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# Higher Order Thinking Skills - Number System Class 9 Notes | EduRev

Created by: Indu Gupta

## Class 9 : Higher Order Thinking Skills - Number System Class 9 Notes | EduRev

``` Page 1

Question 1. Find the value of
32 48
812
+
+
Solution:
Since 32 16 2 =× = 42; 48 16 3 4 3 =×=
8 = 42 × =
22and 12 4 3 2 3 =× =
?
32 48
812
+
+
=
42 4 3
22 2 3
+
+
=
()
()
42 3
2
22 3
+
=
+
Question 2. If
2
= 1.4142 then find the value
â€“ 21
21 +
.
Solution: Rationalising the denominator of
2â€“1
21 +
, we get
2â€“1
21 +
=
2â€“1 2â€“1
21 2â€“1
×
+
=
()
()
()
2
2
2
2â€“1
2â€“1
=
()
2
2â€“1
2â€“1
= ()
2
2â€“1
?
2â€“1
21 +
=
()
2
2â€“1 or
2â€“1
21 +
=
2â€“1
........ (1)
ä
2
= 1.4142
â
2â€“1
21 +
=
2â€“1
= 1.4142 â€“ 1 = 0.4142
Question 3. Find the value of 'a' in
â€“
â€“
35 19
a5
32 5 11
=
+
Solution:
L.H.S. =
3â€“ 5
32 5 +
=
3â€“ 5
32 5 +
×
3â€“ 2 5
3â€“2 5
Rationalising the
denominator

=
()
()
2
2
9â€“6 5 â€“3 5 10
3â€“25
+

22
using (a b) (a â€“ b) a â€“ b +=
=
19 â€“ 9 5 19 â€“ 9 5
9 â€“ 20 â€“11
= =
919
5â€“
11 11
R.H.S. =
19
5â€“
11
a
since,  L.H.S. =  R.H.S.
Page 2

Question 1. Find the value of
32 48
812
+
+
Solution:
Since 32 16 2 =× = 42; 48 16 3 4 3 =×=
8 = 42 × =
22and 12 4 3 2 3 =× =
?
32 48
812
+
+
=
42 4 3
22 2 3
+
+
=
()
()
42 3
2
22 3
+
=
+
Question 2. If
2
= 1.4142 then find the value
â€“ 21
21 +
.
Solution: Rationalising the denominator of
2â€“1
21 +
, we get
2â€“1
21 +
=
2â€“1 2â€“1
21 2â€“1
×
+
=
()
()
()
2
2
2
2â€“1
2â€“1
=
()
2
2â€“1
2â€“1
= ()
2
2â€“1
?
2â€“1
21 +
=
()
2
2â€“1 or
2â€“1
21 +
=
2â€“1
........ (1)
ä
2
= 1.4142
â
2â€“1
21 +
=
2â€“1
= 1.4142 â€“ 1 = 0.4142
Question 3. Find the value of 'a' in
â€“
â€“
35 19
a5
32 5 11
=
+
Solution:
L.H.S. =
3â€“ 5
32 5 +
=
3â€“ 5
32 5 +
×
3â€“ 2 5
3â€“2 5
Rationalising the
denominator

=
()
()
2
2
9â€“6 5 â€“3 5 10
3â€“25
+

22
using (a b) (a â€“ b) a â€“ b +=
=
19 â€“ 9 5 19 â€“ 9 5
9 â€“ 20 â€“11
= =
919
5â€“
11 11
R.H.S. =
19
5â€“
11
a
since,  L.H.S. =  R.H.S.

i.e.
919
5â€“
11 11
=
19
5â€“
11
a
?
9

11
= a
Question 4. Find the value of 'a' and 'b'

75 7â€“5
â€“
7â€“ 5 7 5
+
+
=
7
a5b
11
+
Solution:
L.H.S. =
75 7â€“5
â€“
7â€“ 5 7 5
+
+
=
()()
()()
22
75 â€“7â€“5
7â€“ 5 7 5
+
+
=
49 514 5â€“49â€“514 5
49 â€“ 5
++ +
=
47 5
75
44 11
??
??
=
=
7
05
11
+
R.H.S. =  a +
7
5b
11
Since, L.H.S. = R.H.S
? 0 +
7
5
11
=  a +
7
5b
11
? a = 0 and b = 1
Question 5. If  a =
35
2
+
, then find the value of +
2
2
1
a
a
.
Solution:  a =
35
2
+
?
1
a
=
2
35 +
Now
2
2
1
+ a
a
=
2
1
â€“2
??
+
??
??
a
a
?
2
2
1
+ a
a
=
2
35 2
â€“2
23 5
??
+
+
??
+
??
=
()
()
2
2
2
35 2
â€“2
23 5
??
++
??
??
+
??
??
=
2
95 2 3 5 4
â€“2
23 2 5
??
++ × × +
??
×+
??
=
2
18 6 5
â€“2
62 5
??
+
??
+
??
=
()
()
2
63 5
â€“2
23 5
??
+
??
??
+
??
=
2
6
â€“2
2
??
??
??
= ()
2
3â€“2
=9 â€“ 2 = 7
So,
2
2
1
+ a
a
= 7
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