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.

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 required fraction be x. Then, (1 / x )- x = 9/20
1 - x^(2) / x = 9 / 20  =>  20 - 20 * x^(2) = 9 * x.
20 * x^(2) + 9 *x - 20 = 0.
=> (4 * x + 5) (5 * x - 4) = 0.
=> x = 4 / 5.

As the toys are distinct and not identical,

For each of the 8 toys, we have three choices as to which child will receive the toy. Therefore, there are 58 ways to distribute the toys.

Hence, it is 58 and not 85.

Let their investments be Rs. x for 12 months, Rs. y for 8 months and Rs. z for 6 months respectively.

Then, 12x : 8y : 6z = 4 : 6 : 8

Now, 12x/8y = 4/6   <=>   9x=4y   <=>   y=9x/4

And, 12x/6z = 4/8   <=>   4x=z     <=>    z=4x

Therefore, x : y: z  =  x : 9x/4: 4x  =  4 : 9 : 16

Let the sum be Rs. P. Then,

P1+252x1002-P= 510

P[982- 1] = 510.

Sum = Rs. 1920

So, S.I. = (1920 x 25 x 2) / (2 x 100) = Rs. 480

If t1 and t2 are the upstream and down stream times. Then time taken in still water is given by 

2×t1×t2t1+t2=2×12×2436=16h