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An insurance company sells only one type of health and one type

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An insurance company sells only one type of health and one type

by BTGModeratorVI » Wed Jul 29, 2020 2:54 pm

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An insurance company sells only one type of health and one type of life insurance policy. The monthly premium for a health insurance policy is $80. If the insurance company took in a total $5000 in premiums, what was the monthly premium of a life insurance policy?

(1) The total revenue from health insurance premiums was 4/5 of the total revenue the company received from premiums.

(2) The insurance company sold 2.5 times as many health insurance policies as life insurance policies.

Answer: C
Source: Princeton Review
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Source: — Data Sufficiency |

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Re: An insurance company sells only one type of health and one type

by Brent@GMATPrepNow » Thu Jul 30, 2020 7:32 am

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BTGModeratorVI wrote: ↑
Wed Jul 29, 2020 2:54 pm
 An insurance company sells only one type of health and one type of life insurance policy. The monthly premium for a health insurance policy is $80. If the insurance company took in a total $5000 in premiums, what was the monthly premium of a life insurance policy?

(1) The total revenue from health insurance premiums was 4/5 of the total revenue the company received from premiums.

(2) The insurance company sold 2.5 times as many health insurance policies as life insurance policies.

Answer: C
Source: Princeton Review
Let H = the number of health insurance policies
Let L = the number of life insurance policies
Let p = the monthly premium on a life insurance policy

So, 80H + pL = 5000

Target question: What is the value of p?

Statement 1: The total revenue from health insurance premiums was 4/5 of the total revenue the company received from premiums.
Total revenue = $5000
(4/5)($5000) = $4000
So, 80H = $4000.
Take 80H + pL = 5000 and replace 80H with $4000 to get 4000 + pL = 5000, which simplifies to pL = 1000
Since there are many possible values for p that satisfy this equation, statement 1 is NOT SUFFICIENT

Statement 2: The insurance company sold 2.5 times as many health insurance policies as life insurance policies.
In other words, H = 2.5L
Take 80H + pL = 5000 and replace H with 2.5L to get 80(2.5L) + pL = 5000, which simplifies to be 200L + pL = 5000
Since there are many possible values for p that satisfy this equation, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
From the two statements we get two equations:
200L + pL = 5000
pL = 1000
Subtract the bottom equation from the top equation to get: 200L = 4000, which means L = 20
Now that we know L = 20 and pL = 1000, we can see that p = 50
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer = C

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Brent
Brent Hanneson - Creator of GMATPrepNow.com
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Re: An insurance company sells only one type of health and one type

by deloitte247 » Fri Jul 31, 2020 2:07 am

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Let the no. of insurance policy = x
Let the no. of life policy = y
Let the no. of a monthly premium of a life insurance policy = z
Let the no. of a monthly premium of a health insurance policy = $80

Target question: What was the monthly premium of a life insurance policy?
Given that 80x + zy = 5000; find the value of z.

Statement 1: The total revenue from health insurance premiums was 4/5 of the total revenue the company received from premiums.
Total revenue = 5000
Health insurance policy premiums = 80 * x = 80x
$$Therefore,\ 80x=\frac{4}{5}\cdot5000$$
$$80x=4000$$
From the question stem, 80x + zy = 5000; where 80x = 4000
Therefore, 4000 + zy = 5000
zy = 5000 - 4000 = 1000

Here, the exact value of z and y is unknown. So, the target question cannot be answered with a definite value because there are so many variations of integers that satisfy zy=1000. Therefore, statement 1 is NOT SUFFICIENT.

Statement 2: The insurance company sold 2.5 times as many health insurance policies as life insurance policies.
Therefore, x = 2.5 of y
From question stem , 80x + zy = 5000
80 (2.5y) + zy = 5000
200y + zy =5000
The exact value of z and y is unknown and as such, the target question cannot be answered with certainty. Hence, statement 2 is NOT SUFFICIENT.

Combining both statements together
From statement 1; zy = 1000 eqn (1)
From statement 2; 200y + zy = 5000 eqn (2)
Subtracting equation 1 from equation 2
200y + zy = 5000
- zy = 1000
We have,
200y = 4000
$$y=\frac{4000}{200}=20$$
From eqn (1), zy = 1000, where y=20
20z = 1000
$$z=\frac{1000}{20}=$50$$
Comclusively, both statements combined together ARE SUFFICIENT.
Answer = Option C
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