K.K.

Wagh Education Society’s

K. K. WAGH UNIVERSAL SCHOOL, NASHIK
Annual Exam (2024-2025)
School Code: Subject: Maths Date: ___________________
Name:
Grade: IX Div.: Roll No: _________________
Time: 3:00 Hrs. Marks: 80 Invigilator’s Sign: _________
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ANSWER KEY

Question Part – A Marks

No.
Section-I
Section-I has 16 questions of 1 mark each. Internal choice is provided in 1
5 questions.
1. (c) there are infinitely many rational numbers 1
2. (a) Degree of 4x4 + Ox3 + Ox5 + 5x + 7 is equal to the highest power of variable x. 1
Here, the highest power of x is 4,
Hence, the degree of a polynomial is 4.
3. (a) Every point on the line y = x has same value of x-and y-coordinates i.e., x = a and
y = a. 1
Hence, (a, a) is the required form of the solution of given linear equation.

4. (a) Triangles 1
5. 1

6. 1
7. 1

8. (d) Decimal representation of a rational number cannot be non-terminating non- 1

repeating because the decimal expansion of rational number is either terminating or
non-terminating recurring (repeating).
(c) Since, the graph of linear equation 2x + 3y = 6
9. meets the X-axis. 1
So, we put y = 0 in 2x + 3y = 6 => 2x + 3(0) = 6
=> 2x + 0 = 6
=> x = 6/2 => x = 3
Hence, the coordinate on X-axis is (3, 0).
10. (a) Euclid divided his famous treatise The Elements’ 1
into 13 chapters.
11. Solution: 1
(d) Let the angles of a AABC be ∠A, ∠B and ∠C.
Given, ∠A = ∠B+∠C …(i)
InMBC, ∠A+ ∠B+ ∠C-180° [sum of all angles of atriangle is 180°]…(ii)
From Eqs. (i) and (ii),
∠A+∠A = 180° => 2 ∠A = 180°
=> 180° /2
∠A = 90°
Hence, the triangle is a right triangle.
12. Solution: 1
(c) Surface area of a cube = 96 cm2
Surface area of a cube = 6 (Side)2 = 96 => (Side)2 = 16
=> (Side) = 4 cm
[taking positive square root because side is always a positive quantity]
Volume of cube = (Side)3 = (4)3 = 64 cm3
Hence, the volume of the cube is 64 cm3.
13. Solution: 1
(b) In a given class 90-120, upper class = 120
and lower class = 90
 Solution:
14. 1
(b) Let p (x) = 2x2 + 7x-4
= 2x2 + 8x-x-4 [by splitting middle term]
= 2x(x+ 4)-1(x+ 4)
= (2x-1)(x+ 4)
For zeroes of p(x), put p(x) = 0=> (2x -1) (x + 4) = 0
=> 2x-1 = 0 and x+4 = 0
=> x = ½ and x = -4
Hence, one of the zeroes of the polynomial p(x) is ½.
15. Solution: 1
(d) Given, AD = 34 cm and AB = 30 cm
In figure, draw OL ⊥ AB.
Since, the perpendicular from the centre of a circle to a chord bisects the chord.

16. Solution: 1
(d) In a given data, maximum value = 32 and minimum value = 6
We know, range of the data = maximum value – minimum value
= 32 – 6 = 26
Hence, the range of the given data is 26.
17. (c) Bisect each other 1
18. Solution: 1
Let one of interior angle be x°.
∴ Sum of two opposite interior angles = Exterior angle
∴ x° + x° = 105°
2x° = 105°
x° = 105°/2
x°=52 ½°
Hence, each angle of a triangle is 52 ½°.
19.
 20. Solution:
(a) Let p (x) = 5x – 4x2 + 3 …(i)
On putting x = -1 in Eq. (i), we get
p(-1) = 5(-1) -4(-1)2 + 3= -5-4+3 = -6
21.
Solution: 2
We know that
(x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
(i) (x + 2y + 4z)2
= x2 + (2y)2 + (4z)2 + 2 (x) (2y) + 2 (2y) (4z) + 2(4z) (x)
= x2 + 4y2 + 16z2 + 4xy + 16yz + 8 zx

(ii) (2x – y + z)2 = (2x)2 + (- y)2 + z2 + 2 (2x) (- y)+ 2 (- y) (z) + 2 (z) (2x)
= 4x2 + y2 + z2 – 4xy – 2yz + 4zx
OR
Solution:
(i) We have,
12x2 – 7x + 1 = 12x2 – 4x- 3x + 1
= 4x (3x – 1 ) -1 (3x – 1)
= (3x -1) (4x -1)
Thus, 12x2 -7x + 3 = (2x – 1) (x + 3)

(ii) We have, 2x2 + 7x + 3 = 2x2 + x + 6x + 3

= x(2x + 1) + 3(2x + 1)
= (2x + 1)(x + 3)
Thus, 2×2 + 7x + 3 = (2x + 1)(x + 3)
22. Solution: 2
In ∆BOC and ∆AOD, we have
∠BOC = ∠AOD
BC = AD [Given]
∠BOC = ∠AOD [Vertically opposite angles]
∴ ∆OBC ≅ ∆OAD [By AAS congruency]
⇒ OB = OA [By C.P.C.T.]
i.e., O is the mid-point of AB.
Thus, CD bisects AB.
Solution:
23. (i) In ∆APB and ∆CQD, we have 2
∠APB = ∠CQD [Each 90°]
AB = CD [ ∵ Opposite sides of a parallelogram ABCD are equal]
∠ABP = ∠CDQ
[ ∵ Alternate angles are equal as AB || CD and BD is a transversal]
∴ ∆APB = ∆CQD [By AAS congruency]

(ii) Since, ∆APB ≅ ∆CQD [Proved]

⇒ AP = CQ [By C.P.C.T.]

24. Solution: 2
24. OR
Solution: 2
In 1/17, In the divisor is 17.
Since, the number of entries in the repeating block of digits is less than the divisor,
then the maximum number of digits in the repeating block is 16.
Dividing 1 by 17, we have

The remainder I is the same digit from which we started the division.
∴ 1/17 = 0.0588235294117647
Thus, there are 16 digits in the repeating block in the decimal expansion of 1/17.
Hence, our answer is verified.
25. Solution:
Given: Two congruent circles with centres O & O’ and radii r which have chords AB 2
and CD respectively such that ∠AOB = ∠CO’D.

To Prove: AB = CD
Proof: In ∆AOB and ∆CO’D, we have
OA = O’C [Each equal to r]
OB = O’D [Each equal to r]
∠AOB = ∠CO’D [Given]
∴ ∆AOB ≅ ∆CO’D [By SAS congruence criteria]
Hence, AB = CD [C.P.C.T.]
26. Solution: 3

OR

Solution: 3
27. Let the sides of the wall be
a = 15m, b = 11m, c = 6m
Semi-perimeter,

Thus, the required area painted in colour

= 20√2 m2
28. Solution: 3
We have a circle having a chord AB equal to radius of the circle.
∴ AO = BO = AB
⇒ ∆AOB is an equilateral triangle.
Since, each angle of an equilateral triangle is 60°.
⇒ ∠AOB = 60°
Since, the arc ACB makes reflex ∠AOB = 360° – 60° = 300° at the centre of the
circle and ∠ACB at a point on the minor arc of the circle.

Hence, the angle subtended by the chord on the minor arc = 150°.
Similarly, ∠ADB = 12 [∠AOB] = 12 x 60° = 30°
Hence, the angle subtended by the chord on the major arc = 30°
29. Solution:
3
30. Solution: 3
(i) 2x + y = 7
When x = 0, 2(0) + y = 7 ⇒ y = 7
∴ Solution is (0, 7)
When x =1, 2(1) + y = 7 ⇒ y = 7 – 2 ⇒ y = 5
∴ Solution is (1, 5)
When x = 2, 2(2) + y =7y = 7 – 4 ⇒ y = 3
∴ Solution is (2, 3)
When x = 3, 2(3) + y = 7y = 7 – 6 ⇒ y = 1
∴ Solution is (3, 1).

(ii) πx + y = 9
When x = 0, π(0) + y = 9 ⇒ y = 9 – 0 ⇒ y = 9
∴ Solution is (0, 9)
When x = 1, π(1) + y = 9 ⇒ y = 9 – π
∴ Solution is (1, (9 – π))
When x = 2, π(2) + y = 9 ⇒ y = 9 – 2π
∴ Solution is (2, (9 – 2π))
When x = -1,π(-1) + y = 9 ⇒ y = 9 + π
∴ Solution is (-1, (9 + π))

(iii) x = 4y
When x = 0, 4y = 1 ⇒ y = 0
∴ Solution is (0, 0)
When x = 1, 4y = 1 ⇒ y = 14
∴ Solution is (1,14 )
When x = 4, 4y = 4 ⇒ y = 1
∴ Solution is (4, 1)
When x = 4, 4y = 4 ⇒ y = -1
∴ Solution is (-4, -1)

OR
Solution:
 Solution:
31. Sum of all the angles at a point = 360°
∴ x + y + ⇒ + w = 360° or, (x + y) + (⇒ + w) = 360° 3
But (x + y) = (⇒ + w) [Given]
∴ (x + y) + (x + y) = 360° or,
2(x + y) = 360°
or, (x + y) = 360∘2 = 180°
∴ AOB is a straight line.

32. Solution:
Height (h) of conical tent = 8 m
Radius (r) of base of tent = 6 m

Slant height (l) of tent =

CSA of conical tent = πrl
= (3.14 × 6 × 10) m2
= 188.4 m2 5
Let the length of tarpaulin sheet required be l.
As 20 cm will be wasted, therefore, the effective length will be (l − 0.2 m). Breadth
of tarpaulin = 3 m
Area of sheet = CSA of tent
[(l − 0.2 m) × 3] m = 188.4 m2
l − 0.2 m = 62.8 m
l = 63 m
Therefore, the length of the required tarpaulin sheet will be 63 m.
33. Solution:
AB = AC [Given] …(1)
AB = AD [Given] …(2)
From (1) and (2), we have
AC = AD
Now, in ∆ABC, we have 5
∠ABC + ∠ACB + ∠BAC = 180° [Angle sum property of a A]
⇒ 2∠ACB + ∠BAC = 180° …(3)
[∠ABC = ∠ACB (Angles opposite to equal sides of a A are equal)]
Similarly, in ∆ACD,
∠ADC + ∠ACD + ∠CAD = 180°
⇒ 2∠ACD + ∠CAD = 180° …(4)
[∠ADC = ∠ACD (Angles opposite to equal sides of a A are equal)]
Adding (3) and (4), we have
2∠ACB + ∠BAC + 2 ∠ACD + ∠CAD = 180° +180°
⇒ 2[∠ACB + ∠ACD] + [∠BAC + ∠CAD] = 360°
⇒ 2∠BCD +180° = 360° [∠BAC and ∠CAD form a linear pair]
⇒ 2∠BCD = 360° – 180° = 180°
⇒ ∠BCD = 180∘2 = 90°
Thus, ∠BCD = 90°
33. OR
Solution: 5
Since BE ⊥ AC [Given]

∴ BEC is a right triangle such that ∠BEC = 90°

Similarly, ∠CFB = 90°
Now, in right ∆BEC and ∆CFB, we have
BE = CF [Given]
BC = CB [Common hypotenuse]
∠BEC = ∠CFB [Each 90°]
∴ ∆BEC ≅ ∆CFB [By RHS congruency]
So, ∠BCE = ∠CBF [By C.P.C.T.]
or ∠BCA = ∠CBA
Now, in ∆ABC, ∠BCA = ∠CBA
⇒ AB = AC [Sides opposite to equal angles of a ∆ are equal]
∴ ABC is an isosceles triangle.
34. Solution: 5
(i) We have, (x+ 4) (x + 10)
Using identity,
(x+ a) (x+ b) = x2 + (a + b) x+ ab.
We have, (x + 4) (x + 10) = x2+(4 + 10) x + (4 x 10)
= x2 + 14x+40

(ii) We have, (x+ 8) (x -10)

Using identity,
(x + a) (x + b) = x2 + (a + b) x + ab
We have, (x + 8) (x – 10) = x2 + [8 + (-10)] x + (8) (- 10)
= x2 – 2x – 80

(iii) We have, (3x + 4) (3x – 5)

Using identity,
(x + a) (x + b) = x2 + (a + b) x + ab
We have, (3x + 4) (3x – 5) = (3x)2 + (4 – 5) x + (4) (- 5)
= 9x2 – x – 20
 35 (iii) Wehave -2x + 3y = 6 or (-2)x + (3)y + (-6) = 0
Comparing it with ax – 4 – by + c = 0,we get a = -2, b = 3 and c = -6.

(iv) We have x = 3y or (1)x + (-3)y + (0) = 0 Comparing it with ax + by + c = 0, we 5

get a = 1, b = -3 and c = 0.

(v) We have 2x = -5y or (2)x + (5)y + (0) = 0 Comparing it with ax + by + c = 0, we

get a = 2, b = 5 and c = 0.

(vi) We have 3x + 2 = 0 or (3)x + (0)y + (2) = 0 Comparing it with ax + by + c = 0,

we get a = 3, b = 0 and c = 2.

(vii) We have y – 2 = 0 or (0)x + (1)y + (-2) = 0 Comparing it with ax + by + c = 0,

we get a = 0, b = 1 and c = -2.

36 (i) 10 (ii) 60 4
iii) 50% marks of 60 = 50 x 60/100 = 30 marks Total number of students scored more than
30 marks = 19 + 7 +1 = 27
OR
Range = Highest marks – lowest marks = 59 – 6 = 53
37 (i) (ii)

(iii)
4

OR
38 i) C(10, 6) 4
ii) R(5, 6)
iii)

OR

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