Note:
(i) All questions are compulsory. The marks allotted for each question are given beside the questions.
(ii) Write your name, enrolment numbers, Al name and subject on the first page of the answer sheet.
1. Answer any one of the following question.
(a) (i) Find five rational numbers between 3/5 and 4/5.
Answer: 3/5= 30/50 , 4/5 = 40/50 “Therefore, there are five rational numbers.” 31/50, 32/50, 33/50, 34/50, 35/50
(ii) Are the square roots of all positive integers irrational? If not, give an example of the square root of a number that is a rational number.
Answer: No, the square roots of all positive integers are not necessarily irrational. The square root of a positive integer can be rational in cases where the number is a perfect square. For example, the square root of 4 is 2, and 2 is a rational number. However, for numbers that are not perfect squares, their square roots are typically irrational.
(b) Simplify the following
(i) 3(a4 b3 ) 10 x 5 (a2 b2 ) 3
Answer: 3(a4 b3 ) 10 x 5 (a2 b2 ) 3
= 3a4×10b3×10×5a2×3×b2×3 {∵(am)n=amn , am×an=am+n}
=3a40b30×5a6b6
=3×5×a40+6×b30+6
=15a46b36
(ii) (2x 2 y 3 ) 3
Answer: (2x 2 y 3 ) 3 =23 × x−2×3×y3×3 {∵(am)n=amn am×an=am+n}
=8×x−6×y9=8x−6y9
2. Answer any one of the following question.
(a) (i) Solve: (5m3)/4+ (3m2)/5=(m + 6)/7
Answer: Let’s solve the equation step by step:
To simplify, let’s find a common denominator for 4, 5, and 7, which is 4×5×7=140.
Now, multiply each term by the necessary factor to get the common denominator:
This results in:
Combine the numerators:
Combine like terms:
Now, crossmultiply to eliminate the denominators:
140(259m−161)=140(20m+120)
Simplify:
259m−161=20m+120
Isolate m:
259m−20m=120+161
239m=281
So, the solution to the equation is m =
(ii) The sum of two numbers is 82 and their product is 1456, find the two numbers..
Answer: Let the two numbers be x and y. Given that:
2. xy=1456
We can solve these equations simultaneously to find the values of x and y.
From the first equation, we have y=82−x.
Substitute this into the second equation:
Simplify and solve for x:
82x−x2=1456
x2−82x+1456=0
Factorizing or using the quadratic formula, we find =32 or =45.
If x=32, then y=50.
If x=45, then y=37.
Therefore, the two pairs of numbers are (32,50) and (45,37).
(b) For the polynomial (x3 + 2x + 1) /5 (7/2) x 2 x 6, write
(i) the degree of the polynomial
(ii) the coefficient of x3
(iii) the coefficient of x6
(iv) the constant term
Answer: The given polynomial is,
(i) Degree of the Polynomial is the Highest degree of this terms, which is 6 in this case (degree of )
(ii) Coefficient of is
(iii) Coefficient of is 6
(iv) The constant term in the Polynomial is
3. Answer any one of the following question.
(a) In Fig. 1, lines XY and MN intersect at O. If ZPOY=90° and a: b= 2:3, find c.
Answer:
(b) In Fig. 1, <PQR = <PRQ, then prove that <PQS = <PRT.
Answer:
4. Answer any one of the following questions in about 100150 words.
(a) Find all the angles of a parallelogram if one angle is 80°. In a rectangle, one diagonal is inclined to one of its sides at 25°. Measure the acute angle between the two diagonals.
Answer: The sum of interior angles of a parallelogram is 360°. If one angle is 80°, then the other angle would be:
Angle = 180° – 80° = 100°
In a rectangle, the supplementary angle to any angle is 90°, so the bent angle (supplementary) would be:
Angle = 90° – 25° = 65°
Now, the smallest angle between the two supplementary angles:
Angle = 180° – (100° + 65°) = 15°
(b) Prove that the angle bisectors of a parallelogram form a rectangle. In a trapezium ABCD, AB CD. Calculate ∠C and ∠D if ∠A = 55° and ∠B = 70°
Answer: Consider a parallelogram ABCD,
∠A = 55° (given)
∠B = 70° (given)
Since ABCD is a parallelogram, ∠A = ∠C and ∠B = ∠D.
Therefore, ∠C = ∠A = 55°
and ∠D = ∠B = 70°
5. Answer any one of the following questions.
(a)
(i) Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Answer:
A tangent to a circle is a line that intersects the circle at only one point.
Let’s draw the tangents PQ and RS to the circle at the ends of the diameter AB.
According to Theorem 10.1: The tangent at any point of a circle is perpendicular to the radius through the point of contact.
We know that radius is perpendicular to the tangent at the point of contact.
Thus, OA ⊥ PQ and OB ⊥ RS
Since the tangents are perpendicular to the radius,
∠PAO = 90°, ∠RBO = 90°
and ∠OAQ = 90°, ∠OBS = 90°
Here ∠OAQ is equal to ∠OBR and ∠PAO is equal to ∠OBS, which are two pairs of alternate interior angles.
If the alternate interior angles are equal, then lines PQ and RS should be parallel.
We know that PQ and RS are the tangents drawn to the circle at the ends of the diameter AB.
Hence, it is proved that tangents drawn at the ends of a diameter of a circle are parallel.
(ii) Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.
Answer: Hint: – Perpendicular drawn from the center to the chord of the circle bisects the chord.
(b)
(i) In the figure 3, DEAC and DFAE. Prove that BF/FE = BE/EC.
Answer: We know according to the basic proportionality theorem, if a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct points, the other two sides are divided in the same ratio.
In ΔABC
DE  AC
BD/AD = BE/EC ………(i)
In ΔABE
DF  AE
BD/AD = BF/FE ……..(ii)
From (i) and (ii)
BD/AD = BE/EC = BF/FE
Thus, BE/EC = BF/FE
(ii) AD and BC are equal perpendiculars to a line segment AB. Show that CD bisects AB.
Answer: Given: AD ⊥ AB, BC ⊥ AB, and AD = BC
To Prove: CD bisects AB or OA = OB
We can show that the two triangles OBC and OAD are congruent by using AAS congruency rule and then we can say corresponding parts of congruent triangles will be equal.
Consider two triangles △ BOC and △ AOD,
In △ BOC and △ AOD,
∠BOC = ∠AOD (Vertically opposite angles)
∠CBO = ∠DAO (Each 90º, since AD and BC are ⊥ to AB)
BC = AD (Given)
∴ △BOC ≅ △AOD (AAS congruence rule)
∴ BO = AO (By CPCT)
Thus, CD bisects AB and O is the midpoint of AB.
6. Prepare any one Project out of the following projects given below:
(a) Draw the frequency polygon for the following data
Answer: Construct a histogram for the given data. Mark the mid point of the tops of rectangles of histogram and join these points by line segments. To complete the polygon, mark two classes (−10 to 0) and (80 to 90). Join the end point of the line segments to the mid points of these two classes. The curve so obtained is a frequency polygon.
b) The blood groups of 30 students of Class VIII are recorded as follows:
O. O, AB. O, A, O. B. A. O, B. A, 0, 0.
A, B, A, AB, O, A, A, 0, 0, AB, B. A, O. B. A, B, O.
Represent this data in the form of a frequency distribution table. Which is the most common, and which is the rarest, blood group among these student?
Answer:
NO. of Students 

A 
9 
B 
6 
O 
12 
AB 
3 
Total 
30 
Number of students having blood group AB=3
Total student= 30
OR
(a) The distance (in km) of 40 engineers from their residence to their place of work were found as follows:
Construct a grouped frequency distribution table with class size 5 for the data given above taking the first interval as 0 – 5 (5 not included). What main features do you observe from this tabular representation?
Answer: Mostly engineers who travel from residence to work place travel around 20 km.
(b) A survey conducted by an organisation for the cause of illness and death among the women between the ages 15 – 44 (in years) worldwide, found the following figures (in %) :
S. No. 
Causes 
Female Fatality Rate 
1. 
Reproductive health conditions 
31.8 
2. 
Neuropsychiatric conditions 
25.4 
3. 
Injuries 
12.4 
4. 
Cardiovascular Conditions 
4.3 
5. 
Respiratory conditions 
4.1 
6. 
Other Causes 
22.0 
(i) Represent the information given above graphically.
(ii) Which condition is the major cause of women’s ill health and death worldwide?
(iii) Try to find out, with the help of your teacher, any two factors which play major role in the cause in (ii) above being the major cause.
Answer:
(i) By representing causes on xaxis and family fatality rate on yaxis and choosing an appropriate scale (1 unit = 5% for y axis), the graph of the information given above can be constructed as follows.