The answer is: Â "6 (six) books" .
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Explanation:
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The number of books is: Â "(â…–)*15" .
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Method 1)
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 Â
![\frac{2}{5}](/tpl/images/0307/8223/5255a.png)
* 15 =  [(15 ÷ 5) * 2] = 3 * 2 = 6 .
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Method 2)Â
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![\frac{2}{5}](/tpl/images/0307/8223/5255a.png)
* 15 ;
=
![\frac{2}{5}](/tpl/images/0307/8223/5255a.png)
*Â
![\frac{15}{1}](/tpl/images/0307/8223/a5681.png)
;
=Â
![\frac{2*15}{5*1}](/tpl/images/0307/8223/02f5c.png)
=Â
![\frac{30}{5}](/tpl/images/0307/8223/62728.png)
= 6 .
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Method 3)
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![\frac{2}{5}](/tpl/images/0307/8223/5255a.png)
* 15 ;
=
![\frac{2}{5}](/tpl/images/0307/8223/5255a.png)
*Â
![\frac{15}{1}](/tpl/images/0307/8223/a5681.png)
;
  → At this point, we can "cancel out" the "5" to a "1"; &
              we can "cancel out" the "15" to a "3" ;Â
             → {Since:  "15 ÷ 5 = 3" ;  and since: "5 ÷ 5 = 1"}.
 → And rewrite as: =
![\frac{2}{1}](/tpl/images/0307/8223/40438.png)
*Â
![\frac{3}{1}](/tpl/images/0307/8223/89cb8.png)
;
           → At this point, we can rewrite as:
                 " 2 * 3 " ;  and: "2 * 3 = 6 " .Â
       →  {since: "any value, divided by "1", equals that same value};
As such: Â "
![\frac{2}{1}](/tpl/images/0307/8223/40438.png)
= 2 "; Â AND:Â
![\frac{3}{1}](/tpl/images/0307/8223/89cb8.png)
= 3 .
→  So:
![\frac{2}{1}](/tpl/images/0307/8223/40438.png)
*Â
![\frac{3}{1}](/tpl/images/0307/8223/89cb8.png)
 = {2 * 3} = 6 .
→ Alternately, we can continue as follows:
![\frac{2}{1}](/tpl/images/0307/8223/40438.png)
*Â
![\frac{3}{1}](/tpl/images/0307/8223/89cb8.png)
;
= Â
![\frac{2*3}{1*1}](/tpl/images/0307/8223/1d8dd.png)
 =
![\frac{6}{1}](/tpl/images/0307/8223/121c4.png)
= 6 .
______________________________________________________ Â Â Â Â
Â
Method 4)
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![\frac{2}{5}](/tpl/images/0307/8223/5255a.png)
* 15 ;
      =
![\frac{2}{5}](/tpl/images/0307/8223/5255a.png)
*Â
![\frac{15}{1}](/tpl/images/0307/8223/a5681.png)
;
      =
![\frac{2*15}{5*1}](/tpl/images/0307/8223/02f5c.png)
Â
        →  At this point, we can "cancel out" the "15" in the numerator to a "3"; & we can "cancel out" the "5" in the denominator to a "1" ;
         {Since:  "15 ÷ 5 = 3" ; and since:  "5 ÷ 5 = 1" ;
        → and rewrite as follows:
        → Â
![\frac{2*3}{1*1}](/tpl/images/0307/8223/1d8dd.png)
 ;
      → At this point:  since:  "2 * 3 = 6" ;  &  since "1*1 = 1" ;
 Â
       We can rewrite as:  Â
→ Â
![\frac{2*3}{1*1}](/tpl/images/0307/8223/1d8dd.png)
 = Â
![\frac{6}{1} = 6 .\\Alternately, at the point when we have:\\→ [tex] \frac{2*3}{1*1}](/tpl/images/0307/8223/19b9b.png)
 ; Â
→  We can ELIMINATE the "denominator" completely;Â
Since the denominator, "(1*1)" is equal to "1" ; and since any value (e.g. the "numerator"); divided by "1" (e.g. the value of the denominator); equals the same value (e.g. that same value of the numerator);
As such, we can rewrite; and simplify; our expression (as follows):
  → Â
![\frac{2*3}{1*1}](/tpl/images/0307/8223/1d8dd.png)
  = { 2 * 3 }  = 6 .
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 Â
Method 5)
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![\frac{2}{5}](/tpl/images/0307/8223/5255a.png)
* 15 = ?
→ Convert "
![\frac{2}{5}](/tpl/images/0307/8223/5255a.png)
" to a decimal value:
Note: Â
![\frac{2}{5}](/tpl/images/0307/8223/5255a.png)
  = ? / 10  ???Â
What value belongs in the "question mark" ??
→ Let us examine the denominators.
We have "5" and "10".Â
→ 5 * (what value?) = 10?  ( "2" , by recognition);
→Nonetheless, to get that value:  "10 ÷ 5 = ? " ;  The answer is: "2" ;Â
→ To confirm:  "5 * 2 =? 10?  Yes!
→ As such: Â
![\frac{2}{5}](/tpl/images/0307/8223/5255a.png)
;
     Â
        = Â
![\frac{2*2}{5*2}](/tpl/images/0307/8223/af277.png)
;
            Â
        =
![\frac{4}{10}](/tpl/images/0307/8223/7bcbb.png)
;Â
Convert this value to a decimal value;Â
       Â
   →  4/10 =  4 ÷ 10 ; Â
      → To divide by "10" ; Take the decimal value (The decimal value in "4" is considered the value "directly after the "4"); and move that value backward ONE decimal space; {since we are DIVIDING by "10" ; and "10" has ONE "zero"};  to get: ".4" ;  → Write as:  "0.4"  ;
Â
    →Â
![\frac{2}{5}](/tpl/images/0307/8223/5255a.png)
 =
![\frac{4}{10}](/tpl/images/0307/8223/7bcbb.png)
= Â 0.4 Â .
Alternately, use a calculator to convert "
![\frac{2}{5}](/tpl/images/0307/8223/5255a.png)
" to a decimal value:
    → Â
![\frac{2}{5}](/tpl/images/0307/8223/5255a.png)
 =  2 ÷ 5 = 0.4 .
Now, we can rewrite:
   → Â
![\frac{2}{5}](/tpl/images/0307/8223/5255a.png)
* 15 Â ;
as: Â "(0.4)(15)" ; & calculate:
Â
    → (0.4)(15) = 6 .
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Â
 The answer is:  "6 (six) books" .
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