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1/2 of the work is done by P in 6 days

∴ Full work is done by P in 6/1/2 = 6 × 2 = 12 days

2/3 of work done by Q in 4 days.

∴ Full work done by Q in 4/2/3 = 4 × 3/2 = 6 days

(P + Q) will finish the whole work in ab/(a+b) days = (12×6)/(12+6) = (12×6)/18 = 4 days

(P + Q) will finish 3/4 of the work in 4 × 3/4 = 3 days

p can do a piece of work in 12 days.

∴ p’s 1 day work = 1/12

p’s 1 day work = 3 × 1/12 = 3/12

Q can do a piece of work in 15 days.

∴ Q’s 1 day work = 1/15

Remaining work after 3 days = 1 – 3/12 = 9/12

(P + Q)’s 1 day work

= 1/12 + 1/15 = 5/60 + 4/60 = 9/60

Number of days required to finish the remaining work

= Remaining work/(P+Q)′s 1 day work 

= 9/12/9/60 = 9/12 × 60/9 = 5

Remaining work lasts for 5 days. Total work lasts for 3 + 5 = 8 days.

Given m persons complete a work in n days

(i) Then work measured in terms of Man days = mn

4 m men do the work it will be completed in mn/4m days = n/4 days.

(ii) m/4 persons can complete the same work in mn/m/4 days = 4mn/m = 4n days

(c) 6

Conversion period is the time period after which the interest is added to the principal. If principal is compounded every two months then in a year, there will be 6 12/2 conversation periods.

Principal (P) = ₹ 3200

r = 2.5% p.a

n = 2 years comp, annually

∴ Amount (A) = (1 + r/100)ⁿ = (1 + 2.5/100)²

= 3200 x (1.025)² = 3362

Compound interest (Cl) = Amount – Principal 

= 3362 – 3200 

= ₹ 162

₹ 1272

Compound Interest (Cl) formula is

Cl = Amount – Principal

Amount = A (1 + r/100)ⁿ 

= 5000 (1 + 12/100)²

= 5000 (1 + 112/100)² 

= 6272

∴ Cl = 6272 – 5000 = ₹ 1272

Principal (P) = ₹ 5000

Interest compounded half yearly

r = 12% p.a = 12/2 = 6% for half yearly

t = 1 yr.

Since compounded half yearly, the formula to be used is

Amount A = P (1 + r/100)²ⁿ

A = 5000 (1 + 6/100)^2×1 

= 5000 x (106/100)² 

= ₹ 5618

Case 1: Profit = Selling price (SP) – Cost price (CP)

Case 2: Loss = Cost price (CP) – Selling price (SP)

Given that profit of case 1 = loss of case 2

∴ P = 810 – CP

L = CP – 530

Since profit (P) = loss (L)

810 – CP = CP – 530

∴ 2CP = 810 + 530 = 1340 

⇒ C.P = 1340/2

∴ CP = 670

A = P (1 + r/400)⁴ⁿ

Quarterly means 4 times in a year.

∴ The formula for compound interest is

A = P (1 + r/400)⁴ⁿ

₹ 32

Difference between S.I & C.I is given by the formula

CI – SI = P (r/100)²

Principal (P) = 5000, r = 8% p.a

∴ CI – SI = 5000 (8/100)² 

= 5000 x 8/100 x 8/100 

= ₹ 32

Selling price = Marked price – Discount.

(Loss/C.P x 100)%

I = (1 + x/100)

Over head expenses

(Profit/C.P x 100)%

Comparing quantities easily

Per hundred or out of hundred.

True

Depreciation formula is P (1 – r/100)ⁿ

In one day 15 men earn Rs 900

In one day 15 men earn 900/5 = Rs 180

In one day 1 men earn 180/5 = Rs 12

∴ 1 men earn in 7 days = 12 × 7 = Rs 84

∴ 20 men earn in 7 days = 84 × 20 = 1680

y% of 1000 = 600

y/100 x 1000 = 600

y = 600/10

y = 60

Increase in population = 7,00,000 – 5,00,000

= 2,00,000

Percentage increase in population = 2,00,000/5,00,000 x 100 

= 40%

Let R get x, Q gets 50% of what R gets

∴ Q gets = 50/100 × x = x/2

P gets 50% of what Q gets

∴ P gets = 50/100 x x/2 = x/4

Since 350 is divided among the three

∴ 350 = x + x/2 + x/4

350 = (4x+2x+x)/4 = 7x/4 = 350

x = (350×4)/7 = 200

Q gets = x/2 = 200/2 = 100,

P gets = x/4 = 200/4 = 50

∴ P = 50

Q = 100

R = 200

In a day, there are 24 hours

10 hrs out of 24 hrs is 10/24

As a percentage, we need to multiply by 100

∴ Percentage = 10/24 x 100 = 41.67%

C.I – S.I = P(r/100)² 

= 10,000(4/100)²

= 10,000 x 4/100 x 4/100 

= ₹ 16

P = ₹ 16,000, n = 3 years, r = 5%

S.I = Pnr/100 

= (16000 x 3 x 5)/100

= ₹ 2400

Amount A = P(1 + r/100)ⁿ 

= 8000(1 + 5/100)²

= 8000 x 105/100 x 105/100

= 8000 x 21/20 x 21/20

A = ₹ 8820

Cl = A – P 

= 8820 – 8000 

= 820

False

Let the population ‘n’ yrs ago be ‘x’

Present population (P) = x × (1 + r/100)ⁿ

x = P/(1+r/100)ⁿ

Let the number be ‘x’

Given x – 10/100x = 900

(100x−10x)/100 = 900

90x/100 = 900

x = (900×100)/90 = 1000

D = (1 + x/100)

Let cost price of article be C.P. Let S.P of 1 articles be SP by selling 5 articles at SP he makes a gain of cost price of one article.

Gain on 1 article = SP – CP; 

⇒ Gain% = (SP−CP)/CP x 100

Gain on 2 articles = 2 x (SP – CP)

Gain on 5 articles = 5 x (SP – CP)

Given than gain on 5 articles is CP of 1 article

∴ 5(SP – CP) = CP

(SP−CP)/CP = 1/5

Gain percentage (SP−CP)/CP x 100 

= 1/5 x 100 % = 20%

True

Present value of machine = ₹ 16800

Depreciation rate (n) = 25%

Value after 2 years = P (1 – r/100)ⁿ 

= 16800 (1 – 25/100)²

= 16800 x (1 – 1/4)² 

= 16800 x 3/4 x 3/4 

= 9450

Let Q’s income be 100.

P’s income is 25% more than that of Q

∴ P’s income = 100 + 25/100 x 100 = 125

Q’s income is 25 less than that of P

In percentage terms, Q’s income is less than P’s with respect to P’s income is

(P−Q)/P x 100 

= (125−100)/125 x 100 

= 25/100 x 100 

= 20%

Principal (P) = ₹ 15,000

Time period (n) = 2 yrs.

Rate of interest (r) = 6% p.a compounded annually

Difference between CI and SI given by

CI – SI = P (r/100)ⁿ 

= 15000 (6/100)²

= 15000 x 6/100 x 6/100

= ₹ 54

Principal (P) = ₹ 5000

time period (n) = 1 yr.

Rate of interest (r) = 2% p.a

for half yearly r = 1%

Difference between Cl & SI is given by the formula

CI – SI = P (r/100)²ⁿ [for half yearly compounding]

CI – SI = P (1/100)^2×1

= 5000 x 1/100 x 1/100 

= ₹ 0.50

₹ 820

Compound interest (CI) = Amount – Principal

∴ Amount = P (1 + r/100)²ⁿ [2n as it is compounded half yearly]

r = 10% p.a, for half yearly r = 1 + 10/2 = 5

A = 8000 (1 + 5/100)^2×1 

= 8000 x (105/100)² = 8820

CI = Amount – principal

= 8820 – 8000 

= ₹ 820