Quant Boosters - Maneesh - Set 4



  • P(G/E) = P(GnE)/P(E)
    P(GnE) = 1/2 * 3/7 = 3/14
    P(E) = 1/2
    P(G/E) = 3 * 2/14 = 3/7



  • Q13) The average age of A ,B ,C ,D five years ago was 45 years. By including X, the present average age of all the five is 49 years.The present age of X is
    a) 64 years
    b) 48 years
    c) 45 years
    d) 40 years



  • Total Age 5 years ago = 180
    Total age now = 180 + 20 = 200
    X included,total age = 5 * 49 = 245
    Age of X = 45



  • Q14) There are 42 students in a hostel. Due to admission of 13 new students, the expenses of mess increase by Rs 31 per day while the average expenditure per head diminished by Rs 3. What was the original expenditure of the mess ?



  • Let average expenditure is A
    Total expenditure is 42 A
    After 13 students joined new expenditure = 42A + 31
    Total expenditure 55(A-3)
    55(A-3) = 42A+31
    55A-165 = 42A+31
    13A = 196
    A = 196/13
    original expenditure = 42A = 42 * 196/13 = 633.23



  • Q15) If we divide a given two-digit number by the product of its digits, we obtain 3 as quotient and 9 as a remainder . If we subtract the product of the digits constituting the number, from the square of the sum of its digits, we obtain the given number. Find the number



  • 10x+y = 3xy+9
    (x+y)^2-xy = 10x+y
    x^2+y^2+xy = 10x+y
    x^2+y^2+xy = 3xy+9
    x^2+y^2-2xy = 9
    (x-y)^2 = 9

    x-y = +3
    10y+30+y = 9+3y(y+3)
    11y+30 = 9+3y^2+9y
    3y^2-2y-21 = 0
    y = 2+-16/6 = 18/6 = 3
    x = 6

    or

    x - y = -3
    x = y-3
    11y-30 = 9+3y(y-3)
    11y-30 = 9+3y^2-9y
    3y^2-20y+39 = 0

    b^2- 4ac = 400- 468 = -68 hence no real roots

    hence x= 6, y =3
    number 63



  • Q16) How many numbers from 1000 to 10000 have their digits in AP or GP if we consider the digits from left to right?



  • If digits in AP ...possibilities - common difference 0, 1, 2, -1, -2, -3
    if digits in GP...possibilities - common ratio 1/2, 2

    common difference 0....9 possibilities
    common difference 1....6 possibilities
    common difference 2.....3 possibilities
    common difference -1.....7 possibilities
    common difference -2.....4 possibilities
    common difference -3......1 possibility

    common ratio 1/2 ......1 possibility
    common ratio 2 .....one possibility

    total 32



  • Q17) The product of three numbers in GP is 1000 and the sum of the products of the numbers taken in pairs is 350. Find out the greatest of three numbers.



  • a/r, a, ar
    a^3 = 1000, a = 10
    a^2/r + a^2 + a^2r = 350
    100/r + 100r = 250
    2r^2 - 5r + 2 = 0
    r = 1/2 , 2
    numbers 5, 10, 20
    Greatest is 20



  • Q18) How many three digit numbers increase by 495 when their digits are reversed?
    a) 4
    b) 20
    c) 40
    d) 100



  • 100x + 10y + z -100z - 10y - x = 495
    99x - 99z = 495
    x - z = 5
    (9,4)(8,3)(7,2)(6,1)
    y : 0 to 9 --> 10 values
    4 * 10 = 40



  • Q19) A boy tries to climb a 15m greased pole. Each attempt, the boy climbs 3 m and slips back 2 m. How many attempts will it take to climb 10m of the pole?



  • in one attempt net value of distance boy climb is 3–2 = 1m
    in 6 attempts boy climbs = 6 m
    in the 7th attempt, he climbs = 6+3–2 = 7 m
    in the eighth attempt he reaches 7+3 = 10 m
    Hence answer 8 attempts



  • Q20) What is the remainder when (32^32)^32 is divided by 9?



  • 32^2 mod 9 = 1024 mod 9 = -2mod 9
    (32^32)^32 mod 9 = (-2)^(16 * 32)mod 9 = 2^(512) mod 9
    Since 2^3 mod 9 = -1
    2^512 = 2^510 * 2^2
    2^510 mod 9 = 2^(3 * 170) mod 9 = -1^170 = 1
    2^2 mod 9 = 4 mod 9 = 4
    2^512 mod 9 = 4 * 1 = 4
    Hence answer 4



  • Q21) If (a + 1/a)^2 = 3 then (a^3 + 1/a^3) = ?



  • a+1/a = root3
    (a+1/a)^3 = 3root3
    a^3 + 1/a^3 + 3(a + 1/a) = 3 root3
    a^3 + 1/a^3 + 3root3 = 3root 3
    a^3 + 1/a^3 = 0



  • Q22) If a and b are two-digit prime numbers such that a^2 - b^2 = 2a + 14b + 48. Find the largest possible value of a + b.


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