Linear Equation


  1. A boy agrees to work at the rate of 1 rupees on the first day, 2 rupees on the second day. Four rupees on the third day and so on . how much will the boy get if he starts working on the 1st of February and finishes on the 20th of February ?









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    According to given question,
    First day 1 rs, second day 2 rs, third day 4 rs.....
    1 , 2 , 4............ and so on
    The given series is in geometric progression (GP).
    1, 21, 22,.............
    Sum of first n terms in a geometric progression(GP)
    Sn = a (rn ? 1) /(r ?1)
    if r > 1 and where a= the first term, r = common ratio,n = number of terms

    Correct Option: B

    According to given question,
    First day 1 rs, second day 2 rs, third day 4 rs.....
    1 , 2 , 4............ and so on
    The given series is in geometric progression (GP).
    1, 21, 22,.............
    Sum of first n terms in a geometric progression(GP)
    Sn = a (rn ? 1) /(r ?1)
    if r > 1 and where a= the first term, r = common ratio,n = number of terms

    As per the given question a = 1 , r = 2 and n = 20;
    Sn = 1(220 ? 1) /(2 ?1)
    Sn = (220 ? 1) /1
    Sn = (220 ? 1)
    Sn = 220 ? 1


  1. A number 15 is divided into 3 parts which are in Arithmetic Progression (A.P) and the sum of their squares is 83. What will be the smallest number?









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    Let us assume the second number is a and the difference between consecutive numbers is d.
    According to Arithmetic progression,
    First number = a - d
    Second number = a
    Third number = a + d
    According to question,
    Sum of the all three numbers = 15
    a - d + a + a + d = 15
    Again according to given question,
    sum of square of the 3 numbers = 83
    (a - d) 2 + a 2 + (a + d) 2 = 83
    Solve the equation.

    Correct Option: B

    Let us assume the second number is a and the difference between consecutive numbers is d.
    According to Arithmetic progression,
    First number = a - d
    Second number = a
    Third number = a + d
    According to question,
    Sum of the all three numbers = 15
    a - d + a + a + d = 15
    3a = 15
    a = 5
    Again according to given question,
    sum of square of the 3 numbers = 83
    (a - d) 2 + a 2 + (a + d) 2 = 83
    apply the algebra formula
    a 2 + d 2 - 2ad + a 2 + a 2 + d 2 + 2ad = 83
    3a 2 + 2d 2 = 83
    Put the value of a in above equation.
    3 x 5 2 + 2d2 = 83
    3 x 25 + 2d2 = 83
    75 + 2d2 = 83
    2d 2 = 83 - 75
    2d 2 = 8
    d 2 = 8/2
    d 2 = 4
    d = 2
    Put the value of a and d in below equation.
    First number = a - d = 5 - 2 = 3
    Second number = a = 5
    Third number = a + d = 5 + 2 = 7
    The smallest number is 3.



  1. A man arranges to pay off a debt of 3600 in 40 annual installments which form an Arithmetic Progression (A.P). When 30 of the installments are paid, he dies leaving one-third of the debt unpaid . Find the value of the first installment.









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    Let us assume the first installment is a and difference between two consecutive installments is d.
    According to question,
    Sum of 40 Installments = 3600
    S40 = 40/2[ 2a + (40 ?1)d ] = 3600

    Again according to given question,
    Sum of 30 installments = 2400
    S30 = 30/2[ 2a + (30 ?1)d ] = 3600
    Apply the formula Sum of first n terms in an Arithmetic Progression = Sn = n/2[ 2a + (n?1)d ]
    where a = the first term, d = common difference, n = number of terms.

    Correct Option: C

    Let us assume the first installment is a and difference between two consecutive installments is d.
    According to question,
    Sum of 40 Installments = 3600
    Apply the formula Sum of first n terms in an Arithmetic Progression = Sn = n/2[ 2a + (n?1)d ]
    where a = the first term, d = common difference, n = number of terms.

    n/2[ 2a + (n?1)d ] = 3600
    Put the value of a, n and d from question,
    40/2[ 2a + (40 ?1)d ] = 3600
    20[ 2a + 39d ] = 3600
    [ 2a + 39d ] = 3600/20 = 180
    2a + 39d = 180...........................(1)

    Again according to given question,
    After paying the 30 installments the unpaid amount = 1/3(total unpaid amount) = 3600 x 1/3
    After paying the 30 installments the unpaid amount = 1200
    So After paying the 30 installments the paid amount = 3600 - 1200 = 2400
    Sum of 30 installments = 2400
    30/2[ 2a + (30 ?1)d ] = 2400
    [ 2a + (30 ?1)d ] = 2400 x 2/30
    2a + 29d = 80 x 2 = 160
    2a + 29d = 160..........................(2)
    Subtract the Eq. (2) from Eq. (1), we will get
    2a + 39d - 2a - 29d = 180 - 160
    10d = 20
    d = 2
    Put the value of d in Equation (1), we will get
    2a + 39 x 2 = 180
    2a = 180 - 78
    a = 102/2
    a = 51
    The value of first installment = a = 51


  1. On March 1st 2016 , sherry saved 1. Everyday starting from March 2nd 2016, he save 1 more than the previous day . Find the first date after March 1st 2016 at the end of which his total savings will be a perfect square.









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    According to the question,
    Every day adding 1 rs extra to previous day.
    Let us assume after n day, total saving will become perfect square.
    1 + 2 + 3 + 4 + 5 + 6 +.......................+ tn.
    Apply the algebra A.P formula,
    sum of total rupees after n days = n(n+1)/2
    Hit and trail method, Put the value of n = 2 , 3 , 4, 5 .... and so on to get the perfect square.

    Correct Option: D

    According to the question,
    Every day adding 1 rs extra to previous day.
    Let us assume after n day, total saving will become perfect square.
    1 + 2 + 3 + 4 + 5 + 6 +.......................+ tn.
    Apply the algebra A.P formula,
    sum of total rupees after n days = n(n+1)/2
    Hit and trail method, Put the value of n = 2 , 3 , 4, 5 .... and so on to get the perfect square.
    If n = 2
    n(n+1)/2 = 2 x 3 / 2 = 3 which is not perfect Square.
    If n = 3
    n(n+1)/2 = 3 x 4 / 2 = 6 which is not perfect Square.
    If n = 4
    n(n+1)/2 = 4 x 5 / 2 = 10 which is not perfect Square.
    If n = 5
    n(n+1)/2 = 5 x 6 / 2 = 15 which is not perfect Square.
    If n = 6
    n(n+1)/2 = 6 x 7 / 2 = 21 which is not perfect Square.
    If n = 7
    n(n+1)/2 = 7 x 8 / 2 = 28 which is not perfect Square.
    If n = 8
    n(n+1)/2 = 8 x 9 / 2 = 36 which is perfect Square.

    n(n+1)/2 should be a perfect square . The first value of n when this occurs would be for n = 8. thus , on the 8th of March of the required condition would come true.



  1. How many 3-digits numbers are completely divisible by 6?









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    First 3-digit number divisible by 6 is 102 and last 3-digit number divisible by 6 is 996.
    Difference between two consecutive numbers divisible by 6 is 6.
    So 3-digit numbers divisible by 6 are 102,108,114, ......., 996.
    This is an Arithmetic Progression in which a = 102, d = 6 and l = 996.
    where a = First Number , l = Last Number and d = difference of two consecutive numbers.
    Let the number of terms be n. So Last term = tn
    Then tn = 996
    Use the formula for n term of Arithmetic Progression.

    Correct Option: B

    First 3-digit number divisible by 6 is 102 and last 3-digit number divisible by 6 is 996.
    Difference between two consecutive numbers divisible by 6 is 6.
    So 3-digit numbers divisible by 6 are 102,108,114, ......., 996.
    This is an Arithmetic Progression in which a = 102, d = 6 and l = 996.
    where a = First Number , l = Last Number and d = difference of two consecutive numbers.
    Let the number of terms be n. So Last term = tn
    Then tn = 996
    Use the formula for n terms of arithmetic progression.
    a + ( n - 1) x d = 996
    102 + (n - 1) x 6 = 996
    6(n - 1) = 894
    (n - 1) = 149
    n = 150
    Numbers of terms = 150