Quant Boosters - HandaKaFunda - Set 1


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    If a + b + c + d = 20, how many unique, non-negative integer solutions exist for (a,b,c,d)?

    This is like distributing 20 identical chocolates between 4 kids A, B, C, and D. You arrange these 20 chocolates in a line

    C C C C … C

    Now, you put in partitions in between them. Let me denote the partitions with P

    C C C P C P C C C C C P C C C C C C C C C C C

    A gets the chocolates before the first partition, B gets the chocolates between the first two partitions, C gets the chocolates between the second and the third partition and D gets the chocolate after the third partition. In the arrangement shown above A gets 3 chocolates, B gets 1, C gets 5, and D gets 11. Any rearrangement of the above, will lead to a new distribution of chocolates.

    The above can be rearranged in 23! / 20! 3! We got this because there are a total of 23 entities out of which 20 Cs are identical and 3 Ps are identical.

    Now, to extend this concept, what if we have to distribute ‘n’ chocolates in ‘r’ kids. After putting the n chocolates in a line, we would need r - 1 partitions. This would mean that there will be a total of n+r-1 entities out of which n would be identical (of one type) and the other r - 1 would be identical as well (of another type). So, the number of ways in which that would be possible = (n + r - 1)! / n! (r-1)!

    This, in other words, is (n + r - 1) C (r - 1)

    And now, we can use this formula to solve any similar questions

    a + b + c + d = 20

    Case 1: a, b, c, d are non-negative integers.

    Number of solutions = (20 + 4 - 1) C (4 - 1) = 23 C 3 = 1771

    Case 2: a, b, c, d are positive integers

    We allocate at least a value of 1 to a, b, c, d.

    So, we can say a = a’ + 1, b = b’ + 1, c = c’ + 1, d = d’ + 1 where a’, b’ c’, d’ are non-negative integers

    => a’ + 1 + b’ + 1 + c’ + 1 + d’ + 1 = 20

    => a’ + b’ + c’ + d’ = 16

    => Number of solutions = (16 + 4 - 1) C (4 - 1) = 19 C 3

    Case 3: a, b, c, d are non-negative integers such that a > 5 and b > 2

    We allocate at least a value of 5 to a and 2 to b

    So, a = a’ + 5 and b = b’ + 2

    => a’ + 5 + b’ + 2 + c + d = 20

    => a’ + b’ + c + d = 13

    => Number of solutions = (13 + 4 - 1) C ( 4 - 1) = 16 C 3

    Case 4: a, b, c, d are non-negative integer such that a > b

    Let us first consider the situation where a = b

    If a = b = 0, c + d = 20. This has 21 solutions

    If a = b = 1, c + d = 18. This has 19 solutions

    If a = b = 2, c + d = 16. This has 17 solutions

    .

    .

    If a = b = 10, c + d = 0. This has 1 solution

    So, the total number of a solutions when a = b is 21 + 19 + 17 … + 1 = 11/2*(21 + 1) = 121

    We know that the number of solutions when a, b, c, and d are non-negative integers is 1771. Out of these 1771 cases, in 121 cases a = b.

    So, in 1771 - 121 = 1650 cases a is not equal to b.

    In half of the above cases a will be greater than b whereas in the other half of the cases a will be less than b.

    So, number of solutions where a > b is 1650/2 = 825

    There is a escalator and 2 persons move down it.A takes 50 steps and B takes 75 steps while the escalator is moving down. Given that the time taken by A to take 1 step is equal to time taken by B to take 3 steps.Find the no. of steps in the escalator while it is stationary?

    Let us say that the escalator moves at the rate of n steps per second. Let us also say A takes 1 step per second and B takes 3 steps per second.

    Case 1: When A is coming down
    A will take 50 seconds to complete 50 steps.
    In 50 seconds, escalator would have moved 50n steps.
    Total number of steps on the stationary escalator = 50 + 50n

    Case 2: When B is coming down
    B will take 25 seconds to complete 75 steps
    In 25 seconds, escalator would have moved 25n steps.
    Total number of steps on the stationary escalator = 75 + 25n

    Total number of steps on the stationary escalator is a constant
    => 50 + 50n = 75 + 25n
    => 25n = 25
    => n = 1

    Total number of steps = 50 + 50 = 75 + 25 = 100

    Of 100 people, 86 ate eggs, 75 had bacon, 62 had toast and 82 had coffee. Minimum number who had all 4?

    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.

    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

    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

    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

    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.

    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

    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.

    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


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