Find the value of the first instalment, Mathematics

A man arranges to pay a debt of Rs.3600 in 40 monthly instalments which are in a AP. When 30 instalments are paid he dies leaving one third of the debt unpaid. Find the value of the first instalment.

Ans:    Let the value of I instalment be x S40 = 3600.

⇒ 40/2 [2a + 39d ]=3600

⇒2a + 39d = 180                    -           1

S30 =  30/2 [2a + 29d ]=2400

⇒30a + 435d = 2400

⇒2a + 29d = 160                    -           2

Solve 1 & 2 to get

d = 2 a = 51.

∴ I instalment = Rs.51.

 

Posted Date: 4/8/2013 5:21:03 AM | Location : United States







Related Discussions:- Find the value of the first instalment, Assignment Help, Ask Question on Find the value of the first instalment, Get Answer, Expert's Help, Find the value of the first instalment Discussions

Write discussion on Find the value of the first instalment
Your posts are moderated
Related Questions
What are the other differences between learners that a teacher needs to keep in mind, while teaching?  Let us see an example in which a teacher took the pupil's background into acc

Standardizing a Random Variable       If X is a random variable with E(X) = m and V(X) = s 2 , then Y = (X – m)/ s is a random variable with mean 0 and standard deviatio


uses of maths concept

Strategy for Series Now that we have got all of our tests out of the way it's time to think regarding to the organizing all of them into a general set of strategy to help us

Example of Adding signed numbers: Example: (2) + (-4) =      Solution: Start with 2 and count 4 whole numbers to the left. Thus: (2) + (-4) = -2 Adding

I didn't understand the concept of Technical Coefficients, provide me assistance.

On a piece of machinery, the centers of two pulleys are 3 feet apart, and the radius of each pulley is 6 inches. Determine the size of belt (in feet) is required to wrap around bot

Use the graph of y = x2 - 6x  to answer the following: a)         Without solving the equation (or factoring), determine the solutions to the equation  x 2 - 6x = 0  usi

sin (cot -1 {cos (tan -1 x)}) tan -1 x = A  => tan A =x sec A = √(1+x 2 ) ==>  cos A = 1/√(1+x 2 )    so   A =  cos -1 (1/√(1+x 2 )) sin (cot -1 {cos (tan -1 x)}) = s