Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Mathematics
The value of √10+√25+√108+√154+√225 +(√(225/729)-√(25/144)) ÷ √(16/81)
Q. The value of
10
+
25
+
108
+
154
+
225
+
(
729
225
−
144
25
)
÷
81
16
814
131
Squares and Square Roots and Cubes and Cube Roots
Report Error
A
16
69
B
3
31
C
7
54
D
16
39
Solution:
10
+
25
+
108
+
154
+
225
+
(
729
225
−
144
25
)
÷
81
16
=
(
10
+
25
+
108
+
154
+
15
)
+
(
27
15
−
12
5
)
÷
(
9
4
)
[
As
,
225
=
3
2
×
5
2
729
=
3
6
=
3
3
=
27
]
=
(
10
+
25
+
108
+
13
)
+
(
36
20
−
15
)
×
4
9
[
As,
169
=
1
3
2
=
13
]
=
10
+
25
+
11
+
16
5
[
As,
121
=
1
1
2
=
11
]
=
10
+
36
+
16
5
=
10
+
6
+
16
5
=
4
+
16
5
=
16
69