CAT 2001 | Question: 22

Question

Three math classes; $\text{X, Y},$ and $\text{Z},$ take an algebra test.

• The average score in class $\text{X}$ is $83.$
• The average score in class $\text{Y}$ is $76.$
• The average score in class $\text{Z}$ is $85.$
• The average score of all students in classes $\text{X}$ and $\text{Y}$ together is $79.$
• The average score of all students in classes $\text{Y}$ and $\text{Z}$ together is $81.$

What is the average for all three classes?

• A. $81$
• B. $81.5$
• C. $82$
• D. $84.5$

Answer 1

We are given:

• Average of class X = 83

• Average of class Y = 76

• Average of class Z = 85

• Average of X + Y = 79

• Average of Y + Z = 81

We are to find the average of all three classes X, Y, and Z.

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Step 1: Let’s assume the number of students:

Let the number of students in class X = xxx
Let the number of students in class Y = yyy
Let the number of students in class Z = zzz

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Step 2: Use the given average equations:

From average of X and Y:

83x+76yx+y=79(1)\frac{83x + 76y}{x + y} = 79 \quad \text{(1)}x+y83x+76y​=79(1)

Multiply both sides:

83x+76y=79(x+y)⇒83x+76y=79x+79y⇒4x=3y⇒x=34y83x + 76y = 79(x + y) \Rightarrow 83x + 76y = 79x + 79y \Rightarrow 4x = 3y \Rightarrow \boxed{x = \frac{3}{4}y}83x+76y=79(x+y)⇒83x+76y=79x+79y⇒4x=3y⇒x=43​y​

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From average of Y and Z:

76y+85zy+z=81(2)\frac{76y + 85z}{y + z} = 81 \quad \text{(2)}y+z76y+85z​=81(2)

Multiply both sides:

76y+85z=81(y+z)⇒76y+85z=81y+81z⇒−5y=−4z⇒z=54y76y + 85z = 81(y + z) \Rightarrow 76y + 85z = 81y + 81z \Rightarrow -5y = -4z \Rightarrow \boxed{z = \frac{5}{4}y}76y+85z=81(y+z)⇒76y+85z=81y+81z⇒−5y=−4z⇒z=45​y​

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Step 3: Use weighted average to find final average:

We now know:

• x=34yx = \frac{3}{4}yx=43​y

• z=54yz = \frac{5}{4}yz=45​y

So total students:

x+y+z=34y+y+54y=124y=3yx + y + z = \frac{3}{4}y + y + \frac{5}{4}y = \frac{12}{4}y = 3yx+y+z=43​y+y+45​y=412​y=3y

Total score:

83x+76y+85z=83⋅34y+76y+85⋅54y=2494y+76y+4254y83x + 76y + 85z = 83 \cdot \frac{3}{4}y + 76y + 85 \cdot \frac{5}{4}y = \frac{249}{4}y + 76y + \frac{425}{4}y83x+76y+85z=83⋅43​y+76y+85⋅45​y=4249​y+76y+4425​y

Bring to common denominator:

=(249+4⋅76+4254)y=249+304+4254y=9784y= \left(\frac{249 + 4 \cdot 76 + 425}{4}\right)y = \frac{249 + 304 + 425}{4}y = \frac{978}{4}y=(4249+4⋅76+425​)y=4249+304+425​y=4978​y

So final average:

Total scoreTotal students=9784y3y=97812=81.5\frac{\text{Total score}}{\text{Total students}} = \frac{\frac{978}{4}y}{3y} = \frac{978}{12} = \boxed{81.5}Total studentsTotal score​=3y4978​y​=12978​=81.5​

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✅ Final Answer:

B. 81.5\boxed{B. \ 81.5}B. 81.5​