Mathematical Reasoning & Aptitude Practice Questions 

 Number series

1. Find the missing number in the following series.

 512, 256, 128, ?, 32, 16, 8 

(a) 52 (b) 61  (c) 64 (d) 56  

2. Find the missing number in the following series.

 2, 7, 17, 32, 52, 77,? 

 (a) 107 (b) 91 (c) 101 (d) 92 

3. Find the missing number in the following series. 15, 18, 24, ?, 45 

 (a) 27 (b) 30 (c) 33 (d) 36 

4. Fill in the missing number in the following arrangement based on some principle. 

8

6

4

3

?

7

14

15

18

(a) 4 (b) 5 (c) 6 (d) 8

Answers

 1. (c): Each number is half of its preceding number. 

 2. (a): +5, +10, +15, +20, +25, +30 

3. (c): +3, +6, +9, +12 

4. (b): In first row, it is −2; in second row, it is +2, and in third row, it is +2.

 Letter series

 1. Find the missing letters in the following series.

QAR, RAS, SAT, TAU, __________ 

(a) UAV (b) TAS (c) UAT (d) TAT 

2. Find the missing letters in the following series.

 FTG, GTF, HTI, ITH, __________

 (a) JTK (b) HTL (c) HTK (d) JTI 

3. Find the missing letters in the following series.

 SCD, TEF, UGH, __________, WKL

 (a) CMN (b) UJI (c) VIJ (d) IJT 

4. Find the missing letters in the following series. 

JAK, KBL, LCM, MDN, __________

 (a) OEP (b) NEO (c) MEN (d) PFQ 

Answers 

 1. (a): The third letter is repeated as the first letter of the next segment. The middle letter, A, remains static. The third letters are in alphabetical order beginning with R. 

2. The middle letter T is fixed. First and third letters are swapping their position in second and fourth terms. The first letters are in alphabetical order: F, G, H, I, and J. The missing segment begins with a new letter J. 

3. (c): For first letter of each triplet, the series is STUVW. The remaining two letters of the series goes like CD, EF, GH, IJ, KL. 

4. (b): This is an alternating series in alphabetical order. The middle letters follow the order ABCDE. The first and third letters of the triplets are in alphabetical order beginning with J. The third letter is repeated as a first letter in each subsequent three-letter segment

Codes and Relationships

1. If ‘CERTAIN’ is coded as ‘XVIGZRM’ in a particular code language, then how ‘MUNDANE’ be coded in that language?   

 (a) NFMWZMX  (b) VMZWMFN (c) NFMWZMV (d) MIMXZMV 

2. If ‘EDUCATION’ is coded as NOITACUDE, then ‘RED FORT’ will be coded as

 (a) TROFDER  (b) FORTRED (c) TROFRED  (d) FORTDER 

3. In a certain code, ‘MOTHER’ is written as ‘OMHURF’. How will ‘ANSWER’ be written in that code? 

(a) NBWRRF  (b) MAVSPE (c) NBWTRD  (d) NAWTRF 

Answers 

 1. (c) 2. (a) 3. (d) 

Odd Word

 1. Choose the word that is least like the other words in a group. 

(a) Lion   (b) Cheetah (c) Bear   (d) Tiger

2. Choose the word that is least like the other words in a group.

(a) Sheet   (b) Cot (c) Spain   (d) Pillow 

3. Find the odd word among the following: 

(a) Kiwi   (b) Eagle (c) Penguin  (d) Ostrich

Answers  

1. (c): All, except bear belong to the cat family. 

2. (b): All others are parts of bed spread. 

3. (b): All, except eagle are flightless birds

Odd Pair 

 1. Find the odd number. 

(a) 13 (b) 53 (c) 63 (d) 23 

2. Find the odd number. 

(a) 51 (b) 144 (c) 64 (d) 121 

3. Find the odd number. 

(a) 15 (b) 21 (c) 24 (d) 28 (e) 30  

4. Find the odd number. 

(a) 324 (b) 244 (c) 136 (d) 352 

Answers

1. (c): Each of the numbers except 63 is a prime number. 

2. (a): Each of the number except 51 is a perfect square.

3. (d): Each of the numbers except 28 is divisible by 3. 

4. (a): Sum of the digits in each other number is 10

Speed & Distance 

 1. A person crosses a 1200 m long street in 10 minutes. What is his speed in kmph?

(a) 7.2 kmph  (b) 8 kmph (c) 9 kmph  (d) None of the above 

2. A person completes a journey of 48 km in 2 hours. How much time will he take to cover a distance of 252 km? 

(a) 10 hours  (b) 11 hours (c) 10½ hours (d) None of the above 

3. A train completed half a trip at 30 miles per hour and the other half at 60 miles per hour. If the whole trip was 20 miles, then how much time did the train take to complete the trip? 

(a) 90 minutes  (b) 60 minutes (c) 45 minutes  (d) 30 minutes 

4. A person performs half of his journey by train, onethird by bus and the rest 5 km by auto rickshaw. Find his total journey. 

(a) 30   (b) 36 (c) 40   (d) 45 

5. Excluding stoppages, the speed of a bus is 54 kmph, and including stoppages, it is 45 kmph. For how many minutes does the bus stop per hour? 

(a) 9   (b) 10 (c) 12   (d) 20

Answer

 1. (a): Speed = Distance/Time = 1,200 m/600 s = 2 m/s. Speed in kmph = 2 × 18/5 = 7.2 kmph 

2. (c): Here, Speed = 48/2 = 24 kmph Time taken = 252/24 = 10½ hrs 

3. (d): Average speed = 2 × 30 × 60/(30 + 60) = 2 × 30 × 60/90 = 40 mph Time taken by train to complete the trip = Distance/ Speed = 20/40 = ½ hr = 30 min. 

4. (a): The distance covered by auto-rickshaw = 1 - (1/2 + 1/3) = 1/6 1/6 of total journey = 5 km Total journey = 6 × 5 = 30 km 

5. (b): Due to stoppages, it covers 9 km less. Time taken to cover 9 km = 9/54 × 60 min = 10 min 

Calendar 

 1. It was Wednesday on 15 August 2012. What should be the day on November 15, 2013? 

(a) Wednesday (b) Thursday (c) Friday  (d) None of the above  

2. If it is Sunday today, then what will be the day after 60 days? 

(a) Sunday  (b) Thursday (c) Tuesday  (d) Friday 

3. If 22 April 2013 is Monday, then what was the day of the week on April 22, 2012? 

(a) Sunday  (b) Saturday (c) Tuesday  (d) Wednesday

4-The last day of a century cannot be 

(a) Monday   (b) Wednesday (c) Tuesday   (d) Friday  

5. 10 February, 2005 was Thursday. What was the day of the week on 8 February,

 2004? (a) Tuesday  (b) Monday (c) Sunday  (d) Wednesday 

answer

1. (c): From August 15, 2012, till August 15, 2013, there is one extra day. In the rest of 16 days of August, there are 2 odd days. In September, there are 2 odd days, and in October, 3 odd days. In first 15 days of November, there will be 1 odd day. Thus, the total number of odd days is 1 + 2 + 2 + 3 + 1 = 9; it means 2 extra (odd) days. Hence, November 15, 2013 was Friday. 

2. (b): Each day of the week is repeated after 7 days. Assuming that first day is Sunday, 8, 15, 22, 29, 36, 43, 50 and 57 days will be Sundays. Hence, 58 day is Monday, 59 is Tuesday, 60 is Wednesday. Hence, day after 60 days will be Thursday. 

3. (a): The year 2012 is a leap year. Thus, it has 2 odd days. But as calculation is to be done from April 22, 2012, till April 22, 2013, so it has 1 odd day only. As the calculation is to be done backwards, thus April 22nd was one day less. Thus, it was Sunday. Note: Had the question been about April 22, 2014, the answer would have been Tuesday. If the same question is framed for April 22, 2016, the answer would have been Friday as 2016 is a leap year. 

4. (c): 100 years contain 5 odd days. Last day of 1st century is Friday. 200 years contain (5 × 2) = 3 odd days Last day of 2nd century is Wednesday. 300 years contain (5 × 3) = 15 odd days that is, equal to 1 odd day. Last day of 3rd century is Monday. 400 years contain 0 odd days. Last day of 4th century is Sunday. This cycle is repeated. Last day of a century cannot be Tuesday or Thursday or Saturday. 

5. (c): The year 2004 is a leap year. It has 2 odd days. Hence, February 10, 2004, was Tuesday and February 8, 2004, must be Sunday