Quantitative Aptitude - Arithmetic Ability Questions

perimeter = total cost / cost per m = 10080 /20 = 504m
side of the square = 504/4 = 126m
breadth of the pavement = 3m
side of inner square = 126 - 6 = 120m
area of the pavement = (126 x126) - (120 x 120) = 246 x 6 sq m
cost of pavement = 246*6*50 = Rs. 73800

Here a = 3 and r = 6/3 = 2. Let the number of terms be n.
Then, t = 384 => a * r^(n-1) = 384
=> 3 * 2^(n-1) = 384 => 2^(n-1) = 128 = 2^(7)

=> n-1 = 7 => n = 8.

Let cp= 100,
35 % increase in sp=135
10 % discount in 135((135*10)/100)=13.5
so 1st sp=(135-13.5)=121.5, again 15 % discount in 1st sp((121.5*15)/100)=18.225
2nd sp=(121.5-18.225)=103.275,

so finally cp=100,sp=103.275 ,gain by 3.27%

Assume the weight of alloy a is 100 kg 

Therefore, The weight of alloy B is 400kg

             Gold          silver         copper

    A       40kg          60kg          0kg

    B      140kg         160kg        100kg

  total    180kg          220kg        100kg

Therefore, Ratio of gold and silver in new alloy = 180500:200500= 36% :44%

Speed in downstream = (14 + 4) km/hr = 18 km/hr;

Speed in upstream = (14 – 4) km/hr = 10 km/hr.

Let the distance between A and B be x km. Then,

x/18 + (x/2)/10 = 19 ⇔ x/18 + x/20 = 19 ⇒ x = 180 km.