GRE Questions

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Let the number of sides be n. 

The number of diagonals is given by ⁿC₂ - n  

^{Therefore, n}C₂ - n = 44, n>0 

ⁿC₂ - n = 44

n² - 3n - 88 = 0 

n² -11n + 8n - 88 = 0  

As n>0, n will not be -8. Therefore, n=11.

If we count letters till the end of the alphabet including that letter in reverse order then
Here the values are nothing but the values corresponding to the alphabets when taken in a reverse order,i.e,A=26 ,B=25...,Z=1
G = 20 and O = 12 totaling 32
S = 8, H = 19 and E = 22 totaling 49

So
SOME = 8 + 12 + 14 + 22 = 56.

2500 in 5th year and 3000 in 7th year
So in between 2 years Rs. 500 is increased => for a year 500/2 = 250
So, per year it is increasing Rs.250 then in 5 years => 250 x 5 = 1250
Hence, the initial amount must be 2500 - 1250 = Rs. 1250

If you add all the terms of the LHS of equality, then the last letter of the sum written in words would give you the RHS.

1 + 1 + 1 + 1 = 4 = FOUR

2 + 2 + 2 + 2 = 8 = EIGHT

3 + 3 + 3 + 3 = 12 = TWELVE

4 + 4 + 4 + 4 = 16 = SIXTEEN

So, following the same pattern.

5 + 5 + 5 + 5 = 20 = TWENTY

That gives us our answer to 5 5 5 5 =  "Y".

Here, x = 20 and y = - 20
Therefore, the net % change in value
=x + y + xy100 %
= 20 + -20 + 20 × -20100%  or  - 4%

Since the sign is negative, there is a decrease in value by 4%.

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.

Given number sequence is 28, 16, 28, 76, 274

Here the sequence follows a pattern that

28

28 x 0.5 + 2 = 14 + 2 = 16

16 x 1.5 + 4 = 24 + 4 = 28

28 x 2.5 + 6 = 70 + 6 = 76

76 x 3.5 + 8 = 266 + 8 = 274

274 x 4.5 + 10 = 1233 + 10 = 1243

Hence, the next number in the given number sequence is 1243.

S.I

= Rs.(1200x10x1/100)

= Rs.120.

C.I

= Rs[(1200x1+5/100)² -1200]

= Rs.123. 

Difference

= Rs.[123-120]

= Rs. 3.