Radhika has a total of rs 590 as currency notes in the denominations of rs 50 rs 20 rs 10 The ratio of the number of rs 50 notes and rs 20 notes 3:5. If she has a total of 25 notes. How many notes of each denominations she has
Answers
Radhika has currency notes with value = Rs.590/-
The combinations are 50 rs, 20 rs and 10 rs.
Let us assume that there are x, y and z 50 rs, 20 rs and 10 rs. respectively.
Then value would be
50x+20y+10z = 590 (given)
Also given that ratio of x:y= 3:5
i.e. 5x = 3y and
Total notes = 25
Or x+y+z = 25
Three equations in 3 variables.
50x+20y+10z = 590: substitute 5x=3y
30y+20y+10z = 590 or 50y+10z = 590 :
Multiply by 5, x+y+z = 25 to get
5x+5y+5z =125 now substitute 5x=3y
8y+5z = 125
50y+10z = 590 :
Solving these two, 34y = 340
y=10, and 5x = 30 or x =6
z = 25-x-y = 9
Answer is
Radhika has 6 notes of 50 Rs., 10 notes of Rs.20 and 9 notes of Rs10/-
Answer:
No: of ₹50 notes = 6
No: of ₹20 notes = 10
No: of ₹10 notes = 9
Step-by-step explanation:
No: of ₹50 notes : No of ₹20 notes = 3:5
No: of ₹50 notes = 3x
No: of ₹20 notes = 5x
Total no: of notes = 25
No: of ₹10 notes = 25 - (3x + 5x) = 25 - 8x
Total denomination = ₹590
50(3x) + 20(5x) + 10(25-8x) = 590
150x + 100x + 250 - 80x = 590
170x = 590 - 250 = 340
x = 340/170 = 2
No: of ₹50 notes = 3x =3*2 = 6
No: of ₹20 notes = 5x = 5*2 = 10
No: of ₹10 notes = 25 - 8x = 25 - 8 * 2 = 25 - 16 = 9