Question

bhargan has equal number of rupee 10 ,rupee 20,rupee50,noteand rupee 5 coin in his wallet.If he is carrying rupee 595 in all,find the numberof notes of each type he is carrying and the total number of notes and coins he is carrying.

Answer 1

LET NUMBER OF EACH COIN BE X

5X + 10X + 20X + 50X = 85X = 595

X = 595/85 = 7

NUMBER OF EACH COIN =7

TOTAL NUMBER = 7 X 4 = 28

Answer 2

Answer:

The number of notes of each type he is carrying are 7 and the total number of notes and coins he is carrying are 28.

Step-by-step explanation:

Given,Bhargan has equal number of rupee 10 ,rupee 20,rupee 50 note and rupee 5 coin.

Let he has x number of rupee 10 ,rupee 20,rupee 50 note and rupee 5 coin.

So,total amount with rupee 10 note$= (10 \times x) = 10x \: rupees$

and total amount with rupee 20 note $= (20 \times x) = 20x \: rupees$

and total amount with rupee 50 note $= (50 \times x) = 50x \: rupees$

and total amount with rupee 5 rupee coin $= (5 \times x) = 5x \: rupees$

Therefore he has the total amount $= (10x + 20x + 50x + 5x) \\ = 85x \: rupees$

According to question,he carrying 595 rupees in all.

So,

$85x = 595 \\ x = \frac{595}{85} \\ x = \frac{35}{5} \\ x = 7$

So,the number of notes of each type he is carrying = 7.

And total number of notes and coin = (7×4) = 28

Rupees related two more questions:

https://brainly.in/question/2803140

https://brainly.in/question/35004989

#SPJ2