The definition of temperature says that it is . . .

Case Study Based Question on Linear Equation in Two Variables
The definition of temperature says that it is a measure of the hotness and coldness of a body. The two main units we often use to measure temperature are degree Celsius and degree Fahrenheit. 

The linear equation that converts Fahrenheit (F) to Celsius (C) is given by

°C×1.8=°F-32

Based on the above information, answer the following questions :

(i) Convert the normal body temperature in degree Celsius.

(ii) If the room temperature on a day is 35°C, then what will be its value in Fahrenheit scale ?

(iii) (A) Convert the freezing point and boiling point of water in Fahrenheit scale.

OR

(B) What is the numerical value of the temperature which is same in both the scales ?

Doubt by Parth 

Solution : 

(i) Normal body temperature in Fahrenheit (F) = 98.6 °F

°C×1.8=°F-32
=°C×1.8=98.6-32
=°C×1.8=66.6
=°C=66.6/1.8
°C=666/18
°C=222/6
°C=37

Normal body temperature in degree Celsius is 37°C

(ii) Room Temperature = 35°C

°C×1.8=°F-32
35×1.8=°F-32
63=°F-32
63+32=°F
95=°F
°F=95

Hence, room temperature in Fahrenheit scale will be 95°F.

(iii) (A)
Freezing Point of water = 0°C
Boiling Point of water = 100 °C
°C×1.8=°F-32
°F-32=°C×1.8
°F=[°C×1.8]+32

°F=[0×1.8]+32
°F=0+32
°F=32

°F=[100×1.8]+32
°F=180+32
°F=212

Hence, freezing point and boiling point is 32°F and 212°F respectively.

OR

(B) Let °C=°F=a
°C×1.8=°F-32
a×1.8=a-32
1.8a-a=-32
0.8a=-32
a=-32/0.8
a=-320/8
a=-40

Hence, the numerical value of the temperature which is same in both the scale is -40°.

Case Study Based Question on Statistics

Case Study Based Question on Statistics
Hydroponics is a new farming approach to lay the foundation for future agriculture careers. Class IX students explored hydroponics and discovered its benefits by growing 35 plants without soil (non soil material) in short span of time. The following table shows the record of the lengths of the 35 plants grown in plastic containers following proper guided techniques:

Length of Leaves (in cm)	9.3-9.7	9.8-10.2	10.3-10.7	10.8-11.2
Number of Plants	2	5	12	16

(a) Convert the given frequency distribution table into continuous grouped frequency distribution table. 

(b)How many plants grew in leaf length more than 9 cm.
OR 
How many plants grew in leaf length less than 12 cm?

(c) What is the mid point of the class interval 9.8-10.2

(d) In which field is statistics widely used to analyse and interpret data?

Doubt by Jugal 

Solution : 

(a) 
Here 
9.8-9.7
=0.1
Dividing by 2
0.1/2
=0.05

Now subtracting 0.05 from lower limit and adding 0.05 to upper limit. The continous grouped frequency distribution table is : 

Length of Leaves (in cm)	9.25-9.75	9.75-10.25	10.25-10.75	10.75-11.25
Number of Plants	2	5	12	16

(b) Number of plants which grew in leaf length are more than 9 cm 
= 2+5+12+16
= 35

OR

Number of plants which grew in leaf length are less than 12 cm 
= 2+5+12+16
= 35

(c) Mid Point of 9.8-10.2
Mid Point = (Lower Limit + Upper Limit)/2
Mid Point = (9.8+10.2)/2
Mid Point = 20/2
Mid Point = 10

(d) Business and economics is one of the most common fields where statistics is used, especially for decision-making and forecasting.

Case Study based Question on Coordinate Geometry

Case Study based Question on Coordinate Geometry 

To keep herself fit, Sumati used to walk 5 km daily. On one particular day, Sumati participated in a marathon. The organisers used a coordinate plane to mark the course of the marathon. The path for the marathon is as follows: 

The starting point is A. At B, there is a water stall to keep the participants hydrated. There is a juice stall at point C to keep them energetic. At D, there is a rest point for those who want to take rest in between. E is the final destination point.

Based on the above situation, answer the following questions .

(i) What are the coordinates of the point where juice stall is installed? (1)

(ii) Write the quadrant in which destination point lie. (1)

(iii) (A) Find the distance which each participant has to cover to reach the resting point.

OR

(iv) (B) Find the distance which each participant has to cover to reach the destination point from water stall.

Solution : 

A (Starting point) = (2,1)

B (Water Stall) = (2,3)

C (Juice Stall) = (0,3)

D (Resting point) = (0, 5) 

E (Destination point) = (-3, 5) 

(i) Juice stall is at point C and its coordinates are (0,3) 

(ii) The coordinate of destination point is (-3, 5) and it lies in II Quadrant.

(iii) Distance which each participant has to cover to reach the resting point : 

AB+BC+CD

=(3-1)+(2-0)+(5-3)

=2+2+2

= 6 km

OR 

(iv) Distance covered by each participant to reach the destination point from water stall

= BC+CD+DE

=(2-0)+(5-3)+[0-(-3)]

=2+2+3

= 7 km

Truss bridges are formed with a structure of connected elements . . .

Case Study Based Question on Triangles

Truss bridges are formed with a structure of connected elements that form triangular structures to make up the bridge. Trusses are the triangles that connect to the top and bottom cord and two end posts. You can see that there are some triangular shapes as shown in the picture given below and these are represented as ΔABC, ΔACAD and ΔABE.

Based on the above situation, answer the following questions :

(i) If AB=CD and AD=BC, by which congruency rule ΔABC≅ΔCDA? (1)

(ii) If BE=AC and ∠EBA=∠BAC, by which congruency rule ΔEBA≅ΔCAB? (1)

(iii) (A) If AM⊥BC and AB=AC, prove that AM bisects BC. (2) 

OR

(B) If A is the mid point of DE, BE||CA and AB||DC show that EB=AC. (2)

Doubt by Parth 

Solution : 

(i) AB=CD (Given)
BC=AD (Given)
AC=CA (Common)
Hence ΔABC≅ΔCDA (by SSS) 

(ii) BE=AC (Given)

∠EBA=∠BAC
AB=BA (common)

Hence, ΔEBA≅ΔCAB (by SAS)

(iii) 

Given : AM⊥BC
AB=AC

To prove : AM bisects BC
Proof : 
In ΔAMB and ΔAMC
∠AMB=∠AMC (Each 90°)
AB=AC (Given)
AM=AM (Common)

ΔAMB≅ΔAMC (By RHS)
BM=CM (By CPCT)

Hence, AM bisects BC

OR 

Given : EA=DA
BE||CA
AB||DC

To Prove : EB=AC

Proof : 
In ΔEBA and ΔACD
∠AEB=∠DAC (Corresponding angles)
EA=AD (Given)
∠EAB=∠ADC (Corresponding angles)
ΔEBA≅ΔACD (By ASA)
EB=AC (By CPCT)

Hence Proved ∎

Find the value of x : (4)2x-1-(16)x-1=384 . . .

Question : Find the value of x : (4)^2x-1-(16)^x-1=384  

Doubt by Balvinder

Solution : 

(4)^2x-1 - (16)^x-1 = 384

(4)^2x-1 - (4²)^x-1 = 384
(4)^2x-1 - (4)^2(x-1) = 384

(4)^2x-1 - (4)^2x-2 = 384

(4)^2x-1 - (4)^2x-1-1 = 384
(4)^2x-1 - (4)^2x-1.4^-1 = 384

(4)^2x-1 [1-1/4] = 384
(4)^2x-1 [(4-1)/4] = 384

(4)^2x-1 [3/4] = 384
(4)^2x-1= (384×4)/3
(4)^2x-1= 128×4
(4)^2x-1= 512
(2²)^2x-1=2⁹
2^2(2x-1)=2⁹
2^4x-2=2⁹

equating the power of 2 both sides
4x-2=9
4x=9+2
4x=11
x=11/4

If xyz=1 then simplify (1+x+y-1)-1 . . .

Question : If xyz=1 then simplify (1+x+y^-1)^-1×(1+y+z^-1)^-1×(1+z+x^-1)^-1

Doubt by Balvinder 

Solution : 
xyz=1
xy=1/z
1/z=xy— (1) 

Also 
z=1/xy — (2) 

Now, 

(1+x+y^-1)^-1
=[1+x+(1/y)]^-1
=[(y+xy+1)/y]^-1
=[y/(y+xy+1)]
=[y/(1+y+xy)] — (3)

(1+y+z^-1)^-1
=[1+y+(1/z)]^-1
=[1+y+xy]^-1 [Using equation (1)]
=[1/(1+y+xy)] — (4) 

(1+z+x^-1)^-1
=[1+z+(1/x)]^-1
=[1+(1/xy)+(1/x)]^-1 [Using equation (2)]
=[(xy+1+y)/xy]^-1
=[xy/(xy+1+y)]
=[xy/(1+y+xy)] — (5)

Adding equation (3), (4) and (5) 

(1+x+y^-1)^-1×(1+y+z^-1)^-1×(1+z+x^-1)^-1
= [y/(1+y+xy)] + [1/(1+y+xy)] + [xy/(1+y+xy)] 
=[(y+1+xy)/(1+y+xy)]
=[(1+y+xy)/(1+y+xy)]
= 1 

Hence
(1+x+y^-1)^-1×(1+y+z^-1)^-1×(1+z+x^-1)^-1 = 1

In the figure, we have BX=1/2AB, BY=1/2BC . . .

Question : In the figure, we have BX=½AB, BY=½BC and AB=BC show that BX=BY. Also write the axiom used. 

Doubt by Purvi 

Solution : 

Given : 
BX=½AB
BY=½BC
AB=BC

To Prove : BX=BY

Proof : 

BX=½AB (Given)
2BX=AB
AB=2BX — (1) 

BY=½BC (Given)

2BY=BC
BC=2BY — (2)

AB=BC (Given)
2BX=2BY
BX=BY
[Things which are double of the same things are equal to one another]

Hence Proved. 

[CSBQ] If we look closely at computer screen, we see that the screen is made . . .

Question : If we look closely at computer screen, we see that the screen is made up of millions of tiny squares called pixels. Number of pixels together form a digital image on screen every clickable point represents x-co-ordinates as number of pixels along horizontal axis i.e., x-axis and y co-ordinate as number of pixels along vertical axis i.e. y-axis.

Find

i) What is the perpendicular distance of A (165, 250) from x axis. [1 Marks]

ii) In which quadrant the abscissa of a point is negative. [1 Marks]

iii) What is the sum of abscissa of A and B. [2 Marks] 

Doubt Question by Purvi 

Solution : 
(i) 250 units.
(ii) II and III Quadrant 

(iii) 
A(165, 250) 

B(300, 250)

Sum of abscissa of A and B
=165+300
= 465

Along a path, 100 conical pillars are constructed. Each . . .

 Question : Along a path, 100 conical pillars are constructed. Each pillar has base radius 7 cm and height 24 cm. Find 

(i) Curved surface area of one pillar. [2 Marks]

(ii) Find the total cost of painting 100 pillars at the rate of Rs. 120 per cm².

Doubt by Question by Purvi 

Solution : 

No. of Pillars = 100

radius (r) = 7 cm 
height (h) = 24 cm 

Slant Height (l) 
l=√(r²+h²)
l=√[(7)²+(24)²]
l=√[49+576]
l=√[625]
l=25 cm

(i) CSA  of one pillar
= πrl
= (22/7)×7×25
= 22×25
= 550 cm²

(ii) CSA  of one pillar = 550 cm²
CSA of 100 Pillars
= 550 × 100 
=55000 cm²

Cost of Painting 1 cm² = Rs. 120
Cost of Painting 55000 cm²
= 55000 ×120
= Rs 6600000

If a+b+c=root(42) and a2+b2+c2=16 . . .

Question : If a+b+c=√42 and a²+b²+c²=16, then find the value of ab+bc+ca

Doubt by Parth 

Solution : 

a+b+c=√42 — (1) 

a²+b²+c²=16 — (2) 

We know, 
(a+b+c)²=a²+b²+c²+2ab+2bc+2ca
(√42)²=16+2(ab+bc+ca)
42=16+2(ab+bc+ca)
42-16=2(ab+bc+ca)

26=2(ab+bc+ca)
26/2=(ab+bc+ca)
13=ab+bc+ca

Hence, the value of ab+bc+ca=13