Quant Boosters  Shashank Prabhu, CAT 100 Percentiler  Set 5

Q27) A man, starting from a point P, takes six equal steps. Each step is in one of the four directions – East, West, North and South. What is the total number of ways in which the man ends up at point P after the six steps?
(a) 200
(b) 256
(c) 400
(d) 512

EEEWWW =6!/(3!)^2 = 20
SSSNNN = 6!/(3!)^2 = 20
NNSSEW = 6!/(2!)^2 = 180
EEWWNS=6!/(2!)^2 = 180
Total = 400

Q28) The function f(x) is defined for all positive values of x and y as f(xy)=f(x)+f(y). Also, f(2)=2 and f(3)=3. What is the value of f(32/27)?

f(4)=f(2)+f(2)
f(4)=4
f(16)=f(4)+f(4)=8
f(32)=f(16)+f(2)=10
f(9)=f(3)+f(3)=6
f(3)=f(9)+f(1/3)
f(1/3)=3
f(1/9)=f(1/3)+f(1/3)=6
f(1/27)=f(1/9)+f(1/3)=9
f(32/27)=f(32)+f(1/27)=109=1Short cut: here f(x) = log(x) => log(32/27) = log 32 – log 27 = 5log2 – 3log3 = 5f(2) – 3f(3) = 10 – 9 = 1

Q29) A dealer offers a discount of 20% and still makes a profit of 20%, even when he further allows 16 articles to a dozen to a particular sticky customer. How much % is his items marked up?

Let MP be 1 Rs per article. For a dozen, MP will be 12 Rs but a discount of 20% on this will make it 9.6 Rs. But, he has sold 16 articles. That makes SP of one article as 9.6/16 = 0.6. As he is making 20% profit, 1.2CP = 0.6 which will make CP = 0.5 hence, 100% markup on Cost Price

Q30) The area of triangle ABE is 2sq. cm and that of triangle BEF is3sqcm.What is the area of blue shaded figure ACDE?

AE:FE=2:3
AB:DF=2:3
A(ABE)/A(DEF)=(AB/DF)^2=4/9
A(DEF)=4.5
A(BDF)=7.5
A(FBCD)=15
A(AECD)=5.5

@shashank_prabhu @zabeer mechanic has to check the first two machines only if he is able to identify 2 good or 2 defective machines, he can catch his bus; otherwise he cant. So we have to consider the cases with only first 2 machines; that gives me gg, dd, gd, dg ie 4 total cases and not 6. My answer is 2/4. Can you please explain why we have to consider the arrangements of ddgg ?

@VikrantGarg
Probability of catching the bus = Probability of getting faulty machines in the first 2 tests (this part is clear right?)
This is possible in 2 ways  if we get DD or if we get GG as the result of first 2 tests.
In your doubt, if we get DG or GD we will require to do one more test, which would make the mechanic miss the bus. So DG and GD are not favourable cases.
So total number of ways = 4!/(2! x 2!) = 6 ways.
Favourable cases = 2
Probability = 2/6 = 1/3
Is it clear ?