CMAT Averages, Ratio and Proportion Questions

Let the number of boys be 4x and number of girls be 6x

Given, 4x+6x = 50

                   x = 5

Number of boys = 20 and number of girls = 30

Average marks of girls = 60

Total marks scored by girls = 60$$\times\ $$30 = 1800

Average marks score by batch = 62

Total marks scored by batch = 62$$\times\ $$50 = 3100

Marks scored by boys = 3100 - 1800 = 1300

Average marks of boys = $$\frac{1300}{20}$$ = 65

Answer is option D.

Given average age of three friends is 20 years

Sum of their ages = $$3\times\ 20$$ = 60 years

Given, their ages are in ratio 2:3:5. Let their ages be 2x, 3x and 5x

2x + 3x + 5x = 60

10x = 60

x = 6

Their ages are 12 years, 18 years and 30 years

Age of oldest friend = 30 years

Answer is option C.

Total number of students is given and percentage of students enrolled in different activity is given. Calculating number of students participating in each activity, we get

A) Boys enrolled in singing and handicraft = 140 + 365 = 505 (III)

B) Girls enrolled in swimming and drawing = 245 + 280 = 525 (II)

C) Girls enrolled in singing and drawing = 490 + 280 = 770 (IV)

D) Girls enrolled in dancing and handicraft = 350 + 385 = 735 (I)

Therefore, answer is option D.

Total production = 1500 units

A) A and G production = 240 + 180 = 420 (not greater than 500)

B) B and F production = 165 + 75 = 240 ( is less than 250)

C) C and E production = 315 + 390 = 705 

D) D and G production = 135 +180 = 315 ( not less than 200)

E) E, F and G production = 390 + 75 + 180 = 645 ( not equal to 545)

Only (B) and (C) are true.

Answer is option B.

Total number of votes = 15200

15% of total votes are invalid, this implies 85% are valid.

55% of valid votes are secured by winner, this implies 45% of valid votes are secured by loser

Voted secured by loser = $$\frac{45}{100}\times\ \frac{85}{100}\times\ 15200$$ = 5814

Answer is option C.

Total workforce = 60 lakhs

Male workforce in SA and MA sectors = 450 + 540 = 990

Total workforce is PT sector = 1140

% required = $$\frac{990}{1140}\times\ 100$$ = 86.8% $$\approx\ $$ 87%

Answer is option B.

Total workforce = 60 lakhs

Female workforce in SE and PR = 360 + 630 = 990

Male workforce in CM = 300

% required = $$\frac{990}{300}\times\ 100$$ = 330%

Answer is option A.

Total workforce = 60 lakhs

Female workforce in SA and MA = 2.7 + 7.2 = 9.9 lakhs

Female workforce in PT = 4.275 lakhs

Difference = 9.9 - 4.275 = 5.625 $$\approx\ $$ 5.6 lakhs

Answer is option C.

Number of electronics engineers initially = $$\frac{2}{3}\times\ 252\ =\ 168$$
So number of computer engineers initially = 252-168 =84
Now number of computer engineers recruited to make ratio 1:1 = 168-84 =84
Let average age of computer engineers be x
so total age  will be 168x
Now average age of electronics engineers will be : x+2
So total age will be : (x+2) 168
Now as per given
Total age = 336(22)
Equating we get :
336(22) = 336(x+1)
we get x=21
so B and C are correct

Let girls be x 
Total employees will be 80+x
now total value of money will be same :
i.e Average of 1employee (Number of employees ) will be equal to (Average of boys)* Number of boys +(Average of girls)*Number of girls
we get 48000(80+x) = 50,000(80)+40,000(x)
48(80+x) = 50(80) +40x
we get 8x =160
x=20

Let the ratio be K
so we can say A=BK
and C = Dk
Now $$\frac{A^2+B^2}{C^2+D^2}=\ \frac{A^2\left(1+k^2\right)}{C^2\left(1+k^2\right)}=\ \frac{A^2}{C^2}$$
Now checking the options
A,B,C gets eliminated
And AB/CD = $$\ \frac{A^2}{C^2}$$
so D is the correct

In June 2020 
Let the number of boys be 3x 
So number of girls will be 2x
Now number of boys in September 2020: 3x-80 and girls in September 2020 : 2x-20
Now as per given 3x-80:2x-20 =7:5
we get 15x-400 =14x-140
we get x=260
So number of students in June 2020 = 3x+2x =5x = 1300

$$a=\frac{3}{2}b$$
b=$$\frac{3}{4}c$$ and 
d=$$\frac{c}{4}$$
c=4d
Now substituting we get 
$$a=\frac{3}{2}\ \times\ \frac{3}{4}\times\ 4d$$
so we get a=$$a=\frac{9}{2}d$$
a:d=9:2

Let volume be 36 units 
so X fills at 3 units per hour and Y empties at 2 units per hour
Now Tank was filled in 14 hrs so X filled 14*3 =42 units
Now therefore Y emptied 6 units
so it will empty 6 units in 3 hours
so Y will be closed at 11 am

Let Steve's salary = $$Rs.$$ $$10$$

=> John's salary = $$\frac{80}{100}\times10=Rs.$$ $$8$$

=> Tom's salary = $$\frac{150}{100}\times8=Rs.$$ $$12$$

$$\therefore$$ Ratio of Steve's salary to Tom’s salary = $$\frac{10}{12}=5:6$$

=> Ans - (D)

Sum of ages of A, B, C and D five years ago = $$45\times4=180$$

Sum of their present ages = $$180+(5\times4)=200$$

Sum of present ages of all five (including X) = $$49\times5=245$$

=> Present age of X = $$245-200=45$$ years

=> Ans - (B)

Let $$x=2$$ and $$y=3$$

To find : $$\frac{3x+2y}{2x+5y}$$

= $$\frac{3(2)+2(3)}{2(2)+5(3)}$$

= $$\frac{12}{19}$$

=> Ans - (D)

Let the number of men be 'm' and number of women be 'w' in the class.

Given Average score of men = 70

$$\Rightarrow$$ Total score of men$$\div$$number of men = 70

$$\Rightarrow$$ Total score of men = 70$$\times$$m

Also, Average score of women = 83

$$\Rightarrow$$ Total score of women$$\div$$number of women = 83

$$\Rightarrow$$ Total score of women = 83$$\times$$w

Given the class average = 76

$$\Rightarrow$$ (Total score of men + Total score of women)$$\div$$(number of men + number of women) = 76

$$\Rightarrow \frac{70\times m + 83\times w}{m + w} = 76 $$

Let the ratio of number of men to women $$\frac{m}{w} = k$$

$$\Rightarrow$$ 70k + 83 = 76k + 76

$$\Rightarrow$$ 6k = 7 

$$\Rightarrow$$ k = 7 : 6.

Hence the ratio of men to women in the class is 7 : 6.

Average Salary Expenditure (in Lakh Rupees) per annum = Total salary expenditure in all these years$$\div$$Total number of years

= $$\frac{576+682+648+672+740}{5}$$

= 663.6

Given monthly savings of Amit and Bharat are in the ration 37:18. Let the savings of Amit and Bharat be 37k and 18k, where 'k' is a constant.

$$\Rightarrow$$ Total  monthly  savings  of  Amit  and  Bharat = 55k.

Given Total annual savings of Amit and Bharat = 1,32,000.

$$\Rightarrow$$ Total  monthly  savings = 11,000.

$$\Rightarrow$$ 55k = 11000

$$\Rightarrow$$ k = 200.

Hence monthly savings of Amit and Bharat are 7400 and 3600 respectively.

From the given data,

Let the monthly incomes of Amit and Bharat be 5x and 4x respectively, where 'x' is a constant.

Similalrly, let the monthly expenditures of Amit and Bharat be 19y and 21y respectively, where 'y' is a constant.

Savings =  Income - Expenditure

$$\Rightarrow$$ 5x - 19y = 7400..........................(1)

and  4x - 21y = 3600...............................(2)

Solving both the equations we get x = 3000 and y = 400.

Therefore, the monthly income of Amit is 5x i.e., 15000.

Option A:

In 2011, number of Refrigerators sold =60

        number of Cell phones sold = 336

$$\Rightarrow$$  number of refrigerators sold as a percentage of number of cell phones sold = $$\frac{60}{336}\times 100 = 17.85$$   

Option B:

In 2012, number of Refrigerators sold = 79

             number of Cell phones sold = 404

$$\Rightarrow$$  number of refrigerators sold as a percentage of number of cell phones sold = $$\frac{79}{404}\times 100$$ = 19.5

Option C:

In 2013,number of Refrigerators sold = 93

             number of Cell phones sold = 411

$$\Rightarrow$$  number of refrigerators sold as a percentage of number of cell phones sold = $$\frac{93}{411}\times 100$$ = 22.6

Option D:

In 2014, number of Refrigerators sold = 112

              number of Cell phones sold = 442

$$\Rightarrow$$  number of refrigerators sold as a percentage of number of cell phones sold = $$\frac{112}{442}\times 100$$ = 25.33

Hence Option D is the correct answer.

Given the proportion of ages of three men are 3:5:7.

Let their ages be 3k, 5k, 7k ,where k is any constant.

Given average of ages of three men = 50 

$$\Rightarrow \frac{3k+5k+7k}{3} = 50$$

$$\Rightarrow \frac{15k}{3} = 50$$

$$\Rightarrow 5k = 50$$

$$\Rightarrow k = 10$$

Therefore the ages of three men are 30, 50, and 70 years.

The age of the youngest men is 30 years.

Let the maximum number of marks be 'x' and minimum passing marks be 'y'.

Given, A student who gets 20% marks fails by 20 marks.

$$\Rightarrow \frac{20}{100}\times x = y - 20$$

$$\Rightarrow 20x = 100y - 2000$$.................(1)

Also given that, another student who gets 36% marks gets 44 marks more than minimum passing marks.

$$\Rightarrow \frac{36}{100}\times x = y + 44$$

$$\Rightarrow 36x = 100y + 4400$$....................(2)

(2) - (1) $$\Rightarrow$$ 16x = 6400

$$\Rightarrow$$ x = 400

From (1) or (2), we get y = 100

Hence, maximum number of marks = x = 400

Percentage necessary for passing = $$\frac{y}{x}\times 100$$ = 25%

Let, us assume the total revenue for 2010 as 100.

Since, all the companies except T have generated the same revenue in 2011, the percentage increase in revenue for all companies will be only due to the increase in revenue of T.

Increase in revenue of T(2011) = 20% * 14 = 2.8

Total revenue(2011) = 102.8

$$\therefore $$ % increase in revenue = $$\frac{102.8-100}{100}$$ * 100% = 2.8 %

Hence, option A.

Let say,

The number mobiles sold in last year of the brands LG, SAMSUNG, NOKIA, SONY, MICRO-MAX be A, B, C, D, and E respectively.

Given that A+B+C+D+E = 1 crore.

But from the given data, the values of A, B, C, D, and E cannot be found out.

So the number of LG sets sold last year cannot be determined.

Given X, Y and Z are in the ratio 12: 15: 25

Let say X = 12k, Y = 15k and Z = 25k, where k is any constant.

Difference between Y and X = 15k - 12k = 3k..............................(1)

and Difference between Z and Y = 25k - 15k = 10k.....................(2)

Ratio of (1) and (2) = $$3k\div 10k = 3:10$$.