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CBSE 10th 2007: MATHEMATICS SOLVED PAPER

CBSE CLASS 10th 2007: MATHEMATICS SOLVED PAPER

Time Allowed: 3 Hours                                                                                                    Max Marks: 80  

General Instruction:

  • 1. All the questions are compulsory. Get CBSE class 10 previous year question paper with solution.
  • 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 12x4 + 324x and 36x3 + 90x2 – 54x

Solution:    12x4 + 324x = 12x(x3 + 27) = 6 × 2x(x3 + (3)3) = 6 × 2x (x + 3)(x2 + 9 –3x)

                   36x3 + 90x2 – 54x = 18x(2x2 + 5x – 3) = 6 × 3x (2x2 + 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:

                   frac{2x}{a}+frac{y}{b}=2 and frac{x}{a}-frac{y}{b}=4

Solution:    We are given the equations as

                    frac{2x}{a}+frac{y}{b}=2                                …(1)

                    frac{x}{a}-frac{y}{b}=4                                 …(2)

                    From (1) + (2), we get

Rightarrowfrac{2x}{a}+frac{y}{b}+frac{x}{a}-frac{y}{b}=2+4 Rightarrowfrac{3x}{a}=6 Rightarrow x=2a

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

Rightarrowfrac{2times2a}{a}+frac{y}{b}=2 Rightarrow 4+frac{y}{b}=2 Rightarrow frac{y}{b}=-2 Rightarrow y = -2b

                    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)

Rightarrow n -1 =frac{889}{7}=127

⇒ n = 127 + 1 = 128

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

Rightarrow S=frac{n}{2}[a+l]Rightarrow S=frac{128}{2}[105+994]=70336

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

pq-parallel-ab-pr-parallel-ac-prove-qr-parallel-bc
pq-parallel-ab-pr-parallel-ac-prove-qr-parallel-bc

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

                    frac{OP}{AP}=frac{OQ}{BQ}              …(1)           [ By Basic Proportionally Theorem ]                  

                    Since PR || AC in ∆OAC, we get

                    frac{OP}{AP}=frac{OR}{RC}              …(2)           [ By Basic Proportionally Theorem ]        

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

                    frac{OQ}{BQ}=frac{OR}{RC} Rightarrow 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

incircle-triangle-abc-touches-sides-abbcca-at-def-prove-beec

Solution:

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

incircle-triangle-abc-touches-sides-abbcca-at-def-prove-beec-solution

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:   

CLASSFREQUENCY
0-202
20-406
40-60P
60-805
80-1002

Solution:  

ClassClass-Mark (x)Frequency
(f)
x×f
TOTAL15 + p730 + 50p
0-20 10220
20-40306180
40-6050p50p
60-80705350
80-100902180

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

M=frac{Sigma(xf)}{Sigma(f)}Rightarrow 49= frac{730+50p}{15+p}Rightarrow 49(15+p) = 730 + 50p Rightarrow 735 + 49p= 730+50pRightarrow p=5

Ques [6]: A wrist watch is available for Rs. 1000 cash or Rs. 500 as cash down payment followed by three equal monthly installments 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 = frac{3}{12}=frac{1}{4} years

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

                 We know that SI=frac{PRT}{100}Rightarrow40=frac{1000 times R times 1}{100times 4}Rightarrow R=16% 

                 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

P(AUB)=P(A)+P(B)-P(A intersection B)Rightarrow P(AUB)=frac{3}{6}+frac{2}{6}-frac{1}{6}=frac{2}{3}

 (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 P(A)=frac{2}{6}=frac{1}{3}

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