Quant Boosters - Sagar Gupta - Set 2



  • 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?



  • (a+b)^3 = 10 + 3ab (a+b)
    (a+b)^2 = 60 + 2ab
    a+b = p
    (p^3 - 10 ) / 3p = ab
    p^2 = 60 + 2 [ (p^3 - 10 ) / 3p ]
    3p^3=180p + 2p^3 - 20
    p^3 +20 = 180p
    p=13.36



  • Q12) A and B can build a wall in 4 days and 6 days respectively working alone, while C and D can destroy the wall in 8 days and 12 days respectively working alone. If three of them start working together and the wall was built in 24 days. Which of the four persons did not work?



  • Total work =36 units
    A -> +9 units/day
    B -> +6 units/day
    C- > -4.5 units/day
    D- > -3 units/day
    24 (x+y+z) = 36
    x+y+z =1.5
    9-4.5-3 =1.5
    Hence B did not work



  • Q13) Three persons A, B and C rent the grazing of a park for Rs. 570. A put 126 oxen in the park for 3 months, B puts in 162 oxen for 5 months and C puts in 216 oxen for 4 months. What part of the rent should each person pay?
    a. 105, 220, 245
    b. 125, 205, 245
    c. 105, 225, 240
    d. data inadequate
    d. None of these


Log in to reply