Question

Bansi has 3 times as many two rupee coins as he has five rupee coins. If he has in all a sum of a sum of rs 77 how many coins of each denomination does he have

Answer 1

AnswEr :

• Let the Number of Five Rupee Coins be n.
• Number of Two Rupee Coins be ( 3 times of Five Rupee Coin ) = 3n

• According to the Question Now :

⇒ (Two Rupee + Five Rupee) = Rs. 77

⇒ (No. × Value) + (No. × Value) = Rs. 77

⇒ (3n × Rs. 2) + (n × Rs. 5) = Rs. 77

⇒ Rs. (6n + 5n) = Rs. 77

⇒ 11n = 77

• Dividing both term by 11

⇒ n = 7

• Each Denomination Have :

◗ Five Rupee Coin = n = 7

◗ Two Rupee Coin = 3n = 3(7) = 21

∴ There is 7 Five Rupee Coin, and 21 Two Rupee Coin total sum of Rs. 77.

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• V E R I F I C A T I O N :

↠ (Two Rupee + Five Rupee) = Rs. 77

↠ (No. × Value) + (No. × Value) = Rs. 77

↠ (3n × Rs. 2) + (n × Rs. 5) = Rs. 77

↠ (21 × Rs. 2) + (7 × Rs. 5) = Rs. 77

↠ Rs. (42 + 35) = Rs. 77

↠ Rs. 77 = Rs. 77 ⠀⠀⠀⠀Hence, Verified!

#answerwithquality #BAL

Answer 2

$\bf{\Huge{\underline{\boxed{\bf{\green{ANSWER\::}}}}}}$

$\bf{\Large{\underline{\bf{Given\::}}}}$

Bansi has 3 times as many two rupees coins as he has five rupees coins. If he has in all a sum of Rs.77.

$\bf{\Large{\underline{\bf{To\:find\::}}}}$

The coins of each denomination have.

$\bf{\Large{\underline{\bf{\blue{Explanation\::}}}}}$

Let the number of Rs.5 coins be R

Let the number of Rs.2 coins be 3R

A/q

→ 5R + 2× 3R = 77

→ 5R + 6R = 77

→ 11R = 77

→ R = $\bf{\cancel{\frac{77}{11} }}$

→ R = 7

Thus,

• Number of Rs.5 coins is 7

• Number of Rs.2 coins is (3 × 7) = 21