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Q:

The monthly expenditure of a person is Rs. 6000. The distribution of expenditure on various items is as follows:

Item of expenditure Amount(in Rs.)
Food 2000
Clothing 660
Fuel and rent 1200
Education 480
Miscellaneous 1660

If the above data is represented by a percentage bar diagram of height 15 cm, then what are the lengths of the two segments of the bar diagram corresponding to education and miscellaneous respectively?

A) 1.25 cm and 5 cm	B) 1.2 cm and 4.15 cm
C) 1.2 cm and 3.5 cm	D) 4.15 cm and 6 cm

Answer & Explanation Answer: B) 1.2 cm and 4.15 cm

Explanation:

15 cm corresponds to 6000 rs

Education = 480/6000 * 15 cm = 1.2 cm

Miscellaneous = 1660/6000 * 15 cm = 4.15 cm

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Filed Under: Bar Charts
Exam Prep: Bank Exams

Q:

The data collected from which one of the following methods is not a primary data?

A) By direct personal interviews	B) By indirect personal interviews
C) By schedules sent through enumerators	D) From published thesis

Answer & Explanation Answer: D) From published thesis

Explanation:

Primary data is information that you collect specifically for the purpose of your research project.

An advantage of primary data is that it is specifically tailored to your research needs. A disadvantage is that it is expensive to obtain.

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Filed Under: Quantitative Aptitude - Data Interpretation
Exam Prep: Bank Exams

Q:

Diagrammatic representation of data includes which of the following?

1) Bar diagram

2) Pie-diagram

3) Pictogram

Select the correct answer using the code given below:

A) 1 and 2 only	B) 2 and 3 only
C) 1 and 3 only	D) 1, 2 and 3

Answer & Explanation Answer: D) 1, 2 and 3

Explanation:

All three are types of data representation

Pictogram uses pictures so show different identities with different numbers

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Filed Under: Quantitative Aptitude - Data Interpretation
Exam Prep: Bank Exams

Q:

Consider the following frequency distribution:

x f

8 6

5 4

6 5

10 8

9 9

4 6

7 4

What is the median for the distribution?

A) 6	B) 7
C) 8	D) 9

Answer & Explanation Answer: C) 8

Explanation:

Summation of frequencies = 6+4+5+8+9+6+4 = 42

Median = mid value = average of 21st and 22nd value

Arranging data in increasing order we get

x f

4 6

5 4

6 5

7 4

8 6

9 9

10 8

So mid value i.e 21st and 22nd value = 8

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Filed Under: Average
Exam Prep: Bank Exams

Q:

Consider the following grouped frequency distribution:

x f

0-10 8

10-20 12

20-30 10

30-40 p

40-50 9

If the mean of the above data is 25.2, then what is the value of p?

A) 9	B) 10
C) 11	D) 12

Answer & Explanation Answer: C) 11

Explanation:

Mean = (sum of fixi/ (sum of f) = (8*5 + 12*15 + 10*25 + P*35 +9*45) / (8+12+10+P+9) = 25.2

(875 + 35P)/(39+P) = 25.2

P = 11

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Filed Under: Average
Exam Prep: Bank Exams

Q:

Consider the following statements:

1) The perimeter of a triangle is greater than the sum of its three medinas.

2) In any triangle ABC, if D is any point on BC, then AB + BC + CA > 2AD.

Which of the above statements is/are correct?

A) 1 only	B) 2 only
C) Both 1 and 2	D) Neither 1 nor 2

Answer & Explanation Answer: C) Both 1 and 2

Explanation:

Let ABC be the triangle and D. E and F are midpoints of BC, CA and AB respectively.

Recall that the sum of two sides of a triangle is greater than twice the median bisecting the third side,(Theorem to be remembered)

Hence in ΔABD, AD is a median

⇒

AB + AC > 2(AD)

Similarly, we get

BC + AC > 2CF

BC + AB > 2BE

On adding the above inequations, we get

(AB + AC) + (BC + AC) + (BC + AB )> 2AD + 2CD + 2BE

2(AB + BC + AC) > 2(AD + BE + CF)

∴

AB + BC + AC > AD + BE +CF

2.

To prove: AB + BC + CA > 2AD

Construction: AD is joined

Proof: In triangle ABD,

AB + BD > AD [because, the sum of any two sides of a triangle is always greater than the

third side]

----

1

In triangle ADC,

AC + DC > AD [because, the sum of any two

sides of a tri

angle is always greater than the

third side]

----

2

Adding 1 and 2 we get,

AB + BD + AC + DC > AD + AD

=> AB + (BD + DC) + AC > 2AD

=> AB + BC + AC > 2AD

Hence proved

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Filed Under: Volume and Surface Area
Exam Prep: Bank Exams

Q:

Consider the following inequalities in respect of any triangle ABC:

1) AC –AB < BC

2) BC –AC < AB

3) AB –BC < AC

Which of the above are correct?

A) 1 and 2 only	B) 2 and 2 only
C) 1 and 3 only	D) 1, 2 and 3

Answer & Explanation Answer: D) 1, 2 and 3

Explanation:

AC-AB<BC Or AB+BC>AC

BC-AC<AB Or AB+AC>BC

AB-BC<AC Or AC+BC>AB

Sum of 2 sides of triangle is always greater than the third side

So all three statements are true

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Filed Under: Volume and Surface Area
Exam Prep: Bank Exams

Q:

Consider the following statements:

1) The number of circles that can be drawn through three non-collinear points is infinity.

2) Angle formed in minor segment of a circle is acute.

Which of the above statements is/are correct?

A) 1 only	B) 2 only
C) Both 1 and 2	D) Neither 1 nor 2

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