Quantitative Aptitude - Arithmetic Ability Questions

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

i=j/m

Maturity value
= PV(1 + i)^n

Term
= 5 years - 21 months=
3.25 years

Price paid
= FV(1+ i )^-n

We know that,

$$\begin{gathered} i2 = -1 \\ i3 = i2 x i = -1 x i = -i \\ i4 = i2 x i2 = -1 x -1 = 1 \\ Hence, i233 =i4 x 58 + 1 = i232 x i1 = 1 x i = i \end{gathered}$$

Probability of getting a tail when a single coin is tossed =12
Probability of getting 4 when a die is thrown =16

Required probability  =(12)×(16)
= 1/12

P = Rs. 15225, n = 9 months = 3 quarters, R = 16% p.a. per quarter.

Amount = 15225x1+41003

= (15225 x 26/25 x 26/25 x 26/25) = Rs. 17126.05

=> C.I. = 17126 - 15625 = Rs. 1901.05.

2 * 5 = 10. Sum of decimal places =  4.

.002  * .5 = 0.0010 = 0.001 

Maximum earning will be only when he will won on the maximum yielding table.

     A ----> 10:1

     B ----> 20:1

     C ----> 30:1

 i.e, he won on B and C but lost on A

 20 x 200 + 30 x 200 -1 x 200 = 9800

 minimum earning will be when he won on  table A and B and lose on that table 3.

Therefore, 10 x 200 + 20 x 200 - 1 x 200  =  6000-200 = 5800

Therefore, Difference= 9800 - 5800 = 4000