Contents
Suggestive Question Paper 2025-26 | Set 1
Instructions:
- Section A: 20 MCQs (1 Mark).
- Section B: 5 VSA (2 Marks).
- Section C: 6 SA (3 Marks).
- Section D: 4 LA (5 Marks).
- Section E: 3 Case Study (4 Marks).
Section A (20 Marks)
If two positive integers $a$ and $b$ are written as $a = x^3y^2$ and $b = xy^3$, where $x, y$ are prime numbers, then $HCF(a, b)$ is:
(A) $xy$
(B) $xy^2$
(C) $x^3y^3$
(D) $x^2y^2$
View Answer
Option (B)
Explanation: HCF takes the lowest power. Lowest power of x is $x^1$, lowest power of y is $y^2$. So, $xy^2$.
Explanation: HCF takes the lowest power. Lowest power of x is $x^1$, lowest power of y is $y^2$. So, $xy^2$.
The zeroes of the quadratic polynomial $x^2 + 99x + 127$ are:
(A) both positive
(B) both negative
(C) one positive, one negative
(D) both equal
View Answer
Option (B)
Explanation: In $ax^2+bx+c$, if a, b, c all have the same sign (+ve here), then both zeroes are negative.
Explanation: In $ax^2+bx+c$, if a, b, c all have the same sign (+ve here), then both zeroes are negative.
The value of $k$ for which the system of equations $x + y - 4 = 0$ and $2x + ky - 3 = 0$ has no solution, is:
(A) -2
(B) $\neq 2$
(C) 3
(D) 2
View Answer
Option (D)
Explanation: For no solution $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$.
$\frac{1}{2} = \frac{1}{k} \Rightarrow k = 2$.
Explanation: For no solution $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$.
$\frac{1}{2} = \frac{1}{k} \Rightarrow k = 2$.
Which of the following equations has 2 as a root?
(A) $x^2 - 4x + 5 = 0$
(B) $x^2 + 3x - 12 = 0$
(C) $2x^2 - 7x + 6 = 0$
(D) $3x^2 - 6x - 2 = 0$
View Answer
Option (C)
Explanation: Put $x=2$ in (C). $2(2)^2 - 7(2) + 6 = 8 - 14 + 6 = 0$. Correct.
Explanation: Put $x=2$ in (C). $2(2)^2 - 7(2) + 6 = 8 - 14 + 6 = 0$. Correct.
The 4th term from the end of the AP: -11, -8, -5, ..., 49 is:
(A) 37
(B) 40
(C) 43
(D) 58
View Answer
Option (B)
Explanation: Reverse AP: $a=49$, $d = -(-3) = -3$.
4th term = $a + 3d = 49 + 3(-3) = 49 - 9 = 40$.
Explanation: Reverse AP: $a=49$, $d = -(-3) = -3$.
4th term = $a + 3d = 49 + 3(-3) = 49 - 9 = 40$.
In $\Delta ABC$, $DE \parallel BC$ such that $AD = x$, $DB = x-2$, $AE = x+2$ and $EC = x-1$. The value of $x$ is:
(A) 5
(B) 4
(C) 3
(D) 2
View Answer
Option (B)
Explanation: By BPT, $\frac{x}{x-2} = \frac{x+2}{x-1}$.
$x(x-1) = (x-2)(x+2)$
$x^2 - x = x^2 - 4 \Rightarrow x = 4$.
Explanation: By BPT, $\frac{x}{x-2} = \frac{x+2}{x-1}$.
$x(x-1) = (x-2)(x+2)$
$x^2 - x = x^2 - 4 \Rightarrow x = 4$.
The distance between the points $(0, 5)$ and $(-5, 0)$ is:
(A) 5
(B) $5\sqrt{2}$
(C) $2\sqrt{5}$
(D) 10
View Answer
Option (B)
Explanation: $\sqrt{(-5-0)^2 + (0-5)^2} = \sqrt{25+25} = \sqrt{50} = 5\sqrt{2}$.
Explanation: $\sqrt{(-5-0)^2 + (0-5)^2} = \sqrt{25+25} = \sqrt{50} = 5\sqrt{2}$.
If $\cos A = 4/5$, then the value of $\tan A$ is:
(A) 3/5
(B) 3/4
(C) 4/3
(D) 5/3
View Answer
Option (B)
Explanation: Base=4, Hyp=5. So Perp=$\sqrt{25-16}=3$. $\tan A = P/B = 3/4$.
Explanation: Base=4, Hyp=5. So Perp=$\sqrt{25-16}=3$. $\tan A = P/B = 3/4$.
The value of $(\sin 30^{\circ} + \cos 30^{\circ}) - (\sin 60^{\circ} + \cos 60^{\circ})$ is:
(A) -1
(B) 0
(C) 1
(D) 2
View Answer
Option (B)
Explanation: $(1/2 + \sqrt{3}/2) - (\sqrt{3}/2 + 1/2) = 0$.
Explanation: $(1/2 + \sqrt{3}/2) - (\sqrt{3}/2 + 1/2) = 0$.
From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is:
(A) 7 cm
(B) 12 cm
(C) 15 cm
(D) 24.5 cm
View Answer
Option (A)
Explanation: $r = \sqrt{25^2 - 24^2} = \sqrt{625 - 576} = \sqrt{49} = 7$.
Explanation: $r = \sqrt{25^2 - 24^2} = \sqrt{625 - 576} = \sqrt{49} = 7$.
The area of a quadrant of a circle with circumference 22 cm is:
(A) 77 cm²
(B) 77/8 cm²
(C) 77/2 cm²
(D) 77/4 cm²
View Answer
Option (B)
Explanation: $2\pi r = 22 \Rightarrow r = 3.5 = 7/2$.
Area = $1/4 \pi r^2 = 1/4 \times 22/7 \times 7/2 \times 7/2 = 77/8$.
Explanation: $2\pi r = 22 \Rightarrow r = 3.5 = 7/2$.
Area = $1/4 \pi r^2 = 1/4 \times 22/7 \times 7/2 \times 7/2 = 77/8$.
A solid sphere of radius $r$ is melted and cast into the shape of a solid cone of height $r$. The radius of the base of the cone is:
(A) $2r$
(B) $3r$
(C) $r$
(D) $4r$
View Answer
Option (A)
Explanation: $\frac{4}{3}\pi r^3 = \frac{1}{3}\pi R^2 (r)$.
$4r^2 = R^2 \Rightarrow R = 2r$.
Explanation: $\frac{4}{3}\pi r^3 = \frac{1}{3}\pi R^2 (r)$.
$4r^2 = R^2 \Rightarrow R = 2r$.
If the mean of frequency distribution is 7.5 and $\Sigma f_ix_i = 120 + 3k$, $\Sigma f_i = 30$, then $k$ is:
(A) 40
(B) 35
(C) 50
(D) 45
View Answer
Option (B)
Explanation: $7.5 = \frac{120+3k}{30} \Rightarrow 225 = 120+3k \Rightarrow 3k=105 \Rightarrow k=35$.
Explanation: $7.5 = \frac{120+3k}{30} \Rightarrow 225 = 120+3k \Rightarrow 3k=105 \Rightarrow k=35$.
A card is drawn from a well shuffled deck of 52 cards. The probability that it is a black Queen is:
(A) 1/26
(B) 1/13
(C) 1/52
(D) 2/13
View Answer
Option (A)
Explanation: There are 2 black queens (Spade, Club). $P(E) = 2/52 = 1/26$.
Explanation: There are 2 black queens (Spade, Club). $P(E) = 2/52 = 1/26$.
If the length of shadow of a tower is $\sqrt{3}$ times its height, the angle of elevation of the sun is:
(A) $45^{\circ}$
(B) $30^{\circ}$
(C) $60^{\circ}$
(D) $90^{\circ}$
View Answer
Option (B)
Explanation: $\tan \theta = h / (h\sqrt{3}) = 1/\sqrt{3}$. So $\theta = 30^{\circ}$.
Explanation: $\tan \theta = h / (h\sqrt{3}) = 1/\sqrt{3}$. So $\theta = 30^{\circ}$.
The empirical relationship between the three measures of central tendency is:
(A) 3Median = Mode + 2Mean
(B) Mode = 3Mean - 2Median
(C) Median = 3Mode - 2Mean
(D) Mean = 3Median - 2Mode
View Answer
Option (A)
Explanation: Standard Formula: Mode = 3Median - 2Mean. Rearranging gives 3Median = Mode + 2Mean.
Explanation: Standard Formula: Mode = 3Median - 2Mean. Rearranging gives 3Median = Mode + 2Mean.
The probability of getting a bad egg in a lot of 400 is 0.035. The number of bad eggs in the lot is:
(A) 7
(B) 14
(C) 21
(D) 28
View Answer
Option (B)
Explanation: $0.035 \times 400 = 14$.
Explanation: $0.035 \times 400 = 14$.
If point P(x, y) is equidistant from A(5,1) and B(-1,5), then the relation between x and y is:
(A) 3x = 2y
(B) x = y
(C) 2x = 3y
(D) x + y = 0
View Answer
Option (A)
Explanation: $PA^2 = PB^2$.
$(x-5)^2 + (y-1)^2 = (x+1)^2 + (y-5)^2$.
Solving this gives $3x = 2y$.
Explanation: $PA^2 = PB^2$.
$(x-5)^2 + (y-1)^2 = (x+1)^2 + (y-5)^2$.
Solving this gives $3x = 2y$.
Assertion (A): The product of two distinct irrational numbers is always irrational.
Reason (R): Rational numbers are closed under multiplication.
Reason (R): Rational numbers are closed under multiplication.
(A) Both true, R explains A
(B) Both true, R doesn't explain A
(C) A is true, R is false
(D) A is false, R is true
View Answer
Option (D)
Explanation: Assertion is False because $\sqrt{2} \times \sqrt{2} = 2$ (Rational). Reason is True.
Explanation: Assertion is False because $\sqrt{2} \times \sqrt{2} = 2$ (Rational). Reason is True.
Assertion (A): The value of y is 6, for which the distance between P(2, -3) and Q(10, y) is 10.
Reason (R): Distance formula is $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.
Reason (R): Distance formula is $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.
(A) Both true, R explains A
(B) Both true, R doesn't explain A
(C) A is true, R is false
(D) A is false, R is true
View Answer
Option (D)
Explanation: Solve for y: $(10-2)^2 + (y+3)^2 = 100 \Rightarrow 64 + (y+3)^2 = 100 \Rightarrow (y+3)^2 = 36$. $y+3 = \pm 6$. $y=3$ or $y=-9$. A says only 6, which is wrong.
Explanation: Solve for y: $(10-2)^2 + (y+3)^2 = 100 \Rightarrow 64 + (y+3)^2 = 100 \Rightarrow (y+3)^2 = 36$. $y+3 = \pm 6$. $y=3$ or $y=-9$. A says only 6, which is wrong.
Section B (2 Marks Each)
Given that HCF (306, 657) = 9, find LCM (306, 657).
Show Solution
Formula: $LCM \times HCF = a \times b$
$LCM = \frac{306 \times 657}{9}$
$LCM = 34 \times 657 = 22338$.
$LCM = \frac{306 \times 657}{9}$
$LCM = 34 \times 657 = 22338$.
Find the ratio in which the y-axis divides the line segment joining the points (5, -6) and (-1, -4).
Show Solution
Let ratio be $k:1$. Point on y-axis is $(0, y)$.
$x = \frac{m_1x_2 + m_2x_1}{m_1+m_2} \Rightarrow 0 = \frac{k(-1) + 1(5)}{k+1}$
$-k + 5 = 0 \Rightarrow k = 5$.
Ratio is 5:1.
$x = \frac{m_1x_2 + m_2x_1}{m_1+m_2} \Rightarrow 0 = \frac{k(-1) + 1(5)}{k+1}$
$-k + 5 = 0 \Rightarrow k = 5$.
Ratio is 5:1.
In a right triangle ABC, right-angled at B, if $\tan A = 1$, verify that $2 \sin A \cos A = 1$.
Show Solution
$\tan A = 1 \Rightarrow A = 45^{\circ}$.
LHS = $2 \sin 45^{\circ} \cos 45^{\circ}$
$= 2 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}$
$= 2 \times \frac{1}{2} = 1$ = RHS. Verified.
LHS = $2 \sin 45^{\circ} \cos 45^{\circ}$
$= 2 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}$
$= 2 \times \frac{1}{2} = 1$ = RHS. Verified.
PA and PB are tangents from point P to the circle with centre O. If $\angle APB = 50^{\circ}$, find $\angle OAB$.
Show Solution
In $\Delta PAB$, $PA=PB$, so $\angle PAB = \angle PBA = \frac{180-50}{2} = 65^{\circ}$.
Radius is perpendicular to tangent, so $\angle OAP = 90^{\circ}$.
$\angle OAB = 90^{\circ} - 65^{\circ} = 25^{\circ}$.
Radius is perpendicular to tangent, so $\angle OAP = 90^{\circ}$.
$\angle OAB = 90^{\circ} - 65^{\circ} = 25^{\circ}$.
Find the perimeter of a quadrant of a circle of radius 14 cm.
Show Solution
Perimeter = Arc Length + 2(Radius)
$= \frac{1}{4}(2\pi r) + 2r$
$= \frac{1}{2}(\frac{22}{7} \times 14) + 28$
$= 22 + 28 = 50$ cm.
$= \frac{1}{4}(2\pi r) + 2r$
$= \frac{1}{2}(\frac{22}{7} \times 14) + 28$
$= 22 + 28 = 50$ cm.
Section C (3 Marks Each)
Prove that $\sqrt{3}$ is an irrational number.
Show Solution
Standard text proof: Assume $\sqrt{3} = a/b$. $3b^2 = a^2$. 3 divides $a$. Let $a=3c$. Then $3b^2 = 9c^2 \Rightarrow b^2 = 3c^2$. 3 divides $b$. Contradiction to $a,b$ coprime.
Solve for x and y:
$\frac{2}{x} + \frac{3}{y} = 13$
$\frac{5}{x} - \frac{4}{y} = -2$
$\frac{2}{x} + \frac{3}{y} = 13$
$\frac{5}{x} - \frac{4}{y} = -2$
Show Solution
Let $1/x = u, 1/y = v$.
$2u + 3v = 13$ ...(1)
$5u - 4v = -2$ ...(2)
Multiply (1) by 4 and (2) by 3, add them. $23u = 46 \Rightarrow u=2$.
Put $u=2$ in (1): $4+3v=13 \Rightarrow v=3$.
$x=1/2, y=1/3$.
$2u + 3v = 13$ ...(1)
$5u - 4v = -2$ ...(2)
Multiply (1) by 4 and (2) by 3, add them. $23u = 46 \Rightarrow u=2$.
Put $u=2$ in (1): $4+3v=13 \Rightarrow v=3$.
$x=1/2, y=1/3$.
Find the coordinates of the points of trisection of the line segment joining (4, -1) and (-2, -3).
Show Solution
Case 1 (1:2 ratio): $(\frac{1(-2)+2(4)}{3}, \frac{1(-3)+2(-1)}{3}) = (2, -5/3)$.
Case 2 (2:1 ratio): $(\frac{2(-2)+1(4)}{3}, \frac{2(-3)+1(-1)}{3}) = (0, -7/3)$.
Case 2 (2:1 ratio): $(\frac{2(-2)+1(4)}{3}, \frac{2(-3)+1(-1)}{3}) = (0, -7/3)$.
Prove that: $\frac{\tan \theta}{1-\cot \theta} + \frac{\cot \theta}{1-\tan \theta} = 1 + \sec \theta \text{cosec } \theta$.
Show Solution
Convert all to $\sin/\cos$.
LHS becomes $\frac{\sin^2 \theta}{\cos \theta(\sin \theta - \cos \theta)} + \frac{\cos^2 \theta}{\sin \theta(\cos \theta - \sin \theta)}$.
Combine to $\frac{\sin^3 \theta - \cos^3 \theta}{\sin \theta \cos \theta (\sin \theta - \cos \theta)}$.
Use $a^3-b^3$ formula. Result matches RHS.
LHS becomes $\frac{\sin^2 \theta}{\cos \theta(\sin \theta - \cos \theta)} + \frac{\cos^2 \theta}{\sin \theta(\cos \theta - \sin \theta)}$.
Combine to $\frac{\sin^3 \theta - \cos^3 \theta}{\sin \theta \cos \theta (\sin \theta - \cos \theta)}$.
Use $a^3-b^3$ formula. Result matches RHS.
In the figure, XY and X'Y' are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X'Y' at B. Prove that $\angle AOB = 90^{\circ}$.
Show Solution
Join OC. $\Delta OPA \cong \Delta OCA$ (RHS), so $\angle POA = \angle COA$. Similarly $\angle QOB = \angle COB$.
Sum of angles on line PQ is $180^{\circ}$.
$2\angle COA + 2\angle COB = 180^{\circ} \Rightarrow \angle COA + \angle COB = 90^{\circ} \Rightarrow \angle AOB = 90^{\circ}$.
Sum of angles on line PQ is $180^{\circ}$.
$2\angle COA + 2\angle COB = 180^{\circ} \Rightarrow \angle COA + \angle COB = 90^{\circ} \Rightarrow \angle AOB = 90^{\circ}$.
Two dice are thrown at the same time. Find the probability of getting:
(i) Same number on both dice.
(ii) Different numbers on both dice.
(iii) Product is 12.
(i) Same number on both dice.
(ii) Different numbers on both dice.
(iii) Product is 12.
Show Solution
Total = 36.
(i) (1,1)...(6,6) = 6 outcomes. $P=6/36=1/6$.
(ii) $1 - 1/6 = 5/6$.
(iii) (2,6), (3,4), (4,3), (6,2) = 4 outcomes. $P=4/36=1/9$.
(i) (1,1)...(6,6) = 6 outcomes. $P=6/36=1/6$.
(ii) $1 - 1/6 = 5/6$.
(iii) (2,6), (3,4), (4,3), (6,2) = 4 outcomes. $P=4/36=1/9$.
Section D (5 Marks Each)
The sum of the reciprocals of Rehman's ages (in years) 3 years ago and 5 years from now is $1/3$. Find his present age.
Show Solution
Let present age be $x$.
$\frac{1}{x-3} + \frac{1}{x+5} = \frac{1}{3}$.
$\frac{x+5+x-3}{(x-3)(x+5)} = \frac{1}{3} \Rightarrow \frac{2x+2}{x^2+2x-15} = \frac{1}{3}$.
$x^2 - 4x - 21 = 0 \Rightarrow (x-7)(x+3)=0$.
Age cannot be negative. So, $x=7$ years.
$\frac{1}{x-3} + \frac{1}{x+5} = \frac{1}{3}$.
$\frac{x+5+x-3}{(x-3)(x+5)} = \frac{1}{3} \Rightarrow \frac{2x+2}{x^2+2x-15} = \frac{1}{3}$.
$x^2 - 4x - 21 = 0 \Rightarrow (x-7)(x+3)=0$.
Age cannot be negative. So, $x=7$ years.
A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.
Show Solution
Vol of Cone = $\frac{1}{3}\pi (5)^2(8) = \frac{200\pi}{3}$.
Water flows out = $\frac{1}{4} \times \frac{200\pi}{3} = \frac{50\pi}{3}$.
Vol of 1 shot = $\frac{4}{3}\pi (0.5)^3 = \frac{\pi}{6}$.
Number of shots = $\frac{50\pi/3}{\pi/6} = 100$.
Water flows out = $\frac{1}{4} \times \frac{200\pi}{3} = \frac{50\pi}{3}$.
Vol of 1 shot = $\frac{4}{3}\pi (0.5)^3 = \frac{\pi}{6}$.
Number of shots = $\frac{50\pi/3}{\pi/6} = 100$.
Prove that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. Using this, find x if $DE \parallel BC$ and $AD=4x-3, AE=8x-7, BD=3x-1, CE=5x-3$.
Show Solution
First part: Prove BPT (Standard Proof).
Second part: $\frac{4x-3}{3x-1} = \frac{8x-7}{5x-3}$.
Solve quadratic: $(4x-3)(5x-3) = (8x-7)(3x-1)$.
$20x^2 - 27x + 9 = 24x^2 - 29x + 7$.
$4x^2 - 2x - 2 = 0 \Rightarrow 2x^2 - x - 1 = 0$.
$(2x+1)(x-1) = 0$. $x=1$ or $x=-1/2$.
Length can't be negative, so $x=1$.
Second part: $\frac{4x-3}{3x-1} = \frac{8x-7}{5x-3}$.
Solve quadratic: $(4x-3)(5x-3) = (8x-7)(3x-1)$.
$20x^2 - 27x + 9 = 24x^2 - 29x + 7$.
$4x^2 - 2x - 2 = 0 \Rightarrow 2x^2 - x - 1 = 0$.
$(2x+1)(x-1) = 0$. $x=1$ or $x=-1/2$.
Length can't be negative, so $x=1$.
The median of the following data is 525. Find the values of x and y, if the total frequency is 100.
0-100 (2), 100-200 (5), 200-300 (x), 300-400 (12), 400-500 (17), 500-600 (20), 600-700 (y), 700-800 (9), 800-900 (7), 900-1000 (4).
0-100 (2), 100-200 (5), 200-300 (x), 300-400 (12), 400-500 (17), 500-600 (20), 600-700 (y), 700-800 (9), 800-900 (7), 900-1000 (4).
Show Solution
1. $76 + x + y = 100 \Rightarrow x+y = 24$.
2. Median 525 lies in 500-600 class. $l=500, f=20, h=100, cf=36+x$.
$525 = 500 + \frac{50 - (36+x)}{20} \times 100$.
Solving gives $x=9$.
Then $y = 24 - 9 = 15$.
2. Median 525 lies in 500-600 class. $l=500, f=20, h=100, cf=36+x$.
$525 = 500 + \frac{50 - (36+x)}{20} \times 100$.
Solving gives $x=9$.
Then $y = 24 - 9 = 15$.
Section E (12 Marks)
Orchard Management: A gardener plants apple trees in rows. In the first row, there are 25 trees, in the second row 23, in the third 21, and so on.
(i) Does this arrangement form an AP? If yes, find common difference. (1 M)
(ii) Find the number of trees in the 8th row. (1 M)
(iii) If the last row has 5 trees, find the total number of rows. (2 M)
(i) Does this arrangement form an AP? If yes, find common difference. (1 M)
(ii) Find the number of trees in the 8th row. (1 M)
(iii) If the last row has 5 trees, find the total number of rows. (2 M)
Show Answer
(i) Yes. $23-25 = -2$. $d=-2$.
(ii) $a_8 = 25 + 7(-2) = 11$.
(iii) $5 = 25 + (n-1)(-2) \Rightarrow -20 = -2(n-1) \Rightarrow n=11$.
(ii) $a_8 = 25 + 7(-2) = 11$.
(iii) $5 = 25 + (n-1)(-2) \Rightarrow -20 = -2(n-1) \Rightarrow n=11$.
Kite Festival: A kite is flying at a height of 60m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is $60^{\circ}$.
(i) Find the length of the string, assuming no slack. (2 M)
(ii) If the angle changes to $45^{\circ}$ keeping height same, what is the length of string? (2 M)
(i) Find the length of the string, assuming no slack. (2 M)
(ii) If the angle changes to $45^{\circ}$ keeping height same, what is the length of string? (2 M)
Show Answer
(i) $\sin 60 = 60/L \Rightarrow \sqrt{3}/2 = 60/L \Rightarrow L = 120/\sqrt{3} = 40\sqrt{3}$ m.
(ii) $\sin 45 = 60/L_2 \Rightarrow 1/\sqrt{2} = 60/L_2 \Rightarrow L_2 = 60\sqrt{2}$ m.
(ii) $\sin 45 = 60/L_2 \Rightarrow 1/\sqrt{2} = 60/L_2 \Rightarrow L_2 = 60\sqrt{2}$ m.
Ice Cream Treat: An ice cream seller has a container in the shape of a right circular cylinder having diameter 12 cm and height 15 cm. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top.
(i) Find volume of the cylindrical container. (1 M)
(ii) Find volume of one ice cream cone (including hemisphere). (2 M)
(iii) Find how many such cones can be filled. (1 M)
(i) Find volume of the cylindrical container. (1 M)
(ii) Find volume of one ice cream cone (including hemisphere). (2 M)
(iii) Find how many such cones can be filled. (1 M)
Show Answer
(i) $\pi (6)^2 (15) = 540\pi$.
(ii) Cone Vol + Hemi Vol = $\frac{1}{3}\pi (3)^2(12) + \frac{2}{3}\pi (3)^3 = 36\pi + 18\pi = 54\pi$.
(iii) $540\pi / 54\pi = 10$ cones.
(ii) Cone Vol + Hemi Vol = $\frac{1}{3}\pi (3)^2(12) + \frac{2}{3}\pi (3)^3 = 36\pi + 18\pi = 54\pi$.
(iii) $540\pi / 54\pi = 10$ cones.
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