Summary
Highlights
A question involves a circle with a circumference of 44 cm, and a rhombus inscribed within it, with all vertices on the circle. The goal is to find the area of the rhombus. First, the radius of the circle is determined from the circumference (2πr = 44) to be 7 cm. When a rhombus is inscribed in a circle with all vertices on the circle, its diagonals are equal to the diameter of the circle. Thus, both diagonals (D1 and D2) are 14 cm (2 * radius). The area of a rhombus is (1/2) * D1 * D2, so the area is (1/2) * 14 * 14 = 98 cm².
The video discusses questions from the TCS NQT exams on March 20th and 21st, 2026. The presenter notes that 70-80% of questions are exact recollections from students, while 20-30% are framed based on student-reported patterns. Feedback suggests the numerical ability in the foundation section was more challenging than the advanced section. The speaker advises preparation for all topics and reminds viewers about the on-screen calculator. Key topics to focus on include percentages, profit and loss, data interpretation (pie chart, bar graph, line graph, table graph), ratios, blood relations, syllogisms, averages, mixtures, and partnerships.
A question on ratios involves the salaries of A, B, C, and D in the ratio 1/4:1/8:1/12:1/16. Given a total salary of 50,000, the task is to find the difference between B's and C's salaries. The solution involves converting the inverse ratios to normal ratios by finding the LCM (48) of the denominators. This yields a ratio of 12:6:4:3 for A:B:C:D. The total parts (25) are equated to 50,000, meaning one part is 2,000. The difference between B's (6 parts) and C's (4 parts) salaries is 2 parts, which equals 4,000.
This section explains two crucial applications of LCM and HCF. The first states that the product of two numbers equals their LCM multiplied by their HCF. The second application focuses on when the ratio of numbers is given: LCM = HCF × Product of the Ratios. An example demonstrates finding the LCM when numbers are in the ratio 5:6 and their HCF is 9. Using the second formula, LCM = 9 * (5 * 6) = 270. This method allows for quick calculation in 10-20 seconds.
A problem on profit and loss states that the cost price (CP) of a product is 780 rupees, and the profit is 120%. The selling price (SP) needs to be found. The approach is to consider CP as 100%. If the profit is 120%, the SP is 100% + 120% = 220% of the CP. Therefore, if 100% is 780 rupees, then 220% is (780 * 220) / 100 = 1716 rupees. The presenter highlights the importance of understanding how to frame this percentage relationship.
The video addresses a question on finding the angle between the hands of a clock at 8:24 PM. The formula to use is θ = |30H - (11/2)M|, where H is hours and M is minutes. Substituting 8 for H and 24 for M, the calculation is θ = |30 * 8 - (11/2) * 24| = |240 - 11 * 12| = |240 - 132| = 108°. A quick trick for multiplying by 11 is also demonstrated.
A simple interest problem asks for the total simple interest on a principal amount of 9,000 rupees for 5 years, with the rate of interest being 6% per annum for the first two years and 9% per annum for the next three years. The solution breaks down the calculation into two parts: For the first two years, interest is 6% * 2 years = 12% of 9,000, which is 1,080 rupees. For the next three years, interest is 9% * 3 years = 27% of 9,000, which is 2,430 rupees. The total simple interest for 5 years is the sum of these two amounts, 1,080 + 2,430 = 3,510 rupees.
This section explains a time and work problem using the 'chocolate method'. A and B complete a work in 4 days, B and C in 8 days, and C and A in 6 days. To find how many days all three (A+B+C) take together, the total work (LCM of 4, 8, 6) is taken as 24 chocolates. This means: A+B eat 6 chocolates/day, B+C eat 3 chocolates/day, and C+A eat 4 chocolates/day. By summing these, (A+B) + (B+C) + (C+A) = 2(A+B+C) eat 6 + 3 + 4 = 13 chocolates/day. This implies 2(A+B+C) eat 13 chocolates/day, so A+B+C eat 6.5 chocolates/day. To eat 24 chocolates, they will take 24 / 6.5 = 48/13 days, approximately 3.69 days.
A complex blood relations question is presented with various symbols representing relationships (hashtag for brother, @ for daughter, & for husband, % for wife). A series of symbol relationships is given: X%Y, Y#J, J@W, W&V, V@U. The goal is to determine how Y is related to U. By drawing a family tree: X is wife of Y, Y is brother of J, J is daughter of W, W is husband of V, and V is daughter of U. Tracing the lineage, V is U's daughter, W is V's husband, J is W and V's daughter, and Y is J's brother. Therefore, Y is the son of W and V, making Y the grandson of U.
The final question focuses on finding the unit's digit of a given expression: (27) to the power of 48 + (35) to the power of 125 + (63) to the power of 19. For numbers ending in 0, 1, 5, or 6, the unit's digit remains the same regardless of the power. So, the unit's digit of (35) to the power of 125 is 5. For numbers ending in 7, the cyclicity of unit's digits is 7, 9, 3, 1 (repeating every 4 powers). For (27) to the power of 48, 48 divided by 4 gives a remainder of 0 (or 4), so the unit's digit is 1 (the 4th digit in the cycle). For numbers ending in 3, the cyclicity is 3, 9, 7, 1 (repeating every 4 powers). For (63) to the power of 19, 19 divided by 4 gives a remainder of 3, so the unit's digit is 7 (the 3rd digit in the cycle). Adding the unit digits: 1 + 5 + 7 = 13. The unit's digit of the result is 3. (Correction in video calculation: 1+5+7 = 13, not 21, so unit digit is 3, not 1.)