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RRB JE Aptitude Test - 5
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RRB JE Aptitude Test - 5
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  • Question 1/10
    1 / -0

    If 5 Sin2θ + 3 Cos2θ = 4, find the value of Sinθ and Cosθ:

    Solutions

    ⇒ 5 Sin2θ + 3 Cos2θ = 4

    ⇒ 5 Sin2θ + 3(1 - Sin2θ) = 4

    ⇒ 5 Sin2θ + 3 - 3Sin2θ = 4

    ⇒ 2Sin2θ = 1

    ⇒ Sin θ = ±1/√2

    ⇒ Cosθ = ±√(1 - ½) = ±1/√2

    ∴ Option 1 is correct

     

  • Question 2/10
    1 / -0

    If sin25° = 0.423 and cos10° = 0.985, find the value of (sin35° + sin15°)

    Solutions

    sinA + sinB = 2 sin [(A + B)/2] cos [(A – B)/2]

    Putting A = 35° and B = 15°, we get,

    ⇒ (sin 35° + sin 15°) = 2 sin [(35 + 15)/2] cos [(35 – 15)/2]

    ⇒ (sin 35° + sin 15°) = 2 sin 25° cos 10°

    ⇒ (sin 35° + sin 15°) = 2 × 0.423 × 0.985

    ∴ (sin 35° + sin 15°) = 0.833

     

  • Question 3/10
    1 / -0

    PQRS is a square, M is the mid-point of PQ and N is a point on QR such that NR is two-third of QR. If the area of ΔMQN is 48cm2, then what is the length (in cm) of PR?

    Solutions

    Let sides of square PQRS be a units

    ⇒ MQ = a/2 and QN = a/3

    We know that, area of triangle = 1/2 × base × height

    So, Ar(ΔMQN) = 1/2 × MQ × QN

    ⇒ Ar(ΔMQN) = 1/2 × a/2 × a/3

    ⇒ 48 = a2/12

    ⇒ a2 = 576

    ⇒ a = 24 cm

    Length of side = 24 cm

    So, length of diagonal PR = a√2 = 24√2 cm

     

  • Question 4/10
    1 / -0

    If A(3, 4) is equidistant from P (2, -3) and R (a, 4), find the value of a

    Solutions

     

  • Question 5/10
    1 / -0

    A solid sphere of diameter 7 cm is cut into two equal halves. What will be the increase (in cm2) in the total surface area?

    Solutions

    Given r = 7/2

    ⇒ Initial surface area = 4πr2 = 4 π (3.5)2

    ⇒ Final surface area = 4πr2 + 2 π r2

    ⇒ Final surface area = 6πr2

    ⇒ Increase in total surface area = 6πr2 - 4πr2 = 2πr2

    ⇒ Increase in total surface area = 2π(3.5)2

    ⇒ Increase in total surface area = 76.96 cm2

    ∴ the correct option is 1)

     

  • Question 6/10
    1 / -0

    On reducing the entry fee by 35% in a park, the number of people coming to the park increased by 40%, then the percent increase or decrease in the income from the entry fee is –

    Solutions

    Let the original entry fee be ‘a’ and number of people initially coming to the park be ‘b’.

    Total income = a × b = ab

    Now, reducing the entry fee by 35% in a park, the number of people coming to the park increased by 40%

    ∴ New entry fee = a – (35% of a) = 0.65a

    New number of people = b + (40% of b) = 1.4b

    New total income = 0.65a × 1.4b = 0.91ab

    Decrease in income = ab – 0.91ab = 0.09ab

     

  • Question 7/10
    1 / -0

    Instead of dividing Rs. 468 among A, B, C in the ratio 13 : 14 : 12, by mistake it was divided in the ratio 3 : 4 : 2. Who gained from the transaction?

    Solutions

    Let 13 : 14 :12 

    Actual amount they have to got -

    ⇒ A = (468 × 13)/39 = Rs. 156

    ⇒ B = (468 × 14)/39 = Rs. 168

    ⇒ C = (468 × 12)/39 = Rs. 144

    But mistakenly they got the amount in the ratio = 3 : 4 : 2

    ⇒ A = (468 × 3)/9 = Rs. 156

    ⇒ B = (468 × 4)/9 = Rs. 208

    ⇒ C = (468 × 2)/9 = Rs. 104

    ∴ B gained in the transaction

     

     

  • Question 8/10
    1 / -0

    A man purchased a cycle worth Rs. 4400 and sold it on the profit of 12.5% to another man. If the new man who purchased the cycle sells it after using it for a year at the loss of 33.33%. How much loss incurred to him?

    Solutions

    ⇒ 12.5% = 1/8

    ⇒ Net profit = 4400/8 = 550

    ⇒ Price at which the new man purchased the cycle = 4400 + 550 = 4950

    ⇒ Loss = 33.33% = 1/3

    ∴ The loss incurred = 4950/3 = 1650

     

  • Question 9/10
    1 / -0

    A is 2 times efficient as B. B can finish a work alone in 36 days. If A and B started working together and A left the work 3 days before the completion of work. How much time will it take to complete the work?

    Solutions

    ∵ Efficiency is always inversely proportional to time.

     

    A : B

    Efficiency

    2 : 1

    Time

    1 : 2

    Time taken by B = 2 units = 36 days

    Time taken by A = 1 unit = 18 days

    Let Total work = Lcm of 18 and 36 = 36 units

    ⇒ Efficiency of B = 1 unit/day

    ⇒ Efficiency of A = 2 units/day

    ⇒ Total work done by A in 2 days = 2 × 3 = 6 units

    ⇒ Now New Total work = 36 + 6 = 42 units

    ∴ Total time taken to complete the work = 42/3 = 14 days

     

  • Question 10/10
    1 / -0

    A 100 m long train is running at a speed of 86 km/hr. towards another train, which is running at a speed of 64 km/hr. If the trains are 875 m apart, in how much time will they meet? 

    Solutions

    When two trains are running in opposite directions, i.e., towards each other, their relative speed is the sum of their respective speeds.

    Hence, their relative speed = 86 + 64 = 150 km/hr = 150 × (5/18) = 125/3 m/sec.

    Now, Distance to cover = 875 m

    ∴ They will meet after = 875/(125/3) = 21 sec.

     

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