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$\sf \: Question$


A man has Rs. 49 in coins of denominations pf Rs. 10, Rs. 5 and Re. 1. The ratio between Rs 10 coins and Rs. 5 coins is 1:2 and total coins are 15. How many coins of each denomination have the man?


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Answer 1

Answer:

9  

Step-by-step explanation:

Suppose x, y and z are the coins of Rupees 10 , 5 and 1 respectively;

According to the statements :

10x+5y+1z = 49; eq. 1

x+y+z = 15; eq. 2

x/y = 1/2 or x = 0.5y;

Now putting x = 0.5y in eq. 1 and eq. 2;

10*0.5y+5y+1z = 49;

10y+1z = 49; A

0.5y+y+z = 15;

1.5y+z =15; B

now solving (A-B) both the equations we get :

8.5y = 34;

y = 34/8.5;

y = 4

x = 0.5y = 2;

x+y+z = 15;

2+4+z = 15;

z = 15-6 = 9  

Answer 2

STEP-BY-STEP EXPLANATION:

.

$\sf Let's,$

$\sf The \: number \: of \: Rs. \: 10 \: be \: x.$

$\sf The \: number \: of \: Rs. \: 5 \: will \: be \: 2x.$

$\sf The \: Total \: of \: Coins = 15 \: \: \: ...(Given)$

$\sf The \: number \: of \: Rs. \: 1 + x+ 2x = 15$

$\sf The \: number \: of \: Rs. \: 1 + 3x= 15$

$\sf The \: number \: of \: Rs. \: 1 = 15 - 3x$

$\sf Man \: has \: Rs. \: 49 \: \: \: …(Given)$

$\sf So,$

$\implies \sf 10(x) + 5(2x) + 1(15 - 3x) = 49$

$\implies \sf 10x + 10x + 15 - 3x = 49$

$\implies \sf 17x + 15 = 49$

$\implies \sf 17x = 49 - 15$

$\implies \sf 17x = 34$

$\implies \sf x = 2$

$\bf The \: number \: of \: Rs. \: 10 \: is \: 2.$

$\bf The \: number \: of \: Rs. \: 5 \: is \: 4 \: [ 2 \times 2 = 4 ] .$

$\sf The \: number \: of \: Rs. \: 1 = 15 - 3x$

$\sf The \: number \: of \: Rs. \: 1 = 15 - 3(2)$

$\bf The \: number \: of \: Rs. \: 1 \: is \: 9.$