Question

plss help me to find answer​

Answer 1

Solution :

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(1)

Let, The Present Age Of Anju Be x.

And The Present Age Of Manju Be y.

Then, According To First Case,

➝ x = 2y ____(1)

Now, According To Second Case,

And Six Years Ago,

➝ x - 6 = 5(y - 6)

➝ x - 6 = 5y - 30

➝ x - 5y - 6 + 30 = 0

➝ x - 5y + 24 = 0 ____(2)

Now We Can Put x = 2y in Eq.(2) (According To First Case)

➝ x - 5y + 24 = 0 ____(2)

➝ 2y - 5y + 24 = 0

➝ - 3y + 24 = 0

➝ 3y = 24

➝ y = $\sf \frac{24}{3}$

➝ y = $\sf \frac{\cancel{24}^{\: \: 8 × \cancel3}}{\cancel{3}}$

➝ $\underline{\boxed{\bf y = 8}}$

Now, Put Value Of y In Eq.(1) or Eq.(2)

I Like To Solve x Using By Eq.(1)

[Note : You Can Use Any From Above Eq. For x]

➝ x = 2y ____(1)

➝ x = 2(8)

➝ $\underline{\boxed{\bf x = 16 }}$

Hence,

Present Age Of Anju (x) = 16

Present Age Of Manju (y) = 8

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(2)

Let, The Two Digit Number Be Tens = x And Once = y

Hence, Let The Two Digit Number Be xy.

Now, As Per The Given Question One Of The Two Digit of Two Digit Number Is Two Times The Other Digit,

Therefore,

➝ y = 2x

Now, We Will Break xy In Tens And Once

(xy) ↓

➝ 10x + y (Because Of X is Tens)

Now, Using y = 2x In above Eq.

➝ 10x + y

➝ 10x + 2x

➝ 12x

Now, Interchanging The Digits Of This Two Digit Number,

(yx) ↓

➝ 10y + x

Now, Again Using y = 2x In above Eq.

➝ 10(2x) + x

➝ 20x + x

➝ 21x

Now, According To Last Condition,

Add The Resulting Number To The Original Number, You Get 132

Hence,

➝ xy + yx = 132

➝ 12x + 21x = 132

➝ 33x = 132

➝ x = $\sf \frac{132}{33}$

➝ x = $\sf \frac{\cancel{132}^{\:\: \cancel{33} × 4}}{\cancel{33}}$

➝ x = 4

Now, Putting The Value Of x In y = 2x,

➝ y = 2x

➝ y = 2(4)

➝ y = 8

Now, Finally,

We Have The Number Of Two Digits = xy

So,

➝ xy

➝ (4)(8)

➝ $\underline{\boxed{\bf 32 }}$

Hence,

The Number Of Two Digits (xy) = 32

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I hope it helps you...