Maths-
General
Easy

Question

For x element of R, let [x] denote the greatest integer less or equal than x, then value ofopen square brackets negative fraction numerator 1 over denominator 3 end fraction close square brackets+open square brackets negative fraction numerator 1 over denominator 3 end fraction minus fraction numerator 1 over denominator 100 end fraction close square brackets+ open square brackets negative fraction numerator 1 over denominator 3 end fraction minus fraction numerator 2 over denominator 100 end fraction close square brackets+…+open square brackets negative fraction numerator 1 over denominator 3 end fraction minus fraction numerator 99 over denominator 100 end fraction close square bracketsis -

  1. –100    
  2. –123    
  3. –135    
  4. –153    

The correct answer is: –135


    For 0 less or equal thanless or equal than 66, 0 less or equal than fraction numerator r over denominator 100 end fraction< fraction numerator 2 over denominator 3 end fraction
    rightwards double arrowfraction numerator 2 over denominator 3 end fraction < – fraction numerator r over denominator 100 end fraction less or equal than 0
    rightwards double arrowfraction numerator 1 over denominator 3 end fractionfraction numerator 2 over denominator 3 end fraction< –fraction numerator 1 over denominator 3 end fractionfraction numerator r over denominator 100 end fraction less or equal thanfraction numerator 1 over denominator 3 end fraction
    thereforeopen square brackets negative fraction numerator 1 over denominator 3 end fraction minus fraction numerator r over denominator 100 end fraction close square brackets= –1 for 0 less or equal thanless or equal than 66
    Also, for 67 less or equal thanless or equal than 100, fraction numerator 67 over denominator 100 end fractionless or equal than fraction numerator r over denominator 100 end fraction less or equal than 1
    rightwards double arrow–1 less or equal thanfraction numerator r over denominator 100 end fraction less or equal thanfraction numerator 67 over denominator 100 end fraction
    rightwards double arrowfraction numerator 1 over denominator 3 end fraction – 1 less or equal thanfraction numerator 1 over denominator 3 end fractionfraction numerator r over denominator 100 end fraction less or equal thanfraction numerator 1 over denominator 3 end fractionfraction numerator 67 over denominator 100 end fraction
    thereforeopen square brackets negative fraction numerator 1 over denominator 3 end fraction minus fraction numerator r over denominator 100 end fraction close square brackets= –2 for 67 less or equal thanless or equal than 100
    Hence, not stretchy sum subscript r equals 0 end subscript superscript 100 end superscript open square brackets negative fraction numerator 1 over denominator 3 end fraction minus fraction numerator r over denominator 100 end fraction close square brackets= 67(–1) + 2(–34) = –135.

    Book A Free Demo

    +91

    Grade*

    Related Questions to study

    General
    maths-

    The number of integral solutions of the equation x + y + z = 24 subjected to conditions that 1 less or equal thanless or equal than 5, 12 less or equal thanless or equal than 18, z less or equal than –1

    Let t = z + 1
    Equation reduces to x + y + t = 25
    less or equal thanless or equal than 5, 12 less or equal thanless or equal than 18, t greater or equal than 0
    Required number of ways
    = Coefficient of x25 in [(x + x2 + x3 + x4 + x5) (x12 + x13 +…..+ x18) (1 + x + x2 + …..)]
    = Coefficient of x12 in (1 + x + x2 + x3 + x4) (1 + x + x2 + x3 + x4 + x5 + x6) (1 + x + x2 + …..)
    = Coefficient of x12 in fraction numerator left parenthesis 1 minus x to the power of 6 end exponent right parenthesis over denominator left parenthesis 1 minus x right parenthesis end fraction (1 – x)–1
    = Coefficient of x12 in (1 – x5) (1 – x6) (1 – x)–3
    = Coefficient of x12 in (1 – x5 – x6 + x11) (1 – x)–3
    = 12+3–1C3–17+3–1C3–16+3–1C3–1 + 1+3–1C3–1
    = 14C29C28C2 + 3C2
    = 91 – 36 – 28 + 3 = 30.

    The number of integral solutions of the equation x + y + z = 24 subjected to conditions that 1 less or equal thanless or equal than 5, 12 less or equal thanless or equal than 18, z less or equal than –1

    maths-General
    Let t = z + 1
    Equation reduces to x + y + t = 25
    less or equal thanless or equal than 5, 12 less or equal thanless or equal than 18, t greater or equal than 0
    Required number of ways
    = Coefficient of x25 in [(x + x2 + x3 + x4 + x5) (x12 + x13 +…..+ x18) (1 + x + x2 + …..)]
    = Coefficient of x12 in (1 + x + x2 + x3 + x4) (1 + x + x2 + x3 + x4 + x5 + x6) (1 + x + x2 + …..)
    = Coefficient of x12 in fraction numerator left parenthesis 1 minus x to the power of 6 end exponent right parenthesis over denominator left parenthesis 1 minus x right parenthesis end fraction (1 – x)–1
    = Coefficient of x12 in (1 – x5) (1 – x6) (1 – x)–3
    = Coefficient of x12 in (1 – x5 – x6 + x11) (1 – x)–3
    = 12+3–1C3–17+3–1C3–16+3–1C3–1 + 1+3–1C3–1
    = 14C29C28C2 + 3C2
    = 91 – 36 – 28 + 3 = 30.
    General
    maths-

    Ravish write letters to his five friends and address the corresponding envelopes. In how many ways can the letters be placed in the envelopes so that at least two of them are in the wrong envelopes -

    Required number of ways not stretchy sum subscript r equals 2 end subscript superscript 5 end superscript 5 C subscript 5 minus r end subscript D(r)
    equals sum from r equals 2 to 5 of   fraction numerator 5 factorial over denominator r factorial left parenthesis 5 minus r right parenthesis factorial end fraction r factorial times open curly brackets 1 minus fraction numerator 1 over denominator 1 factorial end fraction plus fraction numerator 1 over denominator 2 factorial end fraction minus fraction numerator 1 over denominator 3 factorial end fraction plus horizontal ellipsis plus fraction numerator left parenthesis negative 1 right parenthesis squared over denominator r factorial end fraction close curly brackets
    equals sum from r equals 2 to 5 of   fraction numerator 5 factorial over denominator left parenthesis 5 minus r right parenthesis factorial end fraction=open curly brackets fraction numerator 1 over denominator 2 factorial end fraction – fraction numerator 1 over denominator 3 factorial end fraction plus.... plus fraction numerator left parenthesis negative 1 right parenthesis to the power of r end exponent over denominator r factorial end fraction close curly brackets
    = 10 + 20 + (60 – 20 + 5) + (60 – 20 + 5 – 1)
    = 10 + 20 + 45 + 44
    = 119.

    Ravish write letters to his five friends and address the corresponding envelopes. In how many ways can the letters be placed in the envelopes so that at least two of them are in the wrong envelopes -

    maths-General
    Required number of ways not stretchy sum subscript r equals 2 end subscript superscript 5 end superscript 5 C subscript 5 minus r end subscript D(r)
    equals sum from r equals 2 to 5 of   fraction numerator 5 factorial over denominator r factorial left parenthesis 5 minus r right parenthesis factorial end fraction r factorial times open curly brackets 1 minus fraction numerator 1 over denominator 1 factorial end fraction plus fraction numerator 1 over denominator 2 factorial end fraction minus fraction numerator 1 over denominator 3 factorial end fraction plus horizontal ellipsis plus fraction numerator left parenthesis negative 1 right parenthesis squared over denominator r factorial end fraction close curly brackets
    equals sum from r equals 2 to 5 of   fraction numerator 5 factorial over denominator left parenthesis 5 minus r right parenthesis factorial end fraction=open curly brackets fraction numerator 1 over denominator 2 factorial end fraction – fraction numerator 1 over denominator 3 factorial end fraction plus.... plus fraction numerator left parenthesis negative 1 right parenthesis to the power of r end exponent over denominator r factorial end fraction close curly brackets
    = 10 + 20 + (60 – 20 + 5) + (60 – 20 + 5 – 1)
    = 10 + 20 + 45 + 44
    = 119.
    General
    maths-

    In how many ways can we get a sum of at most 17 by throwing six distinct dice -

    x1 + x2 + x3 + x4 + x5 + x6  17
    When 1 less or equal than xi less or equal than 6, i = 1, 2, 3, …..6
    Let x7 be a variable such that
    x1 + x2 + x3 + x4 + x5 + x6 + x7 = 17
    Clearly x7 greater or equal than 0 Required number of ways
    = Coefficient of x17 in (x1 + x2 + ….. + x6)6
    (1 + x + x2 + …..)
    = Coefficient of x11 in open parentheses fraction numerator 1 minus x to the power of 6 end exponent over denominator 1 minus x end fraction close parentheses to the power of 6 end exponent open parentheses fraction numerator 1 over denominator 1 minus x end fraction close parentheses
    = Coefficient of x11 in (1– 6C1 x6 + 6C2 x12……) (1 – x)–7
    = Coefficient of x11 in (1 – x)–76C1 × coefficient of x5 in (1 – x)–7
    = 11+7–1C7–16C1 × 7+5–1C7–1
    = 17C6 – 6 × 11C6 = 9604.

    In how many ways can we get a sum of at most 17 by throwing six distinct dice -

    maths-General
    x1 + x2 + x3 + x4 + x5 + x6  17
    When 1 less or equal than xi less or equal than 6, i = 1, 2, 3, …..6
    Let x7 be a variable such that
    x1 + x2 + x3 + x4 + x5 + x6 + x7 = 17
    Clearly x7 greater or equal than 0 Required number of ways
    = Coefficient of x17 in (x1 + x2 + ….. + x6)6
    (1 + x + x2 + …..)
    = Coefficient of x11 in open parentheses fraction numerator 1 minus x to the power of 6 end exponent over denominator 1 minus x end fraction close parentheses to the power of 6 end exponent open parentheses fraction numerator 1 over denominator 1 minus x end fraction close parentheses
    = Coefficient of x11 in (1– 6C1 x6 + 6C2 x12……) (1 – x)–7
    = Coefficient of x11 in (1 – x)–76C1 × coefficient of x5 in (1 – x)–7
    = 11+7–1C7–16C1 × 7+5–1C7–1
    = 17C6 – 6 × 11C6 = 9604.
    General
    maths-

    The number of non negative integral solutions of equation 3x + y + z = 24

    3x + y + z = 24, x greater or equal than 0, y greater or equal than 0, z greater or equal than 0
    Let x = k rightwards double arrow y + z = 24 – 3k …(1)
    rightwards double arrow 24 – 3k greater or equal thanrightwards double arrowless or equal than 8
    rightwards double arrow 0 less or equal thanless or equal than 8
    For fixed value of k the number of solutions of (1) is
    24–3k+2–1C2–1
    = 25–3kC1
    = 25 – 3k
    Hence number of solutions
    not stretchy sum subscript k equals 0 end subscript superscript 8 end superscript left parenthesis 25 minus 3 k right parenthesis= 25 × 9 – fraction numerator 3 cross times 8 cross times 9 over denominator 2 end fraction= 225 – 108 = 117.

    The number of non negative integral solutions of equation 3x + y + z = 24

    maths-General
    3x + y + z = 24, x greater or equal than 0, y greater or equal than 0, z greater or equal than 0
    Let x = k rightwards double arrow y + z = 24 – 3k …(1)
    rightwards double arrow 24 – 3k greater or equal thanrightwards double arrowless or equal than 8
    rightwards double arrow 0 less or equal thanless or equal than 8
    For fixed value of k the number of solutions of (1) is
    24–3k+2–1C2–1
    = 25–3kC1
    = 25 – 3k
    Hence number of solutions
    not stretchy sum subscript k equals 0 end subscript superscript 8 end superscript left parenthesis 25 minus 3 k right parenthesis= 25 × 9 – fraction numerator 3 cross times 8 cross times 9 over denominator 2 end fraction= 225 – 108 = 117.
    General
    maths-

    Sum of divisors of 25 ·37 ·53 · 72 is –

    Any divisor of 25 · 37 · 53 · 72 is of the type of 2l 3m 5n 7p, where 0less or equal than  l less or equal than 5, 0 less or equal thanless or equal than 7, 0  n  3 and 0 less or equal thanless or equal than 2
    Hence the sum of the divisors
    = (1 + 2 + …… + 25) (1 + 3 + ……. + 37) (1 + 5 + 52 + 53) (1 + 7 + 72)
    = open parentheses fraction numerator 2 to the power of 6 end exponent minus 1 over denominator 2 minus 1 end fraction close parentheses open parentheses fraction numerator 3 to the power of 8 end exponent minus 1 over denominator 3 minus 1 end fraction close parentheses open parentheses fraction numerator 5 to the power of 4 end exponent minus 1 over denominator 5 minus 1 end fraction close parentheses open parentheses fraction numerator 7 to the power of 3 end exponent minus 1 over denominator 7 minus 1 end fraction close parentheses
    =fraction numerator left parenthesis 2 to the power of 6 end exponent – 1 right parenthesis left parenthesis 3 to the power of 8 end exponent minus 1 right parenthesis left parenthesis 5 to the power of 4 end exponent minus 1 right parenthesis left parenthesis 7 to the power of 3 end exponent minus 1 right parenthesis over denominator 2 times 4 times 6 end fraction.

    Sum of divisors of 25 ·37 ·53 · 72 is –

    maths-General
    Any divisor of 25 · 37 · 53 · 72 is of the type of 2l 3m 5n 7p, where 0less or equal than  l less or equal than 5, 0 less or equal thanless or equal than 7, 0  n  3 and 0 less or equal thanless or equal than 2
    Hence the sum of the divisors
    = (1 + 2 + …… + 25) (1 + 3 + ……. + 37) (1 + 5 + 52 + 53) (1 + 7 + 72)
    = open parentheses fraction numerator 2 to the power of 6 end exponent minus 1 over denominator 2 minus 1 end fraction close parentheses open parentheses fraction numerator 3 to the power of 8 end exponent minus 1 over denominator 3 minus 1 end fraction close parentheses open parentheses fraction numerator 5 to the power of 4 end exponent minus 1 over denominator 5 minus 1 end fraction close parentheses open parentheses fraction numerator 7 to the power of 3 end exponent minus 1 over denominator 7 minus 1 end fraction close parentheses
    =fraction numerator left parenthesis 2 to the power of 6 end exponent – 1 right parenthesis left parenthesis 3 to the power of 8 end exponent minus 1 right parenthesis left parenthesis 5 to the power of 4 end exponent minus 1 right parenthesis left parenthesis 7 to the power of 3 end exponent minus 1 right parenthesis over denominator 2 times 4 times 6 end fraction.
    General
    maths-

    The length of the perpendicular from the pole to the straight line fraction numerator 6 square root of 2 over denominator r end fraction equals C o s space theta plus S i n space theta is

    The length of the perpendicular from the pole to the straight line fraction numerator 6 square root of 2 over denominator r end fraction equals C o s space theta plus S i n space theta is

    maths-General
    General
    maths-

    The condition for the lines c subscript 1 over r equals a subscript 1 c o s space theta plus b subscript 1 s i n space theta and c subscript 2 over r equals a subscript 2 c o s space theta plus b subscript 2 s i n space theta to be perpendicular is

    The condition for the lines c subscript 1 over r equals a subscript 1 c o s space theta plus b subscript 1 s i n space theta and c subscript 2 over r equals a subscript 2 c o s space theta plus b subscript 2 s i n space theta to be perpendicular is

    maths-General
    General
    Maths-

    If f : R →R; f(x) = sin x + x, then the value of not stretchy integral subscript 0 end subscript superscript pi end superscript blank (f-1 (x)) dx, is equal to

    If f : R →R; f(x) = sin x + x, then the value of not stretchy integral subscript 0 end subscript superscript pi end superscript blank (f-1 (x)) dx, is equal to

    Maths-General
    General
    maths-

    The polar equation of the straight line with intercepts 'a' and 'b' on the rays theta equals 0 and theta equals fraction numerator pi over denominator 2 end fraction respectively is

    The polar equation of the straight line with intercepts 'a' and 'b' on the rays theta equals 0 and theta equals fraction numerator pi over denominator 2 end fraction respectively is

    maths-General
    General
    maths-

    The polar equation of the straight line parallel to the initial line and at a distance of 4 units above the initial line is

    The polar equation of the straight line parallel to the initial line and at a distance of 4 units above the initial line is

    maths-General
    General
    maths-

    The polar equation of x to the power of 3 end exponent plus y to the power of 3 end exponent equals 3 axy is

    The polar equation of x to the power of 3 end exponent plus y to the power of 3 end exponent equals 3 axy is

    maths-General
    General
    maths-

    If x, y, z are integers and x greater or equal than 0, y greater or equal than 1, z greater or equal than 2, x + y + z = 15, then the number of values of the ordered triplet (x, y, z) is -

    Let y = p + 1 and z = q + 2.
    Then x greater or equal than 0, p greater or equal than 0, q greater or equal than 0 and x + y + z = 15
    rightwards double arrow x + p + q = 12
    therefore The reqd. number of values of (x, y, z) and hence of (x, p, q)
    = No. of non-negative integral solutions of x + p + q= 12
    = Coeff. of x12 in (x0 + x1 + x2 + ……)3
    = Coeff. of x12 in (1 – x)–3
    = Coeff. of x12 in [2C0 + 3C1 x + 4C2 x2 + ….]
    = 14C12 = fraction numerator 14 factorial over denominator 2 factorial 12 factorial end fraction = fraction numerator 14 cross times 13 over denominator 2 end fraction = 91.

    If x, y, z are integers and x greater or equal than 0, y greater or equal than 1, z greater or equal than 2, x + y + z = 15, then the number of values of the ordered triplet (x, y, z) is -

    maths-General
    Let y = p + 1 and z = q + 2.
    Then x greater or equal than 0, p greater or equal than 0, q greater or equal than 0 and x + y + z = 15
    rightwards double arrow x + p + q = 12
    therefore The reqd. number of values of (x, y, z) and hence of (x, p, q)
    = No. of non-negative integral solutions of x + p + q= 12
    = Coeff. of x12 in (x0 + x1 + x2 + ……)3
    = Coeff. of x12 in (1 – x)–3
    = Coeff. of x12 in [2C0 + 3C1 x + 4C2 x2 + ….]
    = 14C12 = fraction numerator 14 factorial over denominator 2 factorial 12 factorial end fraction = fraction numerator 14 cross times 13 over denominator 2 end fraction = 91.
    General
    Maths-

    If alpha not equal to beta comma alpha 2 equals 5 alpha minus 3 comma beta 2 times equals 5 beta minus 3 commathen the equation whose roots are alpha divided by beta straight & beta divided by alpha

    If alpha not equal to beta comma alpha 2 equals 5 alpha minus 3 comma beta 2 times equals 5 beta minus 3 commathen the equation whose roots are alpha divided by beta straight & beta divided by alpha

    Maths-General
    General
    Maths-

    Let p, q element of {1, 2, 3, 4}. Then number of equation of the form px2 + qx + 1 = 0, having real roots, is

    Let p, q element of {1, 2, 3, 4}. Then number of equation of the form px2 + qx + 1 = 0, having real roots, is

    Maths-General
    General
    Maths-

    ax2 + bx + c = 0 has real and distinct roots null. Further a > 0, b < 0 and c < 0, then –

    ax2 + bx + c = 0 has real and distinct roots null. Further a > 0, b < 0 and c < 0, then –

    Maths-General