An interior designer, Sana, hired two painters, Manan and Bhima to make paintings for her buildings.

Question

Answer the questions based on the given information:

An interior designer, Sana, hired two painters, Manan and Bhima to make paintings for her buildings. Both painters were asked to make 50 different paintings each. The prices quoted by both the painters are given below: ♦ Manan asked for Rs 6000 for the first painting, and an increment of Rs 200 for each following painting. ♦ Bhima asked for Rs 4000 for the first painting, and an increment of Rs 400 for each following painting.

(i) How much money did Manan get for his 25th painting? Show your work.  

(ii) How much money did Bhima get in all? Show your work.

(iii) If both Manan and Bhima make paintings at the same pace, find the first painting for which Bhima will get more money than Manan. Show your steps.

OR

(iii) Sana's friend, Aarti hired Manan and Bhima to make paintings for her at the same rates as for Sana. Aarti had both painters make the same number of paintings, and paid them the exact same amount in total.

How many paintings did Aarti get each painter to make? Show your work.

Answer 1

(i) Notes that the amounts Manan is paid for each painting forms an AP. 

Takes a = 6000, d = 200 and n = 25 to find the amount as 

6000 + (25 - 1)200 = Rs 10800.

(ii) Finds the total amount earned by Bhima as follows: 

S₅₀ = 50/2 [2(4000) + (50 - 1)(400)] 

Solves the above expression to find the total amount as Rs 6,90,000.

(iii) Frames equation as follows: 

6000 + (n - 1)200 = 4000 + (n - 1) 400 

Solves the above equation to find the value of n as 11. 

Writes that, since they both earn the same amount for the 11th painting, as Bhima's increment is more, Bhima gets more money than Manan for the 12th painting.

OR 

Assumes that the number of paintings required is n. Frames equation as follows: 

S_n(Manan) = S_n(Bhima) 

=> n/2 [2(6000) + (n -1)200] 

= n/2 [2(4000) + (n -1)400] 

Solves the equation from step 1 to find n as 21.