Test Level 1: Ratio & Proportion - 2 - CAT MCQ

Assume a set of values for p, q, r, s such that they are proportional i.e. p/q = r/s.
Suppose we take p:q, as 1:4 and r:s as 3:12 we get the given expression: (p – q)(p – r)/p = – 3 × – 2/1 = 6.
This value is also given by p + s – q – r and hence option (b) is correct.

x = k/(y³ – 1). This gives k = 3 × 7 = 21.
When, y = 4, the equation becomes x = 21/(4³ – 1) = 21/63 = 1/3.

X = K × Y × Z → It is known that when X = 6, Y = 3 and Z = 2.
Thus we get 6 = 6K → K = 1.
Thus, our relationship between X, Y and Z becomes X = Y × Z.
Thus, when Y = 5 and Z = 7 we get X = 35.

Solve this question using options. 1/2 of the first part should equal 1/4th of the second part and 1/8th of the third part. Only, option (b) satisfies these conditions thus this option is correct

Solve using options. It is clear that a ratio of x:y as 1: 3 fits the equation

3 × 10 × 4 = 120 man-hours are required for ‘x’ no. of answer sheets.
So, for ‘3x’ answer sheets we would require 360 man-hours = 4 × 30 × n → n = 3 Hours a day.

1 : 3 : 5 → x, 3x and 5x add up to 18.
So the numbers are: 2, 6 and 10. Ratio of cubes = 8 : 216 : 1000 = 1: 27: 125.

The numbers would be 7x and 11x and their LCM would be 77x. This gives us the values as 14 and 22. The first number is 14.

Since equal quantities are being mixed, assume that both alloys have 18 kgs (18 being a number which is the LCM of 9 and 18).
The third alloy will get, 14 kg of aluminum from the first alloy and 7 kg of aluminum from the second alloy.
Hence, the required ratio: 21:15 = 7:5

The total number of man-days-hours required = 10 × 6 × 15 = 900
20 × 14 × number of days = 900 → number of days = 900/280 = 3.21 days

The given ratio for sines would only be true for a 45 - 45 - 90 triangle.
The sides of such a triangle are in the ratio 1:1: √2.
The square of the longest side is 2 while the sum of the squares of the other two sides is also 2.
Hence, the required ratio is 1:1.

p gets 2/3 of what q gets and q gets 1/4th of what r gets’ means a ratio of 2 : 3 : 12 for p : q : r.
Hence, r’s share = 12/17 × 1360 = 960.
Alternately, you could also solve by using options. Option (a) r = 960 fits perfectly because if r = 960, q = 240 and p = 160.

2x + 20 : 3x + 20 : 5x + 20 = 4 : 5 : 7 → x = 10 and initially the number of students would be 20, 30 and 50 → a total of 100.

The ratio of time would be such that speed × time would be constant for all three. Thus if you take the speeds as x, 2x and 3x respectively, the times would be 6y, 3y and 2y, respectively.

Their ratio being 2: 1, the difference according to the ratio is 11 so the numbers must be 22 and 11 respectively.
Hence, the product is 242.

Assume that 1 cow leap is equal to 3 metres and 1 goat leap is equal to 4 metres. Then the speed of the cow in one unit time = 3 × 5 = 15 meters.
Also, the speed of the goat in one unit time = 4 × 4 = 16 meters.
The required ratio is 15:16.

3x, and 5x are their current ages. According to the problem, 3x – 10 : 5x – 10 = 1:2 → x = 10 and hence the sum total of their present ages is 80 years (30 + 50 = 80).

x + 2x + 4x + 8x = 120 → x = 8 Thus, 8x = 64.

25 × 12 = 300 man-days is required for the job.
If only 20 students turn up, they would require 15 days to complete the task.
The number of days is increasing by 1/4.

The initial amount of water is 9 litres and petrol is 27 litres.
By adding 15 litres of petrol the mixture becomes 42 petrol and 9 water → 14:3 the required ratio.