From bc9c6fe7cb11fb0a59b39ca6f7661cb6013f513c Mon Sep 17 00:00:00 2001
From: Tim Graham <timograham@gmail.com>
Date: Tue, 6 Jun 2017 16:11:48 -0400
Subject: [PATCH] [1.11.x] Fixed #28233 -- Used a simpler example in the
 aggregation "cheat sheet" docs.

Backport of 49b9c89d4094574117c9d5b7a696ce152e02553a from master
---
 docs/topics/db/aggregation.txt | 10 +++++-----
 1 file changed, 5 insertions(+), 5 deletions(-)

diff --git a/docs/topics/db/aggregation.txt b/docs/topics/db/aggregation.txt
index aa56a8b4eb..6f7d3918af 100644
--- a/docs/topics/db/aggregation.txt
+++ b/docs/topics/db/aggregation.txt
@@ -67,11 +67,11 @@ In a hurry? Here's how to do common aggregate queries, assuming the models above
     >>> Book.objects.all().aggregate(Max('price'))
     {'price__max': Decimal('81.20')}
 
-    # Cost per page
-    >>> from django.db.models import F, FloatField, Sum
-    >>> Book.objects.all().aggregate(
-    ...    price_per_page=Sum(F('price')/F('pages'), output_field=FloatField()))
-    {'price_per_page': 0.4470664529184653}
+    # Difference between the highest priced book and the average price of all books.
+    >>> from django.db.models import FloatField
+    >>> Book.objects.aggregate(
+    ...     price_diff=Max('price', output_field=FloatField()) - Avg('price')))
+    {'price_diff': 46.85}
 
     # All the following queries involve traversing the Book<->Publisher
     # foreign key relationship backwards.