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Class 10 Mathematics (Basic) TERM 1 Sample Paper Solution - CBSE | Session- 2021-22

1. A box contains cards numbered 6 to 50. A card is drawn at random from the box. The probability that the drawn card has a number which is a perfect square like 4,9….is  (a) 1/45  (b) 2/15  (c) 4/45  (d) 1/9 SOLUTION 2. In a circle of diameter 42cm ,if an arc subtends an angle of 60 ˚ at the centre where ∏=22/7,then the length of the arc is  (a) 22/7 cm  (c) 22 cm  (d) 44 cm (b) 11cm  SOLUTION 3. If sinƟ = x and secƟ = y , then tanƟ is  (a) xy  (b) x/y  (c) y/x  (d) 1/xy SOLUTION 4. The pair of linear equations y = 0 and y =-5 has  (a) One solution  (b) Two solutions  (c) Infinitely many solutions  (d) No solution SOLUTION 5. A fair die is thrown once. The probability of even composite number is  (a) 0  (b) `1/3`  (c) `3/4`  (d) 1 SOLUTION 6. 8 chairs and 5 tables cost Rs.10500, while 5 chairs and 3 tables cost Rs.6450. The cost of each chair will be  (a) Rs. 750  (b) Rs.600  (c) Rs. 850  (d) Rs. 900 SOLUTION 7. If cosƟ+`"cos"^{2}Ɵ` =1,the value of `"sin"^{2}`Ɵ+`

The HCF of two Numbers is 18 and their Product is 12960. Their LCM will be - Bzziii

The HCF of two numbers is 18 and their product is 12960. Their LCM will be  (a) 420  (b) 600  (c) 720  (d) 800   SOLUTION (c) 720  Explanation: We Know that, HCF `\times` LCM = Product of two numbers 18 `\times` LCM = 12960 LCM = `12960/18` = 720 08 09 10 11 12 CBSE Class 10 Mathematics (Basic) Sample Paper Solution-2022

The LCM of `2^3` × `3^2` and `2^2` × `3^3` is - Bzziii

The LCM of `2^{3}\times 3^{2}`  and `2^{2}\times 3^{3}` is (a) `2^{3}` (b) `3^{3}` (c)   `2^{3}\times 3^{3}` (d) `2^{2}\times 3^{2}`   SOLUTION (c)   `2^{3}\times 3^{3}` Explanation: = `2^{3}\times 3^{2}` = 2 `\times`2   `\times`2   `\times`3   `\times`3 = `2^{2}\times 3^{3}` =  2 `\times`2     3   `\times`3   `\times`3 LCM =  2 `\times`2   `\times`2   `\times` 3   `\times`3   `\times`3 =  `2^{3}\times 3^{3}` 07 08 09 10 11 CBSE Class 10 Mathematics (Basic) Sample Paper Solution-2022

The Decimal Representation of 23/2^3×5^2 will be - Bzziii.com

The decimal representation of `\frac{"23"}{"2"^{"3"}\times "5"^{"2"}}` will be  (a) Terminating  (b) Non-terminating  (c) Non-terminating and repeating  (d) Non-terminating and non-repeating   SOLUTION (a) Terminating  Explanation: ⇒ `\frac{"23"}{"2"^{"3"}\times "5"^{"2"}}` ⇒  `\frac{"23"}{"2"^{"3"}\times "5"^{"2"}}``\times``5/5` ⇒   `\frac{"23"}{"2"^{"3"}\times "5"^{"3"}}` ⇒   ` \frac{115}{(2\times5)^{3}} ` ⇒   ` \frac{115}{10^{3}} ` ⇒   `115/1000` ⇒   0.115 ∴ The decimal expansion is terminating. 06 07 08 09 10 CBSE Class 10 Mathematics (Basic) Sample Paper Solution-2022

If cosƟ+`"cos"^{2}Ɵ` =1,the value of `"sin"^{2}`Ɵ+`"sin"^{4}Ɵ` is - Bzziii

If cosƟ+`"cos"^{2}Ɵ` =1,the value of `"sin"^{2}`Ɵ+`"sin"^{4}Ɵ` is (a) -1 (b) 0 (c) 1 (d) 2 SOLUTION (c) 1 Explanation: Given, cosƟ+`"cos"^{2}Ɵ` =1 cosƟ = 1 - `"cos"^{2}Ɵ` cosƟ = `"sin"^{2}Ɵ`  `"sin"^{2}Ɵ` = cosƟ  Now,  `"sin"^{2}Ɵ` +  `"sin"^{4}Ɵ` = cosƟ  +  `"cos"^{2}Ɵ`  = 1 So, the correct answer is option (c) 05 06 07 08 09 CBSE Class 10 Mathematics (Basic) Sample Paper Solution-2022

8 chairs and 5 tables cost Rs.10500, while 5 chairs and 3 tables cost Rs.6450. The cost of each chair will be - Bzziii

8 chairs and 5 tables cost Rs.10500, while 5 chairs and 3 tables cost Rs.6450. The cost of each chair will be  (a) Rs. 750 (b) Rs.600 (c) Rs. 850 (d) Rs. 900 SOLUTION (a) Rs. 750 Explanation: Let be, Cost of Each Chairs = Rs x Cost of each table = Rs y According to the question, Cost of 8 Chairs and 5 tables = 10,500 8x + 5y = 10,500     .....(1) Cost of 5 Chairs and 3 tables = 6,450 5x + 3y = 6,450 From (1) ⇒ 8x + 5y = 10,500   ⇒ 5y = 10500 - 8x ⇒ y = `1/5`(10,500-8x) Putting value of y in (2) ⇒ 5x + 3y = 6,450 ⇒ 5x + 3 `\times``1/5`(10,500-8x) = 6450 ⇒ 5`\times`[5x + 3 `\times``1/5`(10500-8x)] = 5`\times`6,450 (Multiply both sides by 5) ⇒ 25x + 3 (10500 - 8x) = 32250 ⇒ 25x + 31500 - 24x = 32250 ⇒ 25x - 24x = 32250 - 31500 ⇒ x = 750 Thus, Cost of Chair (x) = Rs. 750 So, the Correct answer is Option (a)  04 05 06 07 08 CBSE Class 10 Mathematics (Basic) Sample Paper Solution-2022

A fair die is thrown once. The probability of even composite number is (A) 0 (B) `1/3` (C) `3/4` (D) 1 - Bzziii

A fair die is thrown once. The probability of even composite number is  (a) 0  (b) `1/3`  (c) `3/4`  (d) 1 SOLUTION Outcomes on thrown of dice = 1, 2, 3, 4, 5, 6 even composite numbers = 4, 6 ∴ Total Numbers of outcomes = 6 and Total even composite Numbers = 2 Thus,  P (Getting an even composite numbers) = `"Number of composite Numbers"/"Total Outcomes"` = `2/6` = `1/3` ∴ The probability of even composite number is `1/3`. 03 04 05 06 07 CBSE Class 10 Mathematics (Basic) Sample Paper Solution-2022

The Pair of Linear Equations y = 0 and y =-5 has - Bzziii

The pair of linear equations y = 0 and y =-5 has  (a) One solution  (b) Two solutions  (c) Infinitely many solutions  (d) No solution SOLUTION (D) No solution Explanation: In this above diagram the lines are never intersect, Thus they have No Solution. 02 03 04 05 06 CBSE Class 10 Mathematics (Basic) Sample Paper Solution-2022

If sinƟ = x and secƟ = y , then tanƟ is (A) xy (B) x/y (C) y/x (D) 1/xy - Bzziii

If sinƟ = x and secƟ = y , then tanƟ is  (a) xy  (b) `"x"/"y"`  (c) `"y"/"x"`   (d) `"1"/"xy"` SOLUTION (A) xy  Explanation: Given, sinƟ = x and, secƟ = y ∴ `"1"/"cosƟ"`= y cosƟ = `"1"/"y"` Now, tanƟ = `"sinƟ"/"cosƟ"` = `\frac{"x"}{\frac{"1"}{"y"}}` = xy 01 02 03 04 05 CBSE Class 10 Mathematics (Basic) Sample Paper Solution-2022

In a Circle of Diameter 42cm ,If an arc Subtends an Angle of 60 ˚ at the Centre Where ∏=22/7,then the Length of the Arc is - Bzziii

In a circle of diameter 42cm ,if an arc subtends an angle of 60 ˚ at the centre where ` `\pi` =`22/7`,then the length of the arc is  (a) 22/7 cm  (b) 22 cm  (c) 44 cm (d) 11cm  SOLUTION (b) 22 cm  Explanation: Length of Arc APB = `\frac{\theta}{360}``\times``(2\pi "r")` = `\frac{60}{360}``\times` 2 `\times` `22/7``\times` 21 = `1/6` `\times` 2`\times` `22/7` `\times` 21 = 22 cm 01 02 03 04 05 CBSE Class 10 Mathematics (Basic) Sample Paper Solution-2022

A Box Contains Cards Numbered 6 to 50. A card is Drawn at random from the Box. The Probability that the Drawn Card has a Number Which is a Perfect - Bzziii

A box contains cards numbered 6 to 50. A card is drawn at random from the box. The probability that the drawn card has a number which is a perfect square like 4,9….is (a) `1/45` (b) `2/15` (c) `4/45` (d) `1/9` SOLUTION (D) `1/9` Explanation: Total Numbers of Cards = 50 - 6 + 1 = 45 (Total Numbers of Outcomes) Perfect Numbers are (between 6 - 50) = 4, 9, 16, 25, 36, 49, 64, 81, 100 Number of Perfect Square (between 6 - 50)= 5 Thus, P ( Getting a Perfect Squares ) = `"Numbers of Perfect Squares"/"Total Numberes"` = `5/45` = `1/9` 01 02 03 04 05 CBSE Class 10 Mathematics (Basic) Sample Paper Solution-2022

Prepare Manufacturing, Trading and Profit and Loss Account from the following figures relating to the year ending 31st march, 2015

Prepare Manufacturing, Trading and Profit and Loss Account from the following figures relating to the year ending 31st march, 2015:   01-04-2014 31-03-2015 Stock: Finished Goods 33,000 27,500 Raw Materials 16,000 18,300 Work-in-progress (at prime cost) 11,100 9,400 Particulars Amoutn (Rs.) Purchase of Materials 1,50,900 Carriage on Purchases 4,100 Wages 65,000 Factory Salaries 26,000 Office Salaries 18,000 Repairs & Maintenance on Machinery 8,300 Repairs & Maintenance on Office Equipment Office Equipment 1,700 Depreciation on Machinery 25,000 Depreciation on Office Equipment 8,100 Sundry Expenses (Factory) 5,300 Sundry Expenses (Office) 17,800 Sales 3,60,000 It is firm's practice to transfer goods from the factory to the sales godown at cost plus 10%. SOLUTION Manufacturing Account for the year ended 31st March, 2015 Particulars Amount Particulars Amount To Material Consumed By Trading A/c (Transfer) 3,12,400 Opening Stock 16,000 Add: Purchases 1,50,900 Carriage on Purch