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Haryana Board Class 10 Maths Half Yearly Question Paper 2024 Answer Key
Instructions :
• All questions are compulsory.
• Questions (1-11) carry 1 mark each.
• Questions (12-14) carry 2 marks each.
• Questions (15-17) carry 3 marks each.
• Questions (18-19) carry 5 marks each.
• Question (20) case study, carry 4 marks.
1. निम्नलिखित में से कौन-सी घटना की प्रायिकता नही हो सकती है?
Which of the following cannot be the Probability of an event?
(a) 0.5
(b) 1
(c) –0.9
(d) 25%
Answer : (c) –0.9
2. प्रथम पाँच प्राकृत संख्याओं का माध्यिका क्या होगा?
What would be the median of the first five natural numbers?
(a) 3
(b) 5
(c) 7
(d) 10
Answer : (a) 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
Answer : (a) 1
α + β = –b/a = –(–1)/1 = 1
4. निम्नलिखित में से कौन सी एक अपरिमेय संख्या नहीं है?
Which of the following is not an irrational number?
(a) 3 + √2
(b) 5 – √5
(c) 3 + √16
(d) 4 – √8
Answer : (c) 3 + √16
5. वर्ग अंतराल 40-80 का वर्ग चिह्न होगा :
The Class Mark of class interval 40-80 will be :
(a) 40
(b) 60
(c) 80
(d) 120
Answer : (b) 60
6. बहुपद p(x) = 4x2 + 4x + 1 के अधिकतम शून्यक होंगे :
Polynomial p(x) = 4x2 + 4x + 1 has maximum number of zeros :
(a) 1
(b) 2
(c) 3
(d) No solution
Answer : (d) 2
7. दो समरूप त्रिभुज की भुजाएं 4 : 9 के अनुपात में हैं। इन त्रिभुजों के क्षेत्रफल का अनुपात …………. होगा।
The sides of two similar triangles are in the ratio 4 : 9. Area of these triangles is in the ratio ………….
Answer : 42 : 92 = 16 : 81
8. बिंदु (–2, –3) ………… चतुर्थांश में स्थित है।
The point (–2, –3) lies in the ………… quadrant.
Answer : 3rd
9. यदि P(E) = 0.03 है, तो P(not E) का मान ज्ञात करे।
If P(E) = 0.03, then find the value of P(not E).
Answer : P(E) + P(not E) = 1
P(not E) = 1 – 0.03 = 0.97
10. संख्या 210 को अभाज्य गुणनखंड के रूप में व्यक्त कीजिए।
Express 210 as a product of its Prime factors.
Answer : 210 = 2 × 3 × 5 × 7
11. अभिकयन (A) : y3 – 7 एक त्रिघात बहुपद है।
कारण (R) : तीन घात वाले बहुपद को त्रिघात बहुपद कहते है।
उत्तर : अभिकथन (A) और कारण (R) दोनों सही है तथा कारण (R), अभिकथन (A) की सही व्याख्या करता है।
Assertion (A) : y3 – 7 is a cubic polynomial.
Reason (R) : A Polynomial of degree three is called a cubic polynomial.
Answer : Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
12. 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
13. Find the distance between the points (–5, 7) and (–1, 3).
Answer : Here, A(–5, 7) and B(–1, 3)
x1 = –5, x2 = –1, y1 = 7, y2 = 3
Using Distance Formula,
AB = √(x2–x1)2+(y2–y1)2
= √(–1+5)2+(3–7)2
= √(4)2+(–4)2
= √16+16
= √32
= 4√2
14. How many three digit numbers are divisible by 7 ?
Answer : The three-digit numbers divisible by 7 are; 105, 112, 119, ………, 994
Here, a = 105, d = 112–105 = 7, last term (an) = 994
nth term of AP, an = a + (n–1)d
105 + (n–1)×7 = 994
105 + 7n – 7 = 994
7n + 98 = 994
7n = 994 – 98
7n = 896
n = 896/7
n = 128
Hence, there are 128 three-digit numbers divisible by 7.
15. द्विघात बहुपद x2 – 9x + 20 के शून्यक ज्ञात करें और शून्य तथा गुणांक के बीच संबंध सत्यापित करें।
Find the zeros of the quadratic polynomial x2 – 9x + 20 and verify the relationship between zeros and coefficients.
Answer : Here a = 1, b = –9, c = 20
x2 – 9x + 20 = 0
x2 – 5x – 4x + 20 = 0
x(x – 5) – 4(x – 5) = 0
(x – 5)(x – 4) = 0
x – 5 = 0 or x – 4 = 0
x = 5, x = 4
so, α = 5 and β = 4
α + β = 5 + 4 = 9 = –b/a = –(–9)/1 = 9
αβ = 5 × 4 = 20 = c/a = 20/1 = 20
Thus, the basic relationship is verified.
16. निम्नलिखित रैखिक समीकरण युग्म को हल कीजिए।
Solve the following pair of linear equations :
3x/2 – 5y/3 = –2 and x/3 + y/2 = 13/6
Answer : Multiply both eqn. by 6,
9x – 10y = –12 ………(i)
2x + 3y = 13 ……….(ii)
Multiply eq.(i) by 2 and multiply eqn.(ii) by 9,
18x – 20y = –24 ……….(iii)
18x + 27y = 117 ………..(iv)
Subtract eqn.(iii) from eqn.(iv), we get
47y = 141
y = 3
Put y = 3 in eqn.(i),
9x – 10(3) = –12
9x = –12 + 30
9x = 18
x = 2
Therefore, x = 2 and y = 3
17. सिद्ध कीजिए कि √3 एक अपरिमेय संख्या है।
Prove that √3 is an irrational number.
Answer : Let us assume that √3 is a rational number.
Now, √3 = p/q where p and q are co-prime integers and q ≠ 0
p = √3q
Squaring both sides, we get
p2 = 3q2 ……..(i)
3 divides p2, then 3 also divides p
Put, p = 3m in eqn.(i),
(3m)2 = 3q2
9m2 = 3q2
3m2 = q2
3 divides q2 then 3 also divides q
Hence p, q have a common factor is 3 . This contradicts our assumption that they are co-primes. Therefore, p/q is not a rational number
Hence √3 is an irrational number.
18. पाँच वर्ष पहले, नूरी की उम्र सोनू से तीन गुना थी। दस साल बाद, नूरी की उम्र सोनू से दोगुनी हो जाएगी। नूरी और सोनू की उम्र कितनी है?
Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?
Answer : Let the present age of Nuri = x
And, the present age of Sonu = y
5 years ago, age of Nuri = x – 5
5 years ago, age of Sonu = y – 5
ATQ,
x – 5 = 3(y – 5)
x – 3y = – 10 ……….(i)
10 years later, age of Nuri = x + 10
10 years later, age of Sonu = y + 10
ATQ,
x + 10 = 2(y + 10)
x – 2y = 10 …….(ii)
Subtract eqn.(i) from eqn.(ii), we get
y = 20
Put y = 20 in eqn.(ii),
x – 2(20) = 10
x – 40 = 10
x = 10 + 40 = 50
Hence, the age of Nuri (x) = 50 years and the age of Sonu (y) = 20 years.
19. निम्नलिखित डेटा का माध्यिका 525 है। यदि कुल आवृत्ति 100 है तो x और y का मान ज्ञात करें।
The median of the following data is 525. Find the values of x and y if the total frequency is 100.
Answer :
n = 100 ⇒ n/2 = 50
76 + x + y = 100
x + y = 24 ……..(i)
Median = 525 (median class is 500-600)
l = 500, f = 20, cf = 36 + x, h = 100
Median = l + (n/2 – cf)/f × h
525 = 500 + (50–36–x)/20 × 100
525 – 500 = (14–x) × 5
25 = 70 – 5x
5x = 70 – 25 = 45
x = 45/5 = 9
From eqn.(i), we get
9 + y = 24
y = 24 – 9 = 15
20.जनशक्ति की कम लागत और उच्च गुणवत्ता वाले उत्पादन में योगदान देने वाली मजबूत तकनीकी और इंजीनियरिंग क्षमताओं के कारण भारत प्रतिस्पर्धी विनिर्माण स्थान है। किसी कारखाने में टीवी सेट का उत्पादन हर साल एक निश्चित संख्या में समान रूप से बढ़ता है। इसने छठे वर्ष में 16000 सेट और 9वें वर्ष में 22600 सेट का उत्पादन किया। उपरोक्त जानकारी के आधार पर निम्नलिखित प्रश्नों के उत्तर दीजिए।
India is competitive manufacturing location due to low cost of manpower and strong technical and engineering capabilities contributing to higher quality production runs. The production of TV sets in a factory increases uniformly by a fixed number every year. It produced 16000 sets in 6th year and 22600 in 9th year. On the basis of above information, answer the following questions.
Questions :
(i) शुरुआत में उत्पादन कितना है।
What is the production at starting.
Answer : nth term of AP, an = a + (n-1)d
a6 = a + 5d = 16000 …….(i)
a9 = a + 8d = 22600 …….(ii)
Subtract eqn.(i) from (ii), we get
3d = 6600
d = 2200
Put d = 2200 in eqn.(i),
a + 5(2200) = 16000
a + 11000 = 16000
a = 16000 – 11000
a = 5000
(ii) आठवें वर्ष के दौरान उत्पादन कितना है?
What is the production during 8th year?
Answer : a8 = a + 7d = 5000 + 7(2200) = 5000 + 15400 = 20400
(iii) पहले तीन वर्षों में (उस दौरान) कुल उत्पादन लिखिए।
Write the total production in (during) the first 3 years.
Answer : Sn = n/2 [2a + (n–1)d]
S3 = 3/2 [2×5000 + 2×2200] = 3/2 [10000 + 4400] = 3/2 × 14400 = 21600
(iv) किस वर्ष में उत्पादन 29200 है?
In which year, the production is 29,200?
Answer : an = a + (n–1)d
5000 + (n–1)2200 = 29200
5000 + 2200n – 2200 = 29200
2200n + 2800 = 29200
2200n = 29200 – 2800
2200n = 26400
n = 12
In 12th year, the production is 29200.