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.
| 30201. |
Find firm’s equilibrium with the help of total revenue and total cost curves. |
|
Answer» A firm is in equilibrium when it produces that amount of output at which the difference between total revenue(TR) and total cost (TC), i.e., total profits, is maximum. |
|
| 30202. |
How do you get Total Revenue? |
|
Answer» It is the total income of a company and is calculated by multiplying the quantity of goods sold by the price of the goods. |
|
| 30203. |
Objective of all the firms is : (a) Selling your item at cost price (b) Earning maximum profit (c) Social service (d) None of these |
|
Answer» (b) Earning maximum profit |
|
| 30204. |
Normal price is fixed in the ………… period: (a) Market (b) Short (c) Long (d) Very long |
|
Answer» Normal price is fixed in the Long period. |
|
| 30205. |
When the production is zero, then the total revenue is equal to : (a) Zero (b) Maximum (c) Equal to Fixed Cost (d) None of these |
|
Answer» Correct answer is (a) Zero |
|
| 30206. |
Slope of supply curve is ………(a) Positive(b) Negative(c) Both(d) Parallel |
|
Answer» Slope of supply curve is Positive. |
|
| 30207. |
Traders enjoy ……… profit at equilibrium price : (a) Normal (b) Abnormal (c) Lesser (d) More |
|
Answer» Traders enjoy Normal profit at equilibrium price. |
|
| 30208. |
What is gross profit? |
|
Answer» The difference between total revenue and total variable cost is called gross profit. It is calculated like this : Gross Profit = TR – TVC |
|
| 30209. |
Secular price is fixed in the ……………Period: (a) Very short (b) Short (c) Long (d) Very long |
|
Answer» Secular price is fixed in the Very long Period. |
|
| 30210. |
At equilibrium price : (a) Demand is more (b) Supply is more (c) Demand and supply are equal (d) None of these |
|
Answer» (c) Demand and supply are equal |
|
| 30211. |
What are the two conditions for equilibrium of a firm by MC = MR method? |
|
Answer» Following are the two conditions for equilibrium of a firm by MC = MR method:
|
|
| 30212. |
What are the conditions for the producer’s equilibrium? |
|
Answer» To maximize his profit, a producer will produce upto that quantity at which the following two conditions are fulfilled:
|
|
| 30213. |
If demand increases at a faster rate than the supply, equilibrium price will: (a) Increase (b) Decrease (c) Neither increase nor decrease (d) None of the above |
|
Answer» (a) Increase |
|
| 30214. |
What do you mean by total revenue and total cost? |
Answer»
|
|
| 30215. |
First condition of equilibrium is :(a) MR = MC(b) MR > MC(c) MR < MC(d) MR ≠ MC |
|
Answer» Correct answer is (a) MR = MC |
|
| 30216. |
………… a dominant role in determining market price : (a) Demand plays (b) Supply plays (c) Demand and supply play (d) None of the above |
|
Answer» Demand plays a dominant role in determining market price . |
|
| 30217. |
How does a firm attain maximum profit position ? |
|
Answer» The maximum benefit to a firm is at the production level where the difference between total revenue and the total cost is highest. It is the also the state of equilibrium of the firm. At this stage, the total cost is the lowest in comparison to total revenue and production level. |
|
| 30218. |
…………… a dominant role in determining equilibrium price in the long period: (a) Demand plays (b) Supply plays (c) Demand and supply play (d) None of the above |
|
Answer» Demand and supply play a dominant role in determining equilibrium price in the long period. |
|
| 30219. |
The concept of equilibrium price is …………(a) Theoretical (b) Practical (c) Both (d) Neither theoretical nor practical |
|
Answer» The concept of equilibrium price is Theoretical. |
|
| 30220. |
If demand and supply change at the same rates, the equilibrium price will: (a) Remain constant (b) Increase (c) Decrease (d) None of these |
|
Answer» (a) Remain constant |
|
| 30221. |
……… a dominant role in determining equilibrium price in the short period: (a) Demand plays (b) Supply plays (c) Demand and supply play (d) None of the above |
|
Answer» Demand plays a dominant role in determining equilibrium price in the short period. |
|
| 30222. |
What do you understand by equilibrium of an industry? |
|
Answer» An industry is in equilibrium at a price and output at which its market demand equals its market supply. |
|
| 30223. |
Firm determines the quantity of output where :(a) MR = MC and MC cuts MR from below(b) MR ≠ MC(c) MR > MC(d) MR < MC |
|
Answer» (a) MR = MC and MC cuts MR from below |
|
| 30224. |
Find firm’s equilibrium with the help of marginal revenue and marginal cost curves. |
|
Answer» Profit will be highest at that level of output where marginal cost coincides with the marginal reveue, i.e., MC = MR. |
|
| 30225. |
What is marginal revenue? |
|
Answer» Marginal revenue means the addition made to the total revenue by selling an additional emit of the commodity. |
|
| 30226. |
Point out the main difference between the equilibrium of a firm and an industry. |
|
Answer» The main difference between the equilibrium of a firm and an industry is the following:
|
|
| 30227. |
What is the point called, where TR = TC ? |
|
Answer» Break Even Point. At this point, the firm neither earns profit nor suffers loss. |
|
| 30228. |
Mention the drawbacks of total revenue – total cost method. |
|
Answer» Following are the drawbacks – |
|
| 30229. |
What is Marginal Revenue and Marginal Cost approach? |
|
Answer» According to marginal revenue and marginal cost approach, a producer will be in equilibrium when two conditions are fulfilled i.e., (i) MC=MR and (ii) MC must cut MR from below. This is a better method to analyse the equilibrium point of a firm, since both optimum production and total profit can be easily determined by it. |
|
| 30230. |
When TR is more than TC of a firm, then firm is getting :(a) Abnormal Profit(b) Loss(c) No Profit, No loss(d) None of these |
|
Answer» (a) Abnormal Profit |
|
| 30231. |
What do you understand by Break-even point in equilibrium of firm? |
|
Answer» Break-even for a firm occurs when it is able to cover all its costs of production. Under this situation, the firm earns only normal profit, i.e. neither super normal profits nor super normal losses. This situation prevails at the point where Total Cost is equal to Total Revenue, i.e. TR = TC or AR = AC. |
|
| 30232. |
What is meant by equilibrium price and production ? |
|
Answer» Equilibirum price and production implies that quantity at which the firm earns highest profit. Above and below this point, the production and price both are not optimum to maximise the firm’s profit. |
|
| 30233. |
Write True or False and justify your answer.An edge of a cube measures r cm. If the largest possible right circular cone is cut out of this cube, then the volume of the cone (in cm3 ) is 1/6πr3. |
|
Answer» False. Height of the cone = r cm. Diameter of the base = r cm. Therefore, volume of the cone = 1/3 π(r/2)2.r = 1/12 πr3 |
|
| 30234. |
Write whether True or False and justify your answer.The volume of the largest right circular cone that can be fitted in a cube whose edge is 2r equals to the volume of a hemisphere of radius r. |
|
Answer» According to the question, Edge of cube, l = 2r Then, diameter of the cone = 2r ⇒ radius of the cone = 2r/2 = r Height of the cone, h = height of the cube = 2r Volume of the cone is given by, Volume of cone = 1/3 πr2h = 1/3 πr2(2r) = 2/3 πr3 = Volume of hemisphere of radius r Hence, the given statement “the volume of the largest right circular cone that can be fitted in a cube whose edge is 2r equals to the volume of a hemisphere of radius r” is true. |
|
| 30235. |
Find the volume largest right circular cone that can be cut out of a cube whose edge is 9 cm. |
|
Answer» Given, The side of the cube = 9 cm The largest cone that can be cut from cube will have the base diameter = side of the cube 2r = 9 r = 9/2 cm = 4.5 cm And, Height of cone = side of cube So, height of cone (h) = 9 cm Thus, Volume of the largest cone to fit in = 1/3 π × r2 × h = 1/3 π × 4.52 × 9 = 190.93 cm3 Therefore, the volume of the largest cone to fit in the cube has a volume of 190.93 cm3. |
|
| 30236. |
A 4 cm cube is cut into 1 cm cubes. Calculate the total surface area of all the small cubes. |
|
Answer» Given, Original edge of cube = 4 cm Original volume of cube = 43 = 64 cm3 After cutting, Edge of cube = 1 cm Hence volume of new cube = 13 = 1 cm3 Number of small cubes thus formed = \(\cfrac{64}1\) = 64 Hence , total surface area of 64 cubes = 64 × 6a2 = 64 × 6 = 384 m2 |
|
| 30237. |
A metal cube of edge 12 cm is melted and formed into three smaller cubes. If the edges of the two smaller cubes are 6 cm and 8 cm, find the edge of the third smaller cube. |
|
Answer» Given details are, Edge of metal cube (volume) = 12cm Edge of smaller two cubes = 6cm, 8cm Let edge of third cube be ‘a’ cm So, Volume of metal cube = sum of volume of three small cubes 123 = 63 + 83 + a3 1728 = 216 + 512 + a3 a3 = 1728 – 216 – 512 = 1000 a = 3√1000 = 10cm ∴ Edge of third smaller cube is 10 cm. |
|
| 30238. |
If two cubes each of side 6 cm are joined face to face, then find the volume of the resulting cuboid. |
|
Answer» Side of two equal cubes = 6 cm When we join, two cubes face to face formed a cuboid. Dimensions of a cuboid are: Length = 6 cm + 6 cm = 12 cm Breadth = 6 cm Height = 6 cm Therefore, volume of cuboid = lbh = 12 x 6 x 6 = 432 cm3 |
|
| 30239. |
Two cubes of each edge ‘a’ is joined end to end then the surface area of the resultant cuboid is ……………. A) 9a2 B) 10a2 C) 18a2 D) 16a2 |
|
Answer» Correct option is (B) \(10a^2\) Two cubes of each edge a is joined end to end then the length of formed cuboid = 2a, breadth of formed cuboid = a and height of formed cuboid = a. \(\therefore\) Surface area of resultant cuboid \(=2\,(2a\times a+a\times a+a\times 2a)\) \(=2\,(2a^2+a^2+2a^2)\) \(=2\times5a^2=10a^2\) Correct option is B) 10a2 |
|
| 30240. |
The number of cubes of side 3 cm that can be cut from a cuboid of dimensions 10 cm × 9 cm × 6 cm, isA. 9B. 10C. 18D. 20 |
|
Answer» Correct option is D. 20 Given, Side of cube = 3 cm Dimension of cuboid = 10 cm × 9cm × 6 cm Volume of cuboid = 10 × 9 × 6 = 540 cm3 Volume of cube = a3 = 33 = 27 cm3 Number of cubes can be formed = \(\cfrac{volume\,of\,cuboid}{vlume\,of\,cube}\) = \(\cfrac{540}{27}\) = 20 |
|
| 30241. |
One cube is cut in 8 equal cubes of each edge 4 cm then the volume of bigger cube is …………… A) 125 cm3 B) 512 cm3 C) 160 cm3 D) None |
|
Answer» Correct option is B) 512 cm3 Remove small word from question volume of cube whose edge length is 4 cm is v = a3 = 43 = 64 cm3. ∴ volume of 8 equal cubes whose edge length are 4 cm = 8 v = 8 × 64 = 512 cm3. Since, given cube is cut in 8 equal cubes of each edge is 4 cm. Therefore, volume of given cube = 8 × volume of one cube whose edge length is 4 cm. = 8 × 43 = 8 × 64 = 512 cm3. |
|
| 30242. |
Two cubes of 6 cm edges are joined end to end then the volume of the resulting figure is ……………….. cm3 A) 432 B) 430 C) 332 D) 440 |
|
Answer» Correct option is (A) 432 Length of resultant cuboid = 6+6 = 12 cm, breadth of resultant cuboid = 6 cm and height of resultant cuboid = 6 cm. \(\therefore\) Volume of resultant cuboid \(=12\times6\times6\,cm^3\) \(=432\,cm^3\) Correct option is A) 432 |
|
| 30243. |
A student was asked to prove a statement P (n) by induction. He proved that P (k + 1) is true whenever P (k) is true for all k > 5 ∈ N and also that P (5) is true. On the basis of this he could conclude that P (n) is true(A) for all n ∈ N (B) for all n > 5(C) for all n ≥ 5 (D) for all n < 5 |
|
Answer» Answer is (C), since P(5) is true and P(k + 1) is true, whenever P (k) is true. |
|
| 30244. |
In a rhombus ABCD, if ∠ACB = 40°, then ∠ADB = A. 70° B. 45° C. 50° D. 60° |
|
Answer» Option : (C) The diagonals in a rhombus are perpendicular, So, ∠BPC = 90° From triangle BPC, The sum of angles is 180° So, ∠CBP = 180° – 40° – 90° = 50° Since, triangle ABC is isosceles We have, AB = BC So, ∠ACB = ∠CAB = 40° Again from triangle APB, ∠PBA = 180° – 40° – 90° = 50° Again, triangle ADB is isosceles, So, ∠ADB = ∠DBA = 50° ∠ADB = 50° |
|
| 30245. |
ABCD is a parallelogram in which diagonal AC bisects ∠BAD. If ∠BAC = 35°, then ∠ABC = A. 70° B. 110° C. 90° D. 120° |
|
Answer» Option : (B) Given : ABCD is a parallelogram AC is a bisector of angle BAD ∠BAC = 35° ∠A = 2∠BAC ∠A = 2(35°) ∠A = 70° ∠A + ∠B = 180° (Adjacent angles of parallelogram are supplementary) 70 + ∠B = 180° ∠B = 110° |
|
| 30246. |
Formula for area of a parallelogram is ……………… sq. units.A) \(\frac{1}{2}\) d (h1 + h2) B) \(\frac{1}{2}\) d1d2C) \(\frac{1}{2}\) bh D) bh |
|
Answer» Correct option is D) bh |
|
| 30247. |
PQRS is a rectangle inscribed in a quadrant of a circle of radius 13 cm. A is any point on PQ. If PS = 5 cm, then find ar(△RAS). |
|
Answer» PQRS is a rectangle with PS = 5 cm and PR = 13 cm (Given) In △PSR: Using Pythagoras theorem, SR2 = PR2 – PS2 = (13)2 – (5)2 = 169 – 25 = 114 or SR = 12 Now, Area of △RAS = 1/2 x SR x PS = 1/2 x 12 x 5 = 30 Therefore, Area of △RAS is 30 cm2. |
|
| 30248. |
Formula for area of a triangle is ………………. sq. units.A) \(\frac{1}{2}\) d (h1 + h2) B) \(\frac{1}{2}\) d1 d2C) \(\frac{1}{2}\) bh D) bh |
|
Answer» Correct option is C) \(\frac{1}{2}\) bh |
|
| 30249. |
The diagonals AC and BD of a rectangle ABCD intersect each other at P. If ∠ABD = 50°, then ∠DPC = A. 70° B. 90° C. 80° D. 100° |
|
Answer» Option : (C) Given that, ∠ABD = ∠ABP = 50° ∠PBC +∠ABP = 90° (Each angle of a rectangle is a right angle) ∠PBC = 40° Now, PB = PC (Diagonals of a rectangle are equal and bisect each other) Therefore, ∠BCP = 40° (Equal sides has equal angle) In triangle BPC, ∠BPC + ∠PBC + ∠BCP = 180° (Angle sum property of a triangle) ∠BPC = 100° ∠BPC + ∠DPC = 180° (Angles in a straight line) ∠DPC = 180° – 100° = 80° |
|
| 30250. |
PQRS is a rectangle inscribed in a quadrant of a circle of radius 13 cm. A is any point on PQ. If PS = 5 cm, then find ar (ΔRAS). |
|
Answer» Given that, PQRS is a rectangle PS = 5 cm PR = 13 cm In triangle PSR, by using Pythagoras theorem SR2 = PR2 – PS2 SR2 = (13)2 – (5)2 SR2 = 169 – 25 SR2 = 114 SR = 12 cm We have to find the area ΔRAS, Area (ΔRAS) = \(\frac{1}{2}\) x Base x Height = \(\frac{1}{2}\) x SR x PS = \(\frac{1}{2}\) x 12 x 5 = 30 cm2 Hence, Area (ΔRAS) is 30 cm2 |
|