Searching for "u"

Q:

The cube root of .000216 is:

A) .6 B) .06
C) 77 D) 87
 
Answer & Explanation Answer: B) .06

Explanation:

.00021613=21610613=6*6*6102*102*10213=6102=0.06

Report Error

View Answer Report Error Discuss

Q:

A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and c in 198 seconds, all starting at the same point. After what time will they again at the starting point ?

A) 26 minutes and 18 seconds B) 42 minutes and 36 seconds
C) 45 minutes D) 46 minutes and 12 seconds
 
Answer & Explanation Answer: D) 46 minutes and 12 seconds

Explanation:

L.C.M. of 252, 308 and 198 = 2772.

 

So, A, B and C will again meet at the starting point in 2772 sec. i.e., 46 min. 12 sec.

Report Error

View Answer Report Error Discuss

Q:

The least number which when divided by 5, 6 , 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is:

A) 1677 B) 1683
C) 2523 D) 3363
 
Answer & Explanation Answer: B) 1683

Explanation:

L.C.M. of 5, 6, 7, 8 = 840.

 

 Required number is of the form 840k + 3

 

Least value of k for which (840k + 3) is divisible by 9 is k = 2.

 

 Required number = (840 x 2 + 3) = 1683.

Report Error

View Answer Report Error Discuss

Filed Under: HCF and LCM

Q:

If the sum of two numbers is 55 and the H.C.F. and L.C.M. of these numbers are 5 and 120 respectively, then the sum of the reciprocals of the numbers is equal to:

A) 55/601 B) 601/55
C) 11/120 D) 120/11
 
Answer & Explanation Answer: C) 11/120

Explanation:

Let the numbers be a and b.

 

Then, a + b = 55 and ab = 5 x 120 = 600.

 

The required sum =1a+1b = a+bab55600=11120

Report Error

View Answer Report Error Discuss

Filed Under: HCF and LCM

Q:

Which of the following has the most number of divisors?

A) 99 B) 101
C) 176 D) 182
 
Answer & Explanation Answer: C) 176

Explanation:

99 = 1 x 3 x 3 x 11

 

101 = 1 x 101

 

176 = 1 x 2 x 2 x 2 x 2 x 11

 

182 = 1 x 2 x 7 x 13

 

So, divisors of 99 are 1, 3, 9, 11, 33, .99

 

Divisors of 101 are 1 and 101

 

Divisors of 176 are 1, 2, 4, 8, 11, 16, 22, 44, 88 and 176

 

Divisors of 182 are 1, 2, 7, 13, 14, 26, 91 and 182.

 

Hence, 176 has the most number of divisors.

Report Error

View Answer Report Error Discuss

Filed Under: HCF and LCM

Q:

The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively, is:

A) 123 B) 127
C) 235 D) 305
 
Answer & Explanation Answer: B) 127

Explanation:

Required number = H.C.F. of (1657 - 6) and (2037 - 5)

= H.C.F. of 1651 and 2032 = 127.

Report Error

View Answer Report Error Discuss

Filed Under: HCF and LCM

Q:

The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:

A) 74 B) 94
C) 184 D) 364
 
Answer & Explanation Answer: D) 364

Explanation:

L.C.M. of 6, 9, 15 and 18 is 90.

 

Let required number be 90k + 4, which is multiple of 7.

 

Least value of k for which (90k + 4) is divisible by 7 is k = 4.

 

=>Required number = (90 x 4) + 4   = 364.

Report Error

View Answer Report Error Discuss

Filed Under: HCF and LCM

Q:

The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is:

A) 1 B) 2
C) 3 D) 4
 
Answer & Explanation Answer: B) 2

Explanation:

Let the numbers 13a and 13b.

Then, 13a x 13b = 2028

=>ab = 12.

Now, the co-primes with product 12 are (1, 12) and (3, 4).

[Note: Two integers a and b are said to be coprime or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1 ]

So, the required numbers are (13 x 1, 13 x 12) and (13 x 3, 13 x 4).

Clearly, there are 2 such pairs.

Report Error

View Answer Report Error Discuss

Filed Under: HCF and LCM