GRE Questions

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9 P in SS  is a short form for 9 Planets i  Solar System.

There are many more of this kind. Some of them are:

1. 7 C in a R        -    7 Colours in a Rainbow

2.  90 D in a R A   -    90 Degrees in a Right Angle

3.  52 W in a Y     -    52 Weeks in a Year

4.  5 C in an OF

5.  12 S of the Z

Let the weight of the empty tin = w

(2 - w) = weight of the cookies; and one quarter of the cookies weigh (2 - w)/4

One quarter of the cookies + tin = 0.8 = w + (2 - w)/4

Multiply through by 4; 3.2 = 4w + 2 - w

1.2 = 3w; w = 0.4

For solving this problem first we would break the whole range in 5 sections

1) From 1 to 9
Total number of zero in this range = 0

2) From 10 to 99
Total possibilities = 9*1 = 9 ( here 9 is used for the possibilities of a non zero integer)

3) From 100 to 999 - three type of numbers are there in this range
a) x00 b) x0x c) xx0 (here x represents a non zero number)
Total possibilities
for x00 = 9*1*1 = 9, hence total zeros = 9*2 = 18
for x0x = 9*1*9 = 81, hence total zeros = 81
similarly for xx0 = 81
total zeros in three digit numbers = 18 + 81 +81 = 180

4) From 1000 to 9999 - seven type of numbers are there in this range
a)x000 b)xx00 c)x0x0 d)x00x e)xxx0 f)xx0x g)x0xx
Total possibilities
for x000 = 9*1*1*1 = 9, hence total zeros = 9*3 = 27
for xx00 = 9*9*1*1 = 81, hence total zeros = 81*2 = 162
for x0x0 = 9*1*9*1 = 81, hence total zeros = 81*2 = 162
for x00x = 9*1*1*9 = 81, hence total zeros = 81*2 = 162
for xxx0 = 9*9*9*1 = 729, hence total zeros = 729*1 = 729
for xx0x = 9*9*1*9 = 729, hence total zeros = 729*1 = 729
for x0xx = 9*1*9*9 = 729, hence total zeros = 729*1 = 729
total zeros in four digit numbers = 27 + 3*162 + 3*729 = 2700
thus total zeros will be 0+9+180+2700+4 (last 4 is for 4 zeros of 10000)
= 2893

Let Cost Price(C.P) = P
gain% = {(S.P-C.P)/C.P} x 100
25 = {(1540-P)/P} x 100
25/100 = (1540-P)/P
=> P = 4(1540)-4P
=> 5P = 4(1540)
=> P = 1232
So, Cost Price = Rs. 1232

Pageant is most similar in meaning to Show.

Total number of female members = 115

Let the number of males = x

Number of females = y

=> x + y = 115  ......(1)

According to the given data,

x + y(0.50) = 69 ......(2)

From (1) & (2), we get

0.5y = 46

=> y = 46 x 2 = 92

=> x = 115 - 92 = 23

Hence, required number of gentlemen in the family = x = 23.

Assuming Sripad has scored the least marks in subject other than science,
Then the marks he could secure in other two are 58 each.
Since the average mark of all the 3 subject is 65.
i.e (58+58+x)/3 = 65
116 + x = 195
x = 79 marks.

Therefore, the maximum marks he can score in maths is 79.

Clearly, we may mark such sets of 3 numbers, in which the middle number is 8 and each of the two numbers on both sides of it is a factor of 8, as shown below :

2 8 3 8 2 4 8 2 4 8 6 8 2 8 2 4 8 3 8 2 8 6

So, there are two such 8's. Hence the answer is C.