5555+ Aptitude and Reasoning Questions, Answers With Explanation

Q:

Akash leaves mumbai at 6 am and reaches Bangalore at 10 am . Prakash leaves Bangalore at 8 am and reaches Mumbai at 11:30 am. At what time do they cross each other?

A) 10 am	B) 8:32 am
C) 8:56 am	D) 9:20 am

Answer & Explanation Answer: C) 8:56 am

Explanation:

Time taken by Akash = 4 h

Time taken by Prakash = 3.5 h

For your convenience take the product of times taken by both as a distance.

Then the distance = 14km

Since, Akash covers half of the distance in 2 hours(i.e at 8 am)

Now, the rest half (i.e 7 km) will be coverd by both prakash and akash

Time taken by them = 7/7.5 = 56 min

Thus , they will cross each other at 8 : 56am.

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71 27331

Q:

In a race of 800m Kaushik gives Piyush a start of 200m and then loses the race by 20 seconds. What is the speed of Piyush and Kaushik respectively ? If the ratio of respective speeds be 3:2.

Answer

Let the speed of Kaushik be Sk and speed of Piyush be Sp and let the time taken by Piyush be t second then,

$\frac{Sk}{Sp} = \frac{\frac{800}{t + 20}}{\frac{600}{t}} = \frac{2}{3} (since given ratio)$

t = 20 s

Speed of Piyush = Sp = 600/20 = 30 m/s

and Speed of Kaushik = 800/40 = 20 m/s

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Subject: Time and Distance Exam Prep: AIEEE , Bank Exams , CAT
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8 5191

Q:

A fast moving superfast express crosses another pasenger train in 20 seconds. The speed of faster train is 72 km/hr and speeds of slower train is 27 km/h. Also the length of faster ntrain is 100m, then find the length of the slower train if they are moving in the same direction.

A) 100 m	B) 125 m
C) 150 m	D) 175 m

Answer & Explanation Answer: C) 150 m

Explanation:

$T i m e = \frac{S u m o f l e n g t h o f t h e t w o t r a i n s}{D i f f e r e n c e i n s p e e d s}$

$20 = \frac{(100 + x)}{\frac{25}{2}}$

$\Rightarrow$ X=150m

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43 7032

Q:

The distance of the college and home of Rajeev is 80km. One day he was late by 1 hour than the normal time to leave for the college, so he increased his speed by 4km/h and thus he reached to college at the normal time. What is the changed (or increased) speed of Rajeev?

A) 28 km/h	B) 30 km/h
C) 40 km/h	D) 20 km/h

Answer & Explanation Answer: D) 20 km/h

Explanation:

Let the normal speed be x km/h, then

$\frac{80}{x} - \frac{80}{(x + 4)} = 1$

$\Rightarrow$ $x^{2} + 4 x - 320 = 0$

$\Rightarrow$ x (x + 20) - 16 (x + 20) = 0

(x + 20 ) (x - 16) =0

x = 16 km/h

Therefore (x + 4) = 20 km/h

Therefore increased speed = 20 km/h

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55 29344

Q:

For $x \in N, x > 1$ , and $p = \log_{x} (x + 1)$ , $q = \log_{x + 1} (x + 2)$ then which one of the following is correct?

A) p < q	B) p = q
C) p > q	D) can't be determined

Answer & Explanation Answer: C) p > q

Explanation:

$\frac{k}{l} > \frac{k + 1}{l + 1}$ for (k,l) > 0 and k > l

Let k = x+1 and l = x

Therefore, $\frac{x + 1}{x} > \frac{(x + 1) + 1}{(x) + 1}$

(x + 1) > x

Therefore, $\frac{\log (x + 1)}{\log (x)} > \frac{\log (x + 2)}{\log (x + 1)}$

$\Rightarrow \log_{x} (x + 1) > \log_{x + 1} (x + 2)$

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33 5765

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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27 5408

Q:

What is the number of digits in $3^{3^{3}}$ ? Given that log3 = 0.47712?

A) 12	B) 13
C) 14	D) 15

Answer & Explanation Answer: B) 13

Explanation:

Let $L e t x = (3^{3^{3}})$ = ${(3^{3})}^{3}$

Then, $\log (x) = 3^{3} \log (3)$

= 27 x 0.47712 = 12.88224

Since the characteristic in the resultant value of log x is 12

$∴$ The number of digits in x is (12 + 1) = 13

Hence the required number of digits in $3^{3^{3}}$ is 13.

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Q:

Find value of $\frac{\log \sqrt{27} + \log 8 + l o g \sqrt{1000}}{\log 120}$

A) 1/2	B) 3/2
C) 2	D) 2/3

Answer & Explanation Answer: B) 3/2

Explanation:

$= \frac{\log \sqrt{3^{3}} + \log 2^{3} + \log \sqrt{10^{3}}}{\log (10 \times 3 \times 2^{2})}$

$= \frac{\log (3^{3})^{\frac{1}{2}} + \log 2^{3} + \log {(10^{3})}^{\frac{1}{2}}}{\log (10 \times 3 \times 2^{2})}$

$= \frac{\frac{1}{2} \log 3^{3} + 3 \log 2 + \frac{1}{2} \log 10^{3}}{\log 10 + \log 3 + \log 2^{2}}$

$= \frac{\frac{3}{2} (\log 3 + 2 \log 2 + \log 10)}{\log 3 + 2 \log 2 + \log 10}$ = $\frac{3}{2}$

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