The perimeter of a triangular field is 420 m
Question: The perimeter of a triangular field is 420 m and its sides are in the ratio 6:7:8. Find the area of the triangular field. Solution: Given, perimeter of a triangular field is 420 m and its sides are in the ratio 6:7:8. Let sides of a triangular field be a = 6x, b = 7x and c = 8x. Perimeter of a triangular field, 2s = a + b + c = 420 = 6x + 7x + 8x = 420 = 21x = x = 420/21 = 20 m $\therefore$ Sides of a triangular field are $a=6 \times 20=120 \mathrm{~m}$ $b=7 \times 20=140 \mathrm{~m}$ ...
Read More →In a corner of a rectangular field with dimensions
Question: In a corner of a rectangular field with dimensions $35 \mathrm{~m} \times 22 \mathrm{~m}$, a well with $14 \mathrm{~m}$ inside diameter is dug $8 \mathrm{~m}$ deep. The earth dug out is spread evenly over the remaining part of the field. Find the rise in the level of the field. Solution: We have, Length of the fiel, $l=35 \mathrm{~m}$, Width of the field, $b=22 \mathrm{~m}$, Depth of the well, $H=8 \mathrm{~m}$ and Radius of the well, $R=\frac{14}{2}=7 \mathrm{~m}$, Let the rise in the...
Read More →A dealer buys a wristwatch for Rs 225 and spends Rs 15 on its repairs.
Question: A dealer buys a wristwatch for Rs 225 and spends Rs 15 on its repairs. If he sells the same for Rs 300, find his profit percent. Solution: A dealer buys a wrist watch for Rs. 225 Money spent on repair ing the watch $=$ Rs. 15 Therefore, C. $\mathrm{P}=$ Rs. $(225+15)=$ Rs. 240 S. $\mathrm{P}=$ Rs. 300 Profit $=\mathrm{SP}-\mathrm{CP}$ $=$ Rs. $(300-240)$ $=$ Rs. 60 Profit $\%=\frac{\text { Profit }}{\text { C.P }} \times 100$ $=\frac{60}{240} \times 100$ $=25 \%$...
Read More →A field in the form of a parallelogram has sides 60 m
Question: A field in the form of a parallelogram has sides 60 m and 40 m and one of its diagonals is 80 m long. Find the area of the parallelogram. Solution: Let $A B C D$ be a parallelogram field with sides $A B=C D=60 \mathrm{~m}, B C=D A=40 \mathrm{~m}$ and diagonal $B D=80 \mathrm{~m}$ Area of parallelogram $A B C D=2$ (Area of $\triangle A B D$ )$\ldots$ (i) In $\triangle A B D$, Semi-perimeter of a triangle $\triangle A B D$, $s=\frac{a+b+c}{2}$ $=\frac{A B+B D+D A}{2}$ $=\frac{60+80+40}{2...
Read More →A retailer buys a cooler for Rs 1200 and overhead expenses on it are Rs 40.
Question: A retailer buys a cooler for Rs 1200 and overhead expenses on it are Rs 40. If he sells the cooler for Rs 1550, determine his profit percent. Solution: Cooler costs $=$ Rs. 1200 Overhead expenses $=$ Rs. 40 C.P $=$ Rs. $(1200+40)=$ Rs. 1240 S. $\mathrm{P}=$ Rs. 1550 Profit $=\mathrm{SP}-\mathrm{CP}$ $=$ Rs. $(1550-1240)$ $=$ Rs. 310 Profit $\%=\frac{\text { Profit }}{\text { C.P }} \times 100$ $=\frac{310}{1240} \times 100$ $=25 \%$...
Read More →In a village, a well with 10 m inside diameter, is dug 14 m deep.
Question: In a village, a well with 10 m inside diameter, is dug 14 m deep. Earth taken out of it is spread all around to a width 5 m to form an embankment. Find the height of the embankment. What value of the villagers is reflected here? Solution: We have, Radius of well, $R=\frac{10}{2}=5 \mathrm{~m}$, Depth of the well, $H=14 \mathrm{~m}$ and Width of the embankment $=5 \mathrm{~m}$, Also, the outer radius of the embankment, $r=R+5=5+5=10 \mathrm{~m}$ And, the inner radius of the embankment $...
Read More →A retailer buys a radio for Rs 225. His overhead expenses are Rs 15.
Question: A retailer buys a radio for Rs 225. His overhead expenses are Rs 15. If he sells the radio for Rs 300, determine his profit percent. Solution: Radio costs $=$ Rs. 225 Overhead expenses $=$ Rs. 15 C. $\mathrm{P}=$ Rs. $(225+15)=$ Rs. 240 S. $\mathrm{P}=$ Rs. 300 Profit $=$ SP $-$ CP $=$ Rs. $(300-240)$ $=$ Rs. 60 Profit $\%=\frac{\text { Profit }}{\text { C.P }} \times 100$ $=\frac{60}{240} \times 100$ $=25 \%$...
Read More →Marbles of diameter 1.4 cm are dropped into a cylindrical beaker of diameter 7 cm, containing some water.
Question: Marbles of diameter 1.4 cm are dropped into a cylindrical beaker of diameter 7 cm, containing some water. Find the number of marbles that should be dropped into the beaker so that the water level rises by 5.6 cm. Solution: Diameter of each marble $=1.4 \mathrm{~cm}$ Radius of each marble $=0.7 \mathrm{~cm}$ Volume of each marble $=\frac{4}{3} \pi r^{3}=\frac{4}{3} \pi \times(0.7)^{3} \mathrm{~cm}^{3}$ The water rises as a cylindrical column. Volume of cylindrical column filled with wat...
Read More →Find the area of a parallelogram given in the figure.
Question: Find the area of a parallelogram given in the figure. Also, find the length of the altitude from vertex A on the side DC. Thinking process (i)Determine the area of ABCD by using Herons formula. (ii)Using relation, area of parallelogram ABCD =2 (Area of ΔBCD) (iii)Also, determine the area of parallelogram by using the formula Base x Altitude. (iv)Further, equating the area of parallelogram in (ii) and (iii). Obtain the required length of the altitude. Solution: Area of parallelogram $\q...
Read More →The cost price of 10 articles is equal to the selling price of 9 articles.
Question: The cost price of 10 articles is equal to the selling price of 9 articles. Find the profit percent. Solution: Let the cost price of one article be Rs. C and the selling price be Rs. S Therefore, $10 \mathrm{C}=9 \mathrm{~S}$ $\mathrm{C}=\frac{9}{10} \mathrm{~S}$ So, the cost price is less than the selling price. S. P. $=\left(\frac{100+P \text { rofit } \%}{100}\right)$ C. P $\mathrm{S}=\left(\frac{100+P \text { rofit } \%}{100}\right) \mathrm{C}$ $\frac{\mathrm{S}}{\mathrm{C}}=\left(\...
Read More →150 spherical marbles, each of diameter 1.4 cm
Question: 150 spherical marbles, each of diameter 1.4 cm, are dropped in a cylindrical vessel of diameter 7 cm containing some water, which are completely immersed in water. Find the rise in the level of water in the vessel. Solution: We have, the radius of spherical marble, $r=\frac{1.4}{2}=0.7 \mathrm{~cm}$ and the radius of the cylindrical vessel, $R=\frac{7}{2} \mathrm{~cm}=3.5 \mathrm{~cm}$ Let the rise in the level of water in the vessel be $H$. Now, Volume of water rised in the cylindrica...
Read More →The perimeter of an isosceles triangle is 32 cm.
Question: The perimeter of an isosceles triangle is 32 cm. The ratio of the equal side to its base is 3 : 2. Find the area of the triangle. Solution: Let $A B C$ be an isosceles triangle with perimeter $32 \mathrm{~cm}$. We have, ratio of equal side to its base is $3: 2$. Let sides of triangle be $\quad A B=A C=3 x, B C=2 x$ $\because$ Perimeter of a triangle $=32 \mathrm{~m}$ Now, $\quad 3 x+3 x+2 x=32$ $\Rightarrow \quad 8 x=32$ $\Rightarrow \quad x=4$ $\therefore \quad A B=A C=3 \times 4=12 \...
Read More →Water running in a cylindrical pipe of inner diameter 7 cm, is collected in a container at the rate of 192.5 litres per minute.
Question: Water running in a cylindrical pipe of inner diameter 7 cm, is collected in a container at the rate of 192.5 litres per minute. Find the rate of flow of water in the pipe in km/hr. Solution: We have, Radius of cylindrical pipe, $r=\frac{7}{2} \mathrm{~cm}$ and The rate of flow of water $=192.5 \mathrm{~L} / \mathrm{min}$ $=\frac{192.5 \mathrm{~L}}{1 \mathrm{~min}}$ $=\frac{192.5 \times 1000 \mathrm{~cm}^{3}}{1 \mathrm{~min}} \quad\left(\right.$ As, $\left.1 \mathrm{~L}=1000 \mathrm{~cm...
Read More →A boy buys 9 apples for Rs 9.60 and sells them at 11 for Rs 12.
Question: A boy buys 9 apples for Rs 9.60 and sells them at 11 for Rs 12. Find his gain or loss percent. Solution: C.P of 9 apples $=$ Rs $9.60$ $\therefore$ C.P of 1 apple $=$ Rs $\frac{9.60}{9}=$ Rs $\frac{16}{15}$ S.P of 11 apples $=$ Rs 12 $\therefore$ S.P of 1 apple $=$ Rs $\frac{12}{11}$ Clearly, S.P of 1 apple $$ C.P of 1 apple So, we get profit on selling apples Gain $\%=\left(\frac{\operatorname{gain} \times 100}{\text { C.P. }}\right)$ $=\left(\frac{\frac{12}{11}-\frac{16}{15}}{\frac{1...
Read More →Rekha bought a saree for Rs 1240 and sold it for Rs 1147.
Question: Rekha bought a saree for Rs 1240 and sold it for Rs 1147. Find her loss and loss percent. Solution: C. P of saree $=$ Rs. 1240 S.P of saree $=$ Rs. 1147 Loss $=\mathrm{CP}-\mathrm{SP}$ Loss $=$ Rs. $(1240-1147)$ $=$ Rs. 93 loss $\%=\left(\frac{\text { loss } \times 100}{\text { C.P. }}\right)$ $=\left(\frac{93 \times 100}{1240}\right)$ $=7.5 \%$...
Read More →A farmer connects a pipe of internal diameter 25 cm from a canal into a cylindrical tank in his field, which is 12 m in diameter and 2.5 m deep.
Question: A farmer connects a pipe of internal diameter 25 cm from a canal into acylindrical tank in his field, which is 12 m in diameter and 2.5 m deep.If water flows through the pipe at the rate of 3.6 km/hr, then in how muchtime will the tank be filled? Also, find the cost of water if the canaldepartment charges at the rate of₹0.07 per m3. Solution: We have, the radius of the cylindrical tank, $R=\frac{12}{2}=6 \mathrm{~m}=600 \mathrm{~cm}$, the depth of the tank, $H=2.5 \mathrm{~m}=250 \math...
Read More →From a point in the interior of an equilateral triangle,
Question: From a point in the interior of an equilateral triangle, perpendiculars are drawn on the three sides. The lengths of the perpendiculars are 14 cm, 10 cm and 6 cm. Find the area of the triangle. Thinking Process (i)From a interior point a triangle, three triangle will be formed Determine area of each triangle by using the formula = (Base x Height). (ii)Determine the area of an equilateral triangle by using the Herons formula i.e., $\sqrt{s(s-a)(s-b)(s-c)}$ (iii) Further, equate the area...
Read More →A student buys a pen for Rs 90 and sells it for Rs 100.
Question: A student buys a pen for Rs 90 and sells it for Rs 100. Find his gain and gain percent. Solution: C. $\mathrm{P}$ of pen $=$ Rs. 90 S.P of pen $=$ Rs. 100 Gain $=\mathrm{SP}-\mathrm{CP}$ Gain $=100-90$ $=$ Rs. 10 Gain $\%=\left(\frac{\text { gain } \times 100}{\text { C.P. }}\right)$ $=\left(\frac{10 \times 100}{90}\right)$ $=11 \frac{1}{9} \%$...
Read More →Water in a canal, 5.4 m wide and 1.8 m deep, is flowing with a speed of 25 km/hr.
Question: Water in a canal, 5.4 m wide and 1.8 m deep, is flowing with a speed of 25 km/hr. How much area can it irrigate in 40 minutes, if 10 cm of standing water is required for irrigation? Solution: Width of the canal $=5.4 \mathrm{~m}$ Depth of the canal $=1.8 \mathrm{~m}$ Height of the standing water needed for irrigation $=10 \mathrm{~cm}=0.1 \mathrm{~m}$ Speed of the flowing water $=25 \mathrm{~km} / \mathrm{h}=\frac{25000}{60}=\frac{1250}{3} \mathrm{~m} / \mathrm{min}$ Volume of water fl...
Read More →Find the inverse of each of the following matrices by using elementary row transformations:
Question: Find the inverse of each of the following matrices by using elementary row transformations: $\left[\begin{array}{ccc}2 0 -1 \\ 5 1 0 \\ 0 1 3\end{array}\right]$ Solution: $A=\left[\begin{array}{ccc}2 0 -1 \\ 5 1 0 \\ 0 1 3\end{array}\right]$ We know $A=I A$ $\Rightarrow\left[\begin{array}{ccc}2 0 -1 \\ 5 1 0 \\ 0 1 3\end{array}\right]=\left[\begin{array}{lll}1 0 0 \\ 0 1 0 \\ 0 0 1\end{array}\right] A$ $\Rightarrow\left[\begin{array}{ccc}1 0 \frac{-1}{2} \\ 5 1 0 \\ 0 1 3\end{array}\ri...
Read More →A garden has 2000 trees.
Question: 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. Solution: Let the number of orange trees be $x$. It is given that there are 2,000 trees. While $12 \%$ of them are mango trees, $18 \%$ are lemon trees and the rest are orange trees. Now, $(12 \%$ of 2000$)+(18 \%$ of 2000$)+x=2000$ $\Rightarrow \frac{12}{100} \times 2000+\frac{18}{100} \times 2000+x=2000$ $\Rightarrow 240+360+x=2000$ $\Rightarrow 600+x=2000$...
Read More →The triangular side walls of a flyover have been
Question: The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 13 m, 14 m and 15 m. The advertisements yield an earning of Rs. 2000 per m2 a year. A company hired one of its walls for 6 months. How much rent did it pay? Solution: Since, the sides of a triangular walls are $a=13 \mathrm{~m}, b=14 \mathrm{~m}$ and $c=15 \mathrm{~m}$ $\therefore$ semi-perimeter of triangular side wall, $s=\frac{a+b+c}{2}=\frac{13+14+15}{2}=\frac{42}{2}=21 \mathrm{~m}$...
Read More →Deepti went to school for 216 days in a full year.
Question: 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. Solution: Let the school opened for $x$ days in the given year. We have : $90 \%$ of $x=216$ $\Rightarrow \frac{90}{100} x=216$ $\Rightarrow x=\frac{216 \times 100}{90}$ $=240$ $\therefore$ The school opened for 240 days in the given year....
Read More →Water is flowing at the rate of 6 km/hr through a pipe of diameter 14 cm
Question: Water is flowing at the rate of 6 km/hr through a pipe of diameter 14 cm into a rectangular tank which is 60 m long and 22 m wide. Determine the time in which the level of water in the tank will rise by 7 cm. Solution: We have, Speed of the water flowing through the pipe, $H=6 \mathrm{~km} / \mathrm{h}=\frac{600000 \mathrm{~cm}}{3600 \mathrm{~s}}=\frac{500}{3} \mathrm{~cm} / \mathrm{s}$, Radius of the pipe, $R=\frac{14}{2}=7 \mathrm{~cm}$, Length of the rectangular $\operatorname{tank}...
Read More →Find the inverse of each of the following matrices by using elementary row transformations:
Question: Find the inverse of each of the following matrices by using elementary row transformations: $\left[\begin{array}{lll}0 1 2 \\ 1 2 3 \\ 3 1 1\end{array}\right]$ Solution: $A=\left[\begin{array}{lll}0 1 2 \\ 1 2 3 \\ 3 1 1\end{array}\right]$ We know $A=I A$ $\Rightarrow\left[\begin{array}{lll}0 1 2 \\ 1 2 3 \\ 3 1 1\end{array}\right]=\left[\begin{array}{lll}1 0 0 \\ 0 1 0 \\ 0 0 1\end{array}\right] A$ $\Rightarrow\left[\begin{array}{ccc}0 1 2 \\ -2 1 2 \\ 3 1 1\end{array}\right]=\left[\b...
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