CBSE 11th Standard Maths Subject HOT Questions 4 Mark Questions 2021 Part - I
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CBSE 11th Standard Maths Subject HOT Questions 4 Mark Questions 2021 Part - I
11th Standard CBSE
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Reg.No. :
Mathematics
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Out of 100 students, 15 passed in English, 12 passed in Mathematics, 8 in Science, 4 in English and Science, 4 in all the three.Find how many students passed in English and Mathematics but not in Science?
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Two trees A and B are on the same side of a river. From a point C in the river the distance of trees A and B are 250m and 300m, respectively of the angle C is 450, find the distance between the trees.[use \(\sqrt { 2 } =1.44\)]
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Find the value of x and y, if \(\frac { \left( 1+i \right) x-2i }{ 3+i } +\frac { \left( 2-3i \right) y+i }{ 3-i } =i\)
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The area of a triangle is 5 sq units and two of its vertices are (2,1) and (3,-2).If third vertex is (x,y), where y=x+3, then find the coordinates of the third vertex.
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A box contains 4 red, 5 white and 6 black balls.A person draws 4 balls from the box at random.Find the probability of selecting atleast one ball of each colour.
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CBSE 11th Standard Maths Subject HOT Questions 4 Mark Questions 2021 Part - I Answer Keys
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n(E)=15,n(M)=12,n(s)=8,\(n(E\cap M)=6\)
\(n(M\cap S)=7,n(E\cap S)=4,n(E\cap M\cap S)=4\)
(i)\(n(E\cap M\cap \overline { S } )=n(E\cap M)-n(E\cap M\cap S)\)
=6-4=2 -
According to the given information,
we have the following
In \(\Delta \)ABC, by cosine rule,
we have
\({ AB }^{ 2 }={ AC }^{ 2 }+{ BC }^{ 2 }-2AC.BC\quad cos\frac { \pi }{ 4 } \)
\(\therefore \quad AB=\sqrt { { (250) }^{ 2 }+{ (300) }^{ 2 }-2\times 250\times 300\times \frac { 1 }{ \sqrt { 2 } } } \)
\(=\sqrt { 62500+90000-75000\sqrt { 2 } } \)
\(=\sqrt { 152500-75000\times 1.44 } \)
\(=\sqrt { 152500-108000 } =\sqrt { 44500 } =210.95m\) -
Given \(\frac { \left( 1+i \right) x-2i }{ 3+i } +\frac { \left( 2-3i \right) y+i }{ 3-i } =i\)
\(\Rightarrow \frac { x+\left( x-2 \right) i }{ 3+i } +\frac { 2y+\left( 1-3y \right) i }{ 3-i } =i\)
\(\Rightarrow \frac { \left[ x+\left( x-2 \right) i \right] \left( 3-i \right) +\left[ 2y+\left( 1-3y \right) i \right] \left( 3+i \right) }{ \left( 3+i \right) \left( 3-i \right) } =i\)
\(\Rightarrow \left( 4x+9y-3 \right) +i\left( 2x-7y-3 \right) =10i\)
\(\Rightarrow 4x+9y-3=0\quad and\quad 2x-7y-3=10\)
\(Ans.\ x=3\ and \ y=-1\) -
\(\frac { 1 }{ 2 } \left| x(1+2)+2(-2-y)+3(y-1) \right| =5 \Rightarrow \left| 3x+y-7 \right| =10\)
\(\Rightarrow \) 3x + y - 7 = 10 or 3x + y - 7 = -10 \(\Rightarrow \) 3x + 7 = 17
or 3x + 7 = -3
Case I When 3x + y = 17
It is given that y = x + 3 \(\Rightarrow \) x - y = -3
On solving Eqs.(i) and (ii) we ,get
\(x=\frac { 7 }{ 2 } and\quad y=\frac { 13 }{ 2 } \)
Case II when 3x + y = -3
On solving Eqs.(ii) and (iii), we get
\(x=\frac { -3 }{ 2 } and\quad y=\frac { 3 }{ 2 } \)
\(\left( \frac { 7 }{ 2 } ,\frac { 13 }{ 2 } \right) or\left( \frac { -3 }{ 2 } ,\frac { 3 }{ 2 } \right) \) -
Let A = event that 1 red, 2 white and 1 black balls are drawn,
B= event that 1 red, 2 white and 1 black balls are drawn and
C= event that 2 red, 1 white and 1 black balls are drawn.
Here A, B and C are munually exclusive events.
Hence, required probability
= P(A\(\cup \)B\(\cup \)C) = P(A) + P(B) + P(C)
Ans. \(\frac { 48 }{ 91 } \)
