Math worksheet with four word problems on systems of equations involving three variables.

Let's solve each of the four problems step by step.

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Problem 1:

The sum of the digits of a three-digit number is 16. If the ten's digit of the number is 3 times the unit's digit and the unit's digit is one-fourth of the hundredth digit, then what is the number?

#### Step 1: Define variables
Let:
- $ h $ = hundreds digit
- $ t $ = tens digit
- $ u $ = units digit

We are given:
1. $ h + t + u = 16 $ → (Equation 1)
2. $ t = 3u $ → (Equation 2)
3. $ u = \frac{1}{4}h $ → (Equation 3)

#### Step 2: Substitute Equation 3 into Equation 2
From (3): $ h = 4u $

Now plug into (2): $ t = 3u $

Now substitute $ h = 4u $, $ t = 3u $ into (1):

$$
h + t + u = 16 \\
4u + 3u + u = 16 \\
8u = 16 \\
u = 2
$$

Now find others:
- $ t = 3u = 3(2) = 6 $
- $ h = 4u = 4(2) = 8 $

So the digits are:
- Hundreds: 8
- Tens: 6
- Units: 2

Thus, the number is 862

✔ Answer: 862

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Problem 2:

If the equation of a parabola is generally expressed as $ y = ax^2 + bx + c $, find the equation of one that passes through the points (-2, 40), (1, 7), (3, 15).

We plug each point into $ y = ax^2 + bx + c $ to get a system of equations.

#### Point (-2, 40):
$$
40 = a(-2)^2 + b(-2) + c = 4a - 2b + c \quad \text{(1)}
$$

#### Point (1, 7):
$$
7 = a(1)^2 + b(1) + c = a + b + c \quad \text{(2)}
$$

#### Point (3, 15):
$$
15 = a(3)^2 + b(3) + c = 9a + 3b + c \quad \text{(3)}
$$

Now solve the system:

From (2): $ a + b + c = 7 $ → $ c = 7 - a - b $ → (2a)

Substitute into (1) and (3):

(1):
$ 4a - 2b + c = 40 $
Substitute $ c $:
$ 4a - 2b + (7 - a - b) = 40 $
$ 4a - 2b + 7 - a - b = 40 $
$ 3a - 3b + 7 = 40 $
$ 3a - 3b = 33 $
Divide by 3:
$ a - b = 11 $ → (4)

(3):
$ 9a + 3b + c = 15 $
Substitute $ c $:
$ 9a + 3b + (7 - a - b) = 15 $
$ 9a + 3b + 7 - a - b = 15 $
$ 8a + 2b + 7 = 15 $
$ 8a + 2b = 8 $
Divide by 2:
$ 4a + b = 4 $ → (5)

Now solve (4) and (5):

From (4): $ a = b + 11 $

Plug into (5):
$$
4(b + 11) + b = 4 \\
4b + 44 + b = 4 \\
5b = -40 \\
b = -8
$$

Then $ a = b + 11 = -8 + 11 = 3 $

Now from (2): $ a + b + c = 7 $
$ 3 + (-8) + c = 7 $
$ -5 + c = 7 $ → $ c = 12 $

So the equation is:
$$
y = 3x^2 - 8x + 12
$$

✔ Answer: $ y = 3x^2 - 8x + 12 $

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Problem 3:

There are 12 pieces of five, ten, and twenty rupee coins whose total value is ₹105. When the number of five and ten rupee coins are interchanged, the total value increases by ₹20. Find the number of coins of each type.

Let:
- $ x $ = number of ₹5 coins
- $ y $ = number of ₹10 coins
- $ z $ = number of ₹20 coins

We are given:

1. Total coins: $ x + y + z = 12 $ → (1)
2. Total value: $ 5x + 10y + 20z = 105 $ → (2)
3. After interchanging ₹5 and ₹10 coins: new total = $ 105 + 20 = 125 $

New configuration: $ 10x + 5y + 20z = 125 $ → (3)

Now solve:

From (1): $ x + y + z = 12 $

From (2): $ 5x + 10y + 20z = 105 $ → divide by 5:
$ x + 2y + 4z = 21 $ → (2a)

From (3): $ 10x + 5y + 20z = 125 $ → divide by 5:
$ 2x + y + 4z = 25 $ → (3a)

Now we have:
- (1): $ x + y + z = 12 $
- (2a): $ x + 2y + 4z = 21 $
- (3a): $ 2x + y + 4z = 25 $

Let’s subtract (1) from (2a):

(2a) – (1):
$ (x + 2y + 4z) - (x + y + z) = 21 - 12 $
$ y + 3z = 9 $ → (4)

Now (3a) – (1):
$ (2x + y + 4z) - (x + y + z) = 25 - 12 $
$ x + 3z = 13 $ → (5)

Now from (5): $ x = 13 - 3z $

From (4): $ y = 9 - 3z $

Now plug into (1): $ x + y + z = 12 $

$$
(13 - 3z) + (9 - 3z) + z = 12 \\
22 - 5z = 12 \\
5z = 10 \\
z = 2
$$

Then:
- $ x = 13 - 3(2) = 7 $
- $ y = 9 - 3(2) = 3 $

Check:
- Total coins: $ 7 + 3 + 2 = 12 $ ✔
- Total value: $ 5(7) + 10(3) + 20(2) = 35 + 30 + 40 = 105 $ ✔
- Interchange: $ 10(7) + 5(3) + 20(2) = 70 + 15 + 40 = 125 $ → increase by 20 ✔

✔ Answer:
- ₹5 coins: 7
- ₹10 coins: 3
- ₹20 coins: 2

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Problem 4:

Rob has a bag containing 35 nickels and some quarters. The total value of these coins is less than $2.50. What is the maximum number of quarters that meets the given condition?

Let:
- Number of quarters = $ q $
- Each nickel = $0.05$
- Each quarter = $0.25$

Total value:
$$
\text{Value} = 35 \times 0.05 + q \times 0.25 = 1.75 + 0.25q
$$

This must be less than $2.50:
$$
1.75 + 0.25q < 2.50
$$

Subtract 1.75:
$$
0.25q < 0.75
$$

Divide by 0.25:
$$
q < 3
$$

Since $ q $ must be an integer, maximum possible $ q = 2 $

✔ Answer: 2 quarters

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✔ Final Answers:

1. 862
2. $ y = 3x^2 - 8x + 12 $
3. 5 rupee: 7, 10 rupee: 3, 20 rupee: 2
4. 2 quarters

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