Quantitative Aptitude - Arithmetic Ability Questions

Let marks obtained by N = x

M's marks = 0.80x

L's marks = 1.25 (0.80x)

K's marks = (0.90) (1.25) (0.80x)

= 0.9x

But 0.9x = 360 ⇒ x = 400

Percentage of Marks obtained by N

= 400 x 100/500 = 80%

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's age 10 years ago be x years 

Then B's age 10 years ago  be 2x yeats

 Total of their present ages =                                        

                                          = 

                                          = 35 years

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)

In the question it is asked for 5% of 1500

5 x 1500/100 = 15 x 5 = 75.

Hence, 5% of 1,500 is equal to 75.

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

Cash realised= Rs. (106.25 - 0.25)
                      = Rs. 106.