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2015-02-20T12:05:10+05:30

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Let the total  number of coins be  N.       N = 36.

Let there be n number of 10 paise coins. Their value     n * 10  paise
let there be m number of 25 paise coins.  Their valie   m * 25 paise

So, there are (N - m - n) number of 5 paise coins.
    Their value   (N-m-n) * 5 paise  = (36 - m - n) * 5 paise = 180 - 5 m - 5 n  paise

The total value:     25 m + 10 n + 180 - 5 m - 5 n = 300 paise
           20 m + 5 n  = 120

           4 m + n = 24
      n = 60 - 4 m = 4 (6 - m)   
     Thus n is a multiple of 4.

    we also say that there is at least one 25 paise, at least one 5 paise and at least one 10 paise coin present.


1) 
    n = 4:   m = 5
    there are  4 of 10 paise, 5 of 25 paise, and  27  of 5 paise coins.

2)  n = 8 ,  m = 4
     so there are  8 of 10 paise, 4 of 25 paise, and 24 of 5 paise coins.

3)  n = 12   ,  m = 3
     so there are  12 of 10 paise,  3 of 25 paise, and 21 of 5 paise coins

4)
    n = 16 ,  m = 2
     so there are 16 of 10 paise, 2 of 25 paise coins, and 18 of 5 paise.
5)
    n = 20  m  = 1
    there are  20 of  10 paise ,  1 of 25 paise coins,  and 15 of 5 paise coins.


2 5 2
2015-02-24T13:44:51+05:30
Let there be x number of 10 paise coins.
y number of 25 paise coins.
that means, the no. of 5 paise coins = (36 - x - y)
The total value:     25x + 10y + 5(36-x-y) = Rs 3
⇒ 25x + 10y + 180 - 5x - 5y = 300
           20x + 5y  = 120
           y = 24 - 4x = 4(6 - x)   
  
As y is the multiple of 4, 
a) y = 4 then x = 5
  10 paise = 4, 25 paise = 5, and  5 paise coins = 27.

b)   y = 8 ,  x = 4
  10 paise = 8, 25 paise = 4, and  5 paise coins = 24

c)  y = 12   , x = 3
10 paise = 12, 25 paise = 3, and  5 paise coins = 21

d)    y = 16 ,  x = 2
10 paise = 16, 25 paise = 2, and  5 paise coins = 18

e)    y = 20  x = 1
 There are  10 paise = 20, 25 paise = 1, and  5 paise coins = 15
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