A bucket is in the form of a frustum of a cone.
Question: A bucket is in the form of a frustum of a cone. Its depth is 15 cm and the diameters of the top and the bottom are 56 cm and 42 cm, respectively. Find how many litres of water can the bucket hold. Solution: Greater diameter of the frustum = 56 cmGreater radius of the frustum =R= 28 cmSmaller diameter of the frustum = 42 cmRadius of the smaller end of the frustum =r= 21 cmHeight of the frustum =h= 15 cmCapacity of the frustum $=\frac{1}{3} \pi h\left(R^{2}+r^{2}+R r\right)$ $=\frac{1}{3...
Read More →After allowing a discount of
Question: After allowing a discount of $7 \frac{1}{2} \%$ on the marked price, an article is sold for Rs 555 . Find its markd price. Solution: Given, SP of the article $=$ Rs. 555 Discount $=7.5 \%$ Let the MP of the article be Rs. $x$. Therefore, Discount $=\frac{\text { Discount } \times \mathrm{MP}}{100}=$ Rs. $\frac{7.5 x}{100}=$ Rs. $0.075 x$ Since SP $=\mathrm{MP}-D$ iscount, $555=x-0.075 x$ $555=0.925 x$ $x=\frac{555}{0.925}$ $=R s .600$ Thus, the MP of the article is Rs. 600 ....
Read More →The radii of the circular ends of a solid frustum of a cone are 33 cm and 27 cm,
Question: The radii of the circular ends of a solid frustum of a cone are 33 cm and 27 cm, and its slant height is 10 cm. Find its capacity and total surface area. Solution: Greater radius =R= 33 cmSmaller radius =r= 27 cmSlant height =l= 10 cmUsing the formula for height of a frustum:Height =h = $=\sqrt{l^{2}-(R-r)^{2}}$ $=\sqrt{10^{2}-(33-27)^{2}}$ $=\sqrt{100-(6)^{2}}$ $=\sqrt{100-36}$ $=\sqrt{64}=8 \mathrm{~cm}$ Capacity of the frustum $=\frac{1}{3} \pi h\left(R^{2}+r^{2}+R r\right)$ $=\frac...
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: Solution: $\left[\begin{array}{lll}3 -3 4 \\ 2 -3 4 \\ 0 -1 1\end{array}\right]$ We know $A=I A$ $\Rightarrow\left[\begin{array}{lll}3 -3 4 \\ 2 -3 4 \\ 0 -1 1\end{array}\right]=\left[\begin{array}{lll}1 0 0 \\ 0 1 0 \\ 0 0 1\end{array}\right] A$ $\Rightarrow\left[\begin{array}{lll}1 -1 \frac{4}{3} \\ 2 -3 4 \\ 0 -1 1\end{array}\right]=\left[\begin{array}{lll}\frac{1}{3} 0 0 \\ 0 1 0 \\ 0 0 1\en...
Read More →The radii of the circular ends of a solid frustum of a cone are 33 cm and 27 cm,
Question: The radii of the circular ends of a solid frustum of a cone are 33 cm and 27 cm, and its slant height is 10 cm. Find its capacity and total surface area. Solution: Greater radius =R= 33 cmSmaller radius =r= 27 cmSlant height =l= 10 cmUsing the formula for height of a frustum:Height =h = $=\sqrt{l^{2}-(R-r)^{2}}$ $=\sqrt{10^{2}-(33-27)^{2}}$ $=\sqrt{100-(6)^{2}}$ $=\sqrt{100-36}$ $=\sqrt{64}=8 \mathrm{~cm}$ Capacity of the frustum $=\frac{1}{3} \pi h\left(R^{2}+r^{2}+R r\right)$ $=\frac...
Read More →On the eve of Gandhi Jayanti a saree is sold for Rs 720 after allowing 20% discount.
Question: On the eve of Gandhi Jayanti a saree is sold for Rs 720 after allowing 20% discount. What is its marked price? Solution: Given, SP of the saree $=$ Rs. 720 Discount on the saree $=20 \%$ We know, Discount $\%=\frac{\text { Discount }}{\text { MP }} \times 100$ Or, Discount $=\frac{\text { Discount } \% \times \text { MP }}{100}$ Let the MP of the saree be Rs. $x$ Therefore, Discount $=\frac{20}{100} x=$ Rs. $0.02 x$ Since S.P $=$ MP $-D$ iscount, $720=x-0.20 x$ $720=0.80 x$ $x=\frac{72...
Read More →A container made of a metal sheet open at the top is of the form of frustum of cone, whose height is 16 cm
Question: A container made of a metal sheet open at the top is of the form of frustum of cone, whose height is 16 cm and the radii of its lower and upper circular edges are 8 cm and 20cm respectively. Find(i) the cost of metal sheet used to make the container if it costs ₹ 10 per 100 cm2 (ii) the cost of milk at the rate of ₹ 35 per litre which can fill it completely. Solution: Let $r=8 \mathrm{~cm}, R=20 \mathrm{~cm}, h=16 \mathrm{~cm}$. $\Rightarrow l=\sqrt{(R-r)^{2}+h^{2}}=\sqrt{(20-8)^{2}+16...
Read More →A container made of a metal sheet open at the top is of the form of frustum of cone, whose height is 16 cm
Question: A container made of a metal sheet open at the top is of the form of frustum of cone, whose height is 16 cm and the radii of its lower and upper circular edges are 8 cm and 20cm respectively. Find(i) the cost of metal sheet used to make the container if it costs ₹ 10 per 100 cm2 (ii) the cost of milk at the rate of ₹ 35 per litre which can fill it completely. Solution: Let $r=8 \mathrm{~cm}, R=20 \mathrm{~cm}, h=16 \mathrm{~cm}$. $\Rightarrow l=\sqrt{(R-r)^{2}+h^{2}}=\sqrt{(20-8)^{2}+16...
Read More →A design is made on a rectangular
Question: A design is made on a rectangular tile of dimensions 50 cm x 17 cm as shown in figure. The design shows 8 triangle, each of sides 26 cm, 17 cm and 25 cm. Find the total area of the design and the remaining area of the tiles. Solution: Given, the dimension of rectangular tile is $50 \mathrm{~cm} \times 70 \mathrm{~cm}$. $\therefore$ Area of rectangular tile $=50 \times 70=3500 \mathrm{~cm}^{2}$ The sides of a design of one triangle be $a=25 \mathrm{~cm}, b=17 \mathrm{~cm}$ and $c=26 \ma...
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: $\left[\begin{array}{lll}2 3 1 \\ 2 4 1 \\ 3 7 2\end{array}\right]$ Solution: $A=\left[\begin{array}{lll}2 3 1 \\ 2 4 1 \\ 3 7 2\end{array}\right]$ We know $A=I A$ $\Rightarrow\left[\begin{array}{lll}2 3 1 \\ 2 4 1 \\ 3 7 2\end{array}\right]=\left[\begin{array}{lll}1 0 0 \\ 0 1 0 \\ 0 0 1\end{array}\right] A$ $\Rightarrow\left[\begin{array}{lll}1 \frac{3}{2} \frac{1}{2} \\ 2 4 1 \\ 3 7 2\end{arr...
Read More →The marked price of a ceiling fan is Rs 720.
Question: The marked price of a ceiling fan is Rs 720. During off season, it is sold for Rs 684. Determine the discount percent. Solution: Given: MP of the ceiling fan $=$ Rs. 720 SP of the ceiling fan $=$ Rs. 684 Since SP $=$ MP $-$ Discount, Discount $=\mathrm{MP}-\mathrm{SP}$ $=$ Rs. $(720-684)$ $=$ Rs. 36 Discount $\%=\left(\frac{\text { Discount }}{\text { MP }}\right) \times 100 \%$ $=\frac{36}{720} \times 100 \%$ $=5 \%$...
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: $\left[\begin{array}{lll}2 3 1 \\ 2 4 1 \\ 3 7 2\end{array}\right]$ Solution: $A=\left[\begin{array}{lll}2 3 1 \\ 2 4 1 \\ 3 7 2\end{array}\right]$ We know $A=I A$ $\Rightarrow\left[\begin{array}{lll}2 3 1 \\ 2 4 1 \\ 3 7 2\end{array}\right]=\left[\begin{array}{lll}1 0 0 \\ 0 1 0 \\ 0 0 1\end{array}\right] A$ $\Rightarrow\left[\begin{array}{lll}1 \frac{3}{2} \frac{1}{2} \\ 2 4 1 \\ 3 7 2\end{arr...
Read More →A shop selling sewing machines offers 3% discount on all cash purchases.
Question: A shop selling sewing machines offers 3% discount on all cash purchases. What cash amount does a customer pay for a sewing machine the price of which is marked as Rs 650. Solution: $D$ iscount $=3 \%$ Marked price $=$ Rs. 650 Now, $3 \%$ of the $\mathrm{MP}=\frac{3}{100} \times 650$ $=$ Rs $19.50$ So, SP $=$ MP $-$ Discount $=650-19.50$ $=$ Rs $630.50$...
Read More →The dimensions of a rectangle ABCD are 51 cm x 25 cm.
Question: The dimensions of a rectangle ABCD are 51 cm x 25 cm. A trapezium PQCD with its parallel sides QC and PD in the ratio 9 : 8, is cut off from the rectangle as shown in the figure. If the area of the trapezium PQCD is 5/6 th part of the area of the rectangle. Find the lengths QC and PD. Solution: We have dimensions of a recstangle $A B C D$ as $51 \mathrm{~cm}$ and $25 \mathrm{~cm}$. Also, in trapezium $P Q C D$, parallel sides $Q C$ and $P D$ are in the ratio $9: 8$. i.e.n$Q C: P D=9: 8...
Read More →Find discount in percent when
Question: Find discount in percent when (i) M.P. = Rs 900 and S.P. = Rs 873 (ii) M.P. = Rs 500 and S.P. = Rs 425 Solution: (i) We know that $\mathrm{SP}=\mathrm{MP}-D$ iscount So, $873=900-D$ iscount Therefore, Discount $=$ Rs. $(900-873)$ $=$ Rs. 27 Discount $\%=\frac{\text { Discount }}{\text { MP }} \times 100 \%$ $=\frac{27}{900} \times 100 \%$ $=3 \%$ (ii) We know that SP $=\mathrm{MP}-D$ iscount So, $425=500-D$ iscount There fore, Discount $=$ Rs. $(500-425)$ $=\mathrm{Rs} .75$ Discount $\...
Read More →Find the M.P. if
Question: Find the M.P. if (i) S.P. = Rs 1222 and Discount = 6% (ii) S.P. = Rs 495 and Discount = 1% Solution: (i) Given, $\mathrm{SP}=\mathrm{Rs} 1222$ Discoun $t=6 \%$ So, $\mathrm{MP}=\left(\frac{100 \times \mathrm{SP}}{100-\text { Discount } \%}\right)$ $=\frac{100 \times 1222}{100-6}$ $=$ Rs. 1300 (ii) Given, $\mathrm{SP}=$ Rs. 495 Discount $=1 \%$ So, $\mathrm{MP}=\left(\frac{100 \times \mathrm{SP}}{100-\text { Discount } \%}\right)$ $=\frac{100 \times 495}{100-1}$ $=$ Rs. 500...
Read More →In figure, ΔABC has sides AB = 7.5 cm,
Question: In figure, ΔABC has sides AB = 7.5 cm, AC = 6.5 cm and BC = 7 cm. On base BC a parallelogram DBCE of same area as that of ΔABC is constructed. Find the height OF of the parallelogram. Thinking process (i) First, determine the area of AABC by using Herons formula i.e., $\sqrt{s(s-a)(s-b)(s-c)}$ (ii) Second, determine the area of parallelogram BCED by using the formula Base x Height. (iii) Further, equate the area of triangle and area of parallelogram, to get the height of the parallelog...
Read More →Find the S.P. if
Question: Find the S.P. if (i) M.P. = Rs 1300 and Discount = 10% (ii) M.P. = Rs 500 and Discount = 15% Solution: (i) We know that $\mathrm{SP}=\mathrm{MP}-D$ iscount Discount $\%=\frac{\text { Discount }}{\text { MP }} \times 100$ Discount $=\frac{\text { Discount } \% \times \mathrm{MP}}{100}$ $=\frac{10 \times 1300}{100}$ So, SP $=$ Rs. $\left(1300-\left(\frac{10}{100} \times 1300\right)\right)$ $=1300-130$ $=$ Rs. 1170 (ii) We know that $\mathrm{SP}=\mathrm{MP}-D$ iscount Discount $\%=\frac{\...
Read More →Toshiba bought 100 hens for Rs 8000 and sold 20 of these at a gain of 5%.
Question: Toshiba bought 100 hens for Rs 8000 and sold 20 of these at a gain of 5%. At what gain percent she must sell the remaining hens so as to gain 20% on the whole? Solution: C.P of 100 hens $=$ Rs. 8000 Cost of one hen $=\frac{8000}{100}=$ Rs. 80 C.P of 20 hens $=$ Rs. $(80 \times 20)=$ Rs. 1600 Gain $\%=5 \%$ $S . P=C . P\left(\frac{100+g \text { ain } \%}{100}\right)$ $S \cdot P=1600 \times \frac{105}{100}=$ Rs. 1680 $C . P$ of 80 hens $=$ Rs. $(80 \times 80)=$ Rs. 6400 Gain on 80 hens $...
Read More →The difference between two selling prices of a shirt at profits of 4% and 5% is Rs 6. Find
Question: The difference between two selling prices of a shirt at profits of 4% and 5% is Rs 6. Find (i) C.P. of the shirt (ii) the two selling prices of the shirt Solution: $L$ et the C.P of both the shirts be Rs. $x$. C. $\mathrm{P}=$ Rs. $x$ For shirt 1 : Profit is $4 \%$ : Profit $\%=\frac{\text { Profit }}{\text { CP }} \times 100$ Profit $=\frac{4}{100} \times$ C.P $=$ Rs. $0.04 x$ S.P $=$ C.P $+$ Profit $=x+0.04 x$ $=$ Rs. $1.04 x$ For shirt 2 : Profit $=5 \%$ : C.P $=$ Rs. $x$ Profit $=\...
Read More →A container, open at the top, is in the form of a frustum of a cone of height 24 cm
Question: A container, open at the top, is in the form of a frustum of a cone ofheight 24 cm with radii of its lower and upper circular ends as 8 cm and20 cm, respectively. Find the cost of milk which can completely fill thecontainer at the rate of₹21 per litre. Solution: We have, Height, $h=24 \mathrm{~cm}$, Upper radius, $R=20 \mathrm{~cm}$ and Lower radius, $r=8 \mathrm{~cm}$ Now, Volume of the container $=\frac{1}{3} \pi h\left(R^{2}+r^{2}+R r\right)$ $=\frac{1}{3} \times \frac{22}{7} \times...
Read More →A field is in the shape of a trapezium
Question: A field is in the shape of a trapezium having parallel sides 90 m and 30 m. These sides meet the third side at right angles. The length of the fourth side is 100 m. If it Rs. 4 costs to plough 1 m2of the field, find the total cost of ploughing the field. Solution: In trapezium ABCD, we draw a perpendicular line CE to the line AB. We have, $D C=A E=30 \mathrm{~m}$ Now, $\quad B E=A B-A E=90-30=60 \mathrm{~m}$ In right angled $\triangle B E C$, $(B C)^{2}=(B E)^{2}+(E C)^{2} \quad$ [usin...
Read More →Shabana bought 16 dozen ball bens and sold them at a loss equal to S.P. of 8 ball pens. Find
Question: Shabana bought 16 dozen ball bens and sold them at a loss equal to S.P. of 8 ball pens. Find (i) her loss percent (ii) S.P. of 1 dozen ball pens, if she purchased these 16 dozen ball pens for Rs 576. Solution: (i) Number of pens bought $=16 \times 12=192$ Let the S. P of one pen be Rs. $x .$ Therefore, $S . P$ of 192 pens $=$ Rs. $192 x$ S. P of 8 pens $=$ Rs. $8 x$ It is given that $S . P$ of 8 pens is equal to the loss on selling 192 pens. Therefore, loss $=$ Rs. $8 x$ C. P of 192 pe...
Read More →A rectangular plot is given for constructing
Question: A rectangular plot is given for constructing a house having a measurement of 40 m long and 15 m in the front. According to the laws, a minimum of 3 m, wide space should be left in the front and back each and 2 m wide space on each of other sides. Find the largest area where house can be constructed. Solution: Let ABCD is a rectangular plot having a measurement of 40 m long and 15 m front. Length of inner-rectangle, EF = 40 3 3 = 34 m and breadth of inner-rectangle, FG =15 2 2 = 11 m An...
Read More →A tricycle is sold at a gain of 16%.
Question: A tricycle is sold at a gain of 16%. Had it been sold for Rs 100 more, the gain would have been 20%. Find the C.P. of the tricycle. Solution: Let the S. P of the tricycle be Rs. $x$ and C.P be Rs. $y$ Gain $\%=16 \%$ $S \cdot P=C \cdot P\left(\frac{100+\text { gain } \%}{100}\right)$ Then we have, $x=y+\left(\frac{y \times 16}{100}\right)$ $x=y+0.16 y$ $x=1.16 y$ When S.P increases by Rs. 100 , we get $\Rightarrow x+100=y+\left(\frac{y \times 20}{100}\right)$ Put ting $x=1.6 y$, we get...
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