Mean weight of 35 students = 45 kg.

Total weight of 35 students = (45 × 35) kg = 1575 kg.

Mean weight of 35 students and the teacher (45 + 0.5) kg = 45.5 kg.

Total weight of 35 students and the teacher = (45.5 × 36) kg = 1638 kg.

Weight of the teacher = (1638 - 1575) kg = 63 kg.

Hence, the weight of the teacher is 63 kg.

Assume x soldiers join the fort. 1200 soldiers have provision for 1200 (days for which provisions last them)(rate of consumption of each soldier)

= (1200)(30)(3) kg.

Also provisions available for (1200 + x) soldiers is (1200 + x)(25)(2.5) k

As the same provisions are available

=> (1200)(30)(3) = (1200 + x)(25)(2.5)

x = [(1200)(30)(3)] / (25)(2.5) – 1200 => x = 528.

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 is 12th letter and 12 * 2 = 24

The 24th letter is X.

Similarly, E is the 5th letter and 5 * 2 = 10

The 10th letter is J.

Calculated average height of 30 boys = 150 cm.

Incorrect sum of the heights of 30 boys

= (150 × 30)cm

= 4500 cm.

Correct sum of the heights of 30 boys

= (incorrect sum) - (wrongly copied item) + (actual item)

= (4500 - 135 + 165) cm

= 4530 cm.

Correct mean = correct sum/number of boys

= (4530/30) cm

= 151 cm.

Hence, the correct mean height is 151 cm.

Let the least number to be subtracted be x, then 11 – x, 15 – x, 21 – x and 30 – x are in proportion.

<=> (11 – x) : (15 – x) = (21 – x) : (30 – x)

=> (11 – x)(30 – x) = (15 – x)(21 – x)

From the options, when x = 3

=> 8 * 27 = 12 * 18

Remaining work = (1 – 7/48) = 41/48

Work done by 1 boy in 1/4 day = ( 1/12 * 1/4) = 1/48

Number of boys required = 41/48 * 41 = 41

Mean weight of 7 boys = 56 kg.

Total weight of 7 boys = (56 × 7) kg = 392 kg.

Total weight of 6 boys = (52 + 57 + 55 + 60 + 59 + 55) kg = 338 kg.

Weight of the 7th boy = (total weight of 7 boys) – (total weight of 6 boys) = (392 – 338) kg = 54 kg.

Mean score of 9 innings = 58 runs.

Total score of 9 innings = (58 x 9) runs = 522 runs.

Required mean score of 10 innings = 61 runs.

Required total score of 10 innings = (61 x 10) runs = 610 runs.

Number of runs to be scored in the 10th innings

= (total score of 10 innings) - (total score of 9 innings)

= (610 -522) = 88.

Hence, the number of runs to be scored in the 10th innings = 88.

mod x1 = mod x2 => x1 & x2 have same value but opposite in sign.

in this case b=0 eqn is ax^2+c = 0 => x=sqrt(-c/a) for x to be real (-c/a) should +ve => c & a are of opposite sign so c>a or a>c for ex x^2-4=0 => x1=-2,

x2=2 & modx1=modx2=2 [a>c] or may -x^2+4=0 => x1=-2,x2=2 & modx1=modx2=2 (d) none

x^3-5x+c=0 if x=1,

1-5+c=0 or c=4 if x=-1,

-1+5+c=0 or c=-4 if x=2, 8-10+c=0 or

c=2 if x=-2, -8+10+c=0 or c=-2 we see that,

for c=-2,2,-4,4

we get x=-2,2,-1,1 etc i.e we get integral roots of x for infinite values of c. d)infinite

Let the given numbers be x1, x2, x3, ….. x14.

Then, the mean of these numbers = x1 + x2 + x3+ ….. x14/14

Therefore, (x1 + x2 + x3 + ….. x14)/14 = 6

⇒ (x1 + x2 + x3 + ….. x14) = 84 ………………. (A)

The new numbers are (x1 + 3), (x2 + 3), (x3 + 3), …. ,(x14 + 3)

Mean of the new numbers

= (x1 + 3), (x2 + 3), (x3 + 3), …. ,(x14 + 3)/14

= (x1 + x2 + x3 + ….. x14) + 42

= (84 + 42)/14, [Using (A)]

= 126/14

= 9

Hence, the new mean is 9.

9/6 remainder is 3 9^2/6 remainder is 3 9^3/6 remainder is 3 9 to the power of any number when divided by 6 ,

the remainder will always be 3.

Now, ( 3+3+3 ……11 times)/6 =(3*11)/6;

therefore the remainder will be 3.

If we take the even multiple of 11, then remainder will be zero.

Therefore answer is cannot be determine

Mean of 5 numbers = 28.

Sum of these 5 numbers = (28 x 5) = 140.

Mean of the remaining 4 numbers = (28 - 2) =26.

Sum of these remaining 4 numbers = (26 × 4) = 104.

Excluded number

= (sum of the given 5 numbers) - (sum of the remaining 4 numbers)

= (140 - 104)

= 36.

Hence, the excluded number is 36.

Let the given numbers be x1, x2, . . ., x8.

Then, the mean of these numbers = (x1 + x2 + ...+ x8)/8.

Therefore, (x1 + x2+...+x8)/8 = 25

⇒ (x1 + x2 + ... + x8) = 200 ……. (A)

The new numbers are (x1 - 5), (x2 - 5), …… ,(x8 - 5)

Mean of the new numbers = {(x1 - 5) + (x1 - 5) + …… + (x8 - 5)}/8

= [(x1 + x2 + ... + x8) - 40]/8

= (200 - 40)/8, [using (A)]

= 160/8

= 20

Hence, the new mean is 20.

Mean of the first 13 observations = 32.

Sum of the first 13 observations = (32 × 13) = 416.

Mean of the last 13 observations = 39.

Sum of the last 13 observations = (39 × 13) = 507.

Mean of 25 observations = 36.

Sum of all the 25 observations = (36 × 25) = 900.

Therefore, the 13th observation = (416 + 507 - 900) = 23.

Hence, the 13th observation is 23.

Calculated mean of 16 items = 30.

Incorrect sum of these 16 items = (30 × 16) = 480.

Correct sum of these 16 items

= (incorrect sum) - (sum of incorrect items) + (sum of actual items)

= [480 - (22 + 18) + (32 + 28)]

= 500.

Therefore, correct mean = 500/16 = 31.25.

Hence, the correct mean is 31.25.

Mean weight of 7 boys = 56 kg.

Total weight of 7 boys = (56 × 7) kg = 392 kg.

Total weight of 6 boys = (52 + 57 + 55 + 60 + 59 + 55) kg

= 338 kg.

Weight of the 7th boy = (total weight of 7 boys) - (total weight of 6 boys)

= (392 - 338) kg

= 54 kg.

Hence, the weight of the seventh boy is 54 kg.

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.

Total expenditure during the year

= $[6240 × 3 + 6780 × 4 + 7236 × 5]

= $ [18720 + 27120 + 36180]

= $ 82020.

Total income during the year = $ (82020 + 7080) = $ 89100.

Average monthly income = (89100/12) = $7425.

Hence, the average monthly income of the family is $ 7425.

= 200 * 5/18 = 500/9 m/sec.

Let the length of the other train be x m.

Then, (x + 270)/9 = 500/9 => x = 230.

Clearly, A beats B by (45-36)=9 seconds

Speed of B = Distance Time=10045 m/sDistance Covered by B in 9 seconds = Speed × Time = 10045×9 = 20 metre i.e., A beats B by 20 metre

So, missing term = 210 + [6 x (6 + 15)] = 210 + 126 = 336.