Quantitative Aptitude - Arithmetic Ability Questions

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.

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

Probability of left handed = 4/5
total student = 40
total left handed = 40 x(4/5)
= 32

There are 12 - 3 = 9 patients.They can be seen in 12P9 = 79833600 ways

The question requires you to find number of the outcomes in which at most 3 coins turn up as heads.

i.e., 0 coins turn heads or 1 coin turns head or 2 coins turn heads or 3 coins turn heads.

The number of outcomes in which 0 coins turn heads is 6C0=1 

The number of outcomes in which 1 coin turns head is =6C1=6 

The number of outcomes in which 2 coins turn heads is6C2=15 

The number of outcomes in which 3 coins turn heads is6C3=20

Therefore, total number of outcomes =1+6+15+20= 42 outcomes

Let their ages one year  ago be 4x and 3x years

 Sum of their present ages =(4x +1+3x+1) = 16 years.

Given principal amount = Rs. 8000

Time = 3yrs

Rate = 5%

C.I for 3 yrs = $$\begin{gathered} 8000 x 105100 x 105100 x 105100 - 8000 \\ = Rs. 1261 \end{gathered}$$

Now, C.I for 2 yrs = $$\begin{gathered} 8000 x 105100 x 105100 \\ = Rs. 820 \end{gathered}$$

Hence, the required difference in C.I is 1261 - 820 = Rs. 441

Speed = 44 kmph x 5/18 = 110/9 m/s

We know that, Time = distance/speed

Time = (360 + 140) / (110/9)

= 500 x 9/110 = 41 sec.