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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.
| 6001. |
When an object is kept at a distance of 30cm from a concave mirror, the image is formed at a distance of 10 cm. If the object is moved with a speed of `9cm s^(-1)` the speed with which the image moves isA. 10 cm/sB. 1 m/sC. 9 m/sD. 0.9 m/s |
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Answer» Correct Answer - B `V_(I)=-m^(2)V_(0)` `=-((10)/(30))^(2)xx9=1m//s" "]` |
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| 6002. |
Two thin lens have a combined power of 10D in contact. When separated by 20 cm their equivalent power is `6.25D`. Find teir in dividual powers in dioptres-A. `3.5` and `6.5`B. 5 and 5C. `7.5` and `2.5`D. 9 and 1 |
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Answer» Correct Answer - C `P = P_(1) +P_(2) rArr 10 = (1)/(f_(1)) +(1)/(f_(2))` `P = P_(1)+P_(2) - dP_(1)P_(2)` `rArr 6.25 =(1)/(f_(1)) +(1)/(f_(2)) -(20)/(f_(1)f_(2))` On solving (3) option is obtained |
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| 6003. |
The following diagram represents an electrical circui containing two uniform resistance wires A and B connected to a single ideal cell. Both wires have same length, but thickness of wire B is twice that of wire A. The conducting wires connecting A and B to the ideal cell are resistanceless. The dependence of electric potential on position along the length of two wires is given in option. A. B. C. D. |
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Answer» Correct Answer - D The electric field in wire A or B is uniform. Hence potential V in wires shall be linear function of x. At x=0 (i.e., at left end ) V=0. `therefore V_(A)=m_(1)x` and `V_(B)=m_(2)x` Since at right end i.e., at x=L potential of both wires is again same `therefore m_(1)=m_(2)` Hence `V_(A)=V_(B)=mx` where m is a positive constant. |
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| 6004. |
Which of the following composition is controlled by purge stream?(a) Feed(b) Product stream(c) Recycle stream(d) None of the mentioned |
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Answer» Correct option is (c) Recycle stream Best explanation: Composition of recycle stream can be controlled by purge stream. |
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| 6005. |
The change of phase from liquid to vapour is(a) Condensation(b) Evaporation(c) Sublimation(d) None of the mentioned |
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Answer» Right answer is (b) Evaporation To elaborate: The change of phase from liquid to vapour is Evaporation. |
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| 6006. |
What is the amount of HCl present in the feed?(a) 80 Kg(b) 160 Kg(c) 240 Kg(d) 320 Kg |
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Answer» The correct choice is (b) 160 Kg Easiest explanation: Mass fraction = Mass of the interested component / Total mass. |
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| 6007. |
What would be the value of D when feed 5 is inverted?(a) 50(b) 100(c) 150(d) 200 |
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Answer» Right option is (a) 50 Explanation: D = Feed (1 + 2 + 3 + 4 – 5). |
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| 6008. |
What would be the value of D when feed 1 is inverted?(a) 100(b) 110(c) 120(d) 130 |
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Answer» Right answer is (d) 130 The best explanation: D = Feed (-1 + 2 + 3 + 4 + 5). |
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| 6009. |
What is the value of D when feed 3 is inverted?(a) 70(b) 80(c) 90(d) 100 |
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Answer» Right option is (c) 90 Easiest explanation: D = Feed (1 + 2 – 3 + 4 + 5). |
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| 6010. |
Saturated liquid or saturated vapour can be found(a) Along the liquid and vapour equilibrium curve(b) Along the liquid and solid equilibrium curve(c) Along the solid and vapour equilibrium curve(d) None of the mentioned |
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Answer» Right answer is (a) Along the liquid and vapour equilibrium curve Easiest explanation: Saturated liquid or saturated vapour can be found along the liquid and vapour equilibrium curve. |
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| 6011. |
Purge stream comes from?(a) Feed stream(b) Product stream(c) Recycle stream(d) None of the mentioned |
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Answer» Correct choice is (c) Recycle stream Explanation: Purge stream is the removal of unwanted material from the recycle stream. |
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| 6012. |
What would be the value of D when feed 2 is inverted?(a) 100(b) 110(c) 120(d) Not possible |
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Answer» Right option is (d) Not possible The best I can explain: It is not a possible case. |
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| 6013. |
If an addition feed of 2 moles are there near the feed 3, what is the value of B?(a) 10(b) 12(c) 13(d) 14 |
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Answer» The correct choice is (b) 12 Explanation: Now B = Initial B + 2. |
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| 6014. |
A vacuum recorded in the condenser of a steam power plant is 740 mm of Hg. Find the absolute pressure in the condenser in Pa. The barometric reading is 760 mm of Hg. |
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Answer» Vacuum recorded in the condenser = 740 mm of Hg Barometric reading = 760 mm of Hg We know that, Absolute pressure in the condenser = Barometric reading – vacuum in the condenser = 760 – 740 = 20 mm of Hg = 20 × 133.4 N/m2 (\(\because\) 1 mm of Hg 133.4 N / m2) = 2668 N/m2 = 2668 Pa. |
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| 6015. |
What is the value of C when feed 3 is inverted?(a) 0(b) 80(c) 90(d) 100 |
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Answer» Right choice is (c) 90 Easy explanation: C = Feed (1 + 2 – 3). |
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| 6016. |
If Feed 5 is not present here, what would be the value of B?(a) 5(b) 10(c) 15(d) 20 |
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Answer» Correct choice is (b) 10 Easy explanation: B is not depended on Feed 5. |
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| 6017. |
If Feed 1 is not present here, what would be the value of D?(a) 5(b) 10(c) 14(d) 16 |
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Answer» Correct choice is (c) 14 Easiest explanation: D = Feed (2 + 3 + 4 + 5). |
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| 6018. |
In a process, two feed A(400Kg) and B(400 Kg) is given. If the conversion is 50% and A & B converts in equal proportion, what is the weight of the product formed?(a) 200 Kg(b) 400 Kg(c) 600 Kg(d) 800 Kg |
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Answer» Right option is (b) 400 Kg For explanation I would say: Product = (400*.05 + 400*0.5). |
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| 6019. |
On a warm day a pool of water transfers energy to the air as heat and freezes. This is a direct violation of: A. the zeroth law of thermodynamics B. the first law of thermodynamics C. the second law of thermodynamics D. the third law of thermodynamics E. none of the above |
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Answer» C. the second law of thermodynamics |
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| 6020. |
A cyclical process that transfers energy as heat from a high temperature reservoir to a low temperature reservoir with no other change would violate: A. the zeroth law of thermodynamics B. the first law of thermodynamics C. the second law of thermodynamics D. the third law of thermodynamics E. none of the above |
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Answer» E. none of the above |
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| 6021. |
A heat engine that in each cycle does positive work and loses energy as heat, with no heat energy input, would violate: A. the zeroth law of thermodynamics B. the first law of thermodynamics C. the second law of thermodynamics D. the third law of thermodynamics E. Newton’s second law |
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Answer» B. the first law of thermodynamics |
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| 6022. |
A heat engine in each cycle absorbs energy from a reservoir as heat and does an equivalent amount of work, with no other changes. This engine violates: A. the zeroth law of thermodynamics B. the first law of thermodynamics C. the second law of thermodynamicsD. the third law of thermodynamics E. none of the above |
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Answer» C. the second law of thermodynamics |
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| 6023. |
For all irreversible processes involving a system and its environment: A. the entropy of the system does not change B. the entropy of the system increases C. the total entropy of the system and its environment does not change D. the total entropy of the system and its environment increases E. none of the above |
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Answer» D. the total entropy of the system and its environment increases |
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| 6024. |
A heat engine: A. converts heat input to an equivalent amount of work B. converts work to an equivalent amount of heat C. takes heat in, does work, and loses energy as heat D. uses positive work done on the system to transfer heat from a low temperature reservoir to a high temperature reservoir E. uses positive work done on the system to transfer heat from a high temperature reservoir to a low temperature reservoir. |
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Answer» C. takes heat in, does work, and loses energy as heat |
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| 6025. |
A heat engine operates between a high temperature reservoir at TH and a low temperature reservoir at TL. Its efficiency is given by 1 − TL/TH : A. only if the working substance is an ideal gas B. only if the engine is reversible C. only if the engine is quasi-static D. only if the engine operates on a Stirling cycle E. no matter what characteristics the engine has |
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Answer» B. only if the engine is reversible |
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| 6026. |
The change in entropy is zero for: A. reversible adiabatic processes B. reversible isothermal processes C. reversible processes during which no work is done D. reversible isobaric processes E. all adiabatic processes |
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Answer» A. reversible adiabatic processes |
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| 6027. |
In a reversible process the system: A. is always close to equilibrium states B. is close to equilibrium states only at the beginning and end C. might never be close to any equilibrium state D. is close to equilibrium states throughout, except at the beginning and end E. is none of the above |
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Answer» A. is always close to equilibrium states |
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| 6028. |
A Carnot cycle: A. is bounded by two isotherms and two adiabats on a p-V graph B. consists of two isothermal and two constant volume processes C. is any four-sided process on a p-V graph D. only exists for an ideal gas E. has an efficiency equal to the enclosed area on a p-V diagram |
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Answer» A. is bounded by two isotherms and two adiabats on a p-V graph |
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| 6029. |
A slow (quasi-static) process is NOT reversible if: A. the temperature changes B. energy is absorbed or emitted as heat C. work is done on the system D. friction is present E. the pressure changes |
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Answer» D. friction is present |
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| 6030. |
In a thermally insulated kitchen, an ordinary refrigerator is turned on and its door is left open. The temperature of the room: A. remains constant according to the first law of thermodynamics B. increases according to the first law of thermodynamics C. decreases according to the first law of thermodynamics D. remains constant according to the second law of thermodynamics E. increases according to the second law of thermodynamics |
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Answer» B. increases according to the first law of thermodynamics |
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| 6031. |
Consider the following processes: The temperature of two identical gases are increased from the same initial temperature to the same final temperature. Reversible processes are used. For gas A the process is carried out at constant volume while for gas B it is carried out at constant pressure. The change in entropy: A. is the same for A and B B. is greater for A C. is greater for B D. is greater for A only if the initial temperature is low E. is greater for A only if the initial temperature is high |
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Answer» C. is greater for B |
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| 6032. |
For all adiabatic processes: A. the entropy of the system does not change B. the entropy of the system increases C. the entropy of the system decreases D. the entropy of the system does not increaseE. the entropy of the system does not decrease |
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Answer» E. the entropy of the system does not decrease |
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| 6033. |
An inventor suggests that a house might be heated by using a refrigerator to draw energy as heat from the ground and reject energy as heat into the house. He claims that the energy supplied to the house as heat can exceed the work required to run the refrigerator. This: A. is impossible by first law B. is impossible by second law C. would only work if the ground and the house were at the same temperature D. is impossible since heat energy flows from the (hot) house to the (cold) ground E. is possible |
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Answer» E. is possible |
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| 6034. |
Consider all possible isothermal contractions of an ideal gas. The change in entropy of the gas: A. is zero for all of them B. does not decrease for any of them C. does not increase for any of them D. increases for all of them E. decreases for all of them |
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Answer» E. decreases for all of them |
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| 6035. |
Fundamental frequency of a stretched sonometer wire is `f_(0)`. When its tension is increased by `96%` and length drecreased by `35%`, its fundamental frequency becomes `eta_(1)f_(0)`. When its tension is decreased by `36%` and its length is increased by `30%`, its fundamental frequency becomes `eta_(2)f_(0)`. Then `(eta_(1))/(eta_(2))` is.A. `4.5`B. `3.5`C. `2`D. `6` |
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Answer» Correct Answer - B `f_(0) = (1)/(2L) sqrt((T)/(mu))` `f = (1)/(2(L + DeltaL))sqrt((T + DeltaT)/(mu)) = etaf_(0)` `:. eta = (L)/(L + DeltaL)sqrt((T + DeltaT)/(T)) = sqrt(1 + (DeltaT)/(T))/(1 + (DeltaT)/(T))` `:. eta_(1) = (sqrt(1 - 0.96))/(1 - 0.35) = (1.4)/(0.65)` and `eta_(2) = (sqrt(1-0.36))/(1+0.3) = (0.8)/(1.3)` `:. (eta_(1))/(eta_(2)) = (1.4)/(0.65) xx (1.3)/(0.8) = 3.5` |
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| 6036. |
A certain mass of a solid exists at its melting temperature of `20^(@)C`. When a heat Q is added `(4)/(5)` of the material melts. When an additional Q amount of heat is added the material transforms to its liquid state at `50^(@)C`. Find the ratio of specific latent heat of fusion (in `J//g`) to the specific heat capacity of the liquid (in `J g^(-1) .^(@)C^(-1)`) for the material.A. `20`B. `40`C. `50`D. `60` |
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Answer» Correct Answer - C `Q = (4)/(5)m.L"………"` `2Q = mL +ms(50 - 20)` `rArr (8)/(5)mL = mL + 30 ms` `rArr (3L)/(5) = 30 s rArr (L)/(s) = 50` |
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| 6037. |
`a_(1), a_(2), a_(3),`…………. are distinct terms of an A.P. We cal `(p,q,r)` an increasing triad if `a_(p),a_(q),a_(r)` are in G.P. where `p,q,repsilonN` such that `pltqltr`. If `(5,9,16)` is an increasing triad, which of the following option is/are correctA. if `a_(1)` is a multiple of 4 then every term of the A.P. is an integerB. `(85,149,261)` is an increasing triadC. if the common difference of the A.P. is `1/4`,then its first term is `1/3`D. ratio of the `(4k+1)th` term and `4k^(th)` term can be 4 |
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Answer» Correct Answer - A::B::C Let `R` be the common ratio of the G.P. and `D` be the common difference of A.P. `a_(5)=a_(5), a_(9)=Ra_(5), a_(16=a_(5)R^(2)` `a_(9)-a_(5)=4Dimplies(R-1)a_(5)=4D`…..(1) `a_(16)-a_(9)=7DimpliesR(R-1)a_(5)=7D`………(2) from equation `(1)//(2)` we get `1/r=4/7=R=7/4` From equation `(2)-(1)` we get `(R-1)^(2)a_(5)=3D=(8a_(5))/16=3D` `3/16(a_(1)=4D)=Dimplies3/16 a_(1)=(1- 3/4)Dimpliesa_(1)=(4D)/3` |
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| 6038. |
`x_(1), x_(2), x_(3)` are three real numbers satisfying the system of equations `x_(1)+3x_(2)+9x_(3)=27, x_(1)+5x_(2)+25x_(3)=125` and `x_(1)+7x_(2)+49x_(3)=343`, then which of the following options are correctA. number of divisor of `x_(1)+x_(3)` is 6B. `(x_(1)+x_(2))/2` is a prime numberC. `x_(3)-x_(2)` is a prme numberD. `x_(1)+x_(2)+x_(3)` is square of an integer |
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Answer» Correct Answer - A::B::D Roots of the equation `y^(3)-x_(3)y^(2)-x_(2)y-x_(1)=0` are 3, 5, and 7 `x_(3)=15, x_(3)=-71, x_(1)=105` |
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| 6039. |
Which of the following options are correct?A. `[(.^(n)C_(0)+.^(n)C_(3)+.^(n)C_(6)+…)-1/2 (.^(n)C_(1)+.^(n)C_(2)+.^(n)C_(4)+.^(n)C_(5)+.)]^(2)+3/4(.^(n)C_(1)-.^(n)C_(2)+.^(n)C_(4)-.^(n)C_(5)+…….)^(2)=1`B. If `a` and `b` are two positive numbers such that `a^(5)b^(2)=4` then the maximum value of `log_(2^(1//5))(a^(2)).log_(2^(1//2))(b^(2))` is equal to 4C. Constant term in `((((((x-2)^(2)-2)^(2)-2)^(2)-2)^(2)-2)^(2)………2)^(2)` is equal to 2D. The coeffocient of `x^(24)` in `((.^(25)C_(1))/(.^(25)C_(0))-x)(x-2^(2)(.^(25)C_(2))/(.^(25)C_(1)))(x-3^(2)(.^(5)C_(3))/(.^(2)C_(2)))(x-4^(2) (.^(25)C_(4))/(.^(25)C_(3)))………..(x-25^(2)(.^(25)C_(25))/(.^(25)C_(24)))` is equal to 2925 |
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Answer» Correct Answer - A::B::D (A) `(1+x)^(n)=.^(n)C_(0)+.^(n)C_(1)x+.^(n)C_(2)x^(2)+.^(n)C_(3)x^(3)+……..+.^(n)C_(n)x^(n)` Put `x=omega` where `omega=-1/2+(isqrt(3))/2` `(.^(n)C_(0)+.^(n)C_(3)+.^(n)C_(6)+……)+….+(.^(n)C_(1)+.^(n)C_(4)+.^(n)C_(7)+……..)omega +(.^(n)C_(2)+.^(n)C_(5)+.^(n)C_(8)+.....)omega^(2)=(1+omega)^(n)` `(.^(n)C_(0)+.^(n)C_(3)+.^(n)C_(6)+.........)+(.^(n)C_(1)+.^(n)C_(4)+.^(n)C_(7)+......)(-1/2+(isqrt(3))/2)+(.^(n)C_(2)+.^(n)C_(5)+.^(n)C_(8)+.)(-1/2-(isqrt(3))/2)=(-omega^(2))^(n)` `=(.^(n)C_(0)+.^(n)C_(3)+.^(n)C_(6)+...)-1/2(.^(n)C_(1)+.^(n)C_(2)+.^(n)C_(4)+.^(n)C_(5)+...)+ (isqrt(3))/2(.^(n)C_(1)-.^(n)C_(2)+.^(n)C_(4)-.^(n)C_(5)+...........)=(-omega^(2))^(n)` Take modulus of both sides (B) `a^(5)b^(2)=4implies5log2(a)+2log_(2)(b)=2` `y=10log_(2)a.4 log_(2)b=40log_(2)a.log_(2)b` A.M. `ge` G.Mgt `implies(5log_(2)(a)+2log_(2)(b))/2 ge sqrt(10 log_(2)a.log_(2)b)` `log_(2)a.log_(2)b le 1/10 impliesyle4, =4` when `5 log_(2)a=2 log_(2)bimpliesa^(5)=b^(2)=2` (C) For constant term put `x=0` and get constant term `=4` (D) Coefficient of `x^(24)=1.25+2.24+3.23+.+25.1=2925` |
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| 6040. |
If `D_1, D_2, D_3, ..... D_1000` are 1000 doors and `P_1, P_2, P_3, ..... P_1000` are 1000 persons. Initially all doorsare closed. Changing the status of doors means closing the door if it is open or opening it if it is closed. `P_1` changes the status of all doors. Then `P_2` changes the status of `D_2, D_4, D_6, .... D_1000` (doors having numbers which are multiples of 2). Then `P_3`; changes the status of `D_3, D_6, D_9, .....D_999` (doors having number which are multiples of 3) and this process is continued till `P_1000` changes the status of `D_1000`, then the doors which are finally open is/areA. `D_(961)`B. `D_(269)`C. `D_(413)`D. `D_(729)` |
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Answer» Correct Answer - A::D `D_(m)` will be open finally if `m` is the perfect square |
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| 6041. |
If in a HP, a5 = 7 and a7 = 5, then what will be the value of a12?1. \(\frac{{28}}{{13}}\)2. \(\frac{{35}}{{12}}\)3. \(\frac{{37}}{{12}}\)4. \(\frac{{29}}{{12}}\) |
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Answer» Correct Answer - Option 2 : \(\frac{{35}}{{12}}\) CONCEPT: If in a Harmonic progression am = n and an = m, then am+n will be given as \({a_{m + n}} = \frac{{mn}}{{m + n}}\) . CALCULATIONS: Given: a5 = 7 and a7 = 5 So \({a_{12}} = {a_{5 + 7}} = \frac{{5 \times 7}}{{5 + 7}} = \frac{{35}}{{12}}\) Therefore, option (2) is the correct answer. |
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| 6042. |
X invested Rs. 48000 for a year. Y joined after 2 months and invested Rs. 36000. Z joined after further 4 months and invested 52000. Find the ratio of their profits after a year?1. 24 : 15 : 132. 24 : 15 : 113. 22 : 15 : 184. 24 : 18 : 13 |
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Answer» Correct Answer - Option 1 : 24 : 15 : 13 Given: X invested Rs. 48000 for a year. Y joined after 2 months and invested Rs. 36000. Z joined after further 4 months and invested 52000. Formula: Ratio of Profit = ratios of product of Amount invested and time Calculation: X : Y : Z 48000 × 12 : 36000× 10 : 52000 × 6 24 : 15 : 13 |
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| 6043. |
A shopkeeper sold a mobile charger at Rs. 864, after giving a discount of 4%. Had he not given the discount, he would have earned a profit of \(11\frac{1}{9}\)% on the cost price. What was the cost price of charger (In Rs.)?1. 8002. 8503. 9104. 810 |
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Answer» Correct Answer - Option 4 : 810 Given: Selling price of charger for shopkeeper = 864 % Discount offered = 4% Profit percent earned = \(11\frac{1}{9}\)% Formula used: Selling price = marked price – discount % Discount = (Discount/M.P) × 100 Cost price = 100/(100 + a%) × selling price Where a% → percent profit Concept used: Profit/loss is always calculated on C.P Discount is always calculated on M.P Selling price of an article is same as marked price of the article, when no discount is being offered Calculation: Let M.P of charger be X % Discount = (Discount/M.P) × 100 ⇒ Discount = 4% of X ⇒ 0.04X S.P = X – 0.04X ⇒ 0.96X = 864 ⇒ X = 864/0.96 = 900 Now, M.P of charger = 900 No discount is offered, M.P of charger becomes equal to S.P of charger S.P = 900 and% Profit = 11 (1/9)% C.P = 100/ (100 + 100/9) × 900 ⇒ (900/1000) × 900 ⇒ 810 ∴ C.P of charger is Rs. 810. |
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| 6044. |
Suresh starts a business with Rs. 36000. After a certain period of time he is joined by Deepak, who invests Rs. 27000. At the end of the year they divide the profit in the ratio of 8 : 3. For what period did Deepak join Suresh?1. 7 months 2. 8 months3. 11 months4. 10 months5. 6 months |
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Answer» Correct Answer - Option 5 : 6 months Given: Suresh starts a business with Rs.36000 , After a certain period of time Deepak invests Rs,27000 Formula used: \(Ratio\;of\;profit = \frac{{Capital\;of\;A \times Times}}{{Capital\;of\;B \times Times\;}}\) Calculation: Let Deepak's investment be for x months. ⇒ \(\frac{8}{3} = \frac{{36000 \times 12}}{{27000 \times X}}\) ⇒ \(X = \frac{{36000 \times 12 \times 3}}{{8 \times 27000}}\) = 6 ∴ Deepak join after 6 months |
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| 6045. |
A shopkeeper sold books at Rs. 340 each after giving discount of 15% on marked price. Had he not given the discount, he would have earned a profit of 25% on cost price. What was the cost price of each book?1. 4002. 3203. 3404. 230 |
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Answer» Correct Answer - Option 2 : 320 Given: Selling price of each book = Rs. 340 Shopkeeper gives a discount of 15% on marked price. Had he not given the discount, he would have earned a profit of 25% on cost price. Formula Used: Profit% = Profit/CP × 100 Loss% = Loss/CP × 100 Calculation: From question, Marked price × 85/100 = 340 MP = 400 Now, Cost price × 125/100 = 400 [As, If discount is not given, profit is 25%. It means book is being sold on Marked Price.] CP = Rs.320 ∴ Cost price of each book is Rs. 320 |
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| 6046. |
Vishnu and Deepak is two business partners. Vishnu invests Rs. 50,000 for 8 months and Deepak invests Rs. 20,000 for 10 months. Find the share of Vishnu from the total profit of Rs. 60,000.1. Rs. 40,0002. Rs. 42,0003. Rs. 42,5004. Rs. 44,5005. None of these |
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Answer» Correct Answer - Option 1 : Rs. 40,000 Given: Vishnu Invests (I1) = Rs.50000 Time (t1) = 8 months Deepak Invests (I2) = Rs.20000 Time (t2) = 10 months Total profit = Rs.60000 Formula: Profit = Investments × time Profit (P1)/Profit (P2) = (I1/I2) × (t1/t2) Calculation: We know that – Profit (P1)/Profit (P2) = (I1/I2) × (t1/t2) …….. (1) Put all the given values in equation (1) then we get (P1/P2) = (50000/20000) × (8/10) ⇒ (5/2) × (8/10) ⇒ 40/20 ⇒ 2/1 Now, Profit shares of Vishnu = (2/3) × 60000 ⇒ 2 × 20000 ⇒ 40000 ∴ The Profit shares of Vishnu will be Rs. 40,000 |
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| 6047. |
Harshad Mehta and his wife invested in the share market. Harshad invests 2/3 of the total sum for 9 months and his wife gets 3/5 of the profits. For how long, her wife invested in the market?1. 18 months2. 16 months3. 27 months4. 12 months5. 29 months |
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Answer» Correct Answer - Option 3 : 27 months Given: Harshad’s investment is 2/3 of the total sum Harshad invest for 9 months Harshad’s wife gets 3/5 of the profit Concept used: Profit ∝ Investment Profit ∝ Time So, the ratio of profit is directly proportional to the product of investment and time. Calculation: Assume that the total sum is P, total profit is M and T is the time for which investment of Harshad's wife was in the market Sum invested by Harshad = 2/3P Sum invested by his wife = 1/3P Profit of his wife = 3/5M ⇒ Profit of Harshad = (1 − 3/5)M = 2/5M For Harshad, 2/3P × 9 = 2/5M ⇒ M/P = 15 For Harshad’s wife, 1/3P × T = 3/5M ⇒ T = 3/5 × 3 × M/P ⇒ T = 27 ∴ The time for which investment of Harshad's wife was in the market is 27 months. |
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| 6048. |
C.P. of two books is same. One sold at 15% profit and the other for ₹ 4800 more than the first. If the net profit is 20% then find the C.P. of each book?A. ₹48000B. ₹480C. ₹4800D. ₹38000 |
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Answer» Correct Answer - A Let `CP_(1)= 100 CP_(2) = 100` overall `CP = 200` `SP_(1) = 115,` overall SP = 240 `SP_(2)` = overall, SP - SPI = 240 -115 = 125` Difference in SP = 125-115= 10 Therefore CP `= 48000 xx 100//10 = ₹ 48000`. Alternate Solution: As price of both books is same one is sold at 15% so other must be sold at 25% to make overall profit of 20% `therefore ` 10%$ = 4800` `rArr 100% = 48000` = Price of 1 book. |
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| 6049. |
P invests ₹ 6,000 for X months while Q invests ₹ 8,000 for 9 months in a scheme. The profit share of Q is t ₹ 24,000 out of total profit ₹ 42,000. Then find the value of X?A. 6 monthsB. 9 monthsC. 8 monthsD. 7 months |
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Answer» Correct Answer - B Profit share ratio of P Q `6000 xx x 8000 xx 9` ATQ `(x)/(12) = (18000)/(24000)` `rArr x = 9` months. |
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| 6050. |
The printed price on a Mathematics book is ₹550. If it is sold at two successive discounts of 20% and 30%, then its selling price will be:1. ₹3122. ₹3053. ₹3104. ₹308 |
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Answer» Correct Answer - Option 4 : ₹308 Given: Two successive discounts = 20% and 30% Formula: Single discount = (a + b) - ab/100 Selling price = Marked price × (100 - discount %)/100 Calculation: Single discount = (20 + 30) - (20 × 30)/100 = 44% Selling price after discount = 550 × (100 - 44)/100 = Rs.308 ∴ Selling price after two successive discounts of 20% and 30% is Rs.308. |
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