This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 14401. |
State one transaction involving a decrease in Liquid ratio and no change in current ratio. |
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Answer» Purchase of goods for cash . |
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| 14402. |
The Debaters turnover Ratio of a company is 6 times. State with reasons whether the ratio will Improve, decrease, or not change due to increases in the value of closing stock by Rs. 50,000? |
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Answer» No change because it will neither affect net credit sales nor average receivable. |
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| 14403. |
State one transaction which result in a decrease in ' debt-equity ratio ' and no change in' current Ratio'. |
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Answer» Conversion of debentures into shares. |
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| 14404. |
Indicate which ratio a shareholders would use who is examining his portfolio and wants to decide Whether he should hold or sell his shareholdings? |
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Answer» Total Assets to Debt Ratio. |
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| 14405. |
How does ratio analysis becomes less effective when the price level changes? |
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Answer» Accounting ratios are calculated from financial statements, which are down on the basis of historical Cost as recorded in the book of accounts. |
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| 14406. |
What is single entry system of book-keeping? |
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Answer» It is a method of Book-keeping where both the aspects are not recorded, and for a few transactions none of the aspects is recorded. |
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| 14407. |
What do you mean by double entry system of Book Keeping? |
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Answer» It is a method of book keeping. Double entry system means “The method of recording of two fold aspects of a transactions". |
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| 14408. |
Write any two disadvantages of double entry system. |
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Answer» The disadvantages of double entry system are : 1. It is a costly 2. It requires special knowledge and skills to maintain the accounts. |
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| 14409. |
State true or False :In this system personal accounts and cash accounts transaction are recorded. |
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Answer» True, In this system personal accounts and cash accounts transactions are recorded. |
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| 14410. |
Single entry system will not be accepted by ……………. (a) Proprietor (b) Partners (c) Tax authorities |
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Answer» (c) Tax authorities |
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| 14411. |
Define of Single Entry System. |
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Answer» According to Kohler, “Single Entry System is a system of Book-keeping in which as a rule, only records of cash and personal accounts are maintained. It is always an incomplete double entry system varying with circumstances. |
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| 14412. |
Single entry system capital is calculated. (a) Capital = Assets – Liabilities (b) Assets = Capital – Liabilities (c) Capital = Assets + Liabilities (d) Assets = Liabilities – Capital |
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Answer» (a) Capital = Assets – Liabilities |
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| 14413. |
A firm has assets worth Rs. 10,00,000 and Capital Rs. 2,25,000 then its liabilities is ……………. |
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Answer» Rs. 7,75,000. |
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| 14414. |
What is the amount of capital of the proprietor, if his assets are Rs. 85,000 and liabilities are Rs. 21,000? (a) Rs. 85,000 (b) Rs. 1,06,000 (c) Rs. 21,000 (d) Rs. 64,000 |
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Answer» (d) Rs. 64,000 |
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| 14415. |
The excess of assets over liabilities is ……………. |
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Answer» The capital. |
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| 14416. |
Single entry system maintains ……………. and ……………. accounts. |
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Answer» Cash, Personal. |
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| 14417. |
Statement of affairs is a ……………. (a) Statement of income and expenditure (b) Statement of assets and liabilities (c) Summary of cash transactions (d) Summary of credit transactions |
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Answer» (b) Statement of assets and liabilities |
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| 14418. |
A statement of affairs resembles a ……………. |
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Answer» Balance Sheet |
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| 14419. |
Opening statement of affairs is usually prepared to find out the ……………. (a) Capital in the beginning of the year (b) Capital at the end of the year (c) Profit made during the year (d) Loss occurred during the year |
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Answer» (a) Capital in the beginning of the year |
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| 14420. |
Under Single Entry System, what is prepared to find out capital ?(a) Statement of Affairs(b) Statement of Profit or Loss(c) Balance sheet(d) Capital account |
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Answer» Correct option is (a) Statement of Affairs |
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| 14421. |
State true or False :Under networth method profit is ascertained by calculating the increase in network after adjusting for drawing and addition to capital. |
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Answer» True, Under networth method profit is ascertained by calculating the increase in the network after adjusting for drawing and addition to capital. |
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| 14422. |
Can Profit or Loss be ascertained under Single Entry System? Why? |
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Answer» As income and expense accounts are not prepared, it is not possible to know the profit or loss at the end of the year like Double Entry System. |
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| 14423. |
How has the rise of coalition politics imposed constraints on the power of the Prime Minister? |
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Answer» The Prime Minister of a coalition government cannot take decisions as he likes. He has to accommodate different groups and factions in his party as well as among alliance partners. He also has need to the views and positions of the coalition partners and other parties, on whose support the survival of the government depends. |
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| 14424. |
The principal cereal crop of India is A. wheat B. rice C. maize D. sorghum |
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Answer» Rice is the principal cereal crop in India. |
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| 14425. |
Given below is a schematic representation of lac operon:(i). Identify i and p.(i). Name the ‘inducer’ for this operon and explain its role. |
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Answer» (i) i is the regulatory gene and p is the promoter gene. (ii) Lactose is the inducer. It is the substrate for the enzyme beta-galactosidase and it regulates switching on and off of the operon. |
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| 14426. |
Main cereal crop of our country (A) Rice (B) Wheat (C) Maize (D) Millet |
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Answer» Main cereal crop of our country is Wheat |
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| 14427. |
How are the DNA fragments separated and isolated for DNA fingerprinting? Explain. |
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Answer» Separation and Isolation of DNA Fragments (Gel Electrophoresis) • Gel electrophoresis is a technique for separating DNA fragments based on their size. • Firstly, the sample DNA is cut into fragments by restriction endonucleases. • The DNA fragments being negatively charged can be separated by forcing them to move towards the anode under an electric field through a medium/matrix. • Commonly used matrix is agarose, which is a natural linear polymer of D-galactose and 3, 6-anhydro-L-galactose which is extracted from sea weeds. • The DNA fragments separate-out (resolve) according to their size because of the sieving property of agarose gel. Hence, the smaller the fragment size, the farther it will move. • The separated DNA fragments are visualised after staining the DNA with ethidium bromide followed by exposure to UV radiation. • The DNA fragments are seen as orange coloured bands. • The separated bands of DNA are cut out and extracted from the gel piece. This step is called elution. • The purified DNA fragments are used to form recombinant DNA which can be joined with cloning vectors. |
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| 14428. |
Forensic department was given three blood samples. Write the steps of the procedure carried to get the DNA fingerprinting done for the above samples. |
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Answer» Methodology and Technique i. DNA is isolated and extracted from the cell or tissue by centrifugation. ii. By the process of polymerase chain reaction (PCR), many copies are produced. This step is called amplification. iii. DNA is cut into small fragments by treating with restriction endonucleases. iv. DNA fragments are separated by agarose gel electrophoresis. v. The separated DNA fragments are visualised under ultraviolet radiation after applying suitable dye. vi. The DNA is transferred from electrophoresis plate to nitrocellulose or nylon membrane sheet. This is called Southern blotting. vii. VNTR probes are now added which bind to specific nucleotide sequences that are complementary to them. This is called hybridisation. viii. The hybridised DNA fragments are detected by autoradiography. They are observed as dark bands on X-ray film. |
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| 14429. |
Explain the steps of DNA fingerprinting that will help in processing of the two blood samples A and B picked up from the crime scene. |
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Answer» DNA Fingerprinting Methodology and Technique i. DNA is isolated and extracted from the cell or tissue by centrifugation. ii. By the process of polymerase chain reaction (PCR), many copies are produced. This step is called amplification. iii. DNA is cut into small fragments by treating with restriction endonucleases. iv. DNA fragments are separated by agarose gel electrophoresis. v. The separated DNA fragments are visualised under ultraviolet radiation after applying suitable dye. vi. The DNA is transferred from electrophoresis plate to nitrocellulose or nylon membrane sheet. This is called Southern blotting. vii. VNTR probes are now added which bind to specific nucleotide sequences that are complementary to them. This is called hybridisation. viii. The hybridised DNA fragments are detected by autoradiography. They are observed as dark bands on X-ray film. |
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| 14430. |
What is budding in plants? |
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Answer» Budding in plants is an artificial method of propagation in which a single bud is joined or grafted on the stock plant. |
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| 14431. |
Explain about artificial methods of vegetative reproduction. |
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Answer» 1. Vegetative propagation is a kind of asexual reproduction which occurs with the help of vegetative plant parts. 2. Cutting and grafting are two methods used to propagate desired varieties of plants. 3. Cutting – small pieces of plant parts having one or more buds are selected for propagation, e.g, Stem cutting – Rose, Root Cutting – Blackberry and Leaf cutting – Sansevieria. 4. Grafting – In this method two plant parts are joined ogether (Stock – rooted plant and Scion-attached plant and they continue their growth as one plant. 5. When a single bud is grafted on stock plant it is called as bud grafting, e.g. Rose, Apple, Pear. |
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| 14432. |
Match the columnsColumn AColumn B(1) Nutritive tissue of embryo(a) Perisperm(2) Remnants of nucellus in seed(b) Cotyledon(3) Nutritive tissue of developing microspores(c) Endosperm(4) First photosynthetic organ of embryo(d) Tapetum |
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| 14433. |
Explain the significance of satellite DNA in DNA fingerprinting technique. |
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Answer» About 3% or so of the human genome has highly repetitive sequences or simple-sequence DNA or simple sequence repeats or satellite DNA or microsatellites which are less than 10 bp long short sequences, repeated in multiples per cell. Satellite DNA, also termed as microsatellite, show relative uniformity within species and great variability between closely related species. Also, different individuals differ in a number of repeats of sDNA. This DNA polymorphism is used in DNA fingerprinting to create DNA profiles of individuals. The length of satellite regions are highly variable between people. These form small peaks during density gradient centrifugation and thus are invaluable for identification purposes. |
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| 14434. |
Why do you think the zygote is dormant for some time in a fertilized ovule ? |
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Answer» The zygote is formed by the fusion of the male gamete with the nucleus of the egg cell. The zygote remains dormant for sometime and waits for the endosperm to form, which develops from the primary endosperm cell resulting from triple fusion. The endosperm provides food for the developing embryo and after the formation of endosperm, further development of the embryo from the zygote starts. |
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| 14435. |
Write the function of coleoptile. |
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Answer» It protects the plumule of the monocot embryo. |
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| 14436. |
Match the columnsColumn A (Mechanism)Column B (Type of pollination)(1) Geitonogamy(a) Thea(2) Herkogamy(b) Gloriosa(3) Self-sterility(c) Cucurbita(4) Protogyny(d) Calotropis |
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| 14437. |
Match the columnsColumn A (Artificial Vegetative Propagation)Column B (Examples)(1) Leaf cutting(a) Blackberry(2) Stem cutting(b) Apple(3) Grafting(c) Bougainvillea(4) Root cutting(d) Sansevieria |
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| 14438. |
Match the columns Column A (Asexual)Column B (Examples)(1) Spore formation(a) Spirogyra(2) Conidia formation(b) Yeast(3) Fragmentation(c) Chlamydomonas(4) Budding(d) Penicillium |
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| 14439. |
Differentiate betweenHypocotyl and epicotyl |
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| 14440. |
Differentiate betweenColeoptile and coleorrhiza |
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| 14441. |
Difference between Epicotyl and Hypocotyl. |
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| 14442. |
Match the columnsColumn A (Adaptation)Column B (Type of pollination)(1) Sticky, spiny pollen grains non-fragrant flowers(a) Anemophily(2) Feathery stigma and vearsatile anther(b) Chiropterophily(3) Presence of nectar glands and sweet smell(c) Ornithophily(4) Dull coloured flowers with strong fragrance(d) Entomophily |
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| 14443. |
Synergids are ……………… (a) haploid (b) triploid (c) diploid (d) tetraploid |
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Answer» Correct answer is (a) haploid |
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| 14444. |
Give Reason or Explain the Statement :When Mendel crossed a tall pea plant with a dwarf pea plant the offspring obtained from this cross were all tall. |
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Answer» 1. The tall habit of the pea plant is dominant over the dwarf habit of the pea plant. 2. Hence, when Mendel crossed a tall pea plant with a dwarf pea plant, the offspring obtained from this cross were all tall. |
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| 14445. |
Endosperm of angiosperm is ……………… (a) haploid (b) diploid (c) triploid (d) tetraploid |
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Answer» Correct answer is (c) triploid |
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| 14446. |
If the number of chromosomes in an endosperm cell is 27, what will be the chromosome number in the definitive nucleus? (a) 9 (b) 18 (c) 27 (d) 36 |
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Answer» Correct answer is (b) 18 |
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| 14447. |
What are lysogenous cavity? |
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Answer» This type of intercellular space arises through dissolution of entire cells, which are therefore called lysigenous intercellular spaces (lysis, loosening, Greek). These cavities of intercellular spaces store up water, gases and essential oils in them. The examples are commonly found in water plants and many monocotyledonous plants. The secretory cavities in Eucalyptus, Citrus and Gossypium are good examples. |
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| 14448. |
Write the short note onEndosperm. |
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Answer» 1. Endosperm is a nutritive tissue. It nourishes the developing embryo. 2. The endosperm develops from the primary endosperm nucleus (PEN). 3. The endosperm is a post fertilization tissue. 4. There are two types of seeds depending upon the presence or absence of endosperm, viz., endospermic and nonendospermic. 5. Castor, coconut, maize, etc. are endospermic seeds, while bean, pea, gram, etc. are non-endospermic seeds. |
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| 14449. |
Differentiate between perisperm and endosperm by giving one example of each. |
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Answer»
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| 14450. |
Write the short note onEntomophily |
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Answer» 1. Pollination with the help of insects is called entomophily. 2. The insect pollinated flowers are called J entomophilous flowers. 3. Entomophilous flowers show the following adaptations: 4. Flowers are large and attractive. 5. Flowers are brightly coloured with i pleasant smell. 6. Flowers produce nectar which is food for the insects. 7. Pollen grains are spiny and sticky for easy adherance to the rough and sticky stigma. 8. Entomophily is seen in plants like rose, Jasmine, Oestrum, Salvia, etc. |
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