The average (arithmetic mean) of a, b, and c… | GMAT Sprint

Free Practice Question

PS · Problem Solving

655-705

Algebra

The average (arithmetic mean) of a, b, and c is 90, while the average (arithmetic mean) of a, b, c, d, and e is 80. Given that $d = \frac{a+b}{2}$ , $e = \frac{b+c}{2}$ , and $a ≥ c$ , what is the minimum possible value of a?

Answer Choices

Correct answer marked below

120

140

Correct

150

160

180

See the full step-by-step explanation

You can see the correct answer above. Sign in for free to unlock the complete worked solution.

Track your performance and improve

Get detailed analytics, unlock full explanations, and move up difficulty tiers as you practice.

Start for free Sign in

Explanation

Preview

Official Explanation

Official

From the first average, we have

a + b + c = 90*3 = 270

…(I). For the second average,

a + b + c + d + e = 80*5 = 400

. This gives us

d + b = 400 - 270 = 130

…(II). We know

d = \frac{(a + b)}{2}

and

e = \frac{(b + c)}{2}

. Therefore,

d + e = \frac{(a + 2b + c)}{2}

. From this, we can write

2*130 = a + 2b + c

,…

Read the full explanation

Create free account Sign in

Comments

No comments yet

Be the first to share your thoughts and help others understand it better.

What this question tests

What it tests

Your ability to manipulate linear and quadratic expressions, substitute unknowns, and solve equations under time pressure.

Common trap

Losing a sign when moving terms across the equals sign, or dividing both sides by a variable that could equal zero.

The average (arithmetic mean) of a, b, and c is 90, while the average (arithmetic mean) of a, b, c, d, and e is 80. Given that d=a+b2d = \frac{a+b}{2}d=2a+b​, e=b+c2e = \frac{b+c}{2}e=2b+c​, and a≥ca ≥ ca≥c, what is the minimum possible value of a?