Question

A person wanted to withdraw AA dollars and BB cents from the bank. But the cashier made a mistake and gave him BB dollars and AA cents. Neither the person nor the cashier noticed this. After spending 20 cents, the person counts the money, and to their surprise, they have double the amount they wanted to withdraw. 1 Dollar equals 100 Cents. 2 What are AA and BB? O AA=26, BB=53 O AA=14,BB=34 O None of the options O AA=34, BB=64?​

Answer 1

Solution :-

→ Person wanted teo withdraw = $A , B cents = (A * 100 + B) cents .

→ Person gets = $B , A cents = (B * 100 + A) cents .

given that, person spends 20 cents from what he gets and amount left is double what of he wanted to withdraw .

So,

→ (100B + A) - 20 = 2(100A + B)

→ 100B + A - 200A - 2B = 20

→ 98B - 199A = 20 -------- Eqn.(1)

Since it is given that, when A and B are interchanged , amount gets double .

So ,

• Either B should be twice of A => B = 2A
• Or, B is one more than twice of A => B = (2A + 1)

Putting B = 2A in Eqn.(1),

→ 98(2A) - 199A = 20

→ 196A - 199A = 20

→ (-3A) = 20

→ A = (-20/3) ≠ Not possible .

Putting B = (2A + 1) in Eqn.(1),

→ 98(2A + 1) - 199A = 20

→ 196A + 98 - 199A = 20

→ 3A = 98 - 20

→ 3A = 78

→ A = 26 .

therefore,

→ B = 2A + 1 = 2*26 + 1 = 52 + 1 = 53 .

Hence, value of A is $26 and B is 53 cents .

Method (2) :- Solving from only one Equation ,

→ 98B = 20 + 199A

→ B = (20 + 199A)/98

Put values of A such that B is also an integer , or now from options put values of A and check which option gives B as an integer .

taking A = 26,

→ B = (20 + 199*26)/98

→ B = (20 + 5174)/98

→ B = (5194/98)

→ B = 53 = An integer .

therefore, value of A is $26 and B is 53 cents .

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Answer 2

Answer:

The value of A is $ $26$ and the value of B is $53$ cents.

Step-by-step explanation:

We have been given that the person here spends 20 cents from the total amount that he gets. The leftover amount is double the one that he wished to withdraw.

He wished to withdraw, $$A$, $B$ cents = $(100A+B)$ cents.

He gets, $$B$, $A$ cents = $(100B +A)$ cents.

Here,

$(100B +A)-20 = 2(100A+B)$

$100B+A-200A-2B=20$

$98B-199A=20 \ ......(1)$

According to the question, when $A$ and $B$ are interchanged, the amount is doubled.

This entails two situations,

1. $B$ is twice $A$, i.e., $B=2A$
2. $B$ is one more than twice $A$, i.e., $B=2A+1$

Now, using $B=2A$ in the equation $(1)$

$98(2A+1)-199A=20\\196+98-199A=20\\3A=98-20\\3A=78\\A=26$

Substituting this value of $A$ in $B=2A+1$ we get,

∴ $B=2A+1\\B= 2 \times 26 +1\\B =52 +1\\B=53$

Hence, the value of $A$ is $ $26$ and the value of $B$ is $53$ cents.

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