Quantitative Aptitude - Arithmetic Ability Questions

We know that am×an=am+n 

∴3131-27 = 314 = 961 x 961.

We know that,

One square meter is equal to 10,00,000 sq mm.
Given, each mesh is 8mm length and 5mm wide
Therefore for each mesh : 8 x 5 = 40 sq mm.
So, number of meshes = 100000040 = 25000.

From March 6th to November 11th,
there are 250 days and
Thus 250 x 24 hours
So from Mar 6th 9AM to Nov 11th 8PM, there will be 250 x 24 + 11 hours = 6011 hrs.

Given, 

25% of three-seventh of 26% of a number is 145.5

Let the required number be 'N'

i.e, 25100x37x26100xN = 145.5

N = 145.5x7x1000075x26

=> N = 5223.07

7 does not occur in 1000. So we have to count the number of times it appears between 1 and 999. Any number between 1 and 999 can be expressed in the form of xyz where 0 < x, y, z < 9.

1. The numbers in which 7 occurs only once. e.g 7, 17, 78, 217, 743 etc

This means that 7 is one of the digits and the remaining two digits will be any of the other 9 digits (i.e 0 to 9 with the exception of 7)

You have 1*9*9 = 81 such numbers. However, 7 could appear as the first or the second or the third digit. Therefore, there will be 3*81 = 243 numbers (1-digit, 2-digits and 3- digits) in which 7 will appear only once.

In each of these numbers, 7 is written once. Therefore, 243 times.

2. The numbers in which 7 will appear twice. e.g 772 or 377 or 747 or 77

In these numbers, one of the digits is not 7 and it can be any of the 9 digits ( 0 to 9 with the exception of 7).

There will be 9 such numbers. However, this digit which is not 7 can appear in the first or second or the third place. So there are 3 * 9 = 27 such numbers.

In each of these 27 numbers, the digit 7 is written twice. Therefore, 7 is written 54 times.

3. The number in which 7 appears thrice - 777 - 1 number. 7 is written thrice in it.

Therefore, the total number of times the digit 7 is written between 1 and 999 is

243 + 54 + 3 = 300

T                 C              B

16              10             15

8                12             12

128            120           180             <------- in one hour

1280          1200         1800            <------- in 10 hours

Since, restriction is imposed by composers i.e,since only 1200 books can be composed i 10 hours so not more than 1200 books can be finally pepared.

Let x, y, z be the number of balls received by the three persons, then

x≥5, y≥5, z≥5 and x+y+z=21

Let u≥0, v≥0, w≥0 then

∴ x + y + z =21

⇒ u + 5 + v + 5 + w + 5 = 21

⇒ u + v + w = 6 

∴Total number of solutions = C3-16+3-1=C28=28

x = y  <=> 1 - q =  2q  + 1  <=> 3q  = 0 <=> q = 0.