Using BODMAS rule we get,

⇒ 5 + (5 × 5) – 5 + 5

⇒ 5 + 25 – 5 + 5

⇒ 30 

Adding digits only after decimal:

⇒ ? = 950 + 016 + 980

= 1946

Hence the number we obtain after addition will have 946 after the decimal.

298.946 is correct option.

⇒ 195.84 + 151.56 = x × 7900 +110.4

⇒ 347.4 = 7900x + 110.4

⇒ 237 = 7900x

⇒ x = 237/7900

⇒ x = 0.03

∴ x = 3% 

Approximating each term to the nearest value, we get,

⇒ 9900 ÷ 25 + 215 –? = 310

⇒ ? = 9900/25 + 215 – 310

⇒ ? = 396 + 215 – 310

⇒ ? = 301 ≈ 300

∴ the approximate value is 300

As we know that BODMAS is used of simplifying a mathematical problem.

According to BODMAS first we have to solve ‘brackets’ followed by ‘of’ then ‘division’ then ‘multiplication’ and later ‘addition’, ‘subtraction’.

L.H.S

24.25 × 108 - 128 × 7

Now, 2619 – 896 = 1723

R.H.S

Let ‘x’ be the final answer

(x/100) × 1723

Now, cross multiplying to find ‘x’

x = (1723 × 100)/1723 = 100

2^{a + 3} = 4^{a + 2} – 48

⇒ 2^{a + 3} = 2^{2a + 4} – 48

⇒ 2^{2a + 4} – 2^{a + 3} = 48

⇒ 2^{a + 3}(2^{a + 1} – 1) = 48

Using option,

Put a = 0;

⇒ 8(1) = 8 which is not equal to 48, thus rejected.

Put a = 1;

⇒ 16(3) = 48 which is equal to 48, thus accepted.

Put a = 2;

⇒ 32(7) = 224 which is not equal to 48, thus rejected.

Put a = 4;

⇒ 64(15) = 960 which is not equal to 48, thus rejected.

∴ a = 1

I. x² – 21x + 104 = 0

⇒ x² – 13x – 8x + 104 = 0

⇒ x(x – 13) – 8(x – 13) = 0

⇒ (x – 13)(x – 8) = 0

Then, x = + 13 or x = + 8

II. y² – 5y – 24 = 0

⇒ y² – 8y + 3y – 24 = 0

⇒ y(y – 8) + 3(y – 8) = 0

⇒ (y – 8)(y + 3) = 0

Then, y = + 8 or y = - 3

So, when x = + 13, x > y for y = + 8 and x > y for y = - 3

And when x = + 8, x = y for y = + 8 and x > y for y = - 3

∴ So, we can observe that x ≥ y.

I. x² + 21x – 72 = 0

⇒ x² + 24x – 3x – 72 = 0

⇒ x(x + 24) – 3(x + 24) = 0

⇒ (x + 24)(x – 3) = 0

Then, x = - 24 or x = + 3

II. y² – 12y + 27 = 0

⇒ y² – 9y – 3y + 27 = 0

⇒ y(y – 9) – 3(y – 9) = 0

⇒ (y – 9)(y – 3) = 0

Then, y = + 9 or y = + 3

So, when x = - 24, x < y for y = + 9 and x < y for y = + 3

And when x = + 3, x < y for y = + 9 and x = y for y = + 3

∴ So, we can observe that x ≤ y.

I. x² – 15x – 184 = 0

⇒ x² – 23x + 8x – 184 = 0

⇒ x(x – 23) + 8(x – 23) = 0

⇒ (x – 23)(x + 8) = 0

Then, x = + 23 or x = - 8

II. y² + 24y + 135 = 0

⇒ y² + 15y + 9y + 135 = 0

⇒ y(y + 15) + 9(y + 15) = 0

⇒ (y + 15)(y + 9) = 0

Then, y = - 15 or y = - 9

So, when x = + 23, x > y for y = - 15 and x > y for y = - 9

And when x = - 8, x > y for y = - 15 and x > y for y = - 9

∴ So, we can observe that x > y.

The series follow the following pattern :

⇒ 7 × 2 – 2 = 14

⇒ 12 × 2 – 2 = 22

⇒ 22 × 2 – 2 = 42

⇒ 42 × 2 – 2 = 82

⇒ 82 × 2 – 2 = 162

∴ Next term = 162 × 2 – 2 = 324

The pattern of the given number series is as:

⇒1788

⇒ (1788/2) – 2 = 894 – 2 = 892

⇒ (892/2) – 2 = 446 – 2 = 444

⇒ (444/2) – 2 = 222 – 2 = 220

⇒ (220/2) – 2 = 110 – 2 = 108

⇒ (108/2) – 2 = 54 – 2 = 52

⇒ (52/2) – 2 = 26 – 2 = 24

Hence, the wrong term of the given number series is 112. The correct term is 108.

The pattern of the given number series is as:

→ 5,           

→5 + (4 × 3) = 5 + 12 = 17,

→17 + (4 × 5) = 17 + 20 = 37,

→ 37 + (4 × 7) = 37 + 28 = 65,

→65 + (4 × 9) = 65 + 36 = 101,

→101 + (4 × 11) = 101 + 44 = 145,

Hence, the required term of the given number series is 101.

The pattern of the given number series is as following:

→ 600

→ 600 ÷ 5 + 5 = 125,

→ 125 ÷ 5 + 5 = 30,

→ 30 ÷ 5 + 5 = 11,

→ 11 ÷ 5 + 5 = 7.2,

→ 7.2 ÷ 5 + 5 = 6.44

→ 6.44 ÷ 5 + 5 = 6.288

Hence, the required number in place of question mark in the given number series would be 11.

The pattern of the given number series is as following:

→ 2³ – 1 = 7,

→ 3³ – 1 = 26,

→ 4³ – 1 = 63,

→ 5³ – 1 = 124,

→ 6³ – 1 = 215,

→ 7³ – 1 = 342,

→ 8³ – 1 = 511

Hence, the missing number in the given number series would be 511.

Let the number of boys be 7x and girls be 5x

Number of boys who are not in dance society = 65% of 7x = 4.55x

Number of girls who are not in dance society = 40% of 5x = 2x

∴ Required percentage = {(4.55x + 2x)/12x} × 100 = 54.58%

Five years ago,

⇒ Sum of Age of Raj and Shruti = 25 × 2 = 50            [∵ Sum of quantity = Average × No. of quantity]

⇒ Sum of present age of Raj and Shruti = 50 + 10 = 60

⇒ Sum of present age of Raj, Shruti and Ram = 29 × 3 = 87

∴ Present age of Ram = 87 - 60 = 27 years

⇒ Cost price of 1 kg of flour = Rs. 35

For 1 kg, he weighs only 750 grams

⇒ Cost price of 750 gram of flour = Rs. (35/1000) × 750 = Rs. 26.25

⇒ Selling price of 750gm of flour = Rs. 30

∴ Profit percentage = {(30 – 26.25) /26.25} × 100 = 14.286%

⇒ P = Rs. 3900

⇒ Amount = A = Rs. 5200

⇒ I = 5200 - 3900 = Rs. 1300

⇒ N = 3 years

From the equation,

⇒ I = PRN/100

⇒ R = 100/9

∴ R = 11.11%

Given the length of the wire is 88 cm

When the wire is bent in the shape of the square, then the total length of the wire is equal to the perimeter of the square

Let us consider the side of the square = ‘x’

⇒ Perimeter of the square = 4x

⇒ 4x = 88

⇒ x = 88/4

⇒ 22 cms

⇒ Area of the square = side × side

⇒ 22 × 22 = 484 cm²

∴ Area = 484 cm²

Let the length be 8x and breadth be 5x

According to question

⇒ 8x = 5x + 39

⇒ 3x = 39

⇒ x = 13

⇒ Length = 8x = 8 × 13 = 104

⇒ Breadth = 5x = 5 × 13 = 65

∴ Perimeter = 2 × (104 + 65) = 338

Let suppose first no as x,

So the series is like,

⇒ x, x + 1, x + 2, ....., x + 5

⇒ Sum of six number = 480

⇒ x + x + 1 + x + 2 + x + 3 + x + 4+ x + 5 = 480

⇒ 6x + 15 = 480

⇒ x =77.5

⇒ Fourth no. = x + 3 = 80.5

⇒ Sixth no. = x + 5 = 82.5

⇒ Sum = 80.5 + 82.5 = 163

∴ Sum of forth and sixth number is 163

⇒ Let the number be x,

⇒ (6x + 4x + 3x)/12 = x + 18

⇒ 13x/12 = x + 18

⇒ 13x = 12x + 216

∴ x = 216

SP = CP(1 + Profit%)

Where, CP = cost price

SP = Selling Price

For Divya, CP = 35 × 1.25 = Rs. 43.75, Profit = 15%

∴ SP = 43.75 (1 + 0.15) = 43.75 × 1.15 = Rs. 50.3125 for 35 apples

For Meha, CP = Rs. 50.3125 for 35 apples

She eats 7 apples.

∴ She has 28 apples now.

Profit % = 20%

∴ SP = 50.3125(1 + 0.2) = Rs. 60.375 for 28 apples.

∴ Mehak sells each banana at a price of Rs. 2.156.

We know that,

Speed = Distance/Time

Speed of the train = 600/16 = 37.5 m/sec

Length of the platform = 37.5 × 24 = 900 m

7 minutes and 5 seconds = (6 × 60) + 15 = 375 sec    (∵ 1 min = 60 sec)

∴ Speed of man = 900/375 = 2.4 m/sec

Given, present height of the tree = 64 cm

The height of the tree increases by 1/8 th every year.

After 1^st year

Height of tree = 64 + 1/8 of 64

⇒ Height of tree = 64 + 8 = 72 cm

After 2^nd year

Height of tree = 72 + 1/8 of 72

⇒ Height of tree 72 + 9 = 81 cm

Another approach, it can be considered a CI problem:

Principal (initial height) = 64 cm, rate = 1/8 × 100 % = 12.5 %, t= 2 years, A= final height

For CI:

Where,

A is the length at the end of time t,

P is the height at the starting of the time,

t is time in years.

r is growth rate in percent

⇒ Final height = 64 × 1.125²

⇒ Final height = 64 × 1.265625 = 81 cm.

Let the Distance travelled by worker on foot = x km

∴ Distance covered by cycle = (56 - x) km.

Time = Distance/Speed

According to the question,

x/4 + (56 – x)/8 = 12

⇒ (2x + 56– x)/8 = 12

⇒ x + 56 = 12 × 8 = 96

⇒ x = 96 – 56 = 40

⇒ x = 40 km.

Total number of possible outcomes when three dice are thrown simultaneously = 6 × 6 × 6 = 216

Three numbers add up to 5 ⇒ numbers can constitute the following sets: {1, 1, 3} or {1, 2, 2}

[No die can show more than 3 and less than 1 for the three numbers to add up to 5]

{1, 1, 3} can be rearranged in 3 ways as (1, 1, 3), (1, 3, 1) and (3, 1, 1)

{1, 2, 2} can be rearranged in 3 ways as (1, 2, 2), (2, 1, 2) and (2, 2, 1)

So, total number of ways in which three numbers on dice add up to 5 is 6.

∴ The required probability = 6/216 = 1/36

1200 ml mixture contains milk and water in ratio of 5:3

∴ Amount of milk in mixture =

∴ Amount of water in mixture = 1200 – 750 = 450 ml

Let’s assume that ‘x’ ml of water is added to 1200 ml of mixture so that the ratio of milk to water becomes 5:4. The total amount of milk does not change.

Total amount of water = 450 + x

∵ the required ratio of milk to water is 5:4,

⇒ 4 × (750) = 5 × (450 + x)

⇒ 3000 = 2250 + 5x

⇒ 750 = 5x

⇒ x = 750/5 = 150 ml.

∴ 150 ml of water has to be added.