Quant Boosters - Ravi Handa - Set 1



  • 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?



  • The total quantity of alloy before the addition of Nickel is 100 kg.
    This 100 kg is composed of 50 kg Copper, 30 kg Zinc, and 20 kg Nickel

    Now we want the new ratio to be 5:3:3 by adding some quantity of Nickel.
    You can just look at it and say that the quantity of Nickel that should be added is 10 kg.

    However, let us try and understand the method for complicated cases.
    Let us say that 'x' kg of Nickel is added so that Nickel becomes 20 + x and the total quantity of the alloy becomes 100 + x

    Now, Nickel is 3/(5 + 3 + 3) = 3/11 th of the total mixture.
    => (20 + x)/(100 + x) = 3/11
    => 220 + 11x = 300 + 3x
    => 8x = 80
    => x = 10
    => Quantity of Nickel that should be added is 10 kg.



  • Q14) A container has 100 liters (mixture of milk and water) in the ratio of 3:2. When 40 liters of mixture is taken out and replaced with the same amount of water, what is the ratio of milk and water left in the container?



  • Total solution in the mixture = 100 liters
    Quantity of milk = (3/5) * 100 = 60 liters
    Quantity of water = (2/5) * 100 = 40 liters

    When 40 liters of mixture is taken out, 40% of everything is taken out
    Quantity of milk = 60% of original = 0.6 * 60 = 36 liters
    Quantity of water = 60% of original = 0.6 * 40 = 24 liters

    Same quantity (40 liters) of water is added
    Quantity of water = 24 + 40 = 64 liters
    New Ratio of milk : water = 36 : 64 = 9 : 16



  • Q15) Walking at 5 kmph I missed my train by 7 min . Walking at 6 kmph I reached the station 5 min early . How far is the station from the house?


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