(a) 32768
(b) 2744
(f) 21952

(a) 1331
(e) 3375
(f) 4913

81

3, 21

6, \(\sqrt { 1764 } \)= 42

No, 242

We have, 2744
Resolving 2744 into prime factors, We get
2744= 2 x 2 x 2 x 7 x 7 x 7 =(2 x 7)³=(14)³
Clearly, prime factor of2744 are grouped into triplets
So,2744 is a perfect cube of 14.

2506

52

5.2

29 and 66

m = \(\frac { 4 }{ 5 } \)

Let the number of Rs. 50 notes and Rs. 20 notes be 9x and 8x, respectively, but he has 61 notes in total.
So, the number Rs.10 notes = 61- (9x + 8x)= 61-17x
Now, the amount he has
From Rs. 50 notes = 9x x 50 = Rs. 450x
From Rs. 20 notes = 8x x 20 = Rs. 160x
From Rs. 10 notes = (61-17x) x 10 = Rs. (610 -170x)
Hence, the total money he has = 450x + 160x + (610 -l70x) = f (610 + 440x)
But he has total of the money = Rs. 1050
\(\therefore\)  610 + 440x = 1050
\(\Rightarrow\) 440x = 1050 - 610  \(\Rightarrow\) 440x = 440 \(\Rightarrow\) x = 440 x\(\left( \frac { 1 }{ 440 } \right) \) = 1
\(\therefore\) The number of \(Rs\).  50 notes he has = 9 x 1 = 9
The number of \(Rs\) 20 notes he has = 8 x 1 = 8
The number \(Rs\) 10 notes he has = (61-17 x 1) = 44

x = 0, it satisfies equation

\(x=\frac { 1 }{ 3 } \)

Let 'x' be the side of the cube.
\(\therefore x^3=64 or \sqrt[3]{x^3}=\sqrt[3]{64}\)
\(\Rightarrow \ x =\sqrt[3]{2\times2\times2\times2\times2\times2}=\sqrt[3]{2^3\times2^3}\)
  =2x2=4
Thus, the required side of the cube is 4 cm.

We have 1008 = 2 x 2 x 2 x 2 x 3 x 3 x 7
As the prime factor 7 has no pair.
\(\therefore\)1008 is not a perfect square.
Obviously, if 7 gets a pair, then the number will become a perfect square.
\(\therefore\)1008 x 7
= [2 x 2 x 2 x 2 x 3 x 3 x 7] x 7
or 7056 = 2 x 2 x 2 x 2 x 3 x 3 x 7 x 7
Thus, 7056 is the required multiple of 1008
which is a perfect square.
Now, \(\sqrt{7056}\) = 2 x 2 x 3 x 7 = 84

We have 31944 = 2 x 2 x 2 x 3
x11x11x11
Since, the prime factors of 31944 do not appear in triples as 3 is left over.
\(\therefore\)31944 is not a perfect cube.
Obviously, 31944 \(\div\) 3 will be a perfect cube
i.e., [31944] \(\div\) 3
= [2 x 2 x 2 x 3 x 11 x 11 x11] \(\div\) 3
or 10648 = 2 x 2 x 2 x 11 x 11 x 11
\(\therefore\) 10648 is a perfect cube.
Thus, the required least number = 3.

2028 \(\times\) 3 = 6084 is a perfect square.
The square root of 6084 is 78.

Let the number of winners be x.
Then, the number of participants who did not win = 36-x
Amount spent on x prizes = Rs. 500 x x = Rs. 500x
Amount spent on (36 - x) prizes = Rs. 100 x (36 - x) =
= Rs. (3600 -100x)
But 500x + (3600 -100x) = 4800
\(\Rightarrow\) 500x + 3600 -100x = 4800
\(\Rightarrow\) 400x = 4800 - 3600
\(\Rightarrow\) x = 1200 x \(\frac { 1 }{ 400 } \Rightarrow x=3\)
So, the number of winners is 3

34 cm, 16cm

23m, 23m,9m

15

Let the numerator be x, then denominator = (x + 2).
New numerator = x +3 ; New denominator = (x + 2) -1 = x +1
\(\therefore\) New Number \(\frac { x+3 }{ x-1 } \)
As per the condition, We have \(\frac { x+3 }{ x-1 } =\frac { 7 }{ 6 } \) \(\Rightarrow\) (x + 3) x 6 = 7 x (x +1)[by cross-multiplication] 
6x + 18 = 7x + 7\(\Rightarrow\) 6x - 7x = 7 - 18 \(\Rightarrow\) x = - 11\(\Rightarrow\) x= - 11\(\Rightarrow\) x-\(\frac { -1 }{ 1 } \) =11
\(\therefore\) Rational number = \(\frac { 11 }{ 11+2 } =\frac { 11 }{ 13 } \)

Given, speed of stream = 2 km/h
Let speed of steamer in still water be x km/h.
Then, speed of downstream = (x + 2) km/h
and speed of upstream = (x - 2) km/h
\(\therefore\)  Distance covered in 7 h while downstream = 7(x + 2)
and distance covered in 8 h while upstream = 8(x - 2)
According to the question, 7(x + 2) = 8(x - 2)
\(\Rightarrow\)  7x + 14 = 8x - 16 ~ x = 30 km/h
Total distance = 7(x + 2) km = 7(30 + 2) km = 7 x 32 km = 224 km
Hence, speed of the steamer in still water is 30 km/h and the distance between the ports is 224 km