## CBSE CLASS 10^{th} 2007: MATHEMATICS SOLVED PAPER

**Time Allowed: 3 Hours Max Marks: 80 **

**General Instruction:**

- 1. All the questions are compulsory.
- 2. The question paper consists of 25 questions divided into three sections- A, B, and C. Section A contains 7 questions of 2 marks each. Section B contains 12 questions of 3 marks each. Section C contains 6 questions of 5 marks each.
- 3. There is no overall choice. However, an internal choice has been provided in two questions of two marks each, two questions of three marks each and two questions of five mark each.
- 4. In questions on construction, the drawing should be neat and exactly as per the given measurement.
- 5. Use of the calculator is not permitted. However, you may ask for Mathematical tables.

**Section A: Question No 1 to 7 carry 2 marks each**.

**Ques [1] : Find the GCD of the following polynomials**** 12x ^{4} + 324x and 36x^{3} + 90x^{2} – 54x**

Solution: 12x^{4} + 324x = 12x(x^{3} + 27) = 6 × 2x(x^{3} + (3)^{3}) = 6 × 2x (x + 3)(x^{2} + 9 –3x)

36x^{3} + 90x^{2} – 54x = 18x(2x^{2} + 5x – 3) = 6 × 3x (2x^{2} + 6x – x – 3)

= 6 × 3x (2x (x+ 3) – 1(x + 3))

= 6 × 3x (x+ 3)(2x – 1)

Thus, the GCD of the given polynomials is 6x (x + 3).

**Ques [2] : Solve for x and y:**

** and **

Solution: We are given the equations as

** ** …(1)

…(2)

From (1) + (2), we get

** **Now, putting the value of x in the equation (1), we get

** **Thus, the solution of the given equations are x = 2a and y = –2b.

**OR**

** Solve for x and y: **** 31x + 29y = 33 and 29x + 31y = 27**

Solution: We are given the equations as

31x + 29y = 33 …(1)

29x + 31y = 27 …(2)** **

From (1) + (2), we get

⇒31x + 29y + 29x + 31y = 33 + 27

⇒60x + 60y = 60

⇒60(x + y) = 60

⇒x + y = 1 …(3)

From (1) – (2), we get

⇒31x + 29y – 29x – 31y = 33 – 27

⇒2x – 2y = 6

⇒2(x – y) = 6

⇒x – y = 3 …(4)

From (3) + (4), we get

⇒2x = 4 x = 2

From (3) – (4), we get

⇒2y = –2 y = –1

Thus, we get x = 2 and y = –1

**Ques [3]: Find the sum of all the three digits whole numbers which are multiple of 7**

Solution: All the three-digits whole numbers are 100, 101, 102,…, 999.

Out of which, the multiples of 7 are 105, 112, 119, …, 994

Now,

⇒994 = 105 + (n – 1)(7)

⇒994 –105 = 7(n –1)

⇒889 = 7(n –1)

⇒ n = 127 + 1 = 128

Thus, the required sum of the whole numbers which are multiples of 7 is

**Ques [4]: In the figure (1), PQ || AB and PR || AC. Prove that QR || BC.**

Solution: Since PQ || AB in ∆OAB, we get

…(1) [ By Basic Proportionally Theorem ] ** **

** **Since PR || AC in ∆OAC, we get

…(2) [ By Basic Proportionally Theorem ]

From the equations (1) and (2), we get

QR | | BC [ By Converse of Basic Proportionally Theorem ]

**OR**

** In the figure (2) incircle of ∆ABC touches its sides AB, BC and CA at D, E and F respectively. ****If AB = AC, prove that BE = EC**

Solution:

We know that the tangents drawn from an external point to the circle are equal in length. Thus, we get

Now, we have AB = AC ⇒ x + y = x + z ⇒ y = z ⇒ BE = EC.

**Ques [5]: If the mean of the following frequency distribution is 49, find the missing ****frequency p:**

CLASS | FREQUENCY |
---|---|

0-20 | 2 |

20-40 | 6 |

40-60 | P |

60-80 | 5 |

80-100 | 2 |

Solution:

Class | Class-Mark (x) | Frequency (f) | x×f |
---|---|---|---|

TOTAL | 15 + p | 730 + 50p | |

0-20 | 10 | 2 | 20 |

20-40 | 30 | 6 | 180 |

40-60 | 50 | p | 50p |

60-80 | 70 | 5 | 350 |

80-100 | 90 | 2 | 180 |

Now, we know that the mean of the grouped frequency is

**Ques [6]: A wrist watch is available for Rs. 1000 cash or Rs. 500 as cash down payment ****followed by three equal monthly instalments of Rs. 180. Calculate the rate of ****interest charged under the installment plan.**

Solution: Here, we have

P = Rs. 1000,

A = Rs. (500 + 180 × 3) = Rs. (500 + 540) = Rs 1040

T = 3 Months = years

SI = A – P = Rs. (1040 – 1000) = Rs. 40. Thus, we need to find the rate of interest R = ?

We know that

Therefore, the rate of the interest is R = 16% per year

**Ques [7]: An unbiased die is tossed once. Find the probability of getting ****(i) ****A multiple of 2 or 3 **

** (ii) ****A prime number greater than 2.**

Solution: In an unbiased die, we have {1,2,3,4,5,6} Thus, we have n(S) = 6

(i) A = multiple of 2 = {2,4,6} ⇒ n(A) = 3

B = multiple of 3 = {3,6} ⇒ n(B) = 2

A ∩ B = multiple of both 2 and 3 = {6} ⇒ n(A ∩ B) = 1

Thus, the required probability of getting a multiple of 2 or 3 is

(ii) A = A prime number greater than 2 = {3,5} n(A) = 2.

Thus, the required probability of getting a prime number greater than 2 is

CBSE 10th 2007: MATHEMATICS SOLVED PAPER PDF

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