Quantitative Aptitude - Arithmetic Ability Questions

Upstream speed = B-S

Downstream speed = B+s

B-S = 15/5 = 3 km/h

Again          B= 4S

Therefore    B-S = 3= 3S

=>             S = 1 and B= 4 km/h

Therefore    B+S = 5km/h

Therefore, Time during downstream = 15/5 = 3h

Let Meena’s age = A.
Then Sivagami’s age = A + 2
After 6 years the total of their ages will be 7 times of what?
Not clear.
So the given data are inadequate.

C.P = Rs.3000 

S.P =Rs. [3600 x 10] / [100+(10 x 2)] = Rs.3000 

Gain =0%

Let breadth = x metres, length = 3x metres, height = H metres. 

Area of the floor=(Total cost of carpeting)/(Rate) = (270/5) sq.m = 54 sq.m 

x×3x2=54⇒x2=54×2⇒x=6  

So, breadth = 6 m and length =362 = 9 m.

Now, papered area = (1720/10) =  172 sq.m 

Area of 1 door and 2 windows = 8 sq.m 

Total area of 4 walls = (172 + 8) sq.m = 180 sq.m 

2×9+6H=180⇒H=6

Let Arun's weight be X kg.

According to Arun, 65 < X < 72.

According to Arun's brother, 60 < X < 70.

According to Arun's mother, X < 68.

The values satisfying all the above conditions are 66 and 67.

Required average = (66 + 67) / 2 = 66.5 kg

NOTE :

Repetition of leap year ===> Add +28 to the Given Year.

Repetition of non leap year

Step 1 : Add +11 to the Given Year. If Result is a leap year, Go to step 2.

Step 2:  Add +6 to the Given Year.

Solution : 

Given Year is 1993, Which is a non leap year.

Step 1 : Add +11 to the given year (i.e 1993 + 11) = 2004, Which is a leap year. 

Step 2 : Add +6 to the given year (i.e 1993 + 6) = 1999

Therfore, The calendar for the year 1993 will be same for the year 1999

             A              :               B
(8000x4)+(4000x8) : (9000x6)+(6000x6)
         64000           :      90000
=>        32 : 45

Let the large number be x.

Then x - 20 = 20x/100

=> x - x/5 = 20  => x = 25.