Quantitative Aptitude - Arithmetic Ability Questions

4/10 + 9/x = 1     => x = 15 

Then both can do in 

1/10 + 1/15 = 1/6     => 6 days

Let the two parts be (54 - x) and x.

Then, 10 (54 - x) + 22x = 780

=> 12x = 240  

=>  x = 20.

Bigger part = (54 - x)  = 34.

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

Let 'd' be the distance and 's' be the speed and 't' be the time
d=sxt
45 mins = 3/4 hr and 48 mins = 4/5 hr
As distance is same in both cases;
s(3/4) = (s-5)(4/5)
3s/4 = (4s-20)/5
15s = 16s-80
s = 80 km.
=> d = 80 x 3/4 = 60 kms.

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.

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 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.

$$\begin{gathered} 23x35x34x?=3174 \\ 3174x10/3 \\ = 10580 \end{gathered}$$