Summary
Highlights
The video introduces the first part of a five-part series, covering 10 math items for college entrance exams such as UPCAT, PUPCET, USTET, and ACET. The full video is available on Facebook and YouTube due to length limitations on TikTok.
The first problem involves solving for 'y' in the equation 2 + (1/x) * y = 1/x. The instructor demonstrates distributing 'y', multiplying by the LCD 'x' to eliminate denominators, factoring out 'y', and isolating 'y' to find the solution: y = 1 / (2x + 1).
The second problem asks '15 is 20% of what number?'. The instructor shows two methods: converting the problem into a mathematical equation (15 = 0.20 * n) and solving for 'n', or using the percentage formula P = R * B and identifying the percentage, rate, and base. Both methods yield the answer 75.
The third problem involves direct variation: 'y varies directly as x squared'. Given that the difference in values of 'y' when x=0 and x=3 is 18, the task is to find 'y' when x=-2. The instructor first sets up the variation equation y = kx², solves for the constant 'k' using the given difference, and then substitutes 'k' and x=-2 to find 'y' = 8.
This problem asks 'Which of the following is not the sum of three consecutive odd integers?'. The key is to understand that the sum of three consecutive odd integers will always be divisible by 3. The instructor shows that if x, x+2, and x+4 are consecutive odd integers, their sum is 3x+6, which is always divisible by 3. By checking the divisibility by 3 of each option, 53 is identified as the answer (5+3=8, not divisible by 3).
The fifth problem is to factorize 2x² + 7x - 15. The instructor uses the 'slide method' by sliding the leading coefficient (2) to multiply the constant term (-15), transforming the expression into x² + 7x - 30. After factoring this simpler quadratic into (x+10)(x-3), the 'slid' coefficient (2) is divided back into the constant terms, simplifying to (x+5)(2x-3).
Given the quadratic equation x² + bx + c = 0, and that its roots are both equal to 6, the task is to find the value of 'c'. The instructor explains that if x=6 is a root (repeated), then (x-6) must be a factor, and since it's a double root, (x-6)² = 0. Expanding (x-6)² gives x² - 12x + 36 = 0, directly revealing that c = 36.
This problem asks to simplify the expression (4-y²) / (y-2). The numerator is recognized as a difference of squares (2² - y²), which factors into (2+y)(2-y). The instructor points out that (2-y) and (y-2) are opposites, so their division results in -1. Thus, the simplified expression is -(2+y) or -2-y.
The eighth problem evaluates the expression cos(60°) + sin(30°). The instructor references a 30-60-90 special right triangle with sides in the ratio 1:√3:2. Using SOH CAH TOA, sin(30°) is determined as opposite/hypotenuse = 1/2, and cos(60°) as adjacent/hypotenuse = 1/2. Adding these values, 1/2 + 1/2, gives the answer 1.
The ninth problem requires solving the equation 3 / (3x + 2) = 2/3. The instructor uses cross-multiplication: 3 * 3 = 2 * (3x + 2). This simplifies to 9 = 6x + 4. Solving for 'x' leads to 5 = 6x, so x = 5/6.
The final problem asks 'How many 1/6s are there in 2 and 1/3?'. This is a division problem: (2 and 1/3) / (1/6). The mixed number 2 and 1/3 is converted to an improper fraction, 7/3. The division is then performed by multiplying 7/3 by the reciprocal of 1/6 (which is 6/1). After cancelling and multiplying, the answer is 14.