Quantitative Aptitude - Arithmetic Ability Questions

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 price of required variety = Rs. P/kg

Then, respective amounts were m kg, m kg and 2m kg

= 126m + 135m + 2pm = 153 x 4m

=> 2p = 351

p = 175.5 / kg

Assume both trains meet after 'p' hours after 7 a.m.
Distance covered by train starting from A in 'p' hours = 20p km
Distance covered by train starting from B in (p-1) hours = 25(p-1)
Total distance = 200
=> 20x + 25(x-1) = 200
=> 45x = 225
=> p= 5
Means, they meet after 5 hours after 7 am, ie, they meet at 12 p.m.

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

Since,each member of the league must meet every other member of the league.If they only played each other once,there would be 8C2 games.Since,each pairing of teams will occur three times,the answer will be triple.

Therfore, 8C2*3=84

let original radius = r and  new radius = (50/100) r = r/2

original area = πr2 and new area = πr/22

decrease in area =  3πr2/4*1πr2 *100 = 75%

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.

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.