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D = 540 + 180 = 720
S = 54 * 5/18 = 15 mps
T = 720/15 = 48 sec

12 * 80 = 16 * x
x = 60 days

Let the four consecutive even numbers be 2(x – 2), 2(x – 1), 2x, 2(x + 1)
Their sum = 8x – 4 = 292 => x = 37
Smallest number is: 2(x – 2) = 70.

Let CP = Rs. 100x
SP = Rs. 120x
MP = 120x/80 * 100 = Rs. 150x
D = Rs. 150x – Rs. 120x = Rs. 30x
D – P = 30x – 20x = Rs. 6, 10x = Rs. 6
120x = 120/10 * 6 = Rs. 72

Area of the path = Area of the outer circle – Area of the inner circle = ∏{4/2 + 25/100}² – ∏[4/2]²
= ∏[2.25² – 2²] = ∏(0.25)(4.25)  { (a² – b² = (a – b)(a + b) }
= (3.14)(1/4)(17/4) = 53.38/16 = 3.34 sq m

4 + 6 + 8 + 9 + 10 + 12 + 14 + 15 + 16 + 18 = 132

Let the monthly incomes of Amar and Bhuvan be 6x and 5x respectively.
Let the monthly expenditure of Amar and Bhuvan be 3y and 2y respectively.
Savings of Bhuvan every month = 1/4(5x)
= (His income) – (His expenditure) = 5x – 2y.
=> 5x = 20x – 8y => y = 15x/8.
Ratio of savings of Amar and Bhuvan
= 6x – 3y : 1/4(5x) = 6x – 3(15x/8) : 5x/4 = 3x/8 : 5x/4
= 3 : 10.

L.C.M of 5, 6, 4 and 3 = 60.
On dividing 2496 by 60, the remainder is 37. Number to be added = 60 – 37 = 23.

64309 – 8703 + 798 – 437 => 65107 – 9140 = 55967

1 to 9     = 9 * 1        =      9
10 to 99 = 90 * 2      =   180
100 to 999 = 900 * 3 =2700
1000                          =       4
———–
2893

Let the marks obtained by the student in Mathematics, Physics and Chemistry be M, P and C respectively.
Given , M + C = 60 and C – P = 20 M + C / 2 = [(M + P) + (C – P)] / 2 = (60 + 20) / 2 = 40.

K⁺²M⁺¹N^-2L, P⁺²R⁺¹S^-2Q, V⁺¹W⁺²Y⁺¹Z, J⁺²L⁺¹M^-2K and W⁺²Y⁺¹Z^-2X.
Except VWYZ, all the other groups follow similar pattern.

Y⁺⁴C^-2A⁺⁴E^-2C, K⁺⁴O^-2M⁺⁴Q^-2O, P⁺⁴T^-2R⁺³U^-1T, G⁺⁴K^-2I⁺⁴M^-2K and D⁺⁴H^-2F⁺⁴J^-2H.
Except PTRUT, all other groups follow similar pattern.

The total number of stations = 20
From 20 stations we have to choose any two stations and the direction of travel (i.e., Hyderabad to Bangalore is different from Bangalore to Hyderabad) in ²⁰P₂ ways.
²⁰P₂ = 20 * 19 = 380.

Ratio of speeds = 3:4
Ratio of times taken = 4:3
Suppose A takes 4x hrs and B takes 3x hrs to reach the destination.
Then, 4x – 3x = 30/60 => x = 1/2
Time taken by A = 4x hrs = 4 * 1/2 = 2 hrs.

Milk = 3/5 * 20 = 12 liters, water = 8 liters
If 10 liters of mixture are removed, amount of milk removed = 6 liters and amount of water removed = 4 liters.
Remaining milk = 12 – 6 = 6 liters
Remaining water = 8 – 4 = 4 liters
10 liters of pure milk are added, therefore total milk = (6 + 10) = 16 liters.
The ratio of milk and water in the new mixture = 16:4 = 4:1
If the process is repeated one more time and 10 liters of the mixture are removed, then amount of milk removed = 4/5 * 10 = 8 liters.
Amount of water removed = 2 liters.
Remaining milk = (16 – 8) = 8 liters.
Remaining water = (4 -2) = 2 liters.
The required ratio of milk and water in the final mixture obtained = (8 + 10):2 = 18:2 = 9:1.

Let the weights of the three boys be 4k, 5k and 6k respectively.
4k + 6k = 5k + 45
=> 5k = 45 => k = 9
Therefore the weight of the lightest boy = 4k = 4 * 9 = 36 kg.

Suppose they meet x hours after 7 a.m.
Distance covered by A in x hours = 20x km.
Distance covered by B in (x – 1) hours = 25(x – 1) km.
Therefore 20x + 25(x – 1) = 110
45x = 135
x = 3.
So, they meet at 10 a.m.

20 = 4² + 4, 42 = 6² + 6, 58 = 7² + 9, 72 = 8² + 8 and 90 = 9² + 9.
20, 42, 72 and 90 can be expressed in n² + n form but not 58.

A can do 1/10 of the work in a day.
B can do 1/12 of the work in a 1 day.
Both of them together can do (1/10 + 1/12) part of work in 1 day = (6 + 5)/60 = 11/60
They take 60/11 days to complete the work together.
Given that they already worked for 2 days.
The number of days required to complete remaining work => 60/11 – 2 = 38/11 = 3 (5/11) days.

Sum of 10 even numbers = 10 * 11 = 110
Average = 110/10 = 11

Sum = (50 * 100) / (2 * 5) = Rs. 500
Amount = [500 * (1 + 5/100)²] = Rs. 551.25
C.I. = (551.25 – 500) = Rs. 51.25.

As numerator is of the form a² – b² = (a + b) (a – b), so the given expression simplifying becomes
(0.9² – 0.8²)/(0.9 + 0.8) (0.9 + 0.8) (0.9 – 0.8)/0.9 + 0.8 = 0.1

Let the present ages of Ajay and Vijay be ‘A’ and ‘V’ years respectively.
A – 8 = 4/3 (V – 8) and A + 8 = 6/5 (V + 8)
3/4(A – 8) = V – 8 and 5/6(A + 8) = V + 8
V = 3/4 (A – 8) + 8 = 5/6 (A + 8) – 8
=> 3/4 A – 6 + 8 = 5/6 A + 20/3 – 8
=> 10 – 20/3 = 10/12 A – 9/12 A
=> 10/3 = A/12 => A = 40.

600 —- 100
100 —- ? => 16 2/3%

x² + 5x – 3x – 15 = 0
x(x + 5) – 3(x + 5) = 0
(x – 3)(x + 5) = 0
=> x = 3 or x = -5.

74330 Largest
30347 Smallest
————-
43983

144) 9999 (69
9936
——–
63
9999 – 63 = 9936

Work done by P alone in one day = 1/6th of the total work done by Q alone in one day = 1/3(of that done by P in one day) = 1/3(1/6 of the total) = 1/18 of the total.
Work done by P and Q, working together in one day = 1/6 + 1/18 = 4/18 = 2/9 of the total
They would take 9/2 days = 4 (1/2) days to complete the work working together.

Given Exp. = 35 + 15 * 3/2
= 35 + 45/2 = 35 + 22.5 = 57.5

36 * 48 / 64 + 36/12 = 27 + 3 = 30

x – 3/8 x = 25
x = 40

1 man’s 1 day work = 1/108
12 men’s 6 day’s work = 1/9 * 6 = 2/3
Remaining work = 1 – 2/3 = 1/3
18 men’s 1 day work = 1/108 * 18 = 1/6
1/6 work is done by them in 1 day.
1/3 work is done by them in 6 * 1/3 = 2 days.

28 * 2 = 56
20 * 3 = 60
———–
4 years

19x + 19y + 17 = -19x + 19y – 21
38x = -38 => x = -1

No two dice show same number would mean all the three faces should show different numbers. The first can fall in any one of the six ways. The second die can show a different number in five ways. The third should show a number that is different from the first and second. This can happen in four ways.
Thus 6 * 5 * 4 = 120 favourable cases.
The total cases are 6 * 6 * 6 = 216.
The probability = 120/216 = 5/9.

2x   3x
8x cube + 27x cube = 945
35x cube = 945
x cube = 27 => x = 3

Number of men in the colony = 4/7 * 70 = 40.
Number of women in the colony = 3/7 * 70 = 40.
Number educated women in the colony = 1/5 * 30 = 6.
Number of uneducated women in the colony = 4/5 * 50 = 24.
Number of educated persons in the colony = 8 /35 * 70 = 16.
As 6 females are educated, remaining 10 educated persons must be men.
Number of uneducated men in the colony = 40 – 10 = 30.
Number of educated men and uneducated men are in the ratio 10 : 30 i.e., 1 : 3.

2 * 22/7 * 14 = 88
88 * 1 1/2 = Rs.132

(25 * 25)/25 + 15 * 40 = 25 + 600 = 625.

Let the length of the stationary train Y be L_Y
Given that length of train X, L_X = 300 m
Let the speed of Train X be V.
Since the train X crosses train Y and a pole in 60 seconds and 25 seconds respectively.
=> 300/V = 25       —> ( 1 )
(300 + L_Y) / V = 60 —> ( 2 )
From (1) V = 300/25 = 12 m/sec.
From (2) (300 + L_Y)/12 = 60
=> 300 + L_Y = 60 (12) = 720
=> L_Y = 720 – 300 = 420 m
Length of the stationary train = 420 m

Except 110, other numbers are divisible by 4.

Let the sides of the rectangle be l and b respectively.
From the given data,
(√l² + b²) = (1 + 108 1/3 %)lb
=> l² + b² = (1 + 325/3 * 1/100)lb
= (1 + 13/12)lb
= 25/12 lb
=> (l² + b²)/lb = 25/12
12(l² + b²) = 25lb
Adding 24lb on both sides
12l² + 12b² + 24lb = 49lb
12(l² + b² + 2lb) = 49lb
but 2(l + b) = 28 => l + b = 14
12(l + b)² = 49lb
=> 12(14)² = 49lb
=> lb = 48
Since l + b = 14, l = 8 and b = 6
l – b = 8 – 6 = 2m.

Let the two-digit number be 10a + b
a = b + 2 — (1)
10a + b = 7(a + b) => a = 2b
Substituting a = 2b in equation (1), we get
2b = b + 2 => b = 2
Hence the units digit is: 2.

The word contains five consonants. Three vowels, three consonants can be selected from five consonants in ⁵C₃ ways, two vowels can be selected from three vowels in ³C₂ ways.
3 consonants and 2 vowels can be selected in ⁵C₂ . ³C₂ ways i.e., 10 * 3 = 30 ways.

30 * 33/2 = 5/2 * 2 * x => x = 99

25% of x = x/4 ; 15% of 1500 = 15/100 * 1500 = 225
Given that, x/4 = 225 – 15 => x/4 = 210 => x = 840.

Original price = 100
CP = 80
S = 80*(140/100) = 112
100 – 112 = 12%