UP Board Class 10 Mathematics Question Paper 2025 PDF (Code 822 BZ) with Answer Key and Solutions PDF is available for download here. UP Board Class 10 exams were conducted between February 24th to March 12th 2025. The total marks for the theory paper were 70. Students reported the paper to be easy to moderate.
UP Board Class 10 Mathematics Question Paper 2025 (Code 822 BZ) with Solutions
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HCF of the numbers 96 and 404 is 4. Value of their LCM will be:
The distance of the point \((2, 5)\) from the origin will be:
The value of \(\sqrt{2} + \tan 45^\circ\) will be:
The value of \(\sec^2 A - \tan^2 A\) will be:
In the equation \(2x + 3y = 11\), if \(x = 1\), the value of \(y\) will be:
The value of a term of an A.P. \(21, 18, 15, \ldots\) is \(-81\). Then the term will be:
There are 2 blue, 3 white, and 4 red balls in a bag. One ball is taken out randomly from this bag. The probability that the ball is not red will be:
The probability that an event will happen surely is:
View Solution
Step 1: Recall probability basics.
The probability of any event lies between 0 and 1.
Step 2: Certain event.
If an event is certain (sure to happen), its probability is the maximum possible value.
Step 3: Conclusion.
Therefore, the probability of a sure event is: \[ P(Sure Event) = 1 \]
Final Answer: \[ \boxed{1} \] Quick Tip: Remember: Impossible event probability = 0, Sure event probability = 1.
In the given figure, if \(DE \parallel BC\), then the value of \(\tfrac{DE}{BC}\) will be:
From a point \(Q\), the length of tangent to a circle is \(24 \, cm\) and the distance of \(Q\) from the centre is \(25 \, cm\). The radius of the circle is:
The angle of a sector of a circle of radius \(6 \, cm\) is \(30^\circ\). The measure of corresponding arc will be:
The height of a solid cylinder of radius \(3 \, cm\) is \(5 \, cm\). A hemisphere of the same radius is placed on its vertex. The total surface area of this solid will be:
The contact point of a tangent to the circle is joined to the centre. The angle at the contact point between tangent and radius will be:
The value of \(\dfrac{1 + \tan^2 A}{1 + \cot^2 A}\) will be:
An arc of a circle of radius \(7 \, cm\) subtends an angle of \(60^\circ\) at the centre. The area of the sector will be:
The surface area of a sphere is \(452 \tfrac{4}{7} \, cm^2\). Its radius will be:
The product of zeroes of the quadratic polynomial \(p(x) = 4x^2 - 4x - 1\) will be:
The discriminant of the quadratic equation \(3x^2 - 4x + \tfrac{4}{3} = 0\) will be:
The mean from the following table will be:
Class-interval & Frequency
0-4 & 4
4-8 & 6
8-12 & 6
12-16 & 5
16-20 & 4
The median class of the following table will be:
Class-interval & Frequency
0-10 & 6
10-20 & 8
20-30 & 10
30-40 & 9
40-50 & 12
Do all the parts:
(a) If a point \((0,1)\) is equidistant from the points \((5,-3)\) and \((x,6)\), then find the values of \(x\).
(a) Find the mode from the following table:
Class-interval & Frequency
0-10 & 6
10-20 & 11
20-30 & 18
30-40 & 21
40-50 & 15
Solve the following pair of equations:
\[ 5x+y=3, 6x-5y=\tfrac{1}{2} \]
View Solution
Step 1: Express \(y\) from first equation.
\[ y=3-5x \]
Step 2: Substitute into second equation.
\[ 6x-5(3-5x)=\tfrac{1}{2} \] \[ 6x-15+25x=\tfrac{1}{2} \] \[ 31x-15=\tfrac{1}{2} \]
Step 3: Simplify.
\[ 31x=\tfrac{1}{2}+15 = \tfrac{1}{2}+ \tfrac{30}{2} = \tfrac{31}{2} \] \[ x=\tfrac{31}{62} = \tfrac{1}{2} \]
Step 4: Find \(y\).
\[ y=3-5\left(\tfrac{1}{2}\right)=3-\tfrac{5}{2}=\tfrac{1}{2} \]
Final Answer: \[ \boxed{x=\tfrac{1}{2}, \; y=\tfrac{1}{2}} \] Quick Tip: For solving pairs of linear equations, substitution method is effective when one equation can be easily expressed in terms of a variable.
The area of a rectangle gets reduced by \(28 \, m^2\), if its length is increased by 2 m and breadth is reduced by 2 m. If the length is reduced by 1 m and breadth is increased by 2 m, the area is increased by \(33 \, m^2\). Find the area of the rectangle.
The angle of elevation of the top of a 20 m high building from a point \(O\) on the horizontal plane is \(30^\circ\). There is a flagstaff on the top of the building. The angle of elevation of the top of flagstaff is \(45^\circ\). Find the height of the flagstaff and measure of the distance of foot of building from the point \(O\).
The angles of depression from a point on the bridge of a river to opposite banks are \(30^\circ\) and \(45^\circ\) respectively. If the height of the bridge from the banks is \(4.5 \, m\), then find the breadth of the river.
View Solution
Step 1: Let breadth of river = \(AB\).
Let point on bridge = \(P\), banks = \(A, B\), with \(PA \perp AB\), height \(PA=4.5\).
Step 2: For angle of depression \(30^\circ\).
\[ \tan 30^\circ = \frac{PA}{AD} = \frac{4.5}{AD} \Rightarrow AD=\frac{4.5}{1/\sqrt{3}}=4.5\sqrt{3} \]
Step 3: For angle of depression \(45^\circ\).
\[ \tan 45^\circ = \frac{PA}{DB} = \frac{4.5}{DB} \Rightarrow DB=4.5 \]
Step 4: Total breadth.
\[ AB=AD+DB=4.5\sqrt{3}+4.5=4.5(\sqrt{3}+1) \] \[ =4.5(1.732+1)=4.5 \times 2.732=12.29 \, m \]
Final Answer: \[ \boxed{AB=4.5(\sqrt{3}+1) \approx 12.29 \, m} \] Quick Tip: Angles of depression are measured from the horizontal, so they form right triangles with the vertical height of the bridge.
(a) A maximum sphere is made by peeling a wooden cube of side 14 cm. Find the volume of the peeled wood.
View Solution
Step 1: Volume of cube.
\[ V_{cube} = a^3 = 14^3 = 2744 \, cm^3 \]
Step 2: Volume of largest sphere inscribed in cube.
Radius of sphere = half of cube’s side = \(\tfrac{14}{2} = 7\) cm.
\[ V_{sphere} = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (7^3) = \frac{4}{3}\pi (343) = \frac{1372}{3}\pi \]
Using \(\pi=\tfrac{22}{7}\): \[ V_{sphere} = \frac{1372}{3}\times \frac{22}{7} = \frac{30184}{21} \approx 1437.33 \, cm^3 \]
Step 3: Volume of peeled wood.
\[ V_{peeled} = V_{cube} - V_{sphere} = 2744 - 1437.33 = 1306.67 \, cm^3 \]
Final Answer: \[ \boxed{1306.67 \, cm^3 \; (approx)} \] Quick Tip: The largest sphere that can fit inside a cube has diameter equal to the side of the cube.
(b) In the figure, two concentric circles of radii 14 cm and 7 cm whose centre is \(O\) and arcs are \(\overset{\frown}{AB}\) and \(\overset{\frown}{CD}\) and \(\angle AOB = 30^\circ\). Find the area of the shaded portion.
View Solution
Step 1: Formula for area of sector.
\[ Area of sector = \frac{\theta}{360^\circ}\pi r^2 \]
Step 2: Outer sector (radius 14).
\[ Outer area = \frac{30}{360}\pi (14^2) = \frac{1}{12}\pi (196) = \frac{196}{12}\pi = \frac{49}{3}\pi \]
Step 3: Inner sector (radius 7).
\[ Inner area = \frac{30}{360}\pi (7^2) = \frac{1}{12}\pi (49) = \frac{49}{12}\pi \]
Step 4: Shaded area (difference).
\[ Shaded area = \frac{49}{3}\pi - \frac{49}{12}\pi \] \[ = \left(\frac{196-49}{12}\right)\pi = \frac{147}{12}\pi = \frac{49}{4}\pi \]
Using \(\pi=\tfrac{22}{7}\): \[ Shaded area = \frac{49}{4}\times \frac{22}{7} = \frac{154}{4} = 38.5 \, cm^2 \]
Final Answer: \[ \boxed{38.5 \, cm^2} \] Quick Tip: For ring-shaped sectors, subtract the area of smaller sector from the larger one.
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