Q1. Find:
(i) 22 % of 120
(ii) 25 % of Rs.1000
(iii) 25 % of 10 kg
(iv) 16.5 % of 5000 metre
(v) 135 % of 80 cm
(vi) 2.5 % of 10000 ml\begin{array}{l} \text {Sol. (i) } \quad 22 \% \text { of } 120 \\ =\frac{22}{100} \times 120 \\ =26.4 \\ \\\text {(ii) } \quad 25 \% \text { of Rs.} 1000 \\ =\frac{25}{100} \times 1000 \\ =\text {Rs.}250 \\ \\\text {(iii) } \quad 25 \% \text { of } 10 \text { kg} \\ =\frac{25}{100} \times 10 \\ =2.5 \text { kg} \\ \\\text {(iv) } \quad 16.5 \% \text { of } 5000 \text { metre } \\=\frac{16.5}{100} \times 5000 \\ =825 \text { metre } \\ \\\text {(v) } \quad 135 \% \text { of } 80 \text { cm} \\ =\frac{135}{100} \times 80 \\ =108 \text { cm} \\ \\\text {(vi) } \quad 2.5 \% \text { of } 10000 \text { ml} \\ =\frac{2.5}{100} \times 10000 \\ =250 \text { ml} \\ \\ \end{array}Q2. Find the number a, if
(i) 8.4 % of a is 42
(ii) 0.5 of a is 3
$$ \text {(iii) } \quad \frac {1}{2} \text { of a is } 50 $$ (iv) 100 % of a is 100\begin{array}{l} \text {Sol. (i) Given that } \quad 8.4 \% \text { of a is } 42 \\ \Rightarrow \quad \frac{8.4}{100} \times a=42 \\ \Rightarrow \quad a=\frac{42 \times 100}{8.4} \\ \Rightarrow \quad a=500 \\ \\\text {(ii) } \quad 0.5 \text { of a is } 3 \\ \Rightarrow \quad \frac{0.5}{100} \times a=3 \\ \Rightarrow \quad a=\frac{3 \times 100}{0.5} \\ \Rightarrow \quad a=600 \\ \\\text {(iii) } \quad \frac {1}{2} \text { of a is } 50 \\ \Rightarrow \quad \frac{1}{200} \times a=50 \\ \Rightarrow \quad a=\frac{50 \times 200}{1} \\ \Rightarrow \quad a=10000 \\ \\\text {(iv) } \quad 100\% \text { of a is } 100 \\ \Rightarrow \quad \frac{100}{100} \times a=100 \\ \Rightarrow \quad a=\frac{100 \times 100}{100} \\ \Rightarrow \quad a=100 \end{array}Q3. x is 5% of y, y is 24% of z, if x=480, find the values of y and z.\begin{array}{l} \text {Sol. Given that x is 5% of y. } \\ \Rightarrow \quad x=\frac{5}{100} \times y \\ \Rightarrow \quad y=\frac{100}{5} \times x \\ \Rightarrow \quad y=20 x \\ \Rightarrow \quad y=20 \times 480 \quad \quad [\text {Given that x = 480}] \\ \Rightarrow \quad y=9600 \\ \text {Also given that y is 24% of z} \\\Rightarrow \quad y=\frac{24}{100} \times z \\ \Rightarrow \quad z=\frac{100}{24} \times y \\ \Rightarrow \quad z=\frac{100}{24} \times 9600 \\ \Rightarrow \quad z=40000 \\ \text {Hence y = 9600 and z = 40000 } \\ \end{array}Q4. A coolie deposits Rs 150 per month in his post office Savings Bank account. If this is 15% of his monthly income, find his monthly income.\begin{array}{l} \text {Sol. Let’s assume the monthly income of the coolie be Rs x.} \\ \text {According to given in question } \\ 15 \% \text { of } x = 150 \\\Rightarrow \quad \frac{15}{100} \times x=150 \\ \Rightarrow \quad x=\frac{150 \times 100}{15} \\ \Rightarrow \quad x=1000 \\\therefore \text {Coolie’s monthly income is Rs. 1000} \\ \end{array}Q5. Asha got 86.875 % marks in the annual examination. If she got 695 marks, find the total number of marks of the examination.\begin{array}{l} \text {Sol. Let’s x be the total number of marks in the examination.} \\ \text {According to given in question } \\ 86.875 \% \text { of } x = 695 \\\Rightarrow \quad \frac{86.875}{100} \times x=695 \\ \Rightarrow \quad x=\frac{695 \times 100}{86.875} \\ \Rightarrow \quad x=800 \\\therefore \text {Total number of marks in the examination are 800.} \\ \end{array}Q6. Deepti went to school for 216 days in a full year. If her attendance is 90%, find the number of days on which the school was opened.\begin{array}{l} \text {Sol. Let’s assume that school was opened for x days.} \\ \text {According to given in question } \\ 90 \% \text { of } x = 216 \\\Rightarrow \quad \frac{90}{100} \times x=216 \\ \Rightarrow \quad x=\frac{216 \times 100}{90} \\ \Rightarrow \quad x=240 \\\therefore \text {The school was opened for 240 days.} \\ \end{array}Q7. A garden has 2000 trees. 12% of these are mango trees, 18 % lemon and the rest are orange trees. Find the number of orange trees.\begin{array}{l} \text {Sol. } \\ \text {Given that total number of trees in the garden are 2000.} \\ \text {No. of mango trees } = 12\% \text { of } 2000 = \frac {12}{100} \times 2000 = 240 \\ \text {No. of lemon trees } = 18\% \text { of } 2000 = \frac {18}{100} \times 2000 = 360 \\ \text {Let’s assume the number of orange trees be x.} \\ \text {According to given in question } \\ x + 240 + 360= 2000 \\ \Rightarrow \quad x = 2000 -240 -360 = 1400 \\\therefore \text {There are 1400 orange trees in the garden.} \\ \end{array}Q8. Balanced diet should contain 12% of proteins, 25% of fats and 63% of carbohydrates. If a child needs 2600 calories in this food daily, find in calories the amount of each of these in his daily food intake.\begin{array}{l} \text {Sol. Given that total 2600 calories a child need daily.} \\ \text {Amount of protien intake } = 12\% \text { of } 2600 = \frac {12}{100} \times 2600 = 312 \\ \text {Amount of fats intake } = 25\% \text { of } 2600 = \frac {25}{100} \times 2600 = 650 \\\text {Amount of carbohydrates intake } = 63\% \text { of } 2600 = \frac {63}{100} \times 2600 = 1638 \\\end{array}
(i) 22 % of 120
(ii) 25 % of Rs.1000
(iii) 25 % of 10 kg
(iv) 16.5 % of 5000 metre
(v) 135 % of 80 cm
(vi) 2.5 % of 10000 ml\begin{array}{l} \text {Sol. (i) } \quad 22 \% \text { of } 120 \\ =\frac{22}{100} \times 120 \\ =26.4 \\ \\\text {(ii) } \quad 25 \% \text { of Rs.} 1000 \\ =\frac{25}{100} \times 1000 \\ =\text {Rs.}250 \\ \\\text {(iii) } \quad 25 \% \text { of } 10 \text { kg} \\ =\frac{25}{100} \times 10 \\ =2.5 \text { kg} \\ \\\text {(iv) } \quad 16.5 \% \text { of } 5000 \text { metre } \\=\frac{16.5}{100} \times 5000 \\ =825 \text { metre } \\ \\\text {(v) } \quad 135 \% \text { of } 80 \text { cm} \\ =\frac{135}{100} \times 80 \\ =108 \text { cm} \\ \\\text {(vi) } \quad 2.5 \% \text { of } 10000 \text { ml} \\ =\frac{2.5}{100} \times 10000 \\ =250 \text { ml} \\ \\ \end{array}Q2. Find the number a, if
(i) 8.4 % of a is 42
(ii) 0.5 of a is 3
$$ \text {(iii) } \quad \frac {1}{2} \text { of a is } 50 $$ (iv) 100 % of a is 100\begin{array}{l} \text {Sol. (i) Given that } \quad 8.4 \% \text { of a is } 42 \\ \Rightarrow \quad \frac{8.4}{100} \times a=42 \\ \Rightarrow \quad a=\frac{42 \times 100}{8.4} \\ \Rightarrow \quad a=500 \\ \\\text {(ii) } \quad 0.5 \text { of a is } 3 \\ \Rightarrow \quad \frac{0.5}{100} \times a=3 \\ \Rightarrow \quad a=\frac{3 \times 100}{0.5} \\ \Rightarrow \quad a=600 \\ \\\text {(iii) } \quad \frac {1}{2} \text { of a is } 50 \\ \Rightarrow \quad \frac{1}{200} \times a=50 \\ \Rightarrow \quad a=\frac{50 \times 200}{1} \\ \Rightarrow \quad a=10000 \\ \\\text {(iv) } \quad 100\% \text { of a is } 100 \\ \Rightarrow \quad \frac{100}{100} \times a=100 \\ \Rightarrow \quad a=\frac{100 \times 100}{100} \\ \Rightarrow \quad a=100 \end{array}Q3. x is 5% of y, y is 24% of z, if x=480, find the values of y and z.\begin{array}{l} \text {Sol. Given that x is 5% of y. } \\ \Rightarrow \quad x=\frac{5}{100} \times y \\ \Rightarrow \quad y=\frac{100}{5} \times x \\ \Rightarrow \quad y=20 x \\ \Rightarrow \quad y=20 \times 480 \quad \quad [\text {Given that x = 480}] \\ \Rightarrow \quad y=9600 \\ \text {Also given that y is 24% of z} \\\Rightarrow \quad y=\frac{24}{100} \times z \\ \Rightarrow \quad z=\frac{100}{24} \times y \\ \Rightarrow \quad z=\frac{100}{24} \times 9600 \\ \Rightarrow \quad z=40000 \\ \text {Hence y = 9600 and z = 40000 } \\ \end{array}Q4. A coolie deposits Rs 150 per month in his post office Savings Bank account. If this is 15% of his monthly income, find his monthly income.\begin{array}{l} \text {Sol. Let’s assume the monthly income of the coolie be Rs x.} \\ \text {According to given in question } \\ 15 \% \text { of } x = 150 \\\Rightarrow \quad \frac{15}{100} \times x=150 \\ \Rightarrow \quad x=\frac{150 \times 100}{15} \\ \Rightarrow \quad x=1000 \\\therefore \text {Coolie’s monthly income is Rs. 1000} \\ \end{array}Q5. Asha got 86.875 % marks in the annual examination. If she got 695 marks, find the total number of marks of the examination.\begin{array}{l} \text {Sol. Let’s x be the total number of marks in the examination.} \\ \text {According to given in question } \\ 86.875 \% \text { of } x = 695 \\\Rightarrow \quad \frac{86.875}{100} \times x=695 \\ \Rightarrow \quad x=\frac{695 \times 100}{86.875} \\ \Rightarrow \quad x=800 \\\therefore \text {Total number of marks in the examination are 800.} \\ \end{array}Q6. Deepti went to school for 216 days in a full year. If her attendance is 90%, find the number of days on which the school was opened.\begin{array}{l} \text {Sol. Let’s assume that school was opened for x days.} \\ \text {According to given in question } \\ 90 \% \text { of } x = 216 \\\Rightarrow \quad \frac{90}{100} \times x=216 \\ \Rightarrow \quad x=\frac{216 \times 100}{90} \\ \Rightarrow \quad x=240 \\\therefore \text {The school was opened for 240 days.} \\ \end{array}Q7. A garden has 2000 trees. 12% of these are mango trees, 18 % lemon and the rest are orange trees. Find the number of orange trees.\begin{array}{l} \text {Sol. } \\ \text {Given that total number of trees in the garden are 2000.} \\ \text {No. of mango trees } = 12\% \text { of } 2000 = \frac {12}{100} \times 2000 = 240 \\ \text {No. of lemon trees } = 18\% \text { of } 2000 = \frac {18}{100} \times 2000 = 360 \\ \text {Let’s assume the number of orange trees be x.} \\ \text {According to given in question } \\ x + 240 + 360= 2000 \\ \Rightarrow \quad x = 2000 -240 -360 = 1400 \\\therefore \text {There are 1400 orange trees in the garden.} \\ \end{array}Q8. Balanced diet should contain 12% of proteins, 25% of fats and 63% of carbohydrates. If a child needs 2600 calories in this food daily, find in calories the amount of each of these in his daily food intake.\begin{array}{l} \text {Sol. Given that total 2600 calories a child need daily.} \\ \text {Amount of protien intake } = 12\% \text { of } 2600 = \frac {12}{100} \times 2600 = 312 \\ \text {Amount of fats intake } = 25\% \text { of } 2600 = \frac {25}{100} \times 2600 = 650 \\\text {Amount of carbohydrates intake } = 63\% \text { of } 2600 = \frac {63}{100} \times 2600 = 1638 \\\end{array}