Maths-
General
Easy

Question

Find the simple interest and amount on Rs 5000 in 3 years at 8% p.a.

Hint:

Use the formula for simple interest and then find the total amount.

The correct answer is: 6200 Rupees


    Complete step by step solution:
    We calculate simple interest by the formula,S I equals fraction numerator P T R over denominator 100 end fraction…(i)
    where P is Principal amount, T is number of years and R is rate of interest
    Here, we have T = 3,R = 8% and P = 5000
    On substituting the known values in (i), we get S I equals fraction numerator 5000 cross times 3 cross times 8 over denominator 100 end fraction
    not stretchy rightwards double arrow S I equals fraction numerator 5000 cross times 3 cross times 8 over denominator 100 end fraction equals 1200
    We have SI = 1200 Rupees.
    We know the formula for total amount = A = P + SI…(ii)
    where A is the total amount, P is the principal amount and SI is simple interest.
    On substituting the known values in (ii), we get A = 5000 + 1200 = 6200
    Hence total amount to be paid after 3 years = A = 6200 Rupees.

    Related Questions to study

    General
    Maths-

    Find the amount to be paid at the end of 3 years, if principal is Rs 1800 at 9% p.a.

    Complete step by step solution:
    We calculate simple interest by the formula,S I equals fraction numerator P T R over denominator 100 end fraction…(i)
    where P is Principal amount, T is number of years and R is rate of interest
    Here, we have T = 3,R = 9% and P = 1800
    On substituting the known values in (i), we get S I equals fraction numerator 1800 cross times 3 cross times 9 over denominator 100 end fraction
    not stretchy rightwards double arrow S I equals fraction numerator 1800 cross times 3 cross times 9 over denominator 100 end fraction equals 486
    We have SI = 486 Rupees.
    We know the formula for total amount = A = P +SI…(ii)
    where A is the total amount, P is the principal amount and SI is simple interest.
    On substituting the known values in (ii), we get A = 1800 + 486 = 2286
    Hence total amount to be paid after 3 years = A = 2286 Rupees.

    Find the amount to be paid at the end of 3 years, if principal is Rs 1800 at 9% p.a.

    Maths-General
    Complete step by step solution:
    We calculate simple interest by the formula,S I equals fraction numerator P T R over denominator 100 end fraction…(i)
    where P is Principal amount, T is number of years and R is rate of interest
    Here, we have T = 3,R = 9% and P = 1800
    On substituting the known values in (i), we get S I equals fraction numerator 1800 cross times 3 cross times 9 over denominator 100 end fraction
    not stretchy rightwards double arrow S I equals fraction numerator 1800 cross times 3 cross times 9 over denominator 100 end fraction equals 486
    We have SI = 486 Rupees.
    We know the formula for total amount = A = P +SI…(ii)
    where A is the total amount, P is the principal amount and SI is simple interest.
    On substituting the known values in (ii), we get A = 1800 + 486 = 2286
    Hence total amount to be paid after 3 years = A = 2286 Rupees.
    General
    Maths-

    Find the compound interest for 3 years on Rs 5000, if the rate of interest for the successive years are 8%, 6% and 10% respectively.

    Complete step by step solution:
    Given that principal amount P = 5000
    Number of years T = 3
    Let R1 = 8%,R2 = 6% and R3 = 10%

    Total amount , A equals P cross times open parentheses 1 plus R subscript 1 over 100 close parentheses cross times open parentheses 1 plus R subscript 2 over 100 close parentheses cross times open parentheses 1 plus R subscript 3 over 100 close parentheses …(i)
    On substituting the known values in (i), we get

    A equals 5000 cross times open parentheses 1 plus 8 over 100 close parentheses cross times open parentheses 1 plus 6 over 100 close parentheses cross times open parentheses 1 plus 10 over 100 close parentheses

    not stretchy rightwards double arrow A equals 5000 cross times 108 over 100 cross times 106 over 100 cross times 110 over 100
    not stretchy rightwards double arrow A equals 6296.4
    We know that, Compound interest ( CI) = total amount (A) - principal amount (P)
    So, Compound interest ( CI) = 6296.4 - 5000 = 1296.4 Rupees

    Find the compound interest for 3 years on Rs 5000, if the rate of interest for the successive years are 8%, 6% and 10% respectively.

    Maths-General
    Complete step by step solution:
    Given that principal amount P = 5000
    Number of years T = 3
    Let R1 = 8%,R2 = 6% and R3 = 10%

    Total amount , A equals P cross times open parentheses 1 plus R subscript 1 over 100 close parentheses cross times open parentheses 1 plus R subscript 2 over 100 close parentheses cross times open parentheses 1 plus R subscript 3 over 100 close parentheses …(i)
    On substituting the known values in (i), we get

    A equals 5000 cross times open parentheses 1 plus 8 over 100 close parentheses cross times open parentheses 1 plus 6 over 100 close parentheses cross times open parentheses 1 plus 10 over 100 close parentheses

    not stretchy rightwards double arrow A equals 5000 cross times 108 over 100 cross times 106 over 100 cross times 110 over 100
    not stretchy rightwards double arrow A equals 6296.4
    We know that, Compound interest ( CI) = total amount (A) - principal amount (P)
    So, Compound interest ( CI) = 6296.4 - 5000 = 1296.4 Rupees

    General
    Maths-

    What annual instalment will discharge a debt of Rs 1092 due in 3 years at 12% simple interest?

    Complete step by step solution:
    Let the principal amount P = 1092
    It is given that T = 2, R = 12%
    We have the formula for annual payment A P equals fraction numerator 100 P over denominator 100 cross times T plus fraction numerator R T left parenthesis T minus 1 right parenthesis over denominator 2 end fraction end fraction…(i)
    On substituting the known values in (i), we get A P equals fraction numerator 100 cross times 1092 over denominator 100 cross times 3 plus fraction numerator 12 cross times 3 left parenthesis 3 minus 1 right parenthesis over denominator 2 end fraction end fraction

    not stretchy rightwards double arrow A P equals fraction numerator 109200 over denominator 300 plus 72 over 2 end fraction
    not stretchy rightwards double arrow A P equals 109200 over 336 equals 325
    So, 325 Rupees is the annual instalment.

    What annual instalment will discharge a debt of Rs 1092 due in 3 years at 12% simple interest?

    Maths-General
    Complete step by step solution:
    Let the principal amount P = 1092
    It is given that T = 2, R = 12%
    We have the formula for annual payment A P equals fraction numerator 100 P over denominator 100 cross times T plus fraction numerator R T left parenthesis T minus 1 right parenthesis over denominator 2 end fraction end fraction…(i)
    On substituting the known values in (i), we get A P equals fraction numerator 100 cross times 1092 over denominator 100 cross times 3 plus fraction numerator 12 cross times 3 left parenthesis 3 minus 1 right parenthesis over denominator 2 end fraction end fraction

    not stretchy rightwards double arrow A P equals fraction numerator 109200 over denominator 300 plus 72 over 2 end fraction
    not stretchy rightwards double arrow A P equals 109200 over 336 equals 325
    So, 325 Rupees is the annual instalment.

    parallel
    General
    Maths-

    What sum of money lent out at 6% for 2 years will produce the same interest as Rs. 1200 lent out at 5% for 3 years.

    Complete step by step solution:
    We calculate simple interest by the formula,S I equals fraction numerator P T R over denominator 100 end fraction …(i)
    where P is Principal amount, T is number of years and R is rate of interest
    Case Ⅰ
    Let the sum of money = P
    Here, we have T equals 2 comma R equals 6 straight percent sign text  and  end text P equals ?
    On substituting the values in (i), we get S I equals fraction numerator P cross times 2 cross times 6 over denominator 100 end fraction equals fraction numerator 12 P over denominator 100 end fraction…(ii)
    Case Ⅱ
    Here, we have T equals 3 comma R equals 5 straight percent sign text  and  end text P equals 1200
    On substituting the values in (i), we get S I equals fraction numerator 1200 cross times 3 cross times 5 over denominator 100 end fraction equals 18000 over 100…(iii)
    It is given that the interest produced in both the cases is the same.
    So, Equate (ii) and (iii)

    On equating, we get fraction numerator 12 P over denominator 100 end fraction equals 18000 over 100

    not stretchy rightwards double arrow P equals 18000 over 12

    not stretchy rightwards double arrow P equals 1500 rupees.
    Hence the sum of money P = 1500 Rupees

    What sum of money lent out at 6% for 2 years will produce the same interest as Rs. 1200 lent out at 5% for 3 years.

    Maths-General
    Complete step by step solution:
    We calculate simple interest by the formula,S I equals fraction numerator P T R over denominator 100 end fraction …(i)
    where P is Principal amount, T is number of years and R is rate of interest
    Case Ⅰ
    Let the sum of money = P
    Here, we have T equals 2 comma R equals 6 straight percent sign text  and  end text P equals ?
    On substituting the values in (i), we get S I equals fraction numerator P cross times 2 cross times 6 over denominator 100 end fraction equals fraction numerator 12 P over denominator 100 end fraction…(ii)
    Case Ⅱ
    Here, we have T equals 3 comma R equals 5 straight percent sign text  and  end text P equals 1200
    On substituting the values in (i), we get S I equals fraction numerator 1200 cross times 3 cross times 5 over denominator 100 end fraction equals 18000 over 100…(iii)
    It is given that the interest produced in both the cases is the same.
    So, Equate (ii) and (iii)

    On equating, we get fraction numerator 12 P over denominator 100 end fraction equals 18000 over 100

    not stretchy rightwards double arrow P equals 18000 over 12

    not stretchy rightwards double arrow P equals 1500 rupees.
    Hence the sum of money P = 1500 Rupees

    General
    Maths-

    What sum of money lent out at 5% for 3 years will produce the same interest as Rs. 900 lent out at 4% for 5 years.

    Complete step by step solution:
    We calculate simple interest by the formula, S I equals fraction numerator P T R over denominator 100 end fraction…(i)
    where P is Principal amount, T is number of years and R is rate of interest
    Case Ⅰ
    Let the sum of money = P
    Here, we have T equals 3 comma R equals 5 straight percent sign text  and  end text P equals ?
    On substituting the values in (i), we get S I equals fraction numerator P cross times 3 cross times 5 over denominator 100 end fraction equals fraction numerator 15 P over denominator 100 end fraction…(ii)
    Case Ⅱ
    Here, we have T equals 5 comma R equals 4 straight percent sign text  and  end text P equals 900
    On substituting the values in (i), we get S I equals fraction numerator 900 cross times 5 cross times 4 over denominator 100 end fraction equals 18000 over 100…(iii)
    It is given that the interest produced in both the cases is the same.
    So, Equate (ii) and (iii)
    On equating, we get fraction numerator 15 P over denominator 100 end fraction equals 18000 over 100
    not stretchy rightwards double arrow P equals 18000 over 15
    not stretchy rightwards double arrow P equals 1200rupees.
    Hence the sum of money P = 1200 Rupees

    What sum of money lent out at 5% for 3 years will produce the same interest as Rs. 900 lent out at 4% for 5 years.

    Maths-General
    Complete step by step solution:
    We calculate simple interest by the formula, S I equals fraction numerator P T R over denominator 100 end fraction…(i)
    where P is Principal amount, T is number of years and R is rate of interest
    Case Ⅰ
    Let the sum of money = P
    Here, we have T equals 3 comma R equals 5 straight percent sign text  and  end text P equals ?
    On substituting the values in (i), we get S I equals fraction numerator P cross times 3 cross times 5 over denominator 100 end fraction equals fraction numerator 15 P over denominator 100 end fraction…(ii)
    Case Ⅱ
    Here, we have T equals 5 comma R equals 4 straight percent sign text  and  end text P equals 900
    On substituting the values in (i), we get S I equals fraction numerator 900 cross times 5 cross times 4 over denominator 100 end fraction equals 18000 over 100…(iii)
    It is given that the interest produced in both the cases is the same.
    So, Equate (ii) and (iii)
    On equating, we get fraction numerator 15 P over denominator 100 end fraction equals 18000 over 100
    not stretchy rightwards double arrow P equals 18000 over 15
    not stretchy rightwards double arrow P equals 1200rupees.
    Hence the sum of money P = 1200 Rupees
    General
    Maths-

    Find the sum which will amount to Rs. 364.80 at 31 half % per annum in 8 years at simple interest

    Complete step by step solution:
    Let the sum of money = P
    We know the formula for total amount = A = P + SI
    where A is the total amount, T is the principal amount and R is simple interest.
    We know that S I equals fraction numerator P T R over denominator 100 end fraction
    where P is Principal amount, T is number of years and R is rate of interest
    So, A equals P plus fraction numerator P T R over denominator 100 end fraction…(i)
    Here, we have A equals 364.80 comma T equals 8 comma R equals 3 1 half equals 3.5 straight percent sign text  and  end text P equals ?
    On substituting these values in (i), we get 364.80 equals P plus fraction numerator p cross times 8 cross times 3.5 over denominator 100 end fraction
    On further simplifications, we get 364.80 equals P plus.28 P
    table attributes columnspacing 1em end attributes row cell not stretchy rightwards double arrow 1.28 P equals 364.80 end cell row cell not stretchy rightwards double arrow P equals R s 285 end cell end table
    Hence the sum of money P = Rs 285.

    Find the sum which will amount to Rs. 364.80 at 31 half % per annum in 8 years at simple interest

    Maths-General
    Complete step by step solution:
    Let the sum of money = P
    We know the formula for total amount = A = P + SI
    where A is the total amount, T is the principal amount and R is simple interest.
    We know that S I equals fraction numerator P T R over denominator 100 end fraction
    where P is Principal amount, T is number of years and R is rate of interest
    So, A equals P plus fraction numerator P T R over denominator 100 end fraction…(i)
    Here, we have A equals 364.80 comma T equals 8 comma R equals 3 1 half equals 3.5 straight percent sign text  and  end text P equals ?
    On substituting these values in (i), we get 364.80 equals P plus fraction numerator p cross times 8 cross times 3.5 over denominator 100 end fraction
    On further simplifications, we get 364.80 equals P plus.28 P
    table attributes columnspacing 1em end attributes row cell not stretchy rightwards double arrow 1.28 P equals 364.80 end cell row cell not stretchy rightwards double arrow P equals R s 285 end cell end table
    Hence the sum of money P = Rs 285.
    parallel
    General
    Maths-

    The simple interest on a sum of money at the end of 3 years is of the sum itself. What rate percent was charged?

    Complete step by step solution:
    Let the sum of money = P
    It is given that SI is 3 over 16 times the sum itself = 3 over 16P.
    We calculate simple interest by the formula, S I equals fraction numerator P T R over denominator 100 end fraction
    where P is Principal amount, T is number of years and R is rate of interest
    Here, we have S I equals 3 over 16 P comma T equals 3 text  and  end text R equals ?
    On substituting the known values we get, 3 over 6 P equals fraction numerator P cross times 3 cross times R over denominator 100 end fraction
    On further simplifications, we have 1 over 16 equals R over 100.
    not stretchy rightwards double arrow R equals 100 over 16 equals 6.25 straight percent sign

    The simple interest on a sum of money at the end of 3 years is of the sum itself. What rate percent was charged?

    Maths-General
    Complete step by step solution:
    Let the sum of money = P
    It is given that SI is 3 over 16 times the sum itself = 3 over 16P.
    We calculate simple interest by the formula, S I equals fraction numerator P T R over denominator 100 end fraction
    where P is Principal amount, T is number of years and R is rate of interest
    Here, we have S I equals 3 over 16 P comma T equals 3 text  and  end text R equals ?
    On substituting the known values we get, 3 over 6 P equals fraction numerator P cross times 3 cross times R over denominator 100 end fraction
    On further simplifications, we have 1 over 16 equals R over 100.
    not stretchy rightwards double arrow R equals 100 over 16 equals 6.25 straight percent sign
    General
    Maths-

    A theatre company uses the revenue function R left parenthesis x right parenthesis equals negative 50 x squared plus 250 x dollars. The cost functions of the production C left parenthesis x right parenthesis equals 450 minus 50 x. What ticket price is needed for the theatre to break even?


    A theatre company uses the revenue function R left parenthesis x right parenthesis equals negative 50 x squared plus 250 x dollars. The cost functions of the production C left parenthesis x right parenthesis equals 450 minus 50 x. What ticket price is needed for the theatre to break even?

    Maths-General

    General
    Maths-

    Rewrite the equation as a system of equations, and then use a graph to solve.
    0.5 x squared plus 4 x equals negative 12 minus 1.5 x

    Hint:
    A graph is a geometrical representation of an equation or an expression. It can be used to find solutions of equation.
    We are asked to rewrite the equation as system of equations and graph them to solve it.
    Step 1 of 3:
    Equate each side of the equation to a new variable, y:




    Here we get two points where both the graphs intersect each other. The points are (-8, 0) and (-3, -7.5). Thus, we can say that the solutions to the given set of equation are the points of intersection.
     Note:
    When you graph a quadratic equation find three coordinate points to get the curve. But when it is a linear equation, just two points would give the path of the line.

    Rewrite the equation as a system of equations, and then use a graph to solve.
    0.5 x squared plus 4 x equals negative 12 minus 1.5 x

    Maths-General
    Hint:
    A graph is a geometrical representation of an equation or an expression. It can be used to find solutions of equation.
    We are asked to rewrite the equation as system of equations and graph them to solve it.
    Step 1 of 3:
    Equate each side of the equation to a new variable, y:




    Here we get two points where both the graphs intersect each other. The points are (-8, 0) and (-3, -7.5). Thus, we can say that the solutions to the given set of equation are the points of intersection.
     Note:
    When you graph a quadratic equation find three coordinate points to get the curve. But when it is a linear equation, just two points would give the path of the line.
    parallel
    General
    Maths-

    Rewrite the equation as a system of equations, and then use a graph to solve.
    x squared minus 6 x equals 2 x minus 16



    Thus, the solutions are (0, 0) and (1, -14)
    Step 3 of 3:
    Plot the points and join them to get the respective graph.

    Here, there is just one point where both the graphs intersect each other. The point is (4, -8). Thus, we can say that the point is the solution of the set of equation.
    Note:
    When you graph a quadratic equation find three coordinate points to get the curve. But when it is a linear equation, just two points would give the path of the line.

    Rewrite the equation as a system of equations, and then use a graph to solve.
    x squared minus 6 x equals 2 x minus 16

    Maths-General


    Thus, the solutions are (0, 0) and (1, -14)
    Step 3 of 3:
    Plot the points and join them to get the respective graph.

    Here, there is just one point where both the graphs intersect each other. The point is (4, -8). Thus, we can say that the point is the solution of the set of equation.
    Note:
    When you graph a quadratic equation find three coordinate points to get the curve. But when it is a linear equation, just two points would give the path of the line.
    General
    Maths-

    Find the simple interest on Rs. 6500 at 14% per annum for 73 days?

    Complete step by step solution:
    We calculate simple interest by the formula, S I equals fraction numerator P T R over denominator 100 end fraction
    where P is Principal amount, T is number of years and R is rate of interest
    Here, we have P equals 6500 comma T equals 73 over 365 equals 1 fifth text  and  end text R equals 14 straight percent sign
    On substituting the known values we get, S I equals fraction numerator 6500 cross times 1 fifth cross times 14 over denominator 100 end fraction
    On further simplifications, we have S I equals 18200 over 100 equals 182 rupees.
    Thus, SI = 182 Rupees.

    Find the simple interest on Rs. 6500 at 14% per annum for 73 days?

    Maths-General
    Complete step by step solution:
    We calculate simple interest by the formula, S I equals fraction numerator P T R over denominator 100 end fraction
    where P is Principal amount, T is number of years and R is rate of interest
    Here, we have P equals 6500 comma T equals 73 over 365 equals 1 fifth text  and  end text R equals 14 straight percent sign
    On substituting the known values we get, S I equals fraction numerator 6500 cross times 1 fifth cross times 14 over denominator 100 end fraction
    On further simplifications, we have S I equals 18200 over 100 equals 182 rupees.
    Thus, SI = 182 Rupees.
    General
    Maths-

    Rewrite the equation as a system of equations, and then use a graph to solve.
    2 x squared plus 3 x equals 2 x plus 1