# EXIM Bank Officer Quantitative Aptitude

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# EXIM Bank Officer Quantitative Aptitude

### Introduction

EXIM Bank Officer Recruitment - Online Exam, conducted in online Mode, has: a duration of 2 hour 30 minutes, a total of 147 questions, a maximum score of 150 marks, and consists of 4 sections, namely - English Language, Quantitative Aptitude, Reasoning Ability and General Knowledge. The 4 sections are separately timed and the questions can be attempted in any order. Candidates must clear the cut-off in all 4 sections to qualify for the EXIM Bank Officer Recruitment Interview. The below sections gives the detailed information about EXIM Bank Officer Quantitative Aptitude.

### Imp Dates

EXIM Bank Officer Quantitative Aptitude - Important Dates
Events Dates
Online Application Starting Date 07.02.2020
Online Application Ending Date 22.02.2020
Written Exam Date 15.03.2020
Interview Date April 2020

### Pattern

Exam Type Subject Names Questions Marks Exam Duration

Objective Type
Quantitative Aptitude 25 25

2 Hours 30 Minutes
Reasoning 25 25
General Knowledge 50 25
English 47 75
Total 147 150

### Syllabus

EXIM Bank Officer Quantitative Aptitude - Administrative Officer and Manager "
EXIM Bank Officer Quantitative Aptitude - Syllabus
Sl.No Subject Syllabus
1. Quantitative Aptitude
• Number Systems. • Discount. • Use of Tables and Graphs. • Averages. • The relationship between Numbers. • Ratio and Proportion. • Time and Work. • Profit and Loss. • Time and Distance. • Percentages. • Ratio and Time. • Computation of Whole Numbers. • Interest • Mensuration. • Decimals and Fractions. • Fundamental arithmetical operations.

### Samples

1. The value of $\frac {\frac{1}{2}\frac{1}{2}of \frac{1}{2}}{\frac{1}{2}\frac{1}{2}of \frac{1}{2}}$
A. 2$\frac{2}{3}$ B. 1 C. 1$\frac{1}{3}$ D. 3

Explanation -
$\frac {\frac{1}{2}\frac{1}{2} * \frac{1}{2}}{\frac{1}{2}\frac{1}{2} * \frac{1}{2}}$
= $\frac {\frac {1}{2} \frac{4}{1}}{\frac{3}{4}}$ = 2 * $\frac{4}{3}$
= $\frac{8}{3}$
= 2$\frac{2}{3}$
2. Taking $\sqrt{2}$ = 1.414, $\sqrt{3}$ = 1.732, $\sqrt{5}$ = 2.236 and $\sqrt{6}$ = 2.449,then the value of $\frac{9 +\sqrt{2}}{\sqrt{5} +\sqrt{3}}$ + $\frac{9 -\sqrt{2}}{\sqrt{5} -\sqrt{3}}$ to three places of decimals is
A. 9.2321 B. 13.716 C. 10.723 D. 15.892

Explanation -
$\frac{9 +\sqrt{2}}{\sqrt{5} +\sqrt{3}}$ + $\frac{9 -\sqrt{2}}{\sqrt{5} -\sqrt{3}}$
= $\frac{1}{2}$[15$\sqrt{5}$- 3$\sqrt{5}$ + 2$\sqrt{6}$]
= $\frac{1}{2}$ [33.540 – 5.196 – 7.898]
= 10.732
1. The monthly income of Shyama and Kamal together is Rs.28000. The income of Shyama and Kamal is increased by 25% and 12.5% respectively. The new income of Kamal becomes 120% of the new salary of Shyama. What is the new income of Shyama?
A. Rs.12000 B. Rs.18000 C. Rs.14000 D. Rs.16000

Explanation:
The monthly income of Shyama and Kamal => S + K = 28000 —(1)
Shyama’s income = x; Kamal’s income = 28000 – x.
K = $\frac{120}{100}$ * S —(2)
S’s new income = (28000 – x)*$\frac{112.5}{100}$
K’s new income = x * $\frac{125}{100}$
(28000 – x)*$\frac{112.5}{100}$ = x * $\frac{125}{100}$
x = 12000
New Income of Shyama = 125% of 12000 = 15000
now percent of water = ($\frac{(80*4 + 60*6)}{1000}$)100 = 68%
2. 500 kg of ore contained a certain amount of iron. After the first blast furnace process, 200 kg of slag containing 12.5% of iron was removed. The percentage of iron in the remaining ore was found to be 20% more than the percentage in the original ore. How many kg of iron were there in the original 500 kg ore?
A. 54.2 B. 58.5 C. 46.3 D. 89.2

Explanation:
Initially ‘x’ kg of iron in 500 kg ore.
Iron in the 200 kg of removed = 200*$\frac{12.5}{100}$= 25 kg.
The percentage of iron in the remaining ore was found to be 20% more than the percentage in the original ore
So $\frac{(x-25)}{300}$ = ($\frac{1205}{100}$)*$\frac{x}{100}$
$\Rightarrow$ x – 25 = $\frac{18x}{25}$
$\Rightarrow$ 7x = 625
$\Rightarrow$ x = 89.2
Direction [1-5]: A school has four sections A, B, C, D of Class IX students.
The results of half yearly and annual examinations are shown in the table given below.
Result No. of Students
Section A Section B Section C Section D
Students failed in both Exams 28 23 17 27
Students failed in half-yearly but passed in Annual Exams 14 12 8 13
Students passed in half-yearly but failed in Annual Exams 6 17 9 15
Students passed in both Exams 64 55 46 76

1. If the number of students passing an examination be considered a criterion for comparison of the difficulty level of two examinations, which of the following statements is true in this context?
A. Half yearly examinations were more difficult. B. Annual examinations were more difficult. C. Both the examinations had almost the same difficulty level. D. The two examinations cannot be compared for difficulty level.

Explanation:
Number of students who passed half-yearly exams in the school
= (Number of students passed in half-yearly but failed in annual exams) + (Number of students passed in both exams)
= (6 + 17 + 9 + 15) + (64 + 55 + 46 + 76)
= 288.
Also, Number of students who passed annual exams in the school
= (Number of students failed in half-yearly but passed in annual exams) + (Number of students passed in both exams)
= (14 + 12 + 8 + 13) + (64 + 55 + 46 + 76)
= 288.
Since, the number of students passed in half-yearly = the number of students passed in annual exams. Therefore, it can be inferred that both the examinations had almost the same difficulty level.
2. How many students are there in Class IX in the school?
A. 336 B. 189 C. 335 D. 430

Explanation:
Since the classification of the students on the basis of their results and sections form independent groups, so the total number of students in the class:
= (28 + 23 + 17 + 27 + 14 + 12 + 8 + 13 + 6 + 17 + 9 + 15 + 64 + 55 + 46 + 76)
= 430.
3. Which section has the maximum pass percentage in at least one of the two examinations?
A. A Section B. B Section C. C Section D. D Section

Explanation:
Pass percentages in at least one of the two examinations for different sections are:
For Section A [$\frac{(14 + 6 + 64)}{(28 + 14 + 6 + 64)}$ x 100] % = [$\frac{84}{112}$ x 100] % = 75%.
For Section B [$\frac{(12 + 17 + 55)}{(23 + 12 + 17 + 55)}$ x 100] % = [$\frac{84}{107}$ x 100] % = 78.5%.
For Section C [$\frac{(8 + 9 + 46)}{(17 + 8 + 9 + 46)}$ x 100] % = [$\frac{63}{80}$ x 100 ] % = 78.75%.
For Section D [$\frac{(13 + 15 + 76)}{(27 + 13 + 15 + 76)}$ x 100] % = [$\frac{104}{131}$ x 100] % = 79.39%.
Clearly, the pass percentage is maximum for Section D.
4. Which section has the maximum success rate in annual examination?
A. A Section B. B Section C. C Section D. D Section

Explanation:
Total number of students passed in annual exams in a section
= [ (No. of students failed in half-yearly but passed in annual exams) + (No. of students passed in both exams) ] in that section
Therefore Success rate in annual exams in Section A
= [$\frac{No. of students of Section A passed in annual exams}{Total number of students in Section A}$ x 100] %
= [$\frac{(14 + 64)}{(28 + 14 + 6 + 64)}$ x 100] %
= [$\frac{78}{112}$ x 100] %
= 69.64%.
Similarly, success rate in annual exams in:
Section B [$\frac{(12 + 55)}{(23 + 12 + 17 + 55)}$ x 100] % = [$\frac{67}{107}$ x 100] % = 62.62%.
Section C [$\frac{(8 + 46)}{(17 + 8 + 9 + 46)}$ x 100] % = [$\frac{54}{80}$ x 100] % = 67.5%.
Section D [$\frac{(13 + 76)}{(27 + 13 + 15 + 76)}$ x 100] % = [$\frac{89}{131}$ x 100] % = 67.94%.
Clearly, the success rate in the annual examination is maximum for Section A.
5. Which section has the minimum failure rate in half yearly examination?
A. A section B. B section C. C section D. D section

Explanation:
Total number of failures in half-yearly exams in a section
= [ (Number of students failed in both exams) + (Number of students failed in half-yearly but passed in Annual exams) ] in that section
Therefore Failure rate in half-yearly exams in Section A
= [$\frac{Number of students of Section A failed in half-yearly}{Total number of students in Section A}$ x 100] %
= [$\frac{(28 + 14)}{(28 + 14 + 6 + 64)}$ x 100] %
= [$\frac{42}{112}$ x 100] %
= 37.5%.
Similarly, failure rate in half-yearly exams in:
Section B [$\frac{(23 + 12)}{(23 + 12 + 17 + 55)}$ x 100] % = [$\frac{35}{107}$ x 100] % = 32.71%.
Section C [$\frac{(17 + 8)}{(17 + 8 + 9 + 46)}$ x 100] % = [$\frac{25}{80}$ x 100] % = 31.25%.
Section D [$\frac{(27 + 13)}{(27 + 13 + 15 + 76)}$ x 100] % = [$\frac{40}{131}$ x 100] % = 30.53%.
Clearly, the failure rate is minimum for Section D
1. Ram and Sunil can finish a work in 8 and 16 hours repectively. If they work at it alternatively for an hour Ram beginning first, the work will be finish in
A. $8 \frac{3}{4}$ B. $10 \frac{2}{3}$ C. $10 \frac{1}{2}$ D. $11 \frac{2}{3}$

Explanation -
$\frac{1}{8} + \frac{1}{16} + \frac{1}{8} + \frac{1}{16}...... = 1 = \frac{2 + 1 + 2+ 1 + ...}{16} = 1$
i.e, $10 \frac{1}{2}$
2. 12 men complete a work in 18 days. 6 days after they had started working, 4 men joined them. How many days will all of them take to complete the remaining work ?
A. 8 days B. 12 days C. 13 days D. 9 days

Explanation -
12 men’s 6 days work = $\frac{6}{18}$$\frac{1}{3}$
i.e, $\frac{1}{\frac{2}{3}}$ = $\frac{12 * 18}{16 * ?}$
? = 9 days
1. A person bought 20 pens at Rs 12 each and sold 5 of them at Rs 10 and other 7 at Rs 14 and remaining at Rs 11. What is the profit/loss percentage ?
A. 1.66% B. 2.33% C. 1.33% D. 2.66%

Explanation:
Total cost price = 20 × 1 = 240
Total selling price = 5 × 10 + 14 × 7 + 8 × 11
= 50 + 98 + 88 = 236
Loss percent = 240 - $\frac{236}{240}$ × 100
= 1.66%
2. Marked price of an article is Rs 300 and a discount of 20% is given. If profit percent is 20% then what is the cost price ?
A. Rs 210 B. Rs 200 C. Rs 180 D. Rs 220

Explanation:
MP = 300
SP = MP - discount
SP = 300 × $\frac{80}{100}$
SP = 240
Cost price = 240 × $\frac{100}{120}$
Cost price = Rs 200
1. A man can reach a certain place in 40 hours. If he reduces his speed by 1/15th, he goes 5 km less in that time. Find the total distance covered by him.
A. 60 km. B. 85 km. C. 75 km. D. 52 km.

Explanation: Let Distance = D km, Speed = S km/h
Time = $\frac{Distance}{Speed}$ - (1)
40 = $\frac{Distance}{Speed}$
After reducing speed,
Distance = Distance – 5 km, Speed = S × $\frac{14}{15}$km/h
Time = $\frac{(Distance – 5)}{\frac{14}{15} ×{Speed}}$ (2)
After solving equation 1 and 2
$\frac{Distance}{Speed}$ = $\frac{(Distance – 5)}{\frac{14}{15} ×{Speed}}$
14 Distance = 15 (Distance – 5)
14 Distance = 15 Distance – 75
Distance = 75 km.
2. A person travelled 132 km by auto, 852 km by train and 248 km by bike. It took 21 hours in all. If the speed of train is 6 times the speed of auto and 1.5 times speed of bike, what is the speed of train?
A. 78 km${h}^{-1}$ B. 104 km${h}^{-1}$ C. 96 km${h}^{-1}$ D. 88 km${h}^{-1}$

Explanation: Let the speed of auto be km${h}^{-1}$ So, the
speed of train will be 6x and that of bike will be
= $\frac{6×}{1.5}$ = 4x
As per the given information, ⇒ $\frac{132}{×}$ + $\frac{852}{6×}$ + $\frac{248}{4×}$ = 21
∴ × = $\frac{336}{21}$ = 16
∴ Speed of the train = 6x = 6 × 16 = 96 km${h}^{-1}$
1. How much should money lender lend at simple rate of interest of 15% in order to have Rs. 3234 at the end of $1 \frac {1} {2}$ years
A. Rs. 1640 B. Rs. 2620 C. Rs. 2610 D. Rs. 2640

Explanation -
Let required money be x
Then $( x + \frac {x * 15} {100} * \frac {3}{2}$ = 3234
$\frac {49 x} {40}$ = 3234
i.e, x = $\frac {3234 * 40} {49}$ = 2640
2. An amount Rs. 8000 becomes Rs. 9200 in 3 years at simple interest. If rate of interest is increased by 3%, it would amount to
A. Rs. 9920 B. Rs. 10560 C. Rs. 11120 D. Rs. 11820

Explanation -
. Principal = Rs. 8000, S.I. = Rs. 1200,
Time = 3 years.
i.e, Rate = $(\frac {100 * 1200} {8000 * 3}) %$ = 5%
New rate = 8%, Principal = Rs. 8000,
Time = 3 years.
S.I = Rs. $(\frac {8000 * 8 * 3} {100})$ = Rs. 1920
i.e, New amount = Rs. (8000 + 1920)
= Rs. 9920
1. A right circular cone is placed over a cylinder of the same radius. Now the combined structure is painted on all sides. Then they are separated now the ratio of area painted on Cylinder to Cone is 3:1. What is the height of Cylinder if the height of Cone is 4 m and radius is 3 m?
A. 5 m B. 6 m C. 8 m D. 10 m
Explanation:
Cylinder painted area = 2 $\pi rh + \pi r²$
Cone painted area = $\pi rl$
$\frac {2h + r}{\sqrt {(r² + h1² )}}$ = 3 : 1
h = 6
2. The diameter of Road Roller is 84 cm and its length is 150 cm. It takes 600 revolutions to level once on a particular road. Then what is the area of that road in m²?
A. 2376
B. 2476 C. 2496 D. 2516
Explanation:
Area
$600 \times 2 \times \frac {22}{7} \times \frac {42}{100} \times \frac {150}{100}$ = 2376
1. A family consists of two grandparents, two parents and three grandchildren. The average age of the grandparents is 67 years, that of the parents is 35 years and that of the grandchildren is 6 years. What is the average age of the family?
A. 28$\frac {4}{7}$ B. 31$\frac {5}{7}$ C. 32$\frac {1}{7}$ D. None of these
Explanation:
Required Average = $\frac {67 \times 2 + 35 \times 2 + 6 \times 3}{2 + 2 + 3}$
$\frac {134 + 170 + 18}{7}$
$\frac {222}{7}$
31$\frac {5}{7}$ years
2. A grocer has a sales of Euro 6435, Euro 6927, Euro 6855, Euro 7230 and Euro 6562 for 5 consecutive months. How much sale must he have in the sixth month so that he gets an average sale of Euro 6500?
A. Euro 4991 B. Euro 5991 C. Euro 6001 D. Euro 6991
Explanation:
Total sale for 5 months = Euro (6435 + 6927 + 6855 + 7230 + 6562) = Euro 34009.
Required sale = Euro [ $(6500 \times 6) - 34009$]
= Euro (39000 - 34009)
= Euro 4991.