SNAP Averages, Ratio and Proportion Questions

This can be answered either by using averages concept or alligations concept.

Alligations :$$p=\frac{\left(p1q1+p2q2\right)}{q1+q2}$$  where q1 and q2 are the number of innings in two group's , p1 and p2 are there respective averages(average runs) , p is the overall average.

Here q1=15 , q2=1 , p1=32 , p= 34 (since increased by 2 runs) 

On solving the equation you get p2=64

Let 100x be the total numbers of voters in the voters list. => 90x voters voted

A=1.4B  (since A received 40% more votes than B)

C=1.2B

Given A-C = 4000

1.4B-1.2B=4000  => B=20000 => A=24000 and C = 28000

So total voter who voted =A+B+C= 72000 which is 90% of the total list.

So when 90%= 72000 , 100%= 80000.

$$\frac{1}{100}\cdot\frac{1}{100}\cdot\frac{25}{100}\cdot1000\ =0.025$$

It is given that B pays 1.5 times that of A

Let B put N number of cows for 3 month

$$\frac{Money\ paid\ by\ B}{Money\ paid\ by\ A}=\ \frac{3}{2}\ =\ \frac{3\times\ N}{80\times\ 7}$$

$$N=280$$

Let $$a_1,\ a_2.....\ a_{11}$$ be the 11 number

$$a_1+a_2+.....\ +a_{11}\ =\ 11\times\ 10.9\ =\ 119.9$$ .... (I)

Average of first six is 10.5 thus $$a_1+a_2+.....\ +a_6\ =\ 6\times\ 10.5\ =\ 63$$ ....(II)

Average of last six is 10.5 thus $$a_6+a_7+.....\ +a_11\ =\ 6\times\ 11.4\ =\ 68.4$$ ....(III)

(II)+(III)-(I) gives $$a_6\ =\ 11.5$$

Let a and b represent salaries of A and B respectively

then $$a+b=40000$$...(I)

After spending 85% of his salary, savings of A will be = (0.15)a

After spending 95% of his salary, savings of A will be = (0.05)b

Their saving are equal, therefore (0.15)a = (0.05b) or 3a=b ...(II)

Solving eqn (I)and (II) we get a = 10,000

Let the five numbers be a, b, c, d, e in the given order.

a+b+c+d+e = 61*5=305    ----- Eq 1

a+c = 69*2 = 138

b+d = 69*2 = 138

From eq 1: e=305-(138+138) =305-276 =29

B is the correct answer.

Let the amount be. Rs x

B's share in the ratio 2:3:4 = $$\ \frac{\ 3}{9}$$ of x=$$\ \frac{\ x}{3}$$

B's share in the ratio 7:2:5 = $$\ \frac{\ 2}{14}$$ of x = $$\ \frac{\ x}{7}$$ , therefore

$$\ \frac{\ x}{3}$$-$$\ \frac{\ x}{7}$$= 40

$$\ \frac{\ 4x}{21}$$ = 40

 x = (40×21)/4 = Rs 210

A is the correct answer.

Number of boys in the school last year = 610

Number of boys in the school in the present year = 610*0.8 (It is given that the boys decreased by 20% this year)

= 488

Number of girls in the class = 1.75* Number of boys 

=1.75*488

=854

A is the correct answer.

Rs.73,689/- are divided between A and B in the ratio 4 :7.

Share of A = $$\ \frac{\ 4}{11}\times\ 73689$$ /- = 26796

Share of B = $$\ \frac{\ 7}{11}\times\ 73689$$ /- = 46893

Now, we have to find the difference between thrice the share of A and twice the share of B 

= 3*26796 - 2*46893 =Rs. 13398

Given, 

Sum of the marks misread = 44+45+61 = 150

Actual marks of the students = 48, 59 and 67

Actual marks sum = 48+59+67 = 174

Difference = 174-150 = 24

Cporrect average = 56+$$\ \frac{\ 24}{24}$$

= 57

D is the correct answer.

Let the present age of Manisha be 5a

The present age of Deepali = Xa

The present age of Praneeta = 5a+9

Praneeta's age after 9 years = 5a+18=33

5a=15

a=3

The difference between Deepali's and Manisha's age is the same as the present age of Parineeta

Xa-5a=5a+9 

3X-15=15+9

3X=39 ==> X=13

D is the correct answer.

Profit received by each officer= Rs.25000
Therefore, Profit received by each clerk= $$\ \frac{\ 3}{5}$$×25000 = Rs.15000
Therefore, Total earned profit
= Rs. (45×25000+80×15000)
= Rs.(1125000+1200000)
= Rs.23.25 lakh

D is the correct answer.

Let the monthly income of Kajal be 100k

It is given that she spent 55% of the monthly income on grocery, clothes and education in the ratio of 4 : 2 : 5 respectively

i.e she spent 55k on grocery, clothes and education in the ratio of 4 : 2 : 5 respectively.

Amount spent on clothes = $$\ \frac{\ 2}{11}\times\ 55k$$

10k = 5540 (given)

So 100k = 55400

Monthly income = 100k = 55400

A is the correct answer.

Given,

Cost of each shirt = Rs. 43

Cost of each tie = Rs. 21

Let x, y be the number of shirts and ties purchased

x*43+ y*21 = 535

y = $$\ \frac{\ 535-43x}{21}$$

when x = 10, y is an integer i.e y =5

ratio of shirts to ties purchased = 10:5 =2:1

B is the correct answer.

Marks obtained by Rohan =75

Marks obtained by Sonia = 65

Marks obtained by Rohit = 65+45=110

Marks obtained by Raman = 110-25 = 85

Marks obtained by Ravi = 85+34 = 119

Maximum Marks = 119+50 = 169

Required percentage = $$\ \frac{\ 119}{169}\times\ 100$$

=70%

B is the correct answer.

The average weight of 19 men is 74 kgs 

The average weight of 38 women is 63 kgs

The average weight of all the men and women together = $$\ \frac{\ 74\times\ 19+63\times\ 38}{19+38}$$

=$$\ \frac{\ 3800}{57}$$

= 66.66 kg 

=approx 67 kg

D is the correct answer.

Let the total monthly income of Mr Giridhar be 100g

The proportion of income he spends on household items = 50% i.e 50g

Of the remaining 50%, the proportion of income he spends on transport, entertainment  and sports are 50%, 25%, 10% 

15% of 50g = Amount saved 

7.5g = 900(given)

100g = 12000

The monthly income of Mr Giridhar = Rs.12000

Total members :240
Males = 2/3 (240) =160
so Females = 240-160 =80
Graduate Males = 160(0.15) =24
Non graduate males = 160-24 = 136
Graduate Females = (0.75) 80 = 60
So non graduate females = 80-60 =20
Now difference between the number of females who are non-graduates and the number of males who are graduates : 24-20 =4

Total members = 240

Number of males = $$\ \frac{\ 2}{3}\times\ 240$$ = 160

Number of males who are non-graduates =  $$\ \frac{\ 85}{100}\times\ 160$$ = 136

Number of females = $$\ \frac{\ 1}{3}\times\ 240$$ = 80

Number of females who are graduates = $$\ \frac{\ 3}{4}\times\ 80$$ = 60

Sum of male non graduates and female graduates = 136+60=196

D is the correct answer.

Total members in the team = 240

Males = $$\ \frac{\ 2}{3}\times\ 240$$ = 160
Females = $$\ \frac{\ 1}{3}\times\ 240$$ = 80

Number of females who are non graduates = $$\ \frac{\ 1}{4}\times\ 80$$ = 20

ratio between the total number of males and the number of females who are non-graduates

= 160:20

=8:1

B is the correct answer.

SP of 1kg rasgulla= Rs.15
Profit= $$\frac{200}{3}$$%
CP of 1 kg rasgulla = ($$\ \frac{\ 15\times100}{\ \frac{\ 500}{3}}$$) = Rs. 9
CP of 1 kg of sugar= 7x
CP of 1 kg of floor= 3x
3x            7x
         9
7x-9       9-3x
$$\ \frac{\ 7x-9}{9-3x}\ =\ \ \frac{\ 5}{3}$$
x=2
CP of sugar= 7x= 14 per kg

D is the correct answer.

Let the initial price of 100 kg Sugar was Rs. 100.
Now, a 20% decrease in rate then rate of 100 kg sugar would be Rs. 80.
So, Rs. 80 = 100 kg sugar.
Rs. 100 = 100× $$\ \ \frac{100}{80}$$ = 125 Kg sugar.
Increment in sugar = 25 kg.
% increment = $$\ \ \frac{125-100}{100}\times\ 100$$ = 25%.
25% increment = 6 kg of sugar

total sugar initially = 6*4 =24 kg

Original price of the sugar = $$\ \ \frac{240}{24}$$

= Rs. 10 per kg

A is the correct answer.

Let the distance covered be 16 km, and the velocities be 1kmph, 4kmph and 16kmph.

Then the time taken: 16h, 4h, and 1h respectively

Ratio of time and velocity for each vehicle: $$\frac{16}{1}:\frac{4}{4}:\frac{1}{16}$$

=$$16:1:\frac{1}{16}$$

=256:16:1.

A+B+C+D = 56  -- Eq 1

B+C+D = 4.6A   -- Eq 2
A+C+D = $$\ \ \frac{11B}{3}$$   --- Eq 3

A+B+D = 2.5C  --Eq 4

Using Eq 1 and 2, we get A = 10

Using Eq 1 and 4, we get C = 16

Using Eq 1 and 3, we get B = 12

On substituting the values of A, B, C in Eq 1, we get D = 18

D is the correct answer.

Let the salary of Saroj be 100x

The salary of Raju = 80x

The salary of Ram =  70x

the percent by which the salary of Raju more than the salary of Ram = $$\ \frac{\ 80x-70x}{70x}\times\ 100$$

=$$\ \frac{\ 1}{7}\times\ 100$$

=14.28%

D is the correct answer.

We know that wire is in the shape of a cylinder.
So, the volume of wire = $$\pi\ r^2h$$

$$\pi\ r^2h_1$$ = $$\pi\ (\frac{r}{3})^2h_2$$

Cancelling π on both sides,

$$\ r^2h_1$$ = $$\frac{\ r}{9}^{^2}h_2$$
$$9h_1$$ = $$h_2$$
length is increased by 9 times

D is the correct answer.

Initial surface area of the cube = 6$$(side)^2$$ 

= 6$$(4)^2$$ 

= 96$$cm^2$$

No. of new cubes = $$\ \ \frac{\ Volume\ of\ older\ cube}{Volume\ of\ 1\ new\ cube}$$

=$$\ \ \frac{\ 4\times\ 4\times\ 4}{1\times\ 1\times\ 1}$$

=64 cubes 

Newer surface area of 64 cubes = 64*6$$(1)^2$$ = 384$$cm^2$$

Percentage increase in surface area = $$\ \ \frac{384-96}{96}$$ *100

=$$\ \ \frac{28800}{96}$$

 = 300%

B is the correct answer

The time spent in school corresponds to $105^0$$

Percentage of time spent in school = $$\ \frac{\ 105}{360}\times\ 100$$

= Approx 30%

B is the correct answer.

Time spent on games corresponds to $$30^0$$ and on sleeping to $$120^0$$

Required ratio = $$\ \frac{\ 30}{120}\times\ 100$$

=25%

C is the correct answer.

The time spent in playing games and in doing homework corresponds to 30° and 45° respectively.

If the time spent in playing games is equal to that spent on doing homework, then there should be an increase of 15° in the time spent in games.

If this 15 ° is obtained by decreasing the time spent in sleeping (with the other activities remaining constant), 

The percentage decrease in the time spent in sleeping = $$\ \frac{\ 15^0}{120^0}\times\ 100$$

= $$12.5^0$$

B is the correct answer.

School corresponds to  $$105^0$$ and homework corresponds to $$45^0$$

Difference =  $$105^0$$ - $$45^0$$

=$$60^0$$

= $$\ \frac{\ 60}{360}\times\ 24$$

= 4 hrs

C is the correct answer.

The total number of hours = 24

Time spent on homework = $$45^0$$

=$$\ \frac{\ 45}{360}\times\ 24$$ = 3 hrs

It is given that she spent $$\frac{1}{3}rd$$ time of homework in Mathematics.

The time she spent on Mathematics = $$\ \frac{\ 1}{3}\times\ 3$$

=1 hr

The time she spent on the rest of the subjects = 3-1 = 2 hr

B is the correct answer.

Total revenue of the 30-day period = Rs.10000 x 30 days = Rs.300000

Total revenue of the weekends (8 days) = Rs.20000 x 8 days = Rs. 160000

Total revenue of weekdays = (300000-160000) = Rs. 140000

Avg. daily revenue on weekdays = 140000/22 = Rs.6363.63 $$\approx\ $$ Rs.6364

They started with 40 dozen rolls. Half of them (=20 dozen) were sold by noon.

40% of the remaining rolls (half of 40 dozen) were left unsold, which is 0.4 x 20 = 8 dozen rolls

Let Sonali's monthly income be 100x 
Now Expenditure on insurance policies = 15x
Expenditure on shopping and household expenses = 55x
Therefore remaining amount = 100x-55x-15x = 30x 
Which is equal to 12,750 
Now therefore 30x=12,750 
x = 425
Therefore total income = 100x = 42,500

Total number of students = 480

Percentage of total students passed

= 75 % of total student = 75×480/100=360student
Now, using the condition from the question,
Let the number of boys be x
Then, 70% of x + 85% of (480 - x) = 360
=> 70×x/100+85×(480−x)/100=360
=> 70x - 85x + 40800 = 36000
=> 40800 - 36000 = 85x - 70x
=> 4800 = 15x
=> x=4800/15=320
There are 320 boys who have appeared for the examination.

The commission amount will be :
15,000 (12.5/100) 
= 15,000(25/200)
= Rs 1875

Gold in 50g of alloy

$$=80\times\frac{50}{100}=40g$$

Let x g gold must be added.

Now, according to the question,

$$40+x=\frac{90}{100}\left(50+x\right)$$

$$x=50g$$

Thus, 50 g of gold must be added to it to get 90%.

Thus, the correct option is A.

Let the incomes of two persons are 7x and 3x
Expenditure of first person = 7x - 300
Expenditure of second person = 3x - 300
According to the question,
7x−300/3x−300=5/2
=> 14x-600 = 15x-1500
x = 900
Income of first person
= 7x = 7×900=RS.6300

Old price of rice = Rs.24/k
New price of rice = Rs.27 per/kg
Therefore, Increase in price of rice = New price - Old price = 27 - 24 = Rs.3/kg
Therefore, Increased percentage in price
= 3/24×100=12.5 %
Now, reduction in consumption, so as to keep the expenditure fixed.
According to the formula,
$$\frac{a}{100+a}\times\ 100$$
= $$\frac{12.5}{100+12.5}\times\ 100$$
=11.1 %

Gas usage per hour of the larger burner = $$\ \frac{\ 14.4}{\ \ 80}$$

Gas usage per hour of the smaller burner = $$\ \frac{\ 14.4}{\ \ 104}$$

The required percentage = $$\ \dfrac{\ \ \frac{14.4\ }{80}-\ \frac{\ 14.4}{104}}{\ \frac{14.4\ }{80}}$$ = 1- $$\ \frac{\ \ 80}{\ 104}$$ = $$\ \frac{\ \ 24}{\ 104}$$ = 23.07

This can be solved easily by using allegations

$$p=\frac{\left(p1q1+p2q2\right)}{q1+q2}$$

Here p = M , p1=P , p2=N

                     q1=3 , q2=6

$$\ \therefore\ M=\frac{\left(3P+6N\right)}{3+6}$$

On solving 3M = P+2N

Average of 5 consecutive number = $$\frac{\left(n\right)+\left(n+1\right)+\left(n+2\right)+\left(n+3\right)+\left(n+4\right)}{5}$$ = n+2

If 2 more consecutive number are added then its average = $$\frac{\left(n\right)+\left(n+1\right)+\left(n+2\right)+\left(n+3\right)+\left(n+4\right)\ +\left(n+5\right)+\left(n+6\right)}{7}$$ = n+3

Hence average is increased by 1

Let the amount Tanya paid be 6x

Amount paid by Veena is 3x

Then Amita paid by 2x

Required fraction $$\frac{3x}{6x+3x+2x}$$ = $$\frac{3}{11}$$

Let the number of student in Physics, Chemistry and Zoology be 4x, 3x and 5x respectively

New number of Physics student = 4x*(1+$$\frac{50}{100}$$) = 6x

New number of Chemistry student = 3x*(1+ $$\frac{25}{100}$$) = $$\frac{15x}{4}$$

New number of Zoology student = 5x * (1 + $$\frac{10}{100}$$) = $$\frac{11x}{2}$$

Required ratio is $$6$$ : $$\frac{15}{4}$$ : $$\frac{11}{2}$$

Multiplying overall by 4, answer is 24:15:22 

For 1000 people, 32 person are born and 11 die.

Overall increase in population = 32-11 = 21

%increase in popoulation = $$\frac{21}{1000}\times\ 100$$% = 2.1%