Data Sufficiency

Hi there,

Could you explain the concept of this DS in simplest way

During a sale, a clothing store sold each shirt at a price of $15 and each sweater at a price of $25.
Did the store sell more sweaters than shirts during the sale?

(1) The average (arithmetic mean) of the prices of all of the shirts and sweaters that the store
sold during the sales was$21.
(2) The total price of all of the shirts and sweaters that the store sold during the sale was $420.

set up equation.

Q asks you whether # Sweater > # Shirts.

W--#of sweater
T--#of Sgirts

stm1.

25W + 15T = 21 (W + T) SUF

stm2.

Not Suf

many thks

25W + 15T = 21 (W + T)

are we gonna apply hit and trial method here to get W and T value?

I have come up with a very easy way to do questions like these.

Basically, remember this as a rule: if you have the arithmetic means for X, Y and X&Y, you can a) know which of them has a higher weight, and b) their ratio. Plot the averages on a number line.

---X----(X&Y)---------Y

Basically, if (X+Y) is closer to X, X has a greater weight.

The ratio X:Y is Y-(X&Y):X-(X&Y)
[NOTE: The values are averages]

IMO D -

Stmt 1 -
Let number of shirts be X and and number of sweaters be Y

therefore, (15x + 25y)/(x+y) = 21
=>15x + 25y = 21x + 21y
=>4y = 6x
=>x/y = 4/6
=>x/y = 2/3

from the above eqn, we can clearly determine that y is greater than x.
Stmt 1 alone is sufficient.

Stmt 2 - is satisfied for the values 15 * 8 + 25 * 12

Hence both the stmts are individually sufficient.

No statement 2 is not sufficient. While 8 shirts and 12 sweaters works so does 13 shirts and 9 sweaters. You can find many values for shirts that leaves a multiple of 25 for sweaters and if you look at those number half sell more sweaters half sell more shirts so 2 is inconclusive. answer is A

Agree....my bad...i inferred the stmt 2 from the result obtained from stmt 1(2:3).

I change my IMO - A