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From a pack of 52 cards, 3 cards are drawn together at random, What is the probability of both the cards are a king?

A.

1/5225

B.

1/5525

C.

5525

D.

1/525

Answer with explanation

Answer: Option BExplanation

Let S be the sample space

n(S) = 52C_{3} = [( 52 x 51 x 50) / (3 x 2 x 1)] = 13600 / 6 = 22100

Let E = event of getting 3 kings out of 4.

n(E) = 4C_{3} = [(4 x 3 x 2 x 1) / (3 x 2 x 1)] = 24/6 = 4

therefore, p(E) = [n(E) / n(S)]

==> p(E) = [ 4 / 22100]

==> p(E) = 1/5525

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16 persons are participated in a party. In how many differentways can they host the seat in a circular table, if the 2particular persons are to be seated on either side of the host?

A.

16! × 2

B.

14! × 2

C.

18! × 2

D.

14!

Answer with explanation

Answer: Option BExplanation

(16 – 2)! × 2 = 14! × 2

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How many 3-digit numbers can be formed from the digit 2,3,5,6,7 and 9, which are divisible by 5 and none of the digits is repeated?

A.

5

B.

10

C.

15

D.

20

Answer with explanation

Answer: Option DExplanation

Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it.

The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.

The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.

Required number of numbers = (1 x 5 x 4) = 20.

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There is meeting of 20 delegates that is to be held in a hotel. In how many ways these delegates can be seated along a round table, if three particular delegates always seat together?

A.

17! 3!

B.

18! 3!

C.

D.

Answer with explanation

Answer: Option AExplanation

Total 20 persons, 3 always seat together, 17 + 1 =18 delegates can be seated in (18 -1)! Ways = 17!

And now that three can be arranged in 3! Ways.

So, 17! 3! is the correct answer.

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Groups each containing 3 boys are to be formed out of 5 boys. A, B, C, D and E such that no group can contain both C and D together. What is the maximum number of such different groups?

A.

5

B.

6

C.

7

D.

8

Answer with explanation

Answer: Option CExplanation

Maximum number of such different groups = ABC, ABD,ABE, BCE,BDE,CEA,DEA =7.

Alternate method:

Total number of way in which 3 boys can be selected out of 5 is 5C_{3}

Number of ways in which CD comes together = 3 (CDA,CDB,CDE)

Therefore, Required number of ways = 5C_{3} -3

= 10-3 =7.

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A and B, there are two companies, selling the packs of cold-drinks. For the same selling price A gives two successive discounts of 10% and 25%. While B sells it by giving two successive discounts of 15% and 20%. What is the ratio of their marked price?

A.

143 : 144

B.

136 : 135

C.

73 : 77

D.

19 : 11

Answer with explanation

Answer: Option BExplanation

A = 90/100*75/100

= .675

B = 85/100*80/100

= .68

680:675

136:135

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The cost price of 20 articles is the same as the selling price of x articles. If the profit is 25%, find out the value of x

A.

13

B.

14

C.

15

D.

16

Answer with explanation

Answer: Option DExplanation

We have 3 elements here CP-cost price, SP-selling price, and profit percentage Profit%.

So the** equations** relating them are:

- Profit=SP-CP (for loss its CP-SP)
- Profit%=((SP-CP)/CP)*100

**Solution:**

Since we don’t know the cost of the article it is ok to assume a value initially.

Let the CP of each article =1 unit.

CP of 20 articles =20 units

SP of x articles=CP of 20 articles= 20 units

SP of 1 article = 20/x

using formula Profit%=((SP-CP)/CP)*100 for 1 article as profit% is a constant for each article.

25=[(20-x)/x)]*100

=> 25x/100 = 20-x

=> x/4 = 20-x

=> 5x=80

solving we get **x=16 units**.

(OR)

Profit%= (purchased qty – selleing qty)/selling qty × 100

25=(20-x)/x *100

1/4=(20-x)/x

x=80–4x

5x=80

x=16

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