GMAT QUANTITATIVE REASONING PRACTICE QUESTIONS - POLISH YOUR MATH SKILLS

The Quantitative section of the Graduate Management Admission Test® measures the candidates’ ability to solve quantitative problems, interpret graphic data and the ability to reason quantitatively.

Are you prepared for the GMAT Quantitative section? Try some practice questions prepared by Jamboree's faculty below!

Ques 1:

In the figure above, AB and CD are diameters of the circle with centre as O. AEB is an arc of the circle with centre as D and AFB is an arc of the circle with centre as C. If AB = 20cm, what is the area of the shaded region?

1. 50

2. 100

3. 150

4. 200

5. 200

Answer:

Considering the figure as under:

If we connect the diameter AB with the point D, it will create right angled triangle ADB with right angled at D {using the property, if we connect the points on the circumference with the diameter it will always create right angled triangle} Now, first we will find the area of the sector AEBD and then from the sector AEBD we will subtract the area of triangle ABD which will give us the area of the segment AEBO.

Since, OB = OD {radius of the circle}

Thus, using Pythagoras theorem in the right angle triangle ?OBD,

BD2 = OB2 + OD2

OR, BD2 = 102 + 102

OR, BD = 10√2

Since, AEB is the arc of the circle with centre at D, thus BD will be the radius of this arc.

Hence, Area of sector AEBD = πr2θ/360

= π(10√2)290/360 = 50π

Now, Area of the triangle ABD = 1/2 × AB × OD

= 12 × 20 × 10 = 100

Thus, area of the segment AEBO = Area of sector AEBD – Area of triangle ABD

=50π - 100

Similarly, we can find the area of segment AFBO which will also be,

= 50π - 100

Thus, total area of the segment AEBF = Area of segment AEBO + Area of segment AFBO

= 50π - 100 + 50π - 100

= 100π - 200

Hence, the Area of the shaded region = Area of the complete circle – Area of the segment AEBF

Thus, Area of shaded region will be,

= π102 - (100π - 200) = π100 - π100 + 200 = 200 

Ques 2.

S is a set of n integers, where 0 < n < 11. If the arithmetic mean of set S is a positive integer b, which of the following could NOT be the median of set S?

1. 0

2. b

3. (-b)

4. n/5

5. 5n/11

Answer:

S is a set of n integers, where 0 < n < 11.

Arithmetic mean of Set S is a positive integer ‘b’.

We have to find which of the following could not be the median of Set S.

Now, We know that if the number of terms are odd, then there will be only one middle term and that will be the median, which will be an Integer value.

(A) And, if the number of terms are even, then there will be two middle terms, and the median will be the average of those two middle terms, in the form of a+b2, where a & b will be those two middle terms.

(F) So, the median can be either an integer or can be in the form of I2, where I will be an integer.

So, option (A), (B) & (C) can be the median because all three are integers.

(A) Option (D), (n/5) can also be the median because 0 < n < 11, so as n can be a multiple of 5, so n/5 can be an integer, hence it can be the median.

But, option (E), 5n11 cannot be either an integer or in the form of I/2, as n is not a multiple of 11. Hence, this cannot be the median.

The correct answer is E.

Ques 3.

In a sequence, an + 2 = an + 1 – an + (n + 2) for all n ? 1, where an + 2 represents the (n + 2)th term. If the second and fourth terms of the

sequence are 2 and 5 respectively then what is the value of the 6th term of the sequence?.

• 5

• 7

• 8

• 9

• None of these

Ans

For n≥1,

an+2 = an+1 - an + n+2 ……Equation(1)

Putting value of n = 2 in the above equation, we get:

a4 = a3 - a2 + 2 + 2

In the question, it’s also mentioned that a2 = 2 , a4 = 5. Putting in the above equation, we get the value of a3 = 3

Now by putting value of n = 3 in equation(1), we get:

a5 = a4 - a3 + 3 + 2

a5 = 5 - 3 + 3 + 2

a5 = 7

Now by putting value of n = 4 in equation(1), we get:

a6 = a5 - a4 + 4 + 2

a6 = 7 - 5 + 4 + 2

a6 = 8

The correct answer is C.

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