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Bunuel
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If n = (2^{-10}+2^{-9})/((7^(-1)) (5^9)) then n is a terminating 27 Nov 2025, 23:28
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65% (hard)

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55% (01:57) correct 45% (02:02) wrong based on 166 sessions

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If \(n =\frac{2^{-10}+2^{-9}}{(7^{-1})(5^9)}\) then n is a terminating decimal with how many zeroes after the decimal point before the first non-zero digit?

(A) 2
(B) 3
(C) 5
(D) 7
(E) 9


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D
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If n = (2^{-10}+2^{-9})/((7^(-1)) (5^9)) then n is a terminating 28 Nov 2025, 00:24
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Bunuel
If \(n =\frac{2^{-10}+2^{-9}}{(7^{-1})(5^9)}\) then n is a terminating decimal with how many zeroes after the decimal point before the first non-zero digit?

(A) 2
(B) 3
(C) 5
(D) 7
(E) 9
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Simplifying the numerator: 2^(-10) + 2^(-9) = 2^(-10) * (1 + 2) = 3 * 2^(-10)

Thus, we have: 3 * 2^(-10)/[7^(-1) * 5^9]
= 3* 7^1 / [2^10 * 5^9]
= 21 * 5/[2^10 * 5^10]
= 105/10^10
= 0.0000000105

Thus, there are 7 zeroes after the decimal point before the first non-zero digit
Ans D
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If n = (2^{-10}+2^{-9})/((7^(-1)) (5^9)) then n is a terminating 28 Nov 2025, 00:28
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Bunuel
If \(n =\frac{2^{-10}+2^{-9}}{(7^{-1})(5^9)}\) then n is a terminating decimal with how many zeroes after the decimal point before the first non-zero digit?

(A) 2
(B) 3
(C) 5
(D) 7
(E) 9


­
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The given expression is

\(n =\frac{2^{-10}+2^{-9}}{(7^{-1})(5^9)}\)

let’s take the numerator alone,

2^-10 + 2^-9

= 2^-9 (2^-1 +1)

= 2^-9 ( 3/2)

= 3* 2^-(9+1)

= 3*2^-10

The Denominator is (7^-1 *5^9)

Combining both we get,

= [3*2^-10 ]/ (7^-1 *5^9)

= (3*7)/(2^10 * 5^9)

= 21 / (2^10 * 5^9)

Multiplying by 5 in the numerator and denominator, we get

= (21 * 5 )/ (2^10 * 5^9*5)

= 105 / (2^10 * 5^10)

= 105 / 10^10

The number of Zeroes after the decimal point = 10 - (Number of digits in the numerator)

= 10 - 3

= 7

Option D
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If n = (2^{-10}+2^{-9})/((7^(-1)) (5^9)) then n is a terminating 30 Nov 2025, 05:46
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The given expression is

n=2^−10+2^−9?(7^−1)(5^9)
when we simply this expression
taking 2^-9(2 ^-1+1)*7/5^9
10.5/2^9*5^9
so we get 10.5* 10^-9
we get seven zeros after which first non zero digit will appear

Hence option D is correct
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If n = (2^{-10}+2^{-9})/((7^(-1)) (5^9)) then n is a terminating 30 Nov 2025, 23:13
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This problem can be solved by simplification
Given n= (2^-10 + 2^-9) / (7^-1 x 5^9)
taking 2^-9 common and bringing it to denominator generates 10^-9
taking 7^-1 to numerator gives 21/2 as value

Hence multiplying with 10^-9 gives 7 zero digits after decimal before the first non zero digit (Option D)
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