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Average Practice Test 4
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© examsnet.com
Question : 8
Total: 40
The average of 5 consecutive integers starting from m is n. What is the average of 6 consecutive integers starting from (m + 2)?
2
n
+
5
2
n
+
3
n
+
2
2
n
+
9
2
None of these
Validate
Solution:
m
+
m
+
1
+
m
+
2
+
m
+
3
+
m
+
4
5
=
n
5
m
+
10
=
5
n
m
+
2
=
n
or
m
=
n
−
2
…(i)
∴ Required average
=
m
+
2
+
m
+
3
+
m
+
4
+
m
+
5
+
m
+
6
+
m
+
7
6
=
6
m
+
27
6
=
2
m
+
9
2
=
2
(
n
−
2
)
9
2
[taking eq (1)]
=
2
n
+
5
2
© examsnet.com
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