Square root and cube root
 Find the cube root of 19683.

 25
 26
 28
 27

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In 19683, 19 lies between 2^{3} and 3^{3}, so left digit is 2 and 683 ends with 3, so right digit is 7.
Correct Option: D
In 19683, 19 lies between 2^{3} and 3^{3}, so left digit is 2 and 683 ends with 3, so right digit is 7.
thus 27 is a cube root of 19683.
 If √y/169 = 54/39, then y is equal to ?

 108
 324
 2916
 4800

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√y/169= 54/39
⇒ y/169 = (54/39) x (54/39)Correct Option: B
√y/169= 54/39
⇒ y/169 = (54/39) x (54/39)
∴ y= (54/39) x (54/39) x 169 = 324
 112/√196 x √579/12 x √256/8 = ?

 8
 12
 16
 32

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Given Expression 112/√196 x √579/12 x √256/8
=(112/14) x (24/12) x (16/8)Correct Option: D
Given Expression 112/√196 x √579/12 x √256/8
=(112/14) x (24/12) x (16/8)
= 32
 Given that √4096 = 64, the value of √4096 + √40.96 + √.004096 is ?

 70.4
 70.464
 71.104
 71.4

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√4096 + √40.96 + √.004096
= √4096 + √4096/100 + √4096/1000000
=√4096 + √4096/10 + √4096/1000Correct Option: B
√4096 + √40.96 + √.004096
= √4096 + √4096/100 + √4096/1000000
=√4096 + √4096/10 + √4096/1000
= 64 + 64/10 + 64/1000
= 70.464
 √.04

 .02
 .2
 .002
 None of these

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√.04 = √4/100
Correct Option: B
√.04 = √4/100
= 2/10
= 0.2