Practice Arithmetical reasoning - verbal reasoning Online Quiz (set-1) For All Competitive Exams
Q-1) A player holds 13 cards of four suits, of which seven are black and six are red. There are twice as many diamonds as spades and twice as many hearts as diamonds. How many clubs does he holds?
(a)
(b)
(c)
(d)
Clearly, the black cards are either clubs or spades while the red cards are either diamonds or hearts.
Let the number of spades be x. Then, number of cubes = (7 – x).
Number of diamonds = 2 x number of spades = 2x
Number of hearts = 2 x number of diamonds = 4x
Total number of cards = x + 2x + 4x + 7 – x = 6x + 7.
Therefore 6x + 7 = 13 ⇔ 6x = 6 ⇔ x – 1.
Hence, number of clubs = (7 – x) = 6.
Q-2) In a family, each daughter has the same number of brothers as she has sisters and each son has twice as many sisters as he has brothers. How many sons are there in the family ?
(a)
(b)
(c)
(d)
Let d and s represent the number of daughters and sons respectively.
Then, we have :
d – 1 = s and 2 (s – 1) = d.
Solving these two equations, we get: d = 4, s = 3.
Q-3) A box contains five sets of balls while there are 3 balls in each set. Each set of balls has one color which is different from every other set, what is the least number of balls that must be removed from the box in order to claim with certainty that a pair of balls of the same colour has been removed?
(a)
(b)
(c)
(d)
| Set | 1 | 2 | 3 | 4 | 5 |
| Balls | 1 | 1 | 1 | 1 | 1 |
Now, any further removal of balls from any set will ensure that removed ball is of the same colour as one of the already removed balls, thus constituting a pair of the removed balls of the same colour.
| Set | 1 | 2 | 3 | 4 | 5 |
| Balls | 1 + 1 | 1 | 1 | 1 | 1 |
Hence, minimum no. of removed balls = 6
Q-4) Six identical cards are placed on a table. Each card has number '1' marked on one side and number '2' marked on its other side. All the six cards are placed in such a manner that the number '1' is on the upper side. In one try, exactly four (neither more nor less) cards are turned upside down. In how many least number of tries can the cards be turned upside down such that all the six cards show number '2' on the upper side ?
(a)
(b)
(c)
(d)
Q-5) The letters L, M, N, O, P, Q, R, S and T in their order are substituted by nine integers 1 to 9 but not in that order. 4 is assigned to P. The difference between P and T is 5. The difference between N and T is 3. What is the integer assigned to N?
(a)
(b)
(c)
(d)
| 6 | 4 | 9 | ||||||
| L | M | N | O | P | Q | R | S | T |
difference between P & T is 5 i.e., T = 5 + 4 = 9
difference between N & T is 3 i.e., N = 9 – 3 = 6
So, integer assigned to N = 6
Q-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)
(b)
(c)
(d)
Clearly, every member except one (i.e. the winner) must lose one game to decide the winner. Thus, minimum number of matches to be played = 30 – 1 = 29.
Q-7) 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)
(b)
(c)
(d)
Generally we may commit mistake of dividing 1200/15. But out of 16 persons there is one captain. so, it will be 1200/16 = 75
Q-8) Mr. X, a mathematician, defines a number as 'connected with 6 if it is divisible by 6 or if the sum of its digits is 6, or if 6 is one of the digits of the number. Other numbers are allenot connected with 6'. As per this definition, the number of integers from 1 to 60 (both inclusive) which are not connected with 6 is
(a)
(b)
(c)
(d)
Numbers from 1 to 60, which are divisible by 6 are : 6, 12, 18, 24, 30, 36, 42, 48, 54, 60.
There are 10 such numbers.
Numbers from 1 to 60, the sum of whose digits is 6 are 6, 15, 24, 33, 42, 51, 60.
There are 7 such numbers of which 4 are common to the above ones. So, there are 3 such uncommon numbers.
Numbers from 1 to 60, which have 6 as one of the digits are 6, 16, 26, 36, 46, 56, 60.
Clearly, there are 4 such uncommon numbers.
So, numbers 'not connected with 6'
= 60 – (10 + 3 + 4) = 43.
Q-9) There are four routes to travel from city A to city B and six routes from city B to city C. How many routes are possible to travel from the city A to city C ?
(a)
(b)
(c)
(d)
Total number of possible routes from the city A to city C = 4 × 6 = 24
Q-10) At a dinner party every two guests used a bowl of rice between them, every three guests used a bowl of daal between them and every four used a bowl of meat between them. There were altogether 65 dishes. How many guests were present at the party ?
(a)
(b)
(c)
(d)
Let the number of guests be x. Then number of bowls of rice = $x/2$; number of bowls of dal = $x/3$; number of bowls of meat = $x/4$.
∴ $x/2 + x/3 + x/4$ = 65
⇔ ${6x + 4x + 3x}/{12}$ = 65 ⇔ 13x = 65 × 12
⇔ x = $({65 × 12}/{13})$ = 60
Q-11) 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)
(b)
(c)
(d)
Since there are socks of only two colours, so two out of any three socks must always be of the same colour.
Q-12) David gets on the elevator at the 11th floor of a building and rides up at the rate of 57 floors per minute. At the same time. Albert gets on an elevator at the 51st floor of the same building and rides down at the rate of 63 floors per minute. If they continue travelling at these rates, then at which floor will their paths cross?
(a)
(b)
(c)
(d)
Suppose their paths cross after x minutes.
Then, 11 + 57x = 51 – 63 x ⇔ 120 x = 10 ⇔ x = $1/3$
Number of floors covered by David in $1/3$ min
= $(1/3 × 57)$ = 19.
So, their paths cross at (11 + 19)th i.e., 30th floor.
Q-13) I have a few sweets to be distributed. If I keep 2, 3 or 4 in a pack, I am left with one sweet. If I keep 5 in a pack, I am left with none. What is the minimum number of sweets I have to pack and distribute ?
(a)
(b)
(c)
(d)
The required number will 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 among options
Q-14) Three ducks can be arranged as shown above to satisfy all the three given conditions. A certain number of horses and an equal number of men are going somewhere. Half of the owners are on their horses' back while the remaining ones are walking along leading their horses. If the number of legs walking on the ground is 70, how many horses are there ?
(a)
(b)
(c)
(d)
Let number of horses = number of men = x.
Then, number of legs = 4x + 2 x (x/2) = 5x.
So, 5x = 70 or x = 14.
Q-15) First bunch of bananas has (1/4) again as many bananas as a second bunch. If the second bunch has 3 bananas less than the first bunch, then the number of bananas in the first bunch is
(a)
(b)
(c)
(d)
Let the number of bananas in the second bunch be x
Then, number of bananas in the first bunch
= x + $1/4 x = 5/4 x$
So, $5/4$ x - x = 3 ⇔ 5x - 4x = 12 ⇔ x = 12
∴ Number of bananas in the first bunch
= $(5/4 × 12)$ = 15
Q-16) In three coloured boxes - red, green and blue, 108 balls are placed. There are twice as many balls in the green and red boxes combined as there are in the blue box and twice as many in the blue box as there are in the red box. How many balls are there in the green box ?
(a)
(b)
(c)
(d)
Let R, G and B represent the number of balls in red, green and blue boxes respectively.
Then, R + G + B = 108 ...(i),
G + R = 2B ...(ii)
B = 2R ...(iii)
From (ii) and (iii), we have G + R = 2 × 2R = 4R or G = 3R. Putting G = 3R and B = 2R in (i), we get:
R + 3R + 2R = 108 6R = 108 R = 18.
Therefore number of balls in green box = G = 3R = (3 × 18) = 54.
Q-17) If every 2 out of 3 ready made shirts need alterations in the sleeves, and every 4 out of 5 need it in the body, how many alterations will be required for 60 shirts?
(a)
(b)
(c)
(d)
Number of alterations required in 1 shirt
= $(2/3 + 3/4 + 4/5) = {133}/{60}$
∴ Number of alterations required in 60 shirts
= $({133}/{60} × 60)$ = 133
Q-18) The total number of digits used in numbering the pages of a book having 366 pages is
(a)
(b)
(c)
(d)
Total number of digits
= (No. of digits in 1- digit page nos. + No. of digits in 2-digit page nos. + No. of digits in 3- digit page nos.)
= (1 × 9 + 2 × 90 + 3 × 267) = (9 + 180 + 801) = 990.
Q-19) Jaideep was given some money by his mother on his birthday. Jaideep spent all of it in five stores. In each store he spent one rupee more than half of what he had when he came in. How much did he get from his mother?
(a)
(b)
(c)
(d)
Q-20) 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)
(b)
(c)
(d)
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 30 feet to touch the flag.