A:B:C ⇒ 12 : 15 : 25

Let 12x, 15x, 25x

Sum=52 x 364

x = 364/52

x = 7

A:B:C=84:105:175

(diff of)AB:CB  = 21:70=3:10

The percentage of students fails in both subjects = 100-[(60+45) -25] = 20%

Let CP be Rs. x

then, 900 - x = 2(x - 450)                                 [profit = 2 Loss]

3x = 1800 ⇒ x= Rs.600

CP = Rs.600,  gain required = 25%

SP = (100+ gain%) x CP/100

SP = (100 + 25) × 600/100 = Rs.750

Let the money in first installment = Rs. x

Second installment = Rs. (35000 - x)

The principal is defined as:

⇒ P = (3600 × 100)/(12 × 2) = Rs.15000

(Using SI in scheme A)

The money in first installment = Rs. x = Rs.15000

Scheme B=Rs. (35000 - 15000) = Rs.20000

⇒ CI = P[(1 + R/100)² - 1]

= 2000[(1 + 10/100)² - 1] = Rs.4200

Total price of all the articles. = 5 × 25 + 10 × 50 + 15 × 35

= 125 + 500 + 525 = 1150Rs .

Average price = 1150/30 = 38.33 Rs.

5men = 7 women

∴ 7 men = 7/5 × 7 = 49/5 women

and 7 men + 13 women women = 49/5 + 13 = 114/5 women

Now,

∴ 7 women get Rs. 5250

∴ 114/5 women get 5250/7 × 114/5 = Rs.17100

Part of the tank that will be filled by both pipe A and B in 1 min = 1/24 + 1/32 = (4 + 3)/96 = 7/96

Let after x min, pipe B should be closed

x × 7/96 + (9 - x)/24 =1 ⇒ 7x+36-4x /96 =1

3x+36=96 ⇒ 3x=60 ⇒ x = 20min

Volume of cylinder = πr²h

Given, r = 25/2 cm

Volume = 22/7 x 25/2 x 25/2 x h = 11000 cm

h = 11000 × 7 × 2 × 2/ 22 × 25 × 25  ⇒ h = 112/5 = 22 2/5 cm

Speed of the train = 110/3 m/sec.

Required time = (165 + 110)/110 × 3

= 275/110 × 3

= 7.5 sec

Let Distance = d

d/5 - d/6 = 25/60

d/30 = 25/60 ⇒ d = 12.5 km

(c); Let the speed of the motorboat = 36x

Speed of current = 5x

let distance = d

d/(41x) = 5 (10/60) = 31/6 ⇒ d/31x = 41/6

Means, time to come back of motor boat = 41/6 hr

= 41/6 × 60min = 410min = 6h 50min

By the rule of alligation

∴ ratio = 4:6 = 2:3

required quantity = 3/5 × 1000 = 600 kg

Length of the outer rectangle = 60m

Breadth of the outer rectangle = 40 m

Area = 60×40= 2400 m².

Width of path = 1 m

Length of the inner rectangle = 60m - (1 + 1) m

= 58 m

Breadth of the inner rectangle = 40-2 = 38 m

Area = 58 × 38 = 2204 m²

Area of path = [Area of outer rectangle] - [Area of inner rectangle]

= 2400 - 2204 = 196m²

By formula

Area= √3/4 × (side)²    ⇒ √3/4 × (side)² = 4√3

(side)² = 16 ⇒ side = 4 cm

∴ perimeter = 3 × side = 3×4 = 12 cm

Let length =1, breadth = b

New length = 1 + 60/100 × 1 =81/5

New breadth = a ⇒ lb = 81/5 × a  ⇒ a=5b/8

Decrease percent =  (b-5b/8) / b × 100 = 3/8 × 100

= 37 × 1/2%

Total no. of children = 6 + 4 = 10

Total no. of ways to select 4 children = 10^ C_4 = 210

No. of ways to select no boys = 4^C_₁ =1

Total no. of ways to select at least one boy = 10^C _4  - 4^C_4 = 210 - 1 = 209

Total no of balls = 15+20 = 35

Total no. of ways to select 2 balls = 35^C_2

Probability of getting both white = 20^C_2 / 35^C_2

= 20 × 19 / 35 × 34

= 38/119

3 × 2 + 1 = 7,  7 × 2 + 1 = 15,  15 × 2 + 1 = 31,

31 × 2 + 1 = 63,  63 × 2 + 1 = 127

Missing term = 31

Ratio → 1:3:2

x + 25 + 3x + 25 + 2x + 25 = 735

6x + 75 = 735 ⇒ 6x = 660  ⇒ x=110

Money received by C = 2x + 25

= 2 × 110 + 25 = Rs. 245

Let B's salary be 100, A's salary be 120

Required pecentage = 20/120 × 100 = 100 / 6 = 16.67%