Home
Class 10
MATHS
Ajay sharma repays the borrowed amount o...

Ajay sharma repays the borrowed amount of Rs 3,25,000 by paying Rs 30500 in the first month and then decreases the payment by Rs 1500 every month. How long will it take to clear his amount?

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem step-by-step, we will use the formula for the sum of an arithmetic progression (AP). Here’s how we can approach the solution: ### Step 1: Identify the parameters of the problem - **Total amount borrowed (S)**: Rs 3,25,000 - **First month's payment (A)**: Rs 30,500 - **Decrease in payment each month (D)**: Rs 1,500 ### Step 2: Write the formula for the sum of the first n terms of an AP The sum of the first n terms (S_n) of an arithmetic progression can be calculated using the formula: \[ S_n = \frac{n}{2} \times (2A + (n - 1)D) \] ### Step 3: Set up the equation We know that Ajay needs to repay a total of Rs 3,25,000, so we can set up the equation: \[ 3,25,000 = \frac{n}{2} \times (2 \times 30,500 + (n - 1)(-1,500)) \] ### Step 4: Simplify the equation 1. Calculate \( 2A \): \[ 2 \times 30,500 = 61,000 \] 2. Substitute into the equation: \[ 3,25,000 = \frac{n}{2} \times (61,000 - 1,500(n - 1)) \] 3. Distribute \( -1,500(n - 1) \): \[ 3,25,000 = \frac{n}{2} \times (61,000 - 1,500n + 1,500) \] \[ 3,25,000 = \frac{n}{2} \times (62,500 - 1,500n) \] ### Step 5: Multiply both sides by 2 to eliminate the fraction \[ 6,50,000 = n(62,500 - 1,500n) \] ### Step 6: Rearrange the equation \[ 6,50,000 = 62,500n - 1,500n^2 \] Rearranging gives: \[ 1,500n^2 - 62,500n + 6,50,000 = 0 \] ### Step 7: Simplify the quadratic equation Divide the entire equation by 100 to simplify: \[ 15n^2 - 625n + 6500 = 0 \] ### Step 8: Use the quadratic formula to find n The quadratic formula is given by: \[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Where \( a = 15 \), \( b = -625 \), and \( c = 6500 \). ### Step 9: Calculate the discriminant 1. Calculate \( b^2 - 4ac \): \[ (-625)^2 - 4 \times 15 \times 6500 \] \[ 390625 - 390000 = 625 \] ### Step 10: Substitute back into the quadratic formula \[ n = \frac{625 \pm \sqrt{625}}{30} \] \[ n = \frac{625 \pm 25}{30} \] ### Step 11: Calculate the two possible values for n 1. First value: \[ n = \frac{650}{30} = \frac{65}{3} \approx 21.67 \] 2. Second value: \[ n = \frac{600}{30} = 20 \] Since n must be a whole number, we take \( n = 20 \). ### Conclusion Ajay will take **20 months** to clear his borrowed amount. ---
Promotional Banner

Topper's Solved these Questions

  • ARITHMETIC PROGRESSION

    TARGET PUBLICATION|Exercise Chapter Assessment|17 Videos

  • ARITHMETIC PROGRESSION

    TARGET PUBLICATION|Exercise Based on Practice Set 3.3|15 Videos

  • BOARD QUESTION PAPER: MARCH 2019

    TARGET PUBLICATION|Exercise (Solve the following questions (Any one):|4 Videos

Similar Questions

Explore conceptually related problems

A man repays a loan of Rs.3250 by paying Rs.20 in the first month and then increases the payment by Rs.15 every month.How lon will it take him to clear the loan?

Reshma repays her total loan of Rs 1,75,000 by paying every month starting with the first installment of Rs 1500. If she increases the installment by Rs 175 every month, what amount will be paid by her in the 21st instalment? What amount of loan does she still have to pay after the 21st instalment?

Your elder brother wants to buy a car and plans to take loan from a bank for his car. He repays his total loan of Rs 1,18,000 by paying every month starting with the first instalment of Rs 1000. If he increases the instalment by Rs 100 every month , answer the following: The amount paid by him in the 30 installments is

Your elder brother wants to buy a car and plans to take loan from a bank for his car. He repays his total loan of Rs 1,18,000 by paying every month starting with the first instalment of Rs 1000. If he increases the instalment by Rs 100 every month , answer the following: The amount paid by him in 30^(th) installment is

A man starts repaying a loan as first of Rs 100. If the increases the instalment by Rs 5 every month, what amount he will pay in the 30th instalment?

Your elder brother wants to buy a car and plans to take loan from a bank for his car. He repays his total loan of Rs 1,18,000 by paying every month starting with the first instalment of Rs 1000. If he increases the instalment by Rs 100 every month , answer the following: What amount does he still have to pay after 30^(th) installment?

Your elder brother wants to buy a car and plans to take loan from a bank for his car. He repays his total loan of Rs 1,18,000 by paying every month starting with the first instalment of Rs 1000. If he increases the instalment by Rs 100 every month , answer the following: If total installments are 40 then amount paid in the last installment?

Your elder brother wants to buy a car and plans to take loan from a bank for his car. He repays his total loan of Rs 1,18,000 by paying every month starting with the first instalment of Rs 1000. If he increases the instalment by Rs 100 every month , answer the following: The ratio of the 1^(st) installment to the last installment is

Jaspal Singh repays his total loan of Rs. 118000 by paying every month starting with the first installment of Rs. 1000. If he increases the installment by Rs. 100 every month, what amount will be paid by him in the 30th installment? What amount of loan does he still have to pay after the 30th installment?

A man starts repaying a loan with first monthly installment of Rs 1000. If he increases the installment by Rs 50 every month , what amount will he pay in the 30^(th) installment ?