Quantitative Aptitude - Arithmetic Ability Questions

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.

If there are total 100 candidates,
40 candidates are really capable and
60 candidates are incapable.
Of the candidates who are really capable = 80%
40*(80/100)=32 pass the test and
of the incapable= 30%
60*(30/100)=18 pass the test.
Total candidates who passed the test = 32+18 = 50
The proportion of the really capable students who can pass the test to the total students who can pass = 32/50 = 16/25.

The maximum area of the triangle will come when the given sides are placed at right angles. If we take 8 as the base and 5 as the height the area = ½ x 8 x 5=20

There are 8 students and the maximum capacity of the cars together is 9.

We may divide the 8 students as follows

Case I: 5 students in the first car and 3 in the second Or

Case II: 4 students in the first car and 4 in the second

Hence,     in Case I: 8 students are divided into groups of 5 and 3 in 8C3 ways.

Similarly, in Case II: 8 students are divided into two groups of 4 and 4 in 8C4 ways.

Therefore, the total number of ways in which 8 students can travel is

8C3+8C4 = 56 + 70 = 126.

S = { HHH, HHT, HTH, HTT, THH, THT, TTH, TTT }

=> n(S) = 8

E = { HHH, HHT, HTH, THH }

=> n(E) = 4

P(E) = 4/8 = 1/2

Let the speed of Kaushik be Sk and speed of Piyush be Sp and let the time taken by Piyush be t second then,

 SkSp = 800t+20600t = 23since given ratio

t = 20 s

Speed of Piyush = Sp = 600/20 = 30 m/s

and    Speed of Kaushik = 800/40 = 20 m/s

Here the number of cogs is inveresly proportional to number of revolutions

=> More cogs less revolution
=> 6 : 14 :: x : 21
=> (21x6)/14 = 9

2(l+b)/b = 5/1 

=> 2l + 2b = 5b  

=> 3b = 2l  

=> b =(2/3) x l

 Then, Area = 216 sq.cm 

l x b = 216  

=> l x [(2/3)x l] =216  

=> l x l = 324 

=> l = 18 cm.