Quantitative Aptitude - Arithmetic Ability Questions

The largest 4 digit number is = 9999
After dividing 9999 with 88, we get 55 as remainder
so largest 4 digit number divisible by 88 = 9999-55 = 9944.

The first letter is E and the last one is R.

Therefore, one has to find two more letters from the remaining 11 letters.

Of the 11 letters, there are 2 Ns, 2Es and 2As and one each of the remaining 5 letters.

The second and third positions can either have two different letters or have both the letters to be the same.

Case 1: When the two letters are different. One has to choose two different letters from the 8 available different choices. This can be done in 8 * 7 = 56 ways.

Case 2: When the two letters are same. There are 3 options - the three can be either Ns or Es or As. Therefore, 3 ways.

Total number of possibilities = 56 + 3 = 59

60 is not a multiple of 8

Less Cogs  more turns and less time  less turns

$$\begin{gathered} Cogs Time Turns \\ A 54 45 80 \\ B 32 8 ? \end{gathered}$$

Number of turns required=80 × 54/32 × 8/45  = 24 times

We know that I = PTR/100
=> P = 20250/4.5 = 4500
Now, new Interest at 5% = 4500x1x5/100 = 225
Now the additional amount = 225 - 202.5 = Rs. 22.5

Work done by the waste pipe in 1 minute =115-120+124=115-11120 = -140 [-ve sign means emptying]

Volume of 140part = 3 gallons

Volume of whole = (3 x 40) gallons = 120 gallons.

Let the filling capacity of the pump be x m3/min. 

Then, emptying capacity of the pump=(x+10)m3/min.  

so,2400x-2400x+10=8 ⇔x2+10x-3000=0 

⇒x-50+x+60=0 ⇔x=50