The probability of guessing the correct answer of a certain test
question is (x/12). If the probability of not guessing the correct
answer is ( ⅚), then find the value of x.
cbse 10 th math old board exam question paper probability question mathematics standard
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At first glance, it may appear simple, but this problem is an excellent example of a question that checks conceptual understanding, numerical accuracy, and knowledge of complementary events in probability. It can appear in numerous exams worldwide, including CBSE, ICSE, IGCSE, GCSE, IB, AP, SAT, ACT, GRE, GMAT, and SOA actuarial examinations.
Step-by-Step Solution
To solve the problem, we start by recalling one of the fundamental rules in probability:
The sum of the probability of an event and the probability of its complement is always equal to one.
Let’s denote:
According to the complementary rule:
P(correct)+P(not correct)=1
Substituting the given values:
x/12+5/6=1
To solve for first express 5/6 as a fraction with denominator 12:
So the equation becomes:
Combine like terms:
Multiply both sides by 12 to eliminate the denominator:
Subtract 10 from both sides:
Therefore, the probability of guessing the correct answer is 2/12, which simplifies to 1/6. This satisfies the given probability of not guessing the correct answer (5/6), since 1/6+5/6=1.
Understanding Complementary Probability
The problem provides a perfect example of the complementary rule. In probability theory, every event has a complement — the set of outcomes where the event does not occur. The sum of the probabilities of an event and its complement always equals one.
This principle is fundamental in mathematics curricula worldwide. It is introduced in CBSE Class 9 as part of Chapter 15: Probability, where students learn about experimental probability using dice, coins, and simple card experiments. ICSE Class 9 and 10 also emphasize complement probability in their algebraic and practical problem sections. In IGCSE and GCSE mathematics, complement rules are a standard part of both foundation and higher-tier probability questions. IB Diploma students encounter similar questions in Analysis and Approaches or Applications and Interpretation courses, where both theoretical and experimental probability are explored.
Even in professional contexts such as AP Statistics, GRE, and GMAT quantitative sections, the complementary rule forms the basis for more complex probability problems, including conditional probability, joint probability, and expected value calculations. SOA actuarial exams often use complement probability as a foundational step before progressing to advanced stochastic models and actuarial risk assessments.
Strategy overview
The most efficient route is to express tangent and cotangent through sine and cosine, then search for a substitution that treats the two square roots in a symmetric way. A natural attempt is to combine sine and cosine linearly because sin x and cos x appear in reciprocal fashion inside tan x and cot x. The expression sin x minus cos x is the key. With that choice you can leverage the identity for sin 2x and a compact derivative that collapses a messy sum into a familiar primitive.
Detailed solution
Start by rewriting the integrand through sine and cosine. Write tan x as sin x over cos x. Write cot x as cos x over sin x. Then the integrand becomes
√(sin x over cos x) + √(cos x over sin x)
This is
(√sin x over √cos x) + (√cos x over √sin x)
Now bring the two terms over a common style of denominator by noting that
(√sin x over √cos x) + (√cos x over √sin x)
equals
( sin x + cos x ) divided by √(sin x cos x)
To see that, write each fraction with denominator √(sin x cos x). The first term becomes sin x divided by √(sin x cos x). The second term becomes cos x divided by √(sin x cos x). Summing yields the expression above. The integrand is now
( sin x + cos x ) / √( sin x cos x )
This is a big step because it reduces the pair of nested roots to a single root. Next recall that sin 2x equals 2 sin x cos x. Thus sin x cos x equals sin 2x over 2. The denominator becomes √( sin 2x over 2 ). That is √(sin 2x) over √2. Therefore the integrand becomes
( sin x + cos x ) divided by ( √(sin 2x) over √2 )
which equals
√2 ( sin x + cos x ) / √( sin 2x )
You want to introduce a substitution whose derivative matches sin x plus cos x. Observe that the derivative of sin x minus cos x is cos x plus sin x. This is exactly the numerator. Let
t = sin x − cos x
Then dt = (cos x + sin x) dx. That means (sin x + cos x) dx equals dt. The integral becomes
∫ √2 dt / √( sin 2x )
We must now express sin 2x in terms of t. Compute t squared:
t² = (sin x − cos x)² = sin² x + cos² x − 2 sin x cos x = 1 − 2 sin x cos x
Hence 2 sin x cos x = 1 − t². But 2 sin x cos x equals sin 2x. Therefore sin 2x = 1 − t². The integrand is now completely in terms of t:
∫ √2 dt / √( 1 − t² )
This is the standard inverse trigonometric integral. The antiderivative is
√2 arcsin t + C
Now substitute back t = sin x − cos x to obtain the final expression
√2 arcsin( sin x − cos x ) + C
This form is accepted across board solutions and university notes because it displays the direct use of the template ∫ dt over √(1 − t²) equals arcsin t plus constant. Some manuals prefer to convert arcsin of a linear combination to an arctan form. That conversion is valid but not necessary for credit in most marking schemes. Present the arcsin form unless your class notes or teacher prefers a different inverse function. The equality of forms follows from standard relations among inverse trigonometric functions and from the identity that links t to sin 2x. For exam presentation keep the answer as shown with the constant of integration included.
Why this method fits widely used rubrics
Board mark schemes value clarity, a correct substitution, and a standard primitive. The steps above check all three boxes. First, the rewrite of the original integrand into a single fraction over √(sin x cos x) is transparent and short. Second, the substitution uses a linear combination of sine and cosine that appears often in these courses, so it looks familiar to graders. Third, the integral of dt over √(1 − t²) is a known template. If you write each of these moves cleanly, you match the expectations in CBSE Class 12 and ISC, in ICSE advanced sets, in Cambridge IGCSE Additional Mathematics extension exercises, in AS and A Level Pure Mathematics across CIE, Edexcel, OCR, and AQA, in IB Mathematics AA at both levels, and in AP Calculus BC. In JEE practice the same sequence gives a speed advantage because the derivative of sin x minus cos x produces the exact numerator, so you avoid detours.
Common slips and how to avoid them
The most frequent error is dropping the factor √2 when converting √( sin 2x over 2 ) to √(sin 2x) over √2. Keep that factor with care. Another slip is to forget the chain of identities that shows sin 2x equals 1 minus t². Build that bridge step by step using t² and the Pythagorean identity for sine and cosine. A third slip is to omit the constant of integration. Examiners look for it.
How to write the solution in a crisp mark scheme style
Begin with the conversion of the integrand to √2 times the fraction (sin x plus cos x) over √( sin 2x ). Introduce t as sin x minus cos x. Replace dt by the numerator times dx. Replace sin 2x by 1 minus t². Complete the integral to yield √2 arcsin t plus constant. Substitute back t. Close with the final expression. This sequence mirrors the solution skeleton provided in many board scoring rubrics and in university course keys.
Why the topic matters for learners moving between programs
Many students switch boards or programs during their senior years. A learner moving from a national board to IGCSE Additional Mathematics or to AS Level Pure will find that the underlying tools remain the same even though notation and pacing shift. The present integral is a friendly point of contact. It uses the identity for sin 2x. It uses a linear substitution. It uses one of the three classic inverse trigonometric primitives. That trio overlaps across curricula. As a result, mastering this single example builds confidence and transfers smoothly to adjacent problems that share the same structure, such as sums and differences of square roots built from sine and cosine ratios.
Connection to classroom assessment and computer based testing
Marking practices in school boards and in university courses place value on clean structure. You can score full credit by showing the three transitions and the final result without long commentary. Computer based practice systems also tend to accept any equivalent antiderivative. If your system presents a different inverse trigonometric function as the model answer, confirm equivalence by differentiation. Different forms that differ by a constant are equally valid. The arcsin form is compact and easy to differentiate, so it is a solid default in time bound conditions.
Tips for revision
Read the solution once without writing. Read it again and copy the key lines by hand. Then differentiate the final answer to check that you recover the original integrand. This last act links the algebra to the geometric meaning of inverse trigonometric functions and strengthens retention for examination day. If you study with a group, agree on the same structure so that peer feedback is consistent with rubric language.
Final answer to present in exams
∫ [ √(tan x) + √(cot x) ] dx = √2 arcsin( sin x − cos x ) + C
That is the complete result. It flows from a single substitution anchored in the identity for sin 2x. It meets the standards of clarity used in the boards and programs listed earlier, and it aligns with university calculus notes that stress method first, pattern recognition second, and presentation third. Place the video above this text in your BlogSpot post so that readers can watch the derivation and then scan the written solution for reinforcement. This pairing helps learners from different curricula follow a consistent set of steps. It also helps teachers and tutors share a uniform explanation that works across classroom settings and practice platforms.
