Directions (1 – 5): Study the following charts carefully and answer the questions that follow:
5 companies – A, B, C, D and E produce TVs and Mobiles. The total production cost of all 5 companies if Rs 4 crores.
Pie chart shows the %age distribution of total production. The table shows the ratio of production of TVs to Mobiles and % profit on them.

1. Find the total production cost (in crores) of TVs by companies C and D together?
   A) Rs 0.95
   B) Rs 0.81
   C) Rs 0.77
   D) Rs 0.52
   E) Rs 0.45
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option B
   Solution:
   TVs by company C = 5/12 * 15/100 * 4 crores = 0.25 crores
   TVs by company D = 7/10 * 20/100 * 4 crores = 0.56 crores
   So total = 0.25+0.56 = 0.81 crores[/su_spoiler]
2. What is the ratio of profit earned by selling Mobiles by company B to that by selling TVs by company C?
   A) 3 : 7
   B) 5 : 7
   C) 6 : 5
   D) 5 : 8
   E) 6 : 11
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option C
   Solution:
   Profit on mobiles by company B = 24/100 * 5/9 * 18/100 * 4 crore
   Profit on TVs by company C = 32/100 * 5/12 * 15/100 * 4 crore
   So ratio is 24 * 5/9 * 18 : 32 * 5/12 * 15 = 6 : 5 [/su_spoiler]
3. Find the total profit earned (in crores) by selling Mobiles by companies A and E.
   A) Rs 0.23
   B) Rs 0.12
   C) Rs 0.28
   D) Rs 0.28
   E) Rs 0.33
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option A
   Solution:
   Profit by company A by selling mobiles = 20/100 * 5/11 * 22/100 * 4 crores = 0.08 crores
   Profit by company E by selling mobiles = 25/100 * 3/5 * 25/100 * 4 crores = 0.15 crores
   So total = 0.23 crores[/su_spoiler]
4. What is the difference in the profits (in crores) earned by companies A and D by selling TVs?
   A) Rs 0.05
   B) Rs 0.14
   C) Rs 0.12
   D) Rs 0.09
   E) Rs 0.06
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option E
   Solution:
   Profit earned by company A by selling TVs = 30/100 * 6/11 * 22/100 * 4 crores = 0.144 crores Profit earned by company D by selling TVs = 15/100 * 7/10 * 20/100 * 4 crores = 0.084 crores
   So difference = 0.144 – 0.084 = 0.06 crores [/su_spoiler]
5. The production cost of Mobiles by company E is what percent more than the production cost of TVs by company B?
   A) 81.5%
   B) 88.5%
   C) 87.5%
   D) 82.5%
   E) None of these
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option C
   Solution:
   Production cost of Mobiles by company E = 3/5 * 25/100 * 4 = 0.6 crores
   Production cost of TVs by company B = 4/9 * 18/100 * 4 = 0.32 crores
   So required % = (0.6 – 0.32)/0.32 * 100 = 87.5%
   [/su_spoiler]

Directions (6 – 10): Study the following charts carefully and answer the questions that follow:

The bar graph shows the number of students in each section – Medical, Non-Medical and Commerce in schools – A, B, C, D and E

The table shows the % of girls in each section in each school.

6. What is the number of girls in Medical Section from schools A, B and D together?
   A) 356
   B) 451
   C) 461
   D) 268
   E) 308
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option D
   Solution:
   54/100 * 200 + 40/100 * 180 + 55/100 * 160 = 108 + 72 + 88 = 268
   [/su_spoiler]
7. What is the total number of boys in Commerce section from all the schools?
   A) 575
   B) 560
   C) 505
   D) 654
   E) 625
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option A
   Solution:
   55/100 * 180 + 50/100 * 240 + 57/100 * 200 + 61/100 * 200 + 48/100 * 250
   = 99 + 120 + 114 + 122 + 120 = 575[/su_spoiler]
8. What is the difference between the average number of girls in Medical and Non-Medical sections from school C and the average number of girls in Non-Medical and Commerce sections from school E?
   A) 45
   B) 23
   C) 43
   D) 39
   E) 35
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option C
   Solution:
   Number of females in Medical and Non-Medical from school C = 45/100 * 120 + 40/100 * 80 = 54 + 32 = 86, so average = 43
   Number of females in Non-Medical and Commerce from school E = 35/100 * 120 + 52/100 * 250 = 42 + 130 = 172, so average = 86
   Required difference = 86 – 43 = 43[/su_spoiler]
9. The number of girls in Commerce section from school D is approximately what percent of number of boys in Non-Medical section from school D?
   A) 137%
   B) 144%
   C) 156%
   D) 149%
   E) 116%
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option A
   Solution:
   Number of girls in Commerce from school D = 39/100 * 200 = 78
   Number of boys in Non-Medical from school D = 38/100 * 150 = 57
   So required % = 78/57 * 100 = 137% [/su_spoiler]
10. The number of girls in Non-Medical section from school B is approximately what percent less than than the number of girls in same section from schools A and C together?
    A) 17%
    B) 34%
    C) 27%
    D) 54%
    E) 22%
    [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option C
    Solution:
    Number of girls in Non-Medical section from school B = 42/100 * 150 = 63
    Number of girls in Non-Medical section from school A and C = 45/100 * 120 +40/100 * 80 = 86
    So required % = (86 – 63)/86 * 100 = 27%[/su_spoiler]

Mixture and Alligation Questions for Banking and Insurance Exams.

1. A container contains some amount of milk. A milkman adds 200 ml of water for each one litre of milk in the container. 6 litres of the mixture is sold from the container and 10 litres of milk is added to the remaining mixture. If now the ratio of milk to water in container is 25 : 3, find the initial quantity of milk in the container.
   A) 26 l
   B) 29 l
   C) 30 l
   D) 20 l
   E) None of these
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option D
   Solution:
   Let initial quantity of milk = 10x litres, For each 1 litre, 200 ml of water is added, so after adding water, quantity of mixture become = 12x litres
   Now 6 l of mixture is sold, and 10 l of milk is added
   So remaining quantity is (12x – 6 +10) = (12x + 4)
   In this final quantity, milk = 10x – (10x/12x * 6) + 10 = (10x + 5)
   So (10x+5)/(12x +4) = 25/(25+3)
   Solve, x = 2
   So initial quantity of milk = 10x = 20 litres[/su_spoiler]
2. A container contains 64 litres of pure milk. One-fourth of the milk is replaced by water. Again the operation is performed, and one-fourth of mixture is replaced by water. Find the final ratio of milk to water in the container.
   A) 11 : 8
   B) 10 : 7
   C) 9 : 7
   D) 10 : 9
   E) 12 : 7
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option C
   Solution:
   After 2 operations, final quantity of milk = 64 (1 – 1/4)² = 36 litres
   So quantity of water is 64  – 36 = 28 l
   So ratio is 36 : 28 = 9 : 7
   [/su_spoiler]
3. In what ratio do the three varieties of rice costing Rs 6, Rs 8 and Rs 9 per 100 grams should be mixed in order to obtain a mixture costing Rs 84 per kg?
   A) 2 : 3 : 4
   B) 1 : 3 : 6
   C) 1 : 2 : 5
   D) 3 : 4 : 2
   E) None of these
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option B
   Explanation:
   Rs 6, Rs 8 and Rs 9 per 100 grams  means Rs Rs 60, Rs 80 and Rs 90 per kg
   84 is middle number between 80 and 90
   So take ratios as:
   60…………………..90
   ……………84
   6…………………..24
   Ratio is 6 : 24 = 1: 4
   AND
   80…………………..90
   ……………84
   6…………………..4
   Ratio is 6 : 4 = 3: 2
   So final ratio is 1 : 3 : (4+2) = 1: 3: 6 [/su_spoiler]
4. Two containers A and B contain mixture of milk and water such that A contains 40% milk and B contains 22% milk. Some part of mixture in container A is replaced by equal quantity of mixture from container B. How much quantity of the mixture was replaced if final mixture contains 32% milk?
   A) 3/7
   B) 2/5
   C) 7/10
   D) 4/7
   E) 5/9
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option E
   Solution:
   By the method allegation:
   Reaming……………….Replaced
   22………………………….…..40
   …………….…..32
   8……………………..………..10
   So ratio is 8 : 10 = 4 : 5
   So replaced part is 5/(4+5) = 5/9 [/su_spoiler]
5. A container filled of milk-water mixture contains 75% milk. 5 litres of this mixture is replaced by water. Next, 10 l of the mixture is replaced by water. If the final percentage of milk in the container is 54%, find the initial quantity of mixture in the container.
   A) 50 l
   B) 40 l
   C) 60 l
   D) 70 l
   E) 55 l
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option A
   Solution:
   Let initial quantity of mixture = x l
   initial quantity of milk = 0.75x l
   So 0.75x (1 – 5/x) (1 – 10/x) = 0.54 x
   Solve, (x-5)(x-10) = 18x²/25
   Use options to check the answer.
    [/su_spoiler]
6. How much milk (in litres) costing Rs 60 per litres should be mixed with 35 litres of milk costing Rs 84 per litres so that there is a profit of 50% on selling the mixture at Rs 111 per litres?
   A) 25 l
   B) 32 l
   C) 17 l
   D) 36 l
   E) 46 l
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option A
   Solution:
   CP of mixture = 100/150 * 111 = Rs 74
   Let x l of milk to be mixed. So by method of allegation:
   (x)……………..…..(35)
   60…………………..84
   ………….74
   10……………………..14
   So ratio is 10 : 14 = 5 : 7
   So x/35 = 5/7
   x = 25 l  [/su_spoiler]
7. A container whose capacity is 60 l contains milk and water in the ratio 3 : 2. How much quantity of the mixture should be replaced with pure milk so that in the final mixture, ratio of milk to water is 7 : 3?
   A) 22 l
   B) 20 l
   C) 15 l
   D) 17 l
   E) 14 l
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option C
   Solution:
   In 60 l of mixture, milk = 3/5 * 60  = 36 l, so water = 24 l
   Let x litres of mixture is replaced
   So
   Remaining Milk after replacement is = 36 – (3/5)*x + x = 36 + 2x/5
   So  (36 + 2x/5)/60 = 7/10
   Solve, x = 15 l[/su_spoiler]
8. 3 containers having capacities in the ratio 2 : 3 : 1 contain mixture of liquids A and B such that the ratio of A to B in them is 2 : 3, 1 : 4 and 3 : 7 respectively. If all the three containers are emptied in a single container, what will be the ratio of A to B in the final mixture?
   A) 13 : 58
   B) 11 : 54
   C) 22: 13
   D) 17 : 43
   E) None of these
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option D
   Solution:
   2+3 = 5, 1+4 = 5, 3+ 7 = 10
   LCM pf 5, 5, 10 = 10 
   Capacities are in the ratio 2 : 3 : 1  
   Suppose the capacities are 20, 30 and 10
   So A in final mixture is 2/5 * 20 + 1/5 * 30 + 3/10 * 10 = 17
   And B in final mixture is (20+30+10) – 17 = 43
   So final ratio = 17 : 43[/su_spoiler]
9. 12 litres of water is drawn out from a container full of water and replaced by milk. Again 12 litres of mixture are drawn and the container is again filled with milk. The ratio of final quantity of water to milk in the container is 11 : 25. How much did the container hold?
   A) 60 litres
   B) 65 litres
   C) 72 litres
   D) 39 litrers
   E) None of these
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option C
   Solution:
   Let x litres is total capacity of container
   Using formula, amount of water left = x [1 – 12/x]²
   [1 – 12/x]²/x = 25/(25+11)
   Solving we get, x = 72 l [/su_spoiler]
10. There are two mixtures such that they contain 75% milk and 80% milk respectively. Some amount from first mixture is mixed with twice the same amount of second mixture. Find the percentage of milk in the resultant mixture?
    A) 90.2%
    B) 75.9%
    C) 84.5%
    D) 76.3%
    E) 78.3%
    [su_spoiler title=” View Answer ” icon=”chevron-circle”]   Option E
    Solution:
    Let x from first mixture, then 2x form second
    So milk from first = (75/100)*x, milk from second = (80/100)*2x
    So milk in resultant mixture is (75x/100) + (160x/100) = 2.35x
    Total mixture in third is x+2x = 3x
    So % of milk is (2.35x/3x)*100 [/su_spoiler]

Type of Quadratic Equations Questions Asked in All Exams – Bank PO, Insurance and Other related Exams.

Directions: In the following questions, two equations numbered are given in variables x and y. You have to solve both the equations and find out the relationship between x and y. Then give answer accordingly-

1. I. 20x² – 31x + 12 = 0
   II. 3y² – 16y + 16 = 0
   A) x > y
   B) x < y
   C) x ≥ y
   D) x ≤ y
   E) x = y or relation cannot be established
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option B
   Solution:
   20x² – 31x + 12 = 0
   20x² – 16x – 15x + 12 = 0
   So x = 3/4, 4/5
   3y² – 16y + 16 = 0
   3y² – 14y – 4y + 16 = 0
   Gives y = 4, 4/3
   [/su_spoiler]
2. I. 3x² + 22 x + 24 = 0
   II. 2y² – 5y – 12 = 0
   A) x > y
   B) x < y
   C) x ≥ y
   D) x ≤ y
   E) x = y or relation cannot be established
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option E
   Solution:
   3x² + 22 x + 24 = 0
   3x² + 18x + 4x + 24 = 0
   So x = -4/3, -6
   2y² – 5y – 12 = 0
   2y² – 8y + 3y – 12 = 0
   Gives y = -3/2, 4
   [/su_spoiler]
3. I. 2x² – 9x + 4 = 0
   II. 4y² – 13y – 12 = 0
   A) x > y
   B) x < y
   C) x ≥ y
   D) x ≤ y
   E) x = y or relation cannot be established
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option E
   Solution:
   Solution:
   2x² – 9x + 4 = 0
   2x² – 8x – x + 4 = 0
   So x = 4 , 1/2
   4y² – 13y – 12 = 0
   4y² – 16y + 3y – 12 = 0
   Gives y = -3/4, 4
   [/su_spoiler]
4. I. 5x² + 23x + 12 = 0
   II. 5y² – 7y – 6 = 0
   A) x > y
   B) x < y
   C) x ≥ y
   D) x ≤ y
   E) x = y or relation cannot be established
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option D
   Solution:
   5x² + 23x + 12 = 0
   5x² + 20x + 3x + 12 = 0
   So x = -4, -3/5
   5y² – 7y – 6 = 0
   5y² – 10y + 3y – 6 = 0
   So y = -3/5, 2
   Put all values on number line and analyze the relationship
   -4….. -3/5…. 2 [/su_spoiler]
5. I. 7x² + 19x – 6 = 0,
   II. 2y² – 7y + 3 = 0
   A) x > y
   B) x < y
   C) x ≥ y
   D) x ≤ y
   E) x = y or relation cannot be established
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option B
   Solution:
   7x² + 19x – 6 = 0
   7x² + 21x – 2x – 6 = 0
   Gives x = -3, 2/7
   2y² – 7y + 3 = 0
   2y² – 6y – y + 3 = 0
   So y = 1/2, 3
   [/su_spoiler]
6. I. 4x² – 12x + 5 = 0,
   II. 2y² – 19y + 35 = 0
   A) x > y
   B) x < y
   C) x ≥ y
   D) x ≤ y
   E) x = y or relation cannot be established
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option D
   4x² – 12x + 5 = 0
   4x² – 2x – 10x + 5 = 0
   x = 1/2, 5/2
   2y² – 19y + 35 = 0
   2y² – 14y – 5y + 35 = 0
   So y = 5/2, 7
   [/su_spoiler]
7. I. 2x² + 5x ¬– 12 = 0,
   II. 4y² + 13y – 12 = 0
   A) x > y
   B) x < y
   C) x ≥ y
   D) x ≤ y
   E) x = y or relation cannot be established
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option E
   2x² + 5x ¬– 12 = 0
   2x² + 8x ¬– 3x – 12 = 0
   So x = -4 , 3/2
   4y² + 13y – 12 = 0
   4y² + 16y – 3y – 12 = 0
   y = -4, 3/4 [/su_spoiler]
8. I. 3x² + 22x + 24 = 0,
   II. 4y² – 9y – 9 = 0
   A) x > y
   B) x < y
   C) x ≥ y
   D) x ≤ y
   E) x = y or relation cannot be established
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option B
   3x² + 22 x + 24 = 0
   3x² + 18x + 4x + 24 = 0
   Gives x = -4/3, -6
   4y² – 9y – 9 = 0
   4y² – 12y + 3y – 9 = 0
   y = -3/4, 3 [/su_spoiler]
9. I. 20x² – 31x + 12 = 0,
   II. 4y² + 9y – 9 = 0
   A) x > y
   B) x < y
   C) x ≥ y
   D) x ≤ y
   E) x = y or relation cannot be established
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option C
   20x² – 31x + 12 = 0
   20x² – 16x – 15x + 12 = 0
   Gives x = 3/4, 4/5
   4y² + 9y – 9 = 0
   4y² + 12y – 3y – 9 = 0
   y = 3/4, -3 [/su_spoiler]
10. I. 6x² – 7x – 3 = 0,
    II. 4y² + 5y – 6 = 0
    A) x > y
    B) x < y
    C) x ≥ y
    D) x ≤ y
    E) x = y or relation cannot be established
    [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option E
    Solution:
    6x² – 7x – 3 = 0
    6x² + 2x – 9x – 3 = 0
    Gives x = -1/3, 3/2
    4y² + 5y – 6 = 0
    4y² + 8y – 3y – 6 = 0
    Gives y = -2, 3/4
    [/su_spoiler]

Quantitative Aptitude Questions: Word Problems for Bank PO Exams. Type of Quantitative Aptitude Questions asked in IBPS PO/SBI PO/NIACL/NICL/BoB and other exams

1. A train starts from station A at 7:00 AM and reaches station B at 9:00 AM. Another train starts from station B at 7:30 AM and reaches station A at 10:30 AM. At what time the two trains will meet?
   A) 8:10 AM
   B) 8:24 AM
   C) 8:48 AM
   D) 9:08 AM
   E) 9:30 AM
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option B
   Solution:
   Let the distance from A to B is ‘d’ km.
   First train reaches destination in 2 hrs, so speed of first train = d/2 km/hr.
   Speed of second train = d/3 km/hr
   Speed of first train = d/2 km/hr, so after 1/2 hr, i.e. at 7:30 AM, the first train has covered d/4 km.
   The distance left = (d – d/4) km = 3d/4 km. relative speed = d/2 + d/3.
   So time they will meet at = 7:30 AM + (3d/4) / (d/2 + d/3)
   = 7:30 AM + (9/10)*60 = 8:24 AM
   OR Use:
   Time = 7:00AM + [(9:00 – 7:00) * (10:30 – 7:00)]/[(9:00 – 7:00) + (10:30 – 7:30)]
   = 7:00 Am + (2 * 7/2)/(2+5) = 8:24 AM[/su_spoiler]
2. An article is marked 10% above the cost price with marked price being Rs 120. The profit gets increased by Rs 14.8, when the selling price of this article is increased by 2%. What is the original selling price of the article?
   A) Rs 740
   B) Rs 720
   C) Rs 785
   D) Rs 735
   E) Rs 715
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option A
   Solution:
   2% of SP= 14.8
   then 100% of SP= 14.8 x 100/2 =Rs 740 [/su_spoiler]
3. The speed of boat A is 3 km/hr less than the speed of boat B. The time taken by boat A to travel a distance of 18 km downstream is 1 hour more than time taken by B to travel the same distance downstream. If the speed of the current is half of the speed of boat A, what is the speed of boat A?
   A) 4 km/hr
   B) 10 km/hr
   C) 9 km/hr
   D) 7 km/hr
   E) None of these
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option A
   Solution:
   Speed of A = x, then of B = x+3
   Speed of current = x/2
   So downstream speed from boat A = x + x/2 = 3x/2
   downstream speed from boat B = x+3 + x/2 = 3x/2 + 3
   so 18/(3x/2) = 18/(3x/2 + 3) + 1
   36/3x = 36/(3x+6) + 1
   12/x = 12/(x+2) + 1
   12(x+2) = 12x + x(x+2)
   x² + 2x – 24 = 0
   solve, x = 4
   [/su_spoiler]
4. The amount that A and B earns is in the ratio 4 : 3. After a year, A’s salary got increased by 20% and B’s salary by 25%. If now the total of their salaries become Rs 1,02,600, What was the initial salary of A?
   A) Rs 36,000
   B) Rs 48,000
   C) Rs 54,000
   D) Rs 45,000
   E) Rs 56,000
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option B
   Solution:
   4x and 3x
   After year, As salary = (120/100)*4x = 480x/100, Bs salary = (125/100)*3x = 375x/100
   So addition of their salaries = (480x/100) +(375x/100) = 855x/100
   So 855x/100 = 1,02,600
   Solve, x = 12,000
   So As initial salary = 4x = 4*12000
   [/su_spoiler]
5. How many 4 letter words starting with vowel can be formed by the letters of word UPSTREAM such that no letter repeats?
   A) 460
   B) 680
   C) 630
   D) 850
   E) 720
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option C
   Solution:
   There are 3 vowels, so we gave 3 choices for the first letter
   There are total 8 letters in UPSTREAM, and one of them has occupied first place, so 7 choices for second, 6 choices for third, and 5 choices for fourth
   So 3*7*6*5 = 630
   [/su_spoiler]
6. A boat can travel 15 km upstream in 30 minutes. If the speed of the current is 1/3 of the speed of the boat in still water, how much distance can the boat travel downstream in 30 minutes?
   A) 25 km
   B) 60 km
   C) 48 km
   D) 30 km
   E) 36 km
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option D
   Solution:
   A boat can travel 15 km upstream in 30 minutes. So Upstream speed = 15 / (30/60) = 30 km/hr
   Let speed of the boat in still water = x
   Speed of the current = x/3
   Upstream speed = x – x/3 = 2x/3
   But from above Upstream speed = 30. So 2x/3 = 30
   Solve, x = 45
   Speed of the current = 45/3 = 15 and speed of boat = 45
   Downstream speed = 45+15 = 60
   Distance travelled in 30min = 60 * (30/60)= 30 km
   [/su_spoiler]
7. Find the probability that a two-digit number is a multiple of 3 given that it is also a multiple of 5.
   A) 1/4
   B) 3/7
   C) 5/6
   D) 3/5
   E) 1/3
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option E
   Solution:
   Given that it is also a multiple of 5, this means that the numbers divisible by 3 are to be taken from these numbers only (which are divisible by 5)
   There are 90 two-digit numbers.
   Out of which 18 are divisible by 5
   LCM of (3 and 5) = 15, Out of 90 numbers, 6 are divisible by 15 (which are also divisible by 3)
   So 6 are divisible by 3 and 18 are total numbers (which are divisible by 5)
   So required probability = 6/18 = 1/3
   [/su_spoiler]
8. After 2 years the respective SI and CI obtained on a certain sum of money are Rs 360 and Rs 396 respectively. Find the amount of money invested.
   A) Rs 1680
   B) Rs 1700
   C) Rs 1800
   D) Rs 1850
   E) None of these
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option C
   Solution:
   Diff = 396-360 = 36
   r = 2*Diff*100/SI
   So r = 2*36*100/360 = 20%
   Now 36 = Pr²/100²
   Solve, P = 900
   [/su_spoiler]
9. The ratio of incomes of P and Q is 6 : 5 respectively. After one year their respective income by same percentage. The difference in their new incomes is Rs 14,300. If the difference in the old income of A and new income of B is Rs 6500, what is the % increase in incomes?
   A) 15%
   B) 12%
   C) 10%
   D) 8%
   E) 13%
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option C
   Solution:
   6x and 5x
   Suppose, y% increased
   So (100+y)/100 * 6x – (100+y)/100 *5x = 14300
   This gives (100+y)*x = 14,30,000 …………..(1)
   Now also given
   6x – [(100+y)/100 * 5x] = 6500
   Solve, x(20-y) = 1,30,000 ………………………(2)
   Divide both equations
   (100+y)/(20-y) = 11
   Solve, y = 10
   [/su_spoiler]
10. Two trains which are 200 m long each are moving in opposite directions. They cross each other in 8 seconds. If one train is moving one and a half times as fast the other train, then find the speed of the faster train.
    A) 88km/hr
    B) 122 km/hr
    C) 118 km/hr
    D) 108 km/hr
    E) 96 km/hr
    [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option D
    Solution:
    Let speed of slower train = x m/s, then speed of faster train = 1.5x m/s
    Their relative speed becomes = x+1.5x = 2.5x m/s
    So (200+200)/8 = 2.5x
    After solving, speed of slower train, x = 20 m/s
    So speed of faster train = 1.5 × 20 = 30 m/s = 30 × 18/5 km/hr [/su_spoiler]

1. A and B started a business. Their investments were in ratio 6 : 5. After 7 months another partner C joined them and the respective ratio between the investments of A and C was 5 : 6. If after a year a total of Rs 63000 is earned as profit, find the total profit shares A and C together?
   A) Rs 31300
   B) Rs 40500
   C) Rs 46900
   D) Rs 30900
   E) Rs 42300
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option B
   Solution:
   B/A = 5/6 and A/C = 5/6
   So B : A : C = 5*5 : 6*5 : 6*6 = 25 : 30 : 36
   So profit ratio is 25*12 : 30*12 : 36*5 [C’s investment for only 5 months]
   which is 5 : 6 : 3
   So total share of A and B = (6+3)/(5+6+3) * 63000 = Rs 40500 [/su_spoiler]
2. The length and the breadth of a rectangular card are increased by 1 m and 2 m respectively and due to this the area of the card increased by 66 sq. m. But if the length and breadth are decreased by 2 m and 1 m respectively, area is decreased by 51 sq. m. Find the perimeter of the original card.
   A) 75 m
   B) 68 m
   C) 72 m
   D) 60 m
   E) 78 m
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option E
   Solution:
   Let original length = l, breadth = b, so area = lb
   When l and b increased:
   (l+1)(b+2) = lb + 66
   Solve, 2l + b = 64
   When both decreased:
   (l-2)(b-1) = lb – 51
   Solve, l + 2b = 53
   Now solve both equations, l = 25, b = 14
   Perimeter = 2(25+14) [/su_spoiler]
3. When Rs 3500 is lent at 8% per annum, simple interest of Rs 1120 is received after some years. In order to receive Rs 2160 as simple interest on Rs 4500 in same time, the sum will have to be invested at how much rate percent per annum?
   A) 10%
   B) 4%
   C) 5%
   D) 8%
   E) 12%
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option E
   Solution:
   1120 = 3500*8*t/100
   So t = 4 years
   So 2160 = 4500*R*4/100
   R = 12% [/su_spoiler]
4. In a business, A, B and C invested Rs 7000, Rs 8000 and Rs 9000 respectively for 4 months, 6 months and 5 months respectively. In the last 4 months, they again invested money as Rs 9000, Rs 7000 and Rs 8000 respectively. If after a year, the total profit earned is Rs 23,870, what did B got as his share?
   A) Rs 5500
   B) Rs 8280
   C) Rs 8360
   D) Rs 6400
   E) RS 9000
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option C
   Solution:
   7000*4 + 9000*4 : 8000*6 + 7000*4 : 9000*5 + 8000*4
   7*4 + 9*4 : 8*6 + 7*4 : 9*5 + 8*4
   64 : 76 : 77
   So B got = [76/217]*23,870 = 8360[/su_spoiler]
5. If 400 coins consists of 25 paise, 50 paise and rupee coins whose values are in the ratio of 1 : 3 : 6. The number of 50 paise coins is
   A) 180
   B) 150
   C) 120
   D) 100
   E) None of these
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option B
   Solution:
   Let coins be a, b, c resp.
   Then a+b+c = 400
   Now values are x, 3x and 6x
   Then, x = (25/100)*a, so a = 4x
   3x = (50/100)*b, so b = 6x
   And 6x = 1*c, so c = 6x
   So now a+b+c = 400, so 4x + 6x + 6x = 400
   Solve, x = 25
   50 paise coins are b which equals 6x = 6*50 = 150
   [/su_spoiler]
6. How many three digit numbers are there each of which is divisible by 5?
   A) 130
   B) 188
   C) 180
   D) 120
   E) None of these
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option C
   Solution:
   For a number to be divisible by 5, the last digit should be 0 or 5, so two choices for last digit, 9 choices (1-9) for first digit and 10 choices (0-9) for second digit.
   So total = 9*10*2 = 180 [/su_spoiler]
7. Four years ago, the ratio of ages of Amit and Bhuvan was 7 : 11. 5 years hence the respective ratio will be 5 : 7. Find the sum of their present ages.
   A) 58
   B) 62
   C) 69
   D) 52
   E) 72
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option B
   Solution:
   7x and 11x
   4+5 years now
   So (7x+9)/(11x+9) = 5/7
   Solve, x = 3
   So Amit’s present age = 7x + 4 = 25 and Bhuvan’s = 11x+4 = 37
   So sum = 25+37 [/su_spoiler]
8. Rs 9404 is divided among A, B, and C, so that if the amounts being invested at 4% simple interest, the amounts received after 2, 3 and 4 yrs by A, B, and C respectively is equal. Find the share of A?
   A) Rs 2,458
   B) Rs 3,248
   C) Rs 2,922
   D) Rs 3,220
   E) None of these
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option B
   Solution:
   9404 will be divided in ratio –
   1/[100+rt1] : 1/[100+rt2] : 1/[100+rt3]
   1/[100+4*2] : 1/[100+4*3] : 1/[100+4*4]
   1/108 : 1/112 : 1/116
   = 812 : 783 : 756
   So A gets 812/(812+783+756) * 9404 [/su_spoiler]
9. A and B are pipes such that A can empty the tank in 40 minutes and B can fill in 15 minutes. The tank is full of water and pipe A is opened. If after 10 minutes, pipe B is also opened, then in how much total time the tank will be full again?
   A) 12 minutes
   B) 6 minutes
   C) 16 minutes
   D) 10 minutes
   E) None of these
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option C
   Solution:
   Emptying pipe A is opened first for 10 minutes, so in 10 minutes the part of tank it has emptied is (1/40)*10 = 1/4
   Now filling pipe is also opened, now since only 1/4 of the tank is empty so 1/4 is only to be filled by both pipes, let it take now x minutes, so
   (1/15 – 1/40)*x = 1/4
   Solve, x= 6
   So total = 10+6 = 16 minutes [total time is asked – 10 minutes when emptying pipe was only opened, 6 minutes when both were operating.] [/su_spoiler]
10. A job that can be completed by 3 men and 18 women in 10 days will be completed by 5 men and 30 women in how many days?
    A) 6
    B) 3
    C) 5
    D) Cannot be determined
    E) None of these
    [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option A
    Solution:
    3m + 18w = 10 D
    Or 3(1m + 6w) = 10D
    Or 1m + 6w = 30D
    So 5(1m + 6w) = 30/5
    Or 5m + 30w = 6D [/su_spoiler]

Data Interpretation Questions with Solutions
Directions (1 – 5): Study the following charts carefully and answer the questions that follow:
The table shows the % of defective items detected out of total items sold.

1. What is the approximate percent decrease in number of defective products in years 2014 as compared to previous year?
   A) 43%
   B) 53%
   C) 40%
   D) 76%
   E) 58%
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option B
   Solution:
   number of defective products in years 2014 = 2/100 * 32,000 = 640
   number of defective products in years 2013 = 4/100 * 34,000 = 1360
   so required % = (1360-640)/1360 * 100 [/su_spoiler]
2. What is the total number of non-defective products in years 2012 and 2014 together?
   A) 52890
   B) 55280
   C) 54320
   D) 43280
   E) 43390
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option B
   Solution:
   92/100 * 26,000 + 98/100 * 32,000 [/su_spoiler]
3. What is the ratio of defective products sold in year 2011 to non-defective products sold in year 2015?
   A) 20 : 441
   B) 13 : 758
   C) 25 : 644
   D) 78 : 239
   E) 23 : 783
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option C
   Solution:
   5/100 * 25,000 : 92/100 * 35,000 [/su_spoiler]
4. In 2013, selling price of each product is Rs.350. If a penalty of Rs.20 each for first 1000 defective products, Rs.40 each for next 1000 defective products, Rs.60 each for next 1000 defective products, and so on is charged, how much is the total penalty which is to be given by company in 2013?
   A) Rs.43,300
   B) Rs.24,400
   C) Rs.34,400
   D) Rs.45,400
   E) Rs.44,300
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option C
   Solution:
   Number of defective products in 2013 = 4/100 * 34,000 = 1360
   So Rs.20 for first 1000 which gives 20*1000 = Rs.20,000
   And Rs.40 for first 360 which gives 40*360 = Rs.14,400
   So total penalty = 20000+14400[/su_spoiler]
5. What is the difference in non-defective products sold in year 2011 and defective products sold in year 2015?
   A) 21,950
   B) 20,590
   C) 29,490
   D) 20,950
   E) None of these
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option D
   Solution:
   95/100 * 25,000 – 8/100 * 35,000 [/su_spoiler]

Directions (6 – 10): Study the following table carefully and answer the questions that follow:
The table shows the dimensions, cost of fencing and cost flooring different fields which are different in shape.

6. Find the cost of flooring triangular field.
   A) Rs 5200
   B) Rs 4900
   C) Rs 5000
   D) Rs 4800
   E) Rs 5300
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option D
   Solution:
   Area of triangle = 1/2 * (16 * 12) = 96 m²
   So Cost of flooring = 96 * 50 = 4800[/su_spoiler]
7. What is the difference between the cost of fencing of rectangular and square fields?
   A) Rs 180
   B) Rs 160
   C) Rs 200
   D) Rs 150
   E) Rs 220
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option A
   Solution:
   Perimeter of rectangular field = 2*(10+20) = 60 m
   So Cost of fencing it = 60 * 15 = Rs 900
   Perimeter of square = 4 *15 = 60 m
   So, cost of fencing it = 60 *18= Rs 1080
   Required difference = 1080 – 900 = 180
   [/su_spoiler]
8. Find the ratio of cost of flooring to that of fencing of parallelogram field?
   A) 5 : 1
   B) 7 : 1
   C) 8 : 1
   D) 9 : 1
   E) 10 : 1
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option C
   Solution:
   Area = Base * Height = 20 * 12 = 240 m²
   So cost of flooring = 240 * 60 = Rs 14400
   Perimeter = 2 * (20+12) = 64 m
   So cost of Fencing = 64 * 25 = Rs 1600
   So ratio = 14400 : 1600 = 9 : 1
   [/su_spoiler]
9. The cost of fencing of circular field is approximately what per cent of the cost of flooring of square field?
   A) 16.59%
   B) 13.44%
   C) 14.56%
   D) 12.45%
   E) 15.36%
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option E
   Solution:
   Perimeter of circle = 2π r = 2 * 22/7 *10 = 440/7 m
   So Cost of Fencing it = 440/7 * 22 = Rs 1382.85
   Area of square = 15² = 225m²
   So Cost of flooring it = 225 * 40 = Rs 9000
   Required % = 1382.85/9000 * 100 = 15.36%
   [/su_spoiler]
10. The cost of fencing of square field is what per cent of the cost of fencing of parallelogram field?
    A) 87.54%
    B) 67.5%
    C) 72.13%
    D) 54.36%
    E) 46.5%
    [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option B
    Solution:
    Cost of fencing square field= Rs 1080
    Cost of fencing parallelogram field= Rs 1600
    Required % = 1080/1600 * 100 = 67.5%
    [/su_spoiler]

Data Interpretation Questions for SBI PO Prelims, NIACL Assistant, NICL AO and other upcoming exams. Data Interpretation for PO Prelims

Direction (1-5): The table shows the number of people visiting six different banks and % of men, women and children visiting these banks.

1. Number of men visiting bank D forms approximately what percent of the total number of people visiting all the banks together?
   A) 6.2%
   B) 7.5%
   C) 8.4%
   D) 5.7%
   E) 9.5%
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option B
   Solution:
   Number of men visiting bank D = 41/100 * 55500 = 22755
   Total number of people visiting all banks = 303600
   So required % = 22755/303600 * 100 [/su_spoiler]
2. Number of children visiting bank C forms approximately what percent of number of children visiting bank F?
   A) 110%
   B) 96%
   C) 102%
   D) 126%
   E) 121%
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option A
   Solution:
   Number of children visiting bank C = 20/100 * 45640 = 9128
   Number of children visiting bank F = 14/100 * 59650 = 8351
   Required % = (9128/8351)*100 = 109.30% [/su_spoiler]
3. What is the total number of children visiting banks B and D together?
   A) 15 : 32
   B) 18 : 37
   C) 12 : 19
   D) 23 : 44
   E) Cannot be determined
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option E
   Solution:
   Children visiting bank B = 20/100 * 65900
   Children visiting bank D = 33/100 * 55500
   So Visiting banks B and D = 13180 + 18315 = 31495 [/su_spoiler]
4. What is the average number of women visiting all the banks together?
   A) 24823.5
   B) 22893.5
   C) 23823.5
   D) 24827.5
   E) 24293.5
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option A
   Solution:
   Total number of women visiting banks is = 55/100 * 34560 + 43/100 * 65900 +   45/100 * 45640 +   26/100 * 55500 + 70/100 * 42350 +   62/100 * 59650 = 148941
   So average = 148941/6 = 24823.5     [/su_spoiler]
5. What is the respective ratio of number of women visiting bank A to those visiting bank C?
   A) 1098 : 1141
   B) 1056 : 1198
   C) 1023 : 1141
   D) 1056 : 1141
   E) None of these
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option D
   Solution:
   Required ratio = 19008 : 20538 = 1056 : 1141 [/su_spoiler]

Direction (6-10): Study the following pie-chart to answer the following questions.

The pie-chart shows the percentage wise distribution of employees in six different organizations

6. The number of employers in organization B is approximately what per cent of the total number of employees in organizations D and E together?
   A) 68%
   B) 49%
   C) 53%
   D) 35%
   E) 45%
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option B
   Solution:
   Required % = 17/(6+29) * 100 = 49% [/su_spoiler]
7. If 25% of the employees in organization C are females, what is the number of male employees in organization C?
   A) 925
   B) 894
   C) 934
   D) 856
   E) 912
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option E
   Solution:
   In C = 19% of 6400 = 1216
   So number of males = 75% of 1216 = 912 [/su_spoiler]
8. The difference between the total number of employees in organizations A, B and C together and the total number of employees in organizations D, E and F together is exactly equal to the number of employees of which organization?
   A) A
   B) B
   C) C
   D) D
   E) E
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option D
   Solution:
   Difference = (D+E+F)% – (A+B+C)% = (53 – 47%) = 6% which is D% [/su_spoiler]
9. If one-thirty sixth of the employees from organization F are post-graduates and salary of each employee is Rs 96,000, what will be the total salary of all the post-graduate employees together from organization F?
   A) 30.72 lakh
   B) 36.56 lakh
   C) 28.60 lakh
   D) 32.64 lakh
   E) None of these
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option A
   Solution:
   Employees in F = 1152
   Post-grads in F = 1/36 * 1152 = 32
   So their total salary = 32*96000 = 30.72 lakh  [/su_spoiler]
10. What is the average number of employees in organizations A, C, D and F together?
    A) 758
    B) 864
    C) 936
    D) 818
    E) None of these
    [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option B
    Solution:
    Average = (704+1216+384+1152)/4 = 3456/4 = 864 [/su_spoiler]

Click here for more Data Interpretation for PO Prelims sets

Direction (1-5): Read the following table to answer the following questions. Some values are missing in the table. Calculate the values as per given data in questions.  [ASKED in IBPS PO Mains 2016]

1. If the total profit in 2014 is Rs 98,000, then find the ratio of the profit of B in 2013 to the profit of A in 2014.
   A) 22 : 15
   B) 25 : 16
   C) 27 : 13
   D) 30 : 17
   E) 28 : 15
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option C
   Solution:
   Profit of A + B = (98000 – 42000) = Rs 56000
   So (A+15000) : 21000 = 56000 : 42000
   Solve A = A’s investment = Rs 13000
   So ratio of investments of A : B is 13000 : 15000 = 13 : 15
   Divide their profit Rs 56000 in this ratio
   So A’s profit = 13/28 * 56000 = Rs 26000
   So required ratio = 54000 : 26000 = 27 : 13 [/su_spoiler]
2. In year 2015 total investment of A and B is Rs 30,000 and A and B invested their amount for 6 months and 5 months respectively then find the number of months for which C invested his amount.
   A) 8 months
   B) 5 months
   C) 7 months
   D) 4 months
   E) 6 months
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option E
   Solution:
   A*6 / B*5 = 30000 / 375000
   So A/B = 2 : 3
   Divide their total 30,000 in ratio 2 : 3
   So A’s investment = Rs 12,000, B’s = Rs 18,000
   So 12000*6/16000*x = 30000 / 40000
   OR
   18000*5/16000*x = 37500/40000
   Solve, x = 6 months [/su_spoiler]
3. The ratio of total profit earned by B in year 2012 to the profit earned by him in the year 2016 is
   A) 15 : 32
   B) 18 : 37
   C) 12 : 19
   D) 23 : 44
   E) Cannot be determined
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option E
   Solution:
   Profit of B in 2016 cannot be determined from the given data. [/su_spoiler]
4. Total investment made by A,B and C in 2013 was Rs53,000. Profit earned by A in 2013 is approximate what % less than the investment made by C in 2013 ?
   A) 35%
   B) 25%
   C) 28%
   D) 32%
   E) 38%
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option B
   Solution:
   Total investment of A and C is 53000 – 18000 = Rs 35000
   So ratio in shres of profit of B : (A+C) is 18 : 35
   Now profit of B is Rs 54000
   So 18/(18+35) *x = 54000
   So x = total profit = Rs 1,59,000
   So that of A = (159000 – (54000+60000)) = Rs 45000
   So required % = (60000-45000)/60000 * 100 = 25%[/su_spoiler]
5. Total profit earned by all in 2016 is Rs 3,50,000 and the ratio of investment made by A and B together and investment made by B and C together is 45 : 49. Then find the difference between the profit made by C and A in 2016.
   A) Rs 22,000
   B) Rs 28,000
   C) Rs 30,000
   D) Rs 20,000
   E) None of these
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option D
   Solution:
   profit earned by all in 2016 is Rs 3,50,000
   (21000+24000/(24000+C) = 45/49
   So, C’s investment is Rs 25,000
   So ratio of their profit sharing is 21 : 24 : 25
   So difference in profits of A and C is (25-21)/(21+24+25) * 350000 = Rs 20000[/su_spoiler]

Direction (6-10): Study the following pie-charts to answer the following questions.

Students from 7 different schools – A, B, C, D, E, F and G were selected to give a competitive exam out of which some students passed the exam as given.

6. What percentage of students passed the exam from school E out of a total number of students enrolled from the same school?
   A) 68.5%
   B) 60.5%
   C) 73.5%
   D) 65.5%
   E) 75.5%
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option B
   Solution:
   [7% of 6400/8% of 9250] * 100 [/su_spoiler]
7. Which schools has the highest percentage of students passed to students enrolled?
   A) E
   B) A
   C) C
   D) F
   E) G
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option D
   Solution:
   A = [13% of 6400/12% of 9250] * 100 = 75%
   B = [16% of 6400/16% of 9250] * 100 = 69%
   C = [18% of 6400/18% of 9250] * 100 = 69%
   D = [12% of 6400/14% of 9250] * 100  = 69%
   E = [7% of 6400/8% of 9250] * 100 = 60.5%
   F = [25% of 6400/22% of 9250] * 100 = 79%
   G = [9% of 6400/10% of 9250] * 100  =   [/su_spoiler]
8. The number of students passed from schools A and D together is what percent less than the number of students enrolled from schools C and E together by
   A) 45%
   B) 27%
   C) 45%
   D) 39%
   E) 33%
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option E
   Solution:
   Students passed from schools A and D = (13+12)% of 6400 = 1600
   Students enrolled from schools C and E = (18+8)% of 9250 = 2405
   The required % is (2405-1600)/2405 * 100 = 33% [/su_spoiler]
9. What is the percentage of students passed to the students enrolled for schools B and C together?
   A) 69%
   B) 78%
   C) 75%
   D) 62%
   E) None of these
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option A
   Solution:
   Total passed from B and C = (16+18)% of 6400 = 2176
   Total enrolled in B and C = (16+18)% of 9250 = 3145
   So required % = 2176/3145 * 100 [/su_spoiler]
10. What is the average of difference between the number of students enrolled from schools D and G together and the number of students passed from schools E and F together?
    A) 75
    B) 86
    C) 93
    D) 81
    E) None of these
    [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option B
    Solution:
    Number of students enrolled from schools D and G = 1295 + 925 = 2220
    Number of students passed from schools E and F = 448+1600 = 2048
    So difference = 2220 – 2048 = 172
    So average = 172/2 = 86 [/su_spoiler]

1. There are five cans containing milk. There is 200l of milk in first can. The second can contains 20% more milk than the third can, which contains 25% more milk than in first can. The fourth can contains 350 kg which is 30% less than the milk in fifth can. What is the average of the weight of milk in all cans?
   A) 260 l
   B) 290 l
   C) 340 l
   D) 320 l
   E) None of these
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option D
   Solution:
   In 1st can = 200 l
   So in 3rd can = (125/100) * 200 = 250
   In 2nd can = (120/100)*250 = 300
   In 4th can = 350
   So in 5th can = (100/70)*350 = 500
   So average = (500+350+300+250+200)/5 = 320 l
   [/su_spoiler]
2. Two stations A and B are 287 km apart. Two trains started their journey from the two stations towards each other at 10 PM. After three and half hours they crossed each other. If the speed of the faster train is 6 km/hr more than the speed of the slower train, then find the speed of the faster train.
   A) 44 km/hr
   B) 38 km/hr
   C) 40 km/hr
   D) 46 km/hr
   E) 35 km/hr
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option A
   Solution:
   Relative speed of both trains = 287/ 3.5 = 82 km/hr
    If they were moving at the same speed, means each travelling at 82/2 = 41 km/hr
   But difference in speed of trains is 6 km/hr.
   So add 6/2 = 3 to 41 for speed of faster train and subtract 3 from 41 for speed of slower train.
   So faster train’s speed = 41+3 = 44 km/hr, and of slower = 41-3 = 38 km/hr
   [/su_spoiler]
3. A shopkeeper keeps the MP of a product at Rs 100. He gives a discount of 10% on the article, and then also gains 35% on it. If the article is sold for Rs 30 less than the MP, what will be the profit%?
   A) 10%
   B) 4%
   C) 5%
   D) 8%
   E) 12%
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option C
   Explanation:
   Using formula: MP = (100+p%)/(100-d%) * CP
   MP = 100, p = 35%, d = 10%,
   So CP = Rs 200/3
   SP of article (at Rs. 30 less than MP) = Rs. 70
   Profit % = (70-200/3)/(200/3) × 100 = 5%  [/su_spoiler]
4. Pipe A and pipe B can fill the tank in 4 hours and 6 hours respectively while pipe C can empty the tank in 4 hours. Both pipes A and B are opened to fill the empty tank. Pipe C was to be opened when ½ of the tank would be filled. But it was opened when 3/4^th of the tank gets filled by pipes A and B. Find the difference in times in which the tank gets filled by the two processes.
   A) 36 minutes
   B) 54 minutes
   C) 45 minutes
   D) 50 minutes
   E) 58 minutes
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option B
   Solution:
   Case 1:
   Time taken to fill the half tank by A and B:
    2 * (1/4 + 1/6) = 5/6 so
   Time taken to fill the half tank by A and B = 6/5 hours
   Time taken by A,B and C to fill rest half of the tank:
   2* (1/4 + 1/6 – 1/4) = 1/3. So 3 hours
   Total time = 6/5 + 3 = 4 hours 12 min
   Case 2:
   Time taken to fill 3/4 tank by A and B:
    4/3 * (1/4 + 1/6) = 5/9 so
   Time taken to fill the half tank by A and B = 9/5 hours
   Time taken by A,B and C to fill rest 1/4 of the tank:
   4/1 * (1/4 + 1/6 – 1/4) = 2/3. So 3/2 hours
   Total time = 9/5 + 3/2 = 3 hours 18 min
   So,  difference in time = 54 minutes [/su_spoiler]
5. The average age of 4 children of a family is 8 years. If the age of their father and mother is also included, then the average age of whole family becomes 21 years. If the age of father is 4 years more than that of the mother, find the age of mother.
   A) 45 years
   B) 43 years
   C) 49 years
   D) 40 years
   E) 48 years
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option A
   Solution:
   Sum of age of children = 4*8 = 32 years
   32 + M + F = 21 × 6 = 126.
   M+F = 94 and F= M+ 4. So, M = 45
    [/su_spoiler]
6. In a company, the number of employees in departments A and B is in the ratio 25 : 4. If 5 more employees join the company and the number of employees in department B was 2 less than earlier, then the ratio of the number of employees in departments A and B becomes 22 : 3. Find the total number of employees in the company originally?
   A) 104
   B) 177
   C) 183
   D) 145
   E) None of these
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option D
   Solution:
   In A = 25x, B = 4x
   If 5 more join so becomes 25x+4x+5 = 29x + 5
   Then A/B = 22/3
   So
   [(29x +5) – (4x-2)]/(4x-2) = 22/3
   Solve, x = 5
   So total employees = 29x = 29*5 = 145  [/su_spoiler]
7. The monthly income of A and B is in the ratio 8 : 9. If the income of A increases by 50% and that of B decreases by 25%, the new ratio of incomes become 16 : 9. Find A’s monthly income.
   A) Rs 38,000
   B) Rs 42,000
   C) Rs 35,000
   D) Rs 40,000
   E) Cannot be determined
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option E
   Solution:
   8x and 9x
   A’s income = 150/100 * 8x = 12x
   B’s income = 75/100 * 9x = 27x/4
   And new ratio is 16 : 9 which we found above too. So cant be found [/su_spoiler]
8. A and B invested Rs 16,000 and Rs 12,000 respectively in a business. After 3 months A withdrew Rs 5000 and B added Rs 5000. After 3 more months, another partner C joins them with Rs 21,000. What is the difference in profit shares of B and C if there is a profit of Rs 26,400 at the end of year?
   A) Rs 3150
   B) Rs 3600
   C) Rs 3200
   D) Rs 3250
   E) None of these
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option B
   Solution:
   Ratio of shares of A : B : C is
   16000*3 + 11000*9 : 12000*3 + 17000*9 : 21000*6
   49 : 63 : 42
   So required difference = (63-42)/154 * 26400 = Rs 3600   [/su_spoiler]
9. Out of the total monthly salary Kirti spends 20% on her food items. Out of the remaining salary she spends 25% on furniture, 15% on transport, 15% on medicine and 10% on entertainment. What is her monthly salary if she is left with Rs 9800?
   A) Rs 35000
   B) Rs 28000
   C) Rs 33000
   D) Rs 36000
   E) None of these
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option A
   Solution:
   Let total salary is Rs 100
   After 20% spent, remaining = 100-20 = 80
   Now 25% spent so 25/100 * 80 = 20
   Now 15% spent so 15/100 * 80 = 12
   Now 15% spent so 25/100 * 80 = 12
   Now 10% spent so 10/100 * 80 = 8
   So remaining is 100 – (20+20+12+12+8) = Rs 28
   So
   when 28 left, income = 100
   when 9800, income = 100/28 * 9800 = 35000  [/su_spoiler]
10. The odds against an event are 2 : 3 and the odds in favor of another independent event are 2 : 5. Find the probability that one of the event will occur.
    A) 17/28
    B) 19/35
    C) 11/25
    D) 9/14
    E) 19/28
    [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option B
     Solution:
    Let 2 events A and B
    Odds against A are 2 : 3
    So probability of occurrence of A = 3/(2+3) = 3/5. And non-occurrence of A = 2/5
    Odds in favor of B are 2 : 5
    So probability of occurrence of B = 2/(2+5) = 2/7. And non-occurrence of B = 5/7
    Case 1: A occurs and B does not occur
    So probability = 3/5 * 5/7 = 3/7
    Case 2: B occurs and A does not occur
    So probability = 2/7 * 2/5 = 4/35
    So probability that one will occur = 3/7 + 4/35 = 19/35 [/su_spoiler]

With the changing pattern in exam, One should be ready for each type of questions. So we are providing you with questions on Cubic Equations.

Directions: In the following questions, two equations numbered are given in variables x and y. You have to solve both the equations and find out the relationship between x and y. Then give answer accordingly –

1. I. x³ + 7x² + 16x + 12 = 0,
   II. y³ – 2y² – 5y + 6 = 0
   A) If x > y
   B) If x < y
   C) If x ≥ y
   D) If x ≤ y
   E) If x = y or relation cannot be established
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option D
   Solution:
   In these equations, you will have to first satisfy the equation with some root.
   Like start with 1: Put x = 1 in x³ + 7x² + 16x + 12 = 0 and see if it equals to 0. It does not. Next move on to -1, again not, next put 2 and then -2
   -2 satisfies the equation: -8 + 28 – 32 + 12 = 0. Since -2 satisfies the equation so (x+2) is a factor of this equation.
   So x = -2, -3
   Next divide x³ + 7x² + 16x + 12 = 0 by (x+2) . It gives x² + 5x + 6 which is a quadratic equation. Its roots are -2 and -3
   Now same with y³ – 2y² – 5y + 6 = 0
   Putting y = 1, satisfies the equation, so (y-1) is a factor of this equation> Now divide y³ – 2y² – 5y + 6 = 0 by (y-1) gives y² – y – 6 = 0 which is a quadratic equation with roots -2 and 3
   So y = 1, -2, 3
   Now put on number line
   -3……-2…..1……3
   so x ≤ y [/su_spoiler]
2. I. x³ – x² – 2x = 0,
   II. y³ – 7y – 6 = 0
   A) If x > y
   B) If x < y
   C) If x ≥ y
   D) If x ≤ y
   E) If x = y or relation cannot be established
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option E
   Solution:
   x³ – x² – 2x = 0 becomes
   x (x² – x – 2) = 0
   So x = 0 and (x² – x – 2) = 0
   Second is a quadratic equation with roots -1 and 2
   So x = 0, -1, 2
   Now y³ – 7y – 6 = 0
   Put y = 1 – Does not satisfy.
   Put y = -1 – Satisfies the equation so (y+1) is a factor
   Divide y³ – 7y – 6 = 0 by (y+1) gives y² – y – 6 = 0
   So y = -1, -2, 3
   Put on number line
   -2……-1..…0……2……3 so no relation [/su_spoiler]
3. I. x³ – 6x² + 11x – 6 = 0,
   II. y³ + 9y² + 26y + 24 = 0
   A) If x > y
   B) If x < y
   C) If x ≥ y
   D) If x ≤ y
   E) If x = y or relation cannot be established
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option A
   Solution:
   For x³ – 6x² + 11x – 6 = 0, x = 1 satisfies the equation so (x-1) is a factor
   Divide x³ – 6x² + 11x – 6 = 0 by (x-1) => gives x² – 5x + 6 = 0
   Gives x = 1, 2, 3
   Similarly for second equation, y = -2, -3, -4
   put on number line
   So x > y
   [/su_spoiler]
4. I. x³ + x² – 4x – 4 = 0,
   II. y³ – 7y² + 14y – 8 = 0
   A) If x > y
   B) If x < y
   C) If x ≥ y
   D) If x ≤ y
   E) If x = y or relation cannot be established
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option E
   Solution:
   x = -1 satisfies x³ + x² – 4x – 4 = 0
   So (x+1) is a factor. On dividing gives x² – 4 = 0
   So x = -1, 2, -2
   Similarly for second equation
   y = 1, 2, 4[/su_spoiler]
5. I. 3x³ – 13x² + 18x – 8 = 0,
   II. 3y³ + 14y² – 32 = 0
   A) If x > y
   B) If x < y
   C) If x ≥ y
   D) If x ≤ y
   E) If x = y or relation cannot be established
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option E
   Solution:
   x = 1 satisfies 3x³ – 13x² + 18x – 8 = 0 so (x-1) is a factor. On dividing gives 3x² – 10x + 8 = 0 which is quadratic equation with roots x = 2, 4/3
   So x = 1, 2, 4/3
   Similarly with second equation, y = -2, -4, 4/3
   [/su_spoiler]
6. I. 2x³ + 15x² + 13x – 30 = 0,
   II. 2y³ + 22y² + 5y – 18 = 0
   A) If x > y
   B) If x < y
   C) If x ≥ y
   D) If x ≤ y
   E) If x = y or relation cannot be established
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option E
   Solution:
   x = -6, -5/2, 1
   y = -9/2, -2, 1
   Put on number line
   -6 … -9/2 …. -5/2 …. -2…. 1
   [/su_spoiler]
7. I. 3x³ + 25x² + 68x + 60 = 0,
   II. 3y³ + 5y² – 4y – 4 = 0
   A) x > y
   B) x < y
   C) x ≥ y
   D) x ≤ y
   E) x = y or relationship cannot be determined
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option D
   Solution:
   x = -10/3, -3, -2
   y = -2, -2/3, 1
   Put on number line
   -10/3….-3…. -2…. -2/3….1 [/su_spoiler]
8. I. x³ + 2x² – 19x – 20 = 0,
   II. 2y³ + 15y² + 28y + 15 = 0
   A) If x > y
   B) If x < y
   C) If x ≥ y
   D) If x ≤ y
   E) If x = y or relation cannot be established
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option E
   Solution:
   x = -5, -1, 4
   y = -5, -3/2, -1
   Put on number line
   -5…. -3/2…-1….. 4[/su_spoiler]
9. I. 5x³ – 12x² + x + 6 = 0,
   II. 5y³ + 28y² + 35y + 12 = 0
   A) If x > y
   B) If x < y
   C) If x ≥ y
   D) If x ≤ y
   E) If x = y or relation cannot be established
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option C
   Solution:
   x = -3/5, 1, 2
   y = -4, -1, -3/5
   Put on number line
   -4…..-1…. -3/5…1. 2[/su_spoiler]
10. I. 2x³ – 13x² + 22x – 8 = 0,
    II. 2y³ + 9y² + 3y – 4 = 0
    A) x > y
    B) x < y
    C) x ≥ y
    D) x ≤ y
    E) x = y or relationship cannot be determined
    [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option C
    Solution:
    x = 2 satisfies 2x³ – 13x² + 22x – 8 = 0. On dividing gives 2x² – 9x + 4 = 0
    x = 4 , 1/2, 2
    y = -1 satisfies 2y³ + 9y² + 3y – 4 = 0. On dividing gives 2y² + 7y – 4 = 0
    y = -4, 1/2, -1
    Put on number line
    -4……-1….. 1/2……2…. 4[/su_spoiler]