Question

A man wishes to buy few masks. There are two types of masks, type-A cost 11 rupees per mask and type-B cost 13 rupees per mask. Given that the man has a total of
369 rupees, in how many ways can he buy masks of both categories?
i)1
ii)2
iii)3
iv)4​

Answer 1

Given  : A man wishes to buy few masks. There are two types of masks, type-A cost 11 rupees per mask and type-B cost 13 rupees per mask

To find : in how many ways can he buy masks of both categories

Solution:

type-A cost 11 rupees per mask  - Quantity = A

type-B cost 13 rupees per mask  - Quantity = B

Assuming all money has to be spend

11A  + 13B =  369

=> 11A + 11B + 2B = 11 * 33  + 6

=> 2B = 11 ( 33 - A - B )  +  6

=> 2(B - 3)  = 11 (33 - A - B)

2 & 11 are prime

Hence  B - 3 must a factor of 11

B - 3  =  0  => B  = 3   , A  = 30

B - 3 =  11 =>  B = 14  ,   A  = 17

B - 3 = 22  => B = 25     A =  4

B - 2  = 33 => B = 36  not possible as B ≤ 33  

Three ways

(30 , 3)  , ( 17 , 14)  & ( 4 , 25)    

option iii

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Answer 2

Step-by-step explanation:

The number of ways the man can buy masks of both categories is 3

Therefore, option (iii) is correct.

Step-by-step explanation:

Cost of the type-A masks = 11 Rs.

Cost of the type-B masks = 13 Rs.

Let the man buys x type-A and y type-B masks

Then

11x+13y=369

$\implies x=\frac{369-13y}{11}$

x and y are whole numbers

If y = 3

$x=\frac{369-39}{11}=\frac{330}{11}=30$

The next value of y will be 11 more than 3

Thus, y = 3 + 11 = 14

If y = 14

$x=\frac{369-182}{11}=\frac{187}{11}=17$

The next value of y will be 11 more than 14

Thus, y = 14 + 11 = 25

If y = 25

$x=\frac{369-325}{11}=\frac{44}{11}=4$

After this we will only get negative values of x

Therefore, there are 3 ways that the man can buy the masks of both categories.

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