111. A sells his goods 30% cheaper than B and 30% dearer than C. By what % is the cost of C’s good cheaper than B’s goods.
Sol:
∴ required % = \(\frac{{1300 – 700}}{{1300}} \times 100\)
= \(\frac{{600}}{{13}}\)
= 46.15% Answer
112. Due to a 20% increase in the price of sugar, a person is able to purchase 30 kg less of sugar for rupees 540. Find the initial price.
Sol:-
∴ original consumption = 6×30 = 180
& new consumption = 5×30 = 150
∴ original price of sugar = \(\frac{{540}}{{180}}\) = 3 Rs/kg Answer
113. A’s salary is 10% lower than B’s salary which is 20% lower than C’s salary. By how much percent is C’s salary more than A’s salary ?
Sol:-
∴ required % = \(\frac{{25 – 18}}{{18}} \times 100\) = \( \boldsymbol{38\frac{8}{9}\% \uparrow} \) Answer
114. A sells his goods 25% cheaper than B and 25% dearer than C. How much % is C’s goods cheaper than B’s ?
Sol:-
∴ required % = \(\frac{{20 – 12}}{{20}} \times 100\) = 40% Answer
115. A has a certain amount with him. Of this he loses 32.5% in gambling. From the balance someone stole the sum of rupees 100000 that he used to keep in his pocket. Of the rest he donated 20% to a charity. Further he purchases a car for Rs. 7.5 lakh. What was the initial amount he has with him if he is left with 2.5 lakh cash.
Sol:-
\(\left( {x \times \frac{{\left( {100 – 32.5} \right)}}{{100}} – 100000} \right)\left( {\frac{{100 – 20}}{{100}}} \right) – 7.5lakh = 2.5lakh\)
\(\left( {x \times \frac{{67.5}}{{100}} – 100000} \right) \times \frac{4}{5} = 1000000\)
\(x \times \frac{{675}}{{1000}} – 100000 = 1250000\)
⟹ x = 20 lakh Answer
116. In an election, the candidate who got 60% of the votes cast won by 240 votes. Find the total number of voters in the voting list if 80% people cast their votes and there were no invalid votes.
Sol:-
∴ total votes casted = 100 × 12 = 1200
now 80% ⟶ 1200
100% ⟶ \(\frac{{100}}{{80}} \times 1200\) = 1500 Answer
117. The population of a village is 1000. The rate of increase is 10% per annum. Find the population at the start of third year and at the end of third year.
Sol:-
∴ population at the start of third year = \(\frac{{1000}}{{10 \times 10}} \times 11 \times 11\) = 1210 Answer
& population at the end of third year = \(\frac{{1000}}{{10 \times 10 \times 10}} \times 11 \times 11 \times 11\) = 1331 Answer
118. The height of a triangle is increased by 20%. What can be the maximum percentage increase in the length of the base so that the increase in Area is restricted to a maximum of 50% ?
Sol:-
Method(1):
50 = 20 + x + \(\frac{{20x}}{{100}}\)
30 = \(\frac{{120x}}{{100}}\)
x = +25% ↑ Answer
119. The population of a village is 7700. If the number of male increases by 11% and the number of female increases by 20% then the population becomes 8862. Then find the population of females in the town.
Sol:-
% ↑ in population = \(\frac{{8862 – 7700}}{{7700}} \times 100 = \frac{{1162}}{{77}}\) = \(\frac{{166}}{{11}}\% \)
120. A’s salary is 75% more than B’s salary. A got a raise of 40% on his salary while B got a raise of 25% on his salary. By what % A’s salary more than B’s salary.
Sol:-
∴ required % = \(\frac{{480}}{{500}} \times 100\) = 96% Answer