# Very important Arithmetical Reasoning Questions and Answers

Verbal Reasoning and Aptitude Questions and Answers for all Exams like CAT,MAT,XAT,GRE,GMAT,MBA,MCA,Bank Exams,Bank PO,SBI,Gate,Nda,Ssc.These questions are all frequently asked in all Exams. If you want to practice in hindi, you can practice **arithmetical reasoning questions in Hindi.**

## Very important Arithmetical Reasoning Questions and Answers

**1. A tailor had a number of shirt pieces to cut from a roll of fabric. He cut each roll of equal length into 10 pieces. He cut at the rate of 45 cuts a minute. How many rolls would be cut in 24 minutes?**

A. 30 rolls

B. 58 rolls

C. 120 rolls

D. 150 rolls

Ans: C. |

Sol.Number of cuts made to cut a roll into 10 pieces = 9. Therefore Required number of rolls = 45 × 24 / 9 = 120. |

**2. A monkey climbs 30 feet at the beginning of each hour and rests for a while when he slips back 20 feet before he again starts climbing in the beginning of the next hour. If he begins his ascent at 8.00 a.m., at what will he first touch a flag at 120 feet from the ground?**

A. 4 p.m

B. 5 p.m.

C. 6 p.m.

D. 8 p.m.

Ans: C. |

Sol.Let ascent of the monkey in 1 hour = ( 30 – 20 ) = 10 feet. So, the monkey ascends 90 feet in 9 hours i.e., 5 p.m. Clearly, in the next 1 hour i.e., till 6 p.m. The monkey ascends remaining 3o feet to touch the flag. |

**3. A is 3 years older to B and 3 years younger to C, while B and D are twins. How many years older is C and D?**

A. 3

B. 6

C. 9

D. 12

Ans: B. |

Sol.Since B and D are twins. So B = D. Now A = B + 3 and A = C – 3. Thus, B + 3 = C – 3 ‹=› D + 3 = C – 3 ‹=› C – D = 6. |

**4. A girl counted in the following way on the fingers of her left hand : She started by calling the thumb 1, the index finger 2, middle finger 3, ring finger 4, little finger 5 and then reversed direction calling the ring finger 6, middle finger 7 and so on. She counted upto 1994. She ended counting on which finger?**

A. Ring finger

B. Middle finger

C. Thumb

D. Index finger

Ans: D. |

Sol.Clearly, while counting the numbers associated to the thumb will be 1,9,17,25,……… i.e., numbers of the form (8n + 1 ). Since 1994 = 249 × 8 + 2, so 1993 shall correspond to the thumb and 1994 to the index finger. |

**5. Between two book ends in your study are displayed your five favourite puzzle books. If you decide to arrange the five books in every possible combination and moved just one book every minute, how long would it take you?**

A. 2 hours

B. 3 hours

C. 4 hours

D. 5 hours

Ans: A. |

Sol.Clearly, number of ways arranging 5 books, = 5! = 5 × 4 × 3 × 2 × 1 = 120. So total time taken = 120 minutes = 2 hours. |

**6. The 30 members of a club decided to play a badminton singles tournament. Every time a member loses a game he is out of the tournament. There are no ties. What is the minimum number of matches that must be played to determine the winner?**

A. 10

B. 15

C. 29

D. 61

Ans: C. |

Sol.Clearly, every member except one must lose one gain to decide the winner. Thus, minimum number of matches to be played = 30 – 1 = 29. |

**7. A man has Rs. 480 in the denominations of one rupee notes, five rupee notes and ten rupee notes. The number of each denomination is equal. What is the total number of notes that he has?**

A. 45

B. 75

C. 90

D. 120

Ans: C. |

Sol.Let number of notes of each denomination be x. Then x + 5x + 10 = 480. 16 x = 480 = x = 30. Hence, total number of notes = 3x = 90. |

**8. In a caravan, in addition to 5o hens, there are 45 goats and 8 camels with some keepers. If the total number of feet be 224 more than the number of heads in the caravan, the number of keepers is**

A. 5

B. 8

C. 10

D. 15

Ans: D. |

Sol.Let the number of keepers be x. Then, Total number of feet = 2 × 50 + 4 × 45 + 4 × 8 + 2x = 2x + 312. Therefore (2x + 312 ) = ( 103 + x) + 224 or x = 15. |

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**9. A father is now three times as old as his son. Five years back, he was four times as old as his son. The age of the son(in years) is**

A. 10

B. 12

C. 15

D. 18

Ans: C. |

Sol.Let son’s age be x years. Then father’s age = (3x) years. Five years ago, father’s age = (3x – 5) years and Son’s age = (x – 5) years. So 3x – 5 = 4 (x – 5)‹=›3x – 5 = 4x – 20 ‹=› x = 5. |

**10. Aruna cut a cake into two halves and cuts one half into smaller pieces of equal size. Each of the small pieces is twenty grams in weight. If she has seven pieces of the cake in all with her, how heavy was the original cake?**

A. 100 grams

B. 150 grams

C. 200 grams

D. 240 grams

Ans: D. |

Sol.The seven pieces consist of 6 smaller equal pieces and one calf cake piece. Weight of each small piece = 20 g. So, total weight of the cake = [2 × (20 × 6)]g = 240g. |

**11. A farmer built a fence around his square plot. He used 27 fence poles on each side of the square. How many poles did he need altogether?**

A. 54

B. 84

C. 104

D. 108

Ans: C. |

Sol.Since each pole at the corner of the plot is common to its two sides. Therefore total number of poles needed = 27 × 4 – 4 = 104. |

**12. The number of boys in a class is three times the number of girls. Which one of the following numbers cannot represent the total number of children in the class?**

A. 40

B. 42

C. 44

D. 48

Ans: B. |

Sol.Let number of girls = x and number of boys = 3x. Then, 3x + x = 4x = total number of students. Thus, to find exact value of x, The total number of students must be divisible by 4. |

**13. Find the number which when added to itself 13 times, gives 112.**

A. 5

B. 6

C. 7

D. 8

Ans: D. |

Sol.Let the number be x. Then x + 13x = 112 ‹=›14x = 112 ‹=› x = 8. |

**14. A bus starts from city X. The number of women in the bus is half of the number of men. In city Y, 10 men leave the bus and five women enter. Now number of men and women is equal. In the beginning, how many passengers entered the bus?**

A. 25

B. 35

C. 45

D. 55

Ans: C. |

Sol.Originally, let number of women = x. Then number of men = 2x. So, in city Y, we have (2x – 10) = (x + 5 ) or x = 15. Total number of passengers in the beginning = ( x + 2x) = 3x = 45. |

**15. A motorist knows four different routes from Bristol to Birmingham. From Birmingham to Sheffield he knows three different routes and from Sheffield to Carlisle he knows two different routes. How many routes does he know from Bristol to Carlisle?**

A. 12

B. 24

C. 48

D. 60

Ans: B. |

Sol.Total number of routes from Bristol to Carlisle = ( 4 × 3 × 2) = 24. |

**16. A placed three sheets with two carbons to get two extra copies of the original. Then he decided to get more carbon copies and folded the paper in such a way that the upper half of the sheets were on top of the lower half. Then he typed. How many carbon copies did he get?**

A. 1

B. 2

C. 3

D. 4

Ans: B. |

Sol.Since the number of carbons is 2, only two copies can be obtained. |

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**17. A group of 1200 persons consisting of captains and soldiers is travelling in a train. For every 15 soldiers there is one captain. The number of captains in the group is**

A. 70

B. 75

C. 80

D. 85

Ans: B. |

Sol.Clearly, out of every 16 persons, there is one captain. So, number of captains = 1200 / 16 = 75. |

**18. There are deer and peacocks in a zoo. By counting heads they are 80. The number of their legs is 200. How many peacocks are there?**

A. 10

B. 20

C. 40

D. 60

Ans: D. |

Sol.Let x and y be the number of deer and peacocks in the zoo respectively. Then, x + y = 80 ——————> (i) 4x + 2y = 200 or 2x + y = 100 ———> (ii) Solving (i) and (ii), we get : x = 20, y = 60. |

**19. When Rahul was born, his father was 32 years older than his brother and his mother was 25 years older than his sister. If Rahul’s brother is 6 years older than him and his mother is 3 years younger than his father, how old was Rahul’s sister when he was born?**

A. 5 years

B. 8 years

C. 10 years

D. 14 years

Ans: C. |

Sol.When Rahul was born, his brother’s age = 6 years, his father’s age = ( 6 + 32 ) = 38 years. his mother’s age = ( 38 – 3) years = 35 years. his sister’s age = ( 35 – 25 ) years = 10 years. |

**20. A man wears socks of two colours ? black and brown. He has altogether 20 black socks and 20 brown socks in a drawer. Supposing he has to take out the socks in the dark, how many must he take out to be sure that he has a matching pair?**

A. 1

B. 3

C. 20

D. 39

Ans: B. |

Sol.Since there are socks of only two colours, so two out of any three socks must always be of the same colour. |

**21. Five bells begin to toll together and toll respectively at intervals of 6, 5, 7, 10 and 12 seconds. How many times will they toll together in one hour excluding the one at the start?**

A. 5 times

B. 6 times

C. 7 times

D. 8 times

Ans: D. |

Sol.L.C.M. of 6,5,7,10 and 12 is 420. So, the bells will toll together after every 420 seconds. i.e. 7 minutes. Now 7 × 8 = 56 and 7 × 9 = 63. Thus, in 1 hour( or 60 minutes), the bells will toll together 8 times, excluding the one at the start. |

**22. At the end of a business conference the ten people present all shake hands with each other once. How many handshakes will there be altogether?**

A. 20

B. 30

C. 40

D. 45

Ans: D. |

Sol.Clearly, total number of handshakes = (9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1) = 45. |

**23. I have a few sweets to be distributed. If I keep 2,3 or 4 in pack, Iam left with one sweet. If I keep 5 in a pack, Iam left with none. What is the minimum number of sweets I have to pack and distribute?**

A. 25

B. 35

C. 56

D. 65

Ans: A. |

Sol.Clearly, the required number would be such that it leaves a remainder of 1 when divided by 2, 3 or 4 and no remainder when divided by 5, Such a number is 25. |

**24. A shepherd had 17 sheep. All but nine died. How many was he left with?**

A. 6

B. 7

C. 8

D. 9

Ans: D. |

Sol.All but nine died means ‘All except nine died’. i.e., 9 sheep remainded alive. |

**25. An enterprising businessman earns an income of Re. 1 on the first day of his business. On every subsequent day, he earns an income which is just double of that made on the previous day. One the 10th day of business. his income is**

A. Rs.10

B. $$ {Rs.10^{9}}$$

C. $$ {Rs.2^{10}}$$

D. $$ {Rs.2^{9}}$$

Ans: D. |

Sol.Income of the first day = Re = 1. Income of the 2nd day = Rs. ( 1 × 2) = Rs. 2 ^{1}.Income of the 3rd day = Rs. ( 2 ^{1} × 2) = Rs. 2^{2} and so on.Thus, Income of the nth day = Rs.2 ^{n – 1}.Therefore Income of the 10th day = Rs. 2 ^{9}. |

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