Test: Average, Ratio & Proportion - CAT MCQ

Let salaries of Ramesh, Ganesh and Rajesh in 2010 be 6x, 5x and 7x respectively.
Also, let salaries of Ramesh, Ganesh and Rajesh in 2015 be 3y, 4y and 3y respectively
Given, 3y = 1.25 × 6x ⇒ = 2.5x
∴ Salary of Rajesh in 2015 = 3y = 3 × 2.5x = 7.5x
∴ Percentage increase in salary of Rajesh 
=  ≈ 7

Let a be  the average of 20 numbers whose average does not exceed 5.
Let b be  the average of rest of the 10 numbers. Clearly,
b > 5 i.e. the average of these numbers exceeds 5
Therefore,
30 × 5 = 20a + 10b
⇒ 2a + b = 15
⇒ b = 15 – 2a
Going by the options, we can say that when a = 4.5, then b = 6 which satisfies all the conditions.
But when a = 5, then b = 5 which does not satisfies b > 5
Hence, highest possible value of the required average = 4.5

Since scores of Anjali, Mohan and Rama after review were in the ratio of 11 : 10 : 3, therefore we can suppose scores of Anjali, Mohan and Rama after review be 11x, 10x and 3x respectively.
Therefore, their scores before review was (11x – 6), (10x – 6) and (3x – 6) respectively.
Since, Rama’s score was one-twelfth of the sum of the scores of Mohan and Anjali

⇒ 12 (3x – 6) = 21x – 12 ⇒ x = 4
Now, Anjali’s score – Rama’s score
= (11x – 6) – (3x – 6) = 8x = 8 × 4 = 32

Initial amount of salt in vessel A = 10 gms per 100 ml. solution. Therefore in 500 ml solution in vessel amount of salt = 50 gms
Similarly, initially in 500 ml solution in vessel B amount of salt = 110 gms
and initially in 500 ml solution in vessel C, amount of salt = 160 gms
When 100 ml is transferred from A to B, the amount of salt now in B = 10 + 110 = 120 gms in 600 ml.
The new concentration of salt in B = 120 / 600 x 100
= 20 gms per 100 ml.
Now, the amount of salt in A = 50 – 10 = 40 gms in 400 ml
Now, when 100 ml is transfered from B to C, the amount of salt now in C = 20 + 160 = 180 gms in 600 ml.
The new concentration of salt C = 180 / 600 x 100
= 30 gms per 100 ml
Finally, when 100 ml is transfered from C to A, the amount of salt now in A = 30 + 40 = 70 gms in 500 ml.
∴ Strength of salt in 70 / 500 x 100 = 14

Weight of liquid 1 per litre = 1000 gm
Weight of liquid 2 per litre = 800 gm
Weight of mixture per litre = 2 × 480 = 960 gm
By alligation rule,
Quantity of liquid 1 / Quantity of liquid 2 =  = 4 / 1
Hence, the liquids are mixed in 4 : 1.
∴ Percentage of liquid 1 = (4 / 4 + 1) x 100 = 80%

Ratio of incomes of Amala, Bina and Gouri = 3 : 4 : 5
Ratio of interests of Amala, Bina and Gouri = 6 : 5 : 4
Therefore, the ratio of their interest income = (3 × 6) : (4 × 5) : (5 × 4) = 18 : 20 : 20
Let the interest incomes of Amala, Bina and Gouri be 18x, 20x and 20x respectively.
Since, Bina’s interest income exceeds Amala’s by Rs 250
Therefore, 20x – 18x = 250 ⇒ x = 125
Total interest incomes = 18x + 20x + 20x = 58x = 58 × 125 = 7250

Let x be the average of all 22 students and Gautam’s score be G.
Hence, total scores of all 22 students = 22x. 
From the first given condition,
22 x – G = 21 × 62 ...(i)
From the second given condition,
22x – 82.5 = 21(x – 1)
⇒ 22x – 21x = 82.5 – 21
⇒ x = 61.5
On putting x = 61.5 in (i), we get
22 × 6 1.5 – G = 21 × 62
⇒ G = 1353 – 1302 = 51

Refer to the below figure.

As we can see, the triangle can be divided into 9 conguruent smaller equilateral triangles. Also, hexagon contains 6 of them.
∴ The required ratio = 6 : 9 = 2 : 3

Let the average score of the aspirant in all the tests be A. Let the number of tests be N.
The aspirant’s average score for the first 10 tests and last 10 tests are 20 and 30 respectively.

⇒    10A – N = 190 and N + 10A = 310
On subtracting,
we get,  – 2N = – 120
⇒ N = 60.

Let the average age of people aged 51 years and above be A₁ years and the average age of people aged below 51 years be A₂ years. Let the number of people aged below 51 years be N₂.
The average age of all the people in the apartment complex is 38 years.

Clearly for A₂ to be maximum, A₁ should be minimum i.e. 51

Clearly for A₂ to be maximum, N₂ should be also maximum i.e. 39 Hence maximum value of A₂

Let the cost prices of A and B be C_a and C_b respectively.
Selling price of the mixture = 40 per kg.
The profit made is 10% if A and B are mixed in the ratio 3:2.
∴ 
⇒ 
The profit made is 5% if A and B are mixed in the ratio 2 : 3.
∴ 
⇒ 
Divide equation (i) by (ii), we get

⇒ 24C_a = 19C_b ⇒ C_a : C_b = 19 : 24

Let the numbers of marbles with Raju and Lalitha be 4x and 9x respectively.
Let Lalitha gave y marbles to Raju.
∴  ⇒  
Fraction of original marbles that Lalitha gave to Raju
= y / 9x = 7 / 33

Let the average score of the boys in the midsemester examination be b
Average score of the girls = b + s
in the exam,
Average score of the girl = b + 5 – 3
= b + 2
Average score of the entire class increased by 2 and is

Let the total no. of popcorn pockets in stock be T
Total no. of chips pockets in stock = T
Required ratio 

C₅ – C₁ = 19, the numbers above are the actual profits
∴ The total profit = 438 crore.

Let the number of Re. 1,50 paise and 25 paise coins be 360, 432 and 576 respectively (ratio 5 : 6 : 8).

Ratio = 512 : 416 =16 : 13.

If all are of equal height. number of handshakes = ⁴⁰C₂.
If all are of different heights, number of handshakes = 0
∴ Required difference = ⁴⁰C₂ – 0 = ⁴⁰C₂.

Let the required number of hour be h.

Going through the options, h = 4 comes out to be the correct answer.

Let the total amount involved in this game in $K.
The first person has (7/18)k in the beginning and (6/15)k in the end. Thus he won something.
Second person has (6/18)k in the beginning and (5/15)k in the end. So he neither gains nor loses. At this point it is very clear that third person loses something.

So, the winner must have started. with $ 420.

Let the number representing A be a, and the common difference be d.
⇒ a + (a + 2d) + (a + 4d) = 36 ...(i)
and (A + C + E + G) – (A + C + E) = 60 – 36
⇒ G = 24
⇒ a + 6d = 24 ...(ii)
From (i) and (ii)
(3a + 6d) – (a + 6d) = 12
⇒ 2a = 12 ⇒ a = 6 and d = 3
So, B + D + F + H = 4a + 16d
= 24 + 48 = 72