Required sum = (80 - 3 $\times$3) years = (80 - 9) years = 71 years.

Let Rs. x be the fare of city B from city A and Rs. y be the fare of city C from city A.

Then, 2x + 3y = 77 ...(i) and

3x + 2y = 73 ...(ii)

Multiplying (i) by 3 and (ii) by 2 and subtracting, we get: 5y = 85 or y = 17.

Putting y = 17 in (i), we get: x = 13

Lets the number of person be x. then

$\frac{96}{x-4}-\frac{96}{x}=4$

$\frac{1}{x-4}- \frac{1}{x}=\frac{4}{96}$

$\frac{x-(x-4)}{x(x-4)}=\frac{1}{24}$

$x^{2}-4x-96=0$

(x - 12) (x + 8) = 0

x = 12

so required number = x - 4 = 8

Clearly, we have :

B - 3 = E ...(i)

B + 3 = D ...(ii)

A + B = D + E+10 ...(iii)

B = C + 2 ...(iv)

A + B + C + D + E= 133 ...(v)

From (i) and (ii), we have : 2 B = D + E ...(vi)

From (iii) and (vi), we have : A = B + 10 ...(vii)

Using (iv), (vi) and (vii) in (v), we get:

(B + 10) + B + (B - 2) + 2B = 133

5B = 125

B = 25.

Lets the Number of pineapple and watermelon  be x and y respectively

Then, 7x + 5y =38

or, 5y =(38 - 7x)

or,  $y=\frac{38-7x}{5}$

Clearly y is whole number, only then(38- 7 x) is divided by 5

This happens when x = 4.

Let x and y be the ten's and unit's digits respectively of the numeral denoting the woman's age.

Then, woman's age = (10x + y) years; husband's age = (10y + x) years.

Therefore (10y + x)- (10x + y) = $\left ( \frac{1}{11} \right )$(10y + x + 10x + y)

(9y-9x) =$\left ( \frac{1}{11} \right )$(11y + 11x) = (x + y)

10x = 8y

x =$\left ( \frac{4}{5} \right )$y

Clearly, y should be a single-digit multiple of 5, which is 5.

So, x = 4, y = 5.

Hence, woman's age = 10x + y = 45 years.

Clearly, while counting, the numbers associated to the thumb will be : 1, 9,17, 25,.....

i.e. numbers of the form (8n + 1).

Since 1994 = 249 x 8 + 2, so 1993 shall correspond to the thumb and 1994 to the index finger.

Let number of notes of each denomination be x.

Then, x + 5x + 10x = 480

$\Leftrightarrow$ 16x = 480

$\Leftrightarrow$ x = 30.

Hence, total number of notes = 3x = 90.

Since one of the numbers on the dial of a telephone is zero, so the product of all the numbers on it is 0.

Since B and D are twins, so B = D.

Now, A = B + 3 and A = C - 3.

Thus, B + 3 = C - 3

$\Leftrightarrow$ D + 3 = C-3

$\Leftrightarrow$ C - D = 6.

Clearly, every member except one (i.e. the winner) must lose one game to decide the winner. Thus, minimum number of matches to be played = 30 - 1 = 29.

Each row contains 12 plants.

There are 11 gapes between the two corner trees (11 x 2) metres and 1 metre on each side is left.

Therefore Length = (22 + 2) m = 24 m.

Manick's present age = 12 years, Rahul's present age = 4 years.

Let Manick be twice as old as Rahul after x years from now.

Then, 12 + x = 2 (4 + x)

$\Leftrightarrow$ 12 + x = 8 + 2x

$\Leftrightarrow$ x = 4.

Hence, Manick's required age = 12 + x = 16 years.

Number of cuts made to cut a roll into 10 pieces = 9.

Therefore Required number of rolls =$\frac{45\times 24}{9}$= 120.

Let the number of boys and girls participating in sports be 3x and 2x respectively.

Then, 3x = 15 or x = 5.

So, number of girls participating in sports = 2x = 10.

Number of students not participating in sports = 60 - (15 + 10) = 35.

Let number of boys not participating in sports be y.

Then, number of girls not participating in sports = (35 -y).

Therefore (35 - y) = y + 5

$\Rightarrow$ 2y

$\Rightarrow$ 30

$\Rightarrow$ y = 15.

So, number of girls not participating in sports = (35 - 15) = 20.

Hence, total number of girls in the class = (10 + 20) = 30.

Since there are socks of only two colours, so two out of any three socks must always be of the same colour.

Total number of routes from Bristol to Carlisle = (4 $\times$3 $\times$ 2) = 24.

Let money with Ken = x. Then, money with Mac = x + £ 3.

Now, 3x = (x + x + £ 3) + £ 2 x = £ 5.

Therefore Total money with Mac and Ken = 2x + £ 3 = £ 13

Let x and y be the number of deer and peacocks in the zoo respectively. Then,

x + y = 80 ...(i) and

4x + 2y = 200 or 2x + y = 100 ...(ii)

Solving (i) and (ii), we get) x = 20, y = 60.

Let the number of boys be x. Then, $\left ( \frac{3}{4} \right )$ x = 18 or x = 18 x$\left ( \frac{4}{3} \right )$ = 24.

If total number of students is y, then $\left ( \frac{2}{3} \right )$y = 24 or y = 24 x$\left ( \frac{3}{2} \right )$ = 36.

Therefore Number of girls in the class = (36 - 24) = 12.

Let son's age be x years. Then, father's age = (3x) years.

Five years ago, father's age = (3x - 5) years and son's age = (x - 5) years.

So, 3x - 5 = 4 (x - 5)

$\Rightarrow$3x - 5 = 4x - 20

$\Rightarrow$x = 15.

Lets Salary Be Rs. x Then tips = Rs$\left ( \frac{5}{4}x\right )$

Total income = $\left ( x+\frac{5}{4}x\right )$= Rs.$\left (\frac{9x}{4}\right )$

$\therefore$ Required Fraction =$\left ( \frac{5x}{4}\times \frac{4}{9x} \right )=\frac{5}{9}$

Clearly, from 1 to 100, there are ten numbers with 3 as the unit's digit- 3, 13, 23, 33, 43, 53, 63, 73, 83, 93; and ten numbers with 3 as the ten's digit - 30, 31, 32, 33, 34, 35, 36, 37, 38, 39.

So, required number = 10 + 10 = 20.

Less Cats more days (indirect proportion)

Less Mice less days  (direct proportion)

Let's the required no of days be x

Cat  4  :  100 mice  100 :  4 } :: x: 100

$\therefore$ 100$\times$4$\times$x =4$\times$100$\times$100

or x = $\left( \frac{4\times 100\times 100}{100\times 4 } \right )=100$

L.C.M. of 6, 5, 7, 10 and 12 is 420.

So, the bells will toll together after every 420 seconds i.e. 7 minutes.

Now, 7 $\times$ 8 = 56 and 7 $\times$ 9 = 63.

Thus, in 1-hour (or 60 minutes), the bells will toll together 8 times, excluding the one at the start.

Originally, let number of women = x. Then, number of men = 2x.

So, in city Y, we have : (2x - 10) = (x + 5) or x - 15.

Therefore Total number of passengers in the beginning = (x + 2x) = 3x = 45.

Clearly, we have :

A = B - 3 ...(i)

D + 5 = E ...(ii)

A + C = 2E ...(iii)

B + D = A + C = 2E ...(iv)

A + B + C + D + E=150 ...(v)

From (iii), (iv) and (v), we get: 5E = 150 or E = 30.

Putting E = 30 in (ii), we get: D = 25.

Putting E = 30 and D = 25 in (iv), we get: B = 35.

Putting B = 35 in (i), we get: A = 32.

Putting A = 32 and E = 30 in (iii), we get: C = 28.

Since each pole at the corner of the plot is common to its two sides, so we have :

Total number of poles needed = 27 $\times$ 4 - 4 = 108 - 4 = 104.

Lets The number of Adult and children be 2x and 3x Respectively

The illiterate population = (100 - 40)% of 2x + 85% of 3x

$\left ( \frac{60}{100}\times 2x \right )+\left ( \frac{85}{100}\times 3x \right )$

$\frac{6x}{5}+ \frac{51x}{20}=\frac{75x}{20}$

$\therefore$ Required Percentage $= \left ( \frac{75x}{20}\times \frac{1}{5x}\times 100\right )\% =75\%$

We have : A = 3B ...(i) and

C - 4 = 2 (A - 4) ...(ii)

Also, A + 4 = 31 or A= 31-4 = 27.

Putting A = 27 in (i), we get: B = 9.

Putting A = 27 in (ii), we get C = 50.

Let annu age today = x years.

Then, Varan's age after 1 year = (x + 1) years.

Therefore x + 1 = 2 (x - 12)

$\Rightarrow$x + 1 = 2x - 24

$\Rightarrow$x = 25.

There were all sparrows but six' means that six birds were not sparrows but only pigeons and ducks.

Similarly, number of sparrows + number of ducks = 6 and number of sparrows + number of pigeons = 6.

This is possible when there are 3 sparrows, 3 pigeons and 3 ducks i.e. 9 birds in all.

Let's the value of holiday be x

Then, pay the seven week's works = £300 + x

so, $\frac{300 + x}{7}\times 4$

$\Rightarrow$ £1200 + 4x = £210 + 7x

$\Rightarrow$ 3x = £990

$\Rightarrow$ x =  £330

Let's Tanya Share = Rs. x Then Veena's Share Rs. $\frac{x}{2}$

Total Bill = Rs. $\left ( x+\frac{x}{2}+\frac{x}{3} \right )$ = Rs. $\left ( \frac{11x}{6} \right )$

Amita's Share = Rs. $\left (\frac{2}{3}+\frac{x}{2} \right )$ =Rs. $\left ( \frac{x}{3} \right )$

$\therefore$ Required Fraction = $\left ( \frac{x}{2}\times \frac{6}{11x} \right ) = \frac{3}{11}$

Let total number of members be 100,

Then, number of members owning only 2 cars = 20.

Number of members owning 3 cars = 40% of 80 = 32.

Number of members owning only 1 car = 100 - (20 + 32) = 48.

Thus, 48% of the total members own one car each.

When Rahul was born, his brother's age = 6 years;

his father's age = (6 + 32) years = 38 years, his mother's age = (38 - 3) years = 35 years;

his sister's age = (35 - 25) years = 10 years.

Let number of horses = number of men = x.

Then, number of legs = 4x + 2x $\frac{x}{2}$ = 5x.

So, 5x = 70 or x = 14.

When Ravi's brother was born, let Ravi's father's age = x years and mother's age = y years.

Then, sister's age = (x - 28) years. So, x - 28 = 4 or x = 32.

Ravi's age = (y - 26) years. Age of Ravi's brother = (y - 26 + 3) years = (y - 23) years.

Now, when Ravi's brother was born, his age = 0 i.e. y - 23 = 0 or y = 23.

Let number of girls = x and number of boys = 3x.

Then, 3x + x = 4x = total number of students.

Thus, to find exact value of x, the total number of students must be divisible by 4.

Let's there be (x + 1) Member.

Then,  Father's Share = $\frac{1}{4}$ Share of each other member =$\frac{3}{4x}$

$\therefore$ 3 $\frac{3}{4x}$ =$\frac{1}{4}$

$\Rightarrow$ 4x = 36

$\Rightarrow$ x = 9

Hence, total number of family Members = 10

Let R, G and B represent the number of balls in red, green and blue boxes respectively.

Then, .

R + G + B = 108 ...(i),

G + R = 2B ...(ii)

B = 2R ...(iii)

From (ii) and (iii), we have G + R = 2x 2R = 4R or G = 3R.

Putting G = 3R and B = 2R in (i), we get:

R + 3R + 2R = 108 

$\Rightarrow$ 6R = 108 

$\Rightarrow$ R = 18.

Therefore Number of balls in green box = G = 3R = (3 x 18) = 54.

Total runs scored = (36 x 5) = 180.

Let the runs scored by E be x.

Then, runs scored by D = x + 5; runs scored by A = x + 8;

runs scored by B = x + x + 5 = 2x + 5;

runs scored by C = (107 - B) = 107 - (2x + 5) = 102 - 2x.

So, total runs = (x + 8) + (2x + 5) + (102 - 2x) + (x + 5) + x = 3x + 120.

Therefore 3x + 120 =180 

$\Rightarrow$ 3x = 60 

$\Rightarrow$ x = 20.

Total number of digits

= (No. of digits in 1- digit page nos. + No. of digits in 2-digit page nos. + No. of digits in 3- digit page nos.)

= (1$\times$ 9 + 2 $\times$ 90 + 3 $\times$267) = (9 + 180 + 801) = 990.

Let d and s represent the number of daughters and sons respectively.

Then, we have :

d - 1 = s and 2 (s - 1) = d.

Solving these two equations, we get: d = 4, s = 3.

Let's the member of guests be x Then,

Number of Bowls of Rice = $\frac{x}{2}$

Number of Bowls of Dal = $\frac{x}{3}$

Number Of Bowls of Meats = $\frac{x}{4}$

$\therefore$ $\frac{x}{2}+\frac{x}{3}+\frac{x}{4}=65$

$\Rightarrow$ $\frac{6x+4x+3x}{12} =65$

$\Rightarrow$ 13x = 65 $\times$ 12

$\Rightarrow$ x =$\left ( \frac{65\times 12}{13} \right )=60$

Ayush's present age = 10 years.

His mother's present age = (10 + 20) years = 30 years.

Ayush's father's present age = (30 + 5) years = 35 years.

Ayush's father's age at the time of Ayush's birth = (35 - 10) years = 25 years.

Therefore Ayush's father's age at the time of marriage = (25 - 2) years = 23 years.

Suppose the boy got x sums right and 2x sums wrong.

Then, x + 2x = 48 

$\Rightarrow$ 3x = 48 

$\Rightarrow$ x = 16.

Support their paths cross after x minutes.

Then, 11 + 57x = 51 - 63x

$\Rightarrow$ 120x = 40

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

Number of Floors covered by David in $\frac{1}{3}$ min =$\left ( \frac{1}{3}\times 57 \right )$ =19

so there paths cross at (11 + 19)th that is 30th floors.

Clearly, the required number would be such that it leaves a remainder of 1 when divided by 2, 3 or 4 and no remainder when divided by 5. Such a number is 25.

Clearly seven strike of clock have 6 intervals while 10 strike have 9 intervals.

$\therefore$ Required Time = $\left ( 9\times \frac{7}{6}\right )$ Seconds = $10\frac{1}{2}$ Seconds.

Let the number of cows be x and the number of hens be y.

Then, 4x + 2y = 2 (x + y) + 14 

$\Rightarrow$ 4x + 2y = 2x + 2y + 14 

$\Rightarrow$ 2x = 14 

$\Rightarrow$ x = 7

Let the father's age be x and the son's age be y.

Then, x - y = y or x = 2y,

Now, x = 36. So, 2y = 36 or y = 18.

Therefore Son's present age = 18 years.

So, son's age 5 years ago = 13 years.

Let's the total number of shots be x. Then,

Shorts Fired By A = $\frac{5}{8}x$

Shorts Fired By A = $\frac{3}{8}x$

Killing Shorts By A = $\frac{1}{3}\, of\, \frac{5}{8}x =\frac{5x}{24}$

Shorts missed By B = $\frac{1}{2}\, of\, \frac{3}{8}x =\frac{3x}{16}$

$\therefore$ $\frac{3x}{16}=27$ or x = $\left ( \frac{27\times 16}{3} \right )=144$

Birds Killed by A = $\frac{5x}{24}=\left ( \frac{5}{24}\times 144 \right )=30$

Girls = $\frac{3}{5}$ , Boys = $\left ( 1-\frac{3}{5} \right )=\frac{2}{5}$

Fraction of student absent = $\frac{2}{9}\, of\, \frac{3}{5}+\frac{1}{4}\, of\, \frac{2}{5}$ =$\frac{6}{45}+\frac{1}{10}$ =$\frac{21}{90}=\frac{7}{30}$

$\therefore$ Fraction of Students Present = $\left ( 1-\frac{7}{30} \right )$ = $\frac{23}{30}$

Let the daughter's age be x years.

Then, father's age = (3x) years.

Mother's age = (3x - 9) years; Son's age = (x + 7) years.

So, x + 7 = $\frac{(3x-9)}{2}$

$\Rightarrow$ 2x + 14 = 3x - 9 

$\Rightarrow$ x = 23

Therefore Mother's age = (3x - 9) = (69 - 9) years = 60 years.

Since the area is halved on folding, so each time the paper is folded in the center that is it's thickness becomes two fold each time. So, we have:

Thickness after 2 folds =$(2\times 2)mm=2^{2}\, mm$

Thickness after 3 folds =$(2^{2}\times 2)mm=2^{3}\, mm$and so on

$\therefore$ The Thickness after 50 folds $=2^{50}mm=\left ( \frac{2^{50}}{1000\times 1000} \right )km.$

Let's x = $\frac{2^{50}}{(1000)^{2}}$Then,

$log\, \, x= 50\, \, log\, \, 2-2\, \, log\, \, 1000=50\times 0.3010-2\times 3=9.050 -9$

So, log = antilog 9 = 1000000000

Hence, Thickness after 50 folds = x km = 1 billion km.

Let's the number of banana in the 2nd bunch be x.

Then, Number of banana in the 1st bunch $= x+\frac{1}{4}x=\frac{5}{4}x$

So,$\frac{5}{4}x-x=3$

$\Rightarrow$ 5x - 4x = 12

$\Rightarrow$ x = 12

$\therefore$ Number of banana in 1st bunch =$\left ( \frac{5}{4}\times 12 \right )=15$

Numbers from 1 to 60, which are divisible by 6 are : 6,12,18, 24, 30, 36,42, 48, 54, 60.

There are 10 such numbers.

Numbers from 1 to 60, the sum of whose digits is 6 are : 6, 15, 24, 33, 42, 51, 60.

There are 7 such numbers of which 4 are common to the above ones. So, there are 3such uncommon numbers.

Numbers from 1 to 60, which have 6 as one of the digits are 6, 16, 26, 36, 46, 56, 60.

Clearly, there are 4 such uncommon numbers.

So, numbers 'not connected with 6' = 60 - (10 + 3 + 4) = 43

Let the number be x. Then, x + 13x = 112 

$\Rightarrow$ 14x = 112 

$\Rightarrow$ x = 8

The seven pieces consist of 6 smaller equal pieces and one half cake piece.

Weight of each small piece = 20 g.

So, total weight of the cake = [2 $\times$ (20 $\times$6)]g= 240 g.

Let the number of 20-paise coins be x. Then, number of 25-paise coins = (324 - x).

Therefore 0.20 x x + 0.25 (324 - x) = 71  20x + 25 (324 - x) = 7100

$\Rightarrow$ 5x= 1000 

$\Rightarrow$x = 200.

Hence, number of 25-paise coins = (324 - x) - 124.

Clearly, the black cards are either clubs or spades while the red cards are either diamonds or hearts.

Let the number of spades be x. Then, number of clubs = (7 - x).

Number of diamonds = 2 x number of spades = 2x;

Number of hearts = 2 x number of diamonds = 4x.

Total number of cards = x + 2x + 4x + 7 - x = 6x + 7.

Therefore 6x + 7 = 13 

$\Rightarrow$ 6x = 6 

$\Rightarrow$ x - 1

Hence, number of clubs = (7 - x) = 6.

Let the fixed charge be Rs. x and variable charge be Rs.y per km. Then,

x + 16y = 156 ...(i) and

x + 24y = 204 ...(ii)

Solving (i) and (ii), we get: x = 60, y = 6.

Therefore Cost of travelling 30 km = 60 + 30 y = Rs. (60 + 30 $\times$ 6) = Rs. 240.

Number of Alterations required in 1 Shirt $=\left ( \frac{2}{3}+\frac{3}{4}+\frac{4}{5}\right )=\frac{133}{60}$

$\therefore$ Number of Alterations required in 50 shirts $=\left ( \frac{133}{60}\times 60 \right )=133$

Clearly, total number of handshakes = (9+ 8 + 7 + 6 + 5 + 4 + 3 + 2+1) = 45.

Let the total number of sweets be (25x + 8).

Then, (25x + 8) - 22 is divisible by 28

$\Rightarrow$(25x - 14) is divisible by 28 

$\Rightarrow$28x - (3x + 14) is divisible by 28

$\Rightarrow$(3x + 14) is divisible by 28 

$\Rightarrow$x = 14.

$\therefore$ Total number of sweets = (25 $\times$ 14 + 8) = 358.

Clearly, out of every 16 persons, there is one captain. So, number of captains$\left ( \frac{1200}{16}\right )$= 75.

Let number of keepers be x. Then,

Total number of feet = 2 x 50 + 4 x 45 + 4 x 8 + 2x = 2x + 312.

Total number of heads = 50 + 45 + 8 + x= 103 + x.

Therefore (2x + 312) = (103 + x) + 224 or x = 15.

Net ascent of the monkey in 1 hour = (30 - 20) feet = 10 feet.

So, the monkey ascends 90 feet in 9 hours i.e. till 5 p.m.

Clearly, in the next 1 hour i.e. till 6 p.m. the monkey ascends remaining 30 feet to touch the flag.

Let the ten's digit be x. Then, unit's digit = (11 - x).

So, number = 10x + (11 - x) = 9x + 11.

Therefore (9x + 11) + 27 = 10 (11 - x) + x 

$\Rightarrow$ 9x + 38 = 110 - 9x 

$\Rightarrow$ 18x = 72 

$\Rightarrow$ x = 4.

Thus, ten's digit = 4 and unit's digit = 7.

Hence, required number = 47.

Income on the first day = Re. 1.

Income on the 2nd day = Rs. (1$\times$ 2) = Rs. 2¹.

Income on the 3rd day = Rs. (2¹ $\times$ 2) = Rs. 2² and so on. Thus, Income on the rath day = Rs. 2^n-1.

Therefore Income on the 10th day = Rs. 2⁹.

Clearly, we have to first find two numbers whose difference is 2 and of which the smaller one is a perfect square and the bigger one a perfect cube.

Such numbers are 25 and 27.

Thus, Nitin is now 26 years old. Since the next perfect cube after 27 is 64,

so required time period = (64 - 26) years = 38 years.

Let's the total number of sweets be x

Then, $\frac{x}{400}-\frac{x}{175}=4$

$\Rightarrow$ 5x - 4x $\times$700

$\Rightarrow$ x = 2800

Clearly, number of ways of arranging 5 books = 5 ! = 5 x 4 x 3 x 2 x 1 = 120.

So, total time taken = 120 minutes = 2 hours.

Since the number of carbons is 2, only two copies can be obtained.

No. of digits in 1-digit page nos. = 1 $\times$9 = 9.

No. of digits in 2-digit page nos. = 2 $\times$90 = 180.

No. of digits in 3-digit page nos. = 3 $\times$900 = 2700.

No. of digits in 4-digit page nos. = 3189 - (9 + 180 + 2700) = 3189 - 2889 = 300.

$\therefore$ No. of pages with 4-digit page nos. = $\left ( \frac{300}{4} \right )$ 75.

Hence, total number of pages = (999 + 75) = 1074.