As a,b ,c ,d , e are consecutive odd numbers than

b = a+2

c=a + 4

d= a+6

e= a + 8

and we know that , average = $$\frac{Sumof Elements}{Number of Elements}$$

= $$\frac{a+a+2+a+4+a+6+a+8}{5}$$

= a+4

Let's assume no. of seats were 100 and price of a ticket was 100 hence total revenue is 10000.
Now acc. to question after increment no. of seats will be 125 and price of a ticket will be 112. Hence total revenue will be 125*112 = 14000.
Increment in revenue = 4000
% increase in revenue = 40%

Given, 90% of A = 30% of B,
so (90/100)*A = (30/100)*B
Solving the above equation gives B=3A,
B as a % of A = (B/A)*100 = 3*100 = 300
So, X = 300
 

Let's say B's income is 100
Hence A's income will be 50
Now B's income is 50 more than A's
Hence it is $$\frac{50}{50}\times100$$ percent ($$\frac{d}{Ai}\times100$$ where d is difference in income and Ai is A's income) more than A's

Given 60% of A = 48% of B
Hence A:B = 48 : 60
or       A:B = 4 : 5

Expenditure on rent = 14%
Expenditure on fuel = 9% 
Let's say total expenditure is 100.
So rent is $$\frac{14}{9} \times 100$$ =156% of fuel.

Total share percentage on food and rent = 45+14 = 59%
Hence price will be = $$25000 \times {59}{100} = 14750$$

After a discount of 40% price will be = $$\frac{500 \times 60}{100}$$=300
After 2 successive discount of 36% and 4%= $$\frac{500\times64}{100}=320 $$ and $$\frac{320\times96}{100}=307.2 $$
Hence difference will be 7.2

Let's say numbers are $$a, a+2, a+4$$
So avg. will be $$\frac{(a+a+2+a+4)}{3} = \frac{a}{3} +12$$
Or a = 15
So numbers will be 15, 17, 19

Let's say 1.14 is $$x$$ percent of 1.9
hence 1.14 = $$\frac{1.9\times x}{100}$$
$$x$$ will be 60

Let's say total number of workers are $$x$$.
Hence above 30 years workers will be $$\frac{60x}{100}$$ 
And male workers will be $$\frac{60x}{100} \times \frac{75}{100}$$ = 1350 (given)
So x will be = 3000.

Percentage of fuel , clothing and others together will be equal to 30%
So when total expense is 100 , fuel, clothing and others together have expense of 30
Now when total expense is $$360^o$$ , clothing , fuel and others will have expense of $$\frac{30}{100} \times 360 = 108^o$$

let the income be Rs 100

As it is given that he spends 75% of his income so his spending is = Rs 75

Saving =100-75=Rs25

Now his income is increased by 20% .So his new income is = 1.2x100 = 120 Rs

His expenditure increased by 10 % .So his new expenditure is = 1.1x 75 = 82.5 Rs

New savings= 120 - 82.5 = Rs 37.5

Increment in savings = $$\frac{37.5-25}{25}$$ * 100 = 50% .

let the retailer buys 110 gram in Rs 100 because of 10 % cheating 

so cost price = 1 gram = Rs 100/110 = Rs 0.909

and retailer sells 90 gram in Rs 100 

and hence selling price = Rs 100 / 90 = Rs 1.111

Profit % = $$\frac{1.111- 0.909}{0.909}$$ = 22.22%

let the ages of A and B 18 years ago be 8x and 13x

So there present ages will be

A= 8x +18

B = 13x + 18

It is given that ,

$$\frac{8x+18}{13x + 18}$$ = $$\frac{5}{7}$$

x = 4

So present age of A = 8x + 18 = 32 +18 = 50 years

Let's say price of article is 100
After first increase its price will be 100(1+ $$\frac{10}{100}$$) = 110
Now second increment will be applied on 110
Hence new price will be 110(1+ $$\frac{10}{100}$$ ) = 121
Which is 21 more than before any increment
Hence total percentage increment = 21

$$\frac{QNR}{total} = \frac{QNR(initial)}{total} (1- \frac{\textrm{Replaced Quantity}}{total})$$

QNR is the quantity which has not entered again

so here QNR is A

$$\frac{A}{5x} = \frac{4x}{5x} (1- \frac{10}{5x})$$

A = 4(x- 2)

B = 5x - 4x + 8 = x+3

$$\frac{4x-8}{x+3} = \frac{2}{3}$$

12x - 24 = 2x + 6

10 x = 30

x = 3

so volume = 5x = 5 x 3 = 15 ltr

Average age of 120 members of the society = 60.7 years

=> Total age of 120 members = $$120\times60.7=7284$$ years

Let the average age of 30 new members = $$x$$ years

=> Total age of the 30 members = $$30x$$ years

According to ques, new average = 56.3 years

=> $$\frac{7284+30x}{(120+30)}=56.3$$

=> $$7284+30x=56.3\times150=8445$$

=> $$30x=8445-7284=1161$$

=> $$x=\frac{1161}{30}=38.7$$

=> Ans - (D)

Runs After 45 overs = 45*5.8 = 261

required runs = 295 - 261 = 34

remaining overs = 50 - 45 = 5

required run rate = 34/5 = 6.8

so the answer is option B.

Let Mohit's salary = Rs. $$100$$

=> Rahul's salary = $$\frac{40}{100}\times100=Rs.$$ $$40$$

Rahul's new salary after 60% increase = $$40\times\frac{160}{100}=Rs.$$ $$64$$

Similarly, Mohit's new salary after 20% decrease = $$100\times\frac{80}{100}=Rs.$$ $$80$$ 

$$\therefore$$ Required % = $$\frac{64}{80}\times100=80\%$$

=> Ans - (A)

Let share of X, Y and Z be Rs. $$x,y$$ and $$z$$ respectively.

Then, $$3x=5y=2z=k$$

=> $$x=\frac{k}{3}$$ , $$y=\frac{k}{5}$$ and $$z=\frac{k}{2}$$ ------------(i)

Sum = $$x+y+z=7750$$

=> $$\frac{k}{3}+\frac{k}{5}+\frac{k}{2}=7750$$

=> $$k\times(\frac{10+6+15}{30})=7750$$

=> $$k=7750\times\frac{30}{31}=7500$$

$$\therefore$$ Share of Z = $$\frac{k}{2}=\frac{7500}{2}=Rs.$$ $$3750$$

=> Ans - (B)

Let 5/- coins = x

50paise coins = 5x

10paise coins = 11x

toal amount = 43

(x)(5)+(5x)(0.5)+(11x)(0.1) = 43

8.6x = 43

x = 5

No.of 50paise coins = 5x = 5(5) = 25

so the answer is option B.

Average = $$\frac{Sumof Elements}{NumberofElements}$$

it is given that average temperature of Monday, Tuesday and Wednesday was 30° C and that of Tuesday, Wednesday and Thursday was 33° C

Sum of temperatures on Monday, Tuesday, Wednesday = 30 x 3 = 90 .....(1)

Sum of temperatures on Thursday, Tuesday, Wednesday = 33 x 3= 99 ....(2)

it is given that temperature on Monday = 32 .......(3)

hence using equation 1 and 3

Tuesday + Wednesday = 90 - 32 = 58 ....(4)

using equation 2 and 4

Thursday = 99 - 58 = 41 Celsius

C does $$\frac{1}{20}$$ amount of work in a day.
And A+B do twice of work as much done by C or A+B=2C or $$\frac{2}{20}$$.
So A+B+C together will do $$\frac{2}{20}+\frac{1}{20}=\frac{3}{20}$$ amount of work in a day.
And complete work will be done in $$\frac{20}{3}$$ days.

Let's initial price is $$x$$, and sale is $$y$$ so revenue will be $$xy$$
After having discount of 20%, initial price will be $$\frac{80x}{100}$$ and sale will be $$\frac{180y}{100}$$ , hence new revenue is $$1.44xy$$.
Increment in revenue will be $$0.44xy$$ i.e. equal to 44%

Having a gain of 12.5%, selling price will be $$800 \times \frac{112.5}{100} = 900$$
So after having a discount of 10% on marked selling price is 900
Hence marked price will be = $$900 \times \frac{100}{90} = 1000$$

After having 20% discount price will be = $$1500 \times \frac{80}{100}$$ = 1200
So for net price of 1104 discount should be 1200-1104 = 96
%discount = $$\frac{96}{1200} \times 100$$ = 8%

Selling price of 20 apples = 1000
gain = 20% 
Cost price of 20 will be = $$\frac{1000 \times 100}{120} = \frac{2500}{3}$$
So on cost price of 1000 amount of apples will be = $$\frac{20 \times 3}{2500} \times 1000$$ =  24
 

Marked price = 480
discount = 10%
Selling price = 480 - $$\frac{480 \times 10}{100}$$ = 432
Gain = 8%
Cost price = $$ 432 \times (\frac{100}{108})$$ = 400
After no discount ,gain will be = 480-400 = 80
Percentage gain = $$\frac{80}{400} \times 100$$ = 20%

cost of 5 kg. rice = 100 rs.
cost of 1 kg. rice = 20 rs.
Before discount cost price will be = $$20 \times \frac{100}{80} = 25$$

Lets assume third number as 100.
As first number is 30% less than third, first number is 70,
And second number is 37% less than third, second number is 63
Difference between first and second number is (70-63) = 7.
Hence the % diff between first and second numbers is (7/70)*100 = 10%
 

Let's say cost price of 15 articles is $$x$$
cost price of 12 articles will be = $$\frac{12x}{15}$$

selling price of 15 articles is = $$x$$

gain = $$\frac{3x}{15}$$

gain% = $$\frac{\frac{3x}{15}}{\frac{12x}{15}} \times 100$$ = 25

Let's say marked price = 100
Then cost price = 64
 and selling price = 88
discount = 24
%discount = $$\frac{24}{64} \times 100$$ = 37.5%

let the marked price of the article be Rs y

Ist case :

Two succesive discounts 40% followed by 30%

So after these two successive discounts the value of article becomes = 0.6 × 0.7 × y = 0.42 y

2nd case

Two successive discounts are 45% followed by 20%

So after these two discounts the price if article becomes = 0.55 × 0.8 × y = 0.44y

It is given that :

0.44y - 0.42y = 12

0.02y = 12

y = Rs 600

let the number added to both numerator and denominator be y

It is given that :

$$\frac{2+y}{5+y}$$ = $$\frac{5}{6}$$

=>12 + 6y = 25 + 5y

=> y = 13

It is given that 10 students are selected

and 3 were of 20 years age, 4 of 21 and 3 of 22 years

As we know that average age = $$\frac{Sum of Ages}{NumberofStudents}$$

=$$\frac{(20x3)+(21x4)+(22x3)}{10}$$

= 21 years

Marked price(M) of article = Rs 975

Selling Price (S) of article = Rs 897

Discount % = $$\frac{M - S}{M}$$x100 = $$\frac{975-897}{975}$$x100

= 8%

$$x^{3}+27x^{2}+243x+631$$ the following equation is equal to $$(x+9)^3$$ - 98

So using value of x = 2 which is given in question

$$x^{3}+27x^{2}+243x+631$$ = $$(x+9)^3$$ - 90 = $$(11^3)$$ - 98

= 1331 - 98

= 1233

Let selling price = Rs. $$100$$

=> Cost price = $$\frac{75}{100}\times100=Rs.$$ $$75$$

$$\therefore$$ Profit % = $$\frac{(100-75)}{75}\times100$$

= $$\frac{100}{3}=33.33\%$$

=> Ans - (C)

In 2nd inning Cheteswar Pujara has scored minimum ,

runs in 1st innings = 70

runs in 2nd innings = 10

Average = $$\frac{SumofElements}{NumberofElements}$$

average runs = $$\frac{70+10}{2}$$ = 40

Average = $$\frac{Sum of Elements}{Number of Elements}$$

Number of subjects = 6

wrong average = 88

sum of marks calculated wrongly in all subjects = 88 x 6 = 528

Correct Marks = 528 - 86 + 68 = 510

Correct Average of marks = $$\frac{510}{6}$$ = 85

As it is mentioned that in a 729 ml mixture of milk and water ,the ratio of Milk : water = 7:2

So amount of milk = $$\frac{7}{9}$$ x 729 ml = 567 ml

amount of water = $$\frac{2}{9}$$ x 729 ml = 162 ml

Let the amount of water added be y ml

So it is given that :

=> $$\frac{567}{162+y}$$ = $$\frac{7}{3}$$

=> $$162 + y = 243$$

=> $$y$$ = 243-162 = 81 ml

Let cost price of article = Rs. $$70$$

=> Marked price = $$\frac{10}{7}\times70=Rs.$$ $$100$$

Similarly, selling price = $$\frac{4}{5}\times100=Rs.$$ $$80$$

$$\therefore$$ Profit % = $$\frac{(80-70)}{70}\times100=14.28\%$$

=> Ans - (C)

Let weight of P, Q and T be $$x,y,z$$ kg respectively.

Average weight of P and Q = $$\frac{x+y}{2}=38$$

=> $$x+y=76$$ ------------(i)

Similarly, $$x+y+z=147$$ -----------(ii)

Subtracting equation (i) from (ii), we get :

=> $$z=147-76=71$$

$$\therefore$$ Weight (in kg) of coach = 71 kg

=> Ans - (A)

Arithmetic Mean = $$\frac{Sumof Elements}{NumberofElements}$$

Sum of first n natural numbers = $$\frac{n(n+1)}{2}$$

Arithmetic Mean of first n natural numbers = $$\frac{n(n+1)}{2\times n}$$ = $$\frac{n+1}{2}$$

let the amount with C be Rs y

so, amount with B = y +6

and amount with A = y + 13

and it is given that total amount is 76 , So

y+y+6+y+13= 76

y= 19

Hence A:B:C = 32:25:19

$$\frac{QNR}{total} = \frac{QNR(initial)}{total} (1- \frac{\textrm{Replaced Quantity}}{total})^3$$

QNR is the quantity which has not entered again

so here QNR is milk ,let milk quantity be A ltr

$$\frac{A}{80} = \frac{80}{80} (1- \frac{8}{80})^3$$

A= 80 $$\frac{729}{1000}$$

A= $$\frac{1458}{25}$$

= 58.32 Ltr

let the total capital be Rs z and total time be y

hence ratio of profit divison for A,B,C will be

$$\frac{z}{6} \times \frac{y}{6} : \frac{y}{4} \times \frac{z}{4} : y \times \frac{7z}{12}$$

A:B:C = 4:9:84

hence profit for B = 9/97 x 19400 = 1800

sum of squares of first n natural numbers are $$\frac{(n)(n+1)(2n+1)}{6}$$

average = $$\frac{SumofElements}{NmberofElements}$$

we are given numbers as 1^2 , 2^2 , 3^2 , 4^2 , 5^2 , 6^2 , 7^2 , 8^2 , 9^2 , 10^2

Sum of 10 elements = $$\frac{10 \times 11 \times 21}{6}$$ = 385

average = $$\frac{385}{10}$$ = 38.5

we know that average =$$\frac {sum of elements}{numberof elements} $$

Number of elements =7

Sum of elements = 7+11+15+14+21+25+x = 93+x

=> Average = $$\frac {93+x}{7} $$

=> 15 = $$\frac {93+x}{7} $$

=> x = 12

Let B's marks = 100

=> A's marks = $$100+(\frac{20}{100}\times100)=120$$

$$\therefore$$ Percent by which B's marks are less than A's = $$\frac{(120-100)}{120}\times100$$

= $$\frac{1}{6}\times100=16.67\%$$

=> Ans - (A)

let the four numbers be a,b,c,d

Using formula average = $$\frac{SumofElements}{NumberofElements}$$

Now, the average of first 3 numbers = 16

=> Sum of first 3 numbers=a+b+c = 16 × 3 = 48.......... (1)

Also, average of last three numbers = 15

=> Sum of last three numbers = b+c+d = 15 × 3 = 45......(2)

Subtracting equation (2) from (1)

=> a - d = 3

Since, d = 20

=> a - 20 = 3

=> a = 23

we need to find the value of $$\sqrt{\frac{(0.064-0.008)(0.16-0.04)}{(0.16+0.08+0.04)(0.4+0.2)^3}}$$

= $$\sqrt{\frac{(0.4^3-0.2^3)(0.4^2-0.2^2)}{(0.16+0.08+0.04)(0.4+0.2)^3}}$$

= $$\sqrt{\frac{(0.4-0.2)(0.4^2+0.2^2+0.4\times 0.2)(0.4-0.2)(0.4+0.2)}{(0.16+0.08+0.04)(0.4+0.2)^3}}$$

= $$\sqrt{\frac{(0.4-0.2)^2}{(0.4+0.2)^2}}$$

= $$\sqrt{\frac{1}{9}}$$

= $$\frac{1}{3}$$

Let the cost price of 1 book be Rs C

and the selling Price of 1 book be Rs S

It is given that cost price of 100 books is equal to the selling price of 60 books

So, 100xC = 60xS

S = $$\frac{5}{3}$$ C

So Profit % = $$\frac{SellingPrice-CostPrice}{CostPrice}$$x100

= $$\frac{\frac{5C}{3}-C}{C}$$x100

= $$\frac{2}{3}$$ * 100 = $$66\frac{2}{3}$$ %

Let assume the initial amount be Rs 100

Now after 10% discount , Rs 100 reduces to = 100 -( $$\frac{10}{100}$$ x 100 )= Rs 90

Now for 2nd discount of 20% the initial amount is Rs 90 .So amount after 2nd discount = 90 - ($$\frac{20}{100}$$ x 90 ) = Rs 72

Now for 3rd discount of 25% the initial amount is Rs 72 .So amount after 3rd discount = 72 - $$\frac{25}{100}$$ x 72 = Rs 54

total percentage change = $$\frac{100 - 54}{100}$$ x 100 = 46 %

hence the value of discount equivalent to the given three discounts of 10%,20%,25% is 46%

Average of 5 numbers is given as 56.4

Total of 5 numbers = Average x Number of Elements

Total of 5 numbers = 56.4 x 5 = 282

Average of 1st and 2nd number is 53

Total of 1st and 2nd number = 53 x 2 = 106

Average of 4th and 5th number is 56

total of 4th and 5th number = 56 x 2 = 112

1st + 2nd + 3rd +4th +5th = 282

106 + 3rd + 112 = 282

3rd = 64

hence option A is correct

Initial price $$x$$ (say) and consumption is $$y$$ , hence expense will be $$xy$$.

After price increment price is $$\frac{120x}{100}$$ and expense should remain $$xy$$.

Hence $$\frac{120x}{100} \times N=xy$$ (where N is new expense)
                                              $$ N=\frac{100y}{120} $$

Decrement in consumption is=$$\frac{20y}{120}$$ or $$\frac{50}{3}$$ %