patterns in Maths

🧠 Class 6 Quiz – Patterns in Mathematics

1. What comes next in the pattern: 2, 4, 8, 16, __?

2. Mirror image of "MATH" is:

3. Identify the 10th shape in 🔵🔵🟢 repeating pattern:
--Select-- 🔵 🟢 🔴

4. How many matchsticks in Step 6 (3, 5, 7, 9...)?

 SAT EXAM PREPARATION 2025-2026

Question:

There are 66 calories in 15 grams of grated Parmesan cheese, and 59% of those calories are from fat.
When measuring Parmesan cheese, 5 grams is equal to 1 tablespoon.

Which of the following is closest to the number of calories from fat per tablespoon of grated Parmesan cheese?

Options:

• A) 3

• B) 8

• C) 9

• D) 13

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Solution:

Step 1: Calculate total fat calories in 15 grams

$$\text{Fat calories} = 59\% \text{ of } 66 = 0.59 \times 66 = 38.94 \approx 39 \text{ calories}$$

Step 2: Find fat calories per gram

$$\frac{39 \text{ calories}}{15 \text{ grams}} = 2.6 \text{ calories per gram}$$

Step 3: Find fat calories per 1 tablespoon (which is 5 grams)

$$2.6 \times 5 = 13 \text{ calories from fat per tablespoon}$$

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✅ Correct Answer: D) 13

This is the closest value to the actual fat calories per tablespoon.

Question:

The base of a tree has 10 mushrooms growing from its roots.
The mushroom population doubles every 5 days.

What type of function best models the relationship between the mushroom population and time?

Options:

• A) Decreasing exponential

• B) Decreasing linear

• C) Increasing exponential

• D) Increasing linear

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Solution:

Let’s understand what’s happening:

• The starting population is 10 mushrooms.

• The population doubles every 5 days, which means it multiplies by 2 repeatedly over time.

This is a classic example of exponential growth, where the population is increasing over time, not decreasing.

Why not linear?

• Linear growth adds a fixed amount each time.

• Exponential growth multiplies (like doubling), so the rate of increase itself increases over time.

---

✅ Correct Answer: C) Increasing exponential

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Bonus (Equation form):

The function could be modeled as:

$$M(t)=10\cdot {\mathrm{2^}}^{t\mathrm{/}5}$$

Where:

• $M(t)$ is the number of mushrooms after $t$ days,

• 10 is the initial count,

• The exponent $t\mathrm{/}5$ reflects doubling every 5 days.

  Question:
  

  Given the quadratic equation:

  $$x^2 + bx + c = 0$$

  where $b$ and $c$ are constants.

  If:

  $$\frac{-b+\sqrt{b^2-4c}}{2}=18\text{and}\frac{-b-\sqrt{b^2-4c}}{2}=10$$

  what is one possible value of $\mathrm{x\; ?}$

  ---

  Solution:

  These two expressions are the quadratic formula results for the roots of the equation:

  $$x = \frac{-b \pm \sqrt{b^2 - 4c}}{2}$$So, the two solutions are:

  $$x_1 = 18, \quad x_2 = 10$$
  

  Thus, one possible value of x is:

  ✅ Answer: $\boxed{10}$ or $\boxed{18}$

  
  

Question:

Solve:

$$3y^2 - 4 = 5y + 1$$

Which of the following is a solution to the equation above?

Options:
A) $\frac{5 - \sqrt{85}}{6}$
B) $\frac{5 - \sqrt{85}}{3}$
C) 5
D) $5 + \sqrt{85}$

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Solution:

Step 1: Start with the given equation

$$3y^2 - 4 = 5y + 1$$

Step 2: Move all terms to one side to set the equation to 0

$$3y^2 - 5y - 5 = 0$$

Now we solve this quadratic using the quadratic formula:

$$y = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(3)(-5)}}{2(3)}$$ $$= \frac{5 \pm \sqrt{25 + 60}}{6} = \frac{5 \pm \sqrt{85}}{6}$$

So the two solutions are:

$$y = \frac{5 + \sqrt{85}}{6} \quad \text{or} \quad y = \frac{5 - \sqrt{85}}{6}$$

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✅ Correct Answer: A) $\frac{5 - \sqrt{85}}{6}$

Here is the question text and a full step-by-step solution based on the image:

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Question:

A rocket is launched vertically from ground level. The rocket reaches a maximum height of 46.36 meters above the ground after 2.4 seconds, and falls back to the ground after 4.8 seconds.

Which equation best represents the height $s$, in meters, of the rocket $u$ seconds after it is launched?

Options:

• A) $s = -8.05u^2 + 38.64u$

• B) $s = -4.8u^2 + 46.36u$

• C) $s = -u^2 + 46.36$

• D) $s = 8.05u^2 - 38.64u$

---

Solution:

We are told:

• The rocket starts at ground level → initial height = 0

• Maximum height is 46.36 meters at u = 2.4 seconds

• It returns to ground at u = 4.8 seconds

• So the vertex of the parabola is at $u = 2.4$, and the parabola opens downward

General form of a quadratic equation:

$$s = au^2 + bu + c$$

Since it starts at ground level: $c = 0$

Let’s use the vertex form:

$$s = a(u - h)^2 + k$$

Where:

• $h = 2.4$ (time of max height)

• $k = 46.36$ (maximum height)

$$s = a(u - 2.4)^2 + 46.36$$

We also know that at $u = 0$, $s = 0$ (ground level). Plug into the equation:

$$0 = a(0 - 2.4)^2 + 46.36 \Rightarrow 0 = a(5.76) + 46.36 \Rightarrow a = -\frac{46.36}{5.76} \approx -8.05$$

Now plug $a$ into the standard form:

$$s = -8.05u^2 + (2 \cdot 8.05 \cdot 2.4)u = -8.05u^2 + 38.64u$$

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✅ Correct Answer: A) $s = -8.05u^2 + 38.64u$

• The rocket is launched from ground level, so the initial height is $0$.

• Using the general form of a quadratic equation for height,

  $$s(t) = -au^2 + bu + c$$

• Since the initial height is 0, $c = 0$

---

At the maximum height:

$$s(2.4) = -a(2.4)^2 + 2.4b = 46.36$$

When it hits the ground:

$$s(4.8) = -a(4.8)^2 + 4.8b = 0$$

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From these conditions, solve the system of equations:

$$-23.04a + 4.8b = 0 \quad \text{(1)}$$

This gives:

$$b = 4.8a$$

---

Substitute $b = 4.8a$ into:

$$-a(2.4)^2 + 2.4(4.8a) = 46.36 \Rightarrow -5.76a + 11.52a = 46.36 \Rightarrow 5.76a = 46.36 \Rightarrow a = 8.05$$

Then:

$$b = 4.8 \times 8.05 = 38.64$$

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Thus, the equation is:

$$s(u)=-8.05u^2+38.64u$$

WORKSHEET ch1 class 6

WORKSHEET - Number pattern

1)  1,3,5,7, ________, ___________,  _______

Rule- ________________

2)  2,4,6,8,________, ___________,  _______

Rule- ________________

3) 24,34,44,54, ________, ___________,  _______

Rule- ________________

4) 3, 6,9,12,________, ___________,  _______

Rule- ________________

5) 35,40,45, ________, ___________,  _______

Rule- ________________

6) 11,22, 33, ________, ___________,  _______

Rule- ________________

7) 9,19,29, ________, ___________,  _______

Rule- ________________

WORKSHEET - Number pattern solutions

1)  1,3,5,7, ___9_____, ___11________,  ____13___
Rule- _________odd number_______

2)  2,4,6,8,____10____, ___12________,  _____14__
Rule- _____even number___________

3)  24,34,44,54, 64 ,74, 84
Rule: By adding 10-10 in each number

4) 36,9,12,15 18 21
Rule-By adding 3 in each number.

5) 35,40,45, 50,55,60
Rule: By adding 5-5 in each number.

6) 11,22, 33, 44, 55
Rule-By adding 11-11 in each number.

7)
9,19,29, 39 49,59

Rule: By adding 10-10. in each number.

My Teaching Philosophy in Mathematics

 

My Teaching Philosophy in Mathematics

I believe mathematics is not just a subject of numbers but a language of logic, patterns, and possibilities that helps learners make sense of the world around them. My teaching philosophy is rooted in the idea that every student can develop mathematical thinking when learning is made meaningful, visual, and connected to real life.

In my classroom, I strive to blend conceptual understanding with hands-on experience. I use activities, visual aids, and real-world contexts to build bridges between abstract concepts and tangible situations—like using number lines for rational numbers, cube models for volume, or surveys for data handling. I believe that math should not be memorized but discovered, discussed, and applied.

I design learning experiences that nurture curiosity, collaboration, and confidence. Each concept is introduced through engaging strategies—story-based equations, interactive geometry, peer-led discussions, and application-oriented tasks—so that students not only understand 'how' but also 'why'.

Assessment, for me, is more than evaluation—it's a way to deepen learning. I use open-ended questions, peer activities, and visual tasks to gauge understanding and guide feedback. When gaps appear, I see them as opportunities to personalize support and reframe learning through remedial teaching.

Ultimately, my goal is to empower students with not just mathematical skills, but mathematical thinking—equipping them to reason logically, make decisions confidently, and appreciate the beauty and power of math in everyday life.

a short and brief version of your teaching philosophy in mathematics:

My Teaching Philosophy in Mathematics

I believe mathematics is best learned through real-life connections, hands-on activities, and visual thinking. My goal is to make math meaningful, engaging, and accessible to all learners. I use interactive strategies and practical examples to build strong conceptual understanding and problem-solving skills. Every student can succeed in math when learning is active, collaborative, and rooted in curiosity.