Quantitative Aptitude - Arithmetic Ability Questions

Let the length of the train = L mts

Relative speed of train and man = 74 - 8 = 66 kmph = 66 x 5/18 m/s

=> 66 x 5/18 = L/9

=> L = 165 mts.

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.

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

Let Pele x km in 1 hour. so Maradona takes (2h-40min)=1 h 20 min to cover x km. Let speed of Maradona and Pele be M and P respectively than 

                            x =M x (4/3)  and x= P x 1

  => M/P = 3/4 

 Again      300M=300P=1

⇒3003k-3004k=1 

 => k= 25

  M= 3k = 75 km/h   and   P= 4k = 100 km/h 

( Through options it is very easy to solve)

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.

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

we can select the 5 member team out of the 8 in 8C5 ways = 56 ways.

The captain can be selected from amongst the remaining 3 players in 3 ways.

Therefore, total ways the selection of 5 players and a captain can be made = 56x3 = 168 ways.

(or)

Alternatively, A team of 6 members has to be selected from the 8 players. This can be done in 8C6 or 28 ways.

Now, the captain can be selected from these 6 players in 6 ways.

Therefore, total ways the selection can be made is 28x6 = 168 ways.