Mathematics

CBSE Class 10

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1.

Given $\u2206ABC~\u2206PQR,if\frac{AB}{PQ}=\frac{1}{3},thenfind\frac{ar\u2206ABC}{ar\u2206PQR}$

The ratio of the area of a similar triangle is equal to the square of their proportional side

$\frac{ar\u2206ABC}{ar\u2206PQR}=\frac{A{B}^{2}}{P{Q}^{2}}\phantom{\rule{0ex}{0ex}}\frac{{\displaystyle ar\u2206ABC}}{{\displaystyle ar\u2206PQR}}={\left(\frac{1}{3}\right)}^{2}=\frac{1}{9}$

2.

Two different dice are tossed together. Find the probability:

(i) of getting a doublet

(ii) of getting a sum 10, of the numbers on the two dice.

The outcomes when two dice are thrown together are

(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)

(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)

(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)

(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)

(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)

(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)

Total number of outcomes = 36

n (s) = 36

i) A = getting a doublet

A = {(1,1), (2,2), (3,3),(4,4), (5,5), (6,6)}

n(A) = 6

$\therefore P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}=\frac{6}{36}=\frac{1}{6}$

B = getting sum of numbers as 10.

B = {(6, 4), (4, 6), (5, 5)}

n(B) =3

$\therefore P\left(B\right)=\frac{n\left(B\right)}{n\left(S\right)}=\frac{3}{36}=\frac{1}{12}$

3.

Find the distance of a point P(x, y) from the origin.

Given point is P(x,y)

Origin point is O (0,0)

Using distance formula

$PO=\sqrt{({x}_{2}-{x}_{1}{)}^{2}+({y}_{2}-{y}_{1}{)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{(x-0{)}^{2}+(y-0{)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{x}^{2}+{y}^{2}}$

4.

If x = 3 is one root of the quadratic equation x^{2}-2kx -6 =0, the find the value of k.

x = 3 is one of the root of x^{2}-2kx -6 =0

(3)^{2} -2k(3) - 6 =0

9-6k - 6 = 0

3 - 6k = 0

3 = 6k

k = 3/6 = 1/2

5.

An integer is chosen at random between 1 and 100. Find the probability that it is :

(i) divisible by 8

(ii) not divisible by 8

An integer is chosen at random from 1 to 100

Therefore n(S) = 100

(i) Let A be the event that number chosen is divisible by 8

∴ A = { 8,16,24,32,40,48,56,64,72,80,88,96}

∴ n (A) = 12

Now, P (that number is divisible by 8)

$P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}\phantom{\rule{0ex}{0ex}}=\frac{12}{100}=\frac{6}{50}=\frac{3}{25}\phantom{\rule{0ex}{0ex}}P\left(A\right)=\frac{3}{25}$

(ii) Let ‘A’ be the event that number is not divisible by 8.

$\therefore P(A\text{'})=1-P\left(A\right)\phantom{\rule{0ex}{0ex}}=1-\frac{3}{25}\phantom{\rule{0ex}{0ex}}P(A\text{'})=\frac{22}{25}$

6.

In an AP, if the common difference (d) = –4, and the seventh term (a_{7}) is 4, then find the first term.

a_{7} = 4

a + 6d = 4

as a_{n} = a + (n-1) d

but d = -4

a + (-24) = 4

a = 4 + 24 = 28

therefore first term a = 28

7.

What is the value of (Cos^{2}67^{o} - sin^{2} 23^{o})?

Cos^{2} 67° - sin^{2}23°

as cos (90° - θ) = sin θ

Let θ = 23°

cos (90° - 23°) = sin 23°

cos 67° = sin 23°

∴ cos^{2}67° = sin^{2}23°

∴ cos^{2}67° = sin^{2}23° = 0

8.

What is the HCF of the smallest prime number and the smallest composite number?

Smallest prime number is 2.

Smallest composite number is 4

therefore, HCF is 2

9.

In figure, ABCD is a rectangle. Find the values of x and y.

Given figure is a rectangle i.e ABCD,

DC = AB and BC = AD

x + y = 30 ...(i)

and

x -y = 14 ...(ii)

Adding (i) and (ii), we get

2x = 44

x = 22

Putting value of x in equation (i)

x + y = 30

22 + y = 30

y = 30-22

y = 8

Thus, x = 22 and y = 8

10.

Find the ratio in which P(4, m) divides the line segment joining the points A(2, 3) and B(6, –3). Hence find m.

Suppose the point P(4, m) divides the line segment joining the points A(2, 3) and B(6, -3) in the ratio K : 1

Co-ordinates of the point,

$P=\left(\frac{6K+2}{K+1},\frac{-3K+3}{K+1}\right)$

But the co-ordinates of point P are given as (4, m)

$\frac{6K+2}{K+1}=4....\left(i\right)\phantom{\rule{0ex}{0ex}}\frac{-3K+3}{K+1}=m.....\left(ii\right)\phantom{\rule{0ex}{0ex}}6K+2=4K+4\phantom{\rule{0ex}{0ex}}2K=2\phantom{\rule{0ex}{0ex}}K=1\phantom{\rule{0ex}{0ex}}PuttingK=1inequ.\left(2\right)\phantom{\rule{0ex}{0ex}}\frac{-3\left(1\right)+3}{1+1}=m\phantom{\rule{0ex}{0ex}}\therefore m=0\phantom{\rule{0ex}{0ex}}Ratiois1:1andm=0\phantom{\rule{0ex}{0ex}}i.ePisthemidpointofAB$

Textbook Solutions | Additional Questions

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