Math, asked by Swarup1998, 7 months ago

2. DOLLARS AND CENTS

A man entered a store and spent one-half ofthe money that was in his pocket. When he came out he found that he had just as many cents as he had dollars when he went in and half as many dollars as he had cents when he went in. How much money did he have on him when he entered?​

Answers

Answered by SujalSirimilla
6

Answer:

  • 1/2 of the money was spent.
  • Let cents be x and dollars be y.
  • And, when he went in, x = 1/2 y. (Well this is obvious)

Now, I hope you know that 100 cents = 1 dollar.

So 1 cent = 1 dollar / 100.

Now, lets see our equations.

When he enters the store, he will have both cents and dollars with him. Therefore, he will have x+y amount of money.

or, we convert cents into dollars and substituting dollars as half cents:

=> x/100 + y/2 ___(1)

I hope its clear till here.

⊱ ──────  ✯  ────── ⊰

Now, he walks out with 1/2 the money. or,

=> 1/2(x+y/100);

=> x/2 + y/200 ___(2)

From (1) & (2):

=> x/100 + y/2 = x/2 + y/200

=> (2x + 100y)/200 = (2y + 200x)/400

=> -49x/100 = -99/200 y

=> x/y = 99/98

Comparing; we get:

x = 99; y = 98.

So his initial amount is x + y/100 => 99 + 98/100 => 99.98.

He had 99.98, which is approximately equal to 100 dollars.

Thanks :D

Answered by studarsani18018
0

Answer:

of Dollars after he came out = half of cents he had before he went in= 98/2 = 49.

Step-by-step explanation:

He spent one-half of the money = $49.99

Money Remains in his pocket= $49.99

Acc. To Statement:

(i) No. of Dollars before he went in = 99

No. of Cents after he came out =No. of Dollars before he went in = 99

(ii) No. of Cents before he went in = 98

No. of Dollars after he came out = half of cents he had before he went in= 98/2 = 49.

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