Quantitative Aptitude - Arithmetic Ability Questions

cost price = (225 + 15) = 240 sell price = 300
gain = (60/240)*100 = 25%

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.

Area of the square =12*diagonal2=12*3.8*3.8=7.22m2

Let last digit is 2
when second last digit is 4 remaining 4 digits can be filled in 120 ways, similarly second last digit is 6 remained 4 digits can be filled in 120 ways.
so for last digit = 2, total numbers=240

Similarly for 4 and 6
When last digit = 4, total no. of ways =240
and last digit = 6, total no. of ways =240
so total of 720 even numbers are possible.

P.W. = Rs. (540 - 90) = Rs. 450.

S.I. on Rs. 450 = Rs. 90.

S.I. on Rs. 540 = Rs.( 90 /450)x 540 = Rs. 108.

B.D. = Rs. 108.

Now as per question, Karan invested for 12 months and Satish invested for 7 months.

So, Karan:Satish = (29000 x 12):(18000 x 7)

= 348:126

= 58:21

Satish Ratio in profit will be= 16970×2179 = Rs. 4511

Here S = {1,2,3,4,5,6}

Let E be the event of getting the multiple of 3

Then, E = {3,6}

P(E) = n(E)/n(S) = 2/6 = 1/3 

Efficiency of P= 100/20= 5% per hour  

Efficiency of Q= 100/25= 4% per hour  

Efficiency of R= 100/40= 2.5% per hour  

Efficiency of S=100/50= 2% per hour 

Cistern filled till 10 am by P, Q and R  

Till 10.00am Pipe P filled 20%Till 10.00am Pipe Q filled 8%Till 10.00am Pipe R filled 2.5%30.5%  

Thus, at 10 am pipe P,Q and R filled 30.5% of the cistern. 

Rest of cistern to be filled = 100 - 30.5 = 69.5%  

Now, the time taken by P,Q,R and S together to fill the remaining capacity of the cistern 

= 69.5 / (5+4+2.5+2) = 5 Hours and 9 minutes(approx).

Therefore, total time =4 hrs + 5hrs 9 mins = 9 hrs and 9 mins

It means cistern will be filled up at 3:09 pm