Let the radius of the base and height of the right circular cylinder be r units and h units respectively.
Then,
Volume of the cylinder = πr²h cubic units
Curved surface area of the cylinder = 2πrh square units
If volume = curved surface area, then πr²h = 2πrh ⇒ r = 2
Hence, it is possible only when radius of the base is 2 units.

In the given figure, number of white keys =10
Number of black keys = 7
Number of total keys = 10 + 7 = 17
(a) Required ratio = \(\frac{Number of white keys}{Number of black keys}=\frac{10}{7}\)=10:7
(b) Required ratio = \(\frac{Number of black keys}{Number of total keys}=\frac{7}{17}=7:17\)
(c) In 1 pattern, number of black keys = 7 In 14 such patterns, number of black keys =7x14=98
(d) On playing musical instruments, we get mental peace and positive energy.

The number of students come by car =\({2\over5}\) of all Number of students come by bus =\({1\over4}\) of all 
\(\therefore\) Number of students walk to school
\(=[1-({2\over5}+{1\over4})]\) of all \(=[1-({8+5\over20})]\)  of all
\(=[1-{13\over20}]\)  of all = \(({20-13\over20})\) of all\(={7\over20}\) of all
Now, number of students walk to school on their own
=\({1\over3}\)of number of students walk to school
=\({1\over3} \ of {1\over7} of \ all\)
Since, 224 students come to school on their own
\(\therefore 224={1\over3}of{7\over20}of \ all \Rightarrow 224={1\over3}\times{7\over20}of \ all \)
So, number of all the students, who study in the school.
\(=224\div{1\over3}\div{7\over20}\\ =224 \times3\times{20\over7}=32\times3\times20\\ = 96\times 20=1920 students\)

(m+n)(m-n)(m²+n²)

15,16

16\(\frac{2}{3}\%\)

(48)²=(50-2)²
Since, (a - b)² = a² - 2ab + b²
(50-2)²=(50)²-2x50x2+(2)²
= 2500 - 200 + 4
[(a-b)²=a²-2ab+b²]
= 2504 - 200 = 2304
(ii) 181² -19² =(181-19)(181 + 19)
where, a = 50 and b = 2
= 162 x 200 = 32400
(iii) 497 x 505 =(500 - 3)(500 + 5)
= 500² + (-3 + 5) x 500 + (-3)(5)
[.: (x + a)(x + b) = x² + (a +b) x + ab]
= 250000 + 1000 -15 = 250985
(iv) 2.07 x1.93 = (2 + 0.07)(2 - 0.07)
= 2² -(0.07)²
[.: where, a = 50 and b = 2]
= 3.9951

Let AC and BD be two sticks, each of the length 7 cm. On joining their end points A, B, C and D, we will get a shape of a rhombus.
[since, in a rhombus diagonals bisect each other all right angles.]
Now, we will see whether it is a square or not.
Let O is the intersection point of AC and BD.
Now, we see that AO = BO = CO = DO
Since, both AC and BD are of equal length and are bisected by O.
Further, in  ΔABO, we see that
ㄥAOB= 90°
and AO =OB
which implies that ㄥOAB = ㄥOBA = x        [say]
∴ 90° + x + x = 180°
[angle sum property of a triangle]
⇒ 2x = 90°  ⇒ x = 45°
Similarly, we see that ㄥOBC, ㄥOCB, ... etc are 45°.
So, finally we can say that the shape, so obtained will be of a square.

It is clear, from the figure that polygon MNOPQR is divided into six parts, out of which four are triangles and two are trapeziums.
Area of polygon MNOPQR =Area of ΔMAN +Area of trapezium ACON + Area of ΔOCP + Area of ΔPDQ+ Area of trapezium DBRQ +Area of ΔRBM      ...(i)
Now, area of ΔMAN = MA AN =\(\frac { 1 }{ 2 } \times \) 2 \(\times\)2.5
 = 2.5 cm²
Area of trapezium ACON = \(\frac { 1 }{ 2 }\)(AN+OC) \(\times\)AC
 = \(\frac { 1 }{ 2 }\)(AN+OC) \(\times\) (MC-MA)            [∵ AC=MC-MA]
=\(\frac { 1 }{ 2 } \times \) (2.5+3) \(\times\)(6-2) =\(\frac { 1 }{ 2 } \times \) 5.5 \(\times\)4
=5.5 \(\times\)2 = 11cm²
Area of ΔOCP =\(\frac { 1 }{ 2 } \times \) CP\(\times\)OC
 =\(\frac { 1 }{ 2 }\) (MP-MC)\(\times\) OC
 = \(\frac { 1 }{ 2 }\)(9-6)\(\times\) 3 =\(\frac { 1 }{ 2 } \times \) 3 \(\times\)3           [∵ CP=MP-MC]
 = \(\frac { 9 }{ 2 } \)=4.5 cm²
Area of ΔPDQ =\(\frac { 1 }{ 2 } \times \) PD\(\times\) DQ
 = \(\frac { 1 }{ 2 } \times \)(MP-MD)\(\times\) DQ
=\(\frac { 1 }{ 2 } \times \) (9-7)\(\times\) 2 =2cm²
Area of trapezium DBRQ =\(\frac { 1 }{ 2 } \times \) (DQ+BR)\(\times\)BD
 =\(\frac { 1 }{ 2 } \times \) (DQ+BR)\(\times\) (MD-MB)        [∵ BD =MD-MB]
=\(\frac { 1 }{ 2 } \times \) (2+2.5)\(\times\) (7-4)
=\(\frac { 1 }{ 2 } \times \) 4.5 \(\times\)3 =\(\frac { 13.5 }{ 2 } \)= 6.75 cm²
Area of ΔRBM =\(\frac { 1 }{ 2 } \times \) MB\(\times\) BR =\(\frac { 1 }{ 2 } \times \) 4\(\times\) 2.5 =5cm²
On putting all these values in Eq. (i), we get
Area of polygon MNOPQR
 =(2.5+11 +4.5+ 2 +6.75 + 5)cm²
= 31.75 cm²
Hence, area of polygon MNOPQR is 31.75 cm²

Here, to represent the given information, we draw the bar graph. So, draw two perpendicular axes OX and OY. Take months on OX and number of watches sold on OY. Choose a suitable scale for OX and OY and drawbars of equal width with equal gaps in between.
Thus, we get the following bar graph

4l + 5m contains two terms.

200 + 200 = 400

Here, ab = 21 and a + b = -10
Possible values of a and b are 7, 3 or -7, -3.
But  7 +3 = 10 \(\neq \) - 10     [not possible]
\(\therefore\) a= -7, b = -3
Now q² - 10q + 21 = q² +(-7-3)q + 21
= q² - 7q - 3q + 21
=q (q  -7)-3(9 - 7) = (q-7)(q -3)
Hence, required factors of given expression are(q - 7) and (q -3).

Rs 1250000

23m, 23m,9m

According to the question, the side along the river is parallel to and twice the side along the road.
Let the length of side along the road be x m
Then, the length of side along the river = 2x m
Given, perpendicular distance between the two parallel sides =100 m
and area of the trapezium shaped field = 10500 m²
⇒\(\frac { 1 }{ 2 } \times \) (Sum of parallel sides) Heigth = 10500
⇒\(\frac { 1 }{ 2 } \times \)  (2x+2) 100= 10500 ⇒ 3\(\times\) 50 =10500
⇒ 3x = \(\frac { 10500 }{ 50 } =\frac { 1050 }{ 5 } \)
∴ x =\(\frac { 1050 }{ 3\times 5 } \) = 70 and 2x = 2 0 =140
Hence, the length of the side along the river is 140 m.

\([(10)\div 2] \div (-5) = (-5) \div (-5) = 1\)
\( (-10) \div [2 \div (-5)] =(-10) \div (- \frac{2}{5})\) = 25
∵ 1 ≠ 25
∴ No ; \([(10)\div 2] \div (-5) \ne (-10) \div [2 \div (-5)].\)

108^o, 72^o, 108^o, 72^o

ax + bx - ay - by
= x[a + b] + (-y)[a + b]
=(x - y)[a + b]

170

10ax - 2by + 2cz

Since, number 24 ends with digit 4, so the number getting from the square of 24, will be the end of the digit 6.