The pattern of number series:
21 × 0.5 -> 10.5, 
10.5 × 1 -> 10.5 (missing)
10.5 × 1.5 -> 15.75
15.75 × 2.0 -> 31.25
31.25 x 2.5 -> 78.75

The pattern of number series
6 + (1 × 13) -> 19,
19 + ( 3 x 13) -> 58
58 + (5 x 13) -> 123 (missing)
123 + (7 x 13) -> 214
214 + (9 x 13) -> 331

The pattern of number series:
14 + ( 1 × 2) -> 16
16 + (3 x 4) -> 28 
28 + ( 5 x 6) -> 58
58 + ( 7 x 8) -> 114
114 + (9 x 10) -> 204

The pattern of number series :
13.76 + (1×1.15) -> 14.91
14.91 + (2×1.15) -> 17.21
17.21 + (3×1.15) -> 20.66
20.66 + (4×1.15) -> 25.26
25.26 + (5×1.15) -> 31.01

The pattern of the number series is :
15 + 12 = 16
16 + 23 = 16 + 8 = 24
24 + 32 = 24 + 9 = 33
33 + 43 = 33 + 64 = 97
97 + 52 = 97 + 25 = 122

Gita's average earning over all the day's = (400 + 420 + 200 + 140)/4 = 1160/4 = Rs 290.

Rahul earned on Tuesday and Thursday = 180 + 340 = Rs 520

Naveen earned on Tuesday and Thursday = 160 + 260 = Rs 420
Total = 520 + 420 = Rs. 940

Gita donated Naveen = Rs. 420
Naveen's total earning on Wednesday = 420 + 120 = Rs. 540.

Rahul's earning on Monday = Rs. 240
Gita's earning on Tuesday = Rs. 200
Diff = 240 – 200 = Rs. 40.

Naveen's earning on Monday, Wednesday and Thursday

= 360 : 120 : 160

= 9 : 3 : 4

? = (13.001)³ = (13)³ = 2197 ≈ 2200.

? = 55 x 55 + 5 = 3025 + 5 = 3030

50.001% of 99.99 ÷ 49.999 = ?

(Take approx. value of each term)

? = (100 x 50)/100 ÷ 50 = 1 .

? = 999 + 900 - 350 = 1549

? = 2³ x (2)^-2 ÷ (4)^-4 = 2/ 4^-4 = 2 x 2⁸ = 2⁹ = 512.

Number of women liking
CSK = 44% of 45525 = 20031
DD = 33% of 36800 = 12144
DC = 30% 0f 56340 = 16902
MI = 28% of 62350 = 17458
RR = 44% of 48300 = 21252
RCB = 35% of 35580 = 12453
KXI = 36% of 56250 = 20250
KKR = 54% of 64000 = 34560
Total = 155050
.'. Required average = 155050/8 = 19381.25 = 19381

Required percentage = Number of women liking RR 100/Number of women likingall teams x 100 = 21252/155050 × 100 = 13.7%

Required number = 45525×36/100 = 16389.

Required ratio = (36800 × 39) : (48300 × 21) = 208 : 147

Number of men liking DC = 45% of 56340 = 25353
Number of men liking RCB = 15% of 355080 = 5337
.'. Required percentage 25353/5337 x 100 = 475.04%

(I) p² – 7p = – 12
=> p² – 7p + 12 = 0
=> p² – 4p – 3p + 12 = 0
=> p (p – 4) –3 (p –4) = 0
=> (p – 4) (p – 3) = 0
=> p = 3 or 4
(II) q² – 3q + 2 = 0
=> q² – 2q – q + 2 = 0
=> q (q – 2) –1 (q – 2) = 0
=> (q – 2) (q – 1) = 0
=> q = 1 or 2
Hence, p > q

(I) 12p² – 7p = – 1
=> 12p² – 7p + 1 = 0
=> 12p² –4p – 3p + 1 = 0
=> 4p (3p –1) –1 (3p –1) = 0
=> (3p –1) (4p – 1) = 0
=> p = 1/4 or 1/3
(II) 6q² – 7q + 2 = 0
=> 6q² – 4q – 3q + 2 = 0
=> 2q (3q – 2) –1 (3q – 2) = 0
=> (3q – 2) (2q – 1) = 0
=> 2/3 or 1/2
Obviously, p <= q

(I) p² + 12p + 35 = 0
=> p² + 7p + 5p + 35 = 0
=> p (p + 7) + 5 (p + 7) = 0
=> (p + 7) (p + 5) = 0
=> p = – 5 or – 7
(II) 2q² + 22q + 56 = 0
=> 2q² + 14q + 8q + 56 = 0
=> 2q (q + 7) + 8 (q + 7) = 0
=> (q + 7) (2q + 8) = 0
=> q = – 7 or – 4
Hence, relation can’t be established

(I) p² – 8p + 15 = 0
=> p² – 3p – 5p + 15 = 0
=> p (p – 3) – 5 (p – 3) = 0
=> (p – 3) (p – 5) = 0
=> p = 3 or 5
(II) q² – 5q = – 6
=> q² – 5q + 6 = 0
=> q² – 3q – 2q + 6 = 0
=> q (q – 3) – 2 (q – 3) = 0
=> q (q – 3) (q – 2) = 0
=> q = 3 or 2
Obviously, p >= q.

(I) 2p² + 20p + 50 = 0
=> p² + 10p + 25 = 0
=> (p + 5)² = 0
=> p + 5 = 0
=> p = – 5
(II) q² = 25
=> q = ±5
Obviously, p <= q.

According to the question, Let the number of boys and girls be x and 2x respectively.
=> (x - 2)/(2x - 2) = 1/3
=> 3x - 6 = 2x - 2
=> x = 4.
Now, number of boys = 4, Girls = 2 x 4 = 8.
Hence , initially number of people were = 12.

Rate of simple interest = (Interest × 100)/ (Principal × time)
= (10800 x 100)/(22500 x 4) = 12%.
Compound interest = Principal [ ( 1 + rate/100)^time - 1]
= 22500[( 1 + 12/100)² - 1]
= 22500[( 1 + 3/25)² - 1]
= 22500[(28/25)² - 1]
= 22500(784/625 - 1)
= 22500 x 159/625 
= Rs 5724

Total age of remaining 40 girls
= (80 × 20 – 20 × 22 – 20 × 24) years
= (1600 – 440 – 480) years
= 680 years
.'. Required average age = 680/40 = 17 years

Given, Speed of a train = 40 km/h = 40 * 5/18 m/s
Speed of another train = 20 m/s
.'.Required ratio = Speed of first train / Speed of second train
= (40 * 5/18 m/s) / 20 
= (2 x 5) / 18 
= 10/18 = 5/9 or 5 : 9.

Distance covered in downstream = 4 km
Time taken in downstream = 3 hours.
Rate of downstream = distance / time = a = 4km / 3 hours = 4/3 km/hr.
Distance covered in upstream = 2 km
Time taken in upstream = 2 hours.
Rate of upstream = distance / time = b = 2 km / 2 hours = 1 km/hr.
Speed in still water = (a + b)/2 = (1/2)(4/3 + 1) km/hr = (1/2)(4 + 3)/3 = 7/6 km/hr.
Time Taken to cover 8 km in still water = distance / speed = 8 x 6/7 = 48 / 7 ≈ 7 hours.

Given, Area of rectangle = Area of circle = 22/7 × 21 × 21 = 1386 sq. cm.
Let the length and breadth of rectangle be 14x and 11x cm respectively. 
Then,
=> 14x * 11x = 1386
=> x² = 1386/(14 x 11) = 9
.'. x = √9 = 3
Hence, Perimeter of rectangle = 2 (14x + 11x) = 50x = 50 × 3 = 150 cm.

Let the total distance be x km.
Then difference in time = 63 min.
x/5 - x/8 = 63/68.
=> (8x - 5x )/ 40 = 63/60
=> 3x = 63/60 x 40 
=> x = 42/3 = 14km

Since, 28 men can complete a piece of work in 15 days.
.'. 1 man can complete it in = 28 x 15 days
30 men can complete it in = (28 x 15)/30 = 14 days.
.'. 1 man 1 day work = 1/14
Again, 
Since, 15 women can complete the same piece od work in 24 days
.'. 1 woman can complete it in = 24 x 15 days
30 women can complete it in = (24 x 15)/24 = 20 days.
.'. 1 man 1 day work = 1/20
Required ratio = 1/14 : 1/20 = 20 : 14 = 10:7

Suppose the cost of one chair = C and the cost of one table = T
Then, 5C + 3T = 3110 .......(i)
and, T - C = 210 
On putting value of T in eq(i), we get :
=> 5C + 3(210 + C) = 3110
=> 5C + 630 + 3C = 3110
=> 8C = 3110 - 630
=> C = 2480/8 = Rs 310.
.'.cost of one table(T) = 210 + 310 = Rs 520
Hence, the cost of two tables and two chairs : 
= 2T + 2C = 2 x 520 + 2 x 310 = 1040 + 620 = Rs 1660.

Let the 8 consecutive odd numbers be x, x + 2, x + 4, x + 6, x + 8, x + 10, x + 12 and x + 14 respectively.
Then, x + (x + 2) + ( x + 6) + ( x + 8) + (x + 10) + (x + 12) + ( x + 14) = 656
=> 8x + 56 = 656
=> x = 600/8 = 75

smallest odd number = x = 75

Let the 4 consecutive even numbers be y, y + 2, y + 4 and y + 6 respectively.
Then, y + (y + 2) + ( y + 4) + ( y + 6) = 4 x 87.
=> 4y + 12 = 348
=> y = 336/4 = 84
.'. Second largest even number = y + 4 = 84 + 4 = 88.

Hence, required sum = 75 + 88 = 163.