TLDR;
This CTM 23 session focuses on most expected questions for selection in railway exams, emphasizing consistent practice and learning. It covers geometry, mensuration, trigonometry, profit and loss, mixtures, partnership, simplification, boat and stream, simple interest, compound interest, algebra, and data interpretation. The session includes problem-solving with tricks and formulas, and discussions on effective study habits and exam strategies.
- Geometry and Similarity
- Trigonometry and Quadratic Equations
- Data Interpretation and Problem Solving
Mirror and Tower Problem [3:34]
A person 1.5 meters tall observes the top of a tower in a mirror placed on the ground. The mirror is 0.5 meters away from the person and 105 meters from the tower. Using the properties of physics, specifically that the angle of incidence equals the angle of reflection, the height of the tower is calculated. By setting up a proportion (1.5/0.5 = H/105), the height (H) of the tower is found to be 315 meters. This problem can also be solved using similarity of triangles.
Quadrilateral Angle Problem [6:48]
In a quadrilateral PQRS, angle bisectors of angles P and Q meet at point T. Given that angle HTR is 42.5 degrees, the sum of angles P and Q is twice this angle, making it 85 degrees. If the ratio of angles P and Q is 5:2, the difference between the angles is 3 parts. Knowing 5 parts equals 85 degrees, 1 part equals 17 degrees. Therefore, the difference (3 parts) is 51 degrees.
Circle Arc Length Problem [9:26]
A circle has an arc that subtends an angle of 84 degrees at the center. The length of the arc is 22 units. To find the radius of the circle, the formula for arc length (angle × radius) is used. The angle is converted from degrees to radians by multiplying by π/180. The equation becomes 22 = (84 × π/180) × r. Solving for r, with π approximated as 22/7, the radius is found to be 15 units.
Trigonometry Problem [10:49]
Given the equation 4sin(θ)cos(θ) = 8cos³(θ) - cos(θ), where θ is an acute angle, the goal is to find sin(θ). By dividing through by cos(θ), the equation simplifies to 4sin(θ) = 8cos²(θ) - 1. Substituting cos²(θ) with 1 - sin²(θ), the equation becomes a quadratic in terms of sin(θ). Letting a = sin(θ), the quadratic equation is 8a² + 4a - 7 = 0. Using the quadratic formula, a = (-4 ± √(4² - 48(-7))) / (2*8). Simplifying, a = (-4 ± √(16 + 224)) / 16 = (-4 ± √240) / 16 = (-4 ± 4√15) / 16 = (-1 ± √15) / 4. Since θ is acute, sin(θ) must be positive, so the solution is sin(θ) = (-1 + √15) / 4.
Profit and Loss Problem [15:30]
The marked price (MRP) of 85 items equals the cost price (CP) of 153 items. Also, the selling price (SP) of 104 items equals the MRP of 65 items. The ratio of MRP to CP is 153:85, which simplifies to 9:5. The ratio of SP to MRP is 65:104, which simplifies to 5:8. To equate the MRP in both ratios, multiply the first ratio by 8 and the second by 9, resulting in MRP:CP = 72:40 and SP:MRP = 45:72. The ratio of CP to SP is therefore 40:45, simplifying to 8:9.
Mixture Problem [17:01]
A mixture contains a chemical X and water in the ratio of 15:9. 48 liters of this mixture are removed and replaced with water, changing the ratio to 11:13. After removing 48 liters, the ratio of X to water remains 15:9. The amount of chemical X remains constant when water is added. Equating the quantity of chemical X in both ratios, the new ratio of water is found. The difference in the water ratio corresponds to the added 48 liters. From this, the quantity of water after replacement is calculated to be 97.5 liters.
Partnership Problem [20:23]
Q and R invest in a partnership in the ratio of 10:12:15. The total profit is 666 rupees. The value of 37 ratios is equal to 666 rupees. Dividing 666 by 37, one ratio is found to be 18 rupees. Q's share is 12 ratios, so Q's share is 18 * 12 = 216 rupees.
Simplification Problem [21:49]
The problem involves simplifying a complex fraction. First, convert mixed fractions to improper fractions. Simplify the numerator and the denominator separately. After simplifying both, divide the simplified numerator by the simplified denominator. The final simplified fraction is 133/8.
Boat and Stream Problem [24:46]
The speed of a boat in still water is 9 km/hr. The time taken to cover a certain distance upstream is twice the time taken to cover the same distance downstream. The relationship between upstream and downstream speeds is used to find the speed of the stream. The speed of the stream is calculated to be 3 km/hr.
Simple Interest Problem [25:43]
A sum of 4500 rupees earns 81 rupees in simple interest at a rate of 4.5% per annum. The time required to earn this interest is calculated using the formula: Simple Interest = (Principal × Rate × Time) / 100. Solving for Time, T = (81 × 100) / (4500 × 4.5) = 4 years.
Algebra Problem [26:40]
Given A + B + C = 13 and A² + B² + C² = 69, find the value of AB + BC + CA. Using the algebraic identity (A + B + C)² = A² + B² + C² + 2(AB + BC + CA), substitute the given values: 13² = 169 = 69 + 2(AB + BC + CA). Solving for AB + BC + CA, we get 2(AB + BC + CA) = 169 - 69 = 100, so AB + BC + CA = 50.
Volume of Cuboid Problem [27:34]
The length and breadth of a cuboidal box are given as 10 cm and 12.5 cm, respectively. The volume of the box is 625 cubic centimeters. To find the height, use the formula for the volume of a cuboid: Volume = Length × Breadth × Height. Substituting the given values, 625 = 10 × 12.5 × Height. Solving for Height, Height = 625 / (10 × 12.5) = 5 cm.
Circle Geometry Problem [28:31]
Two chords, AB and CD, intersect inside a circle at point P. Given that AP = 6, AB = 16, and PD = 12, find PC. Using the intersecting chords theorem, AP × PB = CP × PD. Since AB = 16 and AP = 6, PB = 10. Thus, 6 × 10 = CP × 12. Solving for CP, CP = 60 / 12 = 5.
Simplification Problem with BODMAS Rule [30:56]
The problem requires simplifying an expression using the BODMAS rule. First, solve the 'of' operation, then the brackets, followed by division, multiplication, addition, and subtraction. The simplified expression results in -13.
Average Speed Problem [31:54]
Sudhin cycles for 4 hours at 8.5 km/hr, then travels by auto for 1.5 hours at 20 km/hr, and finally travels for Y hours at 4 km/hr. The overall average speed for the entire journey is 10 km/hr. The total distance is (4 * 8.5) + (1.5 * 20) + (Y * 4). The total time is 4 + 1.5 + Y. Using the formula Average Speed = Total Distance / Total Time, we get 10 = (34 + 30 + 4Y) / (5.5 + Y). Solving for Y, Y = 1.5 hours.
Quadratic Equation Identification [34:12]
The task is to identify which of the given equations is not a quadratic equation. After analyzing the equations, it is determined that option D does not form a quadratic equation because the x² terms cancel out, leaving a linear equation.
Compound Interest Problem [37:27]
A sum is invested at 5% compounded semi-annually. The interest earned over a period is calculated. The interest for the first year is calculated at 2.5% per half-year. The total compound interest earned is 363.
Algebraic Expression Problem [41:07]
Given P = √2 + 1 and Q = √2 - 1, find the value of a complex algebraic expression involving P and Q. Simplify the expression using algebraic identities. The final simplified value of the expression is 198.
Percentage Decrease Problem [47:54]
If the length and breadth of a rectangle are decreased by 12%, find the overall percentage decrease in the area. Use the formula for successive percentage change: -12 - 12 + (12*12)/100 = -24 + 1.44 = -22.56%.
Marked Price Problem [49:42]
A trader gives a discount of 5% on cash payment. To earn a profit of 14%, by what percentage should he mark up his goods above the cost price? Using the formula MP/CP = (100 + Profit%) / (100 - Discount%), MP/CP = 114/95 = 6/5. Therefore, the markup percentage is (6-5)/5 * 100 = 20%.
Circle and Chord Problem [50:31]
In a circle with a radius of 8 cm, a chord is 10 cm long. Find the distance of the chord from the center. Draw a perpendicular from the center to the chord, bisecting it into two 5 cm segments. Using the Pythagorean theorem, the distance from the center is √(8² - 5²) = √39 cm.
Simple Interest Rate Change Problem [51:09]
A sum of money becomes 9 times its original value in 21 years at a certain simple interest rate. The same sum becomes 12 times its original value in 25 years. Find the change in the interest rate. Calculate the initial rate as (8/21)*100 and the new rate as (11/25)*100. The difference between the two rates is approximately 6%. Since the rate increases, the answer is an increase of approximately 6%.
Average Speed Problem [53:32]
A cyclist covers 10 km at 15 km/hr and then 15 km at 10 km/hr. Find the average speed for the entire journey. The total distance is 10 + 15 = 25 km. The total time is (10/15) + (15/10) = (2/3) + (3/2) = 13/6 hours. The average speed is 25 / (13/6) = 150/13 km/hr.
Fraction to P/Q Form Problem [54:37]
Convert the repeating decimal 0.525252... into a fraction in the form of P/Q. The repeating part is 52, so the fraction is 52/99.
Age Ratio Problem [55:26]
The sum of the ages of A, B, and C is 54 years. B's age is 1/3 of C's age, and A's age is twice B's age. Find the sum of the ages of A and B. Let the ages of A, B, and C be 2x, x, and 3x, respectively. The sum of their ages is 2x + x + 3x = 6x = 54. Therefore, x = 9. The sum of the ages of A and B is 2x + x = 3x = 3 * 9 = 27 years.
Data Interpretation - Hockey Preference [56:41]
The problem involves analyzing data from different classes to determine the total number of students who like hockey. Calculate the number of students who like hockey in each class by multiplying the total number of students by the percentage who like hockey. Sum the results from each class to find the total number of students who like hockey. The total number of students who like hockey is 608.
Data Interpretation - Marks Comparison [57:52]
The problem involves comparing marks in different subjects using a pie chart. The combined marks in English and Social Science are 59 + 52 = 111. The combined marks in Hindi and Maths are 60 + 47 = 107. The difference is 111 - 107 = 4. The percentage difference is (4/107) * 100, which is approximately 3.7%.
Data Interpretation - Physics and Maths Ratio [59:45]
The problem involves finding the student whose marks in Physics and Maths are in the ratio of 10:13. By examining the data, student S has marks in Physics and Maths that match this ratio.
Data Interpretation - Sales Below Average [1:00:11]
The problem involves analyzing a bar graph to determine the number of days when sales were below the average. Calculate the average sales by summing the sales for all days and dividing by the number of days. Count the number of days when the sales were below this average. The number of days with sales below average is 2.
Data Interpretation - Highest Temperature [1:01:09]
The problem involves identifying the city with the highest recorded temperature across all 5 months. Sum the temperatures for each city across the 5 months and compare the totals. Hyderabad has the highest total temperature.
