Question

in my elder brother's bag there are rupees 350 with the coins of Rupees 1 and 50 paise Together by sister has put out one third part of 50 paise coin from the bag and it that place equal number of coins of Rupees 1 she has put into the bag and now the total amount of money in the bag is rupees 400 is calculate and write the original number of coins of rupees 1 and 50 paise kept at for separately in my brother's bag.
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Answer 1

Given :

Total sum of money in bag = Rs 350

The number of coins is in Rs 1 and 50 paise

$\dfrac{1}{3}$ of 50 paise coin = Rs 1 coin ion bag, so total amount of money = Rs 400

To Find :

The original number of Rs 1 coin

The original number of 50 paise coin

Solution :

Let The original number of Rs 1 coin = x

The original number of 50 paise coin = y

According to question

  Rs 1 + 50 coin = Rs 350

i.e   x + $\dfrac{y}{2}$ = 350           .........1

Again

  x + $\dfrac{y}{2}$ = 400               ............2

And   $\dfrac{1}{3}$ y = x               ..........3

From eq 1

$\dfrac{y}{3}$ +  $\dfrac{y}{2}$ = 350

taking LCM

   $\dfrac{3y+2y}{6}$ = 350

Or,    $\dfrac{5y}{6}$ = 350

∴     y = 70 × 6

i.e  y = 420

So, The number of 50 paise coin = y = 420

Put the value of y into eq 3

So,    x = $\dfrac{y}{3}$

            = $\dfrac{420}{3}$

            = 140

So, The number of Rs 1 coin = x = 140

Hence, The number of Rs 1 coin is 140

And  The number of 50 paise coin is 420   Answer

Answer 2

Given :-

• Total amount in beg = Rs.350 with the coins of ₹1 and ₹50 paise together.
• Sister took out = (1/3) of 50 paise coins and put equal number od Rs.1 coins .
• Now total amount in beg = Rs.400 .

To Find :-

• original number of coins of ₹1 and ₹50 paise .

Solution :-

Let us assume that, Original number of Rs.1 coins were x and 50 paise coins were y .

so,

→ 1 * x + (1/2) * y = 350

→ 2x + y = 700 -------------- Eqn.(1)

now,

→ coins sister took out = (1/3) of y = (y/3)

and,

→ coins she added of Rs.1 = (y/3)

then,

Total coins of Rs.1 = x + (y/3) = (3x + y)/3

Total coins of 50 paise = y - y/3 = (2y/3)

so,

→ Total amount in beg = Rs.400

→ 1 * (3x + y)/3 + (1/2) * (2y/3) = 400

→ (3x + y)/3 + (y/3) = 400

→ 3x + y + y = 1200

→ 3x + 2y = 1200 -------------- Eqn.(2)

Multiply Eqn.(1) by 2 and subtracting Eqn.(2) from result,

→ 2(2x + y) - (3x + 2y) = 2*700 - 1200

→ 4x - 3x + 2y - 2y = 1400 - 1200

→ x = 200 (Ans.)

putting value of x in Eqn.(1), we get,

→ 2 * 200 + y = 700

→ 400 + y = 700

→ y = 700 - 400

→ y = 300 (Ans.)

Hence, original number of coins of ₹1 and ₹50 paise kept at first separately in my brother's bag were 200 and 300 respectively .

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