(ii) We have:
To find the cube root of 5832, we use the method of unit digits.
Let us consider the number 5832.
The unit digit is 2; therefore the unit digit in the cube root of 5832 will be 8.
After striking out the units, tens and hundreds digits of the given number, we are left with 5.
Now, 1 is the largest number whose cube is less than or equal to 5.
Therefore, the tens digit of the cube root of 5832 is 1.
(iii) We have:
To find the cube root of 2744000, we use the method of factorisation.
On factorising 2744000 into prime factors, we get:
2744000=2×2×2×2×2×2×5×5×5×7×7×72744000=2×2×2×2×2×2×5×5×5×7×7×7
On grouping the factors in triples of equal factors, we get:
2744000={2×2×2}×{2×2×2}×{5×5×5}×{7×7×7}2744000=2×2×2×2×2×2×5×5×5×7×7×7
It is evident that the prime factors of 2744000 can be grouped into triples of equal factors and no factor is left over.
Now, collect one factor from each triplet and multiply; we get:
2×2×5×7=1402×2×5×7=140
This implies that 2744000 is a cube of 140.
Hence,
(iv) We have:
To find the cube root of 753571, we use the method of unit digits.
Let us consider the number 753571.
The unit digit is 1; therefore the unit digit in the cube root of 753571 will be 1.
After striking out the units, tens and hundreds digits of the given number, we are left with 753.
Now, 9 is the largest number whose cube is less than or equal to 753 (93<753<10393<753<103).
Therefore, the tens digit of the cube root 753571 is 9.
(v) We have:
To find the cube root of 32768, we use the method of unit digits.
Let us consider the number 32768.
The unit digit is 8; therefore, the unit digit in the cube root of 32768 will be 2.
After striking out the units, tens and hundreds digits of the given number, we are left with 32.
Now, 3 is the largest number whose cube is less than or equal to 32 (33<32<4333<32<43).
Therefore, the tens digit of the cube root 32768 is 3.
(i) LHS = 3×4=12273×643=3×3×33×4×4×43=3×4=12
RHS =
=3×4=1227×643=3×3×3×4×4×43=3×3×3×4×4×43=3×4=12
Because LHS is equal to RHS, the equation is true.
(ii) LHS =
=4×9=3664×7293=4×4×4×9×9×93=4×4×4×9×9×93=4×9=36
RHS = =4×9=36
Because LHS is equal to RHS, the equation is true.
(iii) LHS =
=
=−5×2×3=−30
RHS =
=−5×(2×3)=−30
Because LHS is equal to RHS, the equation is true.
(iv) LHS =
−5×−10=50
RHS =
=−5×−10=501253×10003=5×5×53×10×10×103=5×10=50
Because LHS is equal to RHS, the equation is true.
(i) From the above property, we have:
=2×5=10
(ii) From the above property, we have:
(For any positive integer x,
Cube root using units digit:
Let us consider the number 1728.
The unit digit is 8; therefore, the unit digit in the cube root of 1728 will be 2.
After striking out the units, tens and hundreds digits of the given number, we are left with 1.
Now, 1 is the largest number whose cube is less than or equal to 1.
Therefore, the tens digit of the cube root of 1728 is 1.
=12
On factorising 216 into prime factors, we get:
216=2×2×2×3×3×3216=2×2×2×3×3×3
On grouping the factors in triples of equal factors, we get:
216={2×2×2}×{3×3×3}216=2×2×2×3×3×3
Now, taking one factor from each triple, we get:
= 2×3 = 6
Thus
=−12×6=−72
(iii) From the above property, we have:
(For any positive integer x,
Cube root using units digit:
Let us consider the number 2744.
The unit digit is 4; therefore, the unit digit in the cube root of 2744 will be 4.
After striking out the units, tens, and hundreds digits of the given number, we are left with 2.
Now, 1 is the largest number whose cube is less than or equal to 2.
Therefore, the tens digit of the cube root of 2744 is 1.
Thus
=−3×14=−42
(iv) From the above property, we have:
(For any positive integer x,
Cube root using units digit:
Let us consider the number 15625.
The unit digit is 5; therefore, the unit digit in the cube root of 15625 will be 5.
After striking out the units, tens and hundreds digits of the given number, we are left with 15.
Now, 2 is the largest number whose cube is less than or equal to 15 ((23<15<33)23<15<33.
Therefore, the tens digit of the cube root of 15625 is 2.
Also
9, because 9×9×9=729
Thus
=−9×−25=225
(i) From the above property, we have:
=4×6=24
(ii) Use above property and proceed as follows:
=2×17=34
(iii) From the above property, we have:
(∵∵ 700=2×2×5×5×7 and 49=7×7)
=2×5×7=70
(iv) From the above property, we have:
=125×a2−(5×a2)
(∵ =a×a=a^{2} and =5)
=125a2−5a2=120a2
 125/729
Now
=  5/9 (∵∵ 729=9×9×9 and 125 = 5×5×5729=9×9×9 and 125 = 5×5×5)
(ii) Let us consider the following rational number:
10648/12167
Now
Cube root by factors:
On factorising 10648 into prime factors, we get:
10648=2×2×2×11×11×1110648=2×2×2×11×11×11
On grouping the factors in triples of equal factors, we get:
10648={2×2×2}×{11×11×11}10648=2×2×2×11×11×11
Now, taking one factor from each triple, we get:
=2×11=22106483=2×11=22
Also
On factorising 12167 into prime factors, we get:
12167=23×23×2312167=23×23×23
On grouping the factors in triples of equal factors, we get:
12167={23×23×23}12167=23×23×23
Now, taking one factor from the triple, we get:
Now
= 22/23
(iii) Let us consider the following rational number:
19683/24389
Now,
Cube root by factors:
On factorising 19683 into prime factors, we get:
19683=3×3×3×3×3×3×3×3×319683=3×3×3×3×3×3×3×3×3
On grouping the factors in triples of equal factors, we get:
19683={3×3×3}×{3×3×3}×{3×3×3}19683=3×3×3×3×3×3×3×3×3
Now, taking one factor from each triple, we get:
=3×3×3=27196833=3×3×3=27
Also
On factorising 24389 into prime factors, we get:
24389=29×29×2924389=29×29×29
On grouping the factors in triples of equal factors, we get:
24389={29×29×29}24389=29×29×29
Now, taking one factor from each triple, we get:
Now
= 27/ 29
(iv) Let us consider the following rational number:
Now
(686 and 3456 are not perfect cubes; therefore, we simplify it as 686/34566863456 by prime factorisation.)
= 7/ 12
(v) Let us consider the following rational number:
Now
Cube root by factors:
On factorising 39304 into prime factors, we get:
39304=2×2×2×17×17×1739304=2×2×2×17×17×17
On grouping the factors in triples of equal factors, we get:
39304={2×2×2}×{17×17×17}39304=2×2×2×17×17×17
Now, taking one factor from each triple, we get:
=2×17=34
Also
On factorising 42875 into prime factors, we get:
42875=5×5×5×7×7×742875=5×5×5×7×7×7
On grouping the factors in triples of equal factors, we get:
42875={5×5×5}×{7×7×7}42875=5×5×5×7×7×7
Now, taking one factor from each triple, we get:
=5×7=35
Now
= 34/35
0.001728=1728/1000000
Now
On factorising 1728 into prime factors, we get:
1728=2×2×2×2×2×2×3×3×31728=2×2×2×2×2×2×3×3×3
On grouping the factors in triples of equal factors, we get:
1728={2×2×2}×{2×2×2}×{3×3×3}
Now, taking one factor from each triple, we get:
=2×2×3=1217283=2×2×3=12
Also
= 100
12/100 = 0.12
(ii) We have:
0.003375= 3375/ 10000000.003375=33751000000
Now
On factorising 3375 into prime factors, we get:
3375=3×3×3×5×5×53375=3×3×3×5×5×5
On grouping the factors in triples of equal factors, we get:
3375={3×3×3}×{5×5×5}3375=3×3×3×5×5×5
Now, taking one factor from each triple, we get:
=3×5=1533753=3×5=15
Also
=10010000003=100×100×1003=100
= 15/100 = 0.15
(iii) We have:
1/10 = 0.1
(iv) We have:
= 11/10 = 1.1
(i) To evaluate the value of the given expression, we need to proceed as follows:
=3+0.2+0.4=3.6 Thus, the answer is 3.6.
(ii) To evaluate the value of the given expression, we need to proceed as follows:
=10+0.2−0.5=9.7
Thus, the answer is 9.7.
(iii) To evaluate the value of the given expression, we need to proceed as follows:
Thus, the answer is 1.
(iv) To evaluate the value of the expression, we need to proceed as follows:
−1=1−1=0
Thus, the answer is 0.
(v) To evaluate the value of the expression, we need to proceed as follows:
=13/10 =1.3
Thus, the answer is 1.3.
9/10
= 9/10
Because LHS is equal to RHS, the equation is true.
(ii)
= 8/7
RHS =
= 8/7
Because LHS is equal to RHS, the equation is true.
(i) 5
3×5
∵
=5×3
=3×5=3×5 (Commutative law)
(ii) 8×8=648×8=64
(iii) 3
12=4×3
(iv) 20
(v) 7×8=567×8=56
(vi) 4×5×6=1204×5×6=120
(vii) 3
(viii) 11
9/11
(ix) 13×13×13=2197
13×13×13=2197
V=s3V=s3, where s = side of the cube
Now
s3=474.552 cubic metres
To find the cube root of 474552, we need to proceed as follows:
On factorising 474552 into prime factors, we get:
474552=2×2×2×3×3×3×13×13×13474552=2×2×2×3×3×3×13×13×13
On grouping the factors in triples of equal factors, we get:
474552={2×2×2}×{3×3×3}×{13×13×13}474552=2×2×2×3×3×3×13×13×13
Now, taking one factor from each triple, we get:
=2×3×13=78
Also
= 7.8
Thus, the length of the side is 7.8 m.
According to the question:
(2x)3+(3x)3+(4x)3=0.334125⇒8x3+27x3+64x3=0.334125⇒8x3+27x3+64x3=0.334125⇒99x3=0.334125
Thus, the numbers are:
2×0.15=0.30 3×0.15=0.454×0.15=0.60 Question 12: Find the side of a cube whose volume is
(By prime factorisation)
= 29/6
for any two integers a and b
(By prime factorisation)
=2×2×2×3=24
Thus, the answer is 24.
(ii) 96 and 122 are not perfect cubes; therefore, we use the following property:
for any two integers a and b
(By prime factorisation)
=2×2×2×3=24 Thus, the answer is 24.
(iii) 100 and 270 are not perfect cubes; therefore, we use the following property:
for any two integers a and b
(By prime factorisation)
=2×3×5=30 Thus, the answer is 30.
(iv) 121 and 297 are not perfect cubes; therefore, we use the following property: for any two integers a and b
(By prime factorisation)
=11×3=33 Thus, the answer is 33.
for two integers a and b
Now
(By the above property)
(By prime factorisation)
=3×5×9=135 Thus, the answer is 135.
(ii) To find the cube root, we use the following property:
for two integers a and b
Now
(By the above property)
(By prime factorisation)
=3×7×13=273 Thus, the answer is 273.
(iii) To find the cube root, we use the following property:
for two integers a and b
Now
(By the above property)
(By prime factorisation)
=5×7×17=595 Thus, the answer is 595.
(iv) To find the cube root, we use the following property:
for two integers a and b
Now
(By the above property)
(By prime factorisation)
=5×11×7=385 Thus, the answer is 385.
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