10 Extremely Important Types to get 45+ in Maths in SSC Selection Post Phase 14 Exams 2026

Main Ideas / Concepts

• The speaker reviews the likely question types for SSC Selection Post Phase 14 Mathematics, noting that similar patterns will appear in SSC CGL shortly after.
• The focus is on “10 extremely important types” to practice for high marks (45+).
• For each topic, the speaker highlights either:
  • Sub-types that commonly appear (with examples), and/or
  • A shortcut/method to solve common formats quickly.

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Methodology / Instruction-Style Content (Organized by Topic)

1) Mensuration (especially “melting”/mixed solids and surface area)

What to expect

• 3–5 questions out of 25 in the upcoming/observed shift.
• Common themes:
  • Melting problems (volume equivalence of transformed shapes)
  • Total surface area (TSA) and curved surface area (CSA)

Key solving hacks

• Melting / volume equivalence approach

  • If different solids “transform” into a new solid, use the shortcut:
    • Equate volumes.

  • Example structure:

    • Given cubes with edges such as 6, 8, 10 (implied).

    • Compute combined volume: [ 6^3 + 8^3 + 10^3 ]

    • Let the equivalent new cube have volume (v^3).

    • Take cube root to get the missing edge.

• Cone + sphere / cone–sphere volume balancing

  • Use volume formulas and set them equal:
    • Sphere: (\frac{4}{3}\pi r^3)
    • Cone: (\frac{1}{3}\pi r^2 h)
  • Then solve after equating.

• TSA/CSA formula practice

  • Memorize formulas for:
    • 2D figures and 3D figures
    • Volume, CSA, TSA
  • Recommendation:
    • Watch an existing dedicated 40–50 minute mensuration video
    • Practice questions after memorizing formulas.

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2) Trigonometry (simplified identities using sec and tan)

Expected pattern

• Questions where values like (\sec\theta + \tan\theta) are given, and you need to find another value such as (\sin\theta).

Core technique

• If: [ \sec\theta + \tan\theta = 3 ] then: [ \sec\theta - \tan\theta = \frac{1}{3} ]

• Add/subtract to eliminate:

  • (2\sec\theta = 3 + \frac{1}{3})
• Then:
  • Find (\sec\theta \Rightarrow \cos\theta)
  • Use (\cos\theta \Rightarrow \sin\theta)

Related shortcut mentioned

• Triplet-style scaling
  • If you get (\sin\theta = \frac{4}{5}), use scaling to obtain integer-friendly answers.
• Triangle/triplet substitution may also be used when applicable:
  • Related expressions like: [ 3\sec\theta,\; 4\tan\theta,\; 2\cot\theta,\; \csc\theta ]

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3) Mixture & Alligation

Expected pattern

• Two/more containers with mixture ratios (e.g., milk and water).
• You must find the mixing ratio so the final ratio matches a target.

Method

• Model each container’s milk-water composition using the given ratio.
• Use an allegation-style difference/combination approach (with fraction-based working).
• Conceptual “pure quantity” trick:
  • Assume pure milk = 100
  • If 20% is replaced by water → milk becomes 80
  • Repeat substitution logic for multiple replacements
  • This effectively multiplies the remaining fractions to compute the final percentage.

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4) Time & Work

What to do

• Practice the basic time-and-work questions that appeared that day.
• The speaker suggests:
  • Screenshot or note those questions
  • Practice using their basic pattern
• Also mentioned:
  • A separate one-shot video covering all Time & Work types.

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5) Interest (CI vs SI; payment frequency changes)

A) Difference between CI and SI

• Remember: [ \text{CI} - \text{SI} = \frac{p\cdot r^2}{100} ]

• Procedure when difference is given:

  • Example: difference (=160), principal (p=25000)

  • Solve: [ 160 = \frac{p\cdot r^2}{100} ]

  • Simplify/cancel zeros to extract (r).

B) CI when payment frequency changes (yearly vs half-yearly)

• Approach mentioned:
  • Treat half-yearly investment growth as separate blocks
  • Example logic:
    • If yearly rate is 8%, then half-yearly periods act like two periods of 4%
  • Use the tree method.

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6) Algebra (values of expressions like (x+\frac{1}{x}), higher powers)

Expected patterns

• Given (x + \frac{1}{x}), find:
  • (x^3 + \frac{1}{x^3}) (via options/identities)
• Also mention:
  • (x^6 + \frac{1}{x^6})

Method guidance

• For cube outcomes:
  • Use option-based checking by comparing candidate cubed values.
• For higher powers:
  • Suggested order:
    1. Start with (x + \frac{1}{x})
    2. Square it to get (x^2 + \frac{1}{x^2})
    3. Proceed further from there (instead of cubing first).

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7) Average (alligation-style; age average “evergreen”)

A) Average with pass/fail

Typical structure

• Total students and averages for:
  • Overall average
  • Passed average
  • Failed average

Method (difference-from-overall logic)

• Use overall average (=50)
• Passed average (=70), Failed average (=30)
• Compute differences:
  • (70 - 50 = 20)
  • (50 - 30 = 20)
• Since differences match:
  • Passed : Failed = 1 : 1
• Then use total count to calculate actual numbers
  • Example: total (=120 \Rightarrow) passed (=60)

B) Evergreen: average ages with teacher

• Average of (students + teacher) is 17
• If teacher is removed:
  • average of 10 students becomes 2 years less (so it becomes 15)
• Teacher’s age is found by distributing the difference across counts (speaker arithmetic leads to 37).

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8) Speed, Distance, Time (average speed via LCM / scaling)

Expected patterns

• Successive segments with different speeds.
• Another type: part distance at one speed and the rest at another.

Method for successive equal-distance segments

• Distances like “12 km” may be used as intentional confusion.
• Use LCM of speeds (example: 20, 30, 60 ⇒ LCM = 60).
• Scale the segment distances so that each segment pattern uses total scaled distance = 60.

• Time: [ \text{Time}=\frac{\text{distance}}{\text{speed}} ]

• Total time → compute average speed: [ \text{Average speed}=\frac{\text{total distance}}{\text{total time}} ]

Method for fractional distance at different speeds

• Example:
  • (2/5) at speed 40
  • (3/5) at speed 60
• Use total time to set up equations:
  • time for each part = (part distance)/(speed)
• Solve for total distance.

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9) Ratio & Proportion

Type A: percentage change in fraction

• Numerator increased by 20%
• Denominator decreased by 10%
• Convert numerator and denominator accordingly, then compute simplified ratio (x/y).

Type B: (P) is 30% of (R); (Q) is 40% of (R)

• Compute: [ \frac{P}{Q}=\frac{30\% \cdot R}{40\% \cdot R}=\frac{30}{40}=0.75 ]

• So (P) is 75% of (Q).

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10) Geometry

Type A: Triangle with (XY \parallel BC) and equal-area division

• Setup:
  • In triangle (ABC), line segment (XY \parallel BC)
  • (X) on (AB), (Y) on (AC)
  • (XY) divides triangle (ABC) into two equal-area regions
• Key inference:
  • Equal areas imply a relationship between the linear ratios.
  • Use:
    • Area ∝ (side)(^2)
  • Compute the required ratio such as (AX:XB).

Type B: Similar triangles area ratio → altitude ratio

• Rule for similar triangles: [ \frac{\text{Area}_1}{\text{Area}_2}=\left(\frac{h_1}{h_2}\right)^2 ]

• Therefore: [ \frac{h_1}{h_2}=\sqrt{\frac{\text{Area}_1}{\text{Area}_2}} ]

• Example mentioned:

  • If area ratio is (5:3), then: [ \frac{h_1}{h_2}=\sqrt{\frac{5}{3}} ]

Other geometry reminders

• Topics likely to appear:
  • common tangents
  • circle segment area questions
• Speaker claims:
  • questions were not very tough
  • the paper was mostly PYQ-based

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Sources / Speakers Featured

• Single speaker: an unnamed instructor/host addressing “hello everyone” and giving solution strategies.
• Additional sources mentioned (reference only):
  • Mensuration video (40–50 minutes)
  • Trigonometry-related video (planned/mentioned)
  • One-shot Time & Work video
  • Arithmetic and Advanced videos totaling ~10:30 hours
  • Planned “top 50 concepts” videos on 21st and 27th (by the same instructor)