CAT Quantitative Aptitude Questions on Ratios with a high difficulty level

Problem 1:

The first six Prime Ministers of the country had an average tenure of 6 years. What is the tenure of the 7th Prime Minister if the average tenure of first nine Prime Ministers is 5.25 years and the duration of the tenure of the 7th, 8th, and 9th Prime Ministers are in the ratio 1:2:3?

[1] 1 year 3 months 18 days

[2] 1 year 10 months 15 days

[3] 2 years

[4] 1 year 6 months

Solution:

This problem involves Averages as well as Ratios.

Total tenure of first six Prime Minsters = 6 × 6 = 36 years.

Total tenure of first nine Prime Ministers = 9 × 5.25 = 47.25 years.

therefore 36 + x + 2x + 3x = 47.25   [where ‘x years’ is the duration of the tenure of 7^th, 8^th and 9^th Prime Ministers]

or 6x = 1125

or x = 1.875 years

= 1 year 10 months and 15 days.

Hence, the correct choice is [2].

(You may have noticed that without getting the application of ratios in this problem, it will be difficult to get a breakthrough in the solution).

Problem 2:

From a plate of the shape of an equilateral triangle, three symmetrical sectors are cut as shown. The shaded portion shown in the plate gets wasted. What is ratio of the area of the plate that is wasted to the total area of the plate?

Solution:

Ar(Sector APQ) =   [because <PAQ 600...angle of an equilateral triangle]

Since the sectors are symmetrical and the plate is in the shape of an equilateral triangle, we can say that the area of each of the other two sectors i.e., Sector BPR and Sector CQR is also equal to.

Πa²/6

therefore Area of the 3 sectors =    ...(i).

Also, side of equilateral triangle ABC = 2a.

Therefore, Area of the triangle =     ...(ii)

Therefore shaded area or wasted area =    ...from (i) & (ii)

Therefore, ratio of wasted area of the plate to total area of the plate will be given as,

  . Hence, choice [1] is the correct answer.

Problem 3:

The figure below shows 2 concentric circles with centre O. PQRS is a square inscribed in the outer circle. It also circumscribes the inner circle touching it at points B, C, D and A. What is the ratio of the circumference of the outer circle to the perimeter of the polygon ABCD?

[1] Π/4

[2] 3Π/4

[3] Π/2

[4] Π

Solution:

                   
Circumference of outer circle = 2Πr = 2ΠK        …(i)

Perimeter of ABCD =?

In ∆OCD, CD2 = k^₂ + k²= 2k²

therefore CD=√2k                       …(ii)

therefore Perimeter of ABCD =            …(iii)

Also, in square OCED, we have

√2k = K

therefore k = k/√2

Perimeter of ABCD in terms of 

                            
                            = Π/2.
Problem 4:    
If |r-6| = 11 and |2q-12| = 8, then what could be the minimum possible value of the ratio of q to r?

[1]  -2/5

[2]  2/17

[3] 3/14

[4] 11/12

|r-6| = 11 therefore r - 6 = 11, therefore (r - 6) > 0 and

therefore r = 17 (since the value of r is greater than 6, this value is accepted)

r-6 = -11 , therefore ( r - 6 ) ≤ 0

therefore r = - 5 (since the value of r is less than 6, this value is accepted)

Similarly , for the second modulus term, we get

(2q - 12) = 8 , therefore (2q - 12) > 8

therefore 2q = 20 therefore q = 10 (this value is rejected because 2q - 12 > 0 and acceptable values of q will be strictly greater than 10)

therefore When 2q = 4 therefore q = 2 (this value of q is accepted because q should be ≤ 10)

therefore r = 17 and - 5 and

q = 2... these are the premissible values of r and q)

therefore minimum of q/r will be -2/5

Therefore, correct option is [1]