Quantitative Aptitude - Arithmetic Ability Questions

Price of 1kg fresh cashew = Rs.120
Therefore, price of 10kg = Rs.120 x 10 = Rs.1200

10% of 10kg which eaten by rodents = 10 x 10/100 = 1kg.

So, End of Year he had 10kg - 1kg = 9kg

So, he sells that 9kg with Rs.20 per kg profit = 9 x 140 = 1260.

He buy 10kg wooden box with cashew for Rs.1200 and sell that for Rs. 1260

So, profit = 1260 - 1200 = Rs.60 more money he get.

let, speed of Ali=X mph and Faizer=Y mph

Now, relative speed when Ali is walking int the same direction with Faizer= (X-Y)mph
hence, according to question, 27/(X-Y)=9 ; solving X-Y=3; --- (1)

Now relative speed when walking in the opposite direction= (X+Y) mph
hence, according to question, 27/(X+y)=3; solving X+Y=9; --- (2)

solving 1 and 2 we get X=6 and Y=3, where X is the speed of Ali.

Let breadth = x metres.

Then, length = (x + 20) metres.

Perimeter = 5300 m = 200 m. 26.50

2[(x + 20) + x] = 200

2x + 20 = 100

2x = 80

x = 40.

Hence, length = x + 20 = 60 m.

Here to say which fraction is greater from the given options, the fraction with greater numerator with same denominator or smaller denominator with same numerator.

To find the greater fraction, among the given options, make the denominators same for the options by finding the LCM of all the denominators.

7/12, 9/16, 13/20, 32/80

Now, LCM of 12, 16, 20, 80 = 240.

Then,

140/240, 135/240, 156/240, 96/240

Here 

$$\begin{gathered} 156240 > 140240 > 135240 > 96240 \\ => 1320 > 712 > 916 > 3280 \end{gathered}$$

Hence, largest fraction = 13/20.

Let the number of males be 'M'
Let the number of females be 'F'
From the given conditions,
(F-15)=2M.....(i)
F=5(M-45).....(ii)
By solving the equations (1) & (2) we get M = 80
Then substitute M = 80 in (1)
F = 2M + 15
F = 2 x 80 + 15
=> F = 175
Therefore, number of males = 80 and females = 175 in the party.