20 litres juice contain 10% Tomato, i.e. 20L juice = $$\frac{{20 \times 10}}{{100}}$$  = 2L Tomato Tomato puree contains 80% of water and 20% tomato. This 80% tomato = 2L (which is contained by 100 puree) So, Now this 2 L Consist 80% in puree Thus, total puree will be $$\frac{2}{{.8}} = 2.5{\text{L}}$$

Let the number of people be x who has been asked for the donations People already solicited = 60% of x = 0.6x Remaining people = 40% of x = 0.4x Amount collected from the people solicited, = 600 × 0.6x = 360x 360x = 75% of the amount collected Remaining amount = 25% = 120x Thus, Average donations from remaining people, $$\eqalign{ & = \frac{{120{\text{x}}}}{{0.4{\text{x}}}} \cr & = 300 \cr} $$

Let the bike's initial cost be x and then car's initial cost be 5x After the increase, Bike price = 1.2x Car price = 5.75x Initial total cost of 5 cars and 10 bikes, = 25x + 10x = 35x New cost, = 28.75x + 12x = 40.75x Change in cost = (40.75x - 35x) = 5.75x % change = $$\frac{{5.75{\text{x}} \times 100}}{{35}} = 16\frac{3}{7}\% $$

YearValue 2000100 (Let) 2001110 2002121 2003133.1 2004119.79 2005107.811 200697.0299 $$\eqalign{ & {\text{Now}}, \cr & \frac{{ {\left( {100 - 97.0299} \right) \times 100} }}{{100}} \approx 3 \cr} $$

$$\eqalign{ & {\text{Let the required number of pages be }}x. \cr & 30 \times 25 \times 35 = x \times 30 \times 28 \cr & x = 31.25 \approx 32 \cr & \% \,{\text{increase}}\,{\text{in}}\,{\text{number}}\,{\text{of}}\,{\text{pages}}, \cr & = {\frac{2}{{30}}} \times 100 \cr & = 6.66\% \cr} $$

100 == 25% ↑ ==> 125 == x%↓ ==> 100 $${\text{x}}\% = \frac{{25 \times 100}}{{100}} = 25\% $$ We don't need the actual value to calculate the answer

Since, Run scored = Over × Run rate If overs is reduced by 25%, run rate will go up by 33.33%. Hence, Australia could have scored any number of runs.

Non-defective products M1 = 25 × 0.98 = 24.5% Non-defective products M2 = 35 × 0.96 = 33.6% Non-defective products M3 = 40 × 0.95 = 38% Percentage of non-defective products
= 24.5 + 33.6 + 38
= 96.1%

$$\eqalign{ & {\text{Reduction}} \cr & = {\frac{7}{{107}}} \times 2568 \cr & = 168 \cr} $$

The average weight of the bars given to Brennan (light) < the average weight of the bars given to Blair < the average weight of the bars given to Maya (heavy). Let the total weight of all the bars be X. The weight of the bars given to Brennan, = 45% of X = 0.45X The weight of the bars given to Maya, = 26% of X = 0.26X The weight of the bars given to Claire = rest = 29% of X = 0.29X The average weight of the bars given to Brennan, $$ = \frac{{{\text{Weight}}}}{{{\text{Number}}\,{\text{of}}\,{\text{bars}}}} = \frac{{0.45{\text{X}}}}{{24}}$$ The average weight of the bars given to Maya = Weight / number of bars = $$\frac{{0.26{\text{X}}}}{{13}}$$ Similarly, if the number of bars given to Blair = B, then the average weight of the bars given to Blair = $$\frac{{0.29{\text{X}}}}{{\text{B}}}$$ As, the average weight of the bars given to Brennan (light) < the average weight of the bars given to Blair < the average weight of the bars given to Maya
So , Option (B) is the right answer

Total student = 100 Boys = 60 Girls = 40 Boys who plays hockey = 40% = 24 There is no information about boys who play badminton. Girls who plays Badminton = 75% = 30 No girls plays hockey. Since, we do not have information that whether the rest of the boys are playing badminton or not. So, we cannot determine the total no. of student who don't play any game.

$$\eqalign{ & {\text{Let the positive number be x}}. \cr & {\text{According}}\,{\text{to}}\,{\text{the}}\,{\text{question}}, \cr & x \times x = x + \frac{{\left( {x \times 2000} \right)}}{{100}} \cr & {x^2} = x + 20x \cr & {x^2} - 21x = 0 \cr & {\text{Either}}, \cr & x = 0,\,\,or \cr & x = 21 \cr & {\text{21}}\,{\text{is}}\,{\text{the}}\,{\text{possible}}\,{\text{value}}{\text{.}} \cr & {\text{Then}}\,{\text{square}}\,{\text{of}}\,21\,{\text{is}}\,441 \cr} $$

Day 1 2 3 Initial mangoes x (let) 0.45x 0.2025x Sales 0.5x 0.225x 0.10125x Over night remaining 0.5x 0.225x 0.10125x Rotten 0.05x 0.0225x 0.010125x Next day stock 0.45x 0.2025x Thus, Total mangoes rotten amount is
⇒ 0.05x + 0.0225x + 0.010125x = 0.082625x = 1983x = 24,000

Let X be the number which is added to 80 80% of X = 0.8X Now, 80 + 0.8X = X 0.2X = 80 X = $$\frac{{80}}{{0.2}} = 400$$

$$\eqalign{ & {\text{Let the total number of applicants be x}}. \cr & {\text{Number of eligible candidates}} \cr & = {\text{ }}95\% {\text{ }}of{\text{ }}x \cr & {\text{Eligible candidates of other categories}}, \cr & = 15\% \,of\,\left( {95\% {\text{ }}of{\text{ }}x} \right) \cr & = {\frac{{15}}{{100}}} \times {\frac{{95}}{{100}}} \times x \cr & = \frac{{57}}{{400}}x \cr & or,\left( {\frac{{57}}{{400}}} \right)x \cr & x = \frac{{ {4275 \times 400} }}{{57}} \cr & \,\,\,\,\,\, = 30000 \cr} $$

$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{fraction}}\,{\text{be}}\,\frac{{100x}}{{100y}} \cr & {\text{Now}}\,{\text{according}}\,{\text{to}}\,{\text{the}}\,{\text{question}}, \cr & {\left( {\frac{{100x}}{{100y}}} \right)^2} = \frac{{125{x^2}}}{{80{y^2}}} = \frac{{25{x^2}}}{{16{y^2}}} \cr & \frac{{25{x^2}}}{{16{y^2}}} = \frac{5}{8}\left( {\frac{{100x}}{{100y}}} \right) \cr & {\frac{{100x}}{{100y}}} = \frac{2}{5} \cr & {\text{hence}}, \cr & {\text{product of numerator and denominator}} \cr & = 2 \times 5 = 10 \cr} $$

$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{original}}\,{\text{price}} = y, \cr & {\text{After}}\,{\text{first}}\,{\text{change,}}\,{\text{it}}\,{\text{becomes}}, \cr & y \times \left( {1 + {\frac{x}{{100}}} } \right) \cr & {\text{After}}\,{\text{second}}\,{\text{change,}}\,{\text{it}}\,{\text{becomes}} \cr & y \times \left( {1 + {\frac{x}{{100}}} } \right)\left( {1 - {\frac{x}{{100}}} } \right) \cr & = y\left( {1 - {{\left( {\frac{x}{{100}}} \right)}^2}} \right) \cr & {\text{Thus}}, \cr & {x^2} \times y = {10^6} - - - - \left( 1 \right) \cr & {x^2} = \frac{{{{10}^6}}}{y} \cr & {\text{Now}}, \cr & y{\left( {1 - {\frac{{{{10}^6}}}{{10000y}}} } \right)^2} \cr & = 2304\left( {{\text{similar}}\,{\text{to}}\,{\text{above}}} \right) \cr & y{\left( {1 - \frac{{100}}{y}} \right)^2} = 2304 \cr & y = 2500 \cr} $$

If we take different values of x then,
Let x = 150, then error%
$$\eqalign{ & = \frac{{ {\frac{{ {200 - 100} }}{{100}}} }}{{100}} \cr & = 100\% \cr} $$
Again if x = 100, then error%
$$\eqalign{ & = \frac{{ {\left( {150 - 50} \right) \times 100} }}{{100}} \cr & = 200\% \cr} $$
If we take different value of x,
we get different answer so we can't determine it.

Female employees = 45% Male employees = 55% Difference = 55 - 45 = 10% → 10% = 72 → 1% = $$\frac{{72}}{{10}}$$ → 100 = 7.2 × 100 = 720

In 2000, the market share was 40%, 30% and 30%, means the ratio is 4 : 3 : 3 In 2001, a new product (A) enters and has 10% market share, 90% of the remaining market is shared by the previous 3. Now, divide 90% in the ratio 4 : 3 : 3 , i.e 36%, 27%, 27%. Now the ratio is 36 : 27 : 27 : 10 In 2002, another new product (B) enters and has 10% market share, now the remaining 90% market share is distributed in the ratio, 36 : 27 : 27 : 10 and hence remaining 32.4%, 24.3%, 24.3%, 9% Thus, the market share of Margo in 2002 is 32.4%