Q.steve gets on the elevator at the 11th floor of a building and rides up at a rate of 57 floors per min. At the same time, joyce gets on an elevator on the 51st floor of the same building and rides down at a rate of 63 floors per min. If they continue travelling at the rates, then at which floor will their paths cross?
a)19 b)28 3)30 4)32
sol: divide the rate with common divisible number
here for 57 and 63 we can divide by 3, then 57/3=19 and 63/3=21
now add with floor i.e., 11+19=30(up) and 51-21=30(down)

Numbers Divisible by 7

Q:When 2 raised to the power of 256 is divided by 17, the remainder would be?
sol:2 raised to 256=16 raised to 64
16 raised to even number divided by 17 always gives a remainder 1
this is from remainder theorem

The codes which when turned upside down gives a number i.e.,
01-10
06-90
08-80
09-60
16-91
18-81
19-61
66-99
68-89
86-98
->so from 0-99 there are 20 such numbers which can create confusion
ex:an intelligence agency decides on a2-digits selected from 0, 1, ....9 so that 1st digit is non-zero.The code, handwritten on a slip, can create confusion when read upside down.How many codes are there for which there is no confusion.
a.80 b. 63
c.71 d.69
sol:The 2-digit number is selected from 10-99(90 codes)
out of these remove 9 codes viz 11,22,33,44,55,66,77,88,99
remaining 81
the codes which can create confusion are 16-91,18-81,19-61,66-99(we have already removed it),68-89,86-98 total 10
therefore ans is 81-10=71.

Guys all of you know the formula n(n+1)/2 for sum of n natural numbers .

Here is some useful info to solve some typical prob quickly

The sum of 1st 10 natural numbers =55
sum of next 10 natural numbers(11-20)=155
sum of next i.e., (21-30)=255.................................it goes on 355, 455, 555, 655 etc

->sum of 1000 natural numbers=(100*101)/2=>5050 i.e.,(55+155+255+355+455+555+655+755+855+955)

ex:A boy starts adding consecutive natural numbers starting with 1, after some time he reaches a total of 1000, and realizes that he has made an error of double counting one number. find the double counted number.
ans:by using the above mentioned info
the sum of 1st 40 natural numbers=55+155+255+355=820
now you go on adding the remaining numbers till you reach near 1000
820+40+41+42+43+44=990
so the required ans is 10 (1000-990=10)
since double count will always have the effect of increasing the sum

If A can finish a work in X time and B can finish the same work in Y time then both of them together can finish that work in (X*Y)/ (X+Y) time.

If A can finish a work in X time and A & B together can finish the same work in S time then B can finish that work in (XS)/(X-S) time.

If A can finish a work in X time and B in Y time and C in Z time then all of them working together will finish the work in (XYZ)/ (XY +YZ +XZ) time

If A can finish a work in X time and B in Y time and A, B & C together in S time then

· C can finish that work alone in (XYS)/ (XY-SX-SY)

· B+C can finish in (SX)/(X-S); and

· A+C can finish in (SY)/(Y-S)