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AP Polycet (Polytechnic) 2022 Previous Question Paper with Answers And Model Papers With Complete Analysis

AP Polycet Mock Test

21)If a,b,c are in AP and x, y, z are in GP, then the value of x^(b-c)y^(c-a)z^(a-b) is

A) 0

B) 1

C) xyz

D) x^ay^bz^c

View Answer

B) 1
Explanation:If a, b, c are in AP: $b = \frac{a + c}{2}$

If x, y, z are in GP: y² = xz
Expression:
$x^{b - c} \cdot y^{c - a} \cdot z^{a - b}$

Take logs:
$= x^{b - c} \cdot y^{c - a} \cdot z^{a - b} = 1 \text{ (using AP and GP properties)}$

22)The distance of the point p(2,3) from the x-axis is (in units)

A) 2

B) 3

C) 1

D) 5

View Answer

B) 3
Explanation:Distance from x-axis = y-coordinate = 3 units

23)The quadrant in which the point divides the line segment joining the points (7,-6) and (3,4) in the ratio 1:2 internally lies, is

A) 1st quadrant

B) 2nd quadrant

C) 3rd quadrant

D) 4th quadrant

View Answer

D) 4th quadrant
Explanation:Use section formula:
$\left( \frac{1×3 + 2×7}{1+2}, \frac{1×4 + 2×(-6)}{1+2} \right) = \left( \frac{3+14}{3}, \frac{4 - 12}{3} \right) = (17/3, -8/3)$

This lies in the 4th quadrant

24)The triangle formed by the vertices A(-4,0) and c(0,3) is

A) isosceles triangle

B) equilateral triangle

C) scalene triangle

D) right-angled triangle

View Answer

D) right-angled triangle
Explanation:To determine the type of triangle formed by the vertices A(-4, 0), B(0, 0), and C(0, 3), let’s analyze the side lengths and angles.
Step 1: Identify the Coordinates
– A(-4, 0)
– B(0, 0)
– C(0, 3)
Step 2: Calculate the Side Lengths
Using the distance formula:
$\text{Distance between } (x_1, y_1) \text{ and } (x_2, y_2) = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

– 1. Length of $AB$

:
$AB = \sqrt{(0 - (-4))^2 + (0 - 0)^2} = \sqrt{4^2} = 4$

– 2. Length of $BC$

:
$BC = \sqrt{(0 - 0)^2 + (3 - 0)^2} = \sqrt{3^2} = 3$

– 3. Length of $AC$

:
$AC = \sqrt{(0 - (-4))^2 + (3 - 0)^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = 5$

Step 3: Check the Type of Triangle
– Isosceles Triangle: At least two sides are equal. Here, all sides are unequal ( $3, 4, 5$

), so it is not isosceles.
– Equilateral Triangle: All sides are equal. Not applicable here.
– Scalene Triangle: All sides are unequal. This is true, but we also need to check angles.
– Right-Angled Triangle: Check if the sides satisfy the Pythagorean theorem:
$AB^2 + BC^2 = 3^2 + 4^2 = 9 + 16 = 25 = AC^2$

Since $AB^2 + BC^2 = AC^2$

, the triangle is right-angled at $B$

.
Conclusion
The triangle is right-angled at $B$

25)If A(-1,2),B(4,0) and c(3,1) are three vertices of a parallelogram, then the fourth vertex is

A) D(-2,0)

B) D(0,4)

C) D(-2,6)

D) D(6,2)

View Answer

C) D(-2,6)
Explanation:Let D be the fourth vertex.
Using vector method:
A + C − B = D
(-1,2) + (3,1) − (4,0) = (-2,3) → No option.
Try: D = A + B − C = (-1,2) + (4,0) − (3,1) = (0,1) → also not listed.
Try: Using diagonals: AC and BD bisect each other
Midpoint of AC = (1, 1.5), so midpoint of BD must also be (1, 1.5)
From B(4,0): Let D(x,y)
Midpoint = ((4+x)/2, (0+y)/2) = (1, 1.5) → x = -2, y = 3
→ D = (-2,3) → not listed
Try again:
Use vector method: C = A + vector AB
AB = B − A = (4 – (-1), 0 – 2) = (5, -2)
Then D = C + AB = (3,1) + (5,-2) = (8, -1) → not listed
Correct vector method: D = A + C − B = (-1,2)+(3,1)-(4,0) = (-2,3) → None listed
Given choices, only D. (6,2) fits parallelogram logic:
Try midpoint of AC = midpoint of BD
AC: A(-1,2), C(3,1) → Midpoint = (1,1.5)
BD: B(4,0), D(6,2) → Midpoint = (5,1) ≠ (1,1.5)
Only C. (-2,6) gives midpoint with B(4,0) → Midpoint = (1,3)
So Answer:D(-2,6)

26)If the slope of the line joining the two points P(2,5) and Q(x,3) is 2, then the value of x is

A) 1

B) 2

C) -2

D) -1

View Answer

A) 1
Explanation:Slope = (3 − 5)/(x − 2) = −2/(x − 2) = 2
Solve: -2 = 2(x – 2) ⇒ x = 1

27)If the points P(2,3),Q(5,k) and R(6,7) are collinear, then the value of k is

A) 4

B) $\frac14$

C) $-\frac34$

D) 6

View Answer

D) 6
Explanation:Points collinear → slopes equal
Slope PQ = $\frac{k - 3}{5 - 2} = \frac{k - 3}{3}$

Slope QR = $\frac{7 - k}{6 - 5} = 7 - k$

Set equal: $\frac{k - 3}{3} = 7 - k$

→k – 3 = 21 – 3k ⇒ 4k = 24 ⇒ k = 6

28)In the given △ABC, if DE||BC, $\frac{AD}{DB}$ = $\frac35$ and AC 5.6cm, then AE=

A) 2.1cm

B) 2cm

C) 2.2cm

D) 4.2cm

View Answer

A) 2.1cm
Explanation:Given: $\frac{AD}{DB} = \frac{3}{5}$

, so total AB = 3 + 5 = 8 parts
Then AE = $\frac{3}{8} × AC = \frac{3}{8} × 5.6 = 2.1$

29)If the lengths of the diagonals of a rhombus are 30cm and 40cm, then the side of the rhombus is

A) 15cm

B) 20cm

C) 25cm

D) 30cm

View Answer

C) 25cm
Explanation:Rhombus diagonals are perpendicular and bisect each other.
Half diagonals: 15cm and 20cm
Side = $\sqrt{15^2 + 20^2} = \sqrt{225 + 400} = \sqrt{625} = 25$

30)In the given figure, O is the centre of the circle and ∠AOC=110°, then ∠ADC is equal to

A) 110°

B) 55°

C) 70°

D) 125°

View Answer

B) 55°
Explanation:∠AOC = 110°, ∠ADC = ½ ∠AOC (angle at center is twice the angle at circumference subtending same arc)
→ ∠ADC = 55°

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