Question
Complete the following activity to solve the simultaneous equations
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A milk centre sold milk to 50 customers. The table below gives the number of customers and the milk they purchased. Find the mean of the milk sold by direct method.
→| Milk Sold (Litre) | $1-2$ | $2-3$ | $3-4$ | $4-5$ | $5-6$ |
| No. of Customers | 17 | 13 | 10 | 7 | 3 |
In an A.P., the first term is -5 and the last term is $4 5$. If the sum of $n$ terms in the A.P. is $1 2 0$, then complete the activity to find $n$.
→Draw a circle with center P. Draw an arc AB of 100° measure. Perform the following steps to draw tangents to the circle from points A and B.
- Draw a circle with any radius and center P.
- Take any point A on the circle.
- Draw ray PB such ∠APB = 100°.
- Draw perpendicular to ray PA from point A.
- Draw perpendicular to ray PB from point B.
Grouped frequency distribution of supply of milk to hotels and the number of hotels is given in the following table. Find the mode of the supply of milk.
Solve the following simultaneous equations by graphical method.
If $x=5$ is a root of equation $\mathrm{kx}^2-14 \mathrm{x}-5=0$, then find the value of $\mathrm{k}$ by completing the following activity.
One of the roots of equation $k x^2-14 x-5=0$ is $\square$.
$\therefore$ Now Let $x=\square$ in the equation.
$ k \square^2-14 \square-5=0$
$\therefore \quad 25 k-70-5=0$
$25 k-\square=0$
$25 k=\square$
$\therefore k=\frac{\square}{\square}=3 $
→One of the roots of equation $k x^2-14 x-5=0$ is $\square$.
$\therefore$ Now Let $x=\square$ in the equation.
$ k \square^2-14 \square-5=0$
$\therefore \quad 25 k-70-5=0$
$25 k-\square=0$
$25 k=\square$
$\therefore k=\frac{\square}{\square}=3 $
Find the correct answer from the alternatives given.
Different expenditures incurred on the construction of a building were shown by a pie diagram. The expenditure
Rs 45,000 on cement was shown by a sector of central angle of 75°. What was the total expenditure of the construction?
A. 2,16,000
B. 3,60,000
C. 4,50,000
D. 7,50,000
Different expenditures incurred on the construction of a building were shown by a pie diagram. The expenditure
Rs 45,000 on cement was shown by a sector of central angle of 75°. What was the total expenditure of the construction?
A. 2,16,000
B. 3,60,000
C. 4,50,000
D. 7,50,000
If the point $P (6,7)$ divides the segment joining $A (8,9)$ and $B (1,2)$ in some ratio, find that ratio
Solution:
Point $P$ divides segment $A B$ in the ratio $m$ : $n$.
$A(8,9)=\left(x_1, y_1\right), B(1,2)=\left(x_2, y_2\right) \text { and } P(6,7)=(x, y)$
Using Section formula of internal division,
$ \therefore 7=\frac{m(\square)-n(9)}{m+n}$
$\therefore 7 m+7 n=\square+9 n$
$\therefore 7 m-\square=9 n-\square$
$\therefore \square=2 n$
$\therefore \frac{m}{n}=\square $
→Solution:
Point $P$ divides segment $A B$ in the ratio $m$ : $n$.
$A(8,9)=\left(x_1, y_1\right), B(1,2)=\left(x_2, y_2\right) \text { and } P(6,7)=(x, y)$
Using Section formula of internal division,
$ \therefore 7=\frac{m(\square)-n(9)}{m+n}$
$\therefore 7 m+7 n=\square+9 n$
$\therefore 7 m-\square=9 n-\square$
$\therefore \square=2 n$
$\therefore \frac{m}{n}=\square $
Complete the following activity to find the sum of all two-digit numbers divisible by 8.
Two-digit numbers divisible by 8 are 16, 24, 32, ..., 88, 96.
Two-digit numbers divisible by 8 are 16, 24, 32, ..., 88, 96.
₹ 1000 is invested at 10 percent simple interest. Check at the end of every year if the total interest amount is in A.P. If this is an A.P. then find interest amount after 20 years. For this complete the following activity.
Simple interest $=\frac{ P \times R \times N }{100}$
Simple interest after 1 year $=\frac{1000 \times 10 \times 1}{100}=$⬜
Simple interest after 2 year$=\frac{1000 \times 10 \times 2}{100}=$ ⬜
Simple interest after 3 year$=\frac{\square \times \square \times \square}{100}=300$
According to this the simple interest for 4, 5, 6 years will be 400,⬜,⬜ respectively.
From this d= ⬜ and a=⬜
Amount of simple interest after 20 years
tn + a + (n – 1) d
t20=⬜+(20-1)⬜
t20=⬜
Amount of simple interest after 20 years is = ⬜
→Simple interest $=\frac{ P \times R \times N }{100}$
Simple interest after 1 year $=\frac{1000 \times 10 \times 1}{100}=$⬜
Simple interest after 2 year$=\frac{1000 \times 10 \times 2}{100}=$ ⬜
Simple interest after 3 year$=\frac{\square \times \square \times \square}{100}=300$
According to this the simple interest for 4, 5, 6 years will be 400,⬜,⬜ respectively.
From this d= ⬜ and a=⬜
Amount of simple interest after 20 years
tn + a + (n – 1) d
t20=⬜+(20-1)⬜
t20=⬜
Amount of simple interest after 20 years is = ⬜