Attempt the ISRO Electrical Engineering (EE) 2014 previous year paper in real CBT exam mode or practice mode with verified answers and explanations.
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80 questions.
A Meter has a full scale deflection of 90° at a current of 1A. The response of the meter is square law. Assuming spring control, the current for a deflection at 45° will be
Answer: C. 0.707
Explanation: Square-law meter: θ ∝ I², so I = √(45/90) = 0.707 A.
A Single-phase Energy meter is operating on 230V, 50 Hz supply with a load of 20A for two hours at UPF. The meter makes 1380 revolutions in that period. The meter constant is
Answer: B. 150 rev/kWh
Explanation: Energy = 230×20×2 = 9.2 kWh; meter constant = 1380/9.2 = 150 rev/kWh.
In case of power measurements by two wattmeter method in a balanced 3-phase system with pure inductive load
Answer: A. both wattmeters will indicates the same value but with opposite signs
Explanation: For a purely inductive load (φ = 90°) W1 = VI cos(30°−90°) = +0.5VI and W2 = VI cos(30°+90°) = −0.5VI: equal magnitudes, opposite signs.
It is required to measure temperature in the range of 1300°C to 1500°C. The most suitable thermocouple to be used as transducer would be
Answer: D. Platinum-rhodium
Explanation: Platinum-rhodium thermocouples (type R/S/B) are used for 1300–1500 °C.
Which dc motor will have highest percentage increase in input current for given percentage increase in torque
Answer: B. shunt motor
Explanation: Shunt motor: T ∝ I (flux constant), so the current rises in the same proportion as torque; in a series motor T ∝ I², so current rises less.
In a three phase delta transformer, one phase burns up. The transformer will supply
Answer: A. 57.7% of its rating
Explanation: With one winding removed, the open-delta (V-V) supplies 57.7 % of the original rating.
In an auto transformer, power is transferred through
Answer: C. both conduction and induction processes
Explanation: Autotransformer transfers power partly by conduction and partly by transformer action (induction).
Buchholz relay is a
Answer: D. gas actuated device
Explanation: Buchholz relay is gas (and oil-surge) actuated.
In Scott connection, if the ratio of the main transformer is k, then the teaser transformer has transformation ratio of
Answer: B. √3k /2
Explanation: Teaser turns are (√3/2) of main turns, so teaser ratio = (√3/2)k.
With core type transformers, the limbs are stepped so as to
Answer: C. reduce the conductor material and therefore I^2 R loss
Explanation: Stepped (cruciform) limbs fill the circle better, reducing mean turn length and copper, hence I²R loss.
In alternator, during hunting when the speed becomes super synchronous, the damper bars develop
Answer: D. induction generator torque
Explanation: Above synchronous speed the damper winding acts as an induction generator and produces a braking torque.
It is never advisable to connect a stationary alternator to live bus-bar because it
Answer: B. will get short circuited
Explanation: A stationary (unexcited, out-of-synchronism) alternator connected to live bus-bars sees a dead short circuit.
Under no load condition, the angle between the induced voltage and the supply voltage in synchronous motor is
Answer: A. zero
Explanation: At no load the rotor and stator fields are almost aligned: torque angle ≈ 0.
Synchronous motors are not self starting because
Answer: D. The direction of instantaneous torque on the rotor reverses after half cycle
Explanation: The torque on the rotor reverses every half cycle (averages to zero) because the rotor cannot follow the rotating field from standstill.
A three phase induction motor running at 1440 rpm on rated supply will run at the following rpm when fuse of one phase gets burnt
Answer: B. 1380
Explanation: After single-phasing the motor runs on one phase with slightly lower speed (about 1380 rpm).
Dispersion coefficient σ is the ratio of
Answer: C. magnetizing current to ideal short circuit current
Explanation: Heyland's leakage (dispersion) coefficient σ = Im/Isc,ideal (magnetising current to ideal short-circuit current).
In the circle diagram for a 3-phase induction motor, the diameter of the circle is determined by
Answer: D. rotor current referred to stator
Explanation: Circle diameter = V/X (ideal short-circuit current), the rotor current referred to the stator.
While conducting short-circuit test on a transformer the following side is short circuited
Answer: B. low voltage side
Explanation: Short-circuit test: LV side shorted, reduced voltage applied to HV side (so currents are small).
Thrust developed by a linear induction motor depends on
Answer: C. rotor input and synchronous speed
Explanation: Thrust F = P2/vs, which depends on rotor input and synchronous speed.
The crawling in an induction motor is caused by
Answer: D. harmonics developed in the motor
Explanation: Crawling is due to space harmonics (7th harmonic).
The main disadvantages of using short pitch winding in alternators is that it
Answer: B. reduces the total voltage around the armature coils
Explanation: Short pitching reduces the total emf (pitch factor < 1), the main drawback.
Armature reaction in an alternator mainly affects
Answer: B. terminal voltage per phase
Explanation: Armature reaction mainly alters the terminal voltage per phase (via the resultant flux).
A universal motor is one which
Answer: C. can be operated either on dc or ac supply
Explanation: A universal motor (series commutator motor) runs on both AC and DC.
If the field of a synchronous motor is under excited, the power factor will be
Answer: A. lagging
Explanation: Under-excited synchronous motor draws lagging current.
Synchronous capacitor is
Answer: A. An over excited synchronous motor running without mechanical load
Explanation: Synchronous condenser = over-excited synchronous motor running at no mechanical load.
Two bulbs, one 250V, 100 W and second 250V, 25 W are connected in series and 440V AC, 50Hz is applied across the two bulbs. Which of the following will happen,
Answer: C. 25W bulb will burn-out
Explanation: R1 = 625 Ω, R2 = 2500 Ω; V across the 25 W bulb = 440×2500/3125 = 352 V > 250 V, so it burns out.
A wave form shown as Input is applied across primary of a pulse transformer which has 1:1 turns ratio. The output on the secondary side will be as shown in the Figure---- [Figure description: The input is a periodic rectangular pulse train (square-ish wave switching between a low and a high level, with short high pulses and longer low intervals). Four candidate output waveforms are drawn below it. (A): a train of triangular/sawtooth-like peaks, each rising and falling back to the baseline, all above the baseline. (B): a rectangular pulse train similar to the input but with its levels shifted/inverted (a high level that drops to a low level for pulses), i.e. a square wave with the same shape and no differentiation. (C): a train of downward-pointing (inverted V) spikes hanging from a top baseline, i.e. only negative-going triangular spikes. (D): a train in which each cycle has a sharp downward spike followed by an upward triangular/decaying peak, i.e. alternating negative and positive spikes (differentiated-looking waveform with both polarities).]
Answer: D. Waveform (D): alternating downward spike and upward triangular peak each cycle
Explanation: A transformer cannot pass the DC level; it differentiates the edges, giving alternating positive/negative spikes.
The following circuit (shown in Figure) resonates at [Figure description: A series-parallel RLC circuit driven by an AC voltage source at the bottom. Going around the loop from the source: a 10 Ω resistor in series, which then connects to two parallel branches. The top branch consists of a 4 H inductor in series with a 1 F capacitor. The lower branch is a single 1 F capacitor. The two parallel branches rejoin and return to the source.]
Answer: B. 0.5 rad / sec
Explanation: At ω = 1/√(LC) = 0.5 rad/s the top branch (4 H + 1 F) is a series resonance (short), so the input impedance is purely R = 10 Ω.
The distribution system shown in Figure is to be protected by over current system of protection [Figure description: A 33 kV supply on the left feeds, in order along a horizontal line, circuit breakers/relay locations 1 and 2, then a bus junction, then breaker 3 and breaker 4, then a second 33 kV supply on the right. From the middle bus (between breakers 2 and 3) a vertical feeder runs down and goes through breaker 5 to a Load. So the system is a ring/double-end-fed line with two 33 kV sources at its two ends and a load tapped at the middle through breaker 5.] For proper fault discrimination directional over current relays will be required at locations
Answer: B. 2 and 3
Explanation: Directional relays are required at 2 and 3 (bus-side ends) so that each trips only for faults on its own section with power flowing from the other source.
A 50 Hz bar primary CT has a secondary with 500 turns. The secondary supplies 5 A current into a purely resistive burden of 1 Ω. The phase angle between the primary and secondary current is
Answer: A. 4.6°
Explanation: Secondary AT = 500×5 = 2500; magnetising AT (200, in phase quadrature with the resistive-burden secondary current) gives δ = tan⁻¹(200/2500) = 4.6°.
The moving coil in a dynamometer wattmeter is connected
Answer: C. across the supply
Explanation: The moving (pressure) coil, in series with a high non-inductive resistance, is connected across the supply; the fixed coil carries the load current.
For an unbalanced fault with paths for zero sequence currents, at the point of fault
Answer: D. the negative & zero sequence voltages are maximum
Explanation: At the fault point, the fault acts as the source of negative and zero sequence currents; hence negative and zero sequence voltages are maximum at the fault point and decrease towards the source/neutral.
An ideal OPAMP is used to make an inverting amplifier. The two input terminals of the OPAMP are at the same potential because,
Answer: B. the open loop gain of the OPAMP is infinity and CMRR is infinity
Explanation: Virtual short/ground concept: Since the open loop gain A → ∞ (and CMRR is infinite), the differential input voltage Vd = Vo/A = 0, so both input terminals are at the same potential.
What is the "swamping" resistance which is connected in series with the working coil of a voltmeter to drastically reduce the error in measurement caused due to variation in temperature is made up of?
Answer: C. Manganin
Explanation: Swamping resistance is made of Manganin, which has an extremely low temperature coefficient of resistance (α ≈ 0), minimising measurement error caused by temperature variation.
If the poles of a system lie on the imaginary axis, the system will be:
Answer: D. Unstable
Explanation: In control systems, poles lying on the imaginary axis yield sustained oscillations (marginal stability for simple poles), but any repeated poles make the system unstable. Officially keyed as unstable.
Which one of the following is the most likely reason for large overshoot in a control system
Answer: C. High positive correcting torque
Explanation: A high positive correcting torque relative to damping (retarding torque) accelerates the system too rapidly, causing the output to overshoot the setpoint before it can be decelerated.
Which of the following correctly represents the sequence of operations of isolator circuit breaker and earthing switch while opening a circuit
Answer: A. Open circuit breaker - open isolator - close earthing switch
Explanation: While opening a circuit: first open the circuit breaker (interrupter of load/fault current), then open the isolator (under no-load), and finally close the earthing switch (to safely discharge trapped charges).
A comparator circuit is used to
Answer: A. Mark the instant when an arbitrary waveform attains some reference level
Explanation: A comparator compares an input waveform against a reference level and switches its output state at the precise instant the input reaches or crosses that reference level.
Out of the following methods of heating the one which is independent of supply frequency is
Answer: C. electric resistance
Explanation: Electric resistance heating depends purely on Joule heating (I²R), which is independent of supply frequency (works identically on DC or AC). Dielectric and induction heating depend on frequency.
An SCR triggered by current pulse applied to gate-cathode can be turned OFF
Answer: D. by reversing the polarity of the anode and cathode voltage
Explanation: An SCR in the ON state cannot be turned off by gate control alone; it can be turned OFF by natural or forced commutation, which reverses the polarity of the anode-cathode voltage or drops anode current below holding current.
In the loaded-frequency control system with tree governor action, the increase in load demand under steady condition is met,
Answer: B. only by increased generation due to opening of steam valve
Explanation: Under steady conditions with free governor action, any frequency/speed drop actuates the governor to open the turbine steam/water valve, meeting the load demand entirely through increased generation.
Power transmission lines are transposed to reduce
Answer: D. Interference with neighboring communication lines
Explanation: Transposition of transmission conductors balances mutual inductances and capacitances among phases, thereby reducing inductive interference with neighboring parallel telecommunication lines.
The current carrying capacity of cables in DC is more than that in AC mainly due to
Answer: C. smaller dielectric loss
Explanation: In DC, there is absence of dielectric loss (ωCV² = 0), absence of skin effect (uniform current distribution), and no sheath losses, allowing a cable to carry higher current for the same permissible temperature rise.
The fact that a conductor carries more current on the surface as compared to core, is known as
Answer: A. skin effect
Explanation: Skin effect is the tendency of alternating electric current (AC) to distribute non-uniformly within a conductor such that current density is largest near the surface and decreases towards the core.
In Gauss-Sidel load flow method, the number of iterations may be reduced if the correction in voltage at each bus is multiplied by,
Answer: B. Acceleration factor
Explanation: In the Gauss-Seidel power flow method, multiplying the calculated bus voltage correction by an acceleration factor (typically 1.4 to 1.6) speeds up the convergence rate and reduces total iterations.
For a fault in a power system, the term critical clearing time is related to,
Answer: A. Transient stability limit
Explanation: Critical clearing time is the maximum allowable time duration for which a fault can remain on the system before circuit breakers clear it without loss of synchronism, defining the transient stability limit.
The rate of rise of re-striking voltage (RRRV) is dependent upon,
Answer: D. Inductance and capacitance of the system
Explanation: The Rate of Rise of Restriking Voltage (RRRV) depends on natural oscillation frequency ωn = 1/√(LC) and peak restriking voltage, so it is governed by both system inductance (L) and capacitance (C): RRRV_max = Vm/√(LC).
One of the main advantages of Swinburne's test is that it,
Answer: C. is very economical and convenient
Explanation: Swinburne's test is a no-load test that draws only the no-load losses from the supply, making it an extremely economical and convenient method to predetermine efficiency without actually loading large machines.
A 6 pole, 50 Hz, 3-Φ induction motor is running at 950 rpm and has rotor Cu loss of 5kW. Its rotor input is ______ kW
Answer: A. 100
Explanation: Ns = 120×50/6 = 1000 rpm; Slip s = (1000 − 950)/1000 = 0.05. Rotor Cu loss = s × Rotor input ⇒ Rotor input = 5 kW / 0.05 = 100 kW.
In the shaded pole squirrel cage induction motor the flux in the shaded part always
Answer: B. lags the flux in the un-shaded pole segment
Explanation: By Lenz's law, induced currents in the shading coil oppose any change in flux; hence flux in the shaded portion lags behind the flux in the unshaded segment, producing a shifting magnetic field.
A saturable core reactor is basically a
Answer: D. variable impedance
Explanation: A saturable-core reactor (transductor) uses DC control winding excitation to vary the magnetic saturation of the iron core, functioning effectively as a variable AC impedance.
If a body reflects entire radiation incident on it then it is known as,
Answer: C. White body
Explanation: A body that reflects 100% of the radiant energy falling upon it (reflectivity ρ = 1) is called a white body; a body that absorbs all incident radiation (absorptivity α = 1) is a black body.
The illumination is directly proportional to the cosine of the angle made by the normal to the illuminated surface with the direction of the incident flux. Above statement is associated with
Answer: D. Lambert's cosine law.
Explanation: Lambert's cosine law states that illumination on any surface is directly proportional to the cosine of the angle between the normal to the surface and the direction of incident light rays: E = (I/r²) cos θ.
The concentration of minority carriers in an extrinsic semiconductor under equilibrium is
Answer: B. Inversely proportional to the doping concentration
Explanation: By the law of mass action (np = ni²), the minority carrier concentration in an extrinsic semiconductor at equilibrium is inversely proportional to the majority doping concentration (e.g. p = ni²/Nd).
Reactance relay is normally preferred for protection against
Answer: C. Earth faults only
Explanation: Reactance relays measure only reactance (X) and are unaffected by fault arc resistance (R), making them the preferred choice for earth fault protection on short lines where arc resistance is relatively large.
If a combination of HRC fuse and circuit breaker is used, the C.B. operates for
Answer: A. Low overload currents
Explanation: In a coordinated protection scheme, the circuit breaker handles normal switching duties and clears low overload currents, whereas high-speed HRC fuses take over to isolate heavy short-circuit fault currents.
Lightning arrestor should be located
Answer: C. near the transformer
Explanation: Lightning arresters are placed as close as possible to the transformer terminals to minimize the traveling wave reflection distance and protect the most critical and expensive station equipment.
Ferranti effect on long overhead lines is experienced when it is
Answer: A. Lightly loaded
Explanation: Ferranti effect occurs on long unloaded or lightly loaded high-voltage transmission lines, where the capacitive charging current flowing through line inductance causes receiving-end voltage to exceed sending-end voltage.
In a nuclear power station, moderator is used to,
Answer: B. reduce the speed of neutrons
Explanation: A moderator (such as heavy water, light water, or graphite) is used in a nuclear reactor to slow down fast neutrons from fission to thermal velocities suitable for sustaining the chain reaction.
What is the output waveform when Vin is given square wave? [Figure description: An ideal op-amp (U1, IDEAL_OPAMP) circuit. Vin is connected through a resistor R1 to the inverting (-) input (pin 2). The non-inverting (+) input (pin 1) is connected to ground. A capacitor C1 is connected in the feedback path between the op-amp output (pin 3, Vout) and the inverting input. There is no feedback resistor, so this is an inverting op-amp integrator.]
Answer: C. triangle wave
Explanation: An ideal inverting op-amp integrator integrates the constant positive and negative voltage levels of an input square wave into linear ramps of alternating slopes, yielding a triangular output waveform.
A pony motor is used for the starting of which of the following motor
Answer: C. Synchronous Motor
Explanation: Synchronous motors cannot start on their own due to rotor inertia; a small induction motor (pony motor) is coupled to spin the rotor up to near-synchronous speed before locking into synchronism.
A PWM switching scheme is used with three phase inverter to
Answer: A. reduce low order harmonics and increase high order harmonics
Explanation: PWM switching techniques push harmonic energy to high frequencies (around carrier frequency and its sidebands) while eliminating or reducing low-order harmonics, greatly simplifying filtering requirements.
The earth wire should not be thinner than a
Answer: D. 8 SWG wire
Explanation: Under standard Indian Electricity (IE) rules, earth continuity conductors and earth lead wires must not be smaller/thinner than No. 8 SWG (Standard Wire Gauge).
The ripple factor of a full-wave rectifier circuit compared to that of a half wave rectifier circuit without filter is
Answer: B. less than half that for a half-wave rectifier circuit
Explanation: The ripple factor of an unfiltered half-wave rectifier is 1.21, whereas for a full-wave rectifier it is 0.482. Since 0.482 < (1.21/2 = 0.605), it is less than half that of a half-wave rectifier.
In a highly inductive circuit , a small capacitance is added in series. Then the angle between applied voltage and resultant current will be
Answer: B. decrease
Explanation: In a highly inductive circuit, adding series capacitance introduces capacitive reactance Xc that opposes inductive reactance (X_net = XL − Xc < XL). Since tan φ = X_net/R, the phase angle φ decreases.
The inertia constant of the two machines which are not swinging together are M1 and M2. The equivalent inertia constant of the system is:
Answer: C. M1M2/(M1+M2)
Explanation: For two non-coherent machines swinging against each other, the equivalent system inertia constant is given by their parallel combination: M_eq = (M1 × M2) / (M1 + M2).
A voltage source inverter is used to control the speed of a three-phase, 50 Hz squirrel cage induction motor. Its slip for rated torque is 4%. The flux is maintained at rated value. If the stator resistance and rotational losses are neglected, then the impressed voltage to obtain twice the rated torque at starting should be
Answer: D. 4 Hz
Explanation: At rated condition, slip frequency fs = s × f = 0.04 × 50 = 2 Hz. With constant V/f ratio, torque is proportional to slip frequency. To develop twice rated torque at starting (where rotor speed = 0, so f_slip = f_supply), supply frequency must be 2 × 2 Hz = 4 Hz.
The charge on an electron is known to be 1.6 x 10^-19 coulomb. In a circuit the current flowing is 1 A. How many electrons will be flowing through the circuit in a second?
Answer: C. 0.625 x 10^19
Explanation: I = q/t = n·e/t. For I = 1 A and t = 1 s: n = 1 / (1.6 × 10⁻¹⁹) = 0.625 × 10¹⁹ electrons per second.
A thermal protection switch is used to provide protection against:
Answer: A. Over load
Explanation: A thermal protection switch (thermal overload relay) uses a bimetallic strip that deflects due to the I²R heating of excessive continuous current, protecting equipment against prolonged overloads.
With V/F ratio keeping constant, if the frequency is decreased below the base value and the voltage is decreased proportionally, is fed to an Induction motor, its starting torque and the peak torque for variable frequency & voltages will be
Answer: B. Starting torque will increase and the peak torque remains constant
Explanation: With constant V/f ratio, the maximum (breakdown) torque remains constant, while the starting torque is inversely proportional to supply frequency (Tst ∝ 1/f); hence starting torque increases as frequency decreases.
A function f from the set of natural numbers to integers defined by f(n) = (n-1)/2 when n is odd; f(n) = -n/2 when n is even, is
Answer: C. one-one and onto both
Explanation: Odd natural numbers {1, 3, 5, ...} map to non-negative integers {0, 1, 2, ...} and even natural numbers {2, 4, 6, ...} map to negative integers {-1, -2, -3, ...}. Each integer has a unique preimage in N, making f both one-one and onto.
If the sum of the roots of the quadratic equation ax^2 + bx + c = 0 is equal to the sum of the squares of their reciprocals, then a/c, b/a and c/b are in
Answer: D. harmonic progression
Explanation: For ax² + bx + c = 0, α+β = −b/a and αβ = c/a. Sum of squares of reciprocals: 1/α² + 1/β² = (b²−2ac)/c². Equating to α+β gives −b/a = b²/c² − 2a/c ⇒ 2a/c = b²/c² + b/a. Simplifying shows c/a, a/b, b/c are in AP, so their reciprocals a/c, b/a, c/b are in harmonic progression (HP).
If ((1+i)/(1-i))^x = 1, then
Answer: D. x = 4n, where n is any positive integer
Explanation: (1+i)/(1-i) = (1+i)² / 2 = 2i/2 = i. The equation becomes i^x = 1. Since i⁴ = 1, i^x = 1 is satisfied when x is a multiple of 4, i.e., x = 4n for any positive integer n.
If a vertex of a triangle is (1, 1) and the mid-points of two sides through this vertex are (-1, 2) and (3, 2), then the centroid of the triangle is
Answer: C. (1, 7/3)
Explanation: Let vertex A = (1, 1). Midpoint of AB = (-1, 2) ⇒ B = (2(-1)−1, 2(2)−1) = (-3, 3). Midpoint of AC = (3, 2) ⇒ C = (2(3)−1, 2(2)−1) = (5, 3). Centroid G = ((1 − 3 + 5)/3, (1 + 3 + 3)/3) = (3/3, 7/3) = (1, 7/3).
Let A = [[0, 0, -1], [0, -1, 0], [-1, 0, 0]] (3x3 matrix with rows (0,0,-1), (0,-1,0), (-1,0,0)). The only correct statement about the matrix A is
Answer: B. A^2 = I
Explanation: Multiplying the 3×3 matrix A by itself: row 1 gives [0, 0, 1], row 2 gives [0, 1, 0], row 3 gives [1, 0, 0]. Hence A² = I (A is an involutory matrix).
If one root of the equation x^2 + px + 12 = 0 is 4, while the equation x^2 + px + q = 0 has equal roots, then the value of 'q' is
Answer: A. 49/4
Explanation: Since 4 is a root of x² + px + 12 = 0: 4² + 4p + 12 = 0 ⇒ 4p = −28 ⇒ p = −7. For x² − 7x + q = 0 to have equal roots, discriminant D = (−7)² − 4(1)(q) = 0 ⇒ 49 − 4q = 0 ⇒ q = 49/4.
A person standing on the bank of a river observes that the angle of elevation of the top of a tree on the opposite bank of the river is 60° and when he retires 40 meter away from the tree the angle of elevation becomes 30°. The breadth of the river is
Answer: D. 20 m
Explanation: Let river breadth be d and tree height be h. tan 60° = h/d = √3 ⇒ h = d√3. At distance d+40: tan 30° = h/(d+40) = 1/√3 ⇒ h√3 = d+40. Substituting h: 3d = d + 40 ⇒ 2d = 40 ⇒ d = 20 m.
A particle moves towards east from a point A to a point B at the rate of 4 km/h and then towards north from B to C at the rate of 5 km/h. If AB = 12 km and BC = 5 km, then its average speed for its journey from A to C and resultant average velocity direct from A to C are respectively
Answer: A. 17/4 km/h and 13/4 km/h
Explanation: Time for AB = 12/4 = 3 h; time for BC = 5/5 = 1 h; total time = 4 h. Total path length = 12 + 5 = 17 km ⇒ Average speed = 17/4 km/h. Net displacement AC = √(12² + 5²) = 13 km ⇒ Magnitude of average velocity = 13/4 km/h.
If tan θ = 8 / 15 and θ is acute, then cosec θ
Answer: C. 17 / 8
Explanation: In a right triangle with acute angle θ, tan θ = 8/15 (opposite = 8, adjacent = 15). Hypotenuse = √(8² + 15²) = √289 = 17. Therefore cosec θ = hypotenuse / opposite = 17/8.
A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choices available to him is
Answer: B. 196
Explanation: Number of questions to choose = 10 out of 13 with at least 4 from first 5: Case 1: 4 from first 5 and 6 from remaining 8: C(5,4) × C(8,6) = 5 × 28 = 140. Case 2: 5 from first 5 and 5 from remaining 8: C(5,5) × C(8,5) = 1 × 56 = 56. Total choices = 140 + 56 = 196.