Question

neha has three times as many rs2 coins as she has rs5 coins if she has a sum rs77 how many coins of denomination does she have​

Answer 1

Given -

• Neha has three times as many ₹2 coins as she has ₹5.

• She has the sum of ₹77

To find -

• Each denomination of coins does she have.

Solution -

Let the ₹5 coins be x

Then, ₹2 coins be 3x

Now,

Sum of all the equations = ₹77

According to the question -

→ 5x + 2(3x) = 77

→ 5x + 6x = 77

→ 11x = 77

→ x = 77/11

$\implies$ x = 7

On substituting the values -

₹5 coins, x = 7 coins

₹2, 3x = 3 × 7 = 21 coins

$\therefore$ The denomination of coins she have is ₹5 = 7 and ₹2 = 21

Verification -

placing 7 in place of x -

→ 5(7) + 2(3 × 7) = 77

→ 35 + 2(21) = 77

→ 35 + 42 = 77

→ 77 = 77

LHS = RHS

Hence, proved!

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Answer 2

Given :-

• Neha has three times as many ₹2 coins as she has ₹5.
• She has the sum of ₹77

To Find :-

Each denomination of coins does she have.

SoluTion :-

Let

The coins be x

As it is given that she have 3 times more than ₹2 coins.

$\sf \: 5x + 2(3x)= 77$

$\sf \: 5x + 6x = 77$

$\sf \: 11x = 77$

$\sf \: x \: = \dfrac{77}{11}$

${\huge \: {\boxed {\mathfrak { \green{ x = 7}}}}} \bigstar$

$\sf \: ₹5 \: coins(x) = 7 \: coins$

$\sf \: ₹2, 3x = 3 × 7 = 21 coins$

Lets verify

$\sf \: 5(x) + 2(3x) = 77$

$\sf \: 5(7) + 2(3 \times 7) = 77$

$\sf \: 5(7) + 2(21) = 77$

$\sf \: 35 + 42 = 77$

$\sf \: 77 = 77$

LHS = RHS