TS Polycet (Polytechnic) 2024 Previous Question Paper with Answers And Model Papers With Complete Analysis

Polycet Mock Test

41)There are exactly _____ tangents to a circle through a point outside the circle.
వృత్త బాహ్యంలో గల ఏదైనా బిందువు గుండా వృత్తానికి ఖచ్చితంగా _____ స్పర్శరేఖలు గీయగలము.

A) two

B) three

C) infinite

D) None

View Answer

A) two

42)The length of the tangent from a point 15 cm away from the centre of a circle of radius 9 cm is ____.
9 సెం.మీ. వ్యాసార్ధముగా గల వృత్తానికి దాని కేంద్రం నుండి 15 సెం.మీ. దూరములో ఒక బిందువు కలదు. అయిన, ఆ బిందువు నుండి వృత్తానికి గీయబడిన స్పర్శరేఖ పొడవు ______.

A) 15 cm

B) 13 cm

C) 11 cm

D) 12 cm

View Answer

D) 12 cm
Explanation:To find the length of the tangent from a point outside the circle, we can use the Pythagorean Theorem.
Given:
– The distance from the point to the center of the circle (the hypotenuse of the right triangle) = 15 cm.
– The radius of the circle (one leg of the right triangle) = 9 cm.
We need to find the length of the tangent, which is the other leg of the right triangle.
Using the Pythagorean theorem:
$\text{(distance from the point to the center)}^2 = \text{(radius)}^2 + \text{(length of the tangent)}^2$

Substitute the values:

*** QuickLaTeX cannot compile formula:
15^2 = 9^2 + \text{(length of the tangent)}^2<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://www.mcqbits.com/wp-content/ql-cache/quicklatex.com-90adcfb5a78f5ebddacbc5a89297bf71_l3.png" height="22" width="273" class="ql-img-displayed-equation quicklatex-auto-format" alt="\[225 = 81 + \text{(length of the tangent)}^2\]" title="Rendered by QuickLaTeX.com"/>\text{(length of the tangent)}^2 = 225 - 81 = 144<pre class="ql-errors">*** QuickLaTeX cannot compile formula:
\[\text{length of the tangent} = \sqrt{144} = 12 \, \text{cm}$
Final Answer:
The length of the tangent is 12 cm. So, the correct option is 4. 12 cm.
[/su_spoiler]
</div>
<div class="mcq-question" data-answer="D"><b>43)If the areas of two similar triangles are 81 cm<sup>2</sup> and 49 cm<sup>2</sup> respectively. If the altitude of the smaller triangle is 3.5 cm, then the corresponding altitude of the bigger triangle Is _____
రెండు సరూప త్రిభుజాల వైశాల్యలు 81 చ.సెం.మీ. మరియు 49 చ.సెం.మీ. చిన్న త్రిభుజములో గీసిన లంబము పొడవు 3.5 సెం.మీ. అయిన, పెద్ద త్రిభుజములో దాని అనురూప లంబము పొడవు ____</b>
<div class="mcq-options">
A) 9.5 cm
B) 9 cm
C) 7 cm
D) 4.5 cm
</div>
[su_spoiler title="View Answer" style="fancy" icon="arrow"]
D) 4.5 cm
Explanation:Given:
- Areas of two similar triangles = 81 cm² and 49 cm²
- Altitude of the smaller triangle = 3.5 cm
Formula:
For two similar triangles, the ratio of their areas is the square of the ratio of their corresponding sides (or altitudes).
$\frac{\text{Area of larger triangle}}{\text{Area of smaller triangle}} = \left(\frac{\text{Corresponding side of larger triangle}}{\text{Corresponding side of smaller triangle}}\right)^2$
Step 1: Find the ratio of the areas$\frac{81}{49} = \left(\frac{\text{Altitude of larger triangle}}{\text{Altitude of smaller triangle}}\right)^2\]
*** Error message:
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leading text: ...he tangent} = \sqrt{144} = 12 \, \text{cm}$
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leading text: [/su_
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leading text: ...ding altitude of the bigger triangle Is __
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leading text: ...ng altitude of the bigger triangle Is ____
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leading text: ...g altitude of the bigger triangle Is _____
Missing { inserted.
leading text: ర
Unicode character ర (U+0C30)
leading text: ర
Unicode character ె (U+0C46)
leading text: రె
Unicode character ం (U+0C02)
leading text: రెం
Unicode character డ (U+0C21)
leading text: రెండ
Unicode character ు (U+0C41)
leading text: రెండు
</pre>\left(\frac{\text{Altitude of larger triangle}}{\text{Altitude of smaller triangle}}\right)^2 = \frac{81}{49} = \left(\frac{9}{7}\right)^2<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://www.mcqbits.com/wp-content/ql-cache/quicklatex.com-4c73d4e2b7525168c50c120b6ae077ef_l3.png" height="137" width="584" class="ql-img-displayed-equation quicklatex-auto-format" alt="\[\frac{\text{Altitude of larger triangle}}{\text{Altitude of smaller triangle}} = \frac{9}{7}$ Step 2: Find the altitude of the larger triangle Now, we can calculate the altitude of the larger triangle by multiplying the altitude of the smaller triangle by the ratio of the corresponding sides. $\text{Altitude of larger triangle} = \text{Altitude of smaller triangle} \times \frac{9}{7}\]" title="Rendered by QuickLaTeX.com"/>\text{Altitude of larger triangle} = 3.5 \times \frac{9}{7} = 3.5 \times 1.2857 \approx 4.5 \, \text{cm}

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leading text: ...5 = 81 + \text{(length of the tangent)}^2\]
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leading text: ...he tangent} = \sqrt{144} = 12 \, \text{cm}$
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Final Answer:
The altitude of the larger triangle is 4.5 cm.

44)A tangent to a circle touches it in _____ point(s).
వృత్తానికి గీయబడిన స్పర్శరేఖ దానిని ____ బిందువు(ల) వద్ద స్పృశిస్తుంది.

A) one

B) two

C) three

D) infinite
అనంత

View Answer

A) one

45)If AP and AQ are the two tangents to a circle with centre ‘O’, so that $\angle POQ\;=110^\circ$ , then $\angle PAQ$ = ______.
క్రింది పటములో ‘O’ కేంద్రముగా గల వృత్తానికి, AP మరియు AQ లు రెండు స్పర్శరేఖలు మరియు $\angle POQ\;=110^\circ$ అయిన, $\angle PAQ$ = ________.

A) 60°

B) 70°

C) 80°

D) 90°

View Answer

B) 70°
Explanation:To find the measure of angle $\angle PAQ$

given that $\angle POQ = 110^\circ$

, we can use the properties of tangents to a circle and the fact that the sum of angles in a quadrilateral is $360^\circ$

.
Given:
– $AP$

and $AQ$

are tangents to the circle with center $O$

.
– $\angle POQ = 110^\circ$

.
Key Properties:
– 1. Tangents are perpendicular to the radius at the point of contact:
– $\angle OAP = 90^\circ$

(since $AP$

is a tangent and $OA$

is the radius).
– $\angle OAQ = 90^\circ$

(since $AQ$

is a tangent and $OA$

is the radius).
– 2. Sum of angles in quadrilateral $APOQ$

:
$\angle OAP + \angle APO + \angle POQ + \angle OAQ = 360^\circ$

Substituting the known values:
$90^\circ + \angle APO + 110^\circ + 90^\circ = 360^\circ$

Simplifying:
$\angle APO = 360^\circ - 290^\circ = 70^\circ$

– 3. Since $AP = AQ$

(tangents from an external point are equal), triangle $APQ$

is isosceles:
$\angle PAQ = 180^\circ - 2 \times \angle APO = 180^\circ - 2 \times 70^\circ = 40^\circ$

However, let’s verify this with a more straightforward approach.
Alternative (Simpler) Approach:
– The angle between two tangents ( $\angle PAQ$

) is related to the angle subtended at the center ( $\angle POQ$

) by:
$\angle PAQ = 180^\circ - \angle POQ$

Substituting the given value:
$\angle PAQ = 180^\circ - 110^\circ = 70^\circ$

Final Answer: $\boxed{70^\circ}$

46)If $\frac{a_1}{a_2}\neq\frac{b_1}{b_2},$ where a₁x+b₁y+c₁=0 and a₂x+b₂y+c₂=0 are two linear equations,
$\frac{a_1}{a_2}\neq\frac{b_1}{b_2},$ అయ్యేటట్లుగాa₁x+b₁y+c₁=0 మరియు a₂x+b₂y+c₂=0 అనేవి రెండు రేఖీయ సమీ కరణాలైతే, ఆ సమీ కరణాలు

A) have a unique solution
ఏకైక సాధన కలిగి ఉంటాయి

B) have infinitely many solutions
అనంతమైన సాధనలు కలిగి ఉంటాయి

C) have finite solutions
పరిమితమైన సాధనలు కలిగి ఉంటాయి

D) have no solution
ఏ సాధనలను కలిగి ఉండవు

View Answer

A) have a unique solution
ఏకైక సాధన కలిగి ఉంటాయి
Explanation:Given the two linear equations:
$a_1x + b_1y + c_1 = 0 \quad \text{and} \quad a_2x + b_2y + c_2 = 0$

And the condition that $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$

.
Solution:
The two equations represent two straight lines in the coordinate plane. The condition $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$

means that the lines are not parallel and will intersect at exactly one point.
– When two lines intersect at exactly one point, the system of linear equations has a unique solution.
– This condition rules out the possibility of the lines being parallel (which would mean no solution) or coincident (which would mean infinitely many solutions).
Therefore, the correct answer is:
– 1. Have a unique solution.

47)The value of p, for which the pair of equations 3x+4y+2=0 and 9x+ py+8=0 represents parallel lines, is
3x +4y+ 2 = 0 మరియు 9x + py + 8 = 0 అను సమీకరణాల జత సమాంతర రేఖలను సూచించిన, P విలువ

A) 2

B) 4

C) 6

D) 12

View Answer

D) 12
Explanation:To determine the value of $p$

for which the pair of equations represents parallel lines, we first examine the general condition for parallel lines.
Given equations:
– 1. $3x + 4y + 2 = 0$

– 2. $9x + py + 8 = 0$

Step 1: Write the equations in slope-intercept form $y = mx + c$

, where $m$

is the slope.
For the first equation 3x + 4y + 2 = 0, solve for $y$

$4y = -3x - 2 <img src="https://www.mcqbits.com/wp-content/ql-cache/quicklatex.com-3bdc84e9d578fdf8b9194fae0629966b_l3.png" height="89" width="582" class="ql-img-displayed-equation quicklatex-auto-format" alt="\[y = -\frac{3}{4}x - \frac{1}{2}$So, the slope $ m_1 $ of the first line is $ -\frac{3}{4} $. For the second equation $ 9x + py + 8 = 0 $, solve for $ y $:$py = -9x - 8\]" title="Rendered by QuickLaTeX.com"/>y = -\frac{9}{p}x - \frac{8}{p}$

So, the slope $m_2$ of the second line is $-\frac{9}{p}$ .
Step 2: Condition for parallel lines
For the lines to be parallel, their slopes must be equal, i.e.,

$m_1 = m_2 <img src="https://www.mcqbits.com/wp-content/ql-cache/quicklatex.com-b13b8b66b6b6bf70949bd31755026aa0_l3.png" height="68" width="328" class="ql-img-displayed-equation quicklatex-auto-format" alt="\[-\frac{3}{4} = -\frac{9}{p}$ Step 3: Solve for $ p\]" title="Rendered by QuickLaTeX.com"/>\frac{3}{4} = \frac{9}{p}$

Cross multiply:

*** QuickLaTeX cannot compile formula:
3p = 36<pre class="ql-errors">*** QuickLaTeX cannot compile formula:
\[p = 12$
Therefore, the value of $ p $ is 12.
The correct answer is:4. 12.
[/su_spoiler]
</div>
<div class="mcq-question" data-answer="B"><b>48)Which of the following equation represent the situation where Kiran bought 5 oranges, 7 apples and Harish bought 2 oranges, 12 apples for same amount of total money?
కిరణ్ 5 నారింజలు, 7 యాపిల్స్ మరియు హరీశ్ 2 నారింజలు, 12 యాపిల్స్ విడివిడిగా ఒకే మొత్తానికి కొన్నారు. కింది సమీకరణాలలో ఈ విషయాన్ని సూచించే సమీకరణము ఏది?</b>
<div class="mcq-options">
A) 5x+12y=2x+7y
B) 5x+7y=2x+12y
C) 5x-7y=2x-12y
D) 5x+2y=7x+12y
</div>
[su_spoiler title="View Answer" style="fancy" icon="arrow"]
B) 5x+7y=2x+12y
Explanation:Let's denote:
- $ x $ as the price of one orange.
- $ y $ as the price of one apple.
Given:
- Kiran bought 5 oranges and 7 apples.
- Harish bought 2 oranges and 12 apples.
- They spent the same amount of money.
Thus, the total money spent by Kiran is:$5x + 7y$And the total money spent by Harish is:$2x + 12y$
Since they spent the same amount, we can set these two expressions equal to each other:$5x + 7y = 2x + 12y$
Simplifying the equation:$5x - 2x = 12y - 7y\]
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leading text: \[p = 12$
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leading text: [/su_
Unicode character క (U+0C15)
leading text: క
Unicode character ి (U+0C3F)
leading text: కి
Unicode character ర (U+0C30)
leading text: కిర
Unicode character ణ (U+0C23)
leading text: కిరణ
Unicode character ్ (U+0C4D)
leading text: కిరణ్
Unicode character న (U+0C28)
leading text: కిరణ్ 5 న
Unicode character ా (U+0C3E)
leading text: కిరణ్ 5 నా
Unicode character ర (U+0C30)
leading text: కిరణ్ 5 నార
Unicode character ి (U+0C3F)
leading text: కిరణ్ 5 నారి
Unicode character ం (U+0C02)
leading text: కిరణ్ 5 నారిం
Unicode character జ (U+0C1C)
</pre>3x = 5y

*** Error message:
Display math should end with $$.
leading text: \[p = 12$
Missing $ inserted.
leading text: [/su_
Unicode character క (U+0C15)
leading text: క
Unicode character ి (U+0C3F)
leading text: కి
Unicode character ర (U+0C30)
leading text: కిర
Unicode character ణ (U+0C23)
leading text: కిరణ
Unicode character ్ (U+0C4D)
leading text: కిరణ్
Unicode character న (U+0C28)
leading text: కిరణ్ 5 న
Unicode character ా (U+0C3E)
leading text: కిరణ్ 5 నా
Unicode character ర (U+0C30)
leading text: కిరణ్ 5 నార
Unicode character ి (U+0C3F)
leading text: కిరణ్ 5 నారి
Unicode character ం (U+0C02)
leading text: కిరణ్ 5 నారిం
Unicode character జ (U+0C1C)

This is the correct equation representing the given situation.
Therefore, the correct answer is:
2. $5x + 7y = 2x + 12y$ .

49)If $\frac2{\sqrt x}+\frac3{\sqrt y}\;=\;2$ and $\frac4{\sqrt x}-\frac9{\sqrt y}\;=\;-1$ , then
$\frac2{\sqrt x}+\frac3{\sqrt y}\;=\;2$ మరియు $\frac4{\sqrt x}-\frac9{\sqrt y}\;=\;-1$ అయిన,

A) x=4, y=3

B) x=2, y=9

C) x=4, y=9

D) x=2, y=3

View Answer

C) x=4, y=9
Explanation:To solve the given system of equations quickly, follow this shortcut method:
Given equations:
– 1. $\frac{2}{\sqrt{x}} + \frac{3}{\sqrt{y}} = 2$

– 2. $\frac{4}{\sqrt{x}} - \frac{9}{\sqrt{y}} = -1$

Step 1: Let’s assume new variables for simplification:
Let:
– $p = \frac{1}{\sqrt{x}}$

– $q = \frac{1}{\sqrt{y}}$

Now the equations become:
– $2p + 3q = 2$

– $4p - 9q = -1$

Step 2: Use substitution or elimination to solve for $p$

and $q$

.
Multiply the first equation by 2 to align coefficients of $p$

for elimination:
$(2p + 3q = 2) \times 2 \quad \Rightarrow \quad 4p + 6q = 4 \quad \text{(Equation 3)}$

Now subtract the second equation from Equation 3:

$(4p + 6q) - (4p - 9q) = 4 - (-1) <img src="https://www.mcqbits.com/wp-content/ql-cache/quicklatex.com-ae6c95d609ac53910b6b37cff2b285cb_l3.png" height="17" width="167" class="ql-img-displayed-equation quicklatex-auto-format" alt="\[4p + 6q - 4p + 9q = 5\]" title="Rendered by QuickLaTeX.com"/>15q = 5 <img src="https://www.mcqbits.com/wp-content/ql-cache/quicklatex.com-e924f2112135ece04c065ad0941a04e8_l3.png" height="69" width="492" class="ql-img-displayed-equation quicklatex-auto-format" alt="\[q = \frac{1}{3}$ Step 3: Substitute $ q = \frac{1}{3} $ into the first equation: $2p + 3 \times \frac{1}{3} = 2\]" title="Rendered by QuickLaTeX.com"/>2p + 1 = 2 <img src="https://www.mcqbits.com/wp-content/ql-cache/quicklatex.com-7843b6c0cc98e1b3cfa8479b62e645c9_l3.png" height="16" width="49" class="ql-img-displayed-equation quicklatex-auto-format" alt="\[2p = 1\]" title="Rendered by QuickLaTeX.com"/>p = \frac{1}{2}$

Step 4: Solve for $x$ and $y$ :
Recall:
– $p = \frac{1}{\sqrt{x}}$ and $q = \frac{1}{\sqrt{y}}$
– So $\sqrt{x} = \frac{1}{p} = 2$ and $\sqrt{y} = \frac{1}{q} = 3$
Thus:
– $x = 4$
– $y = 9$
Final Answer:
– $x = 4$ , $y = 9$

50)The pair of equations x+y=5 and 2x+2y=k has infinitely many solutions if k =
x + y = 5 మరియు 2x + 2y = k అను సమీకరణాల జత అనంతమైన సాధనలను కలిగి ఉండాలి అనిన, k =

A) 4

B) 6

C) 8

D) 10

View Answer

D) 10
Explanation:To determine the value of $k$

for which the system of equations has infinitely many solutions, we need to examine the two equations:
– 1. $x + y = 5$

– 2. $2x + 2y = k$

Step 1: Express both equations in a similar form
The second equation can be simplified by dividing both sides by 2:
$\frac{2x + 2y}{2} = \frac{k}{2} \quad \Rightarrow \quad x + y = \frac{k}{2}$

Now, the system of equations becomes:
– 1. $x + y = 5$

– 2. $x + y = \frac{k}{2}$

Step 2: Condition for infinitely many solutions
For the system to have infinitely many solutions, the two equations must represent the same line. This happens when their right-hand sides are equal. Therefore, we set:
$5 = \frac{k}{2}$

Step 3: Solve for $k$

Multiply both sides by 2 to solve for $k$

:
$10 = k$

Final Answer:
The value of $k$

for which the system of equations has infinitely many solutions is 10.

Your Score: 0 / 0

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