Mathematics (211)
Tutor Marked Assignment
20% of Theory Marks
1. Answer any one of the following questions
(a) By selling a shirt to a customer for Rs. 749 a shopkeeper earns a profit of 7%. Find the cost price of the shirt.
Ans- CP = ( SP × 100 ) /( 100 + percentage profit).
So, Cost Price of Shirt = (749×100)/(100+7) = 700
2. Answer anyone out of the following questions.
(a) In the below-given figure, O is the Centre of a circle, find the value of x.
Ans- Angle PBO = 40
So angle OBQ = 90- 40 = 50
Angle BOQ = 90
So angle X = 180 – angle AOB – angle QBO = 180 – 90-50 = 40 (answer)
3. Answer anyone out of the following questions.
(a) The radius of the two circles are 9cm and 12cm. Find the radius of a circle whose area is equal to the sum of the areas of these two circles.
Ans- 𝜋𝑅² = 𝜋𝐴² + 𝜋𝐵²
= (A² + B²) = 𝜋(9² + 12²) = 𝜋(81 + 144) = 𝜋225
𝜋𝑅² = 𝜋225
R² = 225
𝑅 = 15 𝑖𝑠 𝑡ℎ𝑒 (𝒂𝒏𝒔𝒘𝒆𝒓)
4. Answer any one of the following questions.
(a) In the below given figure. ‘O' is the midpoint of AB and CD, Prove that AC= BD and AC || BD.
Ans-In triangle AOC and BOD
AOC=BOD(VOA)
AO-BO(given)
CO=DO(given)
•AAOC-=ABOD(SAS rule)
•AC=BD(cpct)
•AC|| BD(alternate angles will be equal)
Hence proved (answer)
5. Answer anyone out of the following questions.
Ans-
Let OP intersects AB at point C.
In ∆ PAC and ∆ PBC,
PA = PB (Tangents from an external point are equal)
∠APC = ∠BPC (PA and PB are equally inclined to OP)
PC = PC (common)
→ ∆PAC ≅ ∆ PBC (SAS criterion)
→ AC = BC and ∠ACP = ∠BCP
But, ∠ACP + ∠BCP = 180°
∠ACP = ∠BCP 90°
OP ⊥ AB
6. Prepare any one project out of the following projects given below.
(a) Conduct a survey of at least 50 households from your locality/village, regarding population and family income.
(i) Present the data related to family members in tabular form mentioning frequencies
Ans-
(ii) Calculate the average family size. How many families are above the average family size?
Ans-
(iii) Draw the Bar graph for the top 10 earning families.
Ans-
 








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