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Modern Maths Test - 1
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Modern Maths Test - 1
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  • Question 1/10
    1 / -0.25

    Find 8P6

    Solutions

     

  • Question 2/10
    1 / -0.25

    Find 8C6

    Solutions

     

  • Question 3/10
    1 / -0.25

    Find the number of ways in which the letters of the word BIHAR can be arranged.

    Solutions

    Since there are 5 letters in the word so the total number of arrangement is 5P= 5! that is 120.

     

  • Question 4/10
    1 / -0.25

    Find the number of ways in which the letters of the word AMERICA can be rearranged.

    Solutions

    In the word AMERICA there are 7 letters and the letter ‘A’ is coming twice.

    So the total number of rearrangement

    Note: No. of Rearrangements = Total no. of arrangements – 1(initial arrangement)

     

  • Question 5/10
    1 / -0.25

    Find the number of ways in which the letters of the word CALCUTTA can be rearranged.

    Solutions

    In the word CALCUTTA, there are 8 letters

    Letter C comes twice

    Letter A comes twice

    Letter T comes twice

    So the total number of rearrangements

    Note: No. of Rearrangements = Total no. of arrangements – 1 (initial arrangement)

     

  • Question 6/10
    1 / -0.25

    How many 4-letter different words can be formed by using the letters a, b, c and d?

    Solutions

    Letter a can be placed in all 4 positions.

    Similarly, b can be placed in all 4 positions.

    Similarly, c can be placed in all 4 positions.

    Similarly, d can be placed in all 4 positions.

    So the total number of arrangement is 4 × 4 × 4 × 4 = 256

     

  • Question 7/10
    1 / -0.25

    How many numbers can be formed by using the digits 2, 3, 4 and 5?(repetitions of digits is not allowed).

    Solutions

    The number formed can be one-digit, two-digit, three-digit or four-digit number

    Number of one-digit numbers formed = 4

    Hence, total numbers formed = 4 + 12 + 24 + 24 = 64

     

  • Question 8/10
    1 / -0.25

    How many numbers greater than 4000 can be formed by using the digits 2, 3, 4 and 5? (repetitions of digits is not allowed).

    Solutions

    Thousand’s place can be filled by using two digits i.e. 4 or 5

    Hundred’s place can be filled in 3 ways.

    Ten’s place can’t be filled in 2 ways.

    Unit’s place can be filled in 1 way.

    So total number of possible numbers = 2 × 3 × 2 × 1 = 12.

     

  • Question 9/10
    1 / -0.25

    How many numbers greater than 4000 can be formed by using the digits 2, 3, 4 and 5?

    Solutions

    Thousands place can be filled in two ways.

    Hundred’s place can be filled with any of the 4 digits.

    Ten’s place can be filled with any of the 4 digits.

    Unit’s place can be filled with any of the 4 digits.

    So the total numbers formed = 2 × 4 × 4 × 4 = 128.

     

  • Question 10/10
    1 / -0.25

    There are 3 roads from A to B, 4 roads from B to C and only 1 road from C to D. How many combinations of roads are there from A to D?

    Solutions

     

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