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RRB NTPC Aptitude Test - 78
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RRB NTPC Aptitude Test - 78
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  • Question 1/10
    1 / -0

    15 men can finish a work in 40 days. After some days, 5 men leave the work so that the total work is completed in 50 days. After how many days of working, do the 5 men leave?

    Solutions

    Let the men leave after x days.

    Work done by them in x days = x/40

    Remaining work = 1 – x/40

    Also, M1T1/W1= M2T2/W2

    So, 15×40/1 = 10×(50 – x)/(1 – x/40)

    So, 1-x/40 = 10(50 – x)/(15×40)

    40 – x = 2(50 – x)/3

    120 – 3x = 100 – 2x

    X = 20

     

  • Question 2/10
    1 / -0

    In a 100 m race A beats B by 20m. In a race between B and C of 100 m, B beats C by 25m. If A and C go on a race of 200m, by how much meters does A beat C?

    Solutions

    In 100m race, A = 100m, B = 80m

    Between B and C, B = 100m, C = 75m

    When B = 80m, C = 60m

    So, in a 200m race between A and C, A=200m, C=120m

    So, A beats C by 80m

     

  • Question 3/10
    1 / -0

    If a man can row with a speed of 4km/h in still water and the speed of the stream in 2km/h. If he rows from one bank to the other and returns, find his average speed for the whole journey.

    Solutions

    Let distance between the banks be x km

    Upstream speed = 4 – 2 = 2km/h; Upstream time = x/2

    Downstream speed = 4 + 2 = 6km/h; Downstream time = x/6

    Total distance = 2x

    Total time = x/2 + x/6 = 2x/3

    Average speed for the whole journey = 2x/(2x/3)

    = 3km/h

     

  • Question 4/10
    1 / -0

    A sum of Rs. 10000 amounts to Rs. 20000 in 3 years of compound interest. In how many years will it become Rs. 40000?

    Solutions

    In CI, A=P(1+0.01r)n

    A-Amount, P-Principle/Sum, n- Time period

    20000=10000(1+0.01r)3

    2 = (1+0.01r)3

    Squaring both sides,

    4 = (1+0.01r)6

    Multiplying both sides by 10000,

    40000=10000(1+0.01r)6

    So, the amount will become 40000 in 6 years

     

  • Question 5/10
    1 / -0

    If 3125 × 1250 = 25x × 50, find the value of x.

    Solutions

    5× 5× 2 = 52x × 52 × 2

    58 = 52x + 2

    Equating exponents, 2x = 6

    X = 3

     

  • Question 6/10
    1 / -0

    A ladder of length 15m leans on a straight wall of height 7.5√3m. If the same ladder while leaning against another straight wall of height h, the foot of the ladder makes an inclination that is half of that made by the ladder when it leans against the 7.5√3m wall then Find the distance between the foot of the ladder and the base of the building of height h.

    Solutions

    In dia 1, sin w = 7.5√3/15

    = √3/2

    W = 60°

    According to question, in the next wall, the angle of inclination is half

    So, w/2 = 30°

    If we compare the two triangles, we can see that the second triangle is rotation of the first triangle (If one angle is 30°, the other angle has got to be 60°)

    SO, the height of first wall is equal to the distance between foot of ladder and base of building in second case.

    So, x = 7.5√3m

     

  • Question 7/10
    1 / -0

    Find the approximate value of

    (72.001×6.99×10.001)/(√(64.05)×√(3&0.729))

    Solutions

    √64.05 ≈ 8

    ∛0.729 ≈ 0.9

    So, the given expression becomes

    72×7×10/7.2

    =700

     

  • Question 8/10
    1 / -0

    Two candidates contest in an election in which out of 8500 votes, some votes are declared to be invalid. If one person gets 51% of the valid votes and the other candidate loses by 160 votes, find the number of invalid votes.

    Solutions

    Let total number of invalid votes be x

    Votes for winner = 0.51(8500 - x)

    Votes for loser = 0.49(8500 - x)

    Difference=0.02(8500-x) = 160

    170-0.02x = 160

    0.02x = 10

    X = 500

     

  • Question 9/10
    1 / -0

    In 2012, the ratio of age of A to that of B was 3:4. If in 2017, the ratio of age of A to that of B is 7:9, find the sum of ages of A and B in 2022.

    Solutions

    2012, let the ages be 3x, 4x

    2017, 3x + 5, 4x + 5

    (3x + 5)/(4x + 5) = 7/9

    27x + 45 = 28x + 35

    X = 10

    Ages of A and B in 2012 = 30, 40

    Ages of A and B in 2022 = 40, 50

    Sum = 90

     

  • Question 10/10
    1 / -0

    A card is drawn from a pack of 52 cards. Find the probability that the card drawn is neither a spade nor a number card.

    Solutions

    Number of spade cards = 13

    Remaining cards=52 – 13 = 39

    Number of numbered cards in the remaining three sets = 9 × 3 = 27

    Remaining cards = 39 – 27 = 12

    So, required Probability = 12/52 = 3/13

     

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