Quant Boosters - Nitin Gupta - Set 2


  • QA/DILR Mentor | Be Legend


    Q26) Four persons go to a birthday party. They leave their top-coats and hats in the lounge and pick them while returning back.The number of ways in which none of them picks up his own top-coat as well as his own hat is p. The number of ways in which exactly one of them picks up his own top-coat as well as his own hat is q. The number of ways in which a person picks up someone else’s top-coat and yet someone else’s hat is r. Then p+q+r is
    (a) 117
    (b) 104
    (c) 113
    (d) 108
    (e) none of the foregoing


  • QA/DILR Mentor | Be Legend


    The no. of ways in which the 1st person doesn't pick up his own top-coat is 3. The 2nd person (whose top-coat the 1st one picked)can pick someone else's top-coat in 3 more ways. For the 3rd and 4rth person we have just 1 choice. Thus, in all when none among the 4 pick his/her own top-coat is 9. Thus p = 9 * 9 = 81.
    A person among the 4 who picks his own top-coat as well as hat can be selected in 4 ways. The other 3 falls in the category [not own top-coat or not own hat] = not own top-coat + not-own hat - (not won top-coat and not own hat) = 2 * 6 + 2 * 6 - 2 * 2 = 20. Thus, q = 80.

    As explained in the 1st case, each person can pick someone else's top-coat in 9 ways. To each of these 9 ways we have 2 ways where he picks yet someone else's hat. Thus, r = 9 * 2 = 18
    Hence, choice (e) is the right answer.


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    Let the distance travelled with 40km/hr, 50km/hr, 60km/hr is A, B and C respectively.
    => A/40 + B/50 + C/60 = 3 and A/60 + B/50 + C/40 = 2.
    Eliminating A, we get 2B/5 + 5C/6 = 0 => B = C = 0 => A = 120 km
    Hence, choice (c) is the right answer.


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    Q27) Five containers each have one litre of 10%, 20%, 30%, 40%, 50% milk in them. By mixing the contents of just two of the containers, one litres each of w%, x%, y%, z% and 42% (22 < w < = x < = y < = z < 42) milk is prepared, contents of no two containers are mixed twice in different proportions to get different concentrations. Also each original container is used exactly twice in making new mixtures. If the containers are mixed only in the multiples of 100 ml then z + y - x - w = ?
    a) can not be determined
    (b) 4
    (c) 16
    (d) 20
    (e) none of the foregoing


  • QA/DILR Mentor | Be Legend


    Lets start with 42% milk preparation.

    10% milk can be mixed with 50% milk to produce 42% milk but on doing so we are left with 800 ml of 10% milk which cant give any possible combination with others to produce milk more than 22% milk. Only 30% milk in quantity 400 ml with 50% milk in 600 ml gives us a 42% combination.

    Now we are left with 600 ml of 30% and 400ml of 50%.Now since we are combining 600 and 400 ml of 2 hence all the solution have to be broken in two parts each of 400 ml and 600 ml.

    Proceeding further from here, we find only 1 valid combination so that all the values are satisfied in the given range.

    400ml of C + 600 ml of E ---> [120 + 300] ---> 420 out of 1000 ---> 42%
    400ml of E + 600 ml of A ---> [200 + 60 ] ---> 260 out of 1000 ---> 26%
    400ml of A + 600 ml of D ---> [40 + 240] ---> 280 out of 1000 ---> 28%
    600ml of B + 400 ml of D ---> [120+ 160] ---> 280 out of 1000 ---> 28%
    400ml of B + 600 ml of C ---> [80 + 180] ---> 260 out of 1000 ---> 26%

    Hence, z + y - x - w = 28 + 28 - 26 - 26 = 4
    Hence, choice (b) is the right answer.


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    Q28) Let a, b, c be distinct non-zero integers such that -5 < = a, b, c < = 5. How many solutions (a, b, c) does the equation 1/a + 1/b + 1/c = 1/(a+b+c) have?
    a) 72
    (b) 120
    (c) 192
    (d) 240
    (e) none of the foregoing


  • QA/DILR Mentor | Be Legend


    1/a + 1/b + 1/c = 1/(a+b+c) reduces to (a+b)(b+c)(c+a) = 0;
    If a + b = 0 then we can chose b in 10 ways (integers from -5 to 5 excluding 0) and c in 10 - 2 = 8 ways (since a, b and c are different). So total 80 ways
    Similarly, b + c = 0, and c + a = 0 gives 80 solutions each.
    So total 80 + 80 + 80 = 240
    Hence, choice (d) is the right answer.


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    Q29) A biologist catches a random sample of 60 fish from a lake, tags them and releases them. Six months later she catches a random sample of 70 fish and finds 3 are tagged. She assumes 25% of the fish in the lake on the earlier date have died or moved away and that 40% of the fish on the later date have arrived (or been born) since. What does she estimate as the number of fish in the lake on the earlier date?
    (a) 420
    (b) 560
    (c) 630
    (d) 720
    (e) 840


  • QA/DILR Mentor | Be Legend


    Let no. of Fishes on earlier day = x; 60 fishes are tagged, out of which 25% have died/moved remaining fishes that are tagged = 45.
    No. of fishes on day of second observation = 5/4x
    No. of fishes caught = 70
    No. of Fishes tagged = 3 = 4.2857% (3/70)
    Total No. fishes on Second day of Observation = 45/.042857 (45 * 70/3) = 1050
    No of Fishes on earlier day = x = 4/5 * 1050 = 840
    Hence, choice (e) is the correct option


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    Q30) Consider a regular polygon of p sides .The number of values of p for which the polygon will have angles whose values in degrees can be expressed in integers?
    (a) 24
    (b) 23
    (c) 22
    (d) 20
    (e) 21


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    Each angle is 180(p-2)/p.
    180-{360}/{p} = k
    So 360/p has to be an integer.
    360 = 2^3 * 3^2 * 5^1
    So there are 4 * 3 * 2 = 24 possibilities, but we exclude 1 and 2, because p > = 3
    So , 24 -2 = 22
    Hence, choice (c) is the right answer


 

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