Quant Boosters - Sagar Gupta - Set 2



  • [ 38c2 - 19 ] /2 = 342 , when x > y and x < y
    when x > = y
    342 + 19 = 361



  • Q2) In a group if 80% drink tea and 60% drink coffee,what is the maximum percentage of people drinking either tea /coffee but not both?



  • I + II = 100
    I + 2 II = 140
    II = 40
    I = 60



  • Q3) 1 unit of A is made by mixing 4 units of B and 5 units of C. 1 unit of B is made by mixing 1 unit of X, 4 units of Y and 1 unit of Z. 1 unit of C is made by mixing 2 units of X, 6 units of Y and 1 unit of Z. The weight of 1 unit each of X, Y and Z is 5 kgs, 3 kgs and 8 kgs respectively. What is the total weight of Y required to make 1400 kgs of A?
    (1) 630 kgs
    (2) 720 kgs
    (3) 690 kgs
    (4) 870 kgs
    (5) 570 kgs



  • A = 4B + 5C
    B = X + 4Y + Z
    C = 2X + 6Y + Z
    X = 5 , Y = 3 , Z = 8
    B = 5 + 12 + 8 =25
    C = 10 + 18 + 8 = 36
    A = 100 + 180 = 280
    280 of A with 46Y or 46*3 = 138
    690 kgs for 1400 kgs of A



  • Q4) Find HCF of 2^100 - 1, 2^120- 1 ?



  • HCF (2^a -1, 2^b -1) = 2^(HCF a, b) - 1
    = 2^20 - 1



  • Q5) Find last two digits of 15 * 37 * 63 * 51 * 97 * 17



  • 15 * 37 * 63 * 51 * 97 * 17 mod 100
    15 * 37 * (-37) * 51 * (-3) * 17 mod 100
    45 * 37^2 * 17^2 * 3 mod 100
    45 * 89 * 03 * 69
    35



  • Q6) A five digit number divisible by 3 is to be formed using numbers 0, 1,2,3,4,5 without repetition . The total number of ways in which this can be done is....
    a. 216
    b. 240
    c. 600
    d. 3125



  • 1 + 2 + 3 + 4 + 5 = 15
    -> 5! ways
    0 + 1 + 2 + 4 + 5 =12
    ->5! - [ 5! /5 ] =120 - 24 =96
    total 120 +96 =216



  • Q7) The total number of ways of selecting two numbers from the set {1,2,3,4,......,30}, so that their sum is divisible by 3, is
    a. 95
    b. 145
    c. 190
    d. None of the above.



  • 3k, 3k + 1, 3k + 2
    divisible by 3:
    case 1 : 3k + 3k
    10 * 9 = 90 ways
    case 2 : (3k+2) + (3k+1)
    10 * 10 = 100 ways
    total = 190 ways



  • Q8) Train X started from point A at 9:00 am with a speed 72km/hr towards station Y. 2 hrs after train Y starts from point B towards X at a speed of 90 km/hr. They cross each other at 1:30 pm. But owing to signal problems at 12:00 noon the speeds of both the trains is reduced by same quantity such that they now cross each other at 4:30 pm. Calculate the new speed of train that started from point A.



  • Train X starts at 9:00 am @72 km/hr
    at 11:00 am, it is 144 kms from A
    A...144 kms...A'......d...........B
    Now Y starts from B at 11:00am @90 km/hr
    A' and B is the position at 11:00 am
    :
    They should have met at 1:30pm
    72 * 2.5 + 90 * 2.5 = d =405 kms
    :
    But they meet at 4:30 pm
    from 11:00 to 12:00 , they travel at normal speed
    Distance reduced = 72+90=162
    Distance between them at 12:00 = 243 kms
    Now they reduce their speed by k
    (72-k) * 4.5 + (90-k) * 4.5 =243
    k=54
    72-54 =18km/hr



  • Q9) abcd is a 4 digit perfect square, then if each digit is increased by 3 the number is still a perfect square. Find abcd.



  • abcd = x^2
    1000a +100b +10c +d = x^2
    1000(a+3) +100(b+3) +10(c+3) +d+3 = y^2
    y^2-x^2=3000 +300 + 30 + 3
    y^2 - x^2 = 3 [ 1000 +100 +10 +1 ]
    y^2 - x^2 =3333
    3333=101*33
    y+x =101
    y-x =33
    y = 67
    x = 34
    x^2 =1156



  • Q10) What is the remainder when 72 to the power 7202 is divided by 625?



  • E[625]=500
    7202 mod 500 = 2
    72^2 mod 625 = 5184 mod 625 = 184



  • Q11) If a^3 + b^3 = 10 and a^2 + b^2 = 60 then find a + b?


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