Quant Boosters - Ravi Handa - Set 1



  • 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?



  • Let us say that the actual distance between the house and station is 'd' km and the ideal time to cover the distance is 't' hours.

    When I walk at 5 kmph, I reach 7 minutes (or 7/60 hours) late
    => d/5 = t + 7/60
    => t = d/5 - 7/60

    When I walks at 6 kmph, he reaches 5 minutes (or 1/12 hours) early
    => d/6 = t - 1/12
    => t = d/6 + 1/12

    Using the above equations, we can say
    d/5 - 7/60 = d/6 + 1/12
    => d/5 - d/6 = 7/60 + 1/12
    => d/30 = (7 + 5)/60
    => d = 30*(12/60) = 6 km



  • Q16) The price of a gold nugget is directly proportional to the square of its weight. If a person breaks down the gold nugget in the ratio of 3:2:1 and gets a loss of Rs. 4620, what is the initial price of the gold nugget?



  • We are given that the price of gold is directly proportional to square of its weight.
    => Price = k*weight^2

    Let us say initially the weight of the nugget was 6 units
    => P(original) = k*6^2 = 36k

    After breaking it down into three pieces, we get the weights as 3, 2, and 1
    => P(3) = k * 3^2 = 9k
    => P(2) = k * 2^2 = 4k
    => P(1) = k * 1^2 = k
    => P(new) = 9k + 4k + k = 14k

    Loss = 36k - 14k = 4620 Rs.
    => 22k = 4620 Rs.
    => k = 210
    => Price (original) = 210 * 36 = 7560 Rs.



  • Q17) How can I divide 25000 coins of Rs. 1 each into 15 buckets such that I can obtain any amount just by picking up buckets and without moving the coins?



  • In questions like these the idea is to move in powers of 2 as long as required.

    If the total number of coins was 15, we will first find out the power of 2 it is lesser than
    => 15 = 2^4 - 1
    => We will need 4 buckets with the distribution as 1, 2, 4, 8

    If the total number of coins was 100, we will first find out the power of 2 it is lesser than
    => 100 = 2^7 - 28
    => We will need 7 buckets with the distribution as 1, 2, 4, 8, 16, 32, 37

    Here the total number of coins is 25000
    We know that the number of buckets is 15
    The distribution will be
    1, 2, 4, 8, .... 2^13, and 25000 - (2^14 - 1) = 8617



  • Q18) How many ways are there for 6 men and 7 women to stand in line so that none of the women are next to each other?



  • First, let us arrange the men. We can do that in 6! ways.
    Let us say they are A, B, C, D, E, and F
    The arrangement that we have made is:
    BCFEAD
    Now, we need to place women along with the men so that no two women are together. They can go on the positions indicated by blanks
    _ B _ C _ F _ E _ A _ D _
    7 women can be arranged on 7 positions in 7! ways.
    Total ways = 6!*7!



  • Q19) Which is greater, 70^71 or 71^70?



  • 70^71 (the one with the higher power)
    As a rule, if you are given two natural numbers a > b > 1
    then the one with the higher power will be bigger
    => a^b < b^a
    The exceptions to this are
    a) 3^2 > 2^3
    b) 2^4 = 4^2



  • Q20) What is the remainder when 97! is divided by 101?


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