Question Bank - Modern Math - Hemant Malhotra

  • Q18) In a colony, 100 persons visited U.S., 80 visited U.K. and 60 visited Canada. 100 visited at least two of the above three countries. 30 visited all the three countries. 50 visited none of the three countries. What is the number of persons in the colony?
    a) 180
    b) 210
    c) 160
    d) Cannot be determined.

  • Q19) If 100 items are distributed among X men and Y women (Y > X). Find the probability that the number of items received by men is odd

  • Q20) Some children were standing around a circle. It was observed that the number of distinct pairs in which children were standing was one fifth of the distinct pairs in which children were not standing side by side. Find the number of children

  • Imagine a n sided polygon inside circle
    and all children on vertices of that polygon so number of children =
    now number of diagonals in a polygon of side n
    =nc2 - n
    nc2 (means choose any two vertices and remove n sides )
    it's give n=(nc2-n)/5
    so 5n=nc2-n
    so 6n=nc2
    so n=13

  • Q21) All the possible 5-digit numbers are formed using the digits 1,2,3,4,5 - repetition is allowed. If one of those numbers is selected at random what is the probability that it will have exactly one digit repeated and that too occurring twice

  • Q22)

  • Q23) N students are seated at desks in an m x n array, where m, n >= 3. Each student shakes hands with the students who are adjacent horizontally, vertically or diagonally. If there are 81 handshakes, what is N?

  • students those are in middle will shake hands with 8 students
    those who are in corner will shake hands with 3 students
    and those who are in sides will shake hands with 5 students
    now number of handshakes students who are in corner then will be 4 corners so number of handshakes will be 4 * 3/2
    now students who are in middle (m-2) * (n-2) * 8/2
    becuase let m =5 and n =4 then students who are in middle will be=6 then handshakes wll be(5-2) * (4-2) * 8/2 = 48/2 = 24 same here if m and n then number of handhshakes will be (m-2) * (n-2) * 8/2

    now who are in sides of grid if 5 * 4 then number of studets who are in side will be 10 which is nothing but 2m-4+2n-4
    so number of handshakes wiill be (2m+2n-8) * 5/2
    so total handshakes =81

    Solving, you will get
    16 * m * n - 12m - 12n = 316
    m=4 and n=7
    so N = m * n = 28


  • Q24) There are ten 50 paise coins placed on a table. Six of these show tails and four show heads. A coin is chosen at random and flipped over (not tossed). This operation is performed seven times. One of these coins is then covered. Of the remaining nine coins, five show tails and four show heads. The covered coin shows:
    a) A head
    b) A tail
    c) More likely a head
    d) More likely a tail

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  • Q25) If m is the number of ways of arranging 5 people A, B, C, D and E in 5 seats and n is the number of words comprising of minimum one letter and maximum four letters that can be formed using the letters F, G and H, then what will be the value of |m – n|?
    a) 20
    b) 3005
    c) 39
    d) 0

  • @hemant_malhotra

    Let x students use all three of applications.
    Then,no of students using WhatsA and WeCh but not Line=12(given)
    No of students using WhatsA and Line but not WeCh= 6(given)
    If 22 students use Line and WeCh,then no of students using Line and WeCh but not WhatsA= 22 - x (from Venn Diagram)
    Similarly,we can find from Venn diagram no of students using:
    WhatsA only = 62 - x
    WeCh only = 32
    Line only = 26

    None of three applications= 16(given)

    Sum of all these would be 10x(given)

    So,(62 - x) +32 +26+{12 +(22- x) +6} + x +16 = 10x
    =>176 - x= 10x
    =>x= 16

    So,Total no of students in the class = 10*16 =160

    Answer- 160 students

  • Q26) In a class of 30 students, 12, 16, 15, 22 and 18 students have passed in Phy, Chem, Biology, Maths and English respectively. What is the maximum possible number of students who passed in at most two subjects.

  • Q27) The students of a class appeared for two exams physics and chemistry. In a class, 40% of the students failed in chemistry but 20% of the students passed only in chemistry. Ten students passed in both physics and chemistry. What is the maximum possible number of students, who passed in physics?
    (A) 10
    (B) 15
    (C) 20
    (D) 25

  • Q28) Three judges have to vote on three performers A, B, C and they have to mention their order of preference also. In how many ways they can vote so that two of them agree in their order of preference while third differs ?

  • Q29) A randomly selects two distinct numbers from the set { 1, 2, 3, 4, 5 }, and B randomly selects a number from the set { 1, 2, ..., 10 } . What is the probability that B's number is larger than the sum of the two numbers chosen by A

  • Q30) How many four-digit numbers, which are divisible by 15, are there such that the number 15 occurs in them?

  • The number must be divisible by 3 as well as 5.
    Case 1 : The number looks like (15 _ _ ) or (_15 _).
    a) The last digit is 5 then The other digit is 1, 4 or 7.
    b) The last digit is 0 then The other digit is 3, 6 or 9 (or 0 for 15 _ _ )
    so (3 + 4) ways for 15 _ _ and (3 + 3) ways for _ 15 _ so . 13 ways.

    Case2 - Now, if the number is like _ _ 15 then the first two digits are any two digit multiple of 3
    from 12, 15, 18 ...... 99,so 33 − 4 + 1 = 30 values of which 1 ,5 ,15 is already counted.

    so 13 + 30 − 1 = 42 values.

  • Q31) In Bhavan Vidhalaya , there are total 100 students and each student study at least one subject from HINDI , ENGLISH , MATHS and SCIENCE.

    Subject(s) : Number of Students details given below.
    HINDI : 54
    SCIENCE : 77
    MATH : 64
    ENGLISH : 50

    What could be the maximum number of student who studies all of the subjects?

    a) 47
    b) 48
    c) 49
    d) 50

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