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Let the speeds of the two trains be x m/sec and y m/sec respectively. Then, length of the first train = 27 x meters, and length of the second train = 17 y meters. (27 x + 17 y) / (x + y) = 23 ==> 27 x + 17 y = 23 x + 23 y ==> 4 x = 6 y ==> x/y = 3/2.

100 — 10 DS = 10
? —- 1
75 —- 15 US = 5
? —– 1 S = (10 – 5)/2
= 2 2 ½ kmph

15 CP = 18 SP
18 — 3 CP loss
100 — ? => 16 2/3% loss

(20 * 16) women can complete the work in 1 day.
1 woman’s 1 day work = 1/320
(16 * 15) men can complete the work in 1 day
1 man’s 1 day work = 1/240
So, required ratio = 1/240 : 1/320 = 4:3.

(10x + y) – (10y + x) = 54
x – y = 6

Let the sum be Rs.P. Let the rate of interest be R% p.a.
(P)(8)(R)/100 = [1 + 6/100][ (P)(5)(R) /100] = 1.6{5PR/100]
8PR/100 = 8PR/100 which is anyway true.
R cannot be found.

It all 3 numbers have to be same basically we want triplets. 111, 222, 333, 444, 555 and 666. Those are six in number. Further the three dice can fall in 6 * 6 * 6 = 216 ways.
Hence the probability is 6/216 = 1/36

Let the two successive numbers be n, n + 1.
It is given that n(n + 1) = 4556.
=> n² + n – 4556 = 0
=> n² + 68n – 67n – 4556 = 0
=> n(n + 68) – 67(n + 68) = 0
=> n = 67 and (n + 1) = 68.

Age of the 15th student=
[15 * 15 – (14 * 5 + 16 * 9)] = (225 – 214) = 11 years.

4 πr² : 3 πr² => 4:3

(A + B)’s 1 day work = 1/12;
(B + C)’s 1 day work = 1/15;
(A + C)’s 1 day work = 1/20
Adding, we get: 2(A + B + C)’s 1 day work = (1/12 + 1/15 + 1/20) = 1/5
(A + B + C)’s 1 day work = 1/10
So, A, B and C together can complete the work in 10 days.

Let the cistern be filled by pipe A alone in x hours.
Then, pipe B will fill it in (x + 6) hours.
1/x  + 1/(x + 6) = 1/4
x² – 2x – 24 = 0
(x – 6)(x + 4) = 0 => x = 6.

Amount of Rs. 100 for 1 year when compounded half-yearly = [100 * (1 + 3/100)²] = Rs. 106.09 Effective rate = (106.09 – 100) = 6.09%

2000 + 300 + 40 + 5 = 2345

30 * 20 * x = (8 * 5.5 * 1.5)/2

4x           5x           6x
100        50           25
400x + 350x + 150x = 16000
x = 20
6x = 120

X – (10/100) X = X * ?
? = 90%

Distance covered by Amar = 18/4.8 (1.6km) = 3/8(1600) = 600 m

16 years ago, let T = x years and G = 8x years. After 8 years from now, T = (x + 16 + 8) years and G = (8x + 16 + 8) years.
8x + 24 = 3(x + 24) => 5x = 48
8 years ago, T/G = (x + 8)/(8x + 8) = (48/5 + 8) / (8 * 48/5 + 8) = 88/424 = 11/53

(2*2)/1 = 4

Final answer = (10 1/2 – 4 2/3) * 450
= 5 5/6 * 450 = 5 * 450 + 5/6 * 450
= 2250 + 375 = 2625.

P = 3Q
P + Q = 3Q + Q = 4Q
These 4Q people can do the work in 12 days, which means Q can do the work in 48 days.
Hence, P can do the work in 16 days.

6 – [5/6 + {3 7/8 – 2 1/3} + 1 7/9] => 6 – [5/6 + {3 1/8 – 7/3} + 16/9]
=> 6 – [5/6 + {(279 – 168 + 128)/72}] => 6 – 5/6 + 239/72
=> 6 – (60 + 239)/72 => 6 – 46 – 11/72
=> 2 – 11/72 => 1 + 1 – 11/72 => 1 + 61/72 = 1 61/72

Milk = 3/5 * 125 = 75 liters
Water = 50 liters
75 : (50+P) = 3:4
150 + 3P = 400 => P = 50
50 liters of water are to be added for the ratio become 3:4.

Let the number of papers be x. Then,
63x + 20 + 2 = 65x
= 2x = 22
= x = 11.

X * (110/100) * (95/100) = 1045
X * (11/10) * (1/100) = 11
X = 1000

520 = 26 * 20 = 2 * 13 * 2² * 5 = 2³ * 13 * 5
Required smallest number = 2 * 13 * 5 = 130
130 is the smallest number which should be multiplied with 520 to make it a perfect square.

Sum of the 35 numbers = 35 * 25 = 875
If each number is multiplied by 5, the sum also gets multiplied by 5 and the average also gets multiplied by 5.
Thus, the new average = 25 * 5 = 125.

1/30 + 1/40 = 7/120
120/7 = 17 1/7 days

Work done by A and B in the first two hours, working alternately = First hour A + Second hour B = 1/4 + 1/12 = 1/3.
Total time required to complete the work = 2 * 3 = 6 days.

100x + 400 = 12(x – 5)
x = 50
100 * 50 + 400 = 5400

2x² + 4x + x + 2 = 0
2x(x + 2) + 1(x + 2) = 0
(x + 2)(2x + 1) = 0 => x = -2, -1/2

Let the numbers be x and y.
Then x + y = 25 and x – y = 13.
x = 19; y = 6
xy = 19 * 6 = 114

8 = (?)³
=> ?³ = 2³ => ? = 2

π * 12 * 4 = 48

Speed = 54 * 5/18 = 15 m/sec.
Length of the train = 15 * 20 = 300 m.
Let the length of the platform be x m . Then,
(x + 300)/36 = 15 => x = 240 m.

Let the present ages of A, B and C be a, b and c respectively.
Given, [(a – 5) + (b – 5)] / 2 = 40 => a + b = 90 — (1)
(b + c)/2 = 48 => b + c = 96 — (2)
From (1) and (2), we cannot find b.

y = 3√0.008 = 0.2
y⁵ = (0.2)⁵ = 0.00032.

S = 600/40 = 15 mps
S = 3600/15 = 240 sec = 4 min

1/18 + 1/30 = 8/90 = 4/45
45/4 = 11 ¼ *2 = 22 ½ days

3^4m+1 = 3^7m-5 equating powers of 3 on both sides, 4m+1 = 7m – 5 => 3m = 6 => m = 2.

S = (1 + 2 + 3)/2 = 3
=> No triangle exists

Total time taken = x/40 + 2x/20 hours
= 5x/40 = x/8 hours
Average speed = 3x/(x/8) = 24 kmph

A can do the work in 30/2 i.e., 15 days.
A and B’s one day’s work = 1/15 + 1/30 = (2 + 1)/30 = 1/10
So A and B together can do the work in 10 days.

Let C.P. = Rs. 5x and S.P. = Rs. 7x.
Then, Gain = Rs. 2x
Required ratio = 2x : 5x = 2:5

Except Regiment, all others are caders of army officers.

Kim’s wages : David’s wages = Kim’s 1 day work : David’s 1 day work = 1/3 : 1/2 = 2:3
Kim’s share = 2/5 * 150 = Rs. 60

Given that the area of the square = 1225 sq.units
=> Side of square = √1225 = 35 units
The radius of the circle = side of the square = 35 units Length of the rectangle = 2/5 * 35 = 14 units
Given that breadth = 10 units
Area of the rectangle = lb = 14 * 10 = 140 sq.units