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Answered on 02 Mar CBSE/Class 9/Mathematics

Salman Ahmad Siddiqui

Maths Tutor

It is not a big job to score 100 out of 100 in mathematics. A Good dedicated teacher and a real dedication can help you score 100%.
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Answered on 23 Feb CBSE/Class 9/Mathematics

Salman Ahmad Siddiqui

Maths Tutor

Yes, of course 0 is a rational number. All natural numbers whole numbers and integers are also rational numbers. 0 can be written as 0/1 , 0/2 etc Hence it is natural number
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Lesson Posted on 22 Feb CBSE/Class 9/Mathematics

MAXIMA AND MINIMA

Sujoy D.

1. Strong concept building classes. 2. Weekly tests in weak areas for students to improve their confidence. 3....

QUESTION In a college, where every student follows at least one of the three activities- dance, music, or sports- 65% follow dance, 86% follow music, and 57% follow sports. What can be the maximum and minimum percentage of students who follow A) All three activities B) Exactly two activities ... read more

QUESTION

In a college, where every student follows at least one of the three activities- dance, music, or sports- 65% follow dance, 86% follow music, and 57% follow sports. What can be the maximum and minimum percentage of students who follow

A) All three activities
B) Exactly two activities

 

ANSWER

Here the sum of the three percentage figures is 208%, however since every students is involve in at least one of the three activities, the surplus count is 108%, accounted for by the people engaged in 2 or 3 activities.

A) Exactly 3 activities
Max -> 108% extra is accounted for by 54% ppl, taking 2 extra activities(Total of 3 activities)
Min -> I allocate a 2nd activity to each of the 100%, but I am still left with 8%, which must be born by 8% people having 3 activities

For B)
Max -> 92%, since in worst case scenario 8% ppl must be into all 3 activities
Min -> 0%, In case of 54% ppl being into all 3 activities, as shown in A (Part 1), we end up with 0% ppl being into exactly 2 activities

Max exactly 2 sets cannot be 100%, as total is 208%, we must have a minimum of 8% students who like all three activities, hence only a max of 92% remaining folks can like exactly 2 activities

If you assign all 100% people 2 activities, your total will be 200 and not 208%... Also dint we just conclude that 8% ppl minimum are into 3 activities, So at least these 8% ppl need to be excluded right?

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Answered on 23 Feb CBSE/Class 9/Mathematics Tuition/Class IX-X Tuition

Salman Ahmad Siddiqui

Maths Tutor

2x + 3y=7 Put 2 in place of x and 1 in place of y. 2*2+ 3*1=4+3=7
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Answered on 23 Feb CBSE/Class 9/Mathematics Tuition/Class IX-X Tuition

Salman Ahmad Siddiqui

Maths Tutor

N¹ *N² = LCM * HCF 161 *345 =23 * N² N² =2415
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Answered on 23 Feb CBSE/Class 9/Mathematics Tuition/Class IX-X Tuition

Salman Ahmad Siddiqui

Maths Tutor

2x 3y=8 X=1 and y=8/6 X=8/6 and y=1 X=-1 and y=(-8)/6 X=(-8)/6 and y=-1
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Answered on 22 Feb CBSE/Class 9/Mathematics

Salman Ahmad Siddiqui

Maths Tutor

Sin(360)= sin(0)=0 Cos(360)=cos(0)=1 Formula: Sin((360*x)+y)=sin(y) Cos((360*x)+y)=cos(y) Where x is any integer and y is any real number. Sin(360) +cos (360) =1
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Answered on 22 Feb CBSE/Class 9/Mathematics

Salman Ahmad Siddiqui

Maths Tutor

Volume of hemisphere of radius r=(2πr³)/3 =Half of the volume of full sphere ((4πr³)/3) = 2π/3 times of volume of cube of side r.
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Answered on 24 Feb CBSE/Class 9/Mathematics Tuition/Class IX-X Tuition

In a trapezium ABCD where AB is parallel to CD, E is the mid-point of BC, prove that ?AED = 1/2 trapezium ABCD.

D Mukherjee

Teacher

Draw a line EF joining E, the mid-point of BC and F, the mid-point of AD. Now EF is parallel to AB Area of Triangle AEF = 1/2 (Area of Trapezium ABEF) Area of Triangle DEF = 1/2 (Area of Trapezuim DCEF) Adding we get Area of Triangle AED = 1/2 (Artea of Trapezium ABCD)
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Answered on 27 Feb CBSE/Class 9/Mathematics Tuition/Class IX-X Tuition

Samar Singh

Mathematics Professor

Let?ABC be required triangle AB=AC=5cm BC=8 cm Let AD be the altitude We know that perpendicular drawn from vertex bisect the base. Therefore BD=4 cm Using Pythagoras theorem we get AD= 3 cm Area of ?ABC=1/2 × base×height =1/2×8×3=12 cm2 read more

Let?ABC be required triangle

AB=AC=5cm

BC=8 cm

Let AD be the altitude

We know that perpendicular drawn from vertex bisect the base.

Therefore BD=4 cm

Using Pythagoras theorem we get AD= 3 cm

Area of ?ABC=1/2 × base×height

=1/2×8×3=12 cm2

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