Quantitative Aptitude - Arithmetic Ability Questions

Average speed =  km/hr

                      =   km/hr

6P4 = 6! / (6-4)! = 360

Let
P's capital = p,
Q's capital = q and
R's capital = r.
Then

6p = 8q = 12r
=> 3p = 4q = 6r
=>q = 3p/4
r = 3p/6 = p/2
P : Q : R = p : 3p/4 : p/2
= 4 : 3 : 2
R's share = 4650 * (2/9) = 150*6 = Rs. 1033.33.

Number of questions = 5
Possibilities of choices for each question 1 to 5 respectively = 4, 4, 4, 6, 6

Reuired total number of sequences

= 4 x 4 x 4 x 6 x 6

= 2304.

Here the LCM of 54, 60 is

=> 2 x 3 x 3 x 3 x 10

= 6 x 9 x 10

= 54 x 10

= 540.

S = { 1, 2, 3, 4, .....18 } 

=> n(S) = 18

E1 = {2, 4, 6, 8, 10, 12, 14, 16, 18}

=> n(E1) = 9

E2 = {3, 6, 9, 12, 15, 18 }

=> n(E2) = 6

 E3 =E1∩E2={6, 12, 18} 

=> n(E3) = 3

∴E=E1 ∪ E2 = E1+E2-E3 

=> n(E) = 9 + 6 - 3 =12

 where E = { 2, 3, 4, 6, 8, 9, 10, 12, 12, 14, 15, 16, 18 }

∴P(E)=n(E)n(S)=1218=23

25 Jan, 1981 = (1980 years + Period from 1.1.1981 to 25.1.1981)

Odd days in 1600 years = 0

Odd days in 300 years = 1

80 years = (60 ordinary years + 20 leap year) = (60 x 1 + 20 x 2)= 2 odd days

Jan = 25 days = (3 weeks + 4 days) = 4 odd days.

Total number of odd days = (0 + 1 + 2 + 4) = 7= 0 odd day.

Given day is Sunday.

Let 'A' be the cost price of property P1.

Then from the given data, the selling price of P1 = Rs. 1,00,000

He got 20% loss on selling P1

$$\begin{gathered} ⇒A - 20A100 = 100000 \\ ⇒A = 1,25,000 \end{gathered}$$

Therefore, the amount he lossed on selling P1 = 25,000

As ge he got no loss or gain on sale of P1 and P2, the gain on selling P2 = 25,000

But the selling price of P2 = 1,00,000 => Cost price of P2 = 75,000

Hence, the profit percentage on P2 = GainCPx 100 = 2500075000x100 = 1003= 33.33%