1. How many words can be made out of the letters of the word HEADING such that all vowels occur together?
   A) 120
   B) 144
   C) 124
   D) 136
   E) 720
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option E
   Solution:
   We have H, D, N, G and EAI
   So 5! For these 5 letters (take EAI as single letter because they are to be taken together)
   Now arrangement of EAI is 3!
   So total 5! * 3! = 120 × 6 = 720
   [/su_spoiler]
2. 2 persons same at same time from point A and reach point B at 6 PM and 9 PM respectively. The average speed of one is 25 km/hr and that of another is 15 km/hr respectively. What should be there average speed (approximate) such that both reach at 8 PM?
   A) 15 km/hr
   B) 20 km/hr
   C) 17 km/hr
   D) 16 km/hr
   E) 18 km/hr
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option C
   Solution:
   Let distance from A to B is ‘d’ and time taken to reach at 8 PM is ‘t’
   Then with 6 PM, time becomes ‘t-2’, so d/(t-2) = 25
   With 9 PM, time becomes ‘t+1’, so d/(t+1) = 15
   Solve the equations, t = 6.5 hrs, d = 112.5 km
   So speed to reach at 1 PM = 112.5/6.5 = 17 km/hr
   [/su_spoiler]
3. Rekha’s age is 4 years less than that of Sheetal who age is 8 years less than Ravina. If seven years hence, the ratio of ages of Ravina to Rekha will be 13 : 7, then find the present age of Sheetal.
   A) 11 years
   B) 10 years
   C) 15 years
   D) 18 years
   E) 13 years
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option A
   Solution:
   Sheetal = A, Rekha = B, Ravina = C
   A = B+4
   A = C-8
   B+7 / C+ 7 = 7/13
   Or (A-4)+7 / (A+8)+7 = 7/13
   A+3 / A+15 = 7/13
   A = 11 = Sheetal’s present age
   [/su_spoiler]
4. A bag contains 7 red balls and 5 white balls. 2 balls are drawn at random. What is the probability that both balls are red in color?
   A) 2/33
   B) 7/44
   C) 8/33
   D) 7/22
   E) None of these
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option D
   Solution:
   Total balls = 7+5 = 12
   So required prob. = ⁷C₂ / ¹²C₂ = 7/22
   [/su_spoiler]
5. The incomes of 2 persons A and B is in the ratio 2 : 3. Their expenditures are in the ratio 1 : 2 respectively. If both save Rs 13,000, then what is B’s income?
   A) Rs 39000
   B) Rs 26000
   C) Rs 29000
   D) Rs 32000
   E) Rs 48000
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option A
   Solution:
   Incomes 2x and 3x
   Expenditures y and 2y
   So 2x – y = 13000
   And 3x – 2y = 13000
   Solve both equations, x = 13,000
   So B’s income = 3*13000 = Rs 39000
   [/su_spoiler]
6. A man buys 30 articles at the marked price of 24 articles. If he sells the articles by giving 2% discount, what will be the overall profit% made?
   A) 25.5%
   B) 22.5%
   C) 20.5%
   D) 18.5%
   E) 23.5%
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option B
   Solution:
   Let MP of 1 article = Re 1
   CP of 30 articles = MP of 24 articles = Rs 24
   MP of 30 articles = Rs 30
   So after discount of 2%, SP of 30 articles = 98/100 * 30 = Rs 29.4
   So profit% = (29.4 – 24)/24 * 100 = 22.5%
   [/su_spoiler]
7. 25 men can complete a work in 40 days. They start the work, but because of some reason only one-third work could be completed after 24 days. How many extra men should join them in order to complete the work in 36 days?
   A) 75
   B) 82
   C) 87
   D) 58
   E) 65
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option A
   Solution:
   After 24 days, the work is to be completed in (36-24) = 12 days
   Remaining work = 1 – 1/3 = 2/3
   Let x extra men are needed. So,
   25 * 24 * 2/3 = (x+25) * 12 * 1/3
   Solve, x = 75 [/su_spoiler]
8. If an item is sold at 3/5th of its actual selling price, the profit made is 20%. Now if the same item is sold at 25% less than the actual selling price, what will be the profit%/loss%?
   A) 48%
   B) 52%
   C) 45%
   D) 53%
   E) 50%
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option E
   Solution:
   Profit is 20%. So,
   (3/5 SP – CP)/CP * 100 = 20
   This gives SP = 2*CP
   Now product sold at 25% less S.P. i.e. 75/100 or 3/4 of S.P.
   Profit/loss% = (3/4 SP – CP)/CP * 100
   Put SP = 2CP
   Solve, Profit% is 50% (+50 means profit)
   [/su_spoiler]
9. A solution contains ingredients P, Q, and R in the ratio 2 : 3 : 1 and another solution contains them in the ratio 1 : 3 : 4 respectively. If both solutions are mixed in the ratio 5 : 6. What is the percentage (approximate) of C in the final solution?
   A) 10%
   B) 40%
   C) 20%
   D) 35%
   E) 25%
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option D
   Solution:
   Ratio of C to A+B in 1st solution = 1 : (2+3) = 1 : 5
   Ratio of C to A+B in 2nd solution = 4 : (1+3) = 4 : 4 = 1 : 1
   They are mixed in ratio 5 : 6
   So ratio of C to (A+B) in final solution = 1/6 * 5 + 1/2 * 6 : 5/6 * 5 + 1/2 * 6 = 23 : 43
   So, percentage of C = 23/(23+43) * 100 = 35%
   [/su_spoiler]
10. The total population of a town is 1,20,000. It has 70% males and rest females. If 85% of males and 50% of females are literate, then what is the total number of illiterates in the town?
    A) 32800
    B) 12600
    C) 23400
    D) 30600
    E) 20100
    [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option D
    Solution:
    Males = (70/100)*1,20,000 = 84,000 so females = 1,20,000 – 84,000 = 36,000
    Male illiterates = (15/100)*84,000 = 12,600
    Female illiterates = (50/100)* 36,000 = 18,000
    Total 12600 + 18000
    [/su_spoiler]

Directions (1 – 5):
Study the following pie chart and table to answer the following questions:

1. If the total number of people in department B below the age 40 and years and above 40 years is 1320, find the total number of people in the company altogether?
   A) 6600
   B) 6700
   C) 5400
   D) 6200
   E) 5800
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option A
   Solution:
   20/100 * total people in company = 1320
   So total people in company= 6600 [/su_spoiler]
2. If the total number of people in department C is 160, then what is the number of female employees above 40 years in department D?
   A) 144
   B) 146
   C) 125
   D) 135
   E) 122
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option D
   Solution:
   10/100 * total people in company = 160
   So total people in company= 1600
   So female employees above 40 years in department D = 9/16 * 15/100 * 1600 = 135 [/su_spoiler]
3. If the total number of employee in department D is 375 and total number of males above 40 years of age and females below 40 years of age is 122 in department F, then what is the number of males below 40 years in department F?
   A) 258
   B) 266
   C) 243
   D) 252
   E) 247
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option D
   Solution:
   15/100 * total people in company = 375
   So total people in company= 2500
   And total people in department F is 20/100 * 2500 = 500
   Let total number of employees in dept D below 40 years is x and above 40 years is y. Then
   1/10 * y + 3/10 * x = 122 or
   3x + y = 1220
   Also from above total people in department F is 500 so
   x + y = 500
   Solve the two equations, x = 360
   So number of males below 40 years in department F = 7/10 * 360 = 252 [/su_spoiler]
4. If the number of female employees below 40 years and number of male employees above 40 years in department E are 40 and 75 respectively, then find the number of employees in company A.
   A) 2160
   B) 2090
   C) 2450
   D) 2370
   E) 2280
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option E
   Solution:
   1/5 * employees below 40 years in E = 40
   So employees below 40 years in E = 200
   AND
   5/12 * employees below 40 years in E = 75
   So employees above 40 years in E = 180
   So total employees in E = 200+180 = 380
   So 5/100 * total employees in company = 380
   So total employees in company = 7600
   So total employees in dept A = 30/100 * 7600 = 2280
   [/su_spoiler]
5. If the total number of employee in department A is 255 and total number of males below 40 years of age and females above 40 years of age is 131 in department B, then find the ratio of total number of employees in dept B below 40 years and total number of employees in dept B above 40 years?
   A) 6 : 11
   B) 3 : 8
   C) 5 : 12
   D) 10 : 17
   E) 7 : 11
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option A
   Solution:
   30/100 * total people in company = 255
   So total people in company= 850
   And total people in department B is 20/100 * 850 = 170
   Let total number of employees in dept B below 40 years is x and above 40 years is y. Then
   9/10 * x + 7/10 * y = 131 or
   9x + 7y = 1310
   Also
   x + y = 170
   Solve the two equations, x = 60, y = 110
   So ratio = 60 : 110 = 6 : 11 [/su_spoiler]

Directions (6 – 10): (asked exact same type in SBI PO Main 2016 exam) 
The following table shows the MP, SP, Profit, Profit% and Discount% on 5 different products sold. Some values are missing in the table. Fill in the values and use it to answer the following questions:

6. What is the difference in the MPs of products E and C?
   A) Rs 780
   B) Rs 830
   C) Rs 800
   D) Rs 700
   E) Rs 920
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option C
   Solution:
   MP of C = 300
   For MP of E:
   10/100 * CP = 27
   So CP = 270, then SP = 270 + 27 = Rs 297
   So MP = 100/(100-73) * 297 = Rs 1100
   So difference = 1100 – 300 = Rs 800
   Similarly find all values for questions
   [/su_spoiler]
7. Find the difference in percentage points of discounts given for products E and discounts given for products B and C together.
   A) 49%
   B) 36%
   C) 33%
   D) 45%
   E) 76%
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option B
   Solution:
   Discount% for E = 73%
   Discount% for B:
   10/100 * CP = 22
   So CP = 220, then SP = 220 + 22 = 242
   MP = 1100, so discount % = (1100-242)/1100 * 100 = 78%
   Similarly discount % for C = 31%
   So required percentage points = (31+78) – 73 = 36%
   [/su_spoiler]
8. What is the total of MP of product D and SP of product B?
   A) Rs 642
   B) Rs 532
   C) Rs 628
   D) Rs 568
   E) Rs 544
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option A
   Solution:
   For country D:
   400 + 242 = Rs 642 [/su_spoiler]
9. If a person buys products A and E for their respective selling prices, then what is the resultant discount% given to him?
   A) 50.70%
   B) 60.20%
   C) 48.06%
   D) 67.62%
   E) 54.50%
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option B
   Solution:
   Total MP of A and E = 400+1100 = Rs 1500
   Total SP of A and E = 300+297 = Rs 597
   So discount% = (1500-597)/1500 * 100 = 60.2%
   [/su_spoiler]
10. Find the profit% given for product F.
    A) 16%
    B) 12%
    C) 18%
    D) 15%
    E) Cannot be determined
    [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option D
    Solution:
    Use formula
    MP = (100+p%)/(100-d%) * CP
    So
    600/360 = (100+p%)/(100-31)
    Solve, p% = 15% [/su_spoiler]

1. While filling his age in a form, a student mistakenly filled his age 10% less than his actual age. To correct the age in the form, by how much percent he will have to increase his age, if his actual age is 15 years?
   A) 12 2/3%
   B) 11 1/9%
   C) 16 1/5%
   D) 16 2/3%
   E) 10 2/5%
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option B
   Solution:
   To increase age, use:
   10/(100-10) * 100 = 100/9% [/su_spoiler]
2. Veena, Shweta and Charu started a business by investing Rs 2000, Rs 2500 and Rs 3000 respectively. After 8 months from start of business, due to some problem all of them withdrew Rs 500 for next 2 months. If after this they reinvest the withdrawn amount, then what is the share of Shweta out of a total profit of Rs 66,000 after a year?
   A) Rs 20,900
   B) Rs 33,700
   C) Rs 22,000
   D) Rs 37,000
   E) Rs 23,500
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option C
   Solution:
   Let Veena = A, Shweta = B, Charu = C
   A invested 2000 for total 10 months (first 8 months and last 2 months) and 1500 for 2 months (9th and 10th month). Similarly with others. So A : B : C is
   2000*10 + 1,500*2 : 2500*10 + 2,000*2 : 3000*10 + 2,500*2
   23 : 29 : 35
   So B gets = [29/(23+29+35)] * 66,000 = (29/87) * 66,000 = 66000/3 = 22000[/su_spoiler]
3. A certain sum of money amounts to Rs 1320 in 2 yrs and to Rs 1410 in 3.5 yrs. Find the sum and the rate of interest.
   A) Rs 1360, 9%
   B) Rs 1400, 5%
   C) Rs 1220, 6%
   D) Rs 1200, 5%
   E) Rs 1100, 5%
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option D
   Solution:
   SI for 1.5 yrs = 1410 – 1320 = Rs 90
   So, for 2 yrs SI = (90/1.5) * 2 = Rs 120
   Amount for 2 yrs is Rs 1320, so P = 1320 – 120 = Rs 1200
   This gives, r = 100*120/1200*2 = 5%[/su_spoiler]
4. A bag contains 3 red balls, 5 white balls and 4 green balls. 3 balls are drawn at random. What is the probability that at least 1 ball is white in color?
   A) 21/22
   B) 27/44
   C) 18/33
   D) 37/44
   E) None of these
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option D
   Solution:
   For at least 1 white, we can have prob = 1 – none is white
   None is white means 3 balls from 7 balls (3 red + 4 green), ways are ⁷C₃ = (7*6*5)/(3*2*1) = 35
   There are a total of 12 balls. Ways for drawing 3 balls is ¹²C₃ = (12*11*10)/)3*2*1) = 220
   So prob. of none white = ⁷C₃ / ¹²C₃ = 35/220 = 7/44
   So prob. of at least 1 white = 1 – 7/44 = 37/44 [/su_spoiler]
5. Bhavna’s age 8 years ago was four more than Anjali’s age that time. If the ratio of Bhavna’s age 8 years hence to Anjali’s age 8 years hence will be 7 : 6, what will be the ratio of Anjali’s age 4 years hence to Bhavna’s age that time
   A) 2 : 5
   B) 3 : 4
   C) 5 : 7
   D) 6 : 5
   E) 5 : 6
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option E
   Solution:
   Let Anjali = A, Bhavna = B
   Let the present ages of A and B are x and y respectively.
   8 years ago, x-8 = (y-8) – 4. This gives x – y = -4
   After 8 years, x+8/ y+8 = 6/7. This gives 7x – 6y = -8
   Solve both equations, x = 16, y = 20
   So required ratio = (16+4) : (20+4) = 20 : 24 = 5 : 6 [/su_spoiler]
6. The speed of the stream is 5 km/hr and speed of the boat in still water is 9 km/hr. if a boat takes 55 hrs to travel downstream from point A to B and coming back to point C midway between A and B, the what is the total distance travelled by the boat?
   A) 280 km
   B) 320 km
   C) 420 km
   D) 380 km
   E) 450 km
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option C
   Solution:
   Downstream speed = 9+5 = 14
   Upstream speed = 9-5 = 4
   Let the distance from A to B is x km
   downstream distance = x, upstream distance is half of x because boat goes half way upstream, then
   x/14 + (x/2)/4 = 55
   solve, x = 280
   so boat covered 280 + 280/2[/su_spoiler]
7. Cylindrical vessel A having radius 7 cm ad height 10 cm contains water. Another cylindrical vessel B having radius 14 cm ad height 12 cm contains milk. Half of the water and half of the milk from both vessels is taken out and mixed in another vessel. Find the % of water (approximate) in the third vessel?
   A) 22%
   B) 12%
   C) 17%
   D) 18%
   E) 15%
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option C
   Solution:
   Water in 1st vessel = volume of vessel A = (22/7)*7*7*10 = 1540 cm³
   Milk in 2nd vessel = volume of vessel B = (22/7)*14*14*12 = 7392 cm³
   In third vessel, water taken is 1540/2 = 770 cm³ and milk taken is 7392/2 = 3696 cm³
   So total is 770+3696 = 4516
   So percentage of water in third vessel = (770/4466)*100 = 17% [/su_spoiler]
8. Raman starts driving from city A with speed 60 km/hr. After 2 hours Ravish starts driving from the same place with speed 75 km/hr. Find the distance from city A they have covered when they meet.
   A) 480 km
   B) 800 km
   C) 600 km
   D) 570 km
   E) None of these
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option C
   Solution:
   Raman’s speed = 60, so after 2 hrs when Ravish starts, Raman has covered 2*60 = 120 km
   Now Ravish also starts, in order to meet Raman, he has to cover 120 km with relative speed of 75-60 = 15 [because both travelling in same direction]
   So the time after which Ravish meets Raman is 120/15 = 8 hrs
   So distance that Ravish has travelled in 8 hours = 8*75 = 600 km = distance both have travelled [/su_spoiler]
9. A bag contains 50 paise, Re 1, and Rs 2 coins in the ratio 2 : 5 : 7 respectively. If the total value of money in bag is Rs 200, what is the total number of coins in the bag?
   A) 100
   B) 140
   C) 120
   D) 85
   E) 125
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option B
   Solution:
   Coins – 2x, 5x, 7x, so
   Value of 50 paise coins = (50/100)*2x = x, of Re 1 coin = 1*5x = 5x, of Rs 2 coins = 2*7x = 14x
   So total value is x+5x+14x = 20x which equals 200, so x = 10
   Total coins = 2x+5x+7x = 14x = 140 [/su_spoiler]
10. A is 25% more efficient than B. If A can complete a work in 20 days, then in how many days A and B together can complete twice the work?
    A) 23 1/3 days
    B) 22 2/9 days
    C) 26 2/7 days
    D) 20 1/4 days
    E) 22 2/5 days
    [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option B
    Solution:
    Let efficiency of B is x, so of A = (125/100)*x = 5x/4
    So ratio of efficiencies = 5x/4 : x = 5 : 4
    So ratio of days = 4 : 5
    A can do in 24 days, so 4y = 20, y = 5
    So B can do in 5*5 = 25 days
    A+B can do 1 work in (20*25)/(20+25) = 100/9 days
    So twice work in (100/9)*2 = 200/9 days [/su_spoiler]

Mixture and Alligation Questions for the upcoming exams like IBPS PO, SBI PO, IPPB, LIC, RBI, Manipal University Course exams etc.

1. A 56 litre mixture contains milk and water in the ratio of 5 : 2 . How much water should be added to the mixture so as make the resultant mixture containing 40% water in it?
   A) 35/6 l
   B) 40/3 l
   C) 29/3 l
   D) 27/2 l
   E) 32/3 l
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option E
   Solution:
   In 56 l, milk = 5/(5+2) * 56 = 40 l, so water = 56 – 40 = 16 l
   Final ratio of milk to water will be = 60 : 40 = 3 : 2
   Let x litres of water to be added. So
   40/(16+x) = 3/2
   Solve, x = 32/3 l [/su_spoiler]
2. A mixture of 30 litres contains milk and water in the ratio 7 : 3. 10 litres of the mixture is taken out and replaced with pure milk and the same operation is repeated one more time. Find the final ratio of milk to water in the mixture.
   A) 12 : 7
   B) 9 : 4
   C) 13 : 2
   D) 15 : 7
   E) 11 : 5
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option C
   Solution:
   In 30 l of mixture, milk = 7/10 * 30 = 21l, so water = 9 l
   let x = amount of water after replacement and y = amount of water before replacement, so y = 9
   Now
   x/y = [1 – 10/30]²
   Solve, x = 4 l
   Now since mixture is 30 l only after replacement also. So milk in mixture after replacement = 30 – 4 = 26 l
   So final ratio = 26 : 4 = 13 : 2
   [/su_spoiler]
3. How much milk (in litres) costing Rs 50 per litres should be mixed with 18 litres of milk costing Rs 56 per litres so that there is a profit of 25% on selling the mixture at Rs 65 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 D
   Solution:
   CP of mixture = 100/125 * 65 = Rs 52
   Let x l of milk to be mixed. So by method of allegation:
   (x)……………..…..(18)
   50…………………..56
   ………….52
   4……………………..2
   So x/18 = 4/2
   x = 36 l [/su_spoiler]
4. A 24 litres of milk and water mixture contains milk and water in the ratio 3 : 5. What litres of the mixture should be taken out and replaced with pure milk so that the final mixture contains milk and water in equal proportions?
   A) 22/3 l
   B) 20/3 l
   C) 3 l
   D) 32/5 l
   E) 24/5 l
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option E
   Solution:
   In 24 l of mixture, milk = 3/8 * 24 = 9 l, so water = 15 l
   Now since the mixture is to be replaced with pure milk, the amount of mixture will remain same after replacement too.
   In 24 l mixture, to have 12 l water and 12 l milk, 3 l of water should be taken out, since we are only adding milk.
   Let x l of mixture taken out. So 5/8 * x = 3,
   Solve, x = 24/5 l [/su_spoiler]
5. 25 litres are drawn from a cask full of wine and is then filled with water. This operation is performed one more time. The ratio of the quantity of wine now left in cask to that of the water is 36 : 85. How much wine the cask hold originally?
   A) 66 l
   B) 85 l
   C) 59 l
   D) 55 l
   E) 46 l
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option D
   Solution:
   Let x l wine was there originally. So
   36/(36+85) = (1 – 25/x)²
   Solve, x = 55 l[/su_spoiler]
6. Out of 2100 kg wheat, some part is sold making 10% profit while the remaining part is sold making 16% profit. If there is an overall profit of 14%, what quantity was sold at 16% profit?
   A) 700 kg
   B) 1300 kg
   C) 1400 kg
   D) 1000 kg
   E) 1100 kg
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option C
   Solution:
   By method of Alligation:
   10………………..16
   …………14
   (16-14)…………..(14-10)
   2……………….4
   So 2 : 4 = 1 : 2
   so part at 16% profit = 2/(1+2) * 2100 = 1400 kg [/su_spoiler]
7. Container A and B contains water and alcohol in the ratio 1 : 3 and 3 : 2 respectively. How much amount of mixture from container A should be mixed with 30 l of mixture from container B, so that the resultant mixture contains water and alcohol in the ratio 11 : 12?
   A) 26 l
   B) 16 l
   C) 22 l
   D) 15 l
   E) None of these
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option B
   Solution:
   Water in A = 1/4. Water in B = 3/5 . And in resultant =11/23
   So by allegation method:
   (x)…………….(30)
   1/4………………..3/5
   ………..11/23
   14/(23*5)…………..21/(23*4)
   Take ratio: 14/23*5 : 21/23*4
   Gives 8 : 15
   So x/30 = 8/15
   Solve, x = 16 l [/su_spoiler]
8. The rice sold by a shopkeeper contains 15% low quality rice. What quantity of good quality rice should be added to 70 kg of rice so that percentage of low quality wheat becomes 7%?
   A) 50 kg
   B) 40 kg
   C) 90 kg
   D) 60 kg
   E) 80 kg
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option E
   Solution:
   In good quality rice, there is 0% low quality rice
   So method of allegation:
   (70 kg)……………………(x kg)
   15% …………………………0%
   …………………..7%
   7……………………………..8
   So 7 : 8
   Gives 70/x = 7/8
   Solve, x = 80 kg
   [/su_spoiler]
9. Container A and B contains 25% and 50% water respectively. The rest is milk in both the containers. How much amount should be mixed from container A to some amount in to some amount of container B so as to get 12 litres of new mixture having water to milk ratio 3 : 5?
   A) 6 l
   B) 8 l
   C) 10 l
   D) 7 l
   E) 5 l
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option A
   Solution:
   In resultant mixture, water is 3/8 * 100 = 75/2%
   So by method of allegation:
   25%……………………………50%
   ………………..75/2%
   25/2%……………………….25/2%
   so ratio is 25/2 : 25/2 = 1 : 1
   And the total should be 12 l, so 6 l of mixture from A, and 6 l from B. [/su_spoiler]
10. A mixture contains A and B in the ratio of 5 : 3. 16 litres of this mixture is taken out and 5 litres of A is poured in. the new mixture has ratio of A to B as 11 : 6. Find the total original quantity of mixture.
    A) 80 litres
    B) 96 litres
    C) 98 litres
    D) 84 litres
    E) 92 litres
    [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option B
    Solution:
    A = 5x, B = 3x
    16 l taken out, so let total mixture now = 5x + 3x + 16 = 8x + 16
    Now 5 l of A poured in and then ratio becomes 11 : 6
    So (5x+5)/3x = 11/6
    Solve, x = 10
    So total mixture originally = 8x + 16 = 8*10 + 16 = 96 litres [su_spoiler]

Ages Questions for Bank PO

1. The ratio of ages of Sneha to Bhavna is 6 : 13. Also ratio between Kritika’s age two years after and Bhavna’s age 4 years after will be 2 : 3. If the average age of Sneha and Kritika is 15 years, what will be Kritika’s age three years hence?
   A) 23
   B) 19
   C) 12
   D) 15
   E) 8
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option D
   Solution:
   S/B = 6/13
   (K+2)/(B+4) = 2/3
   S + K = 2*15 = 30
   Solve the equations, K = 12, So K + 3 = 15
   [/su_spoiler]
2. Difference between ages of Raman and Preet is 16 years. If Raman’s age ten years hence will be two times the age of Preet, find Raman’s age.
   A) 26
   B) 19
   C) 42
   D) 38
   E) 46
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option C
   Solution:
   R – P = 16
   (R + 10) = 2P
   Solve, R = 42
   [/su_spoiler]
3. Three years ago, Pihu was thrice old as Ravi that time. How old is Pihu today if ratio of age of Pihu six years hence to Ravi’s age four years ago is 9 : 2?
   A) 25
   B) 30
   C) 33
   D) 22
   E) 28
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option B
   Solution:
   (P – 3) = 3* (R-3)
   (P + 6)/(R – 4) = 9/2
   Solve both equations, P = 30 [/su_spoiler]
4. The average age of Yogita, Kanika and Prachi is 14 years. The ratio of Yogita’s age one year ago to Kanika’s age one years hence to Prachi’s age three years hence is 5 : 6 : 4. Find ratio of Yogita’s age two years hence to Prachi’s age.
   A) 8 : 5
   B) 5 : 3
   C) 1 : 4
   D) 2 : 1
   E) 4 : 7
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option D
   Solution:
   Y + K + P = 42
   (Y – 1) : (K + 1) : (P + 3) = 5 : 6 : 4
   So (5x + 1) + (6x – 1) + (4x – 3) = 42
   Solve, x = 3
   So Yogita’s age 1 ago hence = 5x = 15, so present age = 16
   Prachi’s age 3 years hence = 4x = 12, so present age = 9
   So (Y+2)/P = 18/9 = 2/1
   [/su_spoiler]
5. Sneha’s mother’s age is five years more than twice the age of Sneha. When Sneha was born, her brother Rahul was four years old and her father two years older than her mother. If the average age of her mother and father is 46 years. Find the ratio of age of Rahul to that of Sneha.
   A) 3 : 7
   B) 7 : 4
   C) 6 : 5
   D) 8 : 11
   E) 3 : 10
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option C
   Solution:
   Let age of Sneha = x, So age of Mother = 2x+5, Rahul = x+4, Father = 2x+7
   (2x+5 + 2x+7) = 2*46
   So x = 20
   So (x+4)/x = 24/20 = 6/5
   [/su_spoiler]
6. Ratio of ages of Reena and Prerna is 2 : 3 and the ratio of ages of Tiya and Reena is 3 : 1. If the ratio of age of Prerna four years hence to age of Tiya three years hence is 5 : 9, what is the total of the ages of all three?
   A) 77
   B) 65
   C) 84
   D) 43
   E) 79
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option A
   Solution:
   P/R = 3/2 and R/T = 1/3
   So P : R : T =
   3*1 : 2*1 : 2*3 = 3 : 2 : 6
   Now (3x+4)/(6x+3) = 5/9
   So x = 7
   P = 3x = 21, R = 2x = 14, T = 6x = 42
   So P + R + T = 77
   [/su_spoiler]
7. Richa’s age is thrice of Tisha’s age while Megha’s age is 3 less than that of Richa’s. The ratio of ages of Tisha’s three years hence to Megha’s four years ago is 3 : 5. What is the square root of the total age of Tisha and Richa?
   A) 36
   B) 25
   C) 6
   D) 13
   E) 18
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option C
   Solution:
   Tisha = x, Richa = 3x, Megha = 3x – 3
   (x+3)/(3x-3-4) = 3/5
   Solve, x = 9
   So total age of Tisha + Richa = x + 3x = 4x = 36, So square root = √36 = 6
   [/su_spoiler]
8. The ratio of age of Shaisha one year ago to Piya is 3 : 5. The sum of ages of Piya and Misha after four years will be 50. If difference between the ages of Shaisha and Misha is 7 years, find Misha’s age.
   A) 21
   B) 19
   C) 18
   D) 10
   E) 12
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option E
   Solution:
   (S -1)/P = 3/5
   P + M + 4 + 4 = 50
   So P + M = 42
   S – M = 7
   Solve equations, M = 12[/su_spoiler]
9. Kashish’s age is two-fifth the age of her mother. Nine years hence her age will be half of her mother’s age that time. What is her mother’s present age?
   A) 22
   B) 45
   C) 54
   D) 36
   E) 41
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option B
   Solution:
   K = 2M/5
   (K+9) = (M+9)/2
   Solve, M = 45 [/su_spoiler]
10. Ratio between Rahim’s age four years hence and Meher’s age three years ago is 6 : 5. If the ratio between Rahim’s age two years ago and Meher’s age four years hence is 6 : 11, find the ratio of Rahim’ age to Meher’s age.
    A) 5 : 9
    B) 9 : 4
    C) 2 : 11
    D) 7 : 9
    E) 4 : 9
    [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option D
    Solution:
    (R+4)/(M-3) = 6/5
    (R-2)/(M+4) = 12/22 = 6/11
    Solve R = 14, M = 18 [/su_spoiler]

Click here to go to Quantitative Aptitude Section.

Directions (1 – 5):
Study the following bar graph and table to answer the following:

1. What is the number of students who passed in only Math in the 6 classes together?
   A) 811
   B) 623
   C) 598
   D) 824
   E) 792
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option A
   Solution:
   For class 1: Students passed in at least 1 subject = 50/100 * 1200 = 600
   Students passed in Math = 40/100 * 1200 = 480
   Students passed in English = 29/100 * 1200 = 348
   Similarly for all other classes:
   So the required sum = 252+100+112+75+77+195 = 811 [/su_spoiler]
2. Which class ranks third among the number of students who passed in English when arranged in ascending order?
   A) 40%
   B) 16%
   C) 25%
   D) 18%
   E) 20%
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option C
   Solution:
   See the table in solution of 1st question: Required class is 5th. [/su_spoiler]
3. How many students in class 4 have not passed in Math?
   A) 4:3
   B) 6:5
   C) 2:5
   D) 8:5
   E) 2:3
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option B
   Solution:
   Passed = 20%
   So not passed = 80/100 * 1500 [/su_spoiler]
4. What is the ratio of number of students who have passed in only Math in class 2 to the number of students who have passed in Math in class 3.
   A) 40%
   B) 56%
   C) 2 : 7
   D) 68%
   E) 70%
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option E
   Solution:
   From the table in question 1 above:
   Number of students passed in only math in class 2 = 500-400 = 100
   Number of students passed in math in class 3 = 350
   SO ratio 100 : 350
   [/su_spoiler]
5. Which of the following is true about the number of students, n who have passed in only English in class 5?
   A) 0 < n < 10
   B) 10 < n < 14
   C) 14 < n < 21
   D) 21 < n < 37
   E) 37 < n < 56
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option E
   Solution:
   Number of students who have passed in only English in class 5, n = 440 – 385 = 55[/su_spoiler]

Directions (6 – 10):
The following pie charts show the percentage of people applied for jobs from different countries and people who got the job from each country respectively.
Success rate is defined as the number of people who got the job as a percentage of the number of job seekers.

6. Find the success rate of country A to C.
   A) 125 : 184
   B) 165 : 254
   C) 276 : 123
   D) 255 : 154
   E) None of these
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option D
   Solution:
   Since the total number is same, there is no need to find the exact number of people who got job.
   So required ratio:
   A : C
   15/11 : 14/17 = 255 : 154[/su_spoiler]
7. Find the difference between the number of people who got job from countries F and H.
   A) 4376
   B) 4284
   C) 4128
   D) 4173
   E) None of these
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option B
   Solution:
   Difference = (12-6)% of 71400 = 4284 [/su_spoiler]
8. What is the approximate percentage point difference between the success rate of countries D and G?
   A) 6%
   B) 4%
   C) 8%
   D) 5%
   E) None of these
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option B
   Solution:
   For country D:
   Total job seekers = 9/100 * 2356200 = 212058
   and who got job = 21/100 * 71400 = 14994
   so success rate = 14994/212058 * 100 = 7%
   For county G:
   Total job seekers = 14/100 * 2356200 = 329868
   and who got job = 15/100 * 71400 = 10710
   so success rate = 3%
   So different in % points = 7 – 3 = 4 [/su_spoiler]
9. What is the success rate of country B?
   A) 0.63%
   B) 3.55%
   C) 1.06%
   D) 2.62%
   E) 1.52%
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option C
   Solution:
   Total job seekers = 20/100 * 2356200 = 471240
   and who got job = 7/100 * 71400 = 4998
   so success rate = 4998/471240 * 100 = 1.06%
   [/su_spoiler]
10. Find the success rate of female job seekers if the success rate of male job seekers is 3.14%
    A) 11.23%
    B) 12.17%
    C) 18.45%
    D) 16.24%
    E) Cannot be determined
    [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option E
    Solution:
    Number of females and males is not given so cant be determined. [/su_spoiler]

Profit and Loss Bank PO Questions

1. A loss of 10% is made by selling an article. Had it been sold for Rs 75 more, there would have been a profit of 5%. What would be the selling price of the article if it is sold at 15% profit?
   A) Rs 475
   B) Rs 520
   C) Rs 575
   D) Rs 425
   E) None of these
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option C
   Solution:
   Use shortcut for these type of questions:
   CP of article = 75 × 100/ [5 – (-10)] (+5 for 5% profit, -10 for 10% loss)
   So SP at 15% profit = 115/100 * CP = (115/100) * [75 × 100/15] = Rs 575[/su_spoiler]
2. A profit of 20% is made when an article is sold for Rs 720. What will be the profit % if the same article is sold for Rs 630?
   A) 12%
   B) 9%
   C) 3%
   D) 5%
   E) 8%
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option D
   Solution:
   Let x% is the required profit. So
   630 = [(100+x)/100] * [100/120 * 720]
   Solving, we get, x = 5% profit [/su_spoiler]
3. A trader gets a profit of 30% on buying a kg of wheat. By mistake, he sold 1200 grams of wheat at the price of 1 kg. Find his net profit/loss%.
   A) 25/3% profit
   B) 55/3% profit
   C) 30/11% loss
   D) 125/3% loss
   E) 25/6% profit
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option A
   Solution:
   He sold 1200 grams of wheat at the price of 1 kg, means he incurs a loss here.
   That loss% = (1200-1000)/1200 * 100 = 50/3%
   Now there was a profit of 30% while buying and now a loss of 50/3% while selling. So successive method can be used here:
   Profit/loss% = 30 + (-50/3) + (30)(-50/3)/100 = + 25/3% [/su_spoiler]
4. A trader bought 8 kg of rice at Rs 10 per kg. While selling he uses false weights of 1600 gm instead of 2 kg. What profit did he make by selling 8 kg of rice bought?
   A) Rs 18
   B) Rs 30
   C) Rs 20
   D) Rs 26
   E) Rs 33
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option C
   Solution:
   He uses 1600 grams weight instead of 2 kg, means he makes a profit here.
   That profit% = (2000-1600)/1600 * 100 = 25%
   1 kg or 1000 gm costed him Rs 10. So 1600 gms costs his Rs 16.
   Now on selling each 1600 gms he makes 25% profit. So profit = 25/100 * 16 = Rs 4
   On selling each 1600 gms, he makes a profit of Rs 4, so on selling 5 * 1600 = 8000 gms or 8 kg, he makes a profit of 5 * 4 = Rs 20
   [/su_spoiler]
5. There is a profit of 20% on an article, when a discount of 10% is given. What will be the profit % on the article if discount given is 20%?
   A) 2 1/3%
   B) 7%
   C) 4 4/5%
   D) 8%
   E) 6 2/3%
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option E
   Solution:
   P2% = (100+p%1) [ (100 – d2%)/(100 – d1%)] – 100
   So P2% = (100+20) [ (100 – 20)/(100 – 10)] – 100 = 120 * (80/90) – 100 = 20/3%
   [/su_spoiler]
6. Two mobiles are sold at same price. If on one, a profit of 25% is made and on another, a loss of 10% is incurred, find the net profit/loss%.
   A) 10 62/99%
   B) 15 12/33%
   C) 6 14/23%
   D) 4 28/43%
   E) 15 1/5%
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option D
   Solution:
   profit 25% = 125, loss 10% = 90. Let SP of each mobile = LCM of (125 and 90) = 2250
   So CP1 = 100/125 * 2250 = Rs 1800, and CP2 = 100/90 * 2250 = Rs 2500
   So total SP = 2250+2250 = Rs 4500
   Total CP = 1800+2500 = Rs 4300
   So gain% = (4500-4300)/4300 * 100 = 200/43%
   OR use formula:
   gain/loss% = [100(25 – 10) – 2 (25)(10)] / [(100+25) + (100-90)] = (1500-500)/215 = + 200/43%[/su_spoiler]
7. Cost price of milk is Rs x per litre and it is sold at Rs 1.5x per litre. If the milkman mixes 2 litres water with every 4 litres of pure milk while selling, what is his profit percentage?
   A) 100%
   B) 125%
   C) 80%
   D) 25%
   E) 55%
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option B
   Solution:
   Let CP of 1 l pure milk = Re 1
   Then CP of 4 l of milk = Rs 4, and CP of 2 l of water = Rs 0. So CP of 6 l mixture = Rs 4
   And SP of 6 l of mixture = 6*1.5 = Rs 9
   So profit% = (9-4)/4 * 100 = 125%
   [/su_spoiler]
8. The profit% made by selling an article for Rs 2200 is same as the loss% made by selling the same article for Rs 1700. What would be the selling price of the article if it is sold at 10% profit?
   A) Rs 2145
   B) Rs 2080
   C) Rs 1975
   D) Rs 1900
   E) Rs 2180
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option A
   Solution:
   Let CP of article = Rs x
   So (2200 – x)/x * 100 = (x – 1700)/x * 100
   So 2200 – x = x – 1700
   Solving, we get x = Rs 1950
   So SP = 110/100 * 1950 = Rs 2145 [/su_spoiler]
9. Due to reduction of 20% in price of pens a customer can purchase 5 more pens for Rs. 40. What is the original price of a pen?
   A) Rs 5
   B) Rs 3
   C) Rs 2
   D) Rs 6
   E) Rs 8
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option C
   Solution:
   There is 20% reduction in price. So to keep the total amount same, consumption should increase by 20/(100-20) * 100 = 25%
   For 25% (25/100 = 1/4) increase in consumption, customer can purchase 5 more pens
   So original price of pens = 4*5 = Rs 20
   So original price of 1 pen = 40/20 = Rs 2[/su_spoiler]
10. If the cost price of 200 pens is equal to the selling price of 160 pens, the gain/loss % is
    A) 14% loss
    B) 25% gain
    C) 12% gain
    D) 16% loss
    E) 22% gain
    [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option B
    Solution:
    Here CP is more, so gain%.
    Gain% = (200-160)/160 * 100 = 25%[/su_spoiler]

Click here to go to Quantitative Aptitude Section

1. A student scored 23% of maximum marks and failed by 23 marks. But if he scores 43% of the marks in the same exam, he passes by 17 marks. What is the maximum marks of the exam?
   A) 250
   B) 220
   C) 200
   D) 180
   E) 300
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option C
   Solution –
   Let, maximum marks = x
   (43-23)% of x = (23+17)
   20% of x = 40
   Solving we get, x= 200
   [/su_spoiler]
2. If the price of sugar is increased by 20%, its expenditure gets decreased by 25%. What is the net effect on the total sale?
   A) 10% increase
   B) 10% decrease
   C) 15% increase
   D) 5% decrease
   E) no effect
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option B
   Solution –
   Use successive method
   20 + (-25) + (20)(-25)/100 = -10
   [/su_spoiler]
3. A invested Rs.50000 for starting a venture and B joined his business with a capital of 65000 after 4 months. A get Rs.350 in every 2 monthS for his extra work. Find B’s profit if A receives a total of Rs.62100 as his share.
   A) 52000
   B) 58000
   C) 46500
   D) 50180
   E) 60000
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option A
   Solution –
   (50000*12) : (65000*8)
   15 : 13
   Now, (62100 – 350*6) = 60000
   B’s profit = 60000*13/15 = 52000
   OR
   Let after cutting 350*6 from A’s profit, remaining amount is x. So
   15/28 * x + 350*6 = 62100
   So total x = 112000
   So B’s profit = 13/28 * 112000 = 52000 [/su_spoiler]
4. Two trains having equal speed take 10 seconds and 15 seconds respectively to cross a 250 meter long bridge. If the length of second train is 150 meters more than the first train, then find the speed of the trains?
   A) 30km/h
   B) 75km/h
   C) 108km/h
   D) 115km/h
   E) 45km/h
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option C
   Solution –
   Let, length of first train = x
   (x + 250)/10 = (x + 150 + 250)/15
   Solving, we get x=50
   Speed = 300/10 or 450/15 = 30m/s
   Convert this speed into km/h, 30*18/5 = 108km/h
   [/su_spoiler]
5. Of the two trains A and B, which train is moving faster? (The length of the both trains is same.)
   Statements:
   I. Train B cross a 450m platform in 30 sec.
   II. Train A crossed a 720m platform in 90 sec.
   III. Train A crossed a pole in 30 sec and train B crossed the same pole in 20 sec.
   A) Both I and II are required
   B) Both I and III are required
   C) All I, II and III are required
   D) Only statement III is required
   E) None of these
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option D
   Solution –
   From statement III, speeds of both can be found. Length is same so we can say that train B is faster.
   [/su_spoiler]
6. A train 75 m long overtook a person who was walking at the rate of 6 km/hr, passed him in 7 1/2 seconds. Also it overtook a second person in 6 3/4 seconds. What was the speed of the second person?
   A) 2km/h
   B) 3km/h
   C) 3.5km/h
   D) 5.55 km/h
   E) none of these
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option A
   Solution –
   Let, speed of train in km/h= x
   (x-6) * 5/18 = 75 * 2/15,
   Solving, we get = 42 km/h
   Now, assume speed of second person is y km/hr,
   So, (42-y) * 5/18 = 75 * 4/27
   Solving, we get y = 2km/h
   [/su_spoiler]
7. Thirteen litres are drawn from a cask full of water and then it is filled with milk. Now thirteen litres of mixture are drawn and the cask is again filled with milk. The ratio of quantity of water now left in the cask to that of the milk in it is 16 : 9. How much does the cask hold?
   A) 60 litres
   B) 65 litres
   C) 78 litres
   D) 39 litrers
   E) none of these
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option B
   Solution –
   Water = 16
   Total mixture = 16+9 = 25
   √(16/25) = 4/5
   Difference in ratio= 1
   And this 1 is equal to 13. So total mixture is 13*5= 65
   OR
   Let x litres is total capacity of cask
   Using formula, amount of water left in cask = x [1 – 13/x]²
   [1 – 13/x]²/x = 16/(16+9)
   Solving we get, x = 65 l
   [/su_spoiler]
8. A reduction of 40% in the price of wheat would enable a purchaser to obtain 36 kg more for Rs. 45. What is the reduced price per kg?
   A) 30p
   B) 45p
   C) 50 p
   D) 25p
   E) None of these
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option C
   Solution –
   Assume, purchaser buy 100 kg in Rs.45
   Now the new price is 60/100 * 45 = 27,
   It means in Rs. (45-27) = 18, 36kg more wheat is purchased.
   Rs 18= 1800p
   Now, 1800/36= 50 p
   [/su_spoiler]
9. A dishonest rice seller sells rice at 15% profit of rice CP, and he also uses 800gm weight in place of 1kg.Find his total profit percent.
   A) 30 %
   B) 25%
   C) 33 1/3 %
   D) 38 2/3 %
   E) 43 3/4 %
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option E
   Solution –
   Initial profit on CP = 15%
   Again profit , (1000-800)/800 *100 = 25%
   Use successive method,
   15 + 25 + (15)(25)/100= 43.75% or 43 ¾ %
   [/su_spoiler]
10. Sides of the parallelogram are in the ratio of 4:3, and its area is 1500 sq. units. Altitude on the greater side is 15 units. Find out the Altitude on the smaller side is?
    A) 15 units
    B) 25 units
    C) 20 units
    D) 22.5 units
    E) None of these
    [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option C
    Solution –
    Let the side of parallelogram be = 4x and 3x
    Area of parallelogram = basic * height
    Given area= 1500 units, so,
    4x*15= 1500
    X = 25 units
    Sides = 4*25 and 3* 25= 100 and 75 units,
    Now, height = 1500/75= 20 units

[/su_spoiler]

Directions (1-5): What will come in place in ? in the following questions?

1. 38% of 4500 – 25% of ? = 1640
   A) 240
   B) 265
   C) 280
   D) 320
   E) None of these
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option C
   
   [/su_spoiler]
2. (0.64)⁴ ÷ (0.512)³ × (0.8)⁴ = (0.8)^{? + 3}
   A) 2
   B) 1
   C) 0
   D) 4
   E) None of these
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option C
   
   [/su_spoiler]
3. 3648.24 + 364.824 ÷ ? – 36.4824 = 3794.1696
   A) 2
   B) 6
   C) 4
   D) 5
   E) 8
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option A
   
   [/su_spoiler]
4. 42% of ? = 3147 + 4258 – 7212 ÷ 12
   A) 16200
   B) 15400
   C) 18600
   D) 12800
   E) 14400
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option A
   
   [/su_spoiler]
5. 16121.50 – 48.25 + 3423.75 – 12016 + 512 ÷ 8 = ?
   A) 7545
   B) 7555
   C) 7575
   D) 7495
   E) 7515
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option A
   
   [/su_spoiler]
6. A box contains 6 black and 14 white balls, out of which 3 black and 5 white balls are defective. If we choose two balls at random, what is the probability that either both are white or both are non-defective?
   A) 141/190
   B) 123/190
   C) 121/190
   D) 123/199
   E) None of these
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option C
   Solution –
   Required probability- ¹⁴C₂/²⁰C₂ + ¹²C₂/²⁰C2₂ – ⁹C₂/²⁰C₂ = 121/190 [/su_spoiler]
7. In a class, the average age of some boys is 16 years, and average age of 16 teachers is 56 years. If the average age of the combined group of all the teachers and boys is 20, then the number of students is
   A) 156
   B) 144
   C) 136
   D) 88
   E) 168
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option B
   Solution –
   Use allegation method
   number of boys ‘x’           :             number of teacher’16’
   .      16                                                      56
   .                                      20 
   .    (56-20)=36                                  (20-16)=4
   .     So 36/4 = 9/1
   Now, x/16= 9/1, x=144
   [/su_spoiler]
8. If the CI on a certain sum for 2 yrs at 10% per annum is Rs. 3150, what would be the SI on same rate for same time?
   A) 2500
   B) 3000
   C) 2800
   D) 2200
   E) None of these
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option B
   Solution –
   Let principal = Rs x
   CI for 2years in % = 21% (using successive method)
   21% of x = 3150
   x = 15000
   S.I = 15000*2*10/100 = Rs.3000
   [/su_spoiler]
9. A started a business with initial investment of rs.12000, after 3 month B invest rs.15000 in this business. After 8 month from starting A withdrew one-fourth of his investment and B further invest 1/15 part of his investment. If at the end of one year the difference between the shares of profit of both is 700, what is the B’s profit share?
   A) 14550
   B) 13550
   C) 13900
   D) 42000
   E) 41250
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option C
   Solution –
   (12000*8 + 9000*4) : (15000*5 + 16000*4)
   132 : 139
   Difference in profit sharing ration is= 139x – 132x = 7x
   Given 7x = 700, So x = 100
   B’s profit share = 139x = 13900 [/su_spoiler]
10. There are three taps A, B, and C. A takes thrice as much time as B and C together to fill the tank. B takes twice as much time as A and C to fill the tank. In how much time can the Tap C fill the tank individually, if they would require 10 hours to fill the tank, when opened simultaneously?
    A) 16 hours
    B) 12 hours
    C) 22 hours
    D) 20 hours
    E) 24 hours
    [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option E
    Solution –
    Let A, B, C fills a, b, c units per hour.
    Total units = 10*(a+b+C)
    Now, 3a = b+c and 2b = a+c
    Solving both,
    b = 4c/5 and a = 3c/5
    total units of work = 10c*(4/5 + 3/5 + 1) = 24c
    done by C in 24c/c = 24 hours [/su_spoiler]

1. The average age of Abhilasha and Aadhira is 35 years. If Aaloka replaces Abhilasha, the average age is 31 years, if Aaloka replaces Aadhira average age is 36 years. If the average age of Aditi and Aashirya is half of average age of Abhilasha, Aadhira and Aaloka. then average age of all the five people is
   A) 28.2 years
   B) 32,4 years
   C) 27.2 years
   D) 30.2 years
   E) None of these
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option C
   Solution:
   Abhilasha, Aadhira, Aaloka, Aditi, Aashirya – X, Y, Z, P, Q
   X + Y = 35 * 2 =70 –(1)
   Z + Y = 31 * 2 =62 –(2)
   X + Z = 36 * 2 = 72 –(3)
   From (1) (2) and (3)
   X = 40 ; y =30; Z = 32
   Average age of P and Q =1/2 * [( X + Y + Z)/3]
   = 102/6 = 17
   Sum of the age of P and Q = 34
   Average age of all the five people = (34 + 102)/5 = 27.2
   [/su_spoiler]
2. The average score of a cricket player after 24 innings is 25 and in the 25th innings the player scores 25 runs. In the 26th innings what minimum number of runs will be required to increase his average score by 2 than it was before the 26th innings?
   A) 64
   B) 57
   C) 44
   D) 77
   E) None of these
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option D
   Solution:
   The average score of a cricket player after 25th Innings = (24 * 25 + 25) / 25 = 25
   Required Run = X
   (625 + X)/26 =27
   X = 26 * 27 – 625 = 77
   [/su_spoiler]
3. There are two vessels P and Q filled with cooking oil with different prices and with volumes 160 and 40 liters respectively. Equal quantities are drawn from both P and Q in such a manner that the cooking oil drawn from P is poured in into Q and oil drawn from Q is poured into P. If the price per liter becomes equal in both vessels. What is the (equal) quantity that was drawn from both P and Q?
   A) 52 litre
   B) 32 litre
   C) 42 litre
   D) 62 litre
   E) None of these
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option B
   Solution:
   Vessel (P)                          Vessel (Q)
   Quantity= 160 l                  Quantity= 40 l
   rate – p                                rate – q
   Let quantity taken out from both = a litres
   ‘a’ litres removed from p and ‘a’ litres added from q
   So rate of vessel P after removal and then addition = [(160-a)p + aq]/160
   Similarly rate of vessel Q after removal and then addition = [(40-a)q + ap]/40
   Now equate these equations
   [(160-a)p + aq]/160 = [(40-a)q + ap]/40
   Solving, we get a = 32 l [/su_spoiler]
4. A book seller sold a book at Rs. 56 in such a way that his percentage profit is same as the cost price of the book. If he sells it at twice the percentage profit of its previous percentage profit then new selling price will be?
   A) Rs. 72
   B) Rs. 32
   C) Rs. 42
   D) Rs. 62
   E) None of these
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option A
   Solution:
   CP = x
   SP = x + (x * x)/100 = 56
   x² + 100x – 5600 = 0
   x = 40
   SP = 40 + (40 * 80)/100 = Rs. 72
   [/su_spoiler]
5. A circular road runs round a circular playground. If the difference between the circumferences of the outer circle and the inner circle is 132 metres, then what is the width of the road?
   A) 15 m
   B) 19 m
   C) 17 m
   D) 21 m
   E) None of these
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option D
   Solution:
   Width of the Road = R – r
   2πR – 2πr = 132
   R – r = 132 * (7/44) = 21 m
   [/su_spoiler]
6. There are two concentric circles whose areas are in the ratio of 16:25 and the difference between their diameters is 8 m. Find out the area of the inner circle?
   A) 225πm²
   B) 256πm²
   C) 289πm²
   D) 144πm²
   E) None of these
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option B
   Solution:
   r²/R² = 25/16
   r/R = 5/4
   5x – 4x = 4
   x = 4
   Inner Radius = 16m
   Area of Inner Circle = Π (16 * 16) = 256πm²
   [/su_spoiler]
7. A boat takes 58 hours for travelling downstream from Point X to point Y and coming back to point Z midway between X and Y. If the speed of the stream is 4 kmph and speed of the boat in still water is 11 kmph, then what is the distance between X and Y?
   A) 520 km
   B) 370 km
   C) 480 km
   D) 420 km
   E) None of these
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option D
   Solution:
   Speed downstream = 11 + 4 = 15 kmph.
   Speed upstream = 11 – 4 = 7 kmph.
   Let distance between P and Q be ‘x’ km, then,
   x/15 + (x/2)/7 = 58.
   i.e., x/15 + x/14 = 58.
   Solving we get, x = 420 km.
   [/su_spoiler]
8. A boat takes 4 hours more while going back in upstream than in downstream when the distance between two places is 32 km and the speed of boat in still water is 6kmph. What must be the speed of boat in still water so that it can row downstream, 32km in 4 hours?
   A) 6 kmph
   B) 8 kmph
   C) 4 kmph
   D) 4.5 kmph
   E) None of these
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option A
   Solution:
   32/(6-R) – 32/(6+R) = 4
   R = 2kmph
   (B + 2) = 32/4
   Speed of boat in still water = 6kmph
   [/su_spoiler]
9. Ajay and Bala invest Rs. 4000 and Rs. 5000 in a business. Ajay receives Rs. 20 per month out of the profit as remuneration for running the business and the rest of profit is divided in proportion to the investment. In a year Ajay totally receives Rs. 672. What does Bala receives?
   A) Rs. 630
   B) Rs. 360
   C) Rs. 480
   D) Rs. 380
   E) Rs. 540
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option E
   Solution:
   Annual profit = x
   Ratio of profit share between Ajay and Bala = 4 : 5
   Ajay gets: 20 * 12 + 4/9 * x = 672
   Solving, we get, x = 108*9
   So Bala gets = 5/9 * x = 5/9 * 108*9 = Rs 540 [/su_spoiler]
10. Angel, Beaula and Catherine entered into a partnership in a business. Angel got 5/7 of the profit. Beaula and Catherine distributed the remaining profit equally. If Catherine got Rs.500 less than Angel, then the total profit was?
    A) Rs. 630
    B) Rs. 760
    C) Rs. 680
    D) Rs. 580
    E) Rs. 875
    [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option E
    Solution:
    Total Profit = x
    Angel’s Share = (5x/7)
    Remaining Profit = x – (5x/7) = (2x/7)
    Beaula and Catherine distributed the remaining profit equally- x/7 , x/7
    (5x/7) -(1x/7)= 500
    (4x/7) = 500
    x = 500 * (7/4) = 875
    [/su_spoiler]