Exams Questions and Answers

Q:

All variables in PHP start with which symbol?

A) %	B) !
C) $	D) &

Answer & Explanation Answer: C) $

Explanation:

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Filed Under: Programming
Exam Prep: Bank Exams

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Q:

A merchant commences with a certain capital and gains annually at the rate of 30 per cent. At the end of 2 years he has worth Rs 10,000. What was the original capital?

A) Rs. 4987	B) Rs. 5148
C) Rs. 5917	D) Rs. 6254

Answer & Explanation Answer: C) Rs. 5917

Explanation:

Let the Original capital amount be Rs. P

Given after 2 years the amount is Rs. 10000

Rate of interest = 30%

According to the question,

$P x \frac{130}{100} x \frac{130}{100} = 10000 = > P = \frac{1000000}{169} = 5917$

Therefore, the capital amount P = Rs. 5917

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Filed Under: Percentage
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Q:

When and by whom the PIN coode system was introduced in India ?

A) Savinder Singh	B) Shriram Bhikaji Velankar
C) D.R. Khullar	D) Mahesh Barnwal

Answer & Explanation Answer: B) Shriram Bhikaji Velankar

Explanation:

The PIN code system was introduced on 15 th August 1972 by Shriram Bhikaji Velankar. Currently India has six digit post code and soon it will be converted in eight digits.

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Filed Under: Indian Geography
Exam Prep: Bank Exams , CAT
Job Role: Bank Clerk , Bank PO

29 6750

Q:

The profit earned by selling an article for Rs.753 is equal to the loss incurred when the same article is sold for Rs.455.What should be the sale price for making 50% profit ?

A) Rs.855	B) Rs.955
C) Rs.906	D) Rs.896

Answer & Explanation Answer: C) Rs.906

Explanation:

Let C.P =Rs. x

Then as given, (753 - x ) = (x - 455)

=>2x = 1208 => x = 604

There fore S.P= 150% of 604 => Rs. $\frac{150}{100} \times 604$ = Rs.906

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Filed Under: Profit and Loss
Exam Prep: CAT
Job Role: Bank PO

14 6750

Q:

A tennis ball machine serves a ball vertically into the air from a height of 2 feet, with an initial speed of 110 feet per second. After how many seconds does the ball attain its maximum height?

A) 2.87 sec	B) 3.44 sec
C) 4.65 sec	D) 5.41 sec

Answer & Explanation Answer: B) 3.44 sec

Explanation:

It is launched vertically with an initial speed of 110 ft/sec.

It will reach maximum height when the speed drops to 0.

Vo=110 feet per second

ho=2 feet

So, h(t) = -16t^2 +110t +2

Take the derivative: h'(t) = 110 -32t

The maximum height will be at the inflection when the derivative crosses the x-axis, then h'(t)=0.

So, set h'(t)=0 and solve for t:

0 = 110 -32t

-110 = -32t

t=3.4375

t=3.44 seconds.

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Filed Under: Physics
Exam Prep: AIEEE , Bank Exams , CAT , GATE
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Q:

Solve this logical Math puzzle?

logical_thinking_puzzle1536132282.jpg image

Answer

Let that something be 'x'

From the given data,

x + x/5 = 21

5x + x/5 = 21

6x = 21 x 5

=> x = 17.5

Hence, that something is x = 17.5.

Workspace

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Subject: Math Puzzles Exam Prep: AIEEE , Bank Exams , CAT , GATE , GRE
Job Role: Analyst , Bank Clerk , Bank PO , Database Administration , IT Trainer

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Q:

What is the SI on Rs.2500/- at the rate of 12% per annum for 8 years?

A) 2200	B) 2300
C) 2400	D) 2500

Answer & Explanation Answer: C) 2400

Explanation:

S.I=PNR/100

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Filed Under: Compound Interest
Exam Prep: Bank Exams
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Q:

The Value of $[\log (\tan (1^{0})) + \log (\tan (2^{0})) + \dots \dots + \log (\tan (89^{0}))]$ is

A) -1	B) 0
C) 1/2	D) 1

Answer & Explanation Answer: B) 0

Explanation:

$= [\log \tan (1^{0}) + \log \tan (89^{0})] + [\log \tan (2^{0}) + \log \tan (88^{0})] + \dots \dots + [\log \tan (45^{0})]$

$= \log [\tan (1^{0}) \times \tan (89^{0})] + \log [\tan (2^{0}) \times \tan (88^{0})] + \dots \dots + \log (1)$

$[∵ \tan (90 - θ) = c o t θ a n d \tan (45^{0}) = 1]$

= log 1 + log 1 +.....+log 1

= 0.

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Filed Under: Logarithms
Exam Prep: AIEEE , Bank Exams , CAT , GATE
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