Quantitative Aptitude - Arithmetic Ability Questions

Numbers of students in section A = x
∴ Numbers of students in section B and C = (100 – x)
∴ x 70 + (100 – x) 87.5 = 84 100
=> 70x + 87.5 100 – 87.5x = 8400
=> 8750 – 17.5x = 8400
=> 17.5x = 8750 – 8400 => x = 20.

The year 2006 is an ordinary year. So, it has 1 odd day.
So, the day on 24th Nov, 2007 will be 1 day beyond the day on 24th Nov, 2006.
But, 24th Nov, 2007 is Thursday.
24 - 10 = 14 days.
Therefore, 2 weeks ago it is same day.
Thus, 10th Nov, 2006 is one day before 10th Nov, 2007 i.e. it is Wednesday.

Let initially 'X' ticket has been sold.
So now in 2nd week 20% increases, so
=> X x 120/100
In 3rd week 16% increases, so
=> X x (120/100) x (116/100)
In 4th week 20% decrease, so
=> X x (120/100) x (116/100) x (80/100) = 1392
X = 1250.

Let the number of boys = x

From the given data,

=> [21x + 16(18)]/(x+16) = 19

=> 21x - 19x = 19(16) - 16(18)

=> 2x = 16

=> x = 8

Therefore, the number of boys in the class = 8.

As both A and B invest the same amounts, the ratio of their profits at the end of the year is equal to the ratio of the time periods for which they have invested.

Thus, the required ratio of their profits = A : B = 8 : 12 = 2 : 3.

Hence, share of A in the total profit = 2 x 25000/5 = Rs.10000

Similarly, share of B in the total profit = 3 x 25000/5 = Rs.15000

Let the speed of river and boat be 'r' m/min and 'b' m/min.
so relative speed in upstream (b-r)m/min and in downstream (b+r)m/min.
Now in upstream distnce covered in 5 min is 5(b-r)miles
so total distnce covered => 1 + 5(b-r)miles in upstream
In downstream distance covered in 5min is 5(b+r)miles
Now 1 + 5(b-r) = 5(b+r)
1+5b-5r = 5b+5r
1 = 10r
r = 1/10 mile/min => 6 mile/hr.

Area = 5.5 × 3.75 sq. metre.

Cost for 1 sq. metre. = Rs. 800

Hence total cost = 5.5 × 3.75 × 800 = 5.5 × 3000 = Rs. 16500