Given, 220% of X = 44 Or, X = 20 Thus, 44% of 20 $$ = \frac{{44 \times 20}}{{100}} = 8.8$$

Price after third depreciation, 100 ==25%↓ ==> 75 == 20%↓==>60 == 15% ↓ ==> 51 The price will be, = Rs. 5,10,000 Alternatively : 1000000 × 0.75 × 0.80 × 0.85
= Rs. 5,10,000

$$\eqalign{ & {\text{His bonus}}, \cr & = \frac{{ {20 \times 1000000} }}{{100}} \cr & = 2\, \text{lakh} \cr & {\text{Total}}\,{\text{profit}} = {\text{Net}}\,{\text{profit}} + \frac{{ {10 \times {\text{net}}\,{\text{profit}}} }}{{100}} \cr & 1.32\,\text{lakh} = {\text{Net}}\,{\text{profit}} \times \left[ {1 + {\frac{{10}}{{100}}} } \right] \cr & {\text{Net}}\,{\text{profit}} = \frac{{132000}}{{1.1}} = 120000 \cr & {\text{Commission}}, \cr & = \left( {{\text{Total}}\,{\text{profit}} - {\text{Net}}\,{\text{profit}}} \right) \cr & = 132000 - 120000 \cr & = 12000 \cr & {\text{Hence}}, {\text{his}}\,{\text{total}}\,{\text{earnings}} \cr & = 2\,\text{lakh} + 12000 \cr & = Rs.\,212000 \cr} $$

Result PhysicsChemistry Failed35%45% Passed65%55% Now, 55% of 600 passed in Chemistry = $$\frac{{55 \times 600}}{{100}}$$   = 330
From question, 40% of those who passed in Chemistry(330 Students) also passed in physics
= $$\frac{{40 \times 330}}{{100}}$$
= 132,
so, 132 students passed in both
Therefore % who passed in both = $$\frac{{132 \times 100}}{{600}} = 22\% $$
Passed in both subject = 22% of total students % of students who passed in either Physics or Chemistry or both, = (65 + 55) - 22
= 98% Thus, percentage of students who failed in both subjects = 2% Number of students who failed = 2% of 600 = 12

No. of Machines Output Manuf. cost Est. cost Total cost Profit 12 48, 000 24, 000 10, 000 34, 000 14, 000 11 44, 000 22, 000 10, 000 32, 000 12, 000 Profit, = Output - total cost = 44000 - 32000 = Rs. 12000 Initial value of share holders, = 14000 $$ \times \frac{{10}}{{100}}$$ = Rs. 1400 New value of share holders, = 12000 $$ \times \frac{{10}}{{100}}$$ = Rs. 1200
Decrease in Share holder value = 1400 - 1200 = 200
percentage decrease in the value of shareholders is :
$$\eqalign{ & = \frac{{200 \times 100}}{{1400}} \cr & = 14.28\% \cr} $$

Let the initial cost of raw material be 100. So, initial labor cost was 25 and net cost was 125 Now, 15% increment in raw materials cost and labor cost has gone up to 30% from 25 % Raw material cost = 115 And Labor cost = (115 × 30%) = 34.5 So, New net cost, = 115 + 34.5 = 149.5 Difference of labor cost = 149.5 - 125 = 24.5 % reduction = $$\frac{{24.5 \times 100}}{{149.5}}$$   = 17%(approx.)

Let the initial revenue be 100 100 === 50% ↑ (Ticket up) ===> 150 === 20% ↓ (Visitors down) ===> 120 There is 20% increase in the revenue

Let the fraction be $$\frac{{\text{x}}}{{\text{y}}}$$ When fraction is squared its numerator is reduced by $$33\frac{1}{3}$$ and denominator is reduced by 20% $$\eqalign{ & {\text{According}}\,{\text{to}}\,{\text{question,}} \cr & {\left( {\frac{x}{y}} \right)^2} \times \frac{{33\left( {\frac{1}{3}} \right)\% }}{{20\% }} = 2\left( {\frac{x}{y}} \right) \cr & {\text{Or}},\,{\left( {\frac{x}{y}} \right)^2} \times \frac{{\left( {\frac{2}{3}} \right)}}{{\left( {\frac{1}{5}} \right)}} = 2\left( {\frac{x}{y}} \right) \cr & {\text{Or}},\,\frac{x}{y} = \frac{3}{5} \cr & {\text{Sum of numerator and denominator is}} \cr & \left( {x + y} \right) = 3 + 5 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 8 \cr} $$

$$\eqalign{ & {\text{Daily}}\,{\text{supply}}, \cr & = \left( {100 - z} \right)\% \,{\text{of}}\,y \cr & = \frac{{ {\left( {100 - z} \right)y} }}{{100}} \cr & {\text{Thus,}}\,{\text{required}}\,{\text{number}}\,{\text{of}}\,{\text{days}} \cr & = {\frac{{ {100x} }}{{\left( {100 - z} \right)y}}} \cr} $$

100 === 25% ↑===> 125 ===> 20% ↓===> 100 So there is no change in the area of rectangle Alternatively : Let area be 1 Area of rectangle, = l × b = 0.8 × 1.25 = 1

Cost of fresh mangoes + Cost of packaging = Total cost Let initial Cost of fresh, mangoes = 100 Then, packaging cost = 40 Thus, Initial total cost = 100 + 40 = 140 After increasing in cost of fresh mangoes 30%, Cost of fresh mangoes = 130 And cost of packing go down by 50 % so, Cost of packing = 20 Now Total cost = 130 + 20 = 150 Increased cost = 150 - 140 = 10 % increased = $$ = \frac{{10 \times 100}}{{140}} = 7.14\% $$

Actual result = 10x
By mistake it has been divided by 10 = $$\frac{x}{{10}}$$
%Change = $$ {\frac{{10x - \left( {\frac{{10}}{x}} \right)}}{{10x}}} $$   × 100 = 99 = -99%
Since, actual value is greater than the wrong value.

Alternatively,
Let, x = 10
Actual result = 10 × 10 = 100
Wrong result = $$\frac{{10}}{{10}}$$ = 1
Change = (1 - 100) = -99
% Change = - 99%

Let the percentage marks in QA =(10a + b)% Let the percentage marks in DI = (10b + a)% Let the percentage marks in VA = x% Now, according to the question, we have, $$\eqalign{ & \frac{{\left( {10{\text{a}} + {\text{b}}} \right) + {\text{x}} + \left( {10{\text{b}} + {\text{a}}} \right)}}{3} = {\text{x}} \cr & 11{\text{a}} + 11{\text{b}} + {\text{x}} = 3{\text{x}} \cr & {\text{or, x}} = \frac{{11\left( {{\text{a}} + {\text{b}}} \right)}}{2} \cr} $$ Clearly, we can see that the percentage of the VA section will be a multiple of 11 So, required answer will be 66

Let initially the quantity and rate be 100 each Quantity × rate = Expenditure 100 × 100 = 10000 Now, Increase in price is 25% and new quantity is 70% of original Quantity × rate = Expenditure 70 × 125 = 8750 Decreased expenditure, = 10000 - 8750 = 1250 % decrease $$ = \frac{{1250 \times 100}}{{10000}} = 12.5\% $$

Copper : Aluminum = 7 : 4 Let Copper and Aluminum in the weapon be 7x and 4x respectively Given, Aluminum in weapon = 12 kg So, → 4x = 12 → x = 3 Copper = 7x = 7 × 3 = 21 Kg. Total alloy in the weapon = 12 + 21 = 33 kg But 12% alloy get destroyed in making the weapon, i.e. 88% alloy is used in the weapon, so, → 88 % alloy = 33 kg → 100 % alloy = 37.5 kg

Total employees = 1600 Female employees, 60% of 1600 $$ = \frac{{60 \times 1600}}{{100}} = 960$$ Then male employees = 640 50% of male are computer literate, = 320 male computer literate 62% of total employees are computer literate, $$ = \frac{{62 \times 1600}}{{100}} = 992$$     computer literate Thus, Female computer literate = 992 - 320 = 672 Alternatively : Let 60% employees are female and 40% are male Then, 20% of male are computer literate and 42% are female computer literate Female computer literate $$ = \frac{{1600 \times 42}}{{100}} = 672$$

There is no need of the number of goats given i.e. 34,398. Initially, let there be 100 goats. Then 100 == 40% ↑==> 140 == 30%↓(declined to 70%) ==> 98 == 30%↑ ==> 127.4 == 10%↓(sold) ==> 114.66 Hence, % increase = 14.66% [As 100 becomes 114.66]

Let the initial price = Rs. 10000p
Price after first increment = 10000p + 100xp
Price after first decrement
= 10000p + 100xp - (100px + px2)
= 10000p - px2
Now, total decrement, px2 = 21025 . . . . . (1)
Price after second increment,
= 10000p - px2 + 100xp - $$\frac{{{\text{p}}{{\text{x}}^3}}}{{100}}$$
Price after second increment,
= 10000p - p2 + 100xp - $$\frac{{{{\text{p}}^3}}}{{100}}$$ - 100xp + $$\frac{{{\text{p}}{{\text{x}}^3}}}{{100}}$$ - px2 + $$\frac{{{\text{p}}{{\text{x}}^4}}}{{10000}}$$
= 10000p - 2px2 + $$\frac{{{\text{p}}{{\text{x}}^2}}}{{10000}}$$
= 484416 . . . . . . (2)
On solving equation (1) and (2), We get
x = 20
Substituting back we get,
p = 5,25,625

Let the smaller number be x and larger number be y According to the question, 80% of x + 4 = 40% of y 4y - 8x = 40 y - 2x = 10 --------- (1) y - x = 85 ----------- (2) By using (1) and (2), we get x = 75 y = 160 x + y = 235

Let the price of wheat is x per kg. Then price of wheat will be 5x per kg. Expenditure on rice = 25 × x = 25x Expenditure of wheat = 9 × 5x = 45x Total cost, 25x + 45x = 350 70x = 350 x = 5 Hence, price of Rice = Rs. 5 per kg. Price of wheat = 25 per kg. Now, price of wheat = 25 ---- 20% ↑----> Rs. 30 per kg. Let the new amount of rice is N kg, then N*5 + 9*30 = 350 N = 16 kg. % decrease in the amount of rice
$$\eqalign{ & = \frac{{\left( {25 - 16} \right) \times 100}}{{25}} \cr & = 36\% \cr} $$