Q.6
Triangles
[1 Mark]
The perimeters of two similar triangles are $56\text{ cm}$ and $70\text{ cm}$ respectively. If one side of the first triangle is $14\text{ cm}$, then the corresponding side of the second triangle (in cm) is:
A 5
B 7.5
C 10
D 17.5
💡 Hint: The ratio of the perimeters of two similar triangles is strictly equal to the ratio of their corresponding sides! Set up the fraction: $\frac{P_1}{P_2} = \frac{s_1}{s_2}$. (Ref: www.maheshkumarsir.in)
www.maheshkumarsir.in
We know that for two similar triangles, the ratio of their perimeters is equal to the ratio of their corresponding sides.
$\Rightarrow \frac{\text{Perimeter of } \Delta_1}{\text{Perimeter of } \Delta_2} = \frac{\text{Side of } \Delta_1}{\text{Side of } \Delta_2}$ [½ Mark]
Let the required corresponding side of the second triangle be $x$.
$\frac{56}{70} = \frac{14}{x}$
Cross-multiplying:
$56x = 70 \times 14$
$x = \frac{70 \times 14}{56}$
$x = \frac{70}{4} = 17.5\text{ cm}$ [½ Mark]
$\therefore$ Correct Option is (D)
Q.7
Coordinate Geometry
[1 Mark]
The point which lies on the perpendicular bisector of the line segment joining the points $A(-3, -4)$ and $B(3, 4)$ is:
A $(0, 0)$
B $(0, 3)$
C $(3, 0)$
D $(0, 4)$
💡 Hint: The perpendicular bisector of a line segment always passes exactly through its midpoint. Simply calculate the midpoint of $A$ and $B$!
⚠️ Common Mistake: Trying to find the full equation of the perpendicular bisector and then checking all points. Finding the midpoint is an instant shortcut for this specific MCQ!
www.maheshkumarsir.in
Any point on the perpendicular bisector of a line segment is equidistant from its endpoints. The most obvious point that satisfies this is the midpoint.
Using the Midpoint Formula:
$M(x, y) = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$ [½ Mark]
$M(x, y) = \left( \frac{-3 + 3}{2}, \frac{-4 + 4}{2} \right)$
$M(x, y) = \left( \frac{0}{2}, \frac{0}{2} \right)$
$M(x, y) = (0, 0)$ [½ Mark]
Since the origin $(0,0)$ is the midpoint, it definitely lies on the perpendicular bisector.
$\therefore$ Correct Option is (A)
Q.8
Trigonometry
[1 Mark]
If $\tan (A + B) = \sqrt{3}$ and $\tan (A - B) = \frac{1}{\sqrt{3}}$, $0^\circ < A + B \le 90^\circ$, $A > B$, then the value of $A$ and $B$ respectively are:
A $60^\circ, 30^\circ$
B $60^\circ, 45^\circ$
C $45^\circ, 15^\circ$
D $60^\circ, 15^\circ$
💡 Hint: Ask yourself: "Tan of what angle gives $\sqrt{3}$?" and "Tan of what angle gives $\frac{1}{\sqrt{3}}$?". This creates two simple linear equations in terms of $A$ and $B$.
www.maheshkumarsir.in
Given: $\tan (A + B) = \sqrt{3}$
We know $\tan 60^\circ = \sqrt{3}$
$\Rightarrow A + B = 60^\circ$ ..... (Equation 1)
Given: $\tan (A - B) = \frac{1}{\sqrt{3}}$
We know $\tan 30^\circ = \frac{1}{\sqrt{3}}$
$\Rightarrow A - B = 30^\circ$ ..... (Equation 2) [½ Mark]
Adding Eq 1 and Eq 2:
$(A + B) + (A - B) = 60^\circ + 30^\circ$
$2A = 90^\circ \Rightarrow A = 45^\circ$
Substitute $A = 45^\circ$ in Eq 1:
$45^\circ + B = 60^\circ \Rightarrow B = 15^\circ$ [½ Mark]
$\therefore$ Correct Option is (C)
Q.9
Probability
[1 Mark]
T-shirts marked with numbers 4 to 99 are placed in a box. Gunika is fond of numbers. She randomly takes out a T-shirt from this box. The probability that she gets a T-shirt marked with a number that is either a perfect square or a perfect cube is:
A $\frac{1}{12}$
B $\frac{1}{32}$
C $\frac{11}{96}$
D $\frac{5}{48}$
💡 Hint: First, find the total number of T-shirts correctly using $(Last - First) + 1$. Then, list out the perfect squares and perfect cubes carefully. Don't count $64$ twice (it's both a square and a cube)!
⚠️ Common Mistake: Calculating total outcomes as $99 - 4 = 95$. It should be $99 - 4 + 1 = 96$ because both endpoints are included!
www.maheshkumarsir.in
Total numbers from $4$ to $99 = (99 - 4) + 1 = 96$.
Total possible outcomes $= 96$. [½ Mark]
Let's list the favorable outcomes:
Perfect squares between 4 and 99:
$\{4, 9, 16, 25, 36, 49, 64, 81\}$ $\rightarrow$ ($8$ numbers)
Perfect cubes between 4 and 99:
$\{8, 27, 64\}$
Notice that $64$ ($8^2$ and $4^3$) is already counted in the squares. So, the unique extra perfect cubes are $\{8, 27\}$.
Total favorable outcomes $= 8 (\text{squares}) + 2 (\text{unique cubes}) = 10$.
$\text{Probability} = \frac{\text{Favorable}}{\text{Total}} = \frac{10}{96} = \frac{5}{48}$ [½ Mark]
$\therefore$ Correct Option is (D)
Q.10
Areas Related to Circles
[1 Mark]
The diameter of a car wheel is $21\text{ cm}$. The number of complete revolutions it will make in moving $66\text{ km}$ is:
A $10^4$
B $10^5$
C $10^6$
D $10^7$
💡 Hint: One revolution covers a distance equal to the circumference of the wheel ($\pi d$). Convert the total distance from km to cm before dividing! (Ref: www.maheshkumarsir.in)
⚠️ Common Mistake: Forgetting to convert $66\text{ km}$ into centimeters! $1\text{ km} = 1000\text{ m}$, and $1\text{ m} = 100\text{ cm}$. So, $1\text{ km} = 100,000\text{ cm}$.
www.maheshkumarsir.in
Distance covered in 1 revolution = Circumference of the wheel.
Circumference $C = \pi \times d$
$C = \frac{22}{7} \times 21 = 22 \times 3 = 66\text{ cm}$. [½ Mark]
Total Distance = $66\text{ km}$.
Convert to cm:
$66\text{ km} = 66 \times 1000\text{ m} = 66 \times 1000 \times 100\text{ cm}$
Total Distance = $6,600,000\text{ cm} = 66 \times 10^5\text{ cm}$.
$\text{Number of revolutions} = \frac{\text{Total Distance}}{\text{Circumference}}$
$\text{Revolutions} = \frac{66 \times 10^5}{66} = 10^5$ [½ Mark]
$\therefore$ Correct Option is (B)
