Question

A Sum of Rs.3000 is to be given in the form of 63 prizes . if the prize money is either Rs.100 or Rs.25 .find the number of prize of each type. ​

Answer 1

$\huge{\fbox{\fbox{\bigstar{\mathfrak </p><p>{\pink{Answer:}}}}}}$

Let the number of prizes of money 100 be x, and the number of prizes of cost 25 be y ;

x+y = 63

y = 63-x -------(1)

100x+25y = 3000

4x+y = 120

y = 120-4x ------(2)

63-x = 120-4x

4x-x = 120-63

3x = 57

x = 19

y = 63-x = 63-19

y = 44

Hence the number of prizes of cost 100 each are 19 while other type of prizes are 44

____________________

Answer 2

$\large \bigstar\boxed{\bf \red{ GIVEN}} \bigstar \\ \\ \Longrightarrow \bf Total \: prize = 3000 \: rs \\ \\ \Longrightarrow \bf 100rs \: note = x \\ \\ \Longrightarrow \bf 25rs \: note = y \\ \\ \Longrightarrow \bf Total \: notes = 63$

$\large \bigstar\boxed{\bf \red{TO \: FIND}} \bigstar \\ \\\Longrightarrow \bf No. \: of \: 100rs \: and \: 25rs \: notes$

$\large \bigstar\boxed{\bf \red{SOLUTION}} \bigstar \\ \\\Longrightarrow \bf x + y = 63 \\ \\ \Longrightarrow \bf x = 63 - y \: \: \: \: \: \: eq(1) \\ \\\Longrightarrow \bf 100x + 25y = 3000 \\ \\ \bf Both \: side \: divided \: by \: 25 \\ \\ \Longrightarrow \bf4x + y = 120 \: \: \: \: \: eq(2) \\ \\ \bf Put \: x \:value \: from \: eq(1) \: into \: eq(2) \\ \\ \Longrightarrow \bf4(63 - y) + y = 120 \\ \\ \Longrightarrow \bf252y - 4y + y = 120 \\ \\ \Longrightarrow \bf - 3y = 120 - 252 \\ \\ \Longrightarrow \bf y = \frac{ - 132}{ - 3} \\ \\ \large \boxed{ \bf \blue{y = 44}} \\ \\ \bf Put \: y \: value \: in \: eq(1) \\ \\\Longrightarrow \bf x = 63 - 44 \\ \\ \large \boxed{ \bf \orange{ x = 19}}$

There are 19 100Rs note and 44 25Rs note distributed in competition