AIEEE Questions

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Let sum = S. Then, amount = 7S/6

S.I. = 7S/6 - S = S/6; Time = 3 years.

Rate = (100 x S) / (S x 6 x 3) = 5 5/9 = 50/9 %.

The given number series is 1 8 36 ? 1100 7701

It follows a pattern that,

1

1 x 3 + 5 = 8

8 x 4 + 4 = 36

36 x 5 + 3 = 183

183 x 6 + 2 = 1100

1100 x 7 + 1 = 7701

Hence, the missing number in the given series is 183.

The given number series 11, 12.2, 14.6, 18.2, ?

11

11 + 1.2 = 12.2

12.2 + 2.4 = 14.6

14.6 + 3.6 = 18.2

18.2 + 4.8 = 23

Hence, the next number in the given number series is 23.

The given series is 118, 166, 190, 202, 208, ?

118

118 + 48 = 166

166 + 24 = 190

190 + 12 = 202

202 + 6 = 208

208 + 3 = 211

350 days = (350/7=50 weeks) i.e  No odd days,

 So it will be a Saturday.

I am Prince of a Kingdom. I will be king only if the man who gave me birth i.e, king my father dies.

5 C in an OF denotes 5 Circles in an Olympic Flag.

These are riddles of type

12 S of the Z

52 W in a Y.

Brightness of bulb depends on both voltage and current. The brightness depends on power.

Since we know that  Power (P) = Voltage x Current = V I

So whether you increase the current or the voltage the brightness will increase.

Example Illustration:

 If you connect a long bulb and a round bulb in series, the long bulb glows but the round bulb does not. Both of them have the same current flowing through them (they are in series) but the long bulb has a greater voltage across it because it has more resistance (even though V= IR is not obeyed exactly, voltage drop does depend on the resistance). If you connect a round bulb and a long bulb in parallel, however, it is the round bulb that is brighter because now the voltage across both bulbs is the same but the current is larger through the round bulb (smaller resistance).

Hence, the brightness of a bulb depends on power i.e, both voltage and current.