Hello Aspirants. Welcome to Online Quantitative Aptitude section in AffairsCloud.com. Here we are creating question sample From all topics , which are Important for upcoming IBPS and RRB exams. We have included Some questions that are repeatedly asked in exams !!

**Study the following graph carefully and answer the questions given below.**

**Production of three types of cars by a company over the years(in lakhs).**

**What was the percentage drop in the number of C type cars manufactured from 2006 to 2007**?

A. 10

B. 11

C. 15

D. 25

E. 5

###### Answer

**A.**Required percentage drop=[(25-22.5)/25]x100=10

**What was the difference between the number of B types car manufactured in 2007 and 2008?**

A. 1,00,000

B. 20,00,000

C. 10,00,000

D. 12,50,000

E. None of these

###### Answer

**E.**Required difference=(35-27.5)lakh=7,50,000

**The total numbers of all the three types cars manufactured was the least in which of the following years?**

A. 2008

B. 2009

C. 2006

D. 2007

E. 2005

###### Answer

**B.**Adding all types in all years we can see that 2009 has least numbers of cars manufactured.

**In which of the following years was the percentage production of type B to type C cars the maximum?**

A. 2005

B. 2009

C. 2008

D. 2007

E. 2006

###### Answer

**C.**Production of B type cars is more than the production of C type cars only in 2006,2007 and 2008.We see the largest difference exists in 2008.

**The total production of C type car in 2005 and 2006 together was what percentage of production of B type cars in 2007?**

A. 150

B. 250

C. 300

D. 200

E. 50

###### Answer

**D.**Required percentage=[(30+25)/27.5]x100=200

**A person cross a stationary bus in 21second.The same bus crosses another person standing over there in 9seconds.What is the ratio of the speed of the bus to the first person ?**

A. 3:4

B. 3:7

C. 2:3

D. 7:3

E. Cant be determined

###### Answer

**D.**let the length of the bus be L.

then speed of the first person

=L/21units/sec.

speed of the bus=L/9 units/sec

Ratio=L/9:L/21

=7:3**6 children and 6 women can do an assignment in a duration of 6 days.In how many days 20women can alone complete the given assignment if 9 children alone can complete the assignment in 10 days?**

A. 6.4

B. 8

C. Can’t be determined

D. 7.5

E. None of these

###### Answer

**E.**9children alone do in-10days

6children alone do in- (10×9)/6=15days

6children in 6days do-6/15=2/5 part

Rest 3/5 part will be completed by 6women

in 6days.

6 women do in 1 day-3/(5×6)=1/10part

6women alone do in-10days

=>20women can do in- (6×10)/20=3days

**One of the angles of triangle is three-fifth of the sum of the adjacent angles of a parallelogram.Remaining angles of the triangle are in the ratio 5:4.What is the value of the smallest angle of triangle?**

A. 40

B. 32

C. 108

D. 72

E. None of these

###### Answer

**B.**One of the angles=(3/5)x180=108(sum of the

adjacent angles of a parallelogram=180)

Sum of Remaining angles=180-108=72

Smallest angle=(4/9)x72=32

**Rahul’s salary increases every year by 15% in August.If there is no other increse or reduction in the salary and salary in August 2014 was 30,000 what was his salary(approximately? in August 2011?**

A. 22,685

B. 19,725

C. 21,465

D. 20,895

E. 23,125

###### Answer

**B.**Salary in August 2014=30,000

Salary in August 2011=30,000x

(1.15×1.15×1.15)=19725.48

**3 peons and 4 sweepers can earn 756 in 7days.11peons and 13 sweepers work for 8 days to earn in 3008. For earning 2480 how much time will be required by 7peons and 9 sweepers?**

A. 8Days

B. 11Days

C. 9Days

D. 7Days

E. 10Days

###### Answer

**E.**Let Peons be P and sweepers be S.

(3P+4S) in one day earn 756/7=108

(11P+13S) in one day earn 3008/8=376

By earning 1rs per day we can conclude

(3P+4S)/108=(11P+13S)/376

=>P:S=5:3–(i)

(3P+4S) in 1day earn 108

=>(3+4x(3/5))P in 1 day earn 108

or 27m/5=108

=>1P in 1 days earns (108×5)/27=20

thus a S earns 20x(3/5)=12daily

7P+9S earn (7×20+9×12)=248

So to earn 2480 to have work 10 days.

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