Clock

1. An accurate clock shows 8 o'clock in the morning. Through how may degrees will the hour hand rotate when the clock shows 2 o'clock in the afternoon?
A. 154°	B. 180°
C. 170°	D. 160°

2. A clock is started at noon. By 10 minutes past 5, the hour hand has turned through
A. 155°	B. 145°
C. 152°	D. 140°

  

3. At what time between 7 and 8 o'clock will the hands of a clock be in the same straight line but, not together?
A. 5 minutes past 7	B. 5311 miinutes past 7
C. 5111 minutes past 7	D. 5511 minutes past 7

4. At what time between 5.30 and 6 will the hands of a clock be at right angles?
A. 44 minutes past 5	B. 44711 minutes past 5
C. 43711 minutes past 5	D. 43 minutes past 5

5. At what angle the hands of a clock are inclined at 15 minutes past 5?
A. 6712°	B. 6212°
C. 70°	D. 6334°

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Permutations and Combinations

1. Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?
A. 24400	B. 21300
C. 210	D. 25200

Explanation :

Number of ways of selecting 3 consonants out of 7 = ⁷C₃
Number of ways of selecting 2 vowels out of 4 = ⁴C₂

Number of ways of selecting 3 consonants out of 7 and 2 vowels out of 4 = ⁷C₃ x ⁴C₂

=(7×6×53×2×1)×(4×32×1)=210

It means that we can have 210 groups where each group contains total 5 letters(3 consonants

and 2 vowels).



Number of ways of arranging 5 letters among themselves = 5!

= 5 x 4 x 3 x 2 x 1 = 120



Hence, Required number of ways = 210 x 120 = 25200

2. In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?


A. 159

B. 209


C. 201

D. 212

Explanation :

In a group of 6 boys and 4 girls, four children are to be selected such that
at least one boy should be there.

Hence we have 4 choices as given below

We can select 4 boys ------(Option 1). 
Number of ways to this = 6C4

We can select 3 boys and 1 girl ------(Option 2)
Number of ways to this = 6C3 x 4C1

We can select 2 boys and 2 girls ------(Option 3)
Number of ways to this = 6C2 x 4C2

We can select 1 boy and 3 girls ------(Option 4)
Number of ways to this = 6C1 x 4C3

Total number of ways 
= (6C4) + (6C3 x 4C1) + (6C2 x 4C2) + (6C1 x 4C3)
= (6C2) + (6C3 x 4C1) + (6C2 x 4C2) + (6C1 x 4C1) [Applied the formula nCr = nC(n - r) ] 

=[6×52×1]+[(6×5×43×2×1)×4]+[(6×52×1)(4×32×1)]+[6×4]
= 15 + 80 + 90 + 24

= 209



3. From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?


A. 624

B. 702


C. 756

D. 812




Explanation :

From a group of 7 men and 6 women, five persons are to be selected with at least 3 men.

Hence we have the following 3 choices

We can select 5 men ------(Option 1)
Number of ways to do this = 7C5

We can select 4 men and 1 woman ------(Option 2) 
Number of ways to do this = 7C4 x 6C1

We can select 3 men and 2 women ------(Option 3) 
Number of ways to do this = 7C3 x 6C2 

Total number of ways 
= 7C5 + [7C4 x 6C1] + [7C3 x 6C2]
= 7C2 + [7C3 x 6C1] + [7C3 x 6C2] [Applied the formula nCr = nC(n - r) ]

=[7×62×1]+[(7×6×53×2×1)×6]+[(7×6×53×2×1)×(6×52×1)]
= 21 + 210 + 525 = 756