Quantitative Aptitude - Arithmetic Ability Questions

Profit received by Chinna as working partner = 14.5% of Rs. 19600

= 14.5x19600/100 = Rs. 2842

Balance in profit = 19600-2842 = Rs. 16758

Ratio of investment of Chinna & Munna = 80,000 : 1,40,000 = 4 : 7

Hence share of Chinna in investment = 4x16758/100 = Rs. 6093.85

Therefore, Share of Munna = 19600 - 2842 - 6093.85 = Rs. 10664.15

Total age of players in team = 22 x 11 = 242
When coach is included, total members = 12, Avg age increases by one becomes 23
Total age when coach is included = 23x12 = 276
Age of coach = 276-242 = 34 yrs.

Let the 6 digits of the required 6 digit number be abcdef

Then, the number to be odd number the last digit must be odd digit i.e 3, 5, 7 or 9

The first digit cannot be ‘0’ => possible digits = 3, 5, 7, 2, 6, 8

Remaining 4 places can be of 6 x 5 x 4 x 3 ways

This can be easily understood by

Therefore, required number of ways = 6 x 6 x 5 x 4 x 3 x 4 = 36 x 20 x 12 = 720 x 12

= 8640 ways.

Let the total earnings be Rs.x 

Then 412% of x + 14% of (x-12000) =2400

 ⇒92×x100+14×x-12000100=2400

=>  x=50,589.47

The toys are different; The boxes are identical
If none of the boxes is to remain empty, then we can pack the toys in one of the following ways
a. 2,2,1
b. 3,1,1

Case a. Number of ways of achieving the first option 2−2–1

Two toys out of the 5 can be selected in5C2 5 ways. Another 2 out of the remaining 3 can be selected in3C2 ways and the last toy can be selected in 1C1  way.

However, as the boxes are identical, the two different ways of selecting which box holds the first two toys and which one holds the second set of two toys will look the same. Hence, we need to divide the result by 2.

Therefore, total number of ways of achieving the 2−2–1 option is:

5C2*3C22=10*32=15 ways.

Case b. Number of ways of achieving the second option 3−1–1

Three toys out of the 5 can be selected in 5C3  ways. As the boxes are identical, the remaining two toys can go into the two identical looking boxes in only one way.
Therefore, total number of ways of getting the 3−1–1 option is5C3=10 ways.

Total ways in which the 5 toys can be packed in 3 identical boxes
=number of ways of achieving Case a + number of ways of achieving Case b
=15 + 10 = 25 ways

Total age seven persons = (26 x 7)years
Total age of the first three persons and the last three persons are (19 x 3) years and (32 x 3) years respectively.
Age of the person sitting in the middle of the row = 26 x 7 - 19 x 3 - 32 x 3 = 182 - 57 - 96 = 29 years.

Let the numbers be x, x+1, x+2 

Then x + x+1 + x+2 = 42

           =>  3x + 3 = 42

           => 3x = 39 

          =>  x = 13

The numbers would be 13, 14, 15 

Therefore, Sum of the digits of middle number = 1+4 = 5

Relative speed = 120 + 80 = 200 km/hr.
= 200 x 5/18 = 500/9 m/sec.

Let the length of the other train be L mts.
Then, (L + 260)/9 = 500/9 => L = 240 mts.