Quantitative Aptitude - Arithmetic Ability Questions

let each have base = b and height = h
then p = b*h, R = b*h and T = (1/2) * b*h
so P = R, P = 2T and T = (1/2)*R are all correct statements

Let the the age of the elder boy = E

Let the the age of the younger boy = Y

Given that Y = cube root of EY

=> Y3 = EY => E = Y2 .....(1)
By the condition of number replacement the age of the father is YE

The Mother's age = EY/2

But she is 3 years less than father => EY/2 + 3 = YE
2YE = EY + 6 ......(2)

Then now from the given options we can identify which satisfies the all the conditions.

Here Y =2 and E = 4 satisfies all the conditions.

43 x 616.6/100 + 37 x 217/100

=> 265.138 + 80.29

=>  345.428

Every ball can be distributed in 4 ways. 

Hence the required number of ways = 4 x 4 x 4 x 4 x 4 x 4 = 4096

wine(left)wine(added) = 343169

It means  wine(left)wine(initial amount) = 343512    (since 343 + 169 = 512)

Thus,  343x = 512x1 - 15k3

 343512 = 783 = 1 - 15k3

 1-15k=78=1-18

Thus the initial amount of wine was 120 liters.

Let the salary of Karthik 2 years ago be Rs.100
So, Lucky's salary 2 years ago was Rs.50

Karthik's present salary has seen two 10 percent increases. So as Lucky's salary.

So, Karthik's salary 1 year ago = 100 + 10% of 100 = 100 + 10 = 110.
Karthik's present salary = 110 + 10% of 110 = 110 + 11 = 121.

Lucky's salary 1 year ago = 50 + 10% of 50 = 50 + 5 = 55
Lucky's present salary = 55 + 10% of 55 = 55 + 5.5 = 60.5

Lucky's present salary of 60.5 = 50% of Karthik's present salary of 121.

N = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305)

= H.C.F. of 3360, 2240 and 5600 = 1120.

Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4

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