Quant Capsules by Shashank Prabhu - Set 6


  • CAT 100%iler, 5 times AIR 1, Director - learningroots.in, Ex ITC, Pagalguy, TAS


    A car after traveling 18 km from a point A developed some problem in the engine and the speed became 4/5th of its original speed. As a result, the car reached point B 45 minutes late. If the engine had developed the same problem after traveling 30 km from A, then it would have reached 36 minutes late. The original speed of the car (in km/h) is:
    A. 25
    B. 30
    C. 20
    D. 35
    E. None of these

    By traveling 12 km more at original speed, you save 9 minutes. To save 45 minutes that were overshot in the first case, you need to travel 60 km more at original speed. So, total distance is 60+18=78. Let original speed be 5x. The difference in time is because of the 60 km travel at 4x. So, 60/4x-60/5x=3/4 and x=4 and so, 5x=20.

    Two numbers b and C are chosen at random with replacement from the numbers 1, 2, 3,4,5,6,7,8,9. The probability that x^2 +bx+C>0 for all x€ R is

    Essentially, you have to have a square plus something surplus which will help us eliminate x from the picture.
    x^2+bx+c
    x^2+2.b/2.x+c
    x^2+2.b/2.x+(b^2)/4+c-(b^2)/4
    (x+b/2)^2+c-(b^2)/4
    If this has to be greater than 0, c-(b^2)/4 has to be greater than 0.
    4c > b^2 (which is nothing but the property for a quadratic equation so that it has real roots)
    c=1, b=1
    c=2, b=1,2
    c=3, b=1,2,3
    c=4, b=1,2,3
    c=5, b=1,2,3,4
    c=6, b=1,2,3,4
    c=7, b=1,2,3,4,5
    c=8, b=1,2,3,4,5
    c=9, b=1,2,3,4,5
    Total: 32 out of 81 cases so 32/81

    Find the number of +ve integers n in the range 12 < = n < = 40 such that the product (n-1)(n-2)(n-3).....3.2.1 i s not divisible by n.

    (n-1)! should not be divisible by n. Consider 12 and 11!. 11! will have a 3 and two 2s and so, will be divisible by 12. But 12! will not be divisible by 13 as 13 is prime. 13! will be divisible by 14, 14! by 15 and so on. So, all the prime numbers n will not be divisible by (n-1)! and so, the answer will be 7. This is nothing but an application of the Wilson's theorem which states that when any prime number p divides (p-1)!, it leaves a remainder of (p-1).

    If n is a positive prime number, and n^4 + 3(n^3) is a perfect cube, what is the sum of the lowest two possible values of n?

    n^3 * (n+3) is a perfect cube. So, (n+3) has to be a perfect cube as well. (n+3) can be equal to 1, 8, 27, 64, 125... Now, the trap here is that n is a positive prime number. The corresponding values of n in the above cases will be -2, 5, 24, 61, 122 and so on. The smallest two values of n that are positive and prime are 5 and 61. So, sum = 66.

    There are 2 distinct natural numbers 'p' and 'q'. Each number is first increased by 10 and then by the same percentage as it was increased for the first time respectively to get new numbers p1 and q1. Incidentally, p1 = q1 = 72. What is the value of |p-q|?
    a. 46
    b. 48
    c. 52
    d. 44

    p -> p+10 -> (p+10)*(1+10/p) -> (p+10)^2/p = 72
    p^2 + 20p + 100 = 72p
    p^2 - 52p + 100 = 0
    On factorization, p =2 or p = 50. So, one of the values will be that of p and the other will be q.
    |p-q| = 48.

    Find the remainder when 95! is divided by 101.

    99! mod 101 gives 1 as the remainder from Wilson's theorem.
    95! * 96 *97 * 98 * 99 mod 101 = 1
    95! * -5 * -4 * -3 * -2 mod 101 = 1
    95! * 120 mod 101 = 1
    95! * 19 mod 101 = 1
    Let 95! mod 101 be k
    19k mod 101 = 1
    19k = 101a + 1
    k = (101a + 1)/19
    a=3, k=16

    Let a secret three digit number be cba. If the sum of cab + bac + bca + abc + acb = 2536, what is c+b+a ?

    222(a+b+c)=2536+(cba)
    (a+b+c)>11
    Also, 2536 mod 222 is 94 and so, (cba) mod 222 needs to be 128. Possible values are 128, 350, 572, 794. Only 572 satisfies and so, c+b+a=14.

    If 15x+20y=375, then find the minimum value of (x^2+y^2), where x and y are natural numbers
    Looking for Cauchy Schwarz inequality here:
    (x^2+y^2)(15^2+20^2) >= (15x+20y)^2
    (x^2+y^2) >= 375^2/625
    (x^2+y^2) >= 225

    At what price the shopkeeper sell his article so that if he sells the article at the marked price he makes a profit of 25% and if he increased selling price by 250 he gets a 25 % profit on the selling price find the cost price?

    Marked price: 1.25x
    Increased selling price: 1.25x+250
    Profit: 0.25x+250
    Profit on selling price:
    (0.25x+250)/(1.25x+250)=25/100
    25x+25000=31.25x+6250
    6.25x=18750
    x=3000.

    The cost price of four articles A, B, C and D are a, b, c and d respectively. A, B, C and D are sold at profits of 10%, 20%, 30% and 40% respectively. If the net profit on the sale of these four articles is 25%, a, b, c and d cannot be in the ratio
    (a) 4 : 1 : 4 : 3
    (b) 1 : 2 : 2 : 1
    (c) 2 : 3 : 6 : 1
    (d) 5 : 2 : 7 : 3

    Substitute and check. First option, assume that the cost prices are 400, 100, 400, 300. So, profits will be 40, 20, 120, 120 and total profit will be 300 on total cost price of 1200. So, it is 25%. Similarly for option 2 and 3. The final option will give you profit of 420 on 1700 which is not 25%.


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