Video summary

Ratio and Proportion - Important Questions | CA Foundation May'26 | Hitesh Parmar

Main summary

Key takeaways

Educational

Main ideas / lessons conveyed

  • The session focuses on Advanced Ratio & Proportion (CA Foundation, May’26) with an emphasis on exam-level practice (PYQs).
  • The instructor repeatedly stresses:
    • “Less talk, more study” (solve questions regularly).
    • Use structured reasoning: set variables like (x), set up ratios/equations, then compute.
    • Many questions require careful interpretation of whether the given ratio is A:B or B:A.
    • A timed exam mindset: questions may look simple but become tricky—stay alert.

Methodologies / step-by-step instruction patterns taught

1) Ages ratio problem (using “1 year ago” and “after some years”)

  • Assume ages 1 year ago in the given ratio:
    • If (A/B) one year ago (= 5/4), set:
      • (A) one year ago (= 5x)
      • (B) one year ago (= 4x)
  • Convert to current/future by adding years to both:
    • Age after 1 year: (5x + 1), (4x + 1)
    • Age after additional years (example described: add 4 more, i.e., reach the time point 5 years after the “1 year ago” baseline)
  • Use the given “ratio at a later time” to form an equation:
    • Example: (\dfrac{5x+5}{4x+5} = \dfrac{6}{5})
  • Solve for (x) using cross-multiplication.
  • Compute the required age “after 10 years” by:
    • Finding current age (= (5x+1)),
    • Then adding the remaining years as required by the problem statement.

2) Boys/Girls with added participants (ratio change after admission)

  • Find original counts using the first session ratio:
    • Given boys:girls (= 8:5) among 455
    • Compute boys and girls from the total.
  • Add new admissions:
    • Add the given number (e.g., 50 girls).
  • Update the new ratio condition:
    • If the question requires girls:boys = 3:4, ensure you use the correct order (girls to boys, not boys to girls).
  • Form an equation with unknown additional boys (x):
    • New girls (= (\text{old girls} + \text{new girls}))
    • New boys (= (\text{old boys} + x))
  • Solve using cross-multiplication to get (x).

3) Partnership ratio when time differs (investment + time adjustment)

  • Base concept: partnership profit ratio depends on (investment × time).
  • If partner times differ, adjust contributions using effective time relative to the durations.
  • Instructor’s takeaway:
    • “When time and duration of investment differ, calculate partnership ratio including time/duration.”

4) “Money game” / coin ratio problems (value from quantities)

  • Convert coin distribution ratio into quantities:
    • If coin denominations are in ratio (3:2:1), treat counts as (3x, 2x, x).
  • Convert total money into an equation using denomination × count:
    • Example pattern:
      • (25(3x) + 10(2x) + 5(x) = 40) (after aligning units)
  • Solve for (x).
  • Then compute the required number of coins for a denomination (e.g., number of 5-paise coins (= x)).

5) Ratio after percentage increments

  • New salary = old salary × ((1 + \text{increment}\%)).
  • For ratio update:
    • If salaries are (2:3:5) with increments (15\%, 10\%, 20\%):
      • Multiply each part accordingly: (2(1.15),\ 3(1.10),\ 5(1.20))
  • Convert decimals to integers by multiplying by a common factor (e.g., by 10) to keep the ratio clean.
  • Reduce to the final ratio.

6) “Sandwich technique” for complex ratio expressions

  • Set up fractions using the sandwich method:
    • Write outer parts (A and D) as products of extreme multipliers.
    • Middle parts (B and C) are formed by isolating/removing factors appropriately.
  • Used to compute expressions like (ABC : D) from sequential ratios.

7) “Remaining days” in ration/time-rate with reduced persons

  • Compute total consumption over the original period:
    • Total food for all soldiers for (N) days = (N \times (\text{number of soldiers})).
  • Compute consumption before withdrawal:
    • Remaining food after (k) days = total − ((\text{soldiers} \times k)).
  • Adjust remaining soldiers count after some leave/are called back.
  • Extra days:
    • (\text{Extra days} = \dfrac{\text{remaining food}}{\text{remaining soldiers per day}}).

8) Mixture/acid-water removal & addition (concentration approach)

  • Convert concentration into a ratio:
    • Water is 64% ⇒ acid is 36%
    • So acid:water (= 36:64) (simplify as needed).
  • When:
    • Remove 4 L of solution,
    • Add the same amount of water.
  • Use concentration/amount tracking:
    • Acid removed = (acid proportion × 4).
    • Water added = 4 L (treated as 100% water).
  • Apply final concentration condition (e.g., 30% acid) to form an equation in terms of an unknown initial total.
  • Solve for the variable (x), then compute the required initial total volume.

Workflow emphasized throughout

  • Write in copy / practice together
  • For every question:
    • Identify what is asked (ratio, value, time, future age, etc.).
    • Introduce variables when needed (e.g., (x)).
    • Set up the ratio/equation carefully.
    • Use cross-multiplication or direct proportionality.
    • Compute the final value/ratio option.
  • Re-check ratio direction (example: instructor corrects whether the given ratio is boys:girls vs girls:boys).

Sources / speakers identified

  • Hitesh Parmar (main instructor)
  • Unacademy (platform referenced for results, course/session branding)
  • Unacademy results/topper teams mentioned:
    • Sankalp Team
    • Yoddha Team A
    • (Also mentions “AYoddha Team A”, likely referring to a specific team name)
  • Audience/student speakers/participants referenced by name in chat/comments:
    • Sapna, Abhishek, Rahul, Master Shikhar, Chandan Kumar, Yashwani, Ishika, Manoj Kumar, Ruchi, Gautam, Riya, Anshu, Ashu, Princey, Parth, Sonakshi Rathore, Karan Bhatt, Mr Makwana

Original video