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    Solutions To Math Puzzle-1

    Uploaded by

    Abhijit Kar Gupta
    The document contains solutions to 20 math puzzles or problems. It provides the question, reasoning, and step-by-step working to arrive at the solution. Key concepts covered include geometry, algebra, ratios, percentages, permutations, combinations and trigonometry.

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    Attribution Non-Commercial (BY-NC)
    Download as PDF, TXT or read online from Scribd

    Solutions To Math Puzzle-1

    Uploaded by

    Abhijit Kar Gupta
    1K views5 pages
    The document contains solutions to 20 math puzzles or problems. It provides the question, reasoning, and step-by-step working to arrive at the solution. Key concepts covered include geometry, algebra, ratios, percentages, permutations, combinations and trigonometry.

    Original Description:

    Solutions to the Math puzzle-1; a set of problems prepared to test the general mathematical aptitude of the secondary school students. The problem set can be found in my site at www.scribd.com. Comments are most welcome!

    Original Title

    Solutions to Math puzzle-1

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    The document contains solutions to 20 math puzzles or problems. It provides the question, reasoning, and step-by-step working to arrive at the solution. Key concepts covered include geometry, algebra, ratios, percentages, permutations, combinations and trigonometry.

    Copyright:

    Attribution Non-Commercial (BY-NC)
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    1K views5 pages

    Solutions To Math Puzzle-1

    Uploaded by

    Abhijit Kar Gupta
    The document contains solutions to 20 math puzzles or problems. It provides the question, reasoning, and step-by-step working to arrive at the solution. Key concepts covered include geometry, algebra, ratios, percentages, permutations, combinations and trigonometry.

    Copyright:

    Attribution Non-Commercial (BY-NC)
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    of 5

    1

    Solutions to Math Puzzle-1

    By- Dr. Abhijit Kar Gupta


    E-Mail: kg.abhi@gmail.com

    ---------------------------------------------------------------------------------------------------------

    1. Ans. (ii) 7
    This means you have to divide a circle by three straight lines. Many would answer (i) 6 .
    But after a little thought you find that if you intersect the three lines at three different
    points inside the circle you get 7 segments. This kind of a problem is to examine the
    sense of Geometry.

    2. Ans. (iv) none of these.


    The Answer is 5.
    Suppose the numbers are x and y .
    Therefore, 2 x =3 y
    x 3
    ⇒ =
    y 2
    Add 1 to both sides of the above, you get one equation. Subtract 1 from both sides of the
    above, you get another. Then divide the previous one by the later and you get the
    following.
    ( x + y ) (3 + 2)
    ⇒ =
    ( x − y ) (3 − 2)
    ⇒ ( x + y ) = 5( x − y )
    This problem is to examine the primary knowledge of Algebra and the elementary rules
    of Ratios.

    3. Ans. (ii) 25%


    This is a problem of percentage can be solved arithmetically.
    But applying Algebra will make work easier.
    Let x be the remaining over and y be the asking run rate per over.
    Now x reduces to x − x × 20% = x(1 − 1 / 5) = 4 x / 5 .
    Then y has to be increased by z to have ( y + z ) .
    2

    Therefore, we can write x × y = 4 x / 5 × ( y + z )


    Solving one gets z / y = 1 / 4 = 25% .

    4. Ans. (iii) 2
    Let the digits of the number are x and y .
    Therefore, (10 x + y ) − (10 y + x) = 18
    Solving, ( x − y ) = 2 .

    5. Ans. (iii) 64
    Follow each step carefully and observe the number of pieces goes by 2 n , where n is the
    number of steps executed. The progression follows a power law.
    Therefore, the result would be 2 6 = 64 .

    6. Ans. (iv) none of the above


    Drawing a diagram would make things clear.

    The diagram is self sufficient and one can easily see that the distance between the starting
    point and the end point is 32 + 1 .

    7. Ans. (iv) none


    This may appear as a silly question. However, the given number is quite big so that most
    often it tempts you to fall in the trap of wasting time unless you are alert. But the trick
    lies in the use of Algebra. You can assume that the number is x.
    Therefore, the steps follow:
    2( x + 1) − 2
    x → ( x + 1) → 2( x + 1) → 2( x + 1) − 2 → =x
    2
    So at the end you get the same number.

    8. Ans. (i) Two rectangular pieces of equal size.


    Suppose you take two squares of size a × a and b × b .
    Therefore, you have two square pieces a 2 and b 2 . If you have two more pieces of equal
    size a × b then the four pieces together make a 2 + b 2 + 2ab = (a + b) 2 , a square of size
    (a + b) × (a + b).

    9. Ans. (ii) 100


    Here the information of the population of size 1 billion is not required.
    Suppose the present population is x0 .
    3

    Therefore, the population after one year is x1 =x 0 (1 + 1 / 100)


    After 2 years it is x 2 = x 0 (1 + 1 / 100) 2 and so on.
    Like this the population after 100 years becomes x100 = x0 (1 + 1 / 100)100 .
    x 1
    Thus we have 100 = (1 + 1 / 100)100 ≅ 1 + × 100 = 2 .
    x0 100

    Note that the last approximation can be made in the binomial expression because
    1
    is much less than 1.
    100

    10. Ans. (iii) 14.


    Place 9 squares side by side and form a big square like the following:

    There are 4 number of squares of size 2 × 2 and 1 square of size 3× 3 .


    Therefore, the total number is 9+4+1=14 (= 3 2 + 2 2 + 1) .

    11. Ans. (iv) None of these


    A knock-out tournament is such where one team loses and goes out of tournament. The
    final team survives is the champion. Therefore, total number of matches played will be
    64 − 1 = 63. Try it out with a small number like 3 or 4 and check.

    12. Ans. (iii) b > a > c


    1 1

    You have to compare among three numbers : 3 , 5 and 2.


    2 3

    1 1
    Notice that the powers of the three numbers are , and 1 respectively. Therefore, to
    2 3
    compare properly one has to take HCF of 2 , 3 and 1 which is 6.
    1

    Hence, we check (3 2 ) 6 = 33 = 27,


    1

    (5 3 ) 6 = 5 2 = 25,
    2 6 = 64.

    13. Ans. (iii) 50 years


    Invoking Algebra would give you the answer quickly:
    5( x + 5) − 5( x − 5) = 50

    14. Ans. (iii) 1960


    4

    This is a elementary problem from the chapter of Permutations and Combinations. Note
    that from each station you can go to 30 other stations. Therefore, there are 31× 30
    connections. And if you consider the return journey tickets then the number is
    31 × 30 × 2 = 1960.

    15.Ans. (iv) none of these


    The biggest number you can construct is 5 to the power 5 to the power 5:
    5
    55

    16.Ans. (i) 9
    This is again a problem of permutation and combination. You can think of putting 2
    objects A and B into 3 Boxes which is 3 2 = 9.

    17.Ans. (iv) 24 km/h


    (20 + 30)
    The answer would definitely not be = 25. The reason is the train does not take
    2
    equal time up and down that is why you can not take simple arithmetic average like this.
    Suppose the average speed is x km/h and one way distance is y km.
    We can then write :
    2y y y
    = +
    x 20 30
    Solving one gets x = 24

    18.Ans. (iii) a square

    Here the information of the numerical value of the length of the wire (100 cm) is
    redundant which may only tempt one to manipulate numerically resulting wastage of
    time.
    Suppose we make a rectangle of length x and breadth y so that 2( x + y ) = l , where l is
    the length of the wire (here perimeter of the frame). This means
    l
    x + y = c(= ) =constant.
    2
    The area is A = xy = x(c − x) . Note that the area A would be maximum when you
    c l
    choose x = y = = and which is nothing but a square.
    2 4

    19. Ans. (iv) equal

    Ordinary algebraic manipulation would lead to confusion and wastage of time. Invoking
    ratios (and a simple trick) will make it easy.
    5

    Tapas Mrinal Arun Manju 1 1


    × × × = 2 × 3× × = 1
    Mrinal Arun Manju Kartik 2 3
    Tapas
    ∴ =1
    Kartik
    Therefore, Tapas and Kartik earn equal amount.

    20. Ans. (ii) 60 deg

    Diagram would make things clear.

    The area of the square is a 2 .


    π
    The height of the rhombus is h = a sin( − θ ) = a cos θ .
    2
    1
    The area of the Rhombus is 2 × × a × a cos θ = a 2 cos θ .
    2
    a2 1
    Therefore, it is clear that if θ = 60 deg, the area of the rhombus will be = × area of
    2 2
    square.

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