Quantitative Aptitude - Arithmetic Ability Questions

Given numbers are 1, 2 and 3

Factors of 1 = 1

Factors of 2 = 1, 2

Factors of 3 = 1, 2 and 3

Common factors = 1

Hence, the HCF of 1, 2 and 3 = 1.

Since there are 5 cups of each kind,prepared with milk or tea leaves added first,are identical hence,total number of different people ways of presenting the cups to the expert is 10!/(5! x 5!)= 252

FV=P(1+r/n)^nt

Let the cost price of the item = Rs. 50

Sold at 10% loss => for Rs. 50 loss  S.P = Rs. 45

From the  given data,

25/2 % gain if it is sold at Rs. 9 more

56.25 - 45 = 9

=> 11.25 = 9

=> 22.5 = 18

=> 45 = 36

=> 50 = ?

=> C.P = 50 x 3645 = 40

Hence, the Cost price of the item = Rs. 40.

Total hours worked = 8 x 3 + 6 x 2 = 36

Total earned = 324. Hourly wage = 324 / 36 = 9

Volume of cuboid = (24 x 9 x 8) cm = 1728 cu.cm
Volume of small cube = (3 x 3 x 3) cm = 27 cu.cm

So, No. of small cubes formed = 1728/27 = 64.

Number of positive integral solutions =  n-1Cr-1 = C220 = 190

Probability between z = 0 and z = 3.01 is given by

P(0<z<3.01) = P(z<3.01) - P(z<0)

Reading from the z-table, we have

P(z<0) = 0.5

P(z<3.01) = 0.9987

Hence, P(0<z<3.01) = 0.9987 - 0.5 = 0.4987.