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Find the probability of getting a numbered card when a card is drawn from the pack of 52 cards.

A.

9/13

B.

2/13

C.

4/13

D.

5/13

Answer with explanation

Answer: Option AExplanation

Total Cards = 52. Numbered Cards = (2, 3, 4, 5, 6, 7, 8, 9, 10) 9 from each suit 4 × 9 = 36

P (E) = 36/52 = 9/13

Or

If you count an ace as a number card there are 40 in a pack so p(get a number card) = 40/52 = 10/13.

If aces don’t count then there are 9 x 4 = 36 in a pack so p(get a number card) = 36/52 = 9/13.

Workspace

A.

35

B.

36

C.

37

D.

38

Answer with explanation

Answer: Option CExplanation

If we want to pack the drinks in the least number of cans possible, then each can should contain the maximum numbers of liters possible.As each can contains the same number liters of a drink, the number of liters in each can is a comman factor for 80,144 and 368; and it is also the highest such factor, as we need to store the maximum number of liters in each can.

So, the number of liters in each can = HCF of 80,144 and 368 = 16 liters.

Now, number of cans of Maaza = 80/16 = 5

Number of cans of Pepsi = 144/16 = 9

Number of cans of Sprite = 368/16 = 23

Thus, the total number of cans required = 5 + 9 + 23 = 37

Workspace

A room is 6 meters 24 centimeters in length and 4 meters 32 centimeters in Width. Find the least number of square tiles of equal size required to cover the entire floor of the room.

A.

107

B.

117

C.

127

D.

137

Answer with explanation

Answer: Option BExplanation

Let us calculate both the length and width of the room in centimeters.

Length = 6 meters and 24 centimeters = 624 cm

width = 4 meters and 32 centimeters = 432 cm

As we want the least number of square tiles required, it means the length of each square tile should be as large as possible.Further,the length of each square tile should be a factor of both the length and width of the room.

Hence, the length of each square tile will be equal to the HCF of the length and width of the room = HCF of 624 and 432 = 48

Thus, the number of square tiles required = (624 x 432 ) / (48 x 48) = 13 x 9 = 117

Workspace

Find the Greatest Number that will devide 43, 91 and 183 so as to leave the same remainder in each case

A.

4

B.

7

C.

9

D.

13

Answer with explanation

Answer: Option AExplanation

Required Number = H.C.F of (91- 43), (183- 91) and (183-43)

= H.C.F of 48, 92, and 140 = 4

Workspace

Six bells commence tolling together and toll at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together

A.

4

B.

10

C.

15

D.

16

Answer with explanation

Answer: Option DExplanation

Six bells commence tolling together and toll at intervals of 2, 4, 6, 8, 10, 12 minutes respectively.

Hence, using LCM of the given numbers which is 120 we conclude that bell will toll together after 120 seconds = 2 minutes.

In 30 minutes they will toll together in (30/2) = 15 and 1 (at the starting).

Hence, total number of bells together is 15 + 1 = 16

Workspace

The maximum number of students among them 1001 pens and 910 pencils can be distributed in such a way that each student gets the same number of pens and same number of pencils is:

A.

91

B.

910

C.

1001

D.

1911

Answer with explanation

Answer: Option AExplanation

Required number of students = H.C.F of 1001 and 910 = 91

Workspace

The H.C.F of two numbers is 11 and their L.C.M is 7700. If one of the numbers is 275 , then the other is

A.

279

B.

283

C.

308

D.

318

Answer with explanation

Answer: Option CExplanation

HCF = 11

LCM=7700

ONE NUMBER = 275

We are that,

One number× Other number = HCF×LCM

275× OTHER NUMBER = 11×7700

OTHER NUMBER = 11×7700/275=308

Workspace

L.C.M of two prime numbers x and y (x>y) is 161. The value of 3y-x is :

A.

-2

B.

-1

C.

1

D.

2

Answer with explanation

Answer: Option AExplanation

H. C. F of two prime numbers is 1. Product of numbers = 1 x 161 = 161.

Let the numbers be a and b . Then , ab= 161.

Now, co-primes with product 161 are (1, 161) and (7, 23).

Since x and y are prime numbers and x >y , we have x=23 and y=7.

Therefore, 3y-x = (3 x 7)-23 = -2

Workspace

The ratio of two numbers is 3 : 4 and their H.C.F is 4. Their L.C.M is

A.

12

B.

16

C.

24

D.

48

Answer with explanation

Answer: Option DExplanation

Let the numbers be 3x and 4x . Then their H.C.F = x. So, x=4

Therefore, The numbers are 12 and 16

L.C.M of 12 and 16 = 48

Workspace

The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:

A.

74

B.

94

C.

184

D.

364

Answer with explanation

Answer: Option DExplanation

L.C.M. of 6, 9, 15 and 18 is 90.

Let required number be 90k + 4, which is multiple of 7.

Least value of k for which (90k + 4) is divisible by 7 is k = 4.

=>Required number = (90 x 4) + 4 = 364.

Workspace

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