Practice Question and answers set 2 - verbal reasoning Online Quiz (set-1) For All Competitive Exams
Q-1) 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-2) 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-3) 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-4) 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-5) 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-6) 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-7) 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-8) 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-9) 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-10) 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