Showing the 12 most recent of 109 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.If $\dfrac{{}^{(n-1)}C_{r-1}}{{}^nC_r} = \dfrac{3}{5};\ \dfrac{{}^{(n+1)}C_{r+1}}{{}^nC_r} = \dfrac{11}{7}$, then ${}^nC_{r+3} \div r\ {}^nC_{n/2} =$ (A) $\dfrac{3}{5}$ (B) $12$ (C) $8$ (D) $\dfrac{5}{3}$

   ›Reveal solutionSolution

   The two ratios give $\tfrac{r}{n}=\tfrac35$ and $\tfrac{n+1}{r+1}=\tfrac{11}{7}$, so $n=10,\ r=6$, and the required value is $\tfrac{5}{3}$.

   Simplify the given ratios.

   $\frac{{}^{\,n-1}C_{r-1}}{{}^{\,n}C_{r}}=\frac{r}{n}=\frac35\ \Rightarrow\ 5r=3n,$

   $\frac{{}^{\,n+1}C_{r+1}}{{}^{\,n}C_{r}}=\frac{n+1}{r+1}=\frac{11}{7}\ \Rightarrow\ 7(n+1)=11(r+1).$

   Solve. With $r=\tfrac{3n}{5}$:

   $7(n+1)=11\!\left(\tfrac{3n}{5}+1\right)\Rightarrow 35(n+1)=11(3n+5)\Rightarrow 35n+35=33n+55\Rightarrow n=10,\ r=6.$

   Check: ${}^{9}C_{5}/{}^{10}C_{6}=126/210=3/5$ and ${}^{11}C_{7}/{}^{10}C_{6}=330/210=11/7$. ✓

   Evaluate. With $n=10,\ r=6$: ${}^{\,n}C_{r+3}={}^{10}C_{9}=10$, so

   $\frac{{}^{10}C_{9}}{r}=\frac{10}{6}=\frac53.$ …

   🔒 Start your 14-day free trial to see the full solution →
2. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.Let $[t]$ represent the greatest integer less than or equal to $t$. If $x = (7\sqrt{5} + 15)^9$ and $y = (5\sqrt{7} + 13)^{11}$ then $[x]$ and $[y]$ are (A) even integer and odd integer respectively (B) odd integer and even integer respectively (C) both odd integers (D) both even integers

   ›Reveal solutionSolution

   Pairing each surd with its conjugate gives an integer sequence $a^n+b^n$ (resp. $p^n+q^n$) that is even for all $n$; the conjugate's odd power is a small negative number, so $[x]$ and $[y]$ equal those even integers.

   For $x=(15+7\sqrt5)^{9}$. Let $a=15+7\sqrt5,\ b=15-7\sqrt5$. Then

   $a+b=30,\qquad ab=15^2-(7\sqrt5)^2=225-245=-20,$

   and $b=15-7\sqrt5\approx-0.65$, so $-1<b<0$.

   $S_n=a^n+b^n$ satisfies $S_n=(a+b)S_{n-1}-ab\,S_{n-2}=30S_{n-1}+20S_{n-2}$ with $S_0=2,\ S_1=30$ — an integer for all $n$. Since $b<0$, $b^{9}<0$ and $|b^{9}|<1$, so

   $x=S_9-b^{9}=S_9+|b^{9}|,$

   which lies just above the integer $S_9$; hence $[x]=S_9$. Modulo $2$: $30\equiv0,\ 20\equiv0$, and $S_0,S_1$ are even, so every $S_n$ is even. Thus $[x]=S_9$ is even.

   For $y=(13+5\sqrt7)^{11}$. Let $p=13+5\sqrt7,\ q=13-5\sqrt7$. Then

   $p+q=26,\qquad pq=169-175=-6,$ …

   🔒 Start your 14-day free trial to see the full solution →
3. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.Consider the following Assertion (A): $5^{2n} - 3^{2n-1}$ is divisible by 2 for all $n \in \mathbb{N}$ Reason (R): $5^{2n} + 3^{2n-1}$ is divisible by 7 for all $n \in \mathbb{N}$ The correct answer is (A) Both A and R are correct and R is the correct explanation of A (B) Both A and R are correct and R is not correct explanation of A (C) A is correct, but R is not correct (D) A is not correct, but R is correct

   ›Reveal solutionSolution

   The assertion (A) is true because $5^{2n} - 3^{2n-1}$ is always odd minus odd = even. The reason (R) is false because $5^{2n} + 3^{2n-1}$ is not always divisible by 7 (e.g., fails for $n=1$). So the correct option is (C).

   Concept and intuition:

   Divisibility problems often look like they need induction or modular arithmetic, but sometimes a simple parity check or a single counterexample is enough. Here, Assertion (A) is about divisibility by 2 — that’s just about whether the expression is even or odd. Reason (R) claims divisibility by 7 for all natural numbers $n$; that’s a stronger claim that can be tested with small $n$. The trick is not to overcomplicate: check the simplest cases first.

   ---

   1. Check Assertion (A):

      We need to see if $5^{2n} - 3^{2n-1}$ is always even for $n \in \mathbb{N}$.

      • $5^{2n} = (5^2)^n = 25^n$, which is odd (odd number raised to any power stays odd).
      • $3^{2n-1}$ is odd (odd number to any power is odd).
      • Odd minus odd = even. So the expression is always divisible by 2.
      Tip

      No need for induction — just parity. An odd number minus an odd number is always even.

   2. Check Reason (R):

      We test $5^{2n} + 3^{2n-1}$ for divisibility by 7.

      • For $n=1$: $5^{2} + 3^{1} = 25 + 3 = 28$, which is divisible by 7. …

   🔒 Start your 14-day free trial to see the full solution →
4. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.The number of 3-digit numbers that can be formed by using the digits 2, 3, 5, 7, 9 that are divisible by 3, if repetition of digits is not allowed, is (A) 30 (B) 12 (C) 18 (D) 24

   ›Reveal solutionSolution

   A number is divisible by 3 if the sum of its digits is divisible by 3.

   From the digits {2, 3, 5, 7, 9}, we must choose 3 distinct digits whose sum is a multiple of 3.

   There are 4 such triples, each giving 3! = 6 permutations, so total = 4 × 6 = 24.

   The correct option is (D).

   ---

   The key concept: divisibility by 3

   A number is divisible by 3 if and only if the sum of its digits is divisible by 3.

   This is a classic rule that turns a problem about 3-digit numbers into a problem about choosing digit triples with a certain sum property.

   Since repetition is not allowed, we are simply picking 3 distinct digits from the set {2, 3, 5, 7, 9} and arranging them.

   ---

   Step-by-step reasoning

   1. List the digits and their remainders modulo 3

      • 2 → remainder 2
      • 3 → remainder 0
      • 5 → remainder 2
      • 7 → remainder 1
      • 9 → remainder 0

      So we have:

      • Remainder 0: {3, 9}
      • Remainder 1: {7}
      • Remainder 2: {2, 5}

   2. When does a sum of three digits become divisible by 3?

      The sum of three numbers is divisible by 3 exactly when the sum of their remainders (mod 3) is 0.

      Possible remainder combinations (mod 3) that sum to 0:

      • (0, 0, 0)
      • (1, 1, 1)
      • (2, 2, 2)
      • (0, 1, 2)

   3. Check which combinations are possible with our digits

      • (0, 0, 0): We have only two digits with remainder 0 (3 and 9), so we cannot pick three. ✗
      • (1, 1, 1): Only one digit with remainder 1 (7). ✗
      • (2, 2, 2): We have two digits with remainder 2 (2 and 5), so cannot pick three. ✗
      • (0, 1, 2): We have at least one of each remainder. ✓ This is the only viable pattern.

   4. Find all triples of distinct digits with remainders (0, 1, 2)

      • Choose one from remainder 0: {3, 9} → 2 choices
      • Choose one from remainder 1: {7} → 1 choice
      • Choose one from remainder 2: {2, 5} → 2 choices

      Total triples = $2 \times 1 \times 2 = 4$.

      The triples are: …

   🔒 Start your 14-day free trial to see the full solution →
5. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.If $2x^5 + ax^4 - 12x^3 + bx^2 + cx + c = 0$ is a reciprocal equation of class one, then the sum of all the rational roots of this equation is (A) $-\dfrac{1}{2}$ (B) $-\dfrac{7}{2}$ (C) $\dfrac{1}{2}$ (D) $\dfrac{7}{2}$

   ›Reveal solutionSolution

   A class-one reciprocal equation has palindromic coefficients, fixing it as $2x^5+x^4-12x^3-12x^2+x+2=0$; its rational roots $-1,\,-2,\,-\tfrac12$ sum to $-\tfrac72$.

   A reciprocal equation of class one has palindromic coefficients (coefficient of $x^k$ equals coefficient of $x^{5-k}$). Matching the given $x^5$-coefficient $2$ and $x^3$-coefficient $-12$ determines the equation as

   $2x^5+x^4-12x^3-12x^2+x+2=0.$

   This factorises as

   $(x+1)(x+2)(2x+1)(x^2-3x+1)=0.$ …

   🔒 Start your 14-day free trial to see the full solution →
6. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.The number of numbers between 345 and 543 such that the sum of the digits in each number is 15 is (A) 24 (B) 136 (C) 91 (D) 17

   ›Reveal solutionSolution

   Counting three-digit numbers in $(345,543)$ with digit sum $15$ by hundreds digit gives $6+8+3=17$.

   We need integers $N$ with $346\le N\le 542$ and digit sum $=15$.

   Hundreds digit $=3$ (so $N=3tu$, need $t+u=12$, $N\ge346$): the valid $(t,u)$ are $(4,8),(5,7),(6,6),(7,5),(8,4),(9,3)$ → $348,357,366,375,384,393$. ($339$ is below $346$.) 6 numbers. …

   🔒 Start your 14-day free trial to see the full solution →
7. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.Let X be a discrete random variable. If $P(X = r) = \frac{3}{7} a^r$, $r = 0, 1, 2, \ldots, \infty$ and $0 < a < 1$, then $a.P(X = 2) =$ (A) $P(X = 1)$ (B) $\frac{3}{7} P(X = 2)$ (C) $P(X = 3)$ (D) $\frac{4}{7} P(X = 1)$

   ›Reveal solutionSolution

   The key is to recognize that the given probabilities form a geometric series that must sum to 1, which forces $a = \frac{4}{7}$. Then compute $a \cdot P(X=2)$ and compare with the options; it equals $P(X=3)$, so option (C) is correct.

   We are told $P(X = r) = \frac{3}{7} a^r$ for $r = 0, 1, 2, \dots$ and $0 < a < 1$. Since $X$ is a discrete random variable, the sum of all probabilities must be 1. This is the central constraint that determines $a$.

   1. Sum the probabilities to 1

   $\sum_{r=0}^\infty P(X = r) = \sum_{r=0}^\infty \frac{3}{7} a^r = \frac{3}{7} \sum_{r=0}^\infty a^r = 1.$

   The infinite geometric series $\sum_{r=0}^\infty a^r$ converges to $\frac{1}{1-a}$ because $0 < a < 1$. Thus:

   $\frac{3}{7} \cdot \frac{1}{1-a} = 1.$

   2. Solve for $a$ Multiply both sides by $1-a$:

   $\frac{3}{7} = 1 - a \quad \Rightarrow \quad a = 1 - \frac{3}{7} = \frac{4}{7}.$

   So $a = \frac{4}{7}$, which indeed lies between 0 and 1.

   3. Write the probability mass function explicitly

   $P(X = r) = \frac{3}{7} \left( \frac{4}{7} \right)^r.$

   4. Compute $a \cdot P(X = 2)$

   $a \cdot P(X = 2) = \frac{4}{7} \cdot \left[ \frac{3}{7} \left( \frac{4}{7} \right)^2 \right] = \frac{4}{7} \cdot \frac{3}{7} \cdot \frac{16}{49} = \frac{3 \cdot 64}{7^3} = \frac{192}{343}.$

   5. Compute each option to compare
      • (A) $P(X = 1) = \frac{3}{7} \cdot \frac{4}{7} = \frac{12}{49} = \frac{84}{343}$ — not equal.
      • (B) $\frac{3}{7} P(X = 2) = \frac{3}{7} \cdot \frac{3}{7} \cdot \frac{16}{49} = \frac{144}{343}$ — not equal. …

   🔒 Start your 14-day free trial to see the full solution →
8. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
   Q.Identify the incorrect statements I. Capping = Addition of methyl guanosine triphosphate at $5'$ end of hnRNA II. Tailing = Addition of adenylate residues at the $3'$ end of hnRNA III. Splicing = Joining of introns and removal of exons IV. Ocho’s enzyme = Help in enzymatic synthesis of RNA (A) II and III only (B) I and II only (C) III only (D) III and IV only

   ›Reveal solutionSolution

   This question tests your understanding of post-transcriptional modifications in eukaryotes and the role of Ochoa's enzyme. Statement III incorrectly describes splicing, making it the only incorrect statement. The final answer is (C).

   Concept and Intuition

   After transcription, the primary RNA transcript (heterogeneous nuclear RNA or hnRNA) in eukaryotes undergoes several modifications before it can be translated into protein. These modifications, collectively known as post-transcriptional processing, are crucial for the stability, transport, and proper function of mRNA. The three main processes are capping, tailing, and splicing. Additionally, enzymes play a vital role in synthesizing nucleic acids.

   • Capping protects the 5' end of the mRNA and helps in ribosome binding.
   • Tailing protects the 3' end from degradation and aids in mRNA export from the nucleus.
   • Splicing removes non-coding regions (introns) and joins coding regions (exons) to form a mature mRNA.
   • Ochoa's enzyme (polynucleotide phosphorylase) is known for its ability to synthesize RNA in vitro.

   We need to evaluate each given statement against these established biological facts.

   Step-by-step Evaluation

   1. Evaluate Statement I: Capping = Addition of methyl guanosine triphosphate at $5'$ end of hnRNA

      • Capping is indeed the process where a 7-methylguanosine triphosphate residue is added to the $5'$ end of the hnRNA molecule. This modification occurs early during transcription and is essential for protecting the mRNA from degradation and for its efficient translation.
      • Therefore, Statement I is correct.

   2. Evaluate Statement II: Tailing = Addition of adenylate residues at the $3'$ end of hnRNA

      • Tailing, specifically polyadenylation, involves the addition of approximately 200-300 adenylate residues (adenine ribonucleotides) to the $3'$ end of the hnRNA. This poly-A tail enhances mRNA stability and facilitates its export from the nucleus.
      • Therefore, Statement II is correct.

   3. Evaluate Statement III: Splicing = Joining of introns and removal of exons

      • Splicing is a critical process in eukaryotic gene expression where non-coding sequences called introns are removed from the hnRNA, and the coding sequences called exons are joined together to form a continuous coding sequence. The statement describes the opposite process: "Joining of introns and removal of exons."
      • Therefore, Statement III is incorrect. …

   🔒 Start your 14-day free trial to see the full solution →
9. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
   Q.From the following lists, choose the incorrect combination List-1 (Nutritional type) | List-2 (Source of carbon) | List-3 (Example) A. Photoautotroph | Atmospheric $\mathrm{CO_2}$ | Chlorobium B. Chemoautotroph | Inorganic compounds | Salmonella C. Photoheterotrophs | Organic source | Rhodospirillum D. Chemoheterotrophs | Organic compounds | Xanthomonas (A) A (B) B (C) C (D) D

   ›Reveal solutionSolution

   Nutritional classification depends on energy source (photo/chemo) and carbon source (auto/hetero). Salmonella is a chemoheterotroph, not a chemoautotroph, making option (B) the incorrect combination.

   Organisms are classified nutritionally along two axes: their energy source and their carbon source. Phototrophs harvest light energy; chemotrophs oxidize chemical compounds. Autotrophs fix inorganic carbon (usually $\mathrm{CO_2}$); heterotrophs consume organic carbon. The four combinations give us the major nutritional types.

   Let's verify each pairing:

   1. Photoautotroph – Atmospheric $\mathrm{CO_2}$ – Chlorobium

      Photoautotrophs use light as energy and fix $\mathrm{CO_2}$ as their carbon source. Chlorobium is a green sulfur bacterium that performs anoxygenic photosynthesis, fixing $\mathrm{CO_2}$ via the reverse TCA cycle. This combination is correct.

   2. Chemoautotroph – Inorganic compounds – Salmonella

      Chemoautotrophs derive energy from oxidizing inorganic compounds (like ammonia, sulfur, or ferrous iron) and fix $\mathrm{CO_2}$ for carbon. Classic examples include Nitrosomonas (ammonia oxidizers) and Thiobacillus (sulfur oxidizers).

      Salmonella, however, is a chemoheterotroph. It oxidizes organic compounds for energy and uses organic molecules as its carbon source. It's a pathogenic bacterium that lives in the guts of animals, feeding on organic nutrients. This is the mismatch.

   3. Photoheterotroph – Organic source – Rhodospirillum …

   🔒 Start your 14-day free trial to see the full solution →
10. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
    Q.Match of the following List-1 A Inhibitor of catalytic activity B Possess peptide bonds C Fungi cell wall material D Secondary metabolite List-2 I Ricin II Chitin III Collagen IV Malonate (A) A-IV, B-III, C-I, D-II (B) A-IV, B-III, C-II, D-I (C) A-IV, B-II, C-III, D-I (D) A-IV, B-I, C-III, D-II

    ›Reveal solutionSolution

    This question tests your knowledge of key biological molecules and their functions. We will match inhibitors, proteins, cell wall components, and secondary metabolites to their specific examples. The correct match is A-IV, B-III, C-II, D-I.

    This question requires you to associate specific biological characteristics or roles (List-1) with the correct biomolecules or compounds (List-2). Understanding the fundamental nature and function of each item is key to making the correct connections.

    1. Match A: Inhibitor of catalytic activity

       • An inhibitor of catalytic activity is a substance that reduces or stops the rate of an enzyme-catalyzed reaction. Enzymes are biological catalysts, and their activity can be regulated by various molecules.
       • Malonate (IV) is a classic example of a competitive inhibitor. It structurally resembles succinate, the substrate for the enzyme succinate dehydrogenase in the Krebs cycle. Malonate binds to the active site of succinate dehydrogenase, preventing succinate from binding and thus inhibiting the enzyme's catalytic activity.
       • Therefore, A matches IV.

    2. Match B: Possess peptide bonds

       • Peptide bonds are covalent chemical bonds formed between the carboxyl group of one amino acid and the amino group of another amino acid. These bonds are the fundamental linkages that form polypeptide chains, which fold into proteins.
       • Collagen (III) is the most abundant protein in mammals, forming a major component of connective tissues. As a protein, it is composed of amino acids linked by peptide bonds.
       • Ricin (I) is also a protein, and thus also possesses peptide bonds. However, we need to find the best fit for each category. Let's keep this in mind as we proceed.
       • Therefore, B matches III.

    3. Match C: Fungi cell wall material

       • Cell walls provide structural support and protection to cells. The composition of cell walls varies significantly across different kingdoms of life.
       • Chitin (II) is a long-chain polymer of N-acetylglucosamine, a derivative of glucose. It is the primary structural component of the cell walls of fungi and also forms the exoskeletons of insects and crustaceans. …

    🔒 Start your 14-day free trial to see the full solution →
11. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
    Q.Match the following: List-1 List-2 A. Rhamphotheca I. Mammals B. Precocial hatchings II. Beak of birds C. Absence of renal portal system III. Collumella auris D. Hyomandibula IV. Flight less birds V. Dentary The correct answer is (A) A – II, B – V, C – I, D – III (B) A – II, B – IV, C – I, D – III (C) A – V, B – IV, C – III, D – I (D) A – IV, B – III, C – II, D – I

    ›Reveal solutionSolution

    This question tests your knowledge of specific anatomical and developmental features across different animal groups. We will match the horny beak covering (rhamphotheca) to birds, precocial development to flightless birds, the absence of a renal portal system to mammals, and the evolutionary fate of the hyomandibula to the columella auris. The correct option is (B).

    The ability to correctly match anatomical structures, physiological systems, or developmental patterns to specific animal groups or their evolutionary derivatives is fundamental to understanding comparative anatomy and evolution. Each item in List-1 represents a distinct biological feature, and we need to identify its correct association from List-2. This requires recalling the defining characteristics of various vertebrate classes and their evolutionary adaptations.

    Let's break down each item and find its match:

    1. A. Rhamphotheca

       • The rhamphotheca refers to the horny sheath that covers the bony jaws of birds, forming what we commonly call the beak. This keratinous covering provides protection and shapes the beak for various functions like feeding, preening, and nest building.
       • Looking at List-2, "II. Beak of birds" is the direct and correct match for rhamphotheca.
       • Match: A - II

    2. B. Precocial hatchings

       • Precocial hatchlings are young animals that are relatively mature and mobile from the moment of hatching or birth. They typically have open eyes, are covered in down or fur, and can often stand, walk, and feed themselves shortly after hatching/birth. This contrasts with altricial young, which are helpless and require extensive parental care.
       • Many bird species exhibit precocial development. Among the options in List-2, "IV. Flightless birds" is a strong association. Many flightless birds, such as ostriches, emus, and kiwis (ratites), are known for their highly precocial young. These chicks can often run and forage with their parents almost immediately after hatching, which is crucial for survival in environments where parents might not be able to easily bring food to a nest.
       • Match: B - IV

    3. C. Absence of renal portal system

       • The renal portal system is a venous system found in many vertebrates (fish, amphibians, reptiles, and birds) that carries blood from the caudal (posterior) part of the body, including the hind limbs and tail, to the kidneys before it returns to the heart. This system allows the kidneys to filter blood from these regions directly. …

    🔒 Start your 14-day free trial to see the full solution →
12. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
    Q.Driving force of evolution (A) Disruptive selection (B) Stabilising selection (C) Directional selection (D) Natural selection

    ›Reveal solutionSolution

    The driving force of evolution is natural selection, which acts on heritable variation to cause adaptive change — the correct option is (D).

    The question asks for the "driving force" of evolution — the mechanism that actually pushes populations to change over generations. This is a classic concept from evolutionary biology, and the answer hinges on understanding what each type of selection does versus what powers the whole process.

    Natural selection is the engine. It is the differential survival and reproduction of individuals due to differences in phenotype. Without it, variation alone (from mutation, recombination) would not lead to adaptation — it would just drift. The other options are specific modes of natural selection, not the driving force itself.

    1. Natural selection is the overarching mechanism. It acts on heritable variation in a population, causing traits that improve survival and reproduction to become more common over time. This is the core idea Darwin and Wallace proposed — it is the primary process that drives evolutionary change.

    2. Directional selection is a type of natural selection where one extreme phenotype is favoured, shifting the population mean in one direction. For example, larger beak size in finches during a drought. It is a pattern of selection, not the driving force itself. …

    🔒 Start your 14-day free trial to see the full solution →