Given:

total pipes = 30

filling pipes = 24 hours each

emptying pipes = 18 hours each

Solution:

Let, the filling pipes = 30 - n

emptying pipes = n

(30 - n)/24 - n/18 = 1/1.5

⇒ 90 - 3n - 4n = 72/1.5

⇒ - 7n = 48 - 90

n = 42/7

n = 6 pipes

Hence, the correct option is 4.

Given:

A man saves 30% of his monthly income. If his monthly income increases by 10%, then he saves 20% more than the previous savings.

Concept used:

Income = Saving + Expenditure

Calculation:

Let his income be 100 initially,

So, Saving = 30 and Expenditure= 70

After increasing 10% income, the income becomes

= 110% of 100

= 110

And the savings after 20% increase becomes

= 120% of 30

= 36

So, The final expenditure = 110 - 36 = 74

Increase in expenditure

= 5.714%

Hence, The percentage increase in his expenditure is 5.7%.

Given:

The shadow of a pole 6 m high and 15 m long.

At the same time, the shadow of a tree is 25 m long.

Calculation:

Let the height of the tree be H.

According to the question

⇒ 15H = 150

⇒ H = 10 m

∴ The required height of the tree is 10 m.

Given:

Vendor sells 40% and throws away 10% of the remaining. Next day he sells 50% of the remaining and throws away the rest.

Concept:

To find the percentage of oranges thrown away, calculate the quantity of oranges thrown each day based on the given percentages.

Solution:

⇒ Initial quantity = 100 (assuming)

⇒ After selling 40%, remaining = 60%

⇒ Throwing 10% of remaining, thrown = 10% of 60 = 6%

⇒ New quantity = 60% - 6% = 54%

⇒ Next day, 50% sold of remaining, remaining = 50% of 54 = 27%

⇒ All of remaining is thrown = 27%

⇒ Total percentage of oranges thrown = 6% + 27% = 33%

Hence, the vendor throws away 33 percent of his oranges.

Given:

7sinθ = 5 

sinθ = 5/7

Concept Used:

secθ = 1/cosθ 

tanθ = sinθ/cosθ 

Calculation:

⇒ 12/2 = 6

∴ Correct answer is 6.

Given:

Sugar of Rs. 100/kg and Rs. 120/kg are mixed  in a ratio of 2 : 3 

Concept used:

Allegation and mixture

Calculation:

The average price of variety first and variety second will be

⇒ (100 × 2 + 120 × 3) / (2 + 3)

⇒ 560 / 5 = 112

∴ The price of mixture per kg is Rs. 112

Shortcut Trick

Calculation:

Let price of mixture be x Rs/kg

According to rule of allegation -

According to question -

⇒ (120 - x)/(x - 100) = 2/3

⇒ 120 × 3 - 3x = 2x - 200

⇒ 5x = 560

⇒ x = 112 Rs/kg.

∴ The price of the mixture per kg is Rs. 112.

Given:

Total voters = 12000,

Voters for A = 30%,

Voters for B = 25%,

Voters for C = 40%

Concept:

Number of non-voters = Total voters - (Voters for A + Voters for B + Voters for C)

Calculation:

⇒ Number of non-voters = 12000 - (12000 × 30/100 + 12000 × 25/100 + 12000 × 40/100) = 12000 - (3600 + 3000 + 4800) = 600

Hence, the number of voters who did not vote for any of these three participants is 600.

Let the numbers are X and Y

Given:

(X + Y)² = 784  and XY = 192

Calculation:

(X + Y)² = 784 ⇒ (X + Y) = 28

⇒  X² + Y² + 2XY = 784

⇒ X² + Y² + 2 × 192 = 784

⇒ X² + Y² = 400

So,

⇒ X² + Y² - 2XY = 400 - 2 × 192

⇒  X² + Y² - 2XY = 16

⇒ (X - Y)² = 16

⇒ X - Y = 4

Now,

X² - Y² = (X + Y)(X - Y)

⇒ 28 × 4 = 112

∴ The correct option is 4

Alternate Method

According to the question,

(x + y)² = 784

⇒ x + y = √784 = 28

xy = 192

x - y = √{(x + y)² - 4xy}

⇒ √{28² - 4 × 192} = √16 = 4

Now, x² - y² = (x + y)(x - y) = 28 × 4 = 112  

Let O be the centre of the circle

In quadrilateral AOBX

∠A + ∠O + ∠B + ∠X = 360° [Angle sum property of quadrilateral]

90° + ∠O + 90° + 50° = 360° [∵ Tangent is perpendicular to the radius]

∴ ∠AOB = 130°

∠ACB = ½ × 130 = 65° [∵ Angle subtended by a chord at any point on the circle is half the angle subtended at centre]

cosec²68° + sec²56° – cot²34° – tan²22°

⇒ cosec²68° + sec²56° – cot²(90 – 56)° – tan²(90 – 68)°

⇒ cosec²68° + sec²56° – tan²56° – cot²68°

We know the formulas:

cosec²θ – cot²θ = 1 and sec²θ – tan²θ = 1

∴ cosec²68° + sec²56° – tan²56° – cot²68° = 1 + 1 = 2