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.
| 98701. |
Which order is exclusively hematophagous (blood feeders)? a. Siphonaptera b. Thysanoptera c. Phasmida d. Hymenoptera |
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Answer» Siphonaptera is exclusively hematophagous (blood feeders). |
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| 98702. |
In insect development, the germ band: a. Forms the amnionic membrane. b. Differentiates into three germ layers. c. Gives rise to the cleavage and activation centers. d. Remains undifferentiated throughout the larval stages. |
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Answer» c. Gives rise to the cleavage and activation centres. |
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| 98703. |
Which order is exclusively herbivorous? a. Hemiptera b. Odonata c. Phasmida d. Thysanoptera |
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Answer» Hemiptera order is exclusively herbivorous. |
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| 98704. |
How to find the number of pair of alleles involved in polygenic inheritance using the number of phenotypical forms of the trait they condition? |
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Answer» Considering “p” the number of phenotypical forms and “a” the number of involved alleles of the polygenic inheritance. The formula p = 2a + 1 is then applied. (Many times it is not possible to determine precisely the number of phenotypical forms, p, due to the multigenic feature of the inheritance, since often the observed variation of phenotypes seems to be on a continuum or the trait suffers environmental influence.) |
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| 98705. |
What is the most probable inheritance pattern of a trait with gaussian proportional distribution of phenotypical forms? |
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Answer» If a trait statistically has a normal (gaussian, bell-shaped curve) distribution of its phenotypical forms it is probable that it is conditioned by polygenic inheritance (quantitative inheritance). In quantitative inheritance the effects of several genes add to others making it possible to represent the trait variation of a given population in a gaussian curve with the heterozygous genotypes in the center, i.e., appearing in larger number, and the homozygous in the extremities. |
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| 98706. |
Why is sex-linked inheritance an example of nonmendelian inheritance? |
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Answer» Sex-linked inheritance is a type of nonmendelian inheritance because it opposes Mendel’s first law, which postulates that each trait is always conditioned by two factors (alleles). In nonhomologous regions of the sex chromosomes the genotypes of the genes contain only one allele (even in the case of the XX karyotype, i.e., in women, one of the X chromosomes is inactive). |
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| 98707. |
What is epistasis? What is the difference between dominant epistasis and recessive epistasis? |
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Answer» Epistasis is the gene interaction in which a gene (the epistatic gene) can disallow the phenotypical manifestation of another gene (the hypostatic gene). In dominant epistasis the inhibitor allele is the dominant allele (for example, I) of the epistatic gene so inhibition occurs in dominant homozygosity (II) or in heterozygosity (Ii). In recessive epistasis the inhibitor allele is the recessive allele of the epistatic gene (i) so inhibition occurs only in recessive homozygosity (ii). |
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| 98708. |
What is polygenic inheritance? How does it work? |
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Answer» Polygenic inheritance, also known as quantitative inheritance, is the gene interaction in which a given trait is conditioned by several different genes having alleles that may or may not contribute to increase the phenotype intensity. The alleles may be contributing or noncontributing and there is no dominance among them. Polygenic inheritance is the type of inheritance, for example, of skin color and of stature in humans. Considering a given species of animal in which the length of the individual is conditioned by polygenic inheritance of three genes, for the genotype having only noncontributing alleles (aabbcc) a basal phenotype, for example, 30 cm, would emerge. Considering also that for each contributing allele a 5 cm increase in the length of the animal is added, so in the genotype having only contributing alleles (AABBCC) the animal would present the basal phenotype (30 cm) plus 30 cm more added by each contributing allele, i.e., its length would be 60 cm. In the case of triple heterozygosity, for example, the length of the animal would be 45 cm. That is the way polygenic inheritance works. |
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| 98709. |
What are complementary genes? Does this inheritance pattern obey Mendel’s second law? |
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Answer» Complementary genes are different genes that act together to determine a given phenotypical trait. For example, consider a phenotypical trait conditioned by 2 complementary genes whose alleles are respectively X, x, Y and y. Performing hybridization in F2 4 different phenotypical forms are obtained: X_Y_ (double dominant), X_yy (dominant for the first pair, recessive for the second), xxY_ (recessive for the first pair, dominant for the second) and xxyy (double recessive). This is what happens, for example, regarding the color of budgerigar feathers, in which the double dominant interaction results in green feathers, the dominant for the first pair, recessive for the second interaction results in yellow feathers, the recessive for the first pair, dominant for the second interaction leads to blue feathers and the double recessive interaction leads to white feathers. Each complementary gene segregates independently from the others since they are located in different chromosomes. Therefore the pattern follows Mendel’s second law (although it does not obey Mendel’s first law). |
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| 98710. |
In the hybridization of 2 genes (4 different alleles, 2 of each pair) how does epistasis affect the proportion of phenotypical forms in the F2 generation? |
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Answer» In dihybridism without epistasis double heterozygous parental individuals cross and in F2 4 phenotypical forms appear. The proportion is 9 double dominant to 3 dominant for the first pair, recessive for the second to 3 recessive for the first pair, dominant for the second to 1 double recessive (9:3:3:1). Considering that the epistatic gene is the second pair and that the recessive genotype of the hypostatic gene means lacking of the characteristic, in the F2 generation of the dominant epistasis the following phenotypical forms would emerge: 13 dominant for the second pair or recessive for the first, i.e., the characteristic does not manifest, 3 dominant for the first pair, recessive for the second, i.e., the characteristic manifests. The phenotypical proportion would be 13:3. In the recessive epistasis in F2 the phenotypical forms that would emerge are: 9 double dominant (the characteristic manifests), 7 recessive for the first pair or recessive for the second, i.e., the characteristic does not manifest. So the phenotypical proportion would be 9:7. These examples show how epistasis changes phenotypical forms and proportions, from the normal 9:3:3:1 in F2 to 13:3 in dominant epistasis or to 9:7 in recessive epistasis (note that some forms have even disappeared). (If the recessive genotype of the hypostatic gene is active, not simply meaning that the dominant allele does not manifest, the number of phenotypical forms in F2 changes.) |
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| 98711. |
How is it possible to obtain the probability of emergence of a given genotype formed of more than one pair of different alleles with independent segregation from the knowledge of the parental genotypes? |
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Answer» Taking as example the crossing of AaBbCc with aaBBCc, for each considered pair of allele it is possible to verify which genotypes it can form (as in an independent analysis) and in which proportion. AA x aa: Aa, aa (1:1). Bb x BB: BB, Bb (1:1). Cc x Cc: CC, Cc, cc (1:2:1). The genotype to which the probability is to be determined is for example aaBbcc. For each pair of this genotype the formation probability is determined: to aa, 0.5; to Bb, 0.5; to cc, 0.25. The final result is obtained by multiplication of these partial probabilities, 0.5 x 0.5 x, 0.5, resulting 0.0625. |
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| 98712. |
What are multiple alleles? Is there dominance in multiple alleles? |
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Answer» Multiple alleles is the phenomenon in which the same gene has more than two different alleles (in normal mendelian inheritance the gene has only two alleles). Obviously these alleles combine in pairs to form the genotypes. In multiple alleles relative dominance among the alleles may exist. A typical example of multiple alleles is the inheritance of the ABO blood group system, in which there are three alleles (A, B or O, or IA, IB and i). IA is dominant over i, which is recessive in relation to the other IB allele. IA and IB lack dominance between themselves. Another example is the color of rabbit fur, conditioned by four different alleles (C, Cch, Ch and c). In this case the dominance relations are C > Cch > Ch > c (the symbol > means “dominates over”). |
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| 98713. |
What are gene interactions? What are the three main types of gene interactions? |
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Answer» Gene interaction is the phenomenon in which a given phenotypical trait is conditioned by two or more genes (do not confuse with multiple alleles in which there is a single gene having three or more alleles). The three main types of gene interaction are: complementary genes, epistasis and polygenic inheritance (or quantitative inheritance). |
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| 98714. |
What is pleiotropy? |
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Answer» Pleiotropy (or pliotropy) is the phenomenon in which a single gene conditions several different phenotypical traits. Some phenotypical traits may be sensitive to pleiotropic effects (for example, inhibition) of other genes, even when conditioned by a pair of alleles in simple dominance. In these cases a mixture of pleiotropy and gene interaction is characterized. |
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| 98715. |
What is the genetic condition in which the heterozygous individual has different phenotype from the homozygous individual? |
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Answer» This condition is called lack of dominance and it can happen in two ways: incomplete dominance or codominance. In incomplete dominance the heterozygous presents an intermediate phenotype between the two types of homozygous, as in sickle cell anemia in which the heterozygous produces some sick red blood cells and some normal red blood cells. Codominance occurs, for example, in the genetic determination of the MN blood group system, in which the heterozygous has a phenotype totally different from the homozygous, not being an intermediate form. |
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| 98716. |
According to Mendel’s law phenotypical characteristics would be determined by pair of factors (alleles) that separate independently in gametes. What are the main types of inheritances that are exceptions to Mendel’s rules? |
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Answer» There are many types of inheritance that do not follow the mendelian pattern. Notable among them are: multiple alleles, gene interactions (complementary genes, epistasis and quantitative, or polygenic, inheritance), linkage with or without crossing over and sex-linked inheritance. Pleiotropy, lacking of dominance and lethal genes do not fit as variations of inheritance since genes can have these behaviors and at the same time obey mendelian laws. Mutations and aneuploidies are abnormalities that also alter the mendelian pattern of inheritance as well as mitochondrial inheritance (passage of mitochondrial DNA from the mother through the cytoplasm of the egg cell to the offspring).Mutations and aneuploidies are abnormalities that also alter the mendelian pattern of inheritance as well as mitochondrial inheritance (passage of mitochondrial DNA from the mother through the cytoplasm of the egg cell to the offspring). |
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| 98717. |
Chewing mouthparts are not found in: a. Thrips b. Crayfish c. Millipedes d. Bees |
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Answer» Chewing mouthparts are not found in Thrips. |
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| 98718. |
--------feeds on glue and starchy materials with its chewing mouthparts. a. Cockroach b. Mosquito c. crickets d. Silverfish |
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Answer» d. Silverfish |
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| 98719. |
Ozone depletion would impact the plant community in several ways. These include : I. Increase in photosynthesis II. Decline in water use efficiency III. Decline in yield of plants (a) I & II (b) Only III (c) II & III (d) All the above |
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Answer» (c) Photosynthesis would decrease as increased UV radiations (due to ozone depletion) would make it difficult for leaves of green plants to exchange gases with the atmosphere. Due to increased temperature consequent upon increased UV radiation, the evaporation rate of surface water will increase and soil moisture would decrease leading to drying of agricultural crops and hence reduced yield. |
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| 98720. |
If variance of `1, 3, 5, 7, …….. 215` is `A^(2)` then standard deviation of `1, 5, 9, 13, …..429` in terms of `A` is :A. `2A`B. `2A^(2)`C. `|A|`D. `2|A|` |
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Answer» Correct Answer - D If variance of………. If we use change to origin and scale `u_(1) = 2x_(1)-1` then we get `1, 5, 9, 13`, ……… from `1, 3, 5, 7` ……….. Hence required `S.D. = 2|A|` |
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| 98721. |
Given three statements : `S_(1) : "log"_(0.6) ("log"_(e )(x^(2)+x)/(x^(2)+4))lt 0`, then `x in (8, oo)` `S_(2) : log_(10)(x^(2)-16)le log_(10)(4x-11), then x in (4,5)` `S_(3) : If log_(3)(x+2)(x+4)+log_(((1)/(3)))(x+2)lt(1)/(2)log_(sqrt(3))7` then `x in (-2, -1)uu(0, 1]`, then which of the following option(s) is/are incorrect ?A. `S_(1)` is falseB. `S_(2)` is trueC. `S_(3)` is trueD. `S_(4)` is false |
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Answer» Correct Answer - B::C Given three ………… Obvious |
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| 98722. |
Simplified value of `sqrt(17+12sqrt(2)) - tan ((pi)/(8))` isA. `0`B. `1`C. `-1`D. `2` |
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Answer» Correct Answer - D Simplified value …………. `sqrt(17+12sqrt(2)) = sqrt(3^(2)+(2sqrt(2))^(2)+2.3(2sqrt(2)))` `= (3+2sqrt(2))` `4sqrt(17+12sqrt(12)) = sqrt(3+2sqrt(2)) = sqrt(2) + 1` and we know that `tan ((pi)/(8)) = sqrt(2) -1` ….(ii) from equation (i) and (ii) `=2` |
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| 98723. |
The value of `[(0.16)^(log_(2.5)(1/3+1/3^2+1/3^3+….+oo))]^(1/2)` isA. `1`B. `2`C. `3`D. `-1` |
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Answer» Correct Answer - A If a quadratic………… `0.16^(log_(2.5((1)/(3)+(1)/(3^(2))+(1)/(3^(2))+...oo)))` `= 0.16^(log_(2.5(((1)/(3))/(1-(1)/(3)))))=0.16^(log_(2.5((1)/(2))))` `=0.16^(log_(2//5^(2)))= 1^(2) "root" = 4^((1)/(2))=2` "Roots are" `= 2, A` Sum of root `= 2+A=3` `implies A = 1` |
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| 98724. |
Supose that `alpha lt x lt beta` the solution to the following inequality `((1)/(3))^(x^(2) + 1) gt ((1)/(9))^(x + 2)` then `beta - alpha` is equal toA. `2`B. `-2`C. `5`D. `4` |
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Answer» Correct Answer - D `x^(2) + 1 lt 2x + 4 rArr x^(2) - 2x - 3 lt 0` `(x - 3) (x +1) lt 0` `-1 lt x lt 3` |
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| 98725. |
The sum of all 4 digit numbers formed by using `2, 3, 4` and 7 repetition of digits os not allowed, belongs to the interval :A. `(106000, 106300)`B. `(106300, 106600)`C. `(106600, 106900)`D. `(106900, oo)` |
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Answer» Correct Answer - C The sum of all ……… `(4!)/(4) (2+3+4+7)(10^(0)+10^(1)+10^(2)+10^(3))` `= (24)/(4)(16)(1111) = 106656` |
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| 98726. |
If a and b are any two distinct numbers belonging tothe set {1,2,......100}, then the number of pairs (a, b) such that product ofa and b is divisible by 3 is5478 2.5278 3. 27393. 2639 5. 2837A. `5478`B. `5278`C. `2739`D. `2639` |
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Answer» Correct Answer - A If a and b are …………. `S = {1, 2, 3,…..100}` elements of `S` are of from `3n : 3,6,9,….00 rarr 33` number `3n+1:1,4,7,…..100 rarr 34` `3n+2:2,5,8….98 rarr 33` If ab is to be divisible by 3 either a or b is divisible by 3 or both a and b are to be divisible by 3. The number of ways a and b can be `=(.^(33)C_(2)+^(33)C_(1)xx^(34)C_(1)+^(33)C_(1))xx2` `= 2739xx2=5478` |
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| 98727. |
How many integer values of `x` satisfy the inequality `(32)/(243) lt ((2)/(3))^(x^(2)) lt (9)/(4) ((8)/(27))^(x)`A. `3`B. `4`C. `0`D. `2` |
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Answer» Correct Answer - A `((2)/(3))^(5) lt ((2)/(3))^(x^(2)) lt ((2)/(3))^(-2) lt ((2)/(3))^(3x)` `rArr x^(2) lt 5` and `x^(2) gt -2 + 3x` `sqrt5 lt x lt sqrt(5) cap x^(2) - 3x + 2 gt 0` `(x - 2) ( x- 1) gt 0` `x lt 1 cup x gt 2` so `(-sqrt(5) lt x lt 1) cup (2 lt x lt sqrt5)` |
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| 98728. |
A galvanometer having a coil of resistance 12 Ω gives full scale deflection for a current of 4 mA. How can it be converted into a voltmeter of range 0 to 24V? |
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Answer» Given G = 12 Ω, Ig = 4 x 10-3A. V = 24V We have, R = \(\frac{V}{I_R}\) - G R = \(\frac{24}{4\times 10^{-3}}\) - 12 R = 5988Ω A resistance of 5988 Ω must be connected in series with galvanometer. |
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| 98729. |
1. Why there is a danger of blowing away of thatched roofs in summer?2. How does air move?3. What is a thunderstorm?4. What is a monsoon wind?5. Why does smoke always rise up into the air? |
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Answer» 1. During summer, due to sun rays and heat, water content in thatched roofs will go off and it will become very dry. Due to this, the weight of the thatched leaves will come down drastically and will blowing away even for small wind. 2. Just like water currents in the ocean, the atmosphere has air currents that flow from one place to another. Air is constantly moving around the earth. This moving air is called wind. Winds are created when there are differences in air pressure from one area to another. 3. Thunderstorm, a violent short-lived weather disturbance that is almost always associated with lightning, thunder, dense clouds, heavy rain or hail, and strong gusty winds. On occasion, severe thunderstorms are accompanied by swirling vortices of air that become concentrated and powerful enough to form tornadoes. 4. A monsoon is a seasonal variation in the direction of the predominant, or strongest, winds of a region. 5. As we know that smoke is the combination of hot air and hot air becomes light weight with respect to cold air. Thus, it always rises up. |
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| 98730. |
If the angle between the straight lines represented by `2x^(2)+5xy+3y^(2)+7y+4=0` is `tan^(-1)m` then `[(1)/(pi m)]` is (where `[.]` denote greatest integer function)A. `0`B. `1`C. `2`D. `3` |
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Answer» Correct Answer - B If the angle ……….. `a = 2, b = 3, h = (5)/(2)` `theta = tan^(-1)((2sqrt(h^(2)-ab))/(|a+b|))= tan^(-1) ((2sqrt((25)/(4)-6))/(2+3))` `= tan^(-1)((1)/(5)) :. M = (1)/(5)` `:. [(1)/(pim)] = [(5)/(3.14)] = [1.59] = 1` |
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| 98731. |
A ray of light coming from the point `(1, 2)` is reflected at a point `A` on the line `x+y+1=0` and passes through the point `(7, -4)` then the orthocentre of triangle whose first side is `5x-y+13=0, 2^(nd)` sides is x-axis and `3^(rd)` side lies along the reflected ray is :A. `(-1, 2)`B. `(-2, -3)`C. `(-2, -1)`D. `(-3, -2)` |
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Answer» Correct Answer - D A ray of ligth …………. By using the property of reflection, we can say that reflected ray passes through the image of point `(-1, 2)` in the line `x+y+1=0`, `i.e. (x-1)/(1) =(y-2)/(1) = -2((1(1)+1(2)+1)/(1^(2)+1^(2)))` `implies x-1=y-2= -4` Point `(-3, -2)` Equation of reflected ray (which passes through the points `(7, -4)` and `(-3, -2)` is : `y-(-4)= (-2-(-4))/(-3-7)(x-7)` `implies y+4= (2)/(-10)(x 7)` `implies -5y-20= x-7` `implies x+5y+13 = 0` ....(i) Now the side of triangle are `x+5y+13=0`, slope `= -(1)/(5)` `5x-y+13=0`, slope `= 5` And x-axis, `y= 0` Clearly 2 sides are `_|_` therefore it is right angle triangle whose orthocentre is vertex at which right angled is made i.e. point of intersection of `x+5y+13=0` and `5x-y+13=0` `i.e. (-3,-2)` |
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| 98732. |
Number of proper divisors of `3^(2)5^(3)` which are odd, is `p` then find the value of `(p-5)`A. `6`B. `5`C. `7`D. `8` |
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Answer» Correct Answer - A Number of proper ………… obvious |
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| 98733. |
Rose flowers of equal size are contained in a bag and of those 5 are red, 3 are white and 2 are yellow. If one is taken out from the bag at random, what is the probability of getting a red rose ? (A) 1 / 3 (B) 1 / 5 (C) 3 / 10 (D) 1 / 2 |
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Answer» The correct option is (D) 1 / 2 Explanation: Red = 5 White = 3 Yellow = 2 Total flowers = 10 required probability = 5 / 10 = 1 / 2 |
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| 98734. |
The area of triangle formed by the vertices `(a, 1/a), (b, 1/b), (c, 1/c)` isA. `|((a+b)(b+c)(c+a))/(2abc)|`B. `|((a-b)(b-c)(c-a))/(2abc)|`C. `|((a-(1)/(a))(b-(1)/(b))(c-(1)/(c )))/(2)|`D. `|((a-1)(b-1)(c-1))/(2abc)|` |
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Answer» Correct Answer - B Area of triangle ……….. `Delta=(1)/(2) ||(a,1//a,1),(b,1//b,1),(c,1//c,1)||=(1)/(2)|(1)/(abc)||(a^(2),1,a),(b^(2),1,b),(c^(2),1,c)||` `= (1)/(2)|((a-b)(b-c)(c-a))/(2abc)|` |
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| 98735. |
If `n` is the number of positive integral solution of `x_(1), x_(2)x_(3)x_(4) = 210`, then which of the following is incorrect ?A. n must be a perfect squareB. n must be a perfect `4^(th)` powerC. n must be a perfact `8^(th)` powerD. n must be divisible by an odd prime number |
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Answer» Correct Answer - D If n is the number ………… `x_(1)x_(2)x_(3)x_(4) = 210 = 2xx3xx5xx7` `implies` no.of positive integral solutions `= 4xx4xx4xx4=4^(4)` |
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| 98736. |
The number of solution of the equastion `|x-1|+|x-2|=[{x}]+{[x]}+1` , where `[.]` and `{.}` denotes greatest integer function and fraction part funstion respectively is :A. `0`B. `1`C. `2`D. Infinite |
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Answer» Correct Answer - D The number of ………….. `AA x epsilon R, {x} epsilon[0,1)` Hence `[{x}] = 0` ….(i) Also `AA x epsilon R, [x] epsilon I` Hence `{[x]} = 0` ….(ii) Hence the given equation becomes `|x-1|+|x-2|=1` `implies x epsilon [1, 2]` `:.` Number of solutions = infinite |
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| 98737. |
In a `Delta ABC`, the `AM` and `GM` of ex-radii are 8 asnd 4 respectively, then inradius is :A. `3`B. `2`C. `1`D. `(2)/(3)` |
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Answer» Correct Answer - D In a `Delta aBC`, the ………… Given `AM = 8` `GM = 4` and Let `HM = H` We know that, `G^(2)=AH` `implies H = (G^(2))/(A) = (4^(2))/(8) = 2 implies (3)/((1)/(r_(1))+(1)/(r_(2))+(1)/(r_(3))) = 2` `implies (3)/((s-a)/(Delta)+(s-b)/(Delta)+(s-c)/(Delta)) = 2` `implies (3Delta)/(3s-(a+b+c)) = 2 implies (3Delta)/(s) = 2` `implies 3r = 2 implies r = (2)/(3)` |
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| 98738. |
If `xyz = 1` and `x, y, z gt 0` then the minimum value of the expression `(x+2y)(y+2z)(z+2x)` isA. `81`B. `9`C. `18`D. `27` |
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Answer» Correct Answer - D If `xyz = 1` and ………. apply `AM ge GM` `(x+y+y)/(3)ge(xy^(2))^((1)/(3))` ….(i) `(y+z+z)/(3)ge (yz^(2))^((1)/(3))` ….(ii) `(z+x+x)/(3)ge (zx^(2))^((1)/(3))` ….(iii) From equation (i), (ii), and (iii) `((x+2y)/(3))((y+2z)/(3))((z+2x)/(3))ge xyz` `implies (x+2y) (y+2z) (z+2x) ge 27` |
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| 98739. |
An object is put at a distance of 5cm from the first focus of a convex lens of focal length 10cm. If a real image is formed, its distance from the lens will beA. 15 cmB. 20 cmC. 25 cmD. 30 cm |
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Answer» Correct Answer - 4 `(1)/(v)-(1)/(u)=(1)/(f)` `(1)/(v)-(1)/(-15)=(1)/(10)` `or(1)/(v)=-(1)/(15)+(1)/(10)` `or(1)/(v)=(-2+3)/(30)` or v = 30 cm |
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| 98740. |
If z= 3 – 5i , then show that z3 – 10z2 + 58z – 136 = 0 |
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Answer» z = 3 – 5i ................. (1) ⇒ z2 = (3 – 5i)2 = (3)2 – 2(3) (5i) + (5i)2 = 9 – 30i + 25i2 = 9 – 30i + 25 (–1) = – 16 – 30i ∴ z2 = – 16 – 30i ................. (2) Now z3 = z2 .z = (–16 – 30i) (3 – 5i) = – 48 – 90i + 80i + 150i2 = – 48 – 10i + 150 (–1) = – 198 – 10i ∴ z3 = – 198 – 10i ................. (3) Now z3 – 10z2 + 58z – 136 = (–198 – 10i) – 10 (–16 – 30i) + 58 (3 – 5i) – 136 = – 198 – 10i + 160 + 300i + 174 – 290i – 136 = (–198 + 160 + 174 – 136) + i(–10 + 300 – 290) = 0 + i(0) = 0 ∴ z3 – 10z2 + 58z – 136 = 0 |
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| 98741. |
If `15 cotA = 8`, then find the value of `cosec A`. |
| Answer» `cosec A = 17//15` | |
| 98742. |
If area of quadrant of a circled is `38.5 cm^(2)` then find its diameter (use `pi= (22)/(7)`) |
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Answer» Area of quadrant `= (1)/(4) xx (22)/(7) xx r^(2) = 38.5 `( use `pi= (22)/(7)` ) `rArr r = 7` cm `therefore ` diameter `= 14` cm |
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| 98743. |
Find the common difference of the A.P whose first term is 12 and fifth term is 0. |
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Answer» `A_5 = a_1 + 4d =0` `1^(2) + 4d =0` `d= -3` |
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| 98744. |
A die itthrown once. What is the probability of getting a prime number? |
| Answer» Correct Answer - `(1)/(2)` | |
| 98745. |
A lot of 25 bulbs contain 5 defective ones. One bulb is drawn at random from the lot. What is the probability that the bulb is good. |
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Answer» Good bulbs `= 25- 5= 20` P (good bulb) `= (20)/(25) = (4)/(5)` |
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| 98746. |
In how many ways 9 mathematics papers can be arranged so that the best and the worst i) may come together ii) may not come together ? |
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Answer» (i) If the best and worst papers are treated as one unit, then we have 9 – 2 + 1 = 7 + 1 = 8 papers Now these can be arranged in (7 + 1)! ways and the best and worst papers between themselves can be permuted in 2! ways. Therefore the number of arrangements in which best and worst papers come together is 8!2!. (ii) Total number of ways of arranging 9 mathematics papers is 9!. The best and worst papers come together in 8!2! ways. Therefore the number of ways they may not come together is 9! – 8!2! = 8! (9 – 2) = 8! × 7 |
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| 98747. |
Out of mean, mode, median and range, which is not a measure of central tendency? |
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Answer» Mean is the average. Median is the middle number when the data is arranged in numerical order. Mode is the data item that appears most frequently. But range depends on only its end points. Hence, Range is not a measure of central tendency. |
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| 98748. |
\[\lim _{x \rightarrow 0} \frac{\cosh x-\cos x}{x \sin x}=\](a) 1(b) 0(c) \( -1 \)(d) None |
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Answer» B option will be the correct answer
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| 98749. |
y = sin x + cos x, find its derivative. |
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Answer» y = sin x + cos x The derivative of y is \(\frac{dy}{d\mathrm x}\) = \(\frac{d}{d\mathrm x}\)(sin x + cos x) \(=\frac{d}{d\mathrm x}\)sin x + \(\frac{d}{d\mathrm x}\)cos x = cos x – sin x |
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| 98750. |
tanh\(\left(\frac{\pi i}{4}\right)\) – coth\(\left(\frac{\pi i}{4}\right)\) is equal to |
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Answer» tanh\(\left(\frac{\pi i}{4}\right)\) – coth\(\left(\frac{\pi i}{4}\right)\) \(=\cfrac{\sinh\left(\frac{\pi i}{4}\right)}{\cosh\left(\frac{\pi i}{4}\right)}\) \(-\cfrac{\cosh\left(\frac{\pi i}{4}\right)}{\sinh\left(\frac{\pi i}{h}\right)}\) \(=\cfrac{\sinh^2\left(\frac{\pi i}{4}\right)-\cosh^2\left(\frac{\pi i}{4}\right)}{\sinh\left(\frac{\pi i}{4}\right)\cosh\left(\frac{\pi i}{4}\right)}\) \(=\cfrac{-1}{\frac{1}{2}\times 2\sinh\left(\frac{\pi i}{h}\right)\cosh\left(\frac{\pi i}{h}\right)}\) (\(\because\) 1 + sinh2x = cosh2x ⇒ sinh2x – cosh2x = –1) \(=\cfrac{-2}{\sinh\left(\frac{\pi i}{2}\right)}\) (\(\because\) 2sinh(x) cosh(x) = sinh(2x)) \(=\frac{-2}{i}\) \(\big(\because\) i sinh(x) = sinix ⇒ sinh\(\left(\frac{\pi}{2}i\right)\) \(=\cfrac{\sin\left(i^2\frac{\pi}{2}\right)}{i}\) \(=\cfrac{-\sin \frac{\pi}{2}}{i^2}i = i\big)\) \(=\frac{-2}{i^2}i\) \(=2i\) \(\big(\because\) i2 = –1) |
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