GMAT Question of the day

The number of solution of √(x + 3 )+ √ x = 1, is
(a) two
(b) one
(c) no solution.
(d) None of these

Consider x > 0 ,then

√(x + 3) > √ 3 > 1, so the value of expression will be more than 1,

so no solution is possible.

GMAT Question of the day

If (a + 1/a)^2 = 3, then a^3 + 1/a^3 equals
(a) 0
(b) 3√3
(c) 10√3/3
(d) 6√3

(a + 1/a)^2 = 3 or (a + 1/a) = 3^(1/2) and we know that
(a + 1/a)^3 = a^3 + 1/a^3 + 3(a + 1/a)
⇒ (a3 +1/a3) = 3√ 3 – 3/3 = 0

GMAT Question of the day

What is the smallest possible radius of a circle such that it is possible to
place 6 points on the circumference with an integer distance between any
two?
(a) 1/π
(b) 2/π
(c) 6/π
(d) 3/π

The smallest radius means the smallest circumference.
The circumference is 2πr.
With six points equidistant with distance between them to be integer,
this integer has to be the smallest, i.e., 1.
Therefore, circumference is 6 units.
Therefore, 2πr = 6 or r = 3/π.

GMAT Question of the day

A husband and wife, started painting their house, but husband left painting 5 days before the completion of the work. How many days will it take to complete the work, which the husband alone would have completed in 20 days and wife in 15 days?

a)40/7
b)50/7
c)75/7
d)55/7

HUSBAND takes 20 days to complete one unit of work

So , H take 20 days = 1

W take 15 days = 1

H one day work = 1/20 and W one day work = 1/15

H + W one day work = 1/20 + 1/15 = 7/60.

Since husband left painting 5 days before the completion of the work.

In a total of X day H+W worked for X-5 and W alone for 5

(7/60) (X-5) + 1/15 * 5 = 1 therefore X = 75/7

Hence the answer is option C

GMAT Question of the day

How many positive integers less than 10,000 are such that the product of their digits is 210?
(A) 24
(B) 30
(C) 48
(D) 54
(E) 72

210 = 2*3*5*7.

So the numbers have to be made of above four digits.

If we take all 4 primes as separate digits, then 4*3*2*1 = 24 different numbers.

We can also make numbers from the digits 6 (2*3), 5 and 7 = 3*2*1 = 6 different numbers so total 30 numbers.

Hence the answer is B

GMAT Question of the day

ABCDE is a regular pentagon with F at its center. How many different triangles can be formed by joining 3 of the points A,B,C,D,E and F?
(A) 10
(B) 15
(C) 20
(D) 25
(E) 30

6 points are available to make triangles.
A combination of any 3 will make a unique triangle so:
6C3 = 20
And hence the answer is option C.

GMAT Question of the day

A cone of radius 14 cm height 15 cm is cut in a plane parallel to its base. If
the area of the circle at the intersection is 154 sqm then what is the height
from the base at which the cone is cut?

(a) 10cm
(b) 5 cm
(c) 15 cm
(d) 7.5 cm

Since area of circle with BC as radius is 154 sqcm
Π(BC)2 = 154
BC = 7 cm.
Triangles ABC and ADE are similar triangles.
AB/AD = BC/DE.

As BC = 7 and DE is 14 cm
BC/DE = 1/2
AB/AD must also be 1/2.
As AD = 15, AB must be 7.5 cm.
hence the answer is D

GMAT Question of the day

The chord RS of length 8 cm , of a circle with center C,cuts one of the
diameter PQ in a point T such that CT=TQ, If RT = 6,then the diameter of
the circle is
(a) 14 cm
(b) 8 cm
(c) 16 cm
(d) None of these

Let radius of the circle be r

PT = 3r/2 and TQ = r/2.
(3r/2) (r/2) = 6 x 2 = 12.
3r2/4 = 12
⇒ r2 = 16.
r = + 4.

Hence the diameter of the circle is 8 cm. But it is impossible to draw a chord of length 8 cm (other than the diameter) in a circle of diameter 8 cm. The circle must be imaginary.

GMAT Question of the day.

How many three digit odd nos. can be formed from the digits : 2, 0, 3, 5?(repetition not allowed)

(a) 8
(b) 4
(c) 12
(d) 6

Simple way to solve the issue would be the place that is more restricted should be filled first.
Unit digit place can be filled in two ways (by 3 or 5 only) while the
hundred's place can be filled in three ways (by 2 or 3 or 5).
Therefore starting from the unit's place we can form the nos. in 2 x 2 x 2
ways.

GMAT Question of the Day

Which of the two numbers is greater, 2^300 or 3^200 ?
(a) 2^300
(b) 3^200
(c) Both are equal
(d) Not possible with out calc

We can write 2^300 as (2^3)^100 and 3^200 as (3^2)^100.

So compare 8^100 and 9^100.

Hence answer is B.