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Q:

The cube root of .000216 is:

A) .6	B) .06
C) 77	D) 87

Answer & Explanation Answer: B) .06

Explanation:

${(. 000216)}^{\frac{1}{3}} = {(\frac{216}{10^{6}})}^{\frac{1}{3}} = {[\frac{6 * 6 * 6}{10^{2} * 10^{2} * 10^{2}}]}^{\frac{1}{3}} = [\frac{6}{10^{2}}] = 0.06$

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Filed Under: Square Roots and Cube Roots

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.

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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 = $\frac{1}{a} + \frac{1}{b}$ = $\frac{a + b}{a b}$ = $\frac{55}{600}$ = $\frac{11}{120}$

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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.

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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.

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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.

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Filed Under: HCF and LCM

Q:

Three number are in the ratio of 3 : 4 : 5 and their L.C.M. is 2400. Their H.C.F. is:

A) 40	B) 80
C) 120	D) 200

Answer & Explanation Answer: A) 40

Explanation:

Let the numbers be 3x, 4x and 5x.

Then, their L.C.M. = 60x.

So, 60x = 2400 or x = 40.

The numbers are (3 x 40), (4 x 40) and (5 x 40).

Hence, required H.C.F. = 40.

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Filed Under: HCF and LCM

Q:

The H.C.F. of two numbers is 23 and the other two factors of their L.C.M. are 13 and 14. The larger of the two numbers is:

A) 276	B) 299
C) 322	D) 345

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