Bu'aan kutaa 12ffaa ba'eera; bu'aa kee eessatti akka ilaaltu

Waraqaalee qormaataa

Qormaata galuumsa kutaa 12ffaa Herregaa, bara 2018 A.L.I (2026), maxxansa 2

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 2 qaba; kaan gadi aanee tarreeffamaniiru.

Google Play irraa buufadhaa
  • Damee lamaanuu

    Eenyutu fudhata

  • 2018 A.L.I (2026), maxxansa 2

    Bara qormaataa

  • Gaaffilee 56 deebii waliin

    Kuusaa kana keessa

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

What is the 10th term of Prof. Mulatu's sequence ?

  1. A.309
  2. B.118
  3. C.191
  4. D.89
Deebii fi ibsa ilaalaa

Deebii: B 118

Maaliif

The Mulatu sequence starts with M1=4M_1 = 4 and M2=1M_2 = 1, with subsequent terms defined by Mn=Mn−1+Mn−2M_n = M_{n-1} + M_{n-2}. Generating the terms gives: 4, 1, 5, 6, 11, 17, 28, 45, 73, 118, so the 10th term is 118.

Gaaffii 2

What is the area (in square unit) of the region bounded by y=0 and the graph of f(x)=4-x² on the interval [-2,2]?

  1. A.2/3
  2. B.22/3
  3. C.8
  4. D.32/3
Deebii fi ibsa ilaalaa

Deebii: D 32/3

Maaliif

The area is given by the definite integral ∫−22(4−x2)dx=[4x−x33]−22=(8−83)−(−8+83)=163+163=323\int_{-2}^{2} (4-x^2) dx = [4x - \frac{x^3}{3}]_{-2}^{2} = (8 - \frac{8}{3}) - (-8 + \frac{8}{3}) = \frac{16}{3} + \frac{16}{3} = \frac{32}{3} square units.

Gaaffii 3

Let f be defined on [a, b]. Then which of the following is NOT true?

  1. A.∫ₐᵇ f(x)dx = ∫ₐᶜ f(x)dx - ∫ᶜᵇ f(x)dx for any c in (a,b)
  2. B.∫ᶜᶜ f(x)dx = 0 for any c in (a,b)
  3. C.∫ₐᵇ f(x)dx = -∫ᵇₐ f(x)dx
  4. D.∫ₐᵇ k f(x)dx = k ∫ₐᵇ f(x)dx for any constant k
Deebii fi ibsa ilaalaa

Deebii: A ∫ₐᵇ f(x)dx = ∫ₐᶜ f(x)dx - ∫ᶜᵇ f(x)dx for any c in (a,b)

Maaliif

By the interval union property of definite integrals, ∫abf(x)dx=∫acf(x)dx+∫cbf(x)dx\int_a^b f(x)dx = \int_a^c f(x)dx + \int_c^b f(x)dx for any c∈(a,b)c \in (a,b), which makes the minus sign in option A incorrect.

Gaaffii 4

The sides of an equilateral triangle are increasing at the rate of 2 cm/sec. The rate at which the area increases when side is 10 cm is:

  1. A.10 cm²/sec
  2. B.√3 cm²/sec
  3. C.10√3 cm²/sec
  4. D.10/3 cm²/sec
Deebii fi ibsa ilaalaa

Deebii: C 10√3 cm²/sec

Maaliif

The area of an equilateral triangle is A=34s2A = \frac{\sqrt{3}}{4}s^2. Differentiating with respect to time tt gives dAdt=32sdsdt=32(10)(2)=103 cm2/sec\frac{dA}{dt} = \frac{\sqrt{3}}{2}s\frac{ds}{dt} = \frac{\sqrt{3}}{2}(10)(2) = 10\sqrt{3}\text{ cm}^2/\text{sec}.

Gaaffii 5

A balloon which always remains spherical is being inflated by pumping in 10 cm³ of gas per second. What is the rate at which the radius is increasing when the radius is 15 cm?

  1. A.1/(30π) cm/sec
  2. B.1/(90π) cm/sec
  3. C.1/π cm/sec
  4. D.1/(9π) cm/sec
Deebii fi ibsa ilaalaa

Deebii: B 1/(90π) cm/sec

Maaliif

The volume of a sphere is V=43πr3V = \frac{4}{3}\pi r^3, so dVdt=4πr2drdt\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}. Given dVdt=10\frac{dV}{dt} = 10 and r=15r = 15, we get 10=4π(15)2drdt10 = 4\pi (15)^2 \frac{dr}{dt}, which simplifies to drdt=10900π=190π cm/sec\frac{dr}{dt} = \frac{10}{900\pi} = \frac{1}{90\pi}\text{ cm/sec}.

Gaaffii 6

Suppose Target Population N = 54000 vouchers and sample size n = 6000; following the systematic sampling approach if 4th unit is selected as the first unit in the sample then the next unit selected in the sample would be:

  1. A.8th
  2. B.10th
  3. C.12th
  4. D.13th
Deebii fi ibsa ilaalaa

Deebii: D 13th

Maaliif

The sampling interval is k=N/n=54000/6000=9k = N/n = 54000/6000 = 9. If the first selected unit is the 4th, the next systematic unit is 4+9=134 + 9 = 13th.

Qormaata guutuu PDF dhaan buufadhaa

Gaaffilee 56 kuusaa kana keessa jiran hunda akka waraqaa qormaataa maxxanfameetti, furtuu deebii ibsa 56 waliin immoo akka faayilii lammaffaatti. Maxxansaa yookiin daataa malee dubbisaa.

Bilisa; herregas hin gaafatu.

Qormaata guutuu sa'aatiin daangeffamee hojjedhaa

Herregni Temari bilisaa qormaata Herregaa guutuu fi kan waggoota biroo hunda qormaatilee fakkeenyaa sa'aatiin daangeffaman godhee qabxii battalumaa waliin isiniif kenna. Bilbila kamiyyuu irratti, walitti hidhamiinsa laafaa irrattillee ni hojjeta.

Bilisaan shaakaluu jalqabaa

Temari.et appii Android qaba; Google Play irratti bilisaan argama.

Qormaata kana taa'uu dura Herregaa qayyabadhu

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.

Yaadannoo Herregaa bani
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 56 qaba; tokkoon tokkoon isaanii deebii fi ibsa gara kitaaba barnootaatti qajeelchu qabu.

Qormaata guutuu akkamittan shaakala?

Lakkoofsa bilbilaatiin herrega bilisaa banadhaa. Qormaatichi guutuun qormaata fakkeenyaa sa'aatiin daangeffame ta'ee qabxii battalumaa kenna; fooyya'iinsi gargaaraa AI ji'atti qarshii 199 kan ta'e waan isin jalaa darbe hunda isinii ibsa.

Qormaata Herregaa bara 2018 A.L.I PDF dhaan buufachuu nan danda'aa?

Eeyyee; bilisas. Waraqaan gaaffii faayilii tokko, furtuun deebii ibsa isaa waliin immoo faayilii biraa ti; kanaaf qormaata dursitanii yaaluu dandeessu. Maxxansaa, yookiin yeroo daataan hin jirre irra deebiif bilbila keessan irratti kaa'adhaa.

SeeniHerrega kee bani