Higher Order Thinking Skills - Number System | Extra Documents & Tests for Class 9 PDF Download

 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 
– 21
21 +
.
Solution: Rationalising the denominator of 
2–1
21 +
, we get
2–1
21 +
=
2–1 2–1
21 2–1
×
+
 = 
()
()
()
2
2
2
2–1
2–1
=
()
2
2–1
2–1
 = ()
2
2–1
?  
2–1
21 +
=
()
2
2–1 or  
2–1
21 +
 =  
2–1
........ (1)
ä   
2
= 1.4142
â  
2–1
21 +
=
2–1
= 1.4142 – 1 = 0.4142
Question 3. Find the value of 'a' in 
–
–
35 19
a5
32 5 11
=
+
Solution:
L.H.S. =
3– 5
32 5 +
=
3– 5
32 5 +
 × 
3– 2 5
3–2 5
Rationalising the
denominator
 
=
()
()
2
2
9–6 5 –3 5 10
3–25
+
     
22
using (a b) (a – b) a – b +=
=
19 – 9 5 19 – 9 5
9 – 20 –11
= = 
919
5–
11 11
R.H.S. =
19
5–
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 
– 21
21 +
.
Solution: Rationalising the denominator of 
2–1
21 +
, we get
2–1
21 +
=
2–1 2–1
21 2–1
×
+
 = 
()
()
()
2
2
2
2–1
2–1
=
()
2
2–1
2–1
 = ()
2
2–1
?  
2–1
21 +
=
()
2
2–1 or  
2–1
21 +
 =  
2–1
........ (1)
ä   
2
= 1.4142
â  
2–1
21 +
=
2–1
= 1.4142 – 1 = 0.4142
Question 3. Find the value of 'a' in 
–
–
35 19
a5
32 5 11
=
+
Solution:
L.H.S. =
3– 5
32 5 +
=
3– 5
32 5 +
 × 
3– 2 5
3–2 5
Rationalising the
denominator
 
=
()
()
2
2
9–6 5 –3 5 10
3–25
+
     
22
using (a b) (a – b) a – b +=
=
19 – 9 5 19 – 9 5
9 – 20 –11
= = 
919
5–
11 11
R.H.S. =
19
5–
11
a
since,  L.H.S. =  R.H.S.
  
i.e. 
919
5–
11 11
  =  
19
5–
11
a
  ?
9
 
11
= a
Question 4. Find the value of 'a' and 'b'
 
75 7–5
–
7– 5 7 5
+
+
 =  
7
a5b
11
+
Solution:
L.H.S. = 
75 7–5
–
7– 5 7 5
+
+
=   
()()
()()
22
75 –7–5
7– 5 7 5
+
+
=
49 514 5–49–514 5
49 – 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
–2
??
+
??
??
a
a
? 
2
2
1
+ a
a
=
2
35 2
–2
23 5
??
+
+
??
+
??
 = 
()
()
2
2
2
35 2
–2
23 5
??
++
??
??
+
??
??
=
2
95 2 3 5 4
–2
23 2 5
??
++ × × +
??
×+
??
=
2
18 6 5
–2
62 5
??
+
??
+
??
=
()
()
2
63 5
–2
23 5
??
+
??
??
+
??
=
2
6
–2
2
??
??
??
= ()
2
3–2
=9 – 2 = 7
So, 
2
2
1
+ a
a
= 7