Quantitative Aptitude - Arithmetic Ability Questions

At first glance it might seem that this problem cannot be solved because we do not have enough
information. It can be solved as long as you double whatever amount you start with. If we start with
$100, then P = $100 and FV = $200.

FV=P(1+r/n)^nt

Probability of sending the correct material is 4/5

Probability of the material not being damaged in transit is 1/4

Probability that there is no short shipment = 2/3

∴ Required probability =4/5 x 1/4 x 2/3 = 2/15

Converting each of the given fractions into decimal form, we get

2/3 = 0.66, 3/4 = 0.75, 4/5 = 0.8, 5/6 = 0.833 

Since 0.833>0.8>0.75>0.66 

So, 5/6 > 4/5 > 3/4 > 2/3 

Therefore,  Required Difference = 5/6 - 2/3 = 1/6

At the time of marriage, let Sudeer's age be 4x, then Swetha's age is 3x
12 years after:
Age of Sudeer = 4x+12
Age of Swetha = 3x+12
Now the equation is 3x+12=5/6(4x+12)
Solving equation we get x=6
Hence Swetha got married at the age of 18.

Let the price of an item be 100After 25% decrease, price of item = 75Increase in price to get the original price = 100-75 = 25% increase to get the original price = 25/75 X 100 = 100/3 = 33.33%

Suppose total earnings = Rs. 10000
10% saving = 10000 x (10/100) = Rs. 1000
This year 5% more income than prev year = Rs. 10500
12% saving = 10500 x (12/100) = Rs. 1260
Percentage of amount saved = (1260/1000) x 100 = 126%

% of milk in first bottle = 64%

% of milk in second bottle = 100 - 26 = 74%

Now, ATQ

64%            74%
         

         68%

6                    4

Hence, by using allegation method,

Required ratio = 3 : 2

Volume of water flowing through 1 pipe of diameter x = Volume discharged by 6 pipes of diameter 12 cms.

As speed is same, area of cross sections should be same. 

Area of bigger pipe of diameter x = Total area of 6 smaller pipes of diameter 12 

i.e πR2 = 6πR12   

Here R = R and R1 = 12/2 = 6   

⇒R2 = 6×6×6   

R = 14.696   

=> D = X = 14.696 * 2 = 29.3938 cm.