cmal_s wrote:Data Sufficiey

Martha can buy 2 pencils either 23cents or 21 cents. How many 23 cents pencils did she buy

1) She bought a to total 6 pencils

2) she spent a total of 130cents on the purchase


i know you can go on subtracting but is there a faster method

is the OA D?
if it is,
i would just set up equation
x+y=6
23x+21y=130
2equations 2 unknowns -> D.

OA ? is it b..

For statement 2,

x 23 + y. 21 = 130.

X = 1 means 21.y should end with 7 only possibility is 7 => this doesnot satisfy equation.

X = 2 meands 21.y should end with 4 ==> this satisfies

X = 3 means 21.y should end with 1 ==> Doesnot satisfy.

This is a way that I can think of without substracting.

cmal_s wrote:Data Sufficiey

Martha can buy 2 pencils either 23cents or 21 cents. How many 23 cents pencils did she buy

1) She bought a to total 6 pencils

2) she spent a total of 130cents on the purchase


i know you can go on subtracting but is there a faster method

this is a tricky one I think....

statement 1 is clearly not sufficient.

However, if you consider statement 2 and try to setup an equation, no doubt you'll have two variables.

23 X + 21 Y = 130 .

Notice that 130 is not a multiple of either 23 or 21. hence, she purchased some of each type.

Now, there will be only one set of values of X & Y which will satisfy this equation and that will be (2,4).

Hence, IMO B should be the correct choice.

What's the OA and OE ?

harshavardhanc wrote:

cmal_s wrote:Data Sufficiey

Martha can buy 2 pencils either 23cents or 21 cents. How many 23 cents pencils did she buy

1) She bought a to total 6 pencils

2) she spent a total of 130cents on the purchase


i know you can go on subtracting but is there a faster method

this is a tricky one I think....

statement 1 is clearly not sufficient.

However, if you consider statement 2 and try to setup an equation, no doubt you'll have two variables.

23 X + 21 Y = 130 .

Notice that 130 is not a multiple of either 23 or 21. hence, she purchased some of each type.

Now, there will be only one set of values of X & Y which will satisfy this equation and that will be (2,4).

Hence, IMO B should be the correct choice.

What's the OA and OE ?

Yea actually I think it is B too.

Yes the Answer is B

IMO its B

I would go this way:


u know 1.30 is not divisible by 23 or 21 totally so it has to be a combination of the two


now u want to take the unit digits which are 3 and 1 (from 23 and 21) and make it 0 (to 130)

since we know it is not all 23 cents pencils.... start with one 23 cents pencil and chose the other number for the difference:


so it cld be 1 *3 + 5*1 = 8 --- X
next

2*3 +4*1 = 0 -- Correct answer !

IMO (B).

Here is my approach.

Condition (1) is clearly insufficient.

For condition (2), we can really solve this making equations, however, another approach is to multiply 21 or 23 by certain number to achieve nearby 130.

Taking 21 at first,

21 X 6 = 126 (BTW, I steal number 6 as multiplier from the condition 1)

So, here we lack cents by 4. To add these 4 cents we need 23 cents' pencils.
For each 2 cents, we need to add one pencil costing 23 cents (23 - 21 = 2).

So, 2 pencils of 23 cents are required.

Looks like we have solve the question, but wait is there any other answers could be?

If you have time and want to be sure, check for any other splits otherwise go ahead and choose (B).

I am sure, you won't get any.

cmal_s wrote:Yes the Answer is B

I guess you can just reason it out
if you can prove by statement 2 alone that the total cannot be more than 6.

if you have 7 of 21cents pencils, 147 is more than 130.
therefore the total must be less or equal to 6.
then you can find X and Y by statement 2 alone.