SI = PTR/100

SI = PX100/100

SI = P

So, Amount=P+SI=2P→22500 = 2P→P = 11250

Let larger number be x and smaller number be y.

Also, y = 10

According to question,

⇒(x + y) = 3((x –y)⇒(x + 10) = 3(x –10)

⇒2x = 40⇒x = 20

Let there be x one rupee coin, y 50 paise coin, z 25 paise coinx + y + z = 220......(i)

Total amount = 160

Now, x + (y/2) + z = 160 ......(ii)

⇒4x + 2y + z = 640

Thrice as many 1 re coin as 25 paise

⇒x = 3z ......(iii)

On solving equation (i), (ii) and (iii), we get

z = 40x= 3* 40 = 120y = 220 –(120 + 40 ) = 60

So, 50 paise coin =60

Working efficiency of both typist together = 100/6 = 16.66%

Let the efficiency of first typist be x and second typist be (16.66 –x)

First typist for 4 minute + second tyoistfor 6 minute = 80% work

⇒4x + (16.66 –x) X 6 = 80%

⇒4x + 100% -6x = 80%,

⇒x = 10%

First typist typed 10% per minute, second typst typed (16.66 –10) = 6.66% per minute

Slowest is second typist complete whole job in 100/6.66 = 15.01 = 15 minute

Let the average expenditure of 19 people be x

⇒19x = 13 X 79 + 6 X (x+4)⇒19x = 13 X 79 + 6x +24

⇒x = 80.84

So, total money spent by them = 1536.07

Let the number be x

When 10 is added, number = x + 10

When 7 is multiplied, number = 7 X (x + 10)

According to question,

⇒[7 X (x+10) / 5] –5 = 44

⇒7x + 70 = 49 X 5

⇒x + 10 = 35,

⇒x = 25

Quantity of food = 200 X 31 = 6200 man-days

Till 27 days, quantity consumed = 27 X 200 = 5400 man-days

Remaining quantity = 6200 –5400 = 800 man days

Remaining soilders = 200 –120 = 80

Let remaining food will be sufficient for x days⇒5400 + 80x = 6200⇒x= 10 days

Let a + b = 85→b = 85 –a

LCM(a,b) =102→ab = 102

k→a(85 –a) = 102k, where k is a natural number.