A dance troupe has a total of 50 dancers split into 2 groups. The cost

We can solve this using either one variable or two variables.
Let's use one variable.

Since the question asks "What is the total cost of the costumes worn by Group B?," let's let x = the total number of dancers in Group B

A dance troupe has a total of 50 dancers split into 2 groups
If x = the number of dancers in GROUP B
Then 50 - x = the number of dancers in GROUP A

The costumes worn by Group A cost $80 each, and those worn by Group B cost $90 each. The total cost of all the costumes is $4,270
Cost = (price her costume)(number of dancers)
So, we can write: ($80)(50 - x) + ($90)(x) = $4270
Expand to get: 4000 - 80x + 90x = 4270
Simplify: 4000 + 10x = 4270
So, 10x = 270
Solve: x = 27
So, there are 27 dancers in Group B
If each costume costs $90, then the cost of those costumes = ($90)(27) = $2430

Answer: E

Cheers,
Brent

Dancers in the Group A = x

the total cost of the costumes worn by Group A = 80x

Dancers in the Group B = 50-x

the total cost of the costumes worn by Group B= 90(50-x)

80x+ 90(50-x) = 4500-10x = 4270

x= 23

the total cost of the costumes worn by Group B= 90(50-23)= $ 2430

Let x and y be the number of dancers of Group A and Group B, respectively.
x+y = 50
80x + 90y = 4270
--> 8x +9y = 427
8(50-y) + 9y = 427
400- 8y+9y = 427
y = 27
--> 27*90 = 2430

Answer (E)

Let dancers in A group = x
B group = 50-x

80*x+90(50-x)=4270
80x-90x+4500=4270
230=10x
x=23

Cost of costumes for B= 90*27
B=2430 E

We can create the equation:

A + B = 50

and

80A + 90B = 4270

Multiplying the first equation by 90, we have:

90A + 90B = 4500

Subtracting the second equation from this new equation, we have:

10A = 230

A = 23

Therefore, group B has 50 - 23 = 27 dancers. Since each of their costumes costs $90, the total cost of the costumes worn by Group B is 27 x 90 = $2430.

Answer: E
_________________

BrentGMATPrepNow, thanks for your solution to this problem. I have a question about strategy. When would you choose to use "bottom-up" algebra (like you did here) compared to "top-down" backsolving from the answer choices? The first time I solved this problem, I used backsolving, but I noticed that it was hugely inefficient and took twice as long as when you solve using algebra.

Would appreciate your insight. Thanks!

Backsolving (aka testing the answer choices) is often a valid approach. In fact, when students first encounter a word problem like this, they should automatically consider testing the answer choices. Then they should give themselves 15 to 20 seconds to find a different (and hopefully faster) approach. If they can't come up with some alternate approach, they should start testing the answer choices.

In some cases, testing the answer choices can be very time-consuming.
This is one of those cases.

For example let's test answer choice B ($2,070)
This means the total COST of the costumes worn by Group B = $2,070
Since the costumes worn by Group B cost $90 each, the total NUMBER of costumes = $2,070/90 = 23
So there are 23 dancers in Group B

50 - 23 = 27
So, there are 27 dancers in Group A
The costumes for group a dancer's cost $80 each, so the total costume COST for Group A = (27)($80) = $2160

So the total COST for both groups = $2,070 + $2,160 = $4230
No good. We need the total cost to be $4,270
So, answer choice B is incorrect.

Phew!!! Now we need to test ANOTHER answer choice!

Determining whether testing the answer choices will be too time-consuming can be difficult. You need to be able to quickly predict the number of steps it will take to test each answer choice before wasting a bunch of time.

Cheers,
Brent

Thanks, BrentGMATPrepNow - really appreciate your comments!

MEMBERS OF GROUP A =a
MEMBERS OF GROUP B = b
SO, a+b=50
a= 50-b
80a+90b=4270
80(50-b)+90b=4270
4000-80b+90b=4270
10b=4270-4000=270
b=27

So, the cost for B 27*90=2430

THE ANSWER IS E
_________________

SOLUTION:

Let their be A people in group A and B be the number of people in group B.

Thus A+B=50 .....(1)

Also 80A + 90 B =4270 ....(2) (The cost/costume for group A is $80 and for B its $90)

Solving both (1) and (2), we have A=23 and B=27

Thus, cost for group B = 90x27 = $2630 (OPTION E)

Hope this helps
Devmitra Sen(Math)

Answer: Option E

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*Somehow I took more than 2 mins to solve this with the algebra approach, so I came up with the other approach to consider.


Let's say all the dancers belong to Group B, which uses $10 more on each costume than Group A.
Total cost of the costume would be $90 * 50 = $4500 which is $230 more than the actual total cost.
$230/ $10 (the cost difference between Group A and Group B) = 23, which means that 23 dancers should wear the costumes of $80 to get the total of $4,270.
Then, the rest 27 dancers should wear $90 costumes, so (30 -3) * 90= 2700 - 270 = $2430.


Which solution do you find faster?

Hi BrentGMATPrepNow,

I used back solving and it didn't take me long. I used this approach.
The question asks for the total cost of all the costumes worn by Group B . I know the cost of each costume worn by group B is $90 . So the answer choice has to be a multiple of 90 . Just by looking at the answer options I know that A, C are out (not multiple of 90 )

Left with B,D , and E

Now the total cost of costumes worn by Group B has to be a multiple of 90(since $90 each) and the total cost of costumes worn by Group A has to be a multiple of 80.(since $80 each )

A)1840
D)2160
E)2430

all 3 are multiple of 90 but I need to find out the one where( total cost-cost of group B i.e cost of group A) is multiple of 80

Testing answer option E : 4270(total cost ) -2430(cost of group B )= 1840 (this is a multiple of 80 ) . So keep this answer choice E

Can you please shed some light if it is suggested to use this approach?