Quantitative Aptitude - Arithmetic Ability Questions

Given 5:6 :: 7:10 :: 6:5

We know that compound ratio is calculated as
a:b :: c:d :: e:f
(axcxe):(bxdxf)

=> (5x7x6):(6x10x5)
=> (210):(300)
= 7:10

Let required fraction be  then 

           ............. (i)

and          ............ (ii)

Solving (i) and (ii) we get x=3, y=5

 Required fraction = 

Suppose A, B and C take x, x/2 and x/3 respectively to finish the work.
Then, (1/x + 2/x + 3/x) = 1/2
6/x = 1/2 => x = 12
So, B takes 6 hours to finish the work.

As the win leads to multiplying the amount by 1.5 and loss leads to multiplying the amount by 0.5, we will multiply the initial amount by 1.5 thrice and by 0.5 thrice (in any order).

The overall resultant will remain same.

So final amount with the person will be (in all cases):

= 64(1.5)(1.5)(1.5)(0.5)(0.5)(0.5)= Rs. 27

Hence the final result is: 

64 − 27 = 37

A loss of Rs.37

Given exp . is like this 854*854*854-276*276*276854*854+854*276+276*276=?

so ans = 854 - 276 = 578

Let the age of sushma be x and
the age of her son is y
Then five years before x-5=5(y-5) ...(1)
Five years hence x+5 = 3(y+5)-8 .....(2)

By soving (1) & (2), we get
5y - 15 = 3y + 7
y = 11 => x = 35

Therefore, the age of Sushma = 35 and her son = 11.

Given 30% of 500

$$\begin{gathered} = 30100 x 500 \\ = 30 x 5 \\ = 150 \end{gathered}$$

Hence, 30% of Rs. 500 = Rs. 150.

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