Quant Boosters - Ravi Handa - Set 1



  • We are given that there are 100 people. Let us call them P1, P2, P3 ... P100

    Number of people who ate Egg = 86

    => Those who didn't = 100 - 86 = 14

    => 14 people did not eat eggs.

    Let us number those P1, P2, P3 ... P14

    Similarly, 25 people did not eat bacon

    Let us number those P15, P16, P17 ... P39

    Similarly, 38 did not have toast

    Let us number those people P40, P41, P42 ... P77

    Similarly, 18 people did not have coffee

    Let us number those people P78, P79, P80 ... P95

    Now, we are left with 5 people (P96, P97, P98, P99, P100) who ate / drank all of the four dishes / drink involved. This is the minimum possible configuration.



  • Q4) Find the last 2 digits of (123)^123!?



  • For finding out the last two digits of an odd number raised to a power, we should first try and reduce the base to a number ending in 1.
    After that, we can use the property
    Last two digits of (...a1)^(...b) will be [Last digit of a * b]1
    Let us try and apply this concept in the given question
    Last two digits of 123^123!
    = Last two digits of 23^123!
    = Last two digits of (23^4)^(123!/4)
    = Last two digits of (529^2)^(123!/4)
    = Last two digits of (...41)^(a large number ending in a lot of zeroes)
    = Last two digits of (...01) {Here I have used the concept mentioned above}
    = 01



  • Q5) How many 6 digit numbers can be made using digits 1, 2, 3, 4, 5, and 6 without repetition such that the hundred digit is greater than the ten digit, and the ten is greater than the one digit?



  • Let us assume that the number that we have is abcdef
    The condition given is d > e > f

    Let us first select three digits for a, b, and c.
    From the given 6 digits (1,2,3,4,5,6), 3 digits for a,b, and c can be selected in 6C3 = 6!/3!3! = 20 ways.
    These three selected digits can be arranged on three positions of a, b, and c in 3! = 6 ways

    The remaining 3 digits are automatically selected for d, e, and f.
    So, we have 1 way of selecting them.

    The biggest digit will be allocated to 'd'.
    The second biggest digit will be allocated to 'e'.
    The third biggest digit will be allocated to 'f'
    So, we have 1 way of arranging them on the three positions of d, e, and f.

    Total ways = 20 * 6 * 1 * 1 = 120



  • Q6) What is the units digit of 2! + 4! + 6! ... + 98!?



  • We need to find out the units digits of
    2! + 4! + 6! + .... 98!
    Let us look at the units digit for all of them
    2! = 1 * 2 = 2; unit's digit is 2
    4! = 1 * 2 * 3 * 4 = 24; unit's digit is 4
    6! = 1 * 2 * 3 * 4 * 5 * 6 = 720; unit's digit is 0
    8! = multiple of 6!, divisible by 10; unit's digit is 0
    10! = multiple of 6!, divisible by 10; unit's digit is 0
    12! = multiple of 6!, divisible by 10; unit's digit is 0
    .
    .
    .
    98! = multiple of 6!, divisible by 10; unit's digit is 0

    Unit's digit of 2! + 4! + 6! + .... 98!
    = Sum of unit's digits of individual terms
    = 2 + 4 + 0 + 0 + 0 ..... 0
    = 6



  • Q7) A team has a food stock for N days. After 20 days 1/4th of the team quits and the food stock lasts for another N days. What is the value of N?



  • Let us say that there were '4x' members in the team.
    Total amount of food = N * 4x = 4Nx
    Food consumed in 20 days = 20 * 4x = 80x
    Food left after 20 days = 4Nx - 80x
    The remaining food was consumed by 3x men in N days
    => Food left was = N * 3x = 3Nx
    => 4Nx - 80x = 3Nx
    => Nx = 80x
    => N = 80 days



  • Q8) What is the sum of the series, 1 * (2)^1 + 2 * (2)^2 + 3 * (2)^3 + ... + 100 * (2)^100?



  • S(1) = 1.(2)^1 = 2
    S(2) = 1.(2)^1 + 2.(2)^2 = 2 + 8 = 10
    S(3) = 1.(2)^1 + 2.(2)^2 + 3.(2)^3 = 2 + 8 + 24 = 34 = 2 + 2.(2)^4
    S(4) = 1.(2)^1 + 2.(2)^2 + 3.(2)^3 + 4.(2)^4 = 2 + 8 + 24 + 64 = 98 = 2 + 3.(2)^5
    and so on.
    So, S(n) = 2 + (n-1).2^(n+1)
    So, S(100) = 2 + 99.2^101



  • Q9) Suppose you like a book. It is available on the Flipkart website at a 30% discount and on the Amazon website at a 20% discount. If you buy it via the mobile app, Flipkart offers an additional 20% discount on the reduced price whereas the Amazon app offers an additional 30% discount. So, which app you should use to buy?



  • Let us say the book costs 100 Rs.

    Via Flipkart: Website offers a 30% discount bringing down the price to 70 Rs. The flipkart app offers another 20% discount ( on 70 Rs.) i.e. a discount of 14 Rs. This reduces the price to 56 Rs.

    Via Amazon: Website offers a 20% discount bringing down the price to 80 Rs. The Amazon app offers another 30% discount (on 80 Rs.) i.e. a discount of 24 Rs. This reduces the price to 56 Rs.

    As you can see, both of them are offering it to you at the same price. So go ahead, buy it from Amazon - because it is better.



  • Q10) What is the common ratio in a geometric progression in which: the first term is 7; the last term is 448; and the sum of terms is 889?



  • Let us consider the 'n' terms in the Geometric Progression as a, ar, ar^2 ... ar^(n-1)
    In this question, we know that the first term is 7
    => a = 7
    The last term in the series is 448
    => ar^(n-1) = 448
    => r^(n-1) = 64
    Now, this could be 2^6 or 4^3 or 8^2
    To figure this part out, we also have one more information that the sum of the series is 889
    => a(r^n - 1)/(r - 1) = 889
    Now, we can solve the above equations to get the answer. Or we can just assume some values and solve this out
    Let us assume that the common ratio, r = 2
    => Sum of the GP = 889
    => 7 + 14 + 28 + 56 + 112 + 224 + 448 = 889
    This fits. So, the common ratio is 2



  • Q11) If x + 1/x = 3 then x^5 + 1/x^5 = ?



  • We are given x + 1/x = 3

    The trick to solving these questions is to find out higher powers one by one.
    Basically, we should try and find out the values of x^n + 1/x^n

    When n = 2
    To get the value of n = 2, we need to square the original equation
    (x + 1/x) = 3
    => (x + 1/x)^2 = 3^2
    => x^2 + 1/x^2 + 2 = 9
    => x^2 + 1/x^2 = 7

    When n = 3
    To find the value, we need to multiply the equations where n = 1 and n = 2
    (x + 1/x) * (x^2 + 1/x^2) = 3*7
    => x^3 + 1/x^3 + x + 1/x = 21
    => x^3 + 1/x^3 + 3 = 21
    => x^3 + 1/x^3 = 18

    When n = 4
    To find the value, we can square the equation we got at n = 2
    (x^2 + 1/x^2)^2 = 7^2
    => x^4 + 1/x^4 + 2 = 49
    => x^4 + 1/x^4 = 47

    When n = 5
    To find the value, we can multiply the equations we got at n = 1 and n = 4
    (x + 1/x) * (x^4 + 1/x^4) = 3*47
    => x^5 + 1/x^5 + x^3 + 1/x^3 = 141
    => x^5 + 1/x^5 + 18 = 141
    => x^5 + 1/x^5 = 123

    So, the value we were trying to find out was 123

    Alternatively, you can just multiply the values that you have got at n = 2 and 3 to get the answer.

    (x^2 + 1/x^2) * (x^3 + 1/x^3) = 7*18
    => x^5 + 1/x^5 + x + 1/x = 126
    => x^5 + 1/x^5 +3 = 126
    => x^5 + 1/x^5 = 123



  • Q12) If 5 letters are posted for 5 different addresses, how many ways are there for each of the letters to reach wrong addresses?



  • Number of ways in which 'n' objects can be placed on 'n' positions in such a manner that none of them is correct is given by the Dearrangement formula.

    Dearr(n) = n!(1/0! - 1/1! + 1/2! - 1/3!.... 1/n!)

    In this question, we need to place 5 objects (letters) in 5 positions (addresses) such that none of them is correct. This can be done in

    Dearr(5) = 5! (1/0! - 1/1! + 1/2! - 1/3! + 1/4! - 1/5!)
    => Dearr(5) = 120 (1 - 1 + 1/2 - 1/6 + 1/24 - 1/120)
    => Dearr(5) = 60 - 20 + 5 - 1 = 44

    So, we can send the 5 letters, such that all are delivered at wrong adresses, in 44 different ways.



  • Q13) An alloy contains copper, zinc and nickel in the ratio of 5 : 3 : 2. The quantity of nickel in kg that must be added to 100 kg of this alloy to have the new ratio 5 : 3 : 3 is?


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