Section A: Multiple Choice Questions (1 Mark Each)
Q1. The distance between the points (3,-5) and (11, 10) is
Solution:
Let’s use the distance formula: d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.
d = \sqrt{(11 - 3)^2 + (10 + 5)^2}
d = \sqrt{8^2 + 15^2}
d = \sqrt{64 + 225}
d = \sqrt{289}.
d = 17
Answer: (a).
Q2. The degree of the polynomial x^5+x-4x^3-7 is
Solution:
The degree is the highest power of the variable x. Here, the highest power is 5.
Answer: (d) 5
Q3. What is the value of x? Given \frac{2x+3}{5}+5=12
Solution:
\frac{2x+3}{5} = 12 - 5
\frac{2x+3}{5} = 7
2x + 3 = 35
2x = 32 \implies x = 16
Answer: (d) 16
Q4. The value of 16x^3+10x-9 if x=\frac{1}{2} is _____
Solution:
Substitute x = \frac{1}{2}:
16\left(\frac{1}{8}\right) + 10\left(\frac{1}{2}\right) - 9
2 + 5 - 9 = -2
Answer: (b) -2
Q5. The point A(-3,5) is in which quadrant?
Solution:
The x-coordinate is negative and the y-coordinate is positive. This corresponds to the Second Quadrant.
Answer: (b) Second
Q6. The slope of the line 2y=6x-7 is
Solution:
Convert to y = mx + c format by dividing by 2:
y = 3x - \frac{7}{2}.
The slope (m) is 3.
Answer: (a) 3
Q7. What is the degree of a quadratic polynomial?
Solution:
“Quad” means square.
A quadratic polynomial always has a degree of 2.
Answer: (b) Two
Q8. What angle does a quarter circle subtend at the center
Solution:
A full circle is 360°. A quarter of that is 360 / 4 = 90°.
Answer: (b) 90°
Q9. Which of the following represents linear growth
Solution:
Linear growth requires a positive slope (positive coefficient of x).
In y = -20 + 5x, the slope is +5, showing growth.
Answer: (c) y = -20 + 5x
Q10. Find the 5th term of a GP with common ratio 3, whose 3rd term is 162.
Solution:
For a Geometric Progression, a_n = a r^{n-1}.
Given r = 3 and
a_3 = 162 \implies a(3)^2 = 162 \implies 9a = 162 \implies a = 18.
The 5th term is a_5 = a r^4 = 18(3^4) = 18(81) = 1458.
Answer: (a) 1458
Q11. (-15) x 3 =
Solution:
A negative multiplied by a positive yields a negative.
15 \times 3 = 45, so the answer is -45.
Answer: (b) -45
Q12. A merchant… receives 15 ingots for every 2 bags of spices. If he brings 12 bags… how many copper ingots will he leave with?
Solution:
This is a direct proportion.
Let x be the number of ingots.
\frac{15}{2} = \frac{x}{12} \implies 2x = 180 \implies x = 90
Answer: (b) 90
Q13. Factors of \frac{9x^2}{4}+15xy+25y^2 are
Solution:
Recognize this as a perfect square trinomial a^2 + 2ab + b^2 =(a+b)^2:
a = \frac{3}{2}x and b = 5y.
Check middle term: 2(\frac{3}{2}x)(5y) = 15xy.
Answer: (a) (\frac{3}{2}x+5y)^2
Q14. Calculate the length of the arc of a circle if the radius is 4.2 cm and the angle at the centre is 60°
Solution:
Arc length formula: l = \frac{\theta}{360} \times 2\pi r.
l = \frac{60}{360} \times 2 \times \frac{22}{7} \times 4.2 = \frac{1}{6} \times 2 \times 22 \times 0.6 = 4.4 cm.
Answer: (c) 4.4 cm
Q15. If the diameter of a truck tyre is 98 cm, then how far does the truck need to travel for the tyre to complete one revolution?
Solution:
One revolution equals the circumference of the tyre. C = \pi d.
C = \frac{22}{7} \times 98 = 22 \times 14 = 308 cm.
Answer: (b) 308 cm
Q16. The rational expression \frac{x^2-3x+2}{x^2-1} reduces to
Solution:
Factorize both the numerator and the denominator.
Numerator: x^2 - 3x + 2 = (x-1)(x-2)
Denominator: x^2 - 1 = (x-1)(x+1)
Cancel the common factor (x-1) to get \frac{x-2}{x+1}.
Answer: (c) \frac{x-2}{x+1}
Q17. The distance of the point A(6,3) from the X- axis is
Solution:
The distance from the X-axis is simply the absolute value of the y-coordinate.
Answer: (a) 3
Q18. \frac{5}{12}(\frac{8}{15}-\frac{7}{18})=
Solution:
First, find the LCM of 15 and 18, which is 90.
\frac{5}{12} \left( \frac{48}{90} - \frac{35}{90} \right) = \frac{5}{12} \left( \frac{13}{90} \right)
Cancel 5 and 90 to leave 18 in the denominator: \frac{13}{12 \times 18} = \frac{13}{216}.
Answer: (a) \frac{13}{216}
Q19. Which term of the sequence 11n+32 is 659?
Solution:
Set the general term equal to 659 and solve for n.
11n + 32 = 659
11n = 627 \implies n = 57
Answer: (c) 57
Q20. Assertion & Reason Question:
Assertion: The area of an isosceles trapezium can be calculated using the formula \sqrt{(s-a)(s-b)(s-c)(s-d)}… (True, Brahmagupta’s formula applies to all cyclic quadrilaterals).
Reason: Every Isosceles trapezium is a cyclic quadrilateral. (True)
Solution:
Answer: (a) Both A and R are true but R is the correct explanation of A .
Section B (2 Marks Each)
Q21. Convert 0.\overline{16} in the form of \frac{p}{q}.
Solution:
Step 1: Let x = 0.161616... (Equation 1)
Step 2: Multiply both sides by 100 (since two digits are repeating):
100x = 16.161616... (Equation 2)
Step 3: Subtract Equation 1 from Equation 2:
100x - x = 16.1616... - 0.1616...
99x = 16 \implies x = \frac{16}{99}
Final Answer: \boxed{ \frac{16}{99}}
Q22. The length of a rectangular playground is 4 m more than twice its breadth. If the perimeter… is 56 m, find its length and breadth.
Solution:
Step 1: Let the breadth be b meters.
Then, length l = 2b + 4.
Step 2: Use the perimeter formula P = 2(l + b).
56 = 2(2b + 4 + b)
28 = 3b + 4
24 = 3b \implies b = 8 \text{ m}
Step 3: Find length: l = 2(8) + 4 = 20 m.
Final Answer: Length is 20 m, Breadth is 8 m.
Q23. Plot \frac{1}{4} and \frac{7}{8} on a number line.
Q24. Find 5 rational numbers between \frac{2}{7} and \frac{6}{7}.
Solution:
Step 1: Multiply the numerator and denominator of both fractions by a common integer (like 2) to expand the range.
\frac{2 \times 2}{7 \times 2} = \frac{4}{14} \quad \text{and} \quad \frac{6 \times 2}{7 \times 2} = \frac{12}{14}
Step 2: Pick any 5 numbers between 4 and 12 for your numerators.
Final Answer: \boxed{\frac{5}{14}, \frac{6}{14}, \frac{7}{14}, \frac{8}{14}, \frac{9}{14}}.
Q25. Without performing division, determine whether the decimal expansion of \frac{41}{400} is terminating or non-terminating…
Solution:
Step 1: Find the prime factorization of the denominator (400).
400 = 4 \times 100 = 2^2 \times (2^2 \times 5^2) = 2^4 \times 5^2.
Step 2: Since the prime factors are strictly 2s and 5s, the expansion is terminating.
Step 3: The number of decimal places equals the highest power of 2 or 5. Here, the highest power is 4.
Final Answer: Terminating, after 4 decimal places.
Section C (3 Marks Each)
Q26. Prove that \sqrt{3} is irrational.
Solution:
Assume the opposite.
Let \sqrt{3} be a rational number \frac{p}{q}, where p and q are co-prime integers (no common factors) and q \neq 0.
Square both sides:
3 = \frac{p^2}{q^2} \implies p^2 = 3q^2
This means p^2 is a multiple of 3, so p must also be a multiple of 3.
Let p = 3k.
Substitute p = 3k back into our equation:
(3k)^2 = 3q^2 \implies 9k^2 = 3q^2 \implies q^2 = 3k^2
This means q^2 is a multiple of 3, so q is also a multiple of 3.
Both p and q share a common factor of 3.
This contradiction arises because of our wrong assumption.
Therefore, \sqrt{3} is irrational.
Q27. Expand the following:
(i) (\frac{5}{6}x-\frac{3}{10}y)^2
Use (a-b)^2 = a^2 - 2ab + b^2:
= \left(\frac{5}{6}x\right)^2 - 2\left(\frac{5}{6}x\right)\left(\frac{3}{10}y\right) + \left(\frac{3}{10}y\right)^2
= \frac{25}{36}x^2 - \frac{1}{2}xy + \frac{9}{100}y^2
Final Answer: \boxed{ (\frac{5}{6}x-\frac{3}{10}y)^2 = = \frac{25}{36}x^2 - \frac{1}{2}xy + \frac{9}{100}y^2}
(ii) (3x-5y+4)^2
Use (a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca:
= (3x)^2 + (-5y)^2 + (4)^2 + 2(3x)(-5y) + 2(-5y)(4) + 2(4)(3x)
= 9x^2 + 25y^2 + 16 - 30xy - 40y + 24x
Final Answer: \boxed{ (3x-5y+4)^2 = 9x^2 + 25y^2 + 16 - 30xy - 40y + 24x }
Q28. Use identity and find the product without multiplying directly:
(i) (98)^2
= (100 - 2)^2 = 100^2 - 2(100)(2) + 2^2
= 10000 - 400 + 4 = 9604
Final Answer: \boxed{9604}
(ii) (47 \times 53)
= (50 - 3)(50 + 3) = 50^2 - 3^2
= 2500 - 9 = 2491
Final Answer: \boxed{2491}
Q29. The length of the four sides of a quadrilateral are 8m, 6m, 10m and 12m. If the quadrilateral can be inscribed inside a circle, what is the area…
A quadrilateral inscribed in a circle is a cyclic quadrilateral.
We use Brahmagupta’s formula: Area = \sqrt{(s-a)(s-b)(s-c)(s-d)}.
Calculate the semi-perimeter (s):
s = \frac{8 + 6 + 10 + 12}{2} = \frac{36}{2} = 18 \text{ m}
Apply the formula:
Area = \sqrt{(18-8)(18-6)(18-10)(18-12)}
Area = \sqrt{10 \times 12 \times 8 \times 6} = \sqrt{5760}
Area = \sqrt{576 \times 10} = 24\sqrt{10} \text{ m}^2
Final Answer: Area = \boxed{24\sqrt{10} \text{ m}^2}
Q30. A taxi service charges… y = ax + b find the values of a and b.
Solution:
Create linear equations from the data provided.
For 12 km, fare is Rs. 250: 12a + b = 250 … (Eq 1)
For 20 km, fare is Rs. 370: 20a + b = 370 … (Eq 2)
Subtract Eq 1 from Eq 2 to eliminate b:
(20a + b) - (12a + b) = 370 - 250
8a = 120 \implies a = 15
Substitute a = 15 back into Eq 1:
12(15) + b = 250 \implies 180 + b = 250 \implies b = 70
Final Answer: a = 15, b = 70.
Q31. What is the perimeter of the following shape taking arcs to be semi-circles.
Solution:
The shape consists of one large semi-circle on top and three small semi-circles on the bottom.
The total base length is 4 + 4 + 4 = 12 cm.
Calculate the arc length of the large top semi-circle (diameter = 12, radius = 6):
C_{large} = \pi \times r = \pi \times 6 = 6\pi cm.
Calculate the arc length of the three small semi-circles (diameter = 4, radius = 2):
Length of one small arc = \pi \times 2 = 2\pi cm.
For three small arcs = 3 \times 2\pi = 6\pi cm.
Total perimeter = 6\pi + 6\pi = 12\pi.
Final Answer: \boxed{12\pi} \text{cm}
Section D (5 Marks Each)
Q32. Plot the points A(1,3), B(4,3), C(4,1) and D(1,1) on a graph paper. Is ABCD a rectangle? Can you explain why? What is the area…
Solution:
(Students should draw axes and mark the points).
Calculate the lengths of the sides using coordinates:
- AB (horizontal line, y=3): Length = \vert{}4 - 1\vert{} = 3 units.
- CD (horizontal line, y=1): Length = \vert{}4 - 1\vert{} = 3 units.
- BC (vertical line, x=4): Length = \vert{}3 - 1\vert{} = 2 units.
- DA (vertical line, x=1): Length = \vert{}3 - 1\vert{} = 2 units.
Opposite sides are equal (AB=CD and BC=DA).
Furthermore, the sides are parallel to the x and y axes, meaning the adjacent sides meet at exactly 90° angles.
Area = Length \times Breadth = 3 \times 2 = 6 square units.
Q33. Given the side of an equilateral triangle is ‘a’. Calculate its area using Heron’s formula. Hence calculate the area of an equilateral triangle with side 10cm.
Solution:
Semi-perimeter s = \frac{a + a + a}{2} = \frac{3a}{2}.
Apply Heron’s Formula: Area = \sqrt{s(s-a)(s-b)(s-c)}.
Area = \sqrt{\frac{3a}{2} \left(\frac{3a}{2} - a\right) \left(\frac{3a}{2} - a\right) \left(\frac{3a}{2} - a\right)}
Area = \sqrt{\frac{3a}{2} \times \frac{a}{2} \times \frac{a}{2} \times \frac{a}{2}}
Area = \sqrt{\frac{3a^4}{16}} = \frac{\sqrt{3}}{4}a^2
Substitute a = 10 cm:
Area = \frac{\sqrt{3}}{4}(10)^2 = \frac{\sqrt{3}}{4} \times 100 = 25\sqrt{3} \text{ cm}^2
Final Answer: \boxed{ 25\sqrt{3} \text{ cm}^2 }
Q34. A telecom company charges 900 for a certain recharge scheme. This prepaid balance is reduced by 25 each day…
(i) Write an equation that models the remaining balance b(x):
b(x) = 900 - 25x
(ii) Explain why it represents linear decay:
The balance decreases by a constant, fixed amount (Rs. 25) for every single unit increase in time (1 day).
Graphically, this forms a straight line with a negative slope (-25).
(iii) After how many days will the balance run out?
Set b(x) = 0:
900 - 25x = 0 \implies 25x = 900 \implies x = \frac{900}{25} = 36 days.
(iv) Table of values:
| Day (x) | Balance b(x) (Rs.) |
| 1 | 875 |
| 2 | 850 |
| 3 | 825 |
| 4 | 800 |
| 5 | 775 |
| 6 | 750 |
| 7 | 725 |
Q35. Simplify the following rational expressions:
(i) \frac{x^3-8}{2x^2+x-10}
Numerator: Difference of cubes a^3 - b^3 = (a-b)(a^2+ab+b^2).
x^3 - 2^3 = (x-2)(x^2 + 2x + 4).
Denominator: Split the middle term.
Find factors of -20 that add up to +1 (which are +5 and -4).
2x^2 + 5x - 4x - 10 = x(2x+5) - 2(2x+5) = (x-2)(2x+5).
Result: Cancel (x-2) to get
\frac{x^2+2x+4}{2x+5}.
(ii) \frac{x^3-3x^2-4x+12}{x^2-5x+6}
Numerator: Group terms: x^2(x-3) - 4(x-3) = (x^2 - 4)(x-3).
Expand the difference of squares: (x-2)(x+2)(x-3).
Denominator: Split the middle term: (x-2)(x-3).
Result: Cancel (x-2) and (x-3) to leave just
x+2.
Section E: Case Study Based Questions (6 Marks Each)
Q36. Aditya is a fitness freak… 200 m race. He can currently run that distance in 51 seconds… With each day of practice, it takes him 2 seconds less.
Solution:
This forms an Arithmetic Progression (AP)
AP: 51, 49, 47, 45 ...
where the first term a = 51 (current time) and the common difference d = -2.
(i) In how many days will be able to achieve his target of 31 sec?
Use a_n = a + (n-1)d:
31 = 51 + (n-1)(-2)
-20 = -2(n-1)
10 = n - 1 \implies n = 11
He will achieve his target on the 11th day.
(ii) What will be the difference between the time taken on 5th day and 7th day.
Time on 5th day: a_5 = 51 + 4(-2) = 43 seconds.
Time on 7th day: a_7 = 51 + 6(-2) = 39 seconds.
Difference = 43 - 39 = 4 seconds.
(iii) Which term of the arithmetic progression 3, 15, 27, 39 …. will be 120 more than its 21st term?
Here, a = 3, d = 12.
The 21st term is [/latex]a_{21} = 3 + 20(12) = 3 + 240 = 243[/latex].
The target value is 243 + 120 = 363.
Find n where a_n = 363:
3 + (n-1)12 = 363
(n-1)12 = 360 \implies n-1 = 30 \implies n = 31
It will be the 31st term.
Q37. The figure below shows Esha’s room and an attached bathroom…
(i) What are the coordinates of the four corners O, F, R and P of the bathroom?
Looking at the origin and axes:
- O is the origin: (0, 0)
- P is on the negative x-axis: (-6, 0)
- R is the top-left corner: (-6, 10)
- F is on the positive y-axis, completing the rectangle: (0, 10)
(ii) What is the area of the showering area?
The showering area forms a trapezium with vertices R(-6, 10), W(-2, 10), H(-3, 5), and S(-6, 5).
The top parallel side RW has length \vert{}-2 - (-6)\vert{} = 4 units.
The bottom parallel side SH has length \vert{}-3 - (-6)\vert{} = 3 units.
The height (vertical distance) is 10 - 5 = 5 units.
Area = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}
Area = \frac{1}{2} \times (4 + 3) \times 5 = \frac{1}{2} \times 7 \times 5 = 17.5 \text{ square units}
(iii) Calculate the distance between the corner (W_4) of the wardrobe to the top right corner (B) of the bedroom?
The top-right corner of the bedroom B is given as (12, 10).
The wardrobe sits on the floor starting at W_1(+3, 0) and ending at W_2(+7, 0).
Assuming standard scaling where the wardrobe height spans up to y=2 (since the bed starts at y=4), the top left corner W_4 sits at (3, 2).
Distance W_4 to B:
d = \sqrt{(12 - 3)^2 + (10 - 2)^2}
d = \sqrt{9^2 + 8^2} = \sqrt{81 + 64} = \sqrt{145} \approx 12.04 \text{ units}