Preparation key

CBSE Class 9 Maths Sample Paper Solution Set 1
Answers & Explanations
Section A
1. Solution:

81 𝑦2
We have, 36 𝑥 2 − 25

9 2 𝑦 2
= (6 𝑥) − (5 )

9 𝑦 9 𝑦
= (6 𝑥 + 5 ) (6 𝑥 − 5 ) [∵ 𝑎2 − 𝑏 2 = (𝑎 + 𝑏)(𝑎 − 𝑏)]

81 2 𝑦2 9 𝑦 9 𝑦
Hence, 𝑥 − = ( 𝑥 + )( 𝑥 − )
36 25 6 5 6 5




2. Solution:

Let 𝑃(𝑥) = 𝑥 2 + 9𝑥 − 5 + 𝑘

∵𝑥−1=0⇒𝑥 =1

∴ 𝑃(1) = 0

𝑃(1) = 12 + 9 ∙ 1 − 5 + 𝑘

⇒1+9−5+𝑘 =0

⇒𝑘+5=0

⇒ 𝑘 = −5

Hence, 𝑘 = −5.



3. Solution:
2
93
We have, 1
95

2
2 1
93 𝑎𝑚
= 1 = 93−5 [∵ 𝑎𝑛
= 𝑎𝑚−𝑛 ]
95

10−3
=9 15


7
= 915

, 2
7
93
Hence, 1 = 915 .
95

OR

Solution
49
(343)𝑚 = 𝑚
7

72
(343)𝑚 =
7𝑚

(73 )𝑚 = 72−𝑚

As bases are equal, we can equate the powers

3𝑚 = 2 − 𝑚

4𝑚 = 2
1
𝑚=2



4. Solution:

Given, 𝑃(4,6) and 𝑄(−5, −7)

∴ Abscissa of 𝑃 = 4 and abscissa of 𝑄 = −5

∴ (Abscissa of 𝑃) − (abscissa of 𝑄) = 4 − (−5) = 4 + 5 = 9

Hence, (abscissa of 𝑃) − (abscissa of 𝑄) = 9.



5. Solution:

Given 𝑎 = 25 𝑐𝑚, 𝑏 = 20 𝑐𝑚, 𝑐 = 15 𝑐𝑚

∴ Area of the triangle = √𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐)

To find s:
1
𝑠 = 2 (𝑎 + 𝑏 + 𝑐)

25+20+15 60
𝑠= = = 30
2 2


∴ Area of the triangle = √30(30 − 25)(30 − 20)(30 − 15)

, = √30 × 5 × 10 × 15 = √22500 = 150 𝑐𝑚2

Hence, the area of the triangle = 150 𝑐𝑚2 .

OR

Solution:

Given, interval = 100 − 110.

Here, lower limit = 100 and upper limit = 110

upper limit+lower limit
∴ Class mark = 2

110+100 210
= = = 105
2 2

Hence, the class mark of the interval 100 − 110 is 105.



6. Solution:

Here, length (𝑙) = 20 𝑚, breadth (𝑏) = 20 𝑚 and height (ℎ) = 10 𝑚.

∴ Length of the diagonal in the cuboid = Length of the longest rod

= √𝑙 2 + 𝑏 2 + ℎ2

= √202 + 202 + 102

= √400 + 400 + 100

= √900

= 30 𝑚

Hence, the length of the longest rod = 30 𝑚



Section B

7. Solution:

̅̅̅̅
Let 𝑥 = 0.78

Then, 𝑥 = 0.7878 . . . . .. (1)

Multiplying 100 on both sides in equation (1), we get