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AP Polycet (Polytechnic) 2021 Previous Question Paper with Answers And Model Papers With Complete Analysis

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41)The degree of the equation x²(x²+x+1)=x⁴+x³-x²+3x-1 is

A) 1

B) 2

C) 3

D) 4

View Answer

B) 2
Explanation:To determine the degree of the given equation:
$x^2(x^2 + x + 1) = x^4 + x^3 - x^2 + 3x - 1$

Step 1: Expand the Left Side
$x^2 \cdot x^2 + x^2 \cdot x + x^2 \cdot 1 = x^4 + x^3 + x^2$

So, the equation becomes:
$x^4 + x^3 + x^2 = x^4 + x^3 - x^2 + 3x - 1$

Step 2: Simplify the Equation
Subtract $x^4 + x^3$

from both sides:
$x^2 = -x^2 + 3x - 1$

Bring all terms to one side:
$x^2 + x^2 - 3x + 1 = 0$

Combine like terms:
$2x^2 - 3x + 1 = 0$

Step 3: Determine the Degree
The simplified equation is:
$2x^2 - 3x + 1 = 0$

The degree of an equation is the highest power of $x$

present. Here, the highest power is $x^2$

, so the degree is 2.
Final Answer:
$\boxed{2}$

Correct Option: 2

42)If 18, x, 36 are in arithmetic progression, then x=

A) 9

B) 18

C) 27

D) 26

View Answer

C) 27
Explanation:We are given that 18, x, 36 are in an arithmetic progression (A.P.)
Shortcut method:
In an A.P., the middle term is the average of the first and third terms:
$x = \frac{18 + 36}{2} = \frac{54}{2} = 27$

Final Answer: 27

43)If a, b, c are in arithmetic progression, then a+c=

A) b

B) 2b

C) b-a

D) b+a

View Answer

B) 2b
Explanation:If a, b, c are in arithmetic progression (A.P.), then:
$b = \frac{a + c}{2}$

Multiply both sides by 2:
$2b = a + c$

Final Answer: 2b

44)The common difference of the arithmetic progression 781, 806, 831, … is

A) 26

B) 24

C) 25

D) 23

View Answer

C) 25
Explanation:To find the common difference (d) in an Arithmetic Progression (A.P.), subtract any term from the next term:
d = 806 – 781 = 25
Final Answer: 25

45)The product of two numbers is 91 and their arithmetic mean is 10, then the two numbers are

A) 10, 10

B) 12, 8

C) 13, 7

D) 14, 6

View Answer

C) 13, 7
Explanation:We are given:
Product = 91
Arithmetic Mean = 10 → So,
$\frac{x + y}{2} = 10 \Rightarrow x + y = 20$

We now solve using sum and product shortcut:
We need two numbers whose:
Sum = 20
Product = 91
Try options:
Option C: 13 and 7
13 + 7 = 20 and 13 × 7 = 91
Both conditions are satisfied.
Final Answer: 13, 7

46)The centroid divides each median in the ratio of

A) 1:2

B) 2:1

C) 3:1

D) 1:3

View Answer

B) 2:1

47)If the centroid of the triangle formed with (a, b), (b, c) and (c, a) is O(0, 0) then a³+b³+c³=

A) abc

B) 2abc

C) -3abc

D) 3abc

View Answer

D) 3abc
Explanation:We are given:
Vertices of triangle:
A(a, b), B(b, c), C(c, a)
Centroid is $O(0, 0)$

Step 1: Use centroid formula
The coordinates of the centroid of triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$

is:
$\left( \frac{x_1 + x_2 + x_3}{3},\ \frac{y_1 + y_2 + y_3}{3} \right)$

Plug in:

$x_1 = a,\ x_2 = b,\ x_3 = c<span class="ql-right-eqno"> (1) </span><span class="ql-left-eqno"> </span><img src="https://www.mcqbits.com/wp-content/ql-cache/quicklatex.com-a8fc1c297c69ced1881f80dea60c18fe_l3.png" height="225" width="957" class="ql-img-displayed-equation quicklatex-auto-format" alt="\[y_1 = b,\ y_2 = c,\ y_3 = a$ $\text{Centroid} = \left( \frac{a + b + c}{3},\ \frac{b + c + a}{3} \right)$ Given centroid is $(0, 0)$, so: $a + b + c = 0 $ Step 2: Use identity We want to find: $a^3 + b^3 + c^3 = ?$ Use identity: $a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$ From (1), $a + b + c = 0$, so: $a^3 + b^3 + c^3 - 3abc = 0 \Rightarrow a^3 + b^3 + c^3 = 3abc$ Final Answer: 3abc [/su_spoiler] </div> <div class="mcq-question" data-answer="A"> <b>48)The vertices of a parallelogram are (2,-3),(6,5),(-2,1), (-6,-7) in this order. The point of intersection of the diagonals is</b> <div class="mcq-options"> A) (0,-1) B) (0,0) C) (-1,0) D) (4,1) </div> [su_spoiler title="View Answer" style="fancy" icon="arrow"] A) (0,-1) Explanation:We are given the vertices of a parallelogram in order: $A(2, -3)\]" title="Rendered by QuickLaTeX.com"/>B(6, 5)<span class="ql-right-eqno"> </span><span class="ql-left-eqno"> </span><img src="https://www.mcqbits.com/wp-content/ql-cache/quicklatex.com-b9775757ace816acde4e6577766ba6db_l3.png" height="19" width="67" class="ql-img-displayed-equation quicklatex-auto-format" alt="\[C(-2, 1)\]" title="Rendered by QuickLaTeX.com"/>D(-6, -7)$

Step 1: Diagonals intersect at their midpoint
We can pick opposite vertices (diagonal endpoints):
Diagonal AC: Points $A(2, -3)$ and $C(-2, 1)$
Find midpoint of AC:
$\left( \frac{2 + (-2)}{2}, \frac{-3 + 1}{2} \right) = (0, -1)$
Check diagonal BD just to verify:
Points $B(6, 5)$ and $D(-6, -7)$
Midpoint:
$\left( \frac{6 + (-6)}{2}, \frac{5 + (-7)}{2} \right) = (0, -1)$
Same point! So diagonals intersect at (0, -1)
Final Answer: (0, -1)

49)Distance between the points (0,a) and (0,-a) is

A) a²

B) 2a²

C) 4a²

D) 2a

View Answer

D) 2a
Explanation:We are given two points:
$(0, a)$

$(0, -a)$

Step-by-step shortcut:
These two points lie on the y-axis, so the distance between them is just the difference in y-coordinates.
Distance = |a – (-a)| = |a + a| = 2a
Final Answer: 2a

50)Two poles of height 6 m and 11 m stand on a plain ground and the distance between their feet is 12 m, then the distance between their tops is

A) 11

B) 12

C) 13

D) 14

View Answer

C) 13
Explanation:We are given:
Height of first pole = 6 m
Height of second pole = 11 m
Distance between their feet = 12 m
Step-by-step shortcut (Apply Pythagoras theorem):
Difference in height = 11 – 6 = 5 m
This forms a right triangle with:
vertical side = 5 m
horizontal side = 12 m
hypotenuse (distance between tops) = ?
$\text{Distance}^2 = 12^2 + 5^2 = 144 + 25 = 169 \Rightarrow \text{Distance} = \sqrt{169} = 13$

Final Answer: 13

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