Aptitude and Reasoning Questions

% Profit with respect to A = 40 + 50 + (40 x 50)/100 = 90 + 20 = 110 %.

Speed of the boat upstream = 36/9 = 4 kmph

Speed of the boat in downstream = 36/6 = 6 kmph

Speed of stream = 6-4/2 = 1 kmph

Time taken by 8th tap = 2 x 2 x 2= 8 hours

Time taken by 12th tap = 2x (1/2) x (1/2)  = 1/2 hour

Ratio of time taken by 8th tap and 12th tap = 8 : 1/2 =16:1

Therefore, Ratio of efficiencies of 8th tap and 12th tap =1:16

In these type of fractions, the decimal moves left the number of zeroes in the divisor times. Herein the given question 18/1000, there are 3 zeros in 1000 (divisor). Hence, the quotient is 0.018.

The quotient of 18/1000 = 0.018.

Time taken by one tap to fill half of the tank = 3 hrs.

Part filled by the taps in 1 hour = 4 x 1/6 = 2/3

Remaining part = 1 - 1/2 = 1/2

2/3 : 1/2 :: 1 : p

p = 1/2 x 1 x 3/2 = 3/4 hrs. i.e., 45 min

So, total time taken = 3 hrs 45 min.

As given, we have:

Susmitha > Laxmi > Ramya; Ramya = Hema, Ramya > Avanthi

Thus the sequence of heights becomes:

Susmitha > Laxmi > Ramya = Hema > Avanthi

Clearly, Hema is shorter than Laxmi.

Let the Original capital amount be Rs. P

Given after 2 years the amount is Rs. 10000

Rate of interest = 30%

According to the question,

$$\begin{gathered} P x 130100 x 130100 = 10000 \\ => P = 1000000169 = 5917 \end{gathered}$$

Therefore, the capital amount P = Rs. 5917

Let the number of Rose plants be ‘a’.
Let number of marigold plants be ‘b’.
Let the number of Sunflower plants be ‘c’.
20a+5b+1c=1000; a+b+c=100

Solving the above two equations by eliminating c,
19a+4b=900

b = (900-19a)/4 

b = 225 - 19a/4----------(1)

b being the number of plants, is a positive integer, and is less than 99, as each of the other two types have at least one plant in the combination i.e .:0 < b < 99--------(2)

Substituting (1) in (2),

 0 < 225 - 19a/4 < 99

225 <  -19a/4 < (99 -225)

=> 4 x 225 > 19a > 126 x 4

=> 900/19 > a > 505

a is the integer between 47 and 27 ----------(3)
From (1), it is clear, a should be multiple of 4.

Hence possible values of a are (28,32,36,40,44)

For a=28 and 32, a+b>100
For all other values of a, we get the desired solution:
a=36,b=54,c=10
a=40,b=35,c=25
a=44,b=16,c=40

Three solutions are possible.