AIEEE Questions and Answers with Explanation

Q:

A Shopkeeper sells two articles at Rs.1000 each, making a profit of 20% on the first article and a loss of 20% on the second article. Find the net Profit or loss that he makes?

A) 4%	B) 5%
C) 6%	D) 8%

Answer & Explanation Answer: A) 4%

Explanation:

SP of first article = Rs.1000

Profit = 20%

$C P = S P x \frac{100}{100 + P %} = R s . \frac{2500}{3}$

SP of Second Article = Rs.1000

Loss = 20%

$C P = S P x \frac{100}{100 - L %} = R s . 1250$

So, Total SP = Rs.2000; Total CP = Rs.6250/3

As the CP is more than SP, he makes a loss.

Loss = CP - SP = (6250/3) - 2000 = Rs.(250/3)

So, Loss Percent = Lossx100/CP = 4%

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Q:

Select the related word/letters/number from the given alternatives.

Five Point Someone : Chetan Bhagat : : Swami and Friends : ?

A) William Shakespeare	B) R.K. Narayan
C) Charles Dickens	D) Rabindra Nath Tagore

Answer & Explanation Answer: B) R.K. Narayan

Explanation:

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Q:

11 P in a CT

Answer

The given letter equation implies that 11 P in a CT means 11 Players in a Cricket Team.

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Subject: Logic Puzzles Exam Prep: GRE , GATE , CAT , Bank Exams , AIEEE
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Q:

Develop is to assess as train is to

A) Change	B) Educate
C) Analyze	D) Recruit

Answer & Explanation Answer: C) Analyze

Explanation:

When you develop a product, you assess its performance for proper functioning.

Similarly, when you train a person to do a job, you analyze his/her performance to understand how good he/she is at it.

Hence, Develop : Assess :: Train : Analyze.

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Q:

A school has four sections A, B, C, D of Class IX students.

1. If the number of students passing an examination be considered a criteria for comparision of difficulty level of two examinations, which of the following statements is true in this context?

A. Half yearly examinations were more difficult.

B. Annual examinations were more difficult.

C. Both the examinations had almost the same difficulty level.

D. The two examinations cannot be compared for difficulty level.

2. How many students are there in Class IX in the school?

A. 336 B.189 C. 335 D. 430

3. Which section has the maximum pass percentage in at least one of the two examinations?

A. A Section B. B Section C. C Section D. D Section

4. Which section has the maximum success rate in annual examination?

A. A Section B. B Section C. C Section D. D Section

5. Which section has the minimum failure rate in half yearly examination?

A. A section B. B section C. C section D. D section

Half yearly examinations were more difficult.

If the number of students passing an examination be considered a criteria for comparision of difficulty level of two examinations, which of the following statements is true in this context? - See more at: https://www.theonlinetestcentre.com/table-charts7.html#sthash.slrcbjro.dpuf

Answer

1. ANSWER : C

Explanation - Number of students who passed half-yearly exams in the school

= (Number of students passed in half-yearly but failed in annual exams) + (Number of students passed in both exams)
= (6 + 17 + 9 + 15) + (64 + 55 + 46 + 76)

= 288.

Also, Number of students who passed annual exams in the school

= (Number of students failed in half-yearly but passed in annual exams) + (Number of students passed in both exams)

= (14 + 12 + 8 + 13) + (64 + 55 + 46 + 76)

= 288.

Since, the number of students passed in half-yearly = the number of students passed in annual exams. Therefore, it can be inferred that both the examinations had almost the same difficulty level.

Thus Statements (a), (b) and (d) are false and Statement (c) is true.

2. ANSWER : D

Explanation - Since the classification of the students on the basis of their results and sections form independent groups, so the total number of students in the class:

= (28 + 23 + 17 + 27 + 14 + 12 + 8 + 13 + 6 + 17 + 9 + 15 + 64 + 55 + 46 + 76)

= 430.

3. ANSWER : D

Explanation- Pass percentages in at least one of the two examinations for different sections are:

For Section A = $\frac{14 + 6 + 64}{28 + 14 + 6 + 64} \times 100 = \frac{84}{112} \times 100$ % = 75%

For Section B = $\frac{12 + 17 + 55}{23 + 12 + 17 + 55} \times 100$ % = 78.5%

For Section C = $\frac{8 + 9 + 46}{17 + 8 + 9 + 46} \times 100$ %= 78.75%

For Section D = $\frac{13 + 15 + 76}{27 + 13 + 15 + 76} \times 100$ %= 79.39%

Clearly ,the pass percentage is maximum for Section D

4. ANSWER : A

Explanation - Total number of students passed in annual exams in a section

= [ (No. of students failed in half-yearly but passed in annual exams) + (No. of students passed in both exams) ] in that section

Success rate in annual exams in Section A= $\frac{14 + 64}{112} \times 100$ % = 69.64%

Similarly, success rate in annual exams in:

Section B = $\frac{12 + 55}{107} \times 100$ % = 62.62%

Section C = $\frac{8 + 46}{80} \times 100$ % = 67.5%

Section D = $\frac{89}{131} \times 100$ % = 67.94%

Clearly, the success rate in annual examination is maximum for Section A.

5. ANSWER : D

Explanation - Total number of failures in half-yearly exams in a section

= [ (Number of students failed in both exams) + (Number of students failed in half-yearly but passed in Annual exams) ] in that section

Failure rate in half-yearly exams in Section A %= 37.5 %

Similarly, failure rate in half-yearly exams in:

Section B = 32.71%

Section C = 31.25%

Section D = 30.53%

Clearly, the failure rate is minimum for Section D.

= (Number of students passed in half-yearly but failed in annual exams) + (Number of students passed in both exams)

= (6 + 17 + 9 + 15) + (64 + 55 + 46 + 76)

= 288.

Also, Number of students who passed annual exams in the school

= (Number of students failed in half-yearly but passed in annual exams) + (Number of students passed in both exams)

= (14 + 12 + 8 + 13) + (64 + 55 + 46 + 76)

= 288.

Since, the number of students passed in half-yearly = the number of students passed in annual exams. Therefore, it can be inferred that both the examinations had almost the same difficulty level.

Thus Statements (a), (b) and (d) are false and Statement (c) is true.

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Q:

In a group of persons travelling in a bus, 6 persons can speak Tamil, 15 can speak Hindi and 6 can speak Gujarati. In that group, none can speak any other language. If 2 persons in the group can speak two languages and one person can speak all the three languages, then how many persons are there in the group ?

A) 21	B) 22
C) 23	D) 24

Answer & Explanation Answer: C) 23

Explanation:

Let us assume the two persons who can speak two languages speak Hindi and Tamil. The third person then speaks all the three languages.

Tamil – Number of persons who can speak is 6. Only Tamil 6 – 2 – 1 = 3

Hindi - Number of persons who can speak is 15. Only Hindi 15 – 2 – 1 12

Gujarati – Number of persons who can speak is 6. Only Gujarati 6 – 1 = 5

Thus the number of persons who can speak only one language is 3 + 12 + 5 = 20

Number of persons who can speak two languages = 2

Number of person who an speak all the languages = 1

Total number of persons = 23.

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Q:

Eastern ghats and Western ghats meet at the

A) Annamalai hills	B) Palani hills
C) Nilgiri hills	D) Cardamom hills

Answer & Explanation Answer: C) Nilgiri hills

Explanation:

The Eastern Ghats are a discontinuous range of mountains along India's eastern coast whereas Western Ghats also known as Sahyadri is a mountain range that runs parallel to the western coast of the Indian peninsula.

eastern_ghats_and_western_ghats_meet_at_the1536656083.png image

Eastern ghats and Western ghats meet at the Nilgiri hills in Tamilnadu.

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Q:

A vessel contains 20 liters of a mixture of milk and water in the ratio 3:2. 10 liters of the mixture are removed and replaced with an equal quantity of pure milk. If the process is repeated once more, find the ratio of milk and water in the final mixture obtained ?

A) 5:3	B) 1:4
C) 4:1	D) 9:1

Answer & Explanation Answer: D) 9:1

Explanation:

Milk = 3/5 x 20 = 12 liters, water = 8 liters

If 10 liters of mixture are removed, amount of milk removed = 6 liters and amount of water removed = 4 liters.

Remaining milk = 12 - 6 = 6 liters

Remaining water = 8 - 4 = 4 liters

10 liters of pure milk are added, therefore total milk = (6 + 10) = 16 liters.

The ratio of milk and water in the new mixture = 16:4 = 4:1

If the process is repeated one more time and 10 liters of the mixture are removed,
then amount of milk removed = 4/5 x 10 = 8 liters.

Amount of water removed = 2 liters.

Remaining milk = (16 - 8) = 8 liters.

Remaining water = (4 -2) = 2 liters.

Now 10 lts milk is added => total milk = 18 lts

The required ratio of milk and water in the final mixture obtained

= (8 + 10):2 = 18:2 = 9:1.

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