghaP and
stonecold,
Is there a shortcut-
- it will depend from Q to Q, depending on the properties of the number involved..Should you depend on a short cut--
MAY BE but do not miss out on other combinations which may as well satisfy the equation...
let me explain through this example-
NOTE- I had initially done the Q by taking values one by one, and you too when you get hang of the tables, would be able to do faster. So when now you have written about a shortcut, I can give you one in this Q..\(17x+13y=282\)
=> \(4x+13x+13y=282\)..
\(4x+13(x+y) = 282\)...
now some info you should gather from here
1) 13(x+y) should form most of 282 as 13>4 and x+y>x...
2) (x+y) should be a multiple of 2 but not of 4---
3) if entire 282 was formed from 13(x+y), x+y <22 as 13*22>282..so values of x+y can be 2, 6, 10, 14, 18 only..
Since we are looking for the largest value from 13(x+y), lets try with x+y =18.
\(4x+13(x+y)=4x+13*18=4x+234 = 282\)...
so \(4x= 282-234=48.. x=12\)..
satisfies all conditions written above.. OK x=12, y= 18-12=6
lets try the next possible
\(x+y = 14\)..
\(4x+13(x+y)=4x+13*14=4x+182 = 282\)...
so \(4x= 282-182=100... x=25\)..
is it possible that x= 25 and x+y=14, when both x and y are non negative.... NO
any number below this 10,8,2 can be ruled out straight way as they would make x much larger than x+y..
so our answer is that ONLY one set of number are possible..