Check the solution of CBSE Class 10 Maths Standard Term 2 Question Paper 2022. All solutions are provided by the subject experts. Students can now cross check their answers and calculate the expected marks in their Maths paper.
CBSE Class 10 Maths Answer Key Term 2: CBSE Class 10 students must be waiting for the answer key of Maths paper that was conducted on 5th May, 2022. We have come here with the solutions created by the subject experts. Therefore, students can check now correct answers to all questions given in the CBSE Class 10 Maths (Standard) Term 2 Question Paper 2022. They can also download the question paper in PDF here and check the paper analysis along with the students’ feedback as well.
Check CBSE Class 10 Maths (Standard) Answer Key below:
1. In fig. 1, AB is diameter of a circle centered at O. BC is tangent to the circle at B. If OP bisects the chord AD and ∠AOP = 60°, then find m∠C.
Solution:
Given OP bisect the chord AD.
∴ OP⟂AD
∠P = 90^{0}
and ∠B = 90^{0}
∠BOP = 180^{0} – 60^{0} = 120^{0}
Now, in quad. BOPC, applying angle sum property
∠P + ∠B + ∠O + ∠C = 360^{0}
90^{0} + 90^{0} + 120^{0}+ ∠C = 360^{0}
∠C = 360^{0 }– 30^{0} = 60^{0}
OR
(b) In Fig. 2, XAY is a tangent to the circle centered at O. If ∠ABO = 40°, then find m∠BAY and m∠AOB.
Solution:
Given ∠ABO = 40^{0}
∠XAO = 90^{0}
OA = OB (Radii of same circle)
⇒∠OAB = ∠OBA
∴ ∠OAB = 40^{0}
Now, applying linear pair of angles property
∠BAY + ∠OAB + ∠XAO = 180^{0}
∠BAY + 40^{0 }+ 90^{0} = 180^{0}
∠BAY + 130^{0} = 180^{0}
∠BAY = 50^{0}
Now, in ΔAOB,
∠AOB + ∠OAB + ∠OBA = 180^{0}
∠AOB + 40^{0 }+40^{0} = 180^{0}
∠AOB + 80^{0 }= 180^{0}
∠AOB = 100^{0}
2. If the mode of the following frequency distribution is 55, then find the value of x.
Class |
0-15 |
15-30 |
30-45 |
45-60 |
60-75 |
75-90 |
Frequency |
10 |
7 |
x |
15 |
10 |
12 |
Solution:
Given that mode of frequency distribution is 55.
Here given mode =55
So, mode class is 45−60
lower limit (t) =45
class height (h) =15
then, f_{0} =15
f_{1} =x
f_{2}=10
10(30−x−10)=(15−x)15
10(20−x)=225−15x
200−10x=225−15k
15x−10x=225−200
5x=25
x=5
3. (a) In an AP if some of third and seventh terms is zero, find its 5th term.
Solution:
Let a and d be the first term and common difference of AP
nth term of AP, a_{n}=a+(n−1)d
∴a_{3}=a+(3−1)d=a+2d
a_{7}=a+(7−1)d=a+6d
Given a_{3}+a_{7}=0
∴(a+2d)+(a+6d)=0
⇒2a+8d=0
⇒a+4d=0
Now, 5th term of given AP is,
a_{5}=a+(5−1)d=a + 4d
a_{5}= 0
OR
(b) Determine the AP whose third term is 5 and seventh term is 9.
Solution:
We know that
a_{n}=a+(n−1)d
∴a_{3}=a+(3−1)d=a+2d
⇒ a+2d = 5 …..(i)
a_{7}=a+(7−1)d=a+6d
⇒ a+6d = 9 …..(ii)
Subtracting (ii) from (i), we have
4d = 4
d = 1
substituting d=2 in (i),
a + 2(1) = 5
a = 5-2
a=3
Thus, first term of required AP is
a = 3
second term = first term + common difference
a_{2} = 3+1 =4
Similarly, a_{3} = 4+1 =5, a_{4} = 6 and so on
Thus, the required AP is 3, 4, 5, 6, …….
4. Solve the quadratic equation x2 + 2√2x – 6 = 0.
5. Find the sum of first 20 terms of an AP whose nth term is given as a_{n} = 5 – 2n.
6. A solid piece of metal in the form of a cuboid of dimensions 11 cm x 7 cm x 7 cm is melted to form ‘n’ number of solid spheres of radii 7⁄2 cm each. Find the value of n.
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