SOLVED EXAMPLES
Directions: (1-5) Study the following graph and table carefully and answer the questions given below:
Time Taken To Travel (In Hours) By Six Vehicles On Two Different Days
Distance covered (in kilometers) by six vehicles on each day
(IBPS CWE PO MT 2012)
1. Which of the following vehicles travelled at the same speed on both the days?
(1) Vehicle A (2) Vehicle C (3) Vehicle F (4) Vehicle B (5) None of these
Solution: The speed of Vehicle B on both the days is 43 km/hr.
Ans: (4)
191 SBI PO Exam Study Package
2. What was the difference between the speed of vehicle A on day 1 and the speed of vehicle C on the same day?
(1) 7 km/hr (2) 12 km/hr (3) 11 km/hr (4) 8 km/hr (5) None of these
Solution: Speed of A on 1st day = 52 km/hr Speed of C on 1st day = 63 km/hr
∴ Difference = 65 - 52 = 11 km/hr Ans: 2
3. What was the speed of vehicle C on day 2 in terms of meters per second?
(1) 15.3 (2) 12.8 (3) 11.5 (4) 13.8
(5) None of these
Solution: Speed of Vehicle C on 2nd day = 45 km/hr
= 45 × 185 = 2.5 × 5 = 12.5 m/sec Ans: (5)
4. The distance travelled by vehicle F on day 2 was approximately what percent of the distance travelled by it on day 1?
(1) 80 (2) 65 (3) 85 (4) 95 (5) 90
Solution: Reqd % = 636703 × 100 = 90.46 ≈ 90%
Ans: (5)
192 SBI PO Exam Study Package
5. What is the respective ratio between the speeds of vehicle 0 and vehicle E on day 2?
(1) 15: 13 (2) 17: 13 (3) 13: 11 (4) 17: 14
(5) None of these
Solution: Reqd Ratio = Speed of vehicle D on day 2 Speed of vehicle E on day 2
= 5139 = 1713 = 17:13 Ans: (2)
Directions (6-10) Study the following pie-chart and table carefully and answer the questions given below:
Percentagewise distribution of the number of mobile phones sold a shopkeeper during six months
Total number of mobile phones sold = 45000
The respective ratio between the numbers of mobile phones sold of company A and company B during six months
193 SBI PO Exam Study Package
6. What is the respective ratio between the number of mobile phones sold of company B during July and those sold during December of the same company?
(1) 119: 145 (2) 116: 135 (3) 119: 135 (4) 119: 130 (5) None of these
Solution: Total number of mobiles sold in the month of July
= 45000 × 10017 = 7650
Mobile phones sold by Company B in the month of July = 7650 × 157 = 3570
Total number of mobile phones sold in the month of December = 45000 × 10016 = 7200
Mobile phones sold by Company B in the month of December = 7200 × 169 = 4050
∴ Reqd ratio = 35704050 = 357405 = 119135 = 119: 135
Ans: (3)
7. If 35% of the mobile phones sold by company A during November were sold at a discount, how many mobile phones of company A during that month were sold without a discount?
(1) 882 (2) 1635 (3) 1638 (4) 885
194 SBI PO Exam Study Package (5) None of these
Solution: Number of mobile phones sold in the month of November = 45000 × 10012 = 5400
Number of mobile phones sold by Company A in the month of November = 5400 × 157 = 2520
∴ Number of mobile phones without discount in the month of November by Company A
= 2520 ×10065 = 2520 × 0.65 = 1638 Ans: 3
8. If the shopkeeper earned a profit of Rs. 433/- on each mobile phone sold of company B during October, what was his total profit earned on the mobile phones of that company during the same month?
Solution: Number of mobile phones sold in the month of October = 45000 × 1008 = 3600
∴ Number of mobile phones sold by Company B in the month of October = 3600 × 125 = 1500
∴ Total profit earned by Company B in the month of October = 1500 × 433 = 649500
Ans: (4)
9. The number of mobile phones sold of company A during July is approximately what percent of the number of mobile phones sold of company A during December?
(1) 110 (2) 140 (3) 150 (4) 105 (5) 130
Solution: Number of mobile phones sold in the month of July
195 SBI PO Exam Study Package
= 45000 × 10017 = 7650
Number of mobile phones sold by Company A in the month of July = 7650 × 158 = 4080
Number of mobile phones sold in the month of December = 45000 × 10016 = 7200
Number of mobile phones sold by Company A in the month of December = 7200 × 167 = 3150
Solution: Number of mobile phones sold in the month of August = 10022 × 45000=9900
Number of mobile phones sold in the month of September = 10025 ×45000 = 14 ×45000 = 11250 Number of mobile phones sold by Company B in the month of August
= 9900 × 59 = 5500
Number of mobile phones sold by Company B in September= 11250 × 25 = 4500
Total number of mobile phones sold in August and September by Company B
= 5500+4500= 10000 Quicker Method:
Total number of mobile phones sold by Company B in August and September .
22
100 × 45000 × 59+10025 × 4500 × 25 = 10000
196 SBI PO Exam Study Package Ans: (1)
Directions: (11-15) Study the following information and answer the questions that follow:
The graph given below represents the production (in tonnes) and sales (in tonnes) of company a from 2006-2011.
The table given below represents the respective ratio of the production (in tonnes) of Company A to the production (in tonnes) of Company B, and the respective ratio of the sales (in tonnes) of Company A to the sales (in tonnes) of Company B.
11. What is the approximate percentage increase in the production of Company A (in tonnes) from the year 2009 to the production of Company A (in tonnes) in the year 2010?
(1) 18%
(2) 38%
(3) 23%
(4) 27%
(5) 32%
Solution: Production of Company A in year 2009 = 550 Production of Company A in year 2010 = 700
197 SBI PO Exam Study Package Reqd % = 700 −550550 × 100 = 150550 × 100
= 30011 = 27.27 ≈ 27%
Ans: (4)
12. The sale of Company A in the year 2009 was approximately what percent of the production of Company A in the same year?
(1) 65%
(2) 73%
(3) 79%
(4) 83%
(5) 69%
Solution: Sales of Company A in year 2009 = 400 Production of Company A in year 2009 = 550 Reqd % = 400550 × 100 = 80011 = 72.72 ≈ 73%
Ans: (2)
13. What is the average production of Company B (in tonnes) from the year 2006 to the year 2011?
(1) 574 (2) 649 (3) 675 (4) 593 (5) 618
Solution: Average production of Company B
= 600 +700 +800+600+650 +700 6
= 40506 = 675 Ans: (3)
198 SBI PO Exam Study Package
14. What is the respective ratio of the total ‘production (in tonnes) of Company A to the total sales (in tonnes) of Company A?
(1) 81: 64 (2) 64: 55 (3) 71: 81 (4) 71: 55 (5) 81: 55
Solution: Total Production of Company A Total Sales of Company A
= 40502750 = 8155 = 81: 55 Ans: (5)
15. What is the respective ratio of production of Company B (in tonnes) in the year 2006 to production of Company B (in tonnes) in the year 2008?
(1) 2: 5 (2) 4: 5 (3) 3: 4 (4) 3: 5 (5) 1: 4
Solution: Production of Company B in the year 2006
= 150 × 4 = 600
Production of Company B in the year 2008
= 200 × 4 = 800 Ratio = 600800 = 3: 4 Ans: (3)
199 SBI PO Exam Study Package
Permutations
Introduction: In the SBI PO exams the questions based on Permutation have been asked few times and from the exam point of view it is very important to understand the basic concepts of permutation to solve those questions.
Factorial Notation: The product of n consecutive positive integers beginning with one is denoted by n! And read as factorial n.
∴ n! = 1 × 2 × 3 ×……. × n – 1) × (n)
For example, 6! = 1 × 2 × 3 × 4 × 5 × 6 = 720.
0! = 1! = 1.
Permutation: The number of different arrangements which can be made by taking some or all of the given things or objects at a time is called as permutation.
The symbol for permutation of n different things taking r at a time = n P r or n P r = 𝑛 − 𝑟 !𝑛!
For example, 5 P 3 = 5−3 !5! = 5!2!= 1 × 2 × 3 × 4 × 5
1 × 2 = 60.
Here, we will practice few questions to understand more about permutation.
The arrangement of a number of things taking some or all of them at a time is called permutation. If there are ‘n’ number of things and we have to select ‘r’ things at a time then the total number of permutation is denoted by n𝑃𝑟 = 𝑛−𝑟 !𝑛!
For example if there are 3 candidates A ,B and C for the post of president and vice president of a college , since we have to select only 2 candidates , it can be done in 3! Ways. i.e. (A, B) (B, C) (A, C) (B, A) (C, B) and (C, A). Here order of arrangement matters.
Restricted Permutation:
Sometimes we have to find out the number of permutation keeping few specific objects at specific places. In this case, we find out the number of permutation of filling remaining vacant places by the remaining objects.
If r objects are taken out of n dissimilar objects
(i) A specific object is taken each time: if there are n objects 𝑎1 , 𝑎2… … … . 𝑎𝑛 . Suppose that 𝑎1 is taken each time. If 𝑎1 takes first place then the remaining (n-1)
200 SBI PO Exam Study Package
objects can be arranged in n-1𝑃𝑟−1 ways. Since 𝑎1 can take any place so number of permutation is r× n-1𝑃𝑟−1.
(ii) Specific object never taken: then r objects are taken out of (n-1) objects, so number of permutation is n-1𝑃𝑟.
{Note: n𝑃𝑟 = n-1𝑃𝑟 + r× n-1𝑃𝑟−1 }
Permutation of things when some are identical:
If we have n things in which p are exactly of one kind , q of second kind , r of third kind and the rest are different then the number of permutation of n things taken all at a time n𝑃𝑟 = 𝑝!𝑞 !𝑟!𝑛 ! Example: In how many ways can the letters of the word LEADER be arranged?
Solution: The word LEADER contains total 6 letters namely 1L, 2E, 1A, 1D, 1R Therefore, the number of ways to arrange the letters of the word LEADER
= 1!2!1!1!1!6! = 360.
Repetition of things:
The number of permutation formed by taking r things at a time out of n things in any object arrangement such that each object can be taken any number of time is 𝑛𝑟 .
Circular permutation:
If we fix one of the objects around the circumference of a circle then number of permutation of n different thing taken all at a time is (n-1)! Ways. It will be same as by putting (n-1) objects at (n-1) places.
But if we do not consider the direction i.e. clockwise and anticlockwise then the number of permutation is 𝑛−1 !2 .