5555+ Aptitude and Reasoning Questions, Answers With Explanation

Q:

The average temperature of the town in the first four days of a month was 58 degrees. The average for the second, third, fourth and fifth days was 60 degrees. If the temperatures of the first and fifth days were in the ratio 7 : 8, then what is the temperature on the fifth day ?

A) 62 degrees	B) 64 degrees
C) 65 degrees	D) 66 degrees

Answer & Explanation Answer: B) 64 degrees

Explanation:

Sum of temperatures on 1st, 2nd, 3rd and 4th days = (58 * 4) = 232 degrees ... (1)

Sum of temperatures on 2nd, 3rd, 4th and 5th days - (60 * 4) = 240 degrees ....(2)

Subtracting (1) From (2), we get :

Temp, on 5th day - Temp on 1st day = 8 degrees.

Let the temperatures on 1st and 5th days be 7x and 8x degrees respectively.

Then, 8x - 7x = 8 or x = 8.

Temperature on the 5th day = 8x = 64 degrees.

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Filed Under: Average

56 29228

Q:

A and B can do a piece of work in 40 and 50 days. If they work at it an alternate days with A beginning in how many days, the work will be finished ?

Answer

(A+B)'s two days work = $\frac{1}{40} + \frac{1}{50} = \frac{9}{200}$

Evidently, the work done by A and B duing 22 pairs of days

i.e in 44 days = $22 \times \frac{9}{200} = \frac{198}{200}$

Remaining work = $1 - \frac{198}{200}$ = 1/100

Now on 45th day A will have the turn to do 1/100 of the work and this work A will do in $40 \times \frac{1}{100} = \frac{2}{5}$

Therefore, Total time taken = 44 $\frac{2}{5}$ daya

Workspace

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Subject: Time and Work
Job Role: Bank PO

72 29226

Q:

The base of a triangular field is three times its altitude. If the cost of cultivating the field at Rs. 24.68 per hectare be Rs. 333.18, find its base and height.

A) B=900;H=300	B) B=300;H=900
C) B=600;H=700	D) B=500;H=900

Answer & Explanation Answer: A) B=900;H=300

Explanation:

Area of the field = Total cost/rate = (333.18/25.6) = 13.5 hectares
$13.5 * 10000 m^{2} = 135000 m^{2}$
Let altitude = x metres and base = 3x metres.
Then, $\frac{1}{2} * 3 x * x = 135000 \Rightarrow x^{2} = 90000 \Rightarrow x = 300$

Base = 900 m and Altitude = 300 m.

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Filed Under: Area
Exam Prep: Bank Exams
Job Role: Bank PO

62 29212

Q:

Find the value of 6 + 11 + 16 + 21 + 21 + .... + 71.

A) 539	B) 561
C) 661	D) 639

Answer & Explanation Answer: A) 539

Explanation:

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Filed Under: Number Series
Exam Prep: Bank Exams

10 29189

Q:

A large field of 700 hectares is divided into two parts. The difference of the areas of the two parts is one-fifth of the average of the two areas. What is the area of the smaller part in hectares?

A) 315	B) 385
C) 415	D) 485

Answer & Explanation Answer: A) 315

Explanation:

Let the areas of the two parts be x and (700-x) hectares

therefore, $[x - (700 - x)] = \frac{1}{5} [\frac{x + (700 - x)}{2}]$

$\Rightarrow 2 x - 700 = 70$

$\Rightarrow x = 385$

So, the two parts are 385 and 315.

Hence, Area of the smaller = 315 hectares

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Filed Under: Area
Exam Prep: Bank Exams
Job Role: Bank PO

18 29160

Q:

From a container of wine, a thief has stolen 15 liters of wine and replaced it with same quantity of water.He again repeated the same process. Thus in three attempts the ratio of wine and water became 343:169. The initial amount of wine in the container was:

A) 75 liters	B) 100 liters
C) 150 liters	D) 120 liters

Answer & Explanation Answer: D) 120 liters

Explanation:

$\frac{W i n e (l e f t)}{W a t e r (a d d e d)} = \frac{343}{169}$

It means $\frac{W i n e (l e f t)}{W i n e (i n i t i a l a m o u n t)} = \frac{343}{512}$ $[∵ 343 + 169 = 512]$

Thus ,

$343 x = 512 x {(1 - \frac{15}{K})}^{3}$

$\Rightarrow \frac{343}{512} = {(\frac{7}{8})}^{3} = {(1 - \frac{15}{k})}^{3}$

=> K = 120

Thus the initial amount of wine was 120 liters.

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Filed Under: Alligation or Mixture

40 29149

Q:

Find the appropriate relation for quantity1 and quantity2 in the following question:

Quantity 1: In an examination, Ankita scored 35 marks less than Puneeta. Puneeta scored 65 more marks than Meenakshi. Rakhi scored 115 marks, which is 20 marks more than Meenakshi's. Simpy scored 108 marks less than the maximum marks of the test. What approximate percentage of marks did Simpy score in the examination, if she got 67 marks more than Ankita?

Quantity 2: The length of a rectangle is increased by 60%. By what percent would the width have to be decreased to maintain the same area?

A) Quantity1 < Quantity2	B) Quantity1 ≤ Quantity2
C) Quantity1 ≥ Quantity2	D) Quantity1 > Quantity2

Answer & Explanation Answer: D) Quantity1 > Quantity2

Explanation:

Quantity1-
Rakhi’s marks= 115
Meenakshi’s marks= 115 - 20 = 95
Puneeta’s marks= 95 + 65= 160
Ankita’s marks=160 - 35= 125
Simpy’s marks= 125+ 67= 192
Total maximum marks= 192 + 108= 300
Required percentage marks of Simpy
= $\frac{192}{300} \times 100 = 64 %$
Quantity2-
Let length and breadth be 100.

After increase in length it become 160, then
reduction in breadth be ‘x’
Now, 160*x= 100*100
Hence, x = $\frac{10000}{160} = 62.5$

100-62.5=37.5%

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Filed Under: Quantitative Aptitude - Arithmetic Ability
Exam Prep: Bank Exams , GATE
Job Role: Bank Clerk , Bank PO

2 29140

Q:

Find the missing number in the series.

75, 96, 116, 135, 153, ?

A) 173	B) 170
C) 178	D) 175

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