HBSE Class 10th Maths Half Yearly Question Paper 2022 Answer

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HBSE Class 10th Maths Half Yearly Question Paper 2022 Answer (Exam – Sep 2022) 

विद्यार्थी का नाम ………. रोल नंबर ……….
अर्धवार्षिक आकलन – सितम्बर 2022
कक्षा – 10
विषय – Maths
समय : 90 मिनट         कुल अंक : 40

निर्देश : 1. सभी प्रश्न अनिवार्य हैं।
2. प्रश्न संख्या 1 से 15 तक प्रत्येक प्रश्न 1 अंक का है। प्रश्न संख्या 16 से 18 तक प्रत्येक प्रश्न 2 अंक, प्रश्न संख्या 19 से 21 प्रत्येक प्रश्न 3 अंक तथा प्रश्न संख्या 22 से 23 तक प्रत्येक प्रश्न 5 अंक का है।
Instructions : 1. All questions are compulsory.
2. Question numbers 1 to 15 carry 1 mark each, question numbers 16 to 18 carry 2 marks each, question number 19 to 21 carry 3 marks each and question number 22 to 23 carry 5 marks each.

(1-10) सही उत्तर पर (✓) का निशान लगाएँ।
Mark (✓) on the correct answer.
1. निम्नलिखित में से कौन सी एक अपरिमेय संख्या नहीं है ?
Which of the following is not an irrational number ?
(A) 2.011001…..
(B) 2 + 2√4
(C) 2√2
(D) इनमें से कोई नहीं None of these
Answer : (B) 2 + 2√4
2 + 2√4 = 2 + 2(2) = 2 + 4 = 6

2. निम्नलिखित में से कौन सी संख्या अपरिमेय है ?
Which of these is an irrational number ?
(A) √1
(B) √4
(C) √8
(D) √9
Answer : (C) √8
√8 = 2√2

3. द्विघात बहुपद x² – x – 5 के शून्यकों का योग होगा –
Sum of zeros of quadratic polynomial x² – x – 5 is –
(A) 1
(B) -1
(C) √19
(D) -5
Answer : (A) 1
Compare with ax² + bx + c
Here a = 1, b = -1, c = 5
α + β = -b/a = -(-1/1) = 1

4. 280 और 245 का HCF होगा –
HCF of 280 and 245 is –
(A) 36
(B) 45
(C) 28
(D) 35
Answer : (D) 35
280 = 2 × 2 × 2 × 5 × 7
245 = 5 × 7 × 7
HCF = 5 × 7 = 35

5. A.P. 3, 8, 13, 18….. मे कोनसा पद 78 होगा ?
Which term of the A.P. 3, 8, 13, 18…… is 78 ?
(A) 12th
(B) 13th
(C) 15th
(D) 16th
Answer : (D) 16th
Here a = 3, d = 8-3 = 5, an = 78
nth term of A.P., an = a + (n-1)d
3 + (n-1)(5) = 78
3 + 5n – 5 = 78
5n = 78 + 5 -3
5n = 80
n = 80/5 = 16

6. k के किन मानो के लिए, निम्न रैखिक समीकरणों के युग्म का अद्वितीय हल होगा ?
x – ky + 4 = 0, 2x – 6y – 5=0
What value of k does the pair of linear equations ?
x – ky + 4 = 0, 2x – 6y – 5 = 0 has unique solution
(A) k = 3
(B) k ≠ 3
(C) k = -3
(D) इनमें से कोई नहीं None of these
Answer : (B) k ≠ 3
Here a1 = 1, a2 = 2, b1 = -k, b2 = -6, c1 = 4, c2 = -5
a1/a2 ≠ b1/b2
1/2 ≠ (-k)/(-6)
k ≠ 3

7. बहुपद p(x) = ax² + bx + c के अधिकतम शून्यक होंगे, जहां a ≠ 0
Polynomial p(x) = ax² + bx + c, a ≠0 has maximum number of zeros
(A) 1
(B) 2
(C) 3
(D) 4
Answer : (B) 2

8. द्विघात बहुपद 6x² + x – 2 के शून्यकों का गुणनफल है
Product of zeros of quadratic polynomial 6x² + x – 2 is
(A) ⅓
(B) -⅔
(C) -⅓
(D) ½
Answer : (C) -⅓
Compare with ax² + bx + c
Here a = 6, b = 1, c = -2
αβ = c/a = -2/6 = -⅓

9. एक द्विघात बहुपद ज्ञात कीजिए, जिसके शून्यकों के योगफल तथा गुणनफल क्रमश: 3 और 1 हैं
Find a quadratic polynomial, whose sum and product of zeroes is 3 and 1 respectively
(A) x² + 4x + 1
(B) x² – 4x + 1
(C) x² – 3x – 1
(D) इनमें से कोई नही None of these
Answer : (D) इनमें से कोई नही None of these
α + β = 3, αβ = 1
Quadratic Polynomial = x² – (α+β)x + αβ
= x² – 3x + 1

10. हल करे: x + y = 14, x – y = 4
Solve: x + y = 14, x – y = 4
(A) x = 9, y = -5
(B) x = 5, y = 9
(C) x = 9, y = 5
(D) x = -9, y = -5
Answer : (C) x = 9, y = 5
x + y = 14 …….(i)
x – y = 4 ……..(ii)
Adding eqn.(i) and (ii), we get
2x = 18
x = 18/2 = 9
Put x = 9 in eqn.(i), we get
9 + y = 14
y = 14 – 9 = 5

सही या गलत चुने। Find True or False
11. अनुपात a1/a2, b1/b2 और c1/c2 की तुलना करने पर रैखिक समीकरणों 5x – 3y = 11, -10x + 6y = 22 के युग्म समांतर हैं (सही या गलत )
On comparing the ratios a1/a2, b1/b2 and c1/c2 the pair of linear equations 5x – 3y = 11, -10x + 6y = 22 are parallel. (True or False)
Answer : True
5x – 3y – 11 = 0, -10x + 6y – 22 = 0
Here a1 = 5, a2 = -10, b1 = -3, b2 = 6, c1 = -11, c2 = -22
a1/a2 = 5/(-10) = -½
b1/b2 = (-3)/6 = -½
c1/c2 = (-11)/(-22) = ½
a1/a2 = b1/b2 ≠ c1/c2

12. बिंदुओं (-4, 3) और (8, -5) के मध्य बिंदु के निर्देशांक (2, -1) हैं (सही या गलत)
The coordinates of mid-point between points (-4, 3) and (8, -5) is (2, -1) (True or False)
Answer : True
Coordinates of mid point = [(x1+x2)/2, (y1+y2)/2]
= [(-4+8)/2, (3-5)/2] = (4/2, -2/2) = (2, -1)

खाली स्थान भरो।
Fill in the blanks
13. A.P. 1, 4, 7, ….. का 15वां पद होगा …………
The 15th term of the A.P. 1, 4, 7, ….. will be …………
Answer :
Here a = 1, d = 4-1 = 3, n = 15
nth term of A.P., an = a + (n-1)d
a15 = a + 14d = 1 + 14(3) = 1 + 42 = 43

14. दो समरूप त्रिभुज की भुजाएं 4 : 9 के अनुपात में हैं। इन त्रिभुजों के क्षेत्रफल का अनुपात है ………….
The sides of two similar triangles are in the ratio 4 : 9. Area of these triangles is in the ratio ………….
Answer: 16 : 81
If two triangles are similar to each other, then the ratio of the area of triangles will be equal to the square of the ratio of the corresponding sides of triangle.
Area of triangles is in the ratio = 4² : 9² = 16 : 81

15. मूल बिंदु से (5, -7) की दूरी है ………..
Distance from the origin to the point (5, -7) is ……….
Answer : √74
Here A(0, 0) and B(5, -7)
x1 = 0, x2 = 5, y1 = 0, y2 = -7
Using Distance Formula,
AB = √(x2-x1)²+(y2-y1
= √(5-0)²+(-7-0)²
= √(5)²+(-7)²
= √25+49
= √74

16. k का वह मान ज्ञात कीजिए जिसके लिए द्विघात समीकरण 3x² + kx + 3 = 0 के दो बराबर मूल हैं।
Find value of k for which the quadratic equation 3x² + kx + 3 = 0 has two equal roots.
Answer : k = ±6
Here a = 3, b = k, c = 3
Discriminant (D) = b² – 4ac = 0
k² – 4(3)(3) = 0
k² – 36 = 0
k² = 36
k = √36
k = ±6

17. यदि दो संख्याओं का योग 31 हो और उनका गुणनफल 240 हो तो संख्याएं ज्ञात करो।
If sum of two numbers is 31 and their product is 240 then find numbers.
Answer : 15, 16
Let first number = x
Second number = 31-x
ATQ,
x(31-x) = 240
31x – x² = 240
0 = x² – 31x + 240
x² – 31x + 240 = 0
x² – 16x – 15x + 240 = 0
x(x-16) – 15(x-16) = 0
(x-16)(x-15) = 0
Either x – 16= 0 or x – 15 = 0
x = 16, 15
If first number = 16, then second number = 31-16 = 15
If first number = 15, then second number = 31-15 = 16

18. यदि एक A.P. का 11वाँ पद 38 और 16वाँ पद 73 है तो उसका 22वाँ पद ज्ञात करें।
If in a A.P. 11th term is 38 and 16th term is 73 then find 22th term.
Answer : 115
nth term of A.P., an = a + (n-1)d
11th term of the A.P. = 38
a + 10d = 38 …….(i)
16th term of the A.P. = 73
a + 15d = 73 ……(ii)
Subtract eqn.(i) from eqn.(ii), we get
5d = 35
d = 35/5 = 7
Put d = 7 in eqn.(i),
a + 10(7) = 38
a + 70 = 38
a = 38 – 70 = -32
so, 22th term of A.P. (a22) = a + 21d = – 32 + 21(7) = – 32 + 147 = 115

19. यदि किसी A.P. के प्रथम 7 पदों का योग 49 है और प्रथम 17 पदों का योग 289 है, तो इसके प्रथम पद का मान ज्ञात कीजिए।
If in a A.P. sum of first 7 terms is 49 and sum of first 17 terms is 289, then find first term.
Answer : a = 1
Sum of first n terms of A.P., Sn = n/2 [2a + (n-1)d]
S7 = 7/2[2a + 6d] = 49
2a + 6d = 49 × 2/7 = 14
Dividing both side by 2,
a + 3d = 7 ………(i)
S17 = 17/2[2a + 16d] = 289
2a + 16d = 289 × 2/17 = 34
Dividing both side by 2,
a + 8d = 17 ……..(ii)
Subtract eqn.(i) from (ii), we get
5d = 10
d = 2
Put d = 2 in eqn.(i),
a + 3(2) = 7
a = 7-6 = 1

20. एक सीढ़ी किसी दीवार पर इस प्रकार टिकी हुई है कि इसका निचला सिरा दीवार से 2.5m की दूरी पर है तथा इसका ऊपरी सिरा भूमि से 6m की ऊँचाई पर बनी एक खिड़की तक पहुँचता है। सीढ़ी की लंबाई ज्ञात कीजिए।
A ladder is placed against a wall such that its foot is at a distance of 2.5m from the wall and its top reaches a window 6m above the ground. Find the length of the ladder.
Answer : 6.5m

Using Pythagoras theorem,
H² = P² + B²
AC² = AB² + BC²
AC² = 6² + (2.5)²
AC² = 36 + 6.25 = 42.25
AC = √42.25 = 6.5 m

21. उस त्रिभुज का क्षेत्रफल ज्ञात कीजिए जिसके शीर्ष बिंदुओं के निर्देशांक हैं (1, -1), (-4, 6) और (-3,-5).
Find the area of a triangle whose vertices are (1, -1), (-4, 6) and (-3, -5).
Answer :
Here x1 = 1, x2 = -4, x3 = -3, y1 = -1, y2 = 6, y3 = -5
Area of triangle = ½[x1(y2-y3) + x2(y3-y1) + x3(y1-y2)]
= ½[1(6+5) + (-4)(-5+1) + (-3)(-1-6)]
= ½[11 – 4(-4) + (-3)(-7)]
= ½[11 + 16 + 21]
= ½(48) = 24 sq. unit

22. 6cm त्रिज्या का एक वृत्त खींचिए। केंद्र से 10cm दूर स्थित एक बिंदु से वृत्त पर स्पर्श रेखा युग्म की रचना कीजिए और उनकी लंबाइयाँ मापिए।
Draw a circle of radius 6cm. From a point 10 cm away from its center, construct the pair of tangents to the circle and measure their lengths.
Answer :

Steps of Construction :
Step I- A circle with radius 6 cm is drawn taking O as centre.
Step II- Point P is marked at 10 cm away from centre of circle.
Step III- With the half of compass mark M which is the midpoint of OP.
Step IV- Draw a circle with centre M, taking radius MO or MP which intersects the given circle at Q and R.
Step V- Now join PQ and PR. These are the tangents of the circle.
Given radius = OQ = 6 cm and OP = 10 cm
In ∆OPQ, OQ⊥QP
Applying Pythagoras theorem we get :
OP² = OQ² + PQ²
10² = 6² + PQ²
PQ² = 100 – 36 = 64
PQ = √64 = 8 cm
PQ = PR = 8 cm
Hence, the length of tangents is 8 cm.


Answer : f = 20

 

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