The mean of 25 observations is 36.
Question: The mean of 25 observations is 36. Out of these observations, if the mean of first 13 observations is 32 and that of the last 13 observations is 40, the 13th observation is (a)23 (b)36 (c)38 (d)40 Solution: (b)Given, mean of 25 observations = 36 Sum of 25 observations = 36 x 25 = 900 Now, the mean of first 13 observations = 32 Sum of first 13 observations = 13 x 32 = 416 and the mean of last 13 observations = 40 Sum of last 13 observations = 40 x 13 = 520 So, 13th observation = (Sum of...
Read More →In the class intervals 10-20, 20-30,
Question: In the class intervals 10-20, 20-30, the number 20 is included in (a)10-20 (b)20-30 (c)Both the intervals (d)None of these Solution: (b) Since, the class interval 10-20 is the first interval of frequency distributionand 20-30 is the next one but the number 20 is present in both intervals. We know that, the presence of 20 in the interval 10-20 is not fully 100% while in the next interval 20-30, presence of it fully 100%....
Read More →If m is the mid-point and l is the upper class limit
Question: If m is the mid-point and l is the upper class limit of a class in a continuous frequency distribution, then lower class limit of the class is (a)2 m+l (b)2 m-l (c)m-l (d)m-2l Solution: (b)Let x and y be the lower and upper class limit of a continuous frequency distribution. Now, mid-point of a class = (x + y)/2 = m [given] = x + y = 2 m =x + l = 2m [ y = l = upper class limit (given)] = x = 2 m-l Hence, the lower class limit of the class is 2m l....
Read More →Rohit buys an item at 25% discount on the marked price.
Question: Rohit buys an item at 25% discount on the marked price. He sells it for Rs 660, making a profit of 10%. What is the marked price of the item? Solution: Given, SP of the item $=R$. 660 Discount on the item $=25 \%$ Profit on the item $=10 \%$ We know, $\mathrm{MP}=\left(\frac{100 \times \mathrm{SP}}{100-\text { Discount } \%}\right)$ $=\left(\frac{100 \times 660}{100-25}\right)$ $=$ Rs. 880 Thus, the marked price of the item is Rs. 880 ....
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: $\left[\begin{array}{ccc}1 3 -2 \\ -3 0 -1 \\ 2 1 0\end{array}\right]$ Solution: $A=\left[\begin{array}{ccc}1 3 -2 \\ -3 0 -1 \\ 2 1 0\end{array}\right]$ We know $A=I A$ $\Rightarrow\left[\begin{array}{ccc}1 3 -2 \\ -3 0 -1 \\ 2 1 0\end{array}\right]=\left[\begin{array}{lll}1 0 0 \\ 0 1 0 \\ 0 0 1\end{array}\right] A$ $\Rightarrow\left[\begin{array}{ccc}1 3 -2 \\ 0 9 -7 \\ 0 -5 4\end{array}\righ...
Read More →The width of each of five continuous
Question: The width of each of five continuous classes in a frequency distribution is 5 and the lower class limit of the lowest class is 10. The upper class limit of the highest class is (a)15 (b)25 (c)35 (d)40 Solution: (c)Let x and y be the upper and lower class limit of frequency distribution. Given, width of the class = 5 = x-y= 5 (i) Also, given lower class (y) = 10 On putting y = 10 in Eq.(i),we get x 10= 5 = x = 15 So, the upper class limit of the lowest class is 15. Hence, the upper clas...
Read More →The list price of a table fan is Rs 480 and it is available to a retailer at 25% discount.
Question: The list price of a table fan is Rs 480 and it is available to a retailer at 25% discount. For how much should a retailer sell it to gain 15%? Solution: Marked price of the table fan $=$ Rs. 480 Discount $=25 \%$ Therefore, cost price $=25 \%$ of $R s .480$ $\frac{25}{100} \times 480=$ Rs. 360 It is given that the profit on the table fan is $15 \%$. Gain $\%=\frac{\text { Gain }}{\text { CP }} \times 100$ $15=\frac{\text { Gain }}{360} \times 100$ Gain $=$ Rs. 54 Gain $=$ SP $-$ CP $54...
Read More →In a frequency distribution,
Question: In a frequency distribution, the mid value of a class is 10 and the width of the class is 6. The lower limit of the class is (a)6 (b)7 (c)8 (d)12 Solution: (b)Let x and y be the upper and lower class limit in a frequency distribution. Now, mid value of a class (x + y )/2=10 [given] = x + y = 20 (i) Also, given that, width of class x- y = 6 (ii) On adding Eqs. (i) and (ii), we get 2x =20+ 6 = 2x =26 = x = 13 On putting x = 13 in Eq. (i), we get 13+y = 20= y = 7 Hence, the lower limit of...
Read More →The perimeters of the two circular ends of a frustum of a cone are 48 cm
Question: The perimeters of the two circular ends of a frustum of a cone are 48 cm and 36 cm. If the height of the frustum is 11 cm, then find its volume and curved surface area. Solution: We have, Perimeter of upper end, $C=48 \mathrm{~cm}$, Perimeter of lower end, $c=36 \mathrm{~cm}$ and Height, $h=11 \mathrm{~cm}$ Let the radius of upper end be $R$ and the radius of lower end be $r$. As, $C=48 \mathrm{~cm}$ $\Rightarrow 2 \pi R=48$ $\Rightarrow R=\frac{48}{2 \pi}$ $\Rightarrow R=\frac{24}{\pi...
Read More →The range of the data 25,
Question: The range of the data 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11 and 20 is (a)10 (b)15 (c)18 (d)26 Solution: (d)In a given data, maximum value = 32 and minimum value = 6 We know, range of the data = maximum value minimum value = 32 6 = 26 Hence, the range of the given data is 26....
Read More →After allowing a discount of 20% on the marked price of a lamp, a trader loses 10%.
Question: After allowing a discount of 20% on the marked price of a lamp, a trader loses 10%. By what percentage is the marked price above the cost price? Solution: Let the $\mathrm{CP}$ of the lamp be Rs. 100 . Loss $=10 \%$ of $\mathrm{CP}=$ Rs. 10 So, SP $=\mathrm{CP}-$ Loss $=$ Rs. $100-$ Rs. $10=$ Rs. 90 The trader allows a discount of $20 \%$. This means that when the MP is Rs. 100, the SP will be Rs. $80 .$ Now, If Rs. 80 is the SP, th $e \mathrm{MP}=$ Rs. 100 If Re. 1 is the SP, the M.P ...
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: $\left[\begin{array}{ccc}3 0 -1 \\ 2 3 0 \\ 0 4 1\end{array}\right]$ Solution: $A=\left[\begin{array}{ccc}3 0 -1 \\ 2 3 0 \\ 0 4 1\end{array}\right]$ We know $A=I A$ $\Rightarrow\left[\begin{array}{ccc}3 0 -1 \\ 2 3 0 \\ 0 4 1\end{array}\right]=\left[\begin{array}{lll}1 0 0 \\ 0 1 0 \\ 0 0 1\end{array}\right] A$ $\Rightarrow\left[\begin{array}{ccc}1 0 -\frac{1}{3} \\ 2 3 0 \\ 0 4 1\end{array}\ri...
Read More →A tent consists of a frustum of a cone, surmounted by a cone
Question: A tent consists of a frustum of a cone, surmounted by a cone. If the diameter of the upper and lower circular ends of the frustum be 14 m and 26 m, respectively, the height of the frustum be 8 m and the slant height of the surmounted conical portion be 12 m, find the area of the canvas required to make the tent. (Assume that the radii of the upper circular end of the frustum and the base of the surmounted conical portion are equal.) Solution: For the frustum:Upper diameter = 14 mUpper ...
Read More →A publisher gives 32% discount on the printed price of a book to booksellers.
Question: A publisher gives 32% discount on the printed price of a book to booksellers. What does a bookseller pay for a book whose printed price is Rs 275? Solution: Discount allowed by the publisher $=32 \%$ on the printed price Printed price $=$ Rs. 275 So, $32 \%$ of $275=\frac{32}{100} \times 275$ $=$ Rs. 88 So, the bookseller pay $s=$ Rs. $275-$ Rs. 88 = Rs. 187 for a book...
Read More →A tent consists of a frustum of a cone, surmounted by a cone
Question: A tent consists of a frustum of a cone, surmounted by a cone. If the diameter of the upper and lower circular ends of the frustum be 14 m and 26 m, respectively, the height of the frustum be 8 m and the slant height of the surmounted conical portion be 12 m, find the area of the canvas required to make the tent. (Assume that the radii of the upper circular end of the frustum and the base of the surmounted conical portion are equal.) Solution: For the frustum:Upper diameter = 14 mUpper ...
Read More →30 circular plates, each of radius 14 cm and thickness 3 cm
Question: 30 circular plates, each of radius 14 cm and thickness 3 cm are placed one above the another to form a cylindrical solid.Find the total surface area. volume of the cylinder so formed. Solution: Given, radius of a circular plate, r = 14 cm Thickness of a circular plate = 3 cm Thickness of 30 circular plates = 30 x 3 = 90 cm Since, 30 circular plates are placed one above the another to form a cylindrical solid. Then, Height of the cylindrical solid, h = Thickness of 30 circular plates = ...
Read More →By selling a pair of earings at a discount of 25% on the marked price, a jeweller makes a profit of 16%.
Question: By selling a pair of earings at a discount of 25% on the marked price, a jeweller makes a profit of 16%. If the profit is Rs 48, what is the cost price? What is the marked price and the price at which the pair was eventually bought? Solution: Let the cost price of the pair of earrings be Rs. $\mathrm{x}$. Profit $=16 \%$ Therefore, $\mathrm{SP}=\left(\frac{\text { Profit } \%+100}{100}\right) \times \mathrm{CP}$ $=\left(\frac{16+100}{100}\right) \mathrm{x}$ $=R s . \frac{116 \mathrm{x}...
Read More →A tent is made in the form of a frustum of a cone surmounted by another cone.
Question: A tent is made in the form of a frustum of a cone surmounted by another cone. The diameters of the base and the top of the frustum are 20 m and 6 m, respectively, and the height is 24 m. If the height of the tent is 28 m and the radius of the conical part is equal to the radius of the top of the frustum, find the quantity of canvas required. Solution: For the lower portion of the tent:Diameter of the base= 20 mRadius,R, of the base = 10 mDiameter of the top end of the frustum = 6 mRadi...
Read More →A tent is made in the form of a frustum of a cone surmounted by another cone.
Question: A tent is made in the form of a frustum of a cone surmounted by another cone. The diameters of the base and the top of the frustum are 20 m and 6 m, respectively, and the height is 24 m. If the height of the tent is 28 m and the radius of the conical part is equal to the radius of the top of the frustum, find the quantity of canvas required. Solution: For the lower portion of the tent:Diameter of the base= 20 mRadius,R, of the base = 10 mDiameter of the top end of the frustum = 6 mRadi...
Read More →A right triangle with sides 6 cm,
Question: A right triangle with sides 6 cm, 8 cm and 10 cm is revolved about the side 8 cm. Find the volume and the curved surface of the solid so formed. Solution: When a right triangle with sides 6 cm, 8 cm and 10 cm is revolved about the side 8 cm, then solid formed is a cone whose height of a cone, h = 8 cm and radius of a cone, r = 6 cm. Slant height of a cone, l = 10 cm Volume of a cone = 1/3 r2h = (1/3) x (22/7) x 6 x 6 x 8 = 6336/21 = 301.7cm3 and curved surface of the area of cone = rl ...
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: $\left[\begin{array}{ccc}2 -1 4 \\ 4 0 7 \\ 3 -2 7\end{array}\right]$ Solution: $A=\left[\begin{array}{ccc}2 -1 4 \\ 4 0 2 \\ 3 -2 7\end{array}\right]$ We know $A=I A$ $\Rightarrow\left[\begin{array}{ccc}2 -1 4 \\ 4 0 2 \\ 3 -2 7\end{array}\right]=\left[\begin{array}{lll}1 0 0 \\ 0 1 0 \\ 0 0 1\end{array}\right] A$ $\Rightarrow\left[\begin{array}{ccc}1 -\frac{1}{2} 2 \\ 4 0 2 \\ 3 -2 7\end{array...
Read More →A shopkeeper marks his goods at 40% above the cost price but allows a discount of 5% for cash payment to his customers.
Question: A shopkeeper marks his goods at 40% above the cost price but allows a discount of 5% for cash payment to his customers. What actual profit does he make, if he receives Rs 1064 after paying the discount? Solution: Let the original cost price of the item be Rs. $x$. $\mathrm{MP}=x+\frac{40 x}{100}=1.4 x$ Discount $=M P-S P$ $\mathrm{MP}-\frac{\text { Discount } \% \times \mathrm{MP}}{100}=\mathrm{SP}$ $1.4 x-\frac{5 \times 1.4 x}{100}=1064$ $\frac{1.4 x \times 100-5 \times 1.4 x}{100}=10...
Read More →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.
Question: The radii of the circular ends of a solid frustum of a cone are $33 \mathrm{~cm}$ and $27 \mathrm{~cm}$ and its slant height is $10 \mathrm{~cm}$. Find its total surface area. [Use $\pi=3.14 .$ ] Solution: Greater radius =R= 33 cmSmaller radius =r= 27 cmSlant height =l= 10 cmSurface area of the frustum $=\pi R^{2}+\pi r^{2}+\pi l(R+r)$ $=\pi\left[R^{2}+r^{2}+l(R+r)\right]$ $=\left[33^{2}+27^{2}+10(33+27)\right] \pi$ $=[1089+729+10(60)] \pi$ $=2418 \times 3.14$ $=7592.52 \mathrm{~cm}^{2...
Read More →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.
Question: The radii of the circular ends of a solid frustum of a cone are $33 \mathrm{~cm}$ and $27 \mathrm{~cm}$ and its slant height is $10 \mathrm{~cm}$. Find its total surface area. [Use $\pi=3.14 .$ ] Solution: Greater radius =R= 33 cmSmaller radius =r= 27 cmSlant height =l= 10 cmSurface area of the frustum $=\pi R^{2}+\pi r^{2}+\pi l(R+r)$ $=\pi\left[R^{2}+r^{2}+l(R+r)\right]$ $=\left[33^{2}+27^{2}+10(33+27)\right] \pi$ $=[1089+729+10(60)] \pi$ $=2418 \times 3.14$ $=7592.52 \mathrm{~cm}^{2...
Read More →A storage tank is in the form of a cube.
Question: A storage tank is in the form of a cube. When it is full of water, the volume of water is 15.625 m3. If the present depth of water is 1.3 m, then find the volume of water already used from the tank. Solution: Let side of a cube be = x m Volume of cubical tank = 15.625 m3[given] = x3= 15.625m3 = x = 2.5 m and present depth of water in cubical tank = 1.3 m Height of water used =2.5 1.3 m = 1.2 m Now, volume of water used = 1.2 x 2.5 x 2.5= 7.5m3 = 7. 5 x 1000 = 7500 L [ 1 m3= 1000 L] Hen...
Read More →