Answer keys to CBSE 10th – 2019 – maths question paper Section B)- Answer keys:

                                                   Section B

7. In HCF of 65 and 117 is expressible in the form of 65n-117, then find the value of n.

HCF of 65 and 117

117=65X1 + 52

65=52X1 + 13

52= 13 X 4+0

HCF(65,117)= 13

As given,

65n-117=13

65n=13+117

= 130

n= 130/65= 2

n=2

    OR

On a morning walk, three persons step out together and their steps measure 30 cm, 35 cm and 40 cm respectively. What is the minimum distance each should walk, so that each can cover the same distance in complete steps.

Answer:

Steps sizes are 30cm, 36 cm and 40 cm.

The minimum distance each should walk, should be exactly divisible by 30, 36 and 40.

So the minimum distance = LCM ( 30,36,40)

30=2X3X5

LCM(30,36,40)= 2³X3²X5

=8X9X5

= 360m

Therefore the minimum distance covered by each of them = 360m

8. A die is thrown once. Find the probability of getting (i) a composite number (ii) prime number

Answer:

A die is thrown once, the set of outcomes ={ 1, 2,3,4,5,6}

Composite numbers={ 4,6}

Prime numbers={ 2,3,5}

(i)Probability of getting composite number =2/6=1/3

(ii)Probability of getting a prime number = 3/6=1/2

9. Using the completing the square method, show the equation x²-8x+18  has no solution.

Answer:

x²-8x+18=0

x²-2X x X 4+16-16 +18=0…………………..Adding 16 and Subtracting 16x²

(x-4)²-16 +18=0

(x-4)² +2=0

(x-4)² =-2

(x-4)² cannot be negative , for any real value of x.

Therefore there is no real value of x satisfying the given equation.

10. Cards are numbered 7 to 40 were put in a box. Poonam selects a card at random. What  is the probability that Poonam selects a card which is a multiple of 7?

Outcomes= {7,8,9…….40}

Total number of outcomes= 27

Number of multiples of 7= {7,14,21,28,35}

Number of multiples of 7= 5

Probability of selecting a card which is multiple of 7= 5/27

 

11. Solve the following pair of linear equations

3x+4y=10

2x-2y=2

Answer :

3x+2y=10…………….1

2x-2y=2………………2

Using elimination method

Multiplying equation 1 with 2  and equation 2 with 3

(3x+2y=10)X2

(2x-2y=2)X3

6x     +8y=         20

6x      -6y=           6

(-)      (+)           (-)

________________

0+ 14y = 14

y= 14/14

y=1

Substituting y=1 in equation 1

3x+4X1= 10

3x=10-4

3x=6

x=6/3

x=2

Solution : x=2, y=1

12.Points A(3,1), B(5,1), C(a,b) and D(4,3) are vertices of parallelogram ABCD. Find the values of A and B

Answer:

 

 

OR

Points P and Q trisect the line segment joining the points .A(-2,0) and, B(0,8) such that P is near to A. Find the coordinates of points P and Q.

Answer:

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