Quant Boosters - Sibanand Pattnaik - Set 1


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    Number of Questions - 30
    Topic - Quant Mixed Bag
    Solved ? - Yes
    Source - CAT Preparation With IIMpossible Forum.


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    Q1) Sum of 2 natural numbers & their LCM is 89. How many pairs of numbers satisfy this condition?
    a) 4
    b) 5
    c) 6
    d) 7


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    Let the HCF be H, numbers be HA and HB (A and B re co-prime), therefore LCM= HAB
    Product of numbers and LCM= HA x HB x HAB = H( A+B+AB) =89
    Therefore H=1 and ( A+B+AB)= 89. OR (A+1)(B+1)= 90. Therefore
    i) A= 8, B=9
    ii) A=44, B=1
    iii) A=29, B=2
    iv) A=4, B= 17
    v) A= 5, B=14
    Hence (b)


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    Q2) Difference between LCM & HCF of 2 Numbers is 27. How many pairs of numbers satisfy the condition?
    a) 3
    b) 4
    c) 5
    d) 6


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    Let the HCF and LCM be H and L and the numbers be Ha and Hb. Therefore
    Hab-H= 27 OR
    H(ab-1)=27
    Possibilities
    H ab-1 ab a,b Numbers
    1 27 28 28,1 or 7,4 28,1 or 7,4
    3 9 10 10,1 or 5,2 30,3 or 15,6
    9 3 4 4,1 36,9
    27 1 2 2,1 54,27
    Hence (d)


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    Q3) Sum of HCF and LCM of 2 numbers is 30. How many such pairs of numbers exist?
    a) 6
    b) 4
    c) 7
    d) 8


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    Let the HCF and LCM be H and L and the numbers be Ha and Hb. Therefore
    H+Hab= 30 OR
    H(1+ab)=30

    a and b are co prime to each other
    Possibilities
    H 1+ab ab a,b Numbers
    1 30 29 29, 1 29,1
    2 15 14 14,1 or 7,2 28,2 or 14,4
    3 10 9 9,1 27,3
    5 6 5 5,1 25,5
    6 5 4 4,1 24,6
    10 3 2 2,1 20,10
    15 2 1 15,15 15,15
    30 1 0 Not possible Not possible
    Hence (d)


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    Q4) How many numbers lie between 11 and 1111 which when divided by 9 leave a remainder of 6 and when divide by 21 leave a remainder of 12?


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    Let the possible number be N then it can be expressed as
    N=9k+6
    and N=21l+12
    ∴ 9k+6=21l+12
    ⟹ 9k-21l=6
    or (3k-7l)=6
    or 3k=7l+2 or k=(7l+2)/3
    So put the min. possible value of l such that the value of k an integer or in other words numerator (i.e.,7l+2) will be divisible by 3.
    Thus at l=1, we get k=3 (an integer). So the least possible number N=9×3+6=21×1+12=33.
    Now the higher possible values can be obtained by adding 33 in the multiples of LCM of 9 and 21 i.e., The general form of the number is 63m+33. So the other number in the given range including 33 are 96, 159, 222, 285, 348, …, 1104. Hence there are total 18 numbers which satisfy the given condition.


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    Q5) What is the least possible number which when divided by 13 leaves the remainders 3 and when it is divided by 5 it leaves the remainder 2.


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    Let the required number be N then it can be expressed as follows
    N=13k+3 ….(1)
    and N=5l+2 ….(2)
    where k and l are the quotients belong to the set of integers.
    Thus 5l+2=13k+3
    ⟹ 5l-13k=1
    ⟹ 5l=13k+1
    ⟹ l=(13k+1)/5
    Now we put the value of k such that numerator will be divisible by 5 or l must be integer so considering k=1,2,3,… we find that k=3, l become 8. So the number N=5×8+2=42
    Thus the least possible number = 42
    To get the higher numbers which satisfy the given conditions in the above problem we just add the multiples of the given divisors (i.e., 13 and 5) to the least possible number (i.e., 42)
    NOTE The next higher number =(Multiple of LCM of 13 and 5)+42
    =65m+42
    In the above problem what is the greatest possible number of 4 digits ?
    Since the general form of this number is 65m+42 so by putting m=153 we get (9945+42=)9987, which is required number.
    HINT To get the value of 9945 we can simply divide 9999 (which is the greatest 4 digit number) by 65 and then subtract the remainder from 9999.


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    Q6) Find the least possible 5 digit number which when divided by 2, 4, 6 and 8, it leaves the remainders 1, 3, 5 and 7 respectively


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    The possible value =(LCM of 2,4,6,8)m-1
    = 24m-1
    Since, the least 5 digit number is 10000. So the required number must be atleast 10000. So putting the value of m=417, we get (10008-1)=10007, which is the required number.


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    Q7) What is the least possible number which when divided by 18, 35 or 42 it leaves the 2,19, 26 as the remainders respectively ?


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    Since the difference between the divisors and the respective remainders is same.
    Hence the least possible number
    =(LCM of 18,35 and 42)-16
    =630-16 [∴(18-2)=(35-19)=(42-26)=16]
    =614
    What is the least possible number which when divided by 2, 3, 4, 5, 6 it leaves the remainders 1, 2, 3, 4, 5 respectively ?
    Since the difference is same as
    (2-1)=(3-2)=(4-3)=(5-4)(6-5)=1
    Hence the required number =(LCM of 2,3,4,5,6)-1
    =60-1=59
    In the above problem what is the least possible 3 digit number which is divisible by 11 ?
    Since the form of the number is 60m-1, where m=1,2,3,…but the number (60m-1) should also be divisible by 11 hence at m=9 the number becomes 539 which is also divisible by 11. Thus the required number = 539.


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    Q8) What is the least possible number which when divided by 21,25, 27 and 35 it leaves the remainder 2 in each ?


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    The least possible no.
    =(LCM of 21,25,27 and 35)m + 2
    =4725m+2
    =4727 (for the least possible value we take m=1)


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    Q9) What is the least possible number which when divided by 24, 32 or 42 in each it leaves the remainder 5?


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    Since we know that the LCM of 24, 32 and 42 is divisible by the given numbers. So the required number is
    =LCM of 24,32,42)+(5)=672+5=677
    Hence such a least possible number is 677.
    In the above question how may numbers are possible between 666 and 8888 ?
    Since the form of such a number is 672m +5, where m=1,2,3,…So, the first no.=672×1+5=677 and the highest possible number in the given range
    =672×13+5
    =8736+5=8741
    Thus the total numbers between 666 and 8888 are 13.


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    Q10) 399 and 414 when divided by N gives the same remainder in each case. How many values can N take?


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