Quantitative Aptitude - Arithmetic Ability Questions

Let the length of the train be x metres and its speed by y m/sec.

Then, x/y = 8 => x = 8y

Now, (x+264)/20 = y
8y + 264 = 20y
y = 22.
Speed = 22 m/sec = (22*18/5) km/hr = 79.2 km/hr.

4/10 + 9/x = 1     => x = 15 

Then both can do in 

1/10 + 1/15 = 1/6     => 6 days

let x and y be the sides of the rectangle then 

correct area =105100x×96100y 

Error% = 504500xy-xy=4500xy %

4500xy×1xy×100=45=0.8

Let C1 be the cost price of the first article and C2 be the cost price of the second article.
Let the first article be sold at a profit of 22%, while the second one be sold at a loss of 8%.

We know, C1 + C2 = 600.
The first article was sold at a profit of 22%. Therefore, the selling price of the first article = C1 + (22/100)C1 = 1.22C1
The second article was sold at a loss of 8%. Therefore, the selling price of the second article = C2 - (8/100)C2 = 0.92C2.

The total selling price of the first and second article = 1.22C1 + 0.92C2.

As the merchant did not make any profit or loss in the entire transaction, his combined selling price of article 1 and 2 is the same as the cost price of article 1 and 2.

Therefore, 1.22C1 + 0.92C2 = C1+C2 = 600
As C1 + C2 = 600, C2 = 600 - C1. Substituting this in 1.22C1 + 0.92C2 = 600, we get
1.22C1 + 0.92(600 - C1) = 600
or 1.22C1 - 0.92C1 = 600 - 0.92*600
or 0.3C1 = 0.08*600 = 48
or C1 = 48/(0.3) = 160.
If C1 = 160, then C2 = 600 - 160 = 440.
The item that is sold at loss is article 2. The selling price of article 2 = 0.92*C2 = 0.92*440 = 404.80.

Let the three integers be x, x + 2 and x + 4. Then, 3x = 2 (x + 4) + 3  <=> x = 11.

Third integer = x + 4 = 15.

Minimum number of chocolates are possible when he purchases maximum number of costliest chocolates.

Thus,          2 x 5 + 5 x 2 =Rs.20

Now Rs.100 must be spend on 10 chocolates as 100 = 10 x 10.

Thus minumum number of chocolates = 5 + 2 + 10 = 17