Let the cost price of chair and stool be Rs. x and Rs. y per unit

Cost of 4 chairs and 9 stools = Rs. 1340

⇒ 4x + 9y = 1340       ---- 1

Profit = Profit%/100 × CP

⇒ Profit he makes = 10/100 × 4x + 20/100 × 9y

⇒ 188 = 0.4x + 1.8y

⇒ 10 × 188 = [0.4x + 1.8y] × 10

⇒ 1880 = 4x + 18y

⇒ 1880 = 4x + 9y + 9y

⇒ 1880 = 1340 + 9y

⇒ 9y = 540

y = 60, Substituting in equation 1

⇒ 4x + 540 = 1340

∴ x = Rs. 200

∴ Amount he spends for chairs = 4x = Rs. 800

Solution for the given equation x + 5 = 5 is x = 0

∴ Set of the equations = {0}

Third option cannot be used because it doesn’t represent a set

ϕ means there is no such solution and cannot be used

ab + bc + ca = 0

⇒ -bc = a (b + c)

Similarly -ca = b (c + a) and -ab = c (a + b)

⇒ a²/(a² – bc) + b²/(b² – ca) + c²/(c² – ab) = a²/[a² + a (b + c)] + b²/[b² + b (c + a)] + c²/[c² + c (a + b)]

⇒ a²/[a (a + b + c)] + b²/[b (a + b + c)] + c²/[c (a + b + c)]

⇒ a/(a + b + c) + b/(a + b + c) + c/(a + b + c)

⇒ (a + b + c)/(a + b + c) = 1

Percentage of students failed = Percentage of students who failed in Hindi + Percentage of students who failed in English – Percentage of students who failed in both [∵ Those who failed in both were counted twice]

⇒ 45 + 35 – 20 = 60%

∴ Percentage of students who passed in both subjects = 100 – 60 = 40%

Let (x – y), (y – z) and (z – x) be a, b and c respectively

⇒ a + b + c = x – y + y – z + z – x = 0

If a + b + c = 0, then a³ + b³ + c³ = 3abc

⇒ [(x – y) (y – z) (z – x)]/[(x –y)³ + (y – z)³ + (z – x)³] = [(x – y) (y – z) (z – x)]/3[(x – y) (y – z) (z – x)] = 1/3

log₁₀ 8000 + log₁₀ 600 = log₁₀ (8 × 1000) + log₁₀ (6 × 100)

⇒ log₁₀ 8 + log₁₀ 1000 + log₁₀ 6 + log₁₀ 100 [∵ log(mn) = log m + log n]

⇒ 0.9031 + 3 + 0.7782 + 2 = 6.6813

After 24 days, let ‘x’ men leave

⇒ 16 days of work for 30 men is remaining

⇒ Remaining men needs 40 days to complete the work

⇒ Number of days taken ∝ 1/Number of men

⇒ Number of days taken × Number of men = constant

⇒ 16 × 30 = 40 × (30 – x)

⇒ 16 × 30/40 = 30 – x

⇒ 12 = 30 – x

⇒ x = 18

∴ 18 men leave after 24 days.

6 sheep are equal to 4 goats

⇒ 3 sheep are equivalent to 4/2 = 2 goats

⇒ 2 goats and 3 sheep are equivalent to = 2 + 2 = 4 goats

∴ 4 goats can graze the field in 50 days [Given]

The tank will get filled in 12 hours instead of 10 hours

2 hours of the extra water filled by the tap is emptied by the outlet in 5 hours

∴ 10 hours of the water by tap (Full tub) will be emptied in 10/2 × 5 = 25 hours