Math questions from an image, including algebra, simple interest, and geometry.
Understand the Problem
The image contains several math questions typically found in a math exam for secondary school students. These problems cover topics such as factorization, evaluation of expressions, simple interest calculations, geometry (area of a trapezium, construction of quadrilateral) and algebra.
Answer
10. b. $(3x-1) (3x-1)$ 11. d. $96ab$ 12. c. $5$ 13. d. $4$ years 14. b. $7$ cm 15. a. $₹540$
Answer for screen readers
- b. (3x-1) (3x-1)
- d. 96ab
- c. 5
- d. 4 years
- b. 7 cm
- a. ₹540
Steps to Solve
- Factorize the quadratic expression
The given expression is $9x^2 - 6x + 1$. This is a quadratic expression that can be factored as a perfect square. We look for two terms that multiply to $9x^2$ and $1$, which are $3x$ and $1$. Since the middle term is $-6x$, we use $-1$. Thus we can write the expression as $(3x - 1)^2$.
- Write the expression as a product of two binomials
$(3x - 1)^2 = (3x - 1)(3x - 1)$
- Evaluate the expression using the difference of squares formula
The expression is $(4a + 3b)^2 - (4a - 3b)^2 + 48ab$. Using the difference of squares formula: $A^2 - B^2 = (A + B)(A - B)$, where $A = (4a + 3b)$ and $B = (4a - 3b)$. $(4a + 3b)^2 - (4a - 3b)^2 = [(4a + 3b) + (4a - 3b)][(4a + 3b) - (4a - 3b)]$. $= (4a + 3b + 4a - 3b)(4a + 3b - 4a + 3b) = (8a)(6b) = 48ab$. So, the entire expression becomes $48ab + 48ab = 96ab$.
- Solve for $a - \frac{1}{a}$
Given $a^2 - 5a - 1 = 0$, divide the entire equation by $a$: $\frac{a^2}{a} - \frac{5a}{a} - \frac{1}{a} = 0$ $a - 5 - \frac{1}{a} = 0$ $a - \frac{1}{a} = 5$
- Calculate the time period for simple interest
Simple Interest (SI) $= \frac{P \times R \times T}{100}$, where P is the principal, R is the rate of interest, and T is the time period. Given $P = 4500$, $R = 6%$, and $SI = 1080$. We need to find $T$. $1080 = \frac{4500 \times 6 \times T}{100}$ $1080 = 45 \times 6 \times T$ $1080 = 270T$ $T = \frac{1080}{270} = 4$ years.
- Calculate the radius of the cylinder
The volume of a cylinder is given by $V = \pi r^2 h$, where $V$ is the volume, $r$ is the radius, and $h$ is the height. Given $V = 1570 , cm^3$, $h = 10 , cm$, and $\pi = 3.14$. We need to find $r$. $1570 = 3.14 \times r^2 \times 10$ $1570 = 31.4 \times r^2$ $r^2 = \frac{1570}{31.4} = 50$ $r = \sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2} \approx 7.07$. Since the multiple choice options are integers, we can approximate $r$ to the nearest integer is 7 cm.
- Calculate the simple interest
Simple Interest (SI) $= \frac{P \times R \times T}{100}$, where P is the principal, R is the rate of interest, and T is the time period. Given $P = 1500$, $R = 9%$, and $T = 4$ years. $SI = \frac{1500 \times 9 \times 4}{100} = 15 \times 9 \times 4 = 15 \times 36 = 540$.
- b. (3x-1) (3x-1)
- d. 96ab
- c. 5
- d. 4 years
- b. 7 cm
- a. ₹540
More Information
Here is a breakdown of the answers 10. The factored form of the quadratic expression $9x^2 - 6x + 1$ is $(3x-1)(3x-1)$, which corresponds to option b. 11. Expanding and simplifying the expression $(4a + 3b)^2 - (4a - 3b)^2 + 48ab$ results in $96ab$, which is option d. 12. From the equation $a^2 - 5a - 1 = 0$, we find that $a - \frac{1}{a} = 5$, corresponding to option c. 13. The time period for a principal of ₹4500 at 6% p.a to earn a simple interest of ₹1080 is 4 years which matches with option d. 14. Solving for the radius of the cylinder, the radius is approximately 7 cm, corresponding to option b. 15. The simple interest on ₹1500 for 4 years at 9% p.a is ₹540, which corresponds to option a.
Tips
- When factorizing, not recognizing perfect square trinomials.
- Errors in applying the difference of squares formula.
- Incorrectly rearranging the equation to solve for $a - \frac{1}{a}$.
- Making errors in the simple interest or volume calculation.
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