- (A) -1
- (B) -41
- (C) 41
- (D) 1
उकल:
$$ D = (5 \times -4) - (3 \times -7) $$
$$ D = -20 - (-21) $$
$$ D = -20 + 21 = 1 $$
उत्तर: (D) 1
- (A) \(x^2+4x=11+x^2\)
- (B) \(x^2=4x\)
- (C) \(5x^2=90\)
- (D) \(2x-x^2=x^2+5\)
उकल:
पर्याय (A): \(x^2+4x=11+x^2 \Rightarrow 4x=11\). हे रेषीय समीकरण आहे, वर्गसमीकरण नाही.
उत्तर: (A)
- (A) 0
- (B) 3.5
- (C) 7
- (D) 10.5
उकल:
\(t_2 = a + d = 3.5 + 0 = 3.5\)
उत्तर: (B) 3.5
- (A) 0.6
- (B) 2.0
- (C) 0.15
- (D) 0.75
उकल:
संभाव्यता 0 ते 1 च्या दरम्यान असते. 2.0 ही किंमत 1 पेक्षा जास्त आहे.
उत्तर: (B) 2.0
क्रेमरच्या पद्धतीनुसार: $$ x = \frac{D_x}{D} $$ $$ x = \frac{49}{7} = 7 $$ उत्तर: \(x = 7\)
दिलेले समीकरण: \(2y = 10 - y^2\)
सामान्यरूप \(ax^2 + bx + c = 0\):
उत्तर: \(y^2 + 2y - 10 = 0\)
बाजारभाव (MV) = दर्शनी किंमत (FV) + अधिमूल्य (Premium)
\(MV = 100 + 10 = 110\)
उत्तर: बाजारभाव ₹ 110 आहे.
वर्गमध्य = \(\frac{\text{खालची मर्यादा} + \text{वरची मर्यादा}}{2}\)
वर्गमध्य = \(\frac{6 + 10}{2} = \frac{16}{2} = 8\)
उत्तर: 8
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\(x^2 + 8x - 48 = 0\)
\(x^2 + 8x + 16 - \)16\( - 48 = 0\)
\((x+4)^2 - \)64\( = 0\)
\((x+4)^2 = 64\)
\(x+4 = \)8 किंवा \(x+4 = -8\)
\(x = 4\) किंवा \(x = \)-12
एकूण वस्तू व सेवाकर = केंद्राचा कर + राज्याचा कर
= 45 + 45
= 90
वस्तू व सेवा कराचा दर = \(\frac{90}{500} \times \) 100
\(\therefore\) कुरिअर सेवा देणाऱ्या एजंटने वस्तू व सेवा कराचा दर 18% आकारला.
| बाबी | प्रतिशत खर्च | केंद्रीय कोनाचे माप |
|---|---|---|
| अन्न | 40 | \(\frac{40}{100}\times360^{\circ}=\) 144\(^{\circ}\) |
| कपडे | 20 | \(\frac{20}{100}\times360^{\circ}=\) 72\(^{\circ}\) |
| शिक्षण | 30 | \(\frac{30}{100}\times360^{\circ}=\) 108\(^{\circ}\) |
| इतर खर्च | 10 | \(\frac{10}{100}\times360^{\circ}=\) 36\(^{\circ}\) |
| एकूण | 100 | \(360^{\circ}\) |
\(101x+99y=501\); \(99x+101y=499\)
समीकरणांची बेरीज करून:
\(200x + 200y = 1000 \Rightarrow x+y = 5\)
समीकरण (1) मधून (2) वजा करून:
\(2x - 2y = 2 \Rightarrow x-y = 1\)
उत्तर: \(x+y=5, x-y=1\)
\(x^2 - 9x - 6x + 54 = 0\)
\(x(x-9) - 6(x-9) = 0\)
\((x-9)(x-6) = 0\)
\(x=9\) किंवा \(x=6\)
उत्तर: \(x=9, 6\)
\(a=2, d=9, t_n=560\)
\(t_n = a + (n-1)d\)
\(560 = 2 + (n-1)9\)
\(558 = 9(n-1)\)
\(62 = n-1 \Rightarrow n = 63\)
उत्तर: 63 वे पद 560 आहे.
दलाली = \(0.3\%\) of \(200 = 0.60\)
खरेदी किंमत = बाजारभाव + दलाली
= \(200 + 0.60 = 200.60\)
उत्तर: ₹ 200.60
\(\bar{d} = \frac{\sum f_id_i}{\sum f_i} = \frac{10000}{100} = 100\)
मध्य \(\bar{X} = A + \bar{d}\)
\(\bar{X} = 2000 + 100 = 2100\)
उत्तर: मध्य = 2100
समजा पदे \(a-d, a, a+d\) आहेत.
\(a-d+a+a+d = \) 27
\(3a = 27 \Rightarrow a = \) 9
\((a-d) \times a \times (a+d) = 504\)
\((9^2 - d^2) \times 9 = 504\)
\(d^2 = 81 - 56 = 25 \Rightarrow d = \pm 5\)
जर \(a=9\) आणि \(d=5\) असेल तर, पदे = 4, 9, 14
जर \(a=9\) आणि \(d=-5\) असेल तर, पदे = 14, 9, 4
\(n(S) = \) 52
घटना A: पत्ता लाल असणे.
\(n(A) = \) 26
\(P(A) = \frac{26}{52} = \) \(\frac{1}{2}\)
घटना B: पत्ता चित्रयुक्त असणे.
\(n(B) = \) 12
\(P(B) = \frac{12}{52} = \) \(\frac{3}{13}\)
[येथे आलेख अपेक्षित आहे]
पायऱ्या:
X-अक्ष: गुंतवणूक (वर्ग: 10-20, 20-30, इत्यादी)
Y-अक्ष: कुटुंबांची संख्या (प्रमाण: 1 सेमी = 10 कुटुंबे)
आयतांची उंची: अनुक्रमे 30, 50, 60, 55, 15.
एकूण गुंतवणूक = \(150 \times 120 = 18,000\) रुपये.
एका शेअरवर लाभांश = \(7\%\) of \(100 = 7\) रुपये.
एकूण लाभांश = \(150 \times 7 = 1,050\) रुपये.
परताव्याचा दर = \(\frac{\text{एकूण लाभांश}}{\text{गुंतवणूक}} \times 100\)
\(RoR = \frac{1050}{18000} \times 100 = 5.83\%\)
उत्तर: 5.83%
समजा श्रद्धाचे आजचे वय \(x\) आहे.
\((x-2)(x+3) = 84\)
\(x^2 + x - 6 = 84\)
\(x^2 + x - 90 = 0\)
\((x+10)(x-9) = 0\)
\(x=9\) (वय ऋण असू शकत नाही)
उत्तर: 9 वर्षे.
समीकरण 1: \((0,6), (6,0)\)
समीकरण 2: \((0,-4), (4,0)\)
छेदनबिंदू: \((5,1)\)
उत्तर: \(x=5, y=1\)
समजा कुशल कामगाराचा रोजगार = \(5x\), अकुशल कामगाराचा रोजगार = \(4x\).
\(5x + 4x = 900\)
\(9x = 900 \Rightarrow x = 100\)
कुशल: \(5(100) = 500\)
अकुशल: \(4(100) = 400\)
उत्तर: कुशल: ₹500, अकुशल: ₹400.
\(n(S) = 36\)
(a) बेरीज \(\ge\) 9: \(\{(3,6),(4,5),(4,6),(5,4),(5,5),(5,6),(6,3),(6,4),(6,5),(6,6)\}\). संख्या=10. \(P(A)=10/36 = 5/18\).
(b) बेरजेला 5 ने भाग जातो: \(\{(1,4),(2,3),(3,2),(4,1),(4,6),(5,5),(6,4)\}\). संख्या=7. \(P(B)=7/36\).
(c) पहिला अंक > दुसरा अंक: संख्या=15. \(P(C)=15/36=5/12\).
एकूण \(N=300\), \(N/2 = 150\).
मध्यक वर्ग (cf > 150): 30-40 (f=55, cf=102).
\(L=30, h=10, f=55, cf=102\)
मध्यक \(= L + \left[ \frac{N/2 - cf}{f} \right] \times h\)
\(= 30 + \left[ \frac{150 - 102}{55} \right] \times 10\)
\(= 30 + \frac{480}{55} = 30 + 8.73 = 38.73\) वर्षे.
(a) \(\Delta = 0\) (मुळे वास्तव व समान): उदा., \(x^2 - 4x + 4 = 0\)
(b) \(\Delta > 0\) (मुळे वास्तव व असमान): उदा., \(x^2 - 5x + 6 = 0\)
(c) \(\Delta < 0\) (मुळे वास्तव नाहीत): उदा., \(x^2 + x + 1 = 0\)
येथे, \(a = p\), \(d = q - p\), \(t_n = r\).
सूत्र \(t_n = a + (n-1)d\) वापरून:
\(r = p + (n-1)(q-p)\)
\(r - p = (n-1)(q-p)\)
\(n-1 = \frac{r-p}{q-p}\)
\(n = \frac{r-p}{q-p} + 1 = \frac{r-p+q-p}{q-p} = \frac{q+r-2p}{q-p}\)
बेरीज \(S_n = \frac{n}{2}(t_1 + t_n)\)
\(S_n = \frac{q+r-2p}{2(q-p)} \times (p+r)\)
\(S_n = \frac{(p+r)(q+r-2p)}{2(q-p)}\)
हे सिद्ध झाले.
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