1 x 1 = 1
9 x 9 = 81

3x - 2 = 2x - 1 \(\Rightarrow\) x = 1.

Required measure = \({360^\circ\over 15}=24^\circ\)

100=2\(\times\)2\(\times\)5\(\times\)5

\(\frac{5000}{1000}\) =5 cm is the height of the bar corresponding to food.

6\(\times\)6\(\times\)6=216

100 = 2 \(\times\) 2 \(\times\) 5 \(\times\) 5 = 2²\(\times\) 5².

We have 1372 = 2 x 2 x 7 x 7 x 7
Since, the prime factor 2 does not appear in a group of triples.
\(\therefore\) 1372 is not a perfect cube.
Obviously, to make it a perfect cube we need one more 2 as its factor
i.e., [1372] x 2 = [2 x 2 x 7 x 7 x 7] x 2
or 2744 = 2 x 2 x 2 x 7 x 7 x 7
which is a perfect cube.
Thus, the required smallest number = 2.
 

In ΔABC, ㄥC = 180° - (52° + 64°) =180° -116° =64°
Now, we see that, DG II BC and DG II AE
∴  ㄥACB = ㄥAFG
[∴ FG II BC and FC is a transversal, so corresponding anqles]
⇒ 64°= ㄥAFG
Also,GFD is a straight line.
∴  ㄥGFA + ㄥAFD =180° ⇒ 64° + ㄥAFD =180°
⇒ ㄥAFD = 180° - 64 ° = 116°
Also, FD II AE and AF II ED
So,AEDF is a parallelogram.
∴  ㄥAFD = ㄥAED
[∵ opposite angles in a parallelogram are equal]
or ㄥAED = 116° = x

We can draw the number line accordingly

We have,\(\{ {9\over16}\times {4\over12} \} +\{ {9\over 16}\times {-3\over 9} \} = {9\over16} \times [{4\over12}+({-3\over9})] \)
[by distributivity, taking \( {9\over16}\) as common factor]
\(={9\over16}\times[{4\over12}-{3\over9}]={9\over16}\times[{12-12\over36}]\) 
\([\because LCM \ of\ 12\ and \ 9=36]\) 
\(={9\over16}\times {0\over36}\) 
\(= {9\over16}\times0=0\) 

From the given bar graph, we find that,
i) The information given by the bar graph is the number of students in Class VIII in various academic years in the school.
(ii)In the year 2004-05, increase in the number of students is maximum.
(iii) The number of students is maximum in the year 2007 - 08.
(iv) False, because the number of students during 2005-06 is more than twice that of 2003-04.

If x is a multiple of 11, the next multiple is x + 11. The next to this is x + 11 + 11 or x + 22. So we can take three consecutive multiples of 11 as x, x + 11 and x + 22.
It is given that the sum of these consecutive multiples of 11 is 363. This will give the following equation:
x + (x + 11) + (x + 22) = 363
or x + x + 11 + x + 22 = 363
or 3x + 33 = 363
or 3x = 363 – 33
or 3x = 330
\(x=\frac{330}{3}\)
= 110
Hence, the three consecutive multiples are 110, 121, 132

142^o, 38^o, 142^o, 38^o

4

29

No, 242

We have, 1005
One's digit of 1005 = 5
Now, cube of one's digit of 1005 = (5)³ = 5 x 5 x 5 = 125
Hence,one's digit in the cube of 1005 is 5.

To represent\({-7\over 4}\) on the number line, we make 7 markings each of a distance equal to 1/4 on the left of zero. The 7th point represents the rational number\({-7\over 4}\) as shown below on the number line.

The point P is \(-7\over4\)

On cricket spent money = Rs.125000
On hockey spent money = Rs. 83333.3
On football spent money = Rs. 50000
On tennis sent money = Rs. 41666.7

We have, x =\(11\over15\)
LHS = -(-x) = -\(({11\over15})=​​​​{11\over15}=X=RHS\)
S0, - (-x) = x is verified for x =\(11\over15\)

Since, an equilateral triangle has minimum interior angles, so the maximum exterior angle possible for a regular polygon = 180° - 60° = 120°.

Firstly,find the measure of exterior angle by subtracting measureof interior angle from 180°.

64cm³

Total amount =2209                 [given]
According to the question,
Money Contributed by each student
= Square of the total number of students
\(\therefore\) Total number of students
\(=\sqrt { 2209 } =\sqrt { 47\times 47 } =47\)
The value depicted from student's activities is that all students are very cooperative in nature.

Given, \(\sqrt [ 3 ]{ 27 } +\sqrt [ 3 ]{ 0.008 } +\sqrt [ 3 ]{ 0.064 } \)
Now, \(\sqrt [ 3 ]{ 27 } =\sqrt [ 3 ]{ 3\times 3\times 3 } =3\)
Now \(\sqrt [ 3 ]{ 0.008 } =\sqrt [ 3 ]{ \frac { 8 }{ 1000 } } \)
= \(\sqrt [ 3 ]{ \frac { 2\times 2\times 2 }{ 2\times 2\times 2\times 5\times 5\times 5 } } =\frac { 2 }{ 10 } \)
Now, \(\sqrt [ 3 ]{ 0.064 } =\sqrt [ 3 ]{ \frac { 64 }{ 1000 } } \)
= \(\sqrt [ 3 ]{ \frac { 2\times 2\times 2\times 2\times 2\times 2 }{ 2\times 2\times 2\times 5\times 5\times 5 } } =\frac { 4 }{ 10 } \)
So, \(\sqrt [ 3 ]{ 27 } +\sqrt [ 3 ]{ 0.008 } +\sqrt [ 3 ]{ 0.064 } \)
= \(3+\frac { 2 }{ 10 } +\frac { 4 }{ 10 } \)
LCM of 1 and 10 is 10.
Then, \(\frac { 30+2+4 }{ 10 } =\frac { 36 }{ 10 } \) = 3.6

(a) 120
(b) 37.5%

Let the number of plates on the table be y and the number of puris be x.
Then, according to the first condition of the problem
4(y - 1) = x ...(1)
and, according to the second condition of the
problem
3y + 1= x    ... (2)
From (1) and (2), we have
4(y - 1) = 3y + 1
⇒ 4y - 4 = 3y + 1
⇒ 4y - 3y = 1 + 4
⇒ y=5
Put y = 5 in (1), we get
x = 4(5) - 4 = 20 - 4 = 16
Hence, the number of plates and number of puris on the table are 5 and 16 respectively.