# RBI Assistant Mains Quantitative Aptitude

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# RBI Assistant Mains Quantitative Aptitude

### Introduction

RBI Assistant Mains Quantitative Aptitude - RBI Assistant 2020 Mains Examination, conducted in Online Mode, has: a duration of 2 Hour 15 Minutes, a total of 200 questions, and a maximum score of 200 marks, and, consists of 5 sections, namely: English Language, Numerical Ability, Computer Knowledge, Reasoning Ability and General/Financial Awareness. Candidates must clear the cut-off in all 5 sections to qualify for the RBI Assistant Language Proficiency Test (LPT). The below sections gives detailed information about RBI Assistant Mains Examination.

### RBI Assistant Recruitment Important Dates

Event Date
RBI Assistant Apply Online Start Date 23-12-2019
RBI Assistant 2019 Apply Online Last Date 16-01-2020
Last Date to pay the Application Fee 16-01-2020
Download of call letters for Online examination – Preliminary 1 week prior to the exam
RBI Assistant Exam Date 2019 14th Feb, 15th Feb 2020
RBI Assistant Prelims Result Date 2019 Will Update soon!!!
RBI Assistant Mains Exam Date 2019 March 2020
Download of call letters for Online examination – Mains Will Update Soon!!!
RBI Assistant Mains Result Date Will Update Soon!!!

### Exam Pattern

RBI Assistant Mains Quantitative Aptitude - Mains Exam Pattern
S.No Name Of Test No.of Questions Max Marks Duration
1. General English 40 40 30 Minutes
2. Computer Knowledge 40 40 30 Minutes
3. Quantitative Aptitude 40 40 30 Minutes
4. Reasoning Ability 40 40 25 Minutes
5. General/Financial Awareness 40 40 20 Minutes
Total 200 200 Marks 135 Minutes

### Syllabus

RBI Assistant Mains Quantitative Aptitude - Quantitative Aptitude
RBI Assistant - Quantitative Aptitude
S. No. Name of the Topic
1. Number system
2. Geometry
3. Algebra
4. Permutation and Combination
5. Mixtures and Allegations
6. Ratio & Proportion
7. Logarithm
8. Time & Work
9. Functions
10. Trigonometry
11. Coordinate Geometry
12. Mensuration

### Samples

1. 9, 8, 15, ?, 175, 874
A. 42 B. 24 C. 38 D. 44

Answer - Option D
Explanation - $9 \times 1 – 1$ = 8
$8 \times 2 – 1$ = 15
$15 \times 3 – 1$ = 44
2. 1, 9, 36, 100, 225, ?
A. 441 B. 484 C. 400 D. 289

Answer - Option A
Explanation -
1, 9, 36, 100, 225, (441)
${1}^{2}, {3}^{2}, {6}^{2}, {10}^{2}, {15}^{2}, {21}^{2}$
(3 – 1) = 2
(6 – 3) = 3
(10 – 6) = 4
(15 – 10) = 5
(X – 15) = 6
X = 21
Required term = ${21}^{2}$ = 441
1. In a right angled triangle, two sides are of the same length. Which of the options is one of the angles of that triangle?
A. 30 B. 45 C. 60 D. 75

Answer - Option B
Explanation - If two sides of the right angled triangle are equal, it is a right angled isosceles triangle. So, the angles of the triangle are 45, 45 and 90.
2. A rope makes 125 rounds of a cylinder with base radius 15 cm. How many times can it go round a cylinder with base radius 25 cm?
A. 100 B. 75 C. 80 D. 65

Answer - Option B
Explanation - Let the required number of rounds be x.
More radius, less rounds (Inverse Proportion)
Radius in cm = Round
15 = 125
25 = x $\Rightarrow x = \frac {(15 \times 125)}{25} = 75$ rounds
1. In Champions league, Rohit scored an average of 120 runs per match in the first 3 match and an average of 140 runs per match in the last four match. What is Rohit’s average runs for the first match and the last two match if his average runs per match for all the five match is 122 and total number of matches are 5?
A. 100 B. 200 C. 150 D. 50

Answer - Option A
Explanation - Rohit’s average score in the first 3 exams = 120
Let the scores in the 5 exams be denoted by M1, M2, M3, M4, and M5
M1 + M2 + M3 = 120 × 3 = 360 .......(i)
Average of last 4 match = 140
$\frac{M2 + M3 + M4 + M5}{4}$= 140
M2 + M3 + M4 + M5 = 560 ......(ii)
Average of all the exams
$\frac{M1 + M2 + M3 + M4 + M5}{5}$=122
M1 + M2 + M3 + M4 + M5 = 122 × 5 = 610 .......(iii)
From solving above equation, we get M1 + M4 + M5 = 300
Required average runs =$\frac{300}{3}$= 100
2. A man goes up hill with an average speed at 20 kmph and comes down with an average speed of 30 kmph. The distance traveled in both the cases being the same, the average speed for the entire journey is
A. 18 kmph B. 22 kmph C. 24 kmph D. 26 kmph

Answer - Option C
Explanation - Thus, u = 20kmph and v = 30 kmph
It is given that distance in both the journey is same.
So, average speed for the entire journey
= $\frac{2uv}{u + v}$km/hr
= $\frac{2 * 20* 30}{20 + 30}$kmph
= $\frac{1200}{50}$km/hr = 24 kmph.
1. In how many different ways can the letters of the word "CHARGES" be arranged in such a way that the vowels always come together?
A. 1440 B. 720 C. 360 D. 240

Answer - Option A
Explanation - The arrangement is made in such a way that the vowels always come together.
i.e., "CHRGS(AE)".
Considering vowels as one letter, 6 different letters can be arranged in 6! ways; i.e., 6! = 720 ways.
The vowels "AE" can be arranged themselves in 2! ways; i.e.,2! = 2 ways
Therefore, required number of ways = 720 x 2 = 1440 ways.
2. In how many different ways can the letters of the word "COMPLAINT" be arranged in such a way that the vowels occupy only the odd positions?
A. 1440 B. 43200 C. 1444 D. 5420

Answer - Option B
Explanation - There are 9 different letters in the given word "COMPLAINT", out of which there are 3 vowels and 6 consonants.
Let us mark these positions as under:
[1] [2] [3] [4] [5] [6] [7] [8] [9]
Now, 3 vowels can be placed at any of the three places out of 5 marked 1, 3, 5, 7 and 9.
Number of ways of arranging the vowels = 5P3 = 5x4x3 = 60 ways.
Also, the 6 consonants at the remaining positions may be arranged in 6P6 ways = 6! ways = 720 ways.
Therefore, required number of ways = 60 x 720 = 43200 ways.
1. A jar contains ‘x’ liters of Milk, a seller withdraws 25 liter of it and sells it at Rs.20 per liter. He then replaces it water. He repeated the process total three times. Every time while selling he reduces selling price by Rs.2. After this process Milk left in the mixture is only 108 liters so he decided to sell the entire Mixture at Rs. 15 per liter. Then how much profit did he earned if bought Milk at Rs.20 per liter?
A. 50 B. 70 C. 90 D. 100

Answer - Option B
Explanation - Seller sells Milk at Rs.20,18 and 16 respectively for three times
= 25*(20+18+16) = 1350
108 = x(1-25/100) 3
x =256 liter
He sold entire 256 at Rs.15 =256*15 = 3840
Cost price = 256*20 = 5120
profit = 5190-5120 = 70
2. ‘X’ Liters of the mixture contains Milk and Water in the ratio 4:3. If 13 liters of Water is added then the ratio becomes 1:1. Then what is the final quantity of water in the mixture?
A. 39 B. 52 C. 56 D. 72

Answer - Option B
Explanation - $\frac {4x}{3x+13} = 1$
x = 13
Water = 3x + 13 = 39 + 13 = 52
1. The ratio of Nicotine to Heroine in four types of drugs of equal quantity is 2 : 3, 3 : 7, 4: 11 and 11 :9 respectively. The four drugs are mixed together. What is the ratio of Nicotine to Heroine after mixing?
A. 213 : 91 B. 418 : 189 C. 91 : 149 D. 149 : 81

Answer - Option C
Explanation - Let the quantity of each drugs be ‘a’
Then the quantity of nicotine after mixing is
($\frac{2}{5} + \frac{3}{10} + \frac{4}{15} + \frac{11}{20}$)a
= ($\frac{2 × 12 + 3 × 6 + 4 × 4 + 11 × 3}{60}$)a
= ($\frac{91}{60}$)a
and quantity of heroine after mixing is
($\frac{3}{5}$ + $\frac{7}{10}$ + $\frac{11}{15}$ + $\frac{9}{20}$) a
= ($\frac{3 × 12 + 7 × 6 +11× 4 + 9 × 3}{60}$) a
= ($\frac{149}{60}$) a
⇒ the required ratio of nicotine to heroin after mixing
= $\frac{\frac{91}{60}a}{\frac{149}{60}a}$
= 91 : 149
2. In a certain school, the ratio of boys to girls is 7 : 5. If there are 2400 students in the school, then how many girls are there?
A. 500 B. 700 C. 800 D. 1000

Explanation: Let the number of boys and girls are 7x and 5x, respectively.
Given, the total number of students = 2400
⇒ 7x + 5x = 2400 ⇒ 12x = 2400
∴ x = 200
∴ Required number of girls = 5x = 5 × 200 = 1000.
1. A can do a piece of work in 120 days and B can do it in 150 days. They work together for 20 days. Then A leaves and B continues the work. 12 days after that, C joins the work and the work is completed in 48 more days. In how many days C can do it alone?
A. 225 B. 220 C. 230 D. 160

Answer - Option D
Explanation - A and B per day work = $\frac {1}{120} + \frac {1}{150} = \frac {27}{1800}$
A and B work in 20 days = $\frac {20 \times 27}{1800}$
B work in 12 days = $\frac {12}{150} = \frac {144}{1800}$
Remaining work = 1 - $(\frac {540}{1800} + \frac {144}{1800})$
= $\frac {1116}{1800}$
B and C, 1 day work = $\frac {1116}{\frac {1800}{48}}$
C per day work = $\frac {1116}{1800 \times 48} - \frac {1}{150}$
$\frac {540}{1800 \times 48} = \frac {1}{160}$
So no. of days by C to complete the work = 160
2. A and B can do a piece of work in 10 days, B and C in 15 days and C and A in 20 days. They all work at it for 6 days, and then A leaves, and B and C go on together for 4 days more. If B then leaves, how long will C take to complete the work?
A. 20 B. 25 C. 10 D. 15

Answer - Option C
Explanation - Amount of work done by A, B and C together in aday is $\frac {6 + 4 + 3}{2 \times 60} = \frac {13}{120}$
Work done by all in 6 days = $\frac {13}{120}$
Work done by B and C in 4 days = $\frac {4}{15}$
Remaining work = 1 - $(\frac {13}{20} + \frac {4}{15})$
= $\frac {1}{2}$, which is to be done C
Now, from the question,
C alone can do the whole work in $\frac {\frac {120}{18}\times {10}}{10 - \frac {120}{18}}$ = 120 days
therefore, $\frac {1}{12}$ of the work is done by C $\frac {120}{12}$ = 10 days
1. If tan A – tan B = x and cot B – cot A = y, then cot (A – B) is equal to:
A. $\frac{1}{x} + y$ B. $\frac{1}{xy}$ C. $\frac{1}{x} - \frac{1}{y}$ D. $\frac{1}{x} + \frac{1}{y}$

Answer - Option D
Explanation - Given that
tan A – tan B = x ...(i)
and cot B – cot A = y ...(ii)
Now, cot (A – B) = $\frac{1}{tan (A - B)}$
= $\frac{1 + tan A tan B}{tan A - tan B}$
= $\frac{1}{tan A - tan B)} + \frac{tan A tan B}{tan A - tan B}$
= $\frac{1}{x} + \frac{1}{y}$ [from (i) and (ii)]
2. If cos θ = $\frac{1}{2}(x + \frac{1}{x})$ then $\frac{1}{2}({x}^{2} + \frac{1}{{x}^{2}})$ is equal to:
A. sin 2θ B. cos 2θ C. tan 2θ D. sec 2θ

Answer - Option D
Explanation - Given that cos θ = $\frac{1}{2}(x + \frac{1}{x})$ = $x + \frac{1}{x}$ = 2 cosθ
We know that ${x}^{2} + \frac {1}{{x}^{2}}$ = ${x}^{2} + \frac {1}{{x}^{2}}$
= ${2cos θ}^{2} -2 = {4 cos θ}^{2}$ -2
= 2cos 2θ [from (i)]
i.e, $\frac{1}{2}({x}^{2} + \frac {1}{{x}^{2}})$ =$\frac{1}{2} * {x}^{2} 2cos 2θ$
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. 5m B. 6m C. 8m D. 10m

Answer - Option B
Explanation - Cylinder painted area = 2πrh+πr²
Cone painted area = πrl
2h+r/√ (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. 2516 D. 2496

Answer - Option A
Explanation - 600*2*22/7*42/100*150/100=2376