SOLUTION: Considering \[x\text{ }+\text{ }y\text{ }=\text{ }8,\] The shaded region and the origin both are on the same side of the graph of the line and (0, 0) satisfy the constraint \[x\text{...
SOLUTION: According to the question, Considering \[3x\text{ }+\text{ }2y\text{ }=\text{ }48,\] The shaded region and the origin both are on the same side of the graph of the line and (0, 0) satisfy...
Solution: For above fraction be greater than 0, either both denominator and numerator should be greater than 0 or both should be less than 0. \[\begin{array}{*{35}{l}} \Rightarrow ~6\text{ }\text{...
\[T\text{ }=\text{ }30\text{ }+\text{ }25\left( x\text{ }\text{ }3 \right),\text{ }3\text{ }\le \text{ }x\text{ }\le \text{ }15;\] where, T = temperature and x = depth inside the earth The...
Let the length of shortest side = ‘x’ cm According to the question, The longest side of a triangle is twice the shortest side ⇒ Length of largest side = 2x Also, the third side is 2 cm longer than...
Let temperature in Celsius be C Let temperature in Fahrenheit be F According to the question, Solution should be kept between 40° C and 45°C ⇒ 40 < C < 45 Multiplying each term by 9/5, we get...
According to the question, Let x litres of 3% solution is to be added to 460 liters of the 9% of solution Then, we get, Total solution = \[\left( 460\text{ }+\text{ }x \right)\text{ }litres\] Total...
According to the question, First reading = 8.48 Second reading = 8.35 Now, let the third reading be ‘x’ Average pH should be between 8.2 and 8.5 Average \[pH~=\text{ }\left( 8.48\text{ }+\text{...
Since, Profit = Revenue – cost Requirement is, profit > 0 \[C\left( x \right)\text{ }=\text{ }26,000\text{ }+\text{ }30\text{ }x;\] where x is number of cassettes \[\begin{array}{*{35}{l}}...
\[\begin{array}{*{35}{l}} 4x\text{ }+\text{ }3\text{ }\ge \text{ }2x\text{ }+\text{ }17 \\ \Rightarrow ~4x\text{ }\text{ }-2x\text{ }\ge \text{ }17\text{ }\text{ }-3 \\ \Rightarrow ~2x\text{ }\ge...
Solution: Multiplying each term by 4, we get \[\Rightarrow ~-20\text{ }\le \text{ }2\text{ }\text{ }-3x\text{ }\le \text{ }36\] Adding -2 each term, we get \[\Rightarrow ~-22\text{ }\le \text{...
\[\left| x\text{ }\text{ }-1 \right|\le \text{ }5\] There are two cases, 1:-\[x\text{ }\text{ }-1\text{ }\le \text{ }5\] Adding 1 to LHS and RHS \[\begin{array}{*{35}{l}} \Rightarrow ~x\text{ }\le...
Solution: \[\begin{array}{*{35}{l}} \Rightarrow ~5-\text{ }\text{ }\left| x \right|\text{ }\le \text{ }0\text{ }and\text{ }\left| x \right|\text{ }\text{ }-3\text{ }>\text{ }0\text{ }or\text{...
Solution: Hence, \[\begin{array}{*{35}{l}} 1\text{ }\le \text{ }y\text{ }<\text{ }2 \\ \Rightarrow ~1\text{ }\le \text{ }\left| x-\text{ }\text{ }2 \right|\text{ }<\text{ }2 \\ \end{array}\]...
SOLUTION: Multiplying each term by \[\begin{array}{*{35}{l}} \left( x\text{ }+\text{ }1 \right) \\ \Rightarrow ~4\text{ }\le \text{ }3\left( x\text{ }+\text{ }1 \right)\text{ }\le \text{ }6 \\...
Let z = \[\left| z \right|\text{ }(cos\text{ }{{\theta }_{1}}~+\text{ }I\text{ }sin\text{ }{{\theta }_{1}})\text{ }and\text{ }w\text{ }=\text{ }\left| w \right|\text{ }(cos\text{ }{{\theta...
SOLUTION:
\[z\text{ }+\text{ }\surd 2\text{ }\left| \left( z\text{ }+\text{ }1 \right) \right|\text{ }+\text{ }i\text{ }=\text{ }0\text{ }\ldots \text{ }\left( 1 \right)\] Substituting\[z\text{ }=\text{...
\[\begin{array}{*{35}{l}} Re\text{ }({{z}^{2}})\text{ }=\text{ }0,\text{ }\left| z \right|\text{ }=\text{ }2 \\ Let\text{ }z\text{ }=\text{ }x\text{ }+\text{ }iy. \\ Then,\text{ }\left| z...
Let \[{{z}_{1}}~=\text{ }\left| {{z}_{1}} \right|\text{ }\left( cos\text{ }{{\theta }_{1}}~+\text{ }I\text{ }sin\text{ }{{\theta }_{1}} \right)\text{ }and\text{ }{{z}_{2}}~=\text{ }\left| {{z}_{2}}...
\[\begin{array}{*{35}{l}} \left| {{z}_{1}} \right|\text{ }=\text{ }\left| {{z}_{2}} \right|\text{ }=\text{ }\ldots \text{ }=\text{ }\left| {{z}_{n}} \right|\text{ }=\text{ }1 \\ \Rightarrow...
z1 and z2 are conjugate complex numbers. The negative side of the real axis \[\begin{array}{*{35}{l}} =\text{ }{{r}_{1}}~\left( cos\text{ }{{\theta }_{1}}~\text{ }-i\text{ }sin\text{ }{{\theta...
Let z1 = x + iy Therefore, the real part of z2 is zero.
According to the question, Let \[{{z}_{1}}~=\text{ }|{{z}_{1}}|\text{ }(cos\text{ }{{\theta }_{1}}~+\text{ }I\text{ }sin\text{ }{{\theta }_{1}})\text{ }and\text{ }{{z}_{2}}~=\text{...
\[Let\text{ }z\text{ }=\text{ }x\text{ }+\text{ }iy\] Now, \[\begin{array}{*{35}{l}} \Rightarrow ~{{x}^{2}}~\text{ }-1\text{ }+\text{ }{{y}^{2}}~=\text{ }0 \\ \Rightarrow ~{{x}^{2}}~+\text{...
\[\begin{array}{*{35}{l}} arg(\left| \left( z\text{ }\text{ }-2 \right)\text{ }/\text{ }\left( z-\text{ }\text{ }3 \right) \right|\text{ })=\text{ }2 \\ Substituting\text{ }z\text{ }=\text{...
Let \[z\text{ }=\text{ }x\text{ }+\text{ }iy\] Given that, \[arg\text{ }\left( z-\text{ }\text{ }1 \right)\text{ }=\text{ }arg\text{ }\left( z\text{ }+\text{ }3i \right)\] \[\Rightarrow ~arg\text{...
\[\left| z\text{ }+\text{ }1 \right|\text{ }=\text{ }z\text{ }+\text{ }2\text{ }\left( 1\text{ }+\text{ }i \right)\] Substituting z = x + iy, we get, \[\Rightarrow ~\left| x\text{ }+\text{ }iy\text{...
As per the inquiry, We have, \[\left| z \right|\text{ }=\text{ }z\text{ }+\text{ }1\text{ }+\text{ }2i\] Subbing z = x + iy, we get, \[\begin{array}{*{35}{l}} \Rightarrow \left| x\text{ }+\text{ }iy...
Let \[z\text{ }=\text{ }x\text{ }+\text{ }iy\] arg \[\left( \left( z-1 \right)/\left( z+1 \right) \right)\text{ }=\text{ }\pi /4\] \[\Rightarrow ~arg\text{ }\left( z\text{ }\text{ }1 \right)\text{...
Let z = x + iy Now, \[\Rightarrow ~{{x}^{2}}~+\text{ }x\text{ }\text{ }-2\text{ }+\text{ }{{y}^{2}}~=\text{ }4\text{ }\left( {{x}^{2}}~\text{ }-2x\text{ }+\text{ }1\text{ }+\text{ }{{y}^{2}}...
\[z\text{ }=\text{ }x\text{ }+\text{ }iy\] ⇒ z̅ = x – iy Now, we also have, z z̅ + 2 (z + z̅) + b = 0 \[\Rightarrow ~\left( x\text{ }+\text{ }iy \right)\text{ }\left( x\text{ }\text{ }iy...
SOLUTION: = -iz̅ Hence proved.
SOLUTION: a = cos θ + i sin θ
then find (a, b). Solution: = (i4)25 = 1 Hence, (a, b) = (1, 0)
FIG 1: SOLUTION: We have,
fig 1: SOLUTION: We have,
A/Q = (1 – i)n (1 + i)n = (1 – i2)n = 2n Therefore, (1 – i)n (1 – 1/i)n = 2n
As indicated by the inquiry, \[P\left( n \right)\text{ }is\text{ }n\hat{\ }^2\text{ }<\text{ }2n\text{ }for\text{ }n\ge 5\] Let \[P\left( k \right)\text{ }=\text{ }k^2\text{ }<\text{ }2k\] be...
As indicated by the inquiry, \[P\left( n \right)\text{ }=\text{ }32n\text{ }\text{ }1\] is distinct by 8. Along these lines, subbing various qualities for n, we get, \[P\left( 0 \right)\text{...
As indicated by the inquiry, \[P\left( n \right)\text{ }=\text{ }23n\text{ }\text{ }1\] is detachable by 7. In this way, subbing various qualities for n, we get, \[P\left( 0 \right)\text{ }=\text{...
As indicated by the inquiry, P(n) which is valid for all n. Let P(n) be, ⇒ P(k) is valid for all k. In this manner, P(n) is valid for all n.
As indicated by the inquiry, P(n) which is valid for all n ≥ 4 however P(1), P(2) and P(3) are false Let \[P\left( n \right)\text{ }be\text{ }2n\text{ }<\text{ }n!\] Thus, the instances of the...
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Given Area of the square = \[784\] \[c{{m}^{2}}\] Hence Side of the square = \[\sqrt{Area}\] = \[\sqrt{784}\] = \[28\] cm Given that the four circular plates are congruent, Therefore diameter of...
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solution From the given information, it is given that the four circles are placed such that each piece touches the other two pieces. Now by joining the centers of the circles by a line segment, we...
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solution Given Radius of the circle = r = \[5\] cm Given Arc length of the sector = l = \[3.5\] cm Let us consider the central angle (in radians) be \[\theta \]. As we know that Arc length = Radius...
Solution: Given that the three circles are drawn such that each of them touches the other two. Now, by joining the centers of the three circles, We get, AB = BC = CA = \[2\] (radius) = \[7\] cm...
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Solution Given AB = \[18\] cm, DC = \[32\] cm Given, Distance between AB and DC = Height = \[14\] cm We know that Area of the trapezium = (\[1/2\]) × (Sum of parallel sides) × Height =...
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Solution: Given Diameter of the circular pond = \[17.5\] m Let us consider r be the radius of the park = \[(17.5/2)\] m = \[8.75\] m Given The circular pond is surrounded by a path of width \[2\] m....
Solution: (i) A group of eleven of the world's best cricket batsmen is not a well-defined collection. Because the criteria used to determine a batsman's talent differ from person to person. As a...
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Solution: From the given information, Radius of the circle = r = \[12\] cm ∴ OA = OB = \[12\] cm \[\angle AOB={{60}^{\circ }}\] (given) As triangle OAB is an isosceles triangle, ∴ \[\angle...
Solution: From the given information, Radius of the circle = r = \[12\] cm ∴ OA = OB = \[12\] cm \[\angle AOB={{60}^{\circ }}\] (given) As triangle OAB is an isosceles triangle, ∴ \[\angle...
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Solution From the given question, We got Sides of the triangle are \[15\] m, \[16\] m and \[17\] m. Then, perimeter of the triangle = \[(15+16+17)\] m = \[48\]m Therefore, Semi-perimeter of the...
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solution From the given question, We got, Diameter of front wheels = \[{{d}_{1}}\]= \[80\] cm we got, Diameter of rear wheels = \[{{d}_{2}}\]= \[2\]m = \[200\] cm Let us consider \[{{r}_{1}}\] be...
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From the given question, We got Area of the circular playground = \[22176\] \[{{m}^{2}}\] Let us consider r as the radius of the circle. Therefore, \[\pi {{r}^{2}}=22176\]...
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Solution From the given question, \[AC=6\]cm and \[BC=8\] cm We know that a triangle in a semi-circle with hypotenuse as diameter is right angled triangle. By using Pythagoras theorem in triangle...
Solution: From the given figure, We got that the Length and breadth of the rectangular portion (AFDC) of the flower bed are \[38\] cm and \[10\] cm respectively. We know that, Area of the flower bed...
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Let us consider ABCD be a rectangular field. Given, Length of the field = \[20\] m Given, Breadth of the field = \[16\] m From the given question, A cow is tied with a rope at a point A. Let us...
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From the question Radius of wheel = r = \[35\] cm We know that one revolution of the wheel is equal to Circumference of the wheel i.e., \[2\pi r\] = \[2\times (22/7)\times 35\] = \[220\] cm But,...
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We know that Area of a sector of a circle = \[(1/2){{r}^{2}}\theta \], Here r is the radius and \[\theta \] is the angle in radians subtended by the arc at the center of the circle From the given...
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Let us take a be the side of square. From the question we got, diagonal of square and diameter of circle is \[8\] cm In right angled triangle ABC, By Using Pythagoras theorem we got,...
Given Radius of first circle = \[{{r}_{1}}\] = \[15\] cm Given Radius of second circle = \[{{r}_{2}}\] = \[18\] cm Therefore, Circumference of first circle of radius \[{{r}_{1}}\]= \[2\pi...
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The given statement is True Explanation: The distance travelled by a circular wheel of radius r m in one revolution is equal to the circumference of the circle = \[2\pi r\] So we got, Number of...
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The given statement is False Explanation: We know that, Circumference of the circle = \[2\pi d\](d is the diameter of the circle). Thus, the statement is false
The given statement is False Explanation: In major segment, area is not always greater than the are of its corresponding sector In minor segment, area is always greater than the area of its...
Solution: The given statement is False Explanation: From the fig, Let the Diameter of the circle = d Therefore, Diagonal of inner square (EFGH) = Side of the outer square (ABCD) = Diameter of circle...
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The given statement is true Explanation: Let \[r\] be the radius of circle and is equal \[a\] cm Therefore, Diameter of the circle = d = \[2\times Radius\] = \[2a\] cm From the question we got that...
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The given statement is false Explanation: Let us assume a be the side of square. From the question we got that the circle is inscribed in the square. Therefore, Diameter of circle = Side of square...
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The correct option is (B) \[14:11\] Explanation: Let us take r as the radius of the circle and a as the side of the square. From the given question, Perimeter of a circle of radius r = Perimeter of...
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The correct option is (A) \[{{r}^{2}}\] sq. units Explanation: The largest triangle which can be inscribed in a semi-circle of radius r units is Base of triangle should be equal to the diameter of...
The correction option is (B) Area of the circle > Area of the square Explanation: From the given question, Circumference of a circle of radius r = Perimeter of a square of side a Let us take r...
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The Correct option(A) \[{{R}_{1}}+{{R}_{2}}=R\] Explanation: From the given question, We got sum of the circumferences of two circles with radii \[R1\] and \[R2\] is equal to the circumference of a...
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The Correct option is (B) \[R_{1}^{2}+R_{2}^{2}={{R}^{2}}\] Explanation: From the given question, We got sum of the areas of two circles with radii \[R1\] and \[R2\] is equal to the area of a circle...
Steps of construction: Define a boundary portion \[AB\text{ }=\text{ }5\text{ }cm.\] Develop a right point \[SAB\]at point \[A.\] Draw a circular segment of span \[12\text{ }cm\]with \[B\]as its...
Steps of construction: Define a boundary fragment, \[AB\text{ }=\text{ }7\text{ }cm.\] Draw a beam, \[AX\], making an intense point down ward with \[AB.\] Imprint the focuses\[{{A}_{1}},\text{...
False Support: Allow us to attempt to build the figure as given in the inquiry. Steps of development, Define a boundary section \[BC.\] With \[B\text{ }and\text{ }C\]as focuses, draw two circular...
True Support: As per the inquiry, Ratio\[=\text{ }\surd 3\text{ }:\text{ }\left( \text{ }1/\surd 3 \right)\] On working on we get, \[\surd 3/\left( 1/\surd 3 \right)\text{ }=\text{ }\left( \surd...
![\[\mathbf{3}\text{ }\mathbf{cm}\text{ }\mathbf{and}\text{ }\mathbf{5}\text{ }\mathbf{cm}.\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-e7f0f1ccaef3619b7bf388dc532cfe76_l3.png "Rendered by QuickLaTeX.com")
Steps of construction: Draw a circle with focus \[\mathbf{O}\]and radius \[\mathbf{3}\text{ }\mathbf{cm}.\] Draw one more circle with focus \[\mathbf{O}\]and radius \[\mathbf{5}\text{...
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![\[\mathbf{BC}\text{ }=\text{ }\mathbf{5}\text{ }\mathbf{cm},\text{ }\mathbf{AB}\text{ }=\text{ }\mathbf{3}\text{ }\mathbf{cm}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-96ea4af887c92691519458a43c7e48f9_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{ABC}\text{ }=\text{ }\mathbf{60}{}^\circ ,\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-d97ae2375bc6a0985e2247bc0b2d5ac9_l3.png "Rendered by QuickLaTeX.com")
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Steps of construction: Define a boundary \[\mathbf{AB}=\mathbf{3}\text{ }\mathbf{cm}.\] Draw a beam \[\mathbf{BY}\]making an intense \[\angle \mathbf{ABY}=\mathbf{60}{}^\circ .\] With focus...
![\[\mathbf{AB}\text{ }\mathbf{and}\text{ }\mathbf{AC}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-0f6e61115419a43b9e8a287b541a7701_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{60}{}^\circ \]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-dd2bb0881db8f067898bbb2f484f22b0_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{AB}\text{ }=\text{ }\mathbf{5}\text{ }\mathbf{cm}\text{ }\mathbf{and}\text{ }\mathbf{AC}\text{ }=\text{ }\mathbf{7}\text{ }\mathbf{cm}.\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-b1793bb851dec450a0e88ba1fae151c7_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{P}\text{ }\mathbf{and}\text{ }\mathbf{Q}\text{ }\mathbf{on}\text{ }\mathbf{AB}\text{ }\mathbf{and}\text{ }\mathbf{AC},\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-09badaf99944d9952bc85cef53011334_l3.png "Rendered by QuickLaTeX.com")
![\[\mathbf{AP}\text{ }=\text{ }{\scriptscriptstyle 3\!/\!{ }_4}\text{ }\mathbf{AB}\text{ }\mathbf{and}\text{ }\mathbf{AQ}\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_4}\text{ }\mathbf{AC}.\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-7eacd8c8751e5f1f7360fe85877fd895_l3.png "Rendered by QuickLaTeX.com")
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Steps of construction: 1.Define a boundary portion \[AB\text{ }=\text{ }5\text{ }cm.\] Draw \[\angle BAZ\text{ }=\text{ }60{}^\circ .\] With focus \[A\]and sweep\[7\text{ }cm\], draw a...
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![\[\angle \mathbf{BAX}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-138b66e953a00d08386337b69e9e0974_l3.png "Rendered by QuickLaTeX.com")
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![\[{{\mathbf{A}}_{\mathbf{1}}},\text{ }{{\mathbf{A}}_{\mathbf{2}}},\text{ }{{\mathbf{A}}_{\mathbf{3}}},\text{ }\ldots \]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-922e7c66bc6a00de501ae60c8c9b78ff_l3.png "Rendered by QuickLaTeX.com")
![\[{{\mathbf{B}}_{\mathbf{1}}},\text{ }{{\mathbf{B}}_{\mathbf{2}}},\text{ }{{\mathbf{B}}_{\mathbf{3}}},\text{ }\ldots \]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f8d1704d707366f7df0e55622fa1c93e_l3.png "Rendered by QuickLaTeX.com")
![\[\left( \mathbf{A} \right)\text{ }{{\mathbf{A}}_{\mathbf{5}}}~\mathbf{and}\text{ }{{\mathbf{B}}_{\mathbf{6}}}~\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-89ab77fc48ffa6dcefdea543900ef628_l3.png "Rendered by QuickLaTeX.com")
![\[\left( \mathbf{B} \right)\text{ }{{\mathbf{A}}_{\mathbf{6}}}~\mathbf{and}\text{ }{{\mathbf{B}}_{\mathbf{5}}}~\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-5256cc8ae0ac0b01612bea1f88f4a203_l3.png "Rendered by QuickLaTeX.com")
![\[~\left( \mathbf{C} \right)\text{ }{{\mathbf{A}}_{\mathbf{4}}}~\mathbf{and}\text{ }{{\mathbf{B}}_{\mathbf{5}}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-fcf2583e596b9a81d721035d885acb09_l3.png "Rendered by QuickLaTeX.com")
![\[\left( \mathbf{D} \right)\text{ }{{\mathbf{A}}_{\mathbf{5}}}~\mathbf{and}\text{ }{{\mathbf{B}}_{\mathbf{4}}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-e7fc66a8567a19cef480b32fe735fb48_l3.png "Rendered by QuickLaTeX.com")
\[\left( \mathbf{A} \right)\text{ }{{\mathbf{A}}_{\mathbf{5}}}~\mathbf{and}\text{ }{{\mathbf{B}}_{\mathbf{6}}}\] As per the inquiry, A line portion \[AB\]in the proportion \[5:7\] Along these...
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![\[{{\mathbf{A}}_{\mathbf{1}}},\text{ }{{\mathbf{A}}_{\mathbf{2}}},\text{ }{{\mathbf{A}}_{\mathbf{3}}},\ldots .\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-e8bbd12bc9b45168d39ff2d0bcc0108e_l3.png "Rendered by QuickLaTeX.com")
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![\[\left( \mathbf{A} \right)\text{ }{{\mathbf{A}}_{\mathbf{12}}}~\left( \mathbf{B} \right)\text{ }{{\mathbf{A}}_{\mathbf{11}}}~\left( \mathbf{C} \right)\text{ }{{\mathbf{A}}_{\mathbf{10}}}~\left( \mathbf{D} \right)\text{ }{{\mathbf{A}}_{\mathbf{9}}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-bda43f24e7ebd73ad31bbcc4874f4781_l3.png "Rendered by QuickLaTeX.com")
SOLUTION:- \[\left( \mathbf{B} \right)\text{ }{{\mathbf{A}}_{\mathbf{11}}}\] As per the inquiry, A line section\[~AB\] in the proportion \[4:7\] Thus, \[A:B\text{ }=\text{ }4:7\] Presently, Draw a...
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SOLUTION:- \[\left( D \right)\text{ }12\] As indicated by the inquiry, A line fragment \[AB\]in the proportion \[5:7\] In this way, \[A:B\text{ }=\text{ }5:7\] Presently, Draw a beam \[AX\]making an...
Solution: (i) The number divisible by 9 between 100 and 200 = 108, 117, 126,...198 Let n be the number of terms that are divisible by 9 and are between 100 and 200. ${{a}_{n}}~=\text{ }a\text{...
Solution: We all know that, First term of an AP = a The common difference = d AP’s nth term of an, ${{a}_{n}}~=a+\left( n1 \right)d$ Since, $n\text{ }=\text{ }37$ (odd), Middle term will be $\left(...
Solution: We all know that, AP’s first term = a AP’s common difference = d AP’s nth term, ${{a}_{n}}~=\text{ }a\text{ }+\text{ }\left( n\text{ }\text{ }1 \right)d$ According to the question,...
Solution: We all know that, Multiple of 2 + Multiple of 5 - LCM Multiple (2, 5) = Multiples of 2 or 5 Multiple of 2 + Multiple of 5 – Multiple of LCM (10) = Multiples of 2 or 5 List of multiple of 2...
Solution: (i) It is known to us that, LCM of (2, 5) = 10 for multiples of 2 and 5. Between 1 and 500, multiples of 2 and 5 = 10, 20, 30,..., 490. As a result, We can say that 10, 20, 30,..., 490 is...
Solution: In an A.P, we know that, First term = a The common difference = d An AP’s number of terms of = n According to the question, We have, ${{S}_{5}}~+\text{ }{{S}_{7}}~=\text{ }167$ Using the...
Solution: It is known to us that, AP’s first term = a The common difference = d. According to the question given, 5th term, ${{a}_{5}}~=\text{ }19$ Using the nth term formula, ${{a}_{n}}~=\text{...
Solution: The below given condition needs to be satisfied for a, 7, b, 23, c… to be in AP, ${{a}_{5}}-{{a}_{4}}~=\text{ }{{a}_{4}}-{{a}_{3}}~=\text{ }{{a}_{3}}-{{a}_{2}}~=\text{...
Solution: (i) It is known to us that, AP’s first three terms are : a, a + d, a + 2d √2, √2+1/√2, √2+2/√2 √2, 3/√2, 4/√2
Solution: (i) It is known to us that, AP’s first three terms are : a, a + d, a + 2d ${\scriptscriptstyle 1\!/\!{ }_2},\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\text{ }+\text{ }\left( -1/6...
Solution: Given Here a1 = a ${{a}_{2}}~=\text{ }2a\text{ }+\text{ }1$ ${{a}_{3}}~=\text{ }3a\text{ }+\text{ }2$ ${{a}_{4}}~=~4a\text{ }+\text{ }3$ ${{a}_{2}}-{{a}_{1}}~=\text{ }\left( 2a\text{...
Solution: (i) Given here, ${{a}_{1~}}=\surd 3$ ${{a}_{2}}~=2\surd 3$ ${{a}_{3}}~=3\surd 3$ ${{a}_{4}}~=4\surd 3$ ${{a}_{2}}-{{a}_{1}}=2\sqrt{3}-\sqrt{3}=\sqrt{3}$ ${{a}_{3}}-{{a}_{2}}~=3\surd...
Solution: (i) Given here, ${{a}_{1~}}=0$ ${{a}_{2}}~=\text{ }{\scriptscriptstyle 1\!/\!{ }_4}$ ${{a}_{3}}~=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}$ ${{a}_{4}}~=\text{ }{\scriptscriptstyle 3\!/\!{...
Column A Column B (A1) 2, – 2, – 6, –10,… (B1) 2/3 (A2) a = –18, n = 10, an = 0 (B2) – 5 (A3) a = 0, a10 = 6 (B3) 4 (A4) a2 = 13, a4 =3 (B4) – 4 (B5) 2 (B6) 1/2 (B7) 5 Solution: (A1) AP is 2, – 2, –...
Solution: Consider two APs with the first terms as ‘a’ and ‘A’. The common differences are ‘d’ and ‘D’, respectively. Assume that n is any term. ${{a}_{n}}~=a+\left( n-1 \right)d$ ${{A}_{n}}~=\text{...
Solution: True Provided that The first term, $a=-3$ The common difference, $d=\text{ }{{a}_{2}}~-{{a}_{1}}~=-7\left( -\text{ }3 \right)=-4$ ${{a}_{30}}-{{a}_{20}}~=\text{ }a+29d-\left( a\text{...
Solution: False ${{a}_{1}}~=\text{ }-1,\text{ }{{a}_{2}}~=\text{ }-3/2,\text{ }{{a}_{3}}~=\text{ }-2$and ${{a}_{4}}~=\text{ }5/2$ ${{a}_{2}}-{{a}_{1~}}=-3/2\left( -1 \right)\text{ }=-1/{{}_{2}}$...
Solution: We have, ${{a}_{1~}}=\text{ }\surd 3$ $,{{a}_{2}}~=\text{ }\surd 12,\text{ }{{a}_{3}}~=\text{ }\surd 27$ and ${{a}_{4}}~=\text{ }\surd 48$ ${{a}_{2}}~-{{a}_{1~}}=\text{ }\surd 12-\surd...
Solution: (i) We have ${{a}_{1}}~=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}~,\text{ }{{a}_{2}}~=\text{ }1/3~$and ${{a}_{3}}~=~{\scriptscriptstyle 1\!/\!{ }_4}$ ${{a}_{2}}~-{{a}_{1}}~=-1/6$ We can...
Solution: (i) We have ${{a}_{1}}~=\text{ }1\text{ },\text{ }{{a}_{2}}~=\text{ }1,\text{ }{{a}_{3}}~=\text{ }2$ and ${{a}_{4}}~=\text{ }2$ ${{a}_{2}}~-{{a}_{1}}~=\text{ }0$...
Solution: (i) We have${{a}_{1}}~=-1\text{ },\text{ }{{a}_{2}}~=-1$, ${{a}_{3}}~=-1$and ${{a}_{4}}~=-1$ ${{a}_{2}}~-{{a}_{1}}~=\text{ }0$ ${{a}_{3}}~-{{a}_{2}}~=\text{ }0$...
Solution: Option (C) 25 is the correct answer. Explanation: Provided, the common difference of AP i.e., d = 5 Now, As it is known, an AP’s nth term is ${{a}_{n\text{ }}}=~a+\left( n-1 \right)d$...
Solution: Option (B) 10th is the correct answer. Explanation: Let the given AP’s nth term be 210. According to question, first term, $a=21$ common difference, $d=42-21=21$and ${{a}_{n}}~=210$ We...
Solution: Option (B) 33 is the correct option. Explanation: It is known that the AP’s nth term is ${{a}_{n\text{ }}}=~a\text{ }+\text{ }\left( n-1 \right)d$ In which, first term = a nth term = an...
Solution: Option (B) 137 is the correct option. Explanation: First two terms of an AP are $a=-3$and${{a}_{2\text{ }}}=~4$. It is known, nth term of an AP is ${{a}_{n\text{ }}}=~a\text{ }+\text{...
Solution: Option (C) – 2, – 4, – 6, – 8 is the correct answer. Explanation: First term, a = – 2 Second Term, d = – 2 ${{a}_{1\text{ }}}=~a=\text{ -}2$ It is known that the AP’s nth term is...
Solution: Option (B) 20 is the correct answer. Explanation: First term, a = – 5 Common difference, $d=5-\left( -5/2 \right)=5/2$ $n\text{ }=\text{ }11$ It is known that the AP’s nth term is...
Solution: Option (B) an AP with d = 4 is the correct answer. Explanation: According to the question, ${{a}_{1\text{ }}}=~\text{ }-10$ ${{a}_{2\text{ }}}=\text{ }-6$ ${{a}_{3\text{ }}}=~\text{ }-2$...
Solution: Option (B) 3.5 is the correct answer. Explanation: It is known that nth term of an AP is ${{a}_{n\text{ }=}}~a+\left( n1 \right)d$ In which, first term = a nth term = an common difference...
Solution: According to the given question, DE || AB By the basic proportionality theorem, $CD/AD\text{ }=\text{ }CE/BE$ Therefore, when a line is drawn parallel to one of the triangle's sides and...
Solution: False According to the given question, Let’s assume that, $A\text{ }=\text{ }25\text{ }cm$ $B\text{ }=\text{ }5\text{ }cm$ $C\text{ }=\text{ }24\text{ }cm$ Using the Pythagoras Theorem, We...
Solution: (i) Let (p, q) be the coordinates of a point Q. Provided, The point Q (p, q), Divide the line joining $\mathrm{B}\left(\mathrm{x}{2}, \mathrm{y}{2}\right)$ and...
Solution: According to the given question, A, B and C are the vertices of ΔABC A(x1, y1), B(x2, y2), C(x3, y3) are the coordinates of A, B and C. (i) According to the information provided, D is BC's...
Solution: According to the given question, A (6, 1), B (8, 2) and C (9, 4) are the three vertices of a parallelogram ABCD Let (x, y) be the fourth vertex of parallelogram. It is known to us that,...
Solution: Option (D) 28 is the correct answer. Explanation: It is known that nth term of an AP is ${{a}_{n\text{ }=}}~a+\left( n-1 \right)d$ In which, first term = a nth term = an common difference...
Solution: Let (x,y) be the vertices. The distance between (x,y) & (4,3) is $=\text{ }\surd ({{\left( x-4 \right)}^{2}}~+\text{ }{{\left( y-3 \right)}^{2}})$……(1) The distance between (x,y) &...
Solution: Let 1: k be the ratio in which x-axis divides the line segment joining (–4, –6) and (–1, 7). Therefore, x-coordinate is (-1 – 4k) / (k + 1) y-coordinate is (7 – 6k) / (k + 1) y coordinate...
Solution: The provided vertices are: $({{x}_{1}},\text{ }{{y}_{1}})\text{ }=\text{ }\left( -8,\text{ }4 \right)$ $({{x}_{2}},\text{ }{{y}_{2}})\text{ }=\text{ }\left( -6,\text{ }6 \right)$...
Solution: $A(2,-4), P(3,8)$ and $Q(-10, y)$ are the given points. Now according to the question given, $$ \begin{aligned} P A &=Q A \\ \sqrt{(2-3)^{2}+(-4-8)^{2}} &=\sqrt{(2+10)^{2}+(-4-y)^{2}} \\...
Solution: The points given here i.e., A(5, 1), B(–2, –3) and C(8, 2m) are collinear. Therefore the area of ∆ABC = 0 ${\scriptscriptstyle 1\!/\!{ }_2}\text{ }[{{x}_{1}}~({{y}_{2}}~\text{...
Solution: As the point P lies on the perpendicular bisector of AB, point Q is the midpoint of AB . By the formula for midpoint: $({{x}_{1}}~+\text{ }{{x}_{2}})/2\text{ }=\text{ }\left( -5+4...
Solution: Let P be the point. Now according to the given question, P is at equal distance from A (–5, 4) and B (–1, 6) Then the point P $=\text{ }(({{x}_{1}}+{{x}_{2}})/2,\text{...
Solution: The distance between the two points (x1,y1) ( x2,y2) : $d=\surd ({{x}_{2}}-{{x}_{1}}){}^\text{2}+({{y}_{2}}-{{y}_{1}}){}^\text{2}$ The distance between A (–3, –14) and B (a, –5): $=\surd...
Solution: A (2, –2), B (7, 3), C (11, –1) and D (6, –6) are the given points. Now using the distance formula, $d~=\text{ }\surd \text{ }({{({{x}_{2}}~\text{ }{{x}_{1}})}^{2}}~+\text{...
Solution: (x, 0) = Let coordinates of the point (given that the point lies on x axis) ${{x}_{1}}=7.\text{ }{{y}_{1}}=-4$ ${{x}_{2}}=x.\text{ }{{y}_{2}}=0$ Distance $=\surd...
Solution: A (–5, 6), B (–4, –2) and C (7, 5) are the given points. Now, using the distance formula, $d~=\text{ }\surd \text{ }({{({{x}_{2}}~\text{ }{{x}_{1}})}^{2}}~+\text{ }{{({{y}_{2}}~\text{...
Solution: The statement given in the question is false. Justification: A (4, 3), B (6, 4), C (5, –6) and D (–3, 5) are the points given. We need to find the distance between A and B...
Solution: The statement given in the question is true. Justification: Coordinates of A $=\text{ }({{x}_{1}},\text{ }{{y}_{1}})\text{ }=\text{ }\left( 3,\text{ }1 \right)$ Coordinates of B $=\text{...
Solution: The statement given in the question is false. Justification: We know that the points on the perpendicular bisector of the line segment joining two points are equidistant from the two...
Solution: The statement given in the question is false. Justification: If the area of a triangle formed by its points equals 0, then the points are collinear. Provided, ${{x}_{1}}~=\text{ }0,\text{...
Solution: The statement given in the question is true. Justification: Equation of the line conatining the points A and B using the two-point form is, $\frac{y-6}{x+4}=\frac{-6-6}{-4+4}$...
Solution: The statement given in the question is true. Justification: Distance formula, $d~=\text{ }\surd \text{ }({{({{x}_{2}}~\text{ }{{x}_{1}})}^{2}}~+\text{ }{{({{y}_{2}}~\text{...
Arrangement (i): Leave the two numbers alone x and y individually, to such an extent that y > x. As indicated by the inquiry, \[y\text{ }=\text{ }3x\text{ }\ldots \text{ }\ldots \text{ }\ldots...
Solution: Option (C) 10 is the correct answer. The formula for distance: ${{d}^{2}}~=\text{ }{{({{x}_{2}}~\text{ }{{x}_{1}})}^{2}}~+\text{ }{{({{y}_{2}}~\text{ }{{y}_{1}})}^{2}}$ According to the...
Solution: Option (C) 8 is the correct answer. The vertices of the triangle are, $A\text{ }({{x}_{1}},\text{ }{{y}_{1}})=\left( 3,\text{ }0 \right)$ $B\text{ }({{x}_{2}},\text{ }{{y}_{2}})=\left(...
Solution: Option (B) 12 is the correct answer. (0, 4), (0, 0) and (3, 0) are the vertices of a triangle. The perimeter of triangle AOB = Sum of the length of all its sides: = distance between...
Solution: Option (C) √34 is the correct answer. The three vertices are: $A\text{ }=\text{ }\left( 0,\text{ }3 \right)$, $O\text{ }=\text{ }\left( 0,\text{ }0 \right)$ , $B\text{ }=\text{ }\left(...
Solution: Option (B) 5√ 2 is the correct answer. Distance formula: ${{d}^{2}}~=\text{ }{{({{x}_{2}}~\text{ }{{x}_{1}})}^{2}}~+\text{ }{{({{y}_{2}}~\text{ }{{y}_{1}})}^{2}}$ According to the given...
Solution: Option (B) 8 is the correct answer. The formula for distance : ${{d}^{2}}~=\text{ }{{({{x}_{2}}~\text{ }{{x}_{1}})}^{2}}~+\text{ }{{({{y}_{2}}~\text{ }{{y}_{1}})}^{2}}$ According to the...
Solution: Option (B) 3 is the correct answer. We all know that, On the Cartesian plane (x, y) is a point in first quadrant. Then, Perpendicular distance from Y–axis = x, and Perpendicular distance...
Solution: Let MN the flag pole = 18 m and its shadow LM = 9.6 m. The distance of the top of the pole be LN, N from the far end, L of the shadow. By Pythagoras theorem in right angled ∆LMN,...
Solution: According to the given question, AC⊥CB, $AC\text{ }=\text{ }2x\text{ }km$, $CB=2\left( x+7 \right)km$ and $AB=26\text{ }km$ As a result, we get triangle ACB right angled at C. Now, using...
Solution: Let 5 m be the length of the ladder AC. Let 4m be the height of the wall on which ladder is placed is BC From right angled triangle EBD, Now using the Pythagoras Theorem,...
Solution: According to the given question, The given figure PQRS is a parallelogram, As a result, PQ || SR and PS || QR. It is also given that, AB || PS. To prove: OC || SR From triangles OPS and...
Solution: Let's assume a triangle ABC in which a line DE parallel to BC intersects AB at D and AC at E. To prove: The two sides are divided by DE in the same ratio. $AD/DB\text{ }=\text{ }AE/EC$...
Solution: According to the given question, ∆ABC ∼ ∆EDF By the property of similar triangle, we all know that, corresponding sides of ∆ABC and ∆EDF are in the same ratio. $AB/ED\text{ }=\text{...
Solution: According to the given question, $\angle A\text{ }=~\angle C$, $AB\text{ }=\text{ }6\text{ }cm$, $BP\text{ }=\text{ }15\text{ }cm$, $AP\text{ }=\text{ }12\text{ }cm$ $CP\text{ }=\text{...
Solution: According to the given question, $AB\text{ }=\text{ }4\text{ }cm$, $DE\text{ }=\text{ }6\text{ }cm$ $EF\text{ }=\text{ }9\text{ }cm$ $FD\text{ }=\text{ }12\text{ }cm$ Also, ∆ABC ∼ ∆DEF We...
Solution: Let an equilateral triangle of side 8 cm be ABC. $AB\text{ }=\text{ }BC\text{ }=\text{ }CA\text{ }=\text{ }8\text{ }cm$. (all sides of an equilateral triangle is equal) Construct an...
Solution: According to the given question, At point O, AC and PQ intersect each other and AB||DC. From triangles AOP and COQ, $\angle AOP\text{ }=~\angle COQ$[As they are vertically opposite angles]...
Solution: According to the given question, The given figure, PQRS is a trapezium in which PQ || RS and PQ = 3RS $PQ/RS\text{ }=\text{ }3/1\text{ }=\text{ }3$…(i) In triangles POQ and ROS, $\angle...
Solution: According to the given question, $\text{ }\Delta NSQ~\cong ~\Delta MTR$ $\angle 1\text{ }=~\angle 2$ Since, $\Delta NSQ\text{ }=\text{ }\Delta MTR$ As a result, $SQ\text{ }=\text{...
Solution: According to the given question, In triangle PQR, $P{{R}^{2}}~=\text{ }Q{{R}^{2}}$ and QM⊥PR Using the Pythagoras theorem, we obtain, $P{{R}^{2}}~=\text{ }P{{Q}^{2}}~+\text{ }Q{{R}^{2}}$...
Number of absolute results = 36 (I) When result of the numbers on the highest point of the dice = 6. The potential results = (1, 6), (2,3), (3, 2), (6, 1). Subsequently, number of conceivable ways =...
As indicated by the inquiry, Two dice are tossed all the while. In this way, that number of potential results = 36 (I) Sum of the numbers showing up on the dice is 7. Thus, the potential...
Two dice are tossed simultaneously. Along these lines, absolute number of potential results = 36 (I) Same number on both dice. Potential results = (1,1), (2,2), (3, 3), (4, 4), (5, 5), (6,...
In the given information, the most noteworthy recurrence is 26, which lies in the stretch 201 – 202 Here, l = 201,fm = 26,f1 = 12,f2 = 20 and (class width) h = 1 Subsequently, the modular weight =...
As per the information given, The most elevated recurrence = 41, 41 lies in the stretch 10000 – 15000. Here, l = 10000, fm = 41,f1 = 26,f2 = 16 and h = 5000 \[=\text{ }10000\text{ }+\text{...
First we develop the combined recurrence table Speed ( in km/h) Number of players Cumulative recurrence 85 – 100 11 11 100 – 115 ...
Week by week Income Number of families (fi) Cumulative recurrence (cf) 0-1000 250 250 1000-2000 190 250 + 190 = 400 2000-3000 100 440 + 100 = 540...
The recurrence circulation table for given information. Marks Number of understudies 0 – 20 12 20 – 40 22 – 17 = 5 40 – 60 29 – 22 = 7 60 – 80 37 – 29 = 8 80 – 100 50 – 37 =...
(I) Less than type Age (in year) Number of patients Under 10 0 Under 20 60 + 0 = 60 Under 30 60 + 42 = 102 Under 40 102 + 55 = 157 Under...
Tallness (in cm) Frequency Cumulative recurrence given Cumulative recurrence 150 – 155 12 a 12 155 – 160 b ...
The recurrence dissemination table for the given information is: Class Interval Number of understudies 0-10 34 – 32 = 2 10-20 32 – 30 = 2 20-30 30 – 27 = 3 30-40 27 – 23 = 4...
The recurrence circulation table for the given information is: Class Interval Number of understudies 0-10 10 10-20 50 – 10 = 40 20-30 130 – 50 = 80 30-40 270 – 130 = 140...
Weight (in kg) Cumulative recurrence Under 45 4 Under 50 4 + 4 = 8 Under 55 8 + 13 = 21 Under 60 21 + 5 = 26 Under 65 26 + 6 = 32 Under...
Mileage (km L-1) Class – Marks (xi) Number of vehicles (fi) fixi 10 – 12 11 7 77 12 – 14 13 12 156 14 – 16 15 ...
Weight (in kg) Number of Wrestlers (fi) Class Marks (xi) Deviation (di = xi – a) fidi 100 – 110 4 105 –20 –80 110 – 120 14 ...
Class Interval Class Marks (xi) Frequency (fi) Deviation (di = xi – a) fidi 100 – 104 102 15 –8 –120 104 – 108 106 ...
C.I xi di = (xi – a) Fi fidi 1 – 200 100.5 –200 14 –2800 201 – 400 300.5 0 15 0 401 – 600 ...
Class Marks Mid – Value (xi) Number of days (fi) fixi 15.5 – 18.5 17 1 17 18.5 – 21.5 20 3 60 21.5 – 24.5 ...
The given information isn't constant. Thus, we deduct 0.5 from as far as possible and add 0.5 in the maximum furthest reaches of each class. Class Class Marks (xi) Frequency (fi) fixi 3.5...
We first, discover the class mark xi of each class and afterward continue as follows Class Class Marks (xi) Frequency (fi) fixi 10-20 15 2 30 20-30 ...
We first, discover the class mark xi of each class and afterward continue as follows. Class Class Marks (xi) Frequency (fi) fixi 1-3 2 9 18 3-5 ...
No, the assertion isn't right. It isn't required that expected mean ought to be the mid – mark of the class span. a can be considered as any worth which is not difficult to work on it.
To ascertain the middle of an assembled information, the recipe utilized depends with the understanding that the perceptions in the classes are consistently disseminated or similarly divided....
No, the upsides of mean, mode and middle of gathered information can be equivalent to well, it relies upon the sort of information given.
The middle class and modular class of assembled information isn't generally unique, it relies upon the information given.
No it isn't right that in a family having three youngsters, there might be no young lady, one young lady, two young ladies or three young ladies, the likelihood of each is ¼. . Let young men be B...
All out no. of result = 360 \[p\left( 1 \right)=\text{ }90/360\text{ }=1/4\] \[p\left( 2 \right)\text{ }=\text{ }90/360\text{ }=\text{ }1/4\] \[p\left( 3 \right)\text{ }=\text{ }180/360\text{...
Apoorv toss two dice on the double. Thus, the all out number of results = 36 Number of results for getting item 36 = 1(6×6) ∴ Probability for Apoorv = 1/36 Peehu tosses one kick the bucket, Thus,...
(D)17/16 Clarification: Likelihood of an occasion consistently lies somewhere in the range of 0 and 1. Likelihood of any occasion can't be mutiple or negative as (17/16) > 1 Consequently, choice...
(D) 0 Clarification: The occasion which can't happen is supposed to be incomprehensible occasion. The likelihood of incomprehensible occasion = zero. Thus, choice (D) is right
(A) 3 Clarification: Imprints Obtained Number of students Cumulative Frequency 0-10 (63 – 58) = 5 5 10-20 (58 – 55) = 3 3 20-30 (55 – 51) =...
(C) 82 Clarification: The quantity of competitors who finished the race in under 14.6 second= 2 + 4 + 5 + 71 = 82 Subsequently, choice (C) is right
(C) 20 Clarification: Class Frequency Cumulative Frequency 65-85 4 4 85-105 5 9 105-125 13 22 ...
(C) 30-40 Clarification: Marks Number of students Cumulative Frequency Underneath 10 3=3 3 10-20 (12 – 3) = 9 12 20-30 (27 – 12) = 15 27...
(B) 17.5 Clarification: As per the inquiry, Classes are not constant, henceforth, we make the information nonstop by taking away 0.5 from lower limit and adding 0.5 to furthest reaches of each...
(B) 25 Clarification: Class Frequency Cumulative Frequency 0-5 10 10 5-10 15 25 10-15 12 37 15-20 ...
(B) Median Clarification: Since, the convergence point of not as much as ogive and more than ogive gives the middle on the abscissa, the abscissa of the mark of convergence of the not as much...
(C) (xi – a)/h Clarification: As indicated by the inquiry, \[x\text{ }=\text{ }a\text{ }+\text{ }h\left( fiui/fi \right),\] Above equation is a stage deviation recipe. In the above recipe, xi is...
(A) 0 Clarification: Mean (x) = Sum of the relative multitude of perceptions/Number of perceptions \[x\text{ }=\text{ }\left( f1x1\text{ }+\text{ }f2x2\text{ }+\text{ }\ldots \text{ }..+\text{ }fnxn...
(B) Centered at the class characteristics of the classes Clarification: In figuring the mean of assembled information, the frequencies are focused at the class signs of the classes. Subsequently,...
(C) Mid marks of the classes Clarification: We know, \[di\text{ }=\text{ }xi\text{ }\text{ }a\] Where, xi are information and 'a' is the expected to be mean In this way, di are the deviations from...
As indicated by the inquiry, Normal dislodging by an individual = 0.04 m3 Normal uprooting by 500 people = 500 × 0.04 = 20 m3 Subsequently, the volume of water brought up in lake = 20 m3 It is...
Thinking about cuboidal square Length, l = 4 m Expansiveness, b = 2.6 m Stature, h = 1 m We realize that, Volume of tank = lbh Volume of cuboid = 4.4(2.6)(1) = 11.44 m3 We realize that, The volume...
Let the time taken by line to fill lake = t hours Water streams 15 km in 60 minutes, in this way, it will stream 15t meters in t hours. We realize that, Volume of cuboidal lake up to tallness 21 cm...