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

TS Polycet (Polytechnic) 2024 Previous Question Paper with Answers And Model Papers With Complete Analysis
TS Polycet (Polytechnic) Previous Year Question Papers And Model Papers:
While preparing for TS Polycet (Polytechnic), candidates must also refer to the previous year question papers of the same. Scoring well in TS Polycet (Polytechnic) and understanding weaknesses and strengths in the respective sections.

TS Polycet (Polytechnic) Previous Year Question Papers can be found on this page in PDF format. Students taking the exam to get into some of the best Polytechnic colleges/institutes in the state of Andhra Pradesh may practice these papers to get a clear idea of the structure of the exam, marking scheme, important topics, etc.

TS Polycet Mock Test

Section — A
MATHEMATICS

1) The centroid of the triangle with vertices (1,-1), (0, 6) and (-3, 0) is బిందువులు (1, −1), (0, 6) మరియు (-3, 0) లు శీర్షాలుగా గల త్రిభుజము యొక్క గురుత్వ కేంద్రము

A) $\left(\frac23,\frac53\right)$
B) $\left(\frac{-2}3,\frac{-5}3\right)$
C) $\left(\frac{-2}3,\frac53\right)$
D) $\left(\frac23,\frac{-5}3\right)$

View Answer

C) $\left(\frac{-2}3,\frac53\right)$

Explanation:Sure! Let’s find the **centroid** of the triangle with the given vertices:
Vertices:
– A = (1, -1)
– B = (0, 6)
– C = (-3, 0)
Centroid Formula:
The centroid $G$ of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is:
$G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)$
Applying the values:
x-coordinate:
$\frac{1 + 0 + (-3)}{3} = \frac{-2}{3}$
y-coordinate:
$\frac{-1 + 6 + 0}{3} = \frac{5}{3}$
Final Answer: $G = \left( \frac{-2}{3}, \frac{5}{3} \right)$

2) In ΔABC, if DE || BC, $\frac{AE}{CE}=\frac35$ and AB 5.6 cm, then AD = _________. ΔABC లో DE || BC, $\frac{AE}{CE}=\frac35$ మరియు AB = 5.6 సెం.మీ. అయిన, AD = __________.

A) 2.8 cm
B) 2.1 cm
C) 3 cm
D) 2.4 cm

View Answer

B) 2.1 cm

Explanation:We are given:
– In triangle $\Delta ABC$ , line DE || BC
– $\frac{AE}{EC} = \frac{3}{5}$
– Total length AB = 5.6 cm
– We are to find AD
—
Since DE || BC, by Basic Proportionality Theorem (Thales’ Theorem), we know:
$\frac{AD}{DB} = \frac{AE}{EC} = \frac{3}{5}$
Let’s assume:
– $AD = 3x$
– $DB = 5x$
So total length of $AB = AD + DB = 3x + 5x = 8x$
Given: $AB = 5.6$
So: $8x = 5.6 \Rightarrow x = \frac{5.6}{8} = 0.7$
Now, $AD = 3x = 3 \times 0.7 = \boxed{2.1 \text{ cm}}$

3) A Kiddy bank contains hundred 50 paise coins, fifty ₹ 1 coins, twenty ₹ 2 coins and ten ₹ 5 coins. If one of the coins will fall out when the bank is turned upside down, what is the probability that the coin is ₹ 5 coin? ఒక కిడ్డీ బ్యాంక్ డబ్బాలో వంద 50 పైసల నాణెములు, యాభై కౌ ₹ 1 నాణెములు, ఇరవై ₹ 2 నాణెములు మరియు పది ₹ 5 నాణెములు ఉన్నాయి. డబ్బాను తలక్రిందులు చేసినప్పుడల్లా యాదృచ్ఛికంగా ఒక నాణెం పడుతుంటే, అది ₹ 5 నాణెము కావడానికి సంభావ్యత ఎంత?

A) 5/9
B) 5/18
C) 1/9
D) 1/18

View Answer

D) 1/18

Explanation:We are given the number of each type of coin:
– 50 paise coins = 100
– ₹1 coins = 50
– ₹2 coins = 20
– ₹5 coins = 10
Let’s find the total number of coins:
$100 + 50 + 20 + 10 = 180$
We are asked to find the probability that the coin falling out is a ₹5 coin.
There are 10 ₹5 coins, and total coins are 180.
So, the probability is:
$\frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{10}{180} = \frac{1}{18}$

4) In a grouped frequency distribution, the formula to find median is వర్గీకృత పౌనఃపున్య విభాజనానికి, మధ్యగతము సూత్రము

A) $l+\left(\frac{{\displaystyle\frac n2}+cf}f\right)\times h$
B) $l-\left(\frac{{\displaystyle\frac n2}+cf}f\right)\times h$
C) $l+\left(\frac{{\displaystyle\frac n2}-cf}f\right)\times h$
D) $l-\left(\frac{{\displaystyle\frac n2}-cf}f\right)\times h$

View Answer

C) $l+\left(\frac{{\displaystyle\frac n2}-cf}f\right)\times h$

Explanation:To find the median in a grouped frequency distribution, we use the following formula:
$\text{Median} = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h$
Where:
– l = lower boundary of the median class
– n = total frequency
– cf = cumulative frequency before the median class
– f = frequency of the median class
– h = class width
—

5) In a G. P. the 3rd term is 24 and 6th is 192, then the 10th term is ఒక గుణశ్రేఢిలో 3 వ పదము ’24’ మరియు 6 వ పదము ‘192’ అయిన, 10 వ పదము

A) 2072
B) 3072
C) 1072
D) 1672

View Answer

B) 3072

Explanation:Sure! Here’s a shortcut method using the property of G.P. terms:
—
We are given:
– $T_3 = 24$
– $T_6 = 192$
We are asked to find: $T_{10} = ?$
—
Shortcut Idea:
In G.P., $T_n = T_m \cdot r^{n - m}$
So we use: $T_{10} = T_6 \cdot r^{10 - 6} = T_6 \cdot r^4$
First, find $r$ from: $T_6 = T_3 \cdot r^{6-3} = 24 \cdot r^3 = 192 \Rightarrow r^3 = \frac{192}{24} = 8 \Rightarrow r = 2$
Now: $T_{10} = 192 \cdot 2^4 = 192 \cdot 16 = \boxed{3072}$