TLB-Memory-access-live-test-2

Question

A system uses $2$ level paging scheme.

A regular memory access takes $100$ ns and servicing a page fault takes $10$ millisecond.

An avg instruction takes $100$ ns of CPU time and $2$ memory access.

The TLB hit ratio is $95\%$.

and page fault is $1$ in every $10,000$ memory access.

Avg instruction execution time.$?$

Options given are 

$1300$ns

$1150$ns

$2320$ns

$1275$ns

Comments

Here is my work :
$\{0.95 \times (\frac{1}{10^4}\times 10\times 10^6+(1-\frac{1}{10^4})\times 100)\}\\ \{+0.05\times (200+\frac{1}{10^4}\times 10\times 10^6+(1-\frac{1}{10^4})\times 100)\}$

$\Rightarrow1109.99$ ns.

Is something wrong with this work?

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This is the given solution .

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= {

100(CPU) +

2*100(2 mandatory mem access) +

(for 2 memory access, so mul by 2; TLB access time not given so considering it as 0; 2 level page table)

2* [ 0.95*0 + 0.05*( 2*100 ) ]

+ ( page fault rate* page fault service time )1000

} ns

= 1320 ns.

I think they have done a typo.

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ref: https://gateoverflow.in/318/gate2004-47