Quantitative Aptitude - Arithmetic Ability Questions

Let the numerator be x Then, denominator =  x + 3.

Now. (x + 4)/(x + 3) +4 = 4/5  <=> 5 (x + 4) =  4(x + 7)

=>  x = 8.

The fraction is  8/11.

Let the total population be 'P'

P x (96/100) = 23040

P = 240 * 100

P = 24000.

150 men in 25 days do = 14 work  

Let 1 man in 1 day does = x  work  

Total work done by 150 men in 25 days = 150x *  25 = 14   work => x = 115000   

Therefore, 100 men in 60 days do = 100 * 60x = 6000x work = 6/15 = 2/5

Total work done =14 + 25 = 1320 

Therefore, Remaining work = 1 - 1320 = 720  

Remaining time = 130 - (25+60+10) = 35 days 

Therefore, work is done in 25 days by 150 men.

Therefore, Work is done in 35 days by 150 men.

Hence, he should employ 50 more men.

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.

Given 40% minimum passing marks for boys = 483 + 117 = 600
=> 1% = 600/40
=> 100% = 600 x 100/40 = 1500
Minimum passing marks for girls = 35% of 1500 = 35 x 1500/100 = 525.