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Bunuel
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Let present age be x

1 year from today => x + 1 , equal to
twice his age 10 yrs ago => 2(x - 10)

x+1 = 2(x - 10)
x = 21
Choice C

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IMO:- C

Let us assume the present age of Saurabh be "y"

so y +1 = 2 (y-10)
y = 21

Check:- 22 will be 2 times of 11
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Bunuel
Today is Saurabh's birthday. One year from today he will be twice as old as he was 10 years ago. How old is Saurabh today?

A. 19 years
B. 20 years
C. 21 years
D. 22 years
E. 23 years

Solution


    • Let us assume that Saurabh is x years old today.
    • One year from today, his age will be (x + 1) years.
    • Also, ten years ago his age would have been (x - 10) years.
      o So, \((x +1) = 2*(x -10) \)
      \(⟹ x + 1 = 2x – 20\)
      \(⟹ x = 21\)
Thus, the correct answer is Option C.
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Bunuel
Today is Saurabh's birthday. One year from today he will be twice as old as he was 10 years ago. How old is Saurabh today?

A. 19 years
B. 20 years
C. 21 years
D. 22 years
E. 23 years

Solution:

Let’s let x = Saurabh’s current age. One year from today he will be (x + 1) years old, and 10 years ago, he was (x - 10) years old. We can create the equation:

x + 1 = 2(x - 10)

x + 1 = 2x - 20

21 = x

Answer: C
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Bunuel
Today is Saurabh's birthday. One year from today he will be twice as old as he was 10 years ago. How old is Saurabh today?

A. 19 years
B. 20 years
C. 21 years
D. 22 years
E. 23 years
S+1=2(S-10)
S=21
C:)
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