# HBSE Class 10 Maths Half Yearly Question Paper 2023 Answer Key

Haryana Board (HBSE) Class 10 Maths Half Yearly Question Paper 2023 Answer Key. Haryana Board Class 10th Half Yearly Question Paper Pdf Download 2023. Haryana Board Class 10th Half Yearly Solved Question Paper 2023. HBSE Class 10th Half Yearly Exam 2023. HBSE Maths Half Yearly Question Paper 2023 Answer Key.

HBSE Class 10 Maths Half Yearly Question Paper 2023 Answer Key

Instructions :
• All questions are compulsory.
• Questions (1-15) carry 1 mark each.
• Questions (16-18) carry 2 marks each.
• Questions (19-21) carry 3 marks each.
• Questions (22-23) carry 5 marks each.

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) 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
√1 = 1, √4 = ±2, √8 = 2√2, √9 = ±3

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
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
280 = 2 × 2 × 2 × 5 × 7
245 = 5 × 7 × 7
HCF = 5 × 7 = 35

5. A.P. 3, 8, 13, 18, ….. मे कोन-सा पद 78 होगा?
Which term of the A.P. 3, 8, 13, 18, …… is 78?
(a) 12th
(b) 13th
(c) 15th
(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?
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) 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

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
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) None of these
Answer : (d) 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

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)
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)
Coordinates of mid point = [(x1+x2)/2, (y1+y2)/2]
= [(–4+8)/2, (3–5)/2] = (4/2, –2/2) = (2, –1)

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
a15 = 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 : 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 : 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)2
= √(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 : 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 : 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 : 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 : 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 2.5 m की दूरी पर है तथा इसका ऊपरी सिरा भूमि से 6 m 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 2.5 m from the wall and its top reaches a window 6 m above the ground. Find the length of the ladder.

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, x22 = –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. 6 cm त्रिज्या का एक वृत्त खींचिए। केंद्र से 10 cm दूर स्थित एक बिंदु से वृत्त पर स्पर्श रेखा युग्म की रचना कीजिए और उनकी लंबाइयाँ मापिए।
Draw a circle of radius 6 cm. From a point 10 cm away from its center, construct the pair of tangents to the circle and measure their lengths.

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.