Quantitative Aptitude - Arithmetic Ability Questions

NOTE: For an century year,it is divided by 400,when leaves a remainder 0 then it is a leap year.

So, 1600

Total number of female members = 115

Let the number of males = x

Number of females = y

=> x + y = 115  ......(1)

According to the given data,

x + y(0.50) = 69 ......(2)

From (1) & (2), we get

0.5y = 46

=> y = 46 x 2 = 92

=> x = 115 - 92 = 23

Hence, required number of gentlemen in the family = x = 23.

Let C's salary be Rs. x

Then B's salary = 80x/100

=> A's salary = (50/100) * (80x/100) = 40x/100

But given total earnings in a month = 22000

=> x + 80x/100 + 40x/100 = 22000

=> x + 120x/100 = 22000

=> x = 10000

=> A's salary = 40 x 10000/100 = Rs. 4000 

x = y  <=> 1 - q =  2q  + 1  <=> 3q  = 0 <=> q = 0.

Given year 1856, when divided by 4 leaves a remainder of 0.

NOTE: When remainder is 0, 28 is added to the given year to get the result.

So, 1856 + 28 = 1884

CP of 1000gm = Rs. 10
SP of 800gm = Rs. 12
SP of 1000gm =12x1000/800 = Rs. 15

Now take 1000gm as reference to calculate profit.
Profit=SP-CP=15-10=Rs. 5
Profit % = 5x100/10 = 50%

(10 * 6) women can complete the work in 1 day.

 Therefore, 1 woman's  1 day's work = 1/60

(6 * 5) men can complete the work in 1 day.

 Therefore, 1 man's  1 day's work = 1/30

so, required ratio =1/30 : 1/60 = 2:1

Let number of boys in the school = x

Let number of girls in the school = y

Number of students participated in the sports = 300

Out of which boys = 100

but given number of boys participated in sports = x/3

=> x/3 = 100

=> x = 300

Remaining girls = 300 - 100 = 200

=> y/2 = 200

=> y = 400

Therefore, total number of students in the school = x + y = 300 + 400 = 700