71. A student multiplied a number by \( \boldsymbol{\frac{2}{5}} \) instead of \( \boldsymbol{\frac{5}{2}} \). What is the % error in calculation ?
Sol:
In such type of question let a number which is exactly divisible by denominator of both the fractions.
72. In an examination a student got 25% marks and failed by 5 marks. While another student got 28% marks and got 7 marks more than pass marks. Then find the maximum marks in the examination ?
Sol:
73. In an examination the first student got 25% marks and failed by 18 marks. While in the same exam the second student got 27% marks and failed by 12 marks. Then find the maximum marks and minimum pass marks in the examination.
Sol:
∴ max marks = 100% = 100 × 3 = 300 Answer
& min pass marks = 25% + 18
= 25 × 3 + 18
= 93 Answer
74. The marks of Ankit in physics are 60% of the marks in chemistry and marks in chemistry are 60% of the marks in biology. How many marks he got in physics. If marks in these three subjects are 196 in all.
Sol:
∴ 9x + 15x + 25x = 196
x = 4
∴ marks in physics = 9 × 4 = 36 Answer
75. In an examination a student got 25% marks and failed by 5 marks while another student got 28% marks and got 7 marks more than pass marks. Then find the minimum marks % required to pass in the exam.
Sol:
∴ minumum marks % required to pass in the exam
= 25% + \(\frac{5}{4}\)%
= 31.25% Answer
76. A student has to secure 30% marks to pass the exam. If he gets 70 marks and fails by 50 marks. Then find the maximum marks set for the examination.
Sol:
30% ⟶ (70 + 50) ⟹ 1% ⟶ 4
∴ max marks = 100% = 100 × 4 = 400
Answer
77. A company gives 10% commission to his salesman on total sales and 1% bonus on the sales over Rs. 20000. If the salesman deposite Rs. 44700 after deducting his earning from total sales. Then find total sales.
Sol:
Let total sales = x
∴ earning = x × 10% + (x – 20000) × 1%
= \(\frac{x}{{10}} + \frac{x}{{10}} – 200\)
= \(\frac{{11x}}{{100}} – 200\)
then as per question,
x – \(\left( {\frac{{11x}}{{100}} – 200} \right)\) = 44700
\(\frac{{89x}}{{100}}\) = 44500
x = 50000 Answer
78. If the annual increase in the population of a town is \( \boldsymbol{14\frac{2}{7}} \)% and the present number of population is 2401. What will the population be in 4 years ?
Sol:
So after 4 years the population will be 4096 Answer
79. The population of a village is 14000. If the males and females are increased by 5% and 12% respectively then population will become 14770. Find the number of females at present.
Sol:
Increased population = 14770 – 14000 = 770
increased % = \(\frac{{770}}{{14000}} \times 100 = \frac{{77}}{{14}} = \frac{{11}}{2} = 5\frac{1}{2}\% \)
80. The population of town is 10000. It increases by 10% during the first year. During the second year it decreases by 20% and increased by 30% during the third year. What will the population be after 3 years ?
Sol: