Quant with Kamal Lohia - Part 8

  • Faculty and Content Developer at Tathagat | Delhi College of Engineering

    Q1) A certain city has a circular wall around it, and this wall has four gates pointing north, south, east and west. A house stands outside the city, 3 km north of the north gate, and it can just be seen from a point 9 km east of the south gate. What is the diameter of the wall that surrounds the city?
    a. 6 km
    b. 9 km
    c. 12 km
    d. None of these

    Ans: b. 9 km
    Sol: This can be solved easily by using similarity of triangles.

    Q2) Two sides of a quadrilateral plot measure 32 m and 24 m and the angle between them is a perfect right angle. The other two sides measure 25 m each and the other three angles are not right angles. What is the area of the plot?
    a. 768 m²
    b. 534 m²
    c. 696.5 m²
    d. 684 m²

    Sol: This is a simple application of primitive Pythagorean triplets only.
    Required area = (1/2)[8² (3 * 4) + 5² (3 * 4) + 5² (3 * 4)] = 6 * 114 = 684 m².

    Q3) Euclid has a triangle in mind. Its longest side has length 20 and another of its sides has length 10. Its area is 80. What is the exact length of its third side?
    a. rt(260)
    b. rt(250)
    c. rt(240)
    d. rt(270)

    Let θ be the angles between lengths 20 and 10.
    So Area = 80 = (1/2) * 20 * 10 * Sinθ
    i.e. Sinθ = 4/5, i.e. Cosθ = 3/5
    Using cosine rule, we have third side = rt(20² + 10² - 2 * 20 * 10 * Cosθ) = rt(260)

    Q4) The area of the triangle whose vertices are (a, a), (a + 1, a + 1) and (a + 2, a) is
    a. a³
    b. 1
    c. 2a
    d. rt(2)

    Just consider a = 0 and solve to get the base as 2 and height 1, so area = (1/2) * 2 * 1 = 1.

    Q5) Neeraj has agreed to mow a lawn, which is a 20 m × 40 m rectangle. He mows it with 1 m wide strip. If Neeraj starts at one corner and mows around the lawn toward the centre, about how many times would he go round as he mowed complete the lawn?
    a. 9.5
    b. 10
    c. 19.5
    d. 20

    Ans. Option b ( 10 ). It is simply half of smaller side. Draw the diagram and observe yourself.

    Q6) Let A and B be two solid spheres such that the surface area of B is 300% higher than the surface area of A. The volume of A is found to be k% lower than the volume of B. The value of k must be
    a. 85.5
    b. 92.5
    c. 90.5
    d. 87.5

    Equating surface areas we get that 2r(A) = r(B)
    So, volume of A is 1/8 that of B i.e. 87.5% lesser.

    Q7) x^2 – 3y^2 = 1376
    How many integer solutions exist for the given equation?

    RHS is 2 mod 3 but LHS is either 0 or 1 mod 3. So NO integral solution exists for the equation.

    Q8) The number 3 can be written as 3, 2+1, 1+2, 1+1+1 in four ways. In how many ways the number "n" can be written?

    Consider it as 1_1_1_1....n times. Now you just need to put + signs in place of _. There are total n-1 times _ is used and for each one you have two choices whether to put a + or not. That's why answer is 2^(n-1)

    Q9) Find the unit digit of 1^5 + 2^5 + 3^5 + … + 99^5

    Unit digit of each number repeats after every fourth power i.e. unit digit of 29⁵ = unit digit of 29 = 9 and so on.
    So unit digit of 1⁵ + 2⁵ + 3⁵ + ... + 99⁵ = unit digit of (1 + 2 + 3 + ... + 99) = unit digit of 99(99+1)/2 = unit digit of 99(50) = 0

    Q10) A military general needs to take his troops of 100 soldiers across a river from bank A to bank B. He engages a boat with two boys, both of whom can row, at the bank A. But the boat can take only up to two boys, or only one soldier. What is the min. number of round trips that the boat has to make, to transfer all the 100 soldiers and the general to bank B and come to bank A?
    a) 400
    b) 200
    c) 202
    d) 403

    Answer is (c) 202. See, here we are assuming that each soldier and the general too can row the boat. Initially all 100 soldiers, one general, 2 boys are at bank A. In first trip both boys go from A to B and one boy stays at B and other comes back to A in second trip. That completes 1 round trip. In next trip, one soldier moves from A to B and in the return journey, other boy returns from B to A and thus completes 2nd round trip. Thus in 2 round trips, one soldier is transferred from A to B and the boat is back at A. Thus it'll take exactly 2 * 101 = 202 round trips to take all 100 soldiers and a general to take from A to B and back the boat at A.

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