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Qormaata galuumsa kutaa 12ffaa Herregaa, bara 2018 A.L.I (2026), maxxansa 5

Gaaffilee dhugaa qormaata galuumsa yunivarsiitii kutaa 12 kan Herregaa, akkuma barattoonni bara 2018 A.L.I itti fudhatanitti. Tokkoon tokkoon gaaffii deebii sirrii fi ibsa qaba. Qormaanni bara 2018 A.L.I maxxansa adda addaatiin bahe. Fuulli kun maxxansa 5 qaba; kaan gadi aanee tarreeffamaniiru.

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  • 2018 A.L.I (2026), maxxansa 5

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Gaaffilee qormaata kanaa yaalaa

Deebii banuun dura tokkoon tokkoon gaaffii ofii keessaniin deebisaa. Ibsi gara kutaa kitaaba barnootaa sirriitti waan qajeelchuuf, yeroo gaaffii dogoggortan fuula kam irra deebi'uu akka qabdan ni beektu.

Gaaffii 1

Which term of the following arithmetic sequence -6, 1, 8, 15, ... is 414?

  1. A.31
  2. B.62
  3. C.61
  4. D.32
Deebii fi ibsa ilaalaa

Deebii: C 61

Maaliif

The n-th term of an arithmetic sequence is given by an=a1+(n−1)da_n = a_1 + (n-1)d. Substituting a1=−6a_1 = -6, d=7d = 7, and an=414a_n = 414, we get 414=−6+(n−1)7414 = -6 + (n-1)7, which simplifies to n=61n = 61.

Gaaffii 2

∑_{n=4}^{17} 1/(k²+5k+6) is equal to: (Note: the index likely n, but given as k)

  1. A.11/102
  2. B.7/60
  3. C.12/85
  4. D.10/119
Deebii fi ibsa ilaalaa

Deebii: B 7/60

Maaliif

Using partial fractions, the term can be written as 1(k+2)(k+3)=1k+2−1k+3\frac{1}{(k+2)(k+3)} = \frac{1}{k+2} - \frac{1}{k+3}. This forms a telescoping sum from k=4k=4 to 1717, which simplifies to 16−120=760\frac{1}{6} - \frac{1}{20} = \frac{7}{60}.

Gaaffii 3

Which of the following Geometric sequence is divergent?

  1. A.a_n = (-1)^{2n+1}(3/7)^n
  2. B.a_n = (-3/5)^{2n+1}
  3. C.a_n = (1/10)(5/2)^n
  4. D.a_n = 2/5^n
Deebii fi ibsa ilaalaa

Deebii: C a_n = (1/10)(5/2)^n

Maaliif

A geometric sequence an=a1rn−1a_n = a_1 r^{n-1} is divergent if the absolute value of its common ratio ∣r∣≥1|r| \ge 1. Here, option C has r=5/2=2.5>1r = 5/2 = 2.5 > 1, making it divergent, while the others have ∣r∣<1|r| < 1 and are convergent.

Gaaffii 4

An auditorium has 30 seats in first row, 34 in second, 38 in third. If pattern continues, total seating capacity with 20 rows?

  1. A.106
  2. B.3880
  3. C.580
  4. D.1360
Deebii fi ibsa ilaalaa

Deebii: D 1360

Maaliif

The seating capacity forms an arithmetic series with a1=30a_1 = 30, d=4d = 4, and n=20n = 20. The sum of the first 20 terms is S20=202[2(30)+(20−1)4]=10[60+76]=1360S_{20} = \frac{20}{2}[2(30) + (20-1)4] = 10[60 + 76] = 1360.

Gaaffii 7

A manufacturer wants to design an open box with square base and surface area 75 square units. What height gives maximum volume?

  1. A.5 units
  2. B.2.5 units
  3. C.7.5 units
  4. D.15 units
Deebii fi ibsa ilaalaa

Deebii: B 2.5 units

Maaliif

An open box with a square base xx and height hh has surface area S=x2+4xh=75S = x^2 + 4xh = 75, so h=75−x24xh = \frac{75-x^2}{4x}. The volume V=x2h=75x−x34V = x^2h = \frac{75x - x^3}{4} is maximized when V′(x)=75−3x24=0V'(x) = \frac{75-3x^2}{4} = 0, giving x=5x = 5 and a height of h=75−2520=2.5h = \frac{75-25}{20} = 2.5 units.

Gaaffii 8

What is area of region enclosed between f(x)=x²-4 over [1,5]?

  1. A.5/3
  2. B.65/3
  3. C.16/3
  4. D.86/3
Deebii fi ibsa ilaalaa

Deebii: D 86/3

Maaliif

The function f(x)=x2−4f(x) = x^2 - 4 is negative on [1,2][1, 2] and positive on [2,5][2, 5]. The total area is ∫12(4−x2)dx+∫25(x2−4)dx=[4x−x33]12+[x33−4x]25=53+27=863\int_{1}^{2} (4-x^2) dx + \int_{2}^{5} (x^2-4) dx = [4x - \frac{x^3}{3}]_1^2 + [\frac{x^3}{3} - 4x]_2^5 = \frac{5}{3} + 27 = \frac{86}{3}.

Qormaata guutuu PDF dhaan buufadhaa

Gaaffilee 51 kuusaa kana keessa jiran hunda akka waraqaa qormaataa maxxanfameetti, furtuu deebii ibsa 51 waliin immoo akka faayilii lammaffaatti. Maxxansaa yookiin daataa malee dubbisaa.

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Qormaatilee Herregaa biroo bara 2018 A.L.I

2018 A.L.I (2026)2018 A.L.I (2026), damee hawaasaa2018 A.L.I (2026), maxxansa 12018 A.L.I (2026), maxxansa 22018 A.L.I (2026), maxxansa 32018 A.L.I (2026), maxxansa 42018 A.L.I (2026), maxxansa 6

Qormaatilee Herregaa kan waggoota biroo

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Yaadannoon gabaabaan Herregaa kutaa 12 kaardii 32 qaba. Tokkoon tokkoon boqonnaa boqonnaadhaan kitaaba biyyaalessaa irraa qophaa'ee seericha, mallattoowwan isaa maal bakka akka bu'an fi dogoggora qabxii nama dhabsiisu ni ibsa.

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Kunneen gaaffilee dhugaa qormaata galuumsa Herregaa bara 2018 A.L.I ti?

Eeyyee. Qormaata galuumsa yunivarsiitii biyyaalessaa kan Herregaa barattoonni bara 2018 A.L.I (2026) fudhatan irraa fudhataman. Kuusaan keenya gaaffilee isaa 51 qaba; tokkoon tokkoon isaanii deebii fi ibsa gara kitaaba barnootaatti qajeelchu qabu.

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