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And, the direction against the stream is called upstream.\n\nIf the speed of a boat in still water is u km/hr and the speed of the stream is v km/hr, then:\n\nSpeed downstream = (u + v) km/hr.\n\nSpeed upstream = (u - v) km/hr.\n\nIf the speed downstream is a km/hr and the speed upstream is b km/hr, then:\n\nSpeed in still water =\t1/2*(a + b) km/hr.\n\nRate of stream =\t1/2*(a - b) km/hr.\n\n\n## Profit \u0026 Loss\n\n\n- Profit, P = SP – CP; SP\u003eCP\n\n- Loss, L = CP – SP; CP\u003eSP\n\n- P% = (P/CP) x 100\n\n- L% = (L/CP) x 100\n\n- SP = {(100 + P%)/100} x CP\n\n- SP = {(100 – L%)/100} x CP\n\n- CP = {100/(100 + P%)} x SP\n\n- CP = {100/(100 – L%)} x SP\n\n- Discount = MP – SP\n\n- SP = MP -Discou\n\n- For *false weight*, profit percentage will be **P% = (True weight – false weight/ false weight) x 100**.\n\n- When there are two successful profits say m% and n%, then the net percentage *profit* equals to **(m+n+mn)/100**\n\n- When the profit is m% and loss is n%, then the net % profit or loss will be: **(m-n-mn)/100**\n\n- If a product is sold at m% profit and then again sold at n% *profit* then the actual cost price of the product will be:\n    **CP = [100 x 100 x P/(100+m)(100+n)]**. In case of *loss*, **CP =[100 x 100 x P/(100-m)(100-n)]**\n\n- If P% and L% are equal then, P = L and **%loss = P^2/100**\n\n## Divisiblty Rule\n\n![](https://github.com/Psingh12354/Apptitude/blob/main/Divisibility%20Rule.jpeg)\n\n### Divisible by 7\n\nTo check multiply the number by 5 or 2 and than that number add(if mul by 5) else if mul by 2(subtract) through left number\n\n**eg :** 532 div by 7\n\n**Step 1-**\nmul 2x5=10\n**Step 2-** \nneglect the unit digit and add to left number\n 53|2+10=63 \n \ncheck number divisible if yes than original number is also divisible.\n\nSimilary multiply it by 2 and instead addition subtract it.\n\n### Divisible by 17\n\nMultiply unit digit by 5 and substract that multiplied value with rest digit if this multile than the number is divisible by 17.\n\n**eg :** 272 iv by 17\n\n**Step 1-**\n\nmul 2*5=10\n**Step 2- **\n\n27-10=17\n17 is divisble by 17 hence 272 is divisble by 17.\n\n## Age Base Questions\n\n![](https://github.com/Psingh12354/Apptitude/blob/main/AgeBase.jpeg)\n\n## Blood Relation\n\n\u003cimg src=\"https://github.com/Psingh12354/Apptitude/blob/main/jagran.jpg\" width=\"500\" height=\"400\" /\u003e\n\n## Clock\n\n1. Minute Spaces\n\nThe face or dial of clock is a circle whose circumference is divided into 60 equal parts, named minute spaces.\n\n2. Hour hand and minute hand, A clock has two hands. The smaller hand is called the hour hand or short hand and the larger one is called minute hand or long hand.\n\n3. In 60 minutes, minute hand gains 55 minute spaces over the hour hand.\n\n(In 60 minutes, hour hand will move 5 minute spaces while the minute hand will move 60 minute spaces. In effect the space gain of minute hand with respect to hour hand will be 60 - 5 = 55 minutes.)\n\n4. Both the hands of a clock coincide once in every hour.\n\n5. The hands of a clock are in the same straight line when they are coincident or opposite to each other.\n\n6. When the two hands of a clock are at *right angles*, they are *15 minute* spaces apart.\n\n7. When the hands of a clock are in opposite directions, they are 30 minute spaces apart.\n\n8. Angle traced by hour hand in **12 hrs = 360°**\n\n9. Angle traced by minute hand in **60 min. = 360°**.\n\n10. Theta( aka Degree) = 30 X H- (11/2) X M\n\n11. Reflax Angle = 360 - Theta\n## Probability\n\n![](https://github.com/Psingh12354/Apptitude/blob/main/probability.PNG)\n\n![](https://github.com/Psingh12354/Apptitude/blob/main/Probability%204.PNG)\n\n##  AP \u0026 GP\n\n![](https://github.com/Psingh12354/Apptitude/blob/main/ap1.jpg)\n\n![](https://github.com/Psingh12354/Apptitude/blob/main/ap2.png)\n\n\n## Number System\n\n### TYPES OF NUMBERS :\n**1.** Natural numbers (N) = 1, 2, 3, . . . .\n**2.** Whole numbers (W) = 0, 1, 2, 3, . . . .\n**3.** Intezers (Z) = −∞ . . . −2, −1, 0, 1, 2, 3, . . .\n**4.** Rational numbers (Q) = The numbers of the form p⁄q  where q ≠ 0.  Eg: 1⁄5 , 0.46, 0.333333 they are **terminating, repeating**.\n**5.** Irrational numbers (I) = The numbers of the form x1⁄n  ≠ Intezer.  Also π and e also irrational numbers they are  **non-terminating, non-repeating**.\n\nOther types of numbers:\n**a.** Even numbers : Numbers which are exactly divisible by 2.  These numbers are in the format of 2n.\n**b.** Odd numbers: Numbers which gives remainder 1 when divided by 2. These numbers are in the format of 2n ± 1.\n**c.** Prime numbers : The numbers which are divisible by 1 and the number itself are primes.  The least prime is 2.\n**d.** Composite numbers : The numbers of which are divisible by more than 2 numbers.\n\n### IMPORTANT RULES RELATED TO EVEN AND ODD NUMBERS:\n\nodd ± odd = even;\n\neven ± even = even;\n\neven ± odd = odd\n\n\nodd × odd = odd;\n\neven × even = even;\n\neven × odd = even.\n\nodd(any number) = odd\n\neven(any number) = even\n\n## BODMAS\n\n1) (a - b)2 = (a2 + b2 - 2ab)\n\n2) (a + b)2 = (a2 + b2 + 2ab)\n\n3) (a + b) (a – b) = (a2 – b2 )\n\n4) (a3 + b3) = (a + b) (a2 – ab + b2)\n\n5) (a3 - b3) = (a - b) (a2 – ab + b2)\n\n6) (a + b + c)2 = a2 + b2 + c2 + 2 (ab + bc + ca)\n\n7) (a3 + b3 + c3 – 3abc) = (a + b + c) (ar\u003e2 + b2 + c2 – ab – bc – ac)\n\nQuick Tips and Tricks\n\n1) Virnaculum (Bar ‘¬¬–’): If a given expression contains a virnaculum (bar), then first simplify the given expression under virnaculum and then follow the rule of ‘BODMAS’.\n\nStep1: When simplifying the given expressions, first brackets are to be removed in the order: ‘––’, ‘( )’, ‘{ }’, ‘[ ]’\n\nStep2: The operations are to be performed strictly in the order: Division, Multiplication, Addition and Subtraction\n\n2) ‘BODMAS Rule’: This rule is used to solve and find out the value of given expressions by performing the operations in a correct sequence.\n\nBODMAS is the shortcut used to remember the procedure of simplification.\n\nB: Bracket\n\nO: Order (i.e. power, square root etc.)\n\nD: Division (Left to right)\n\nM: Multiplication (Left to right)\n\nA: Addition (Left to right)\n\nS: Subtraction (Left to right)\n\nStep1: When simplifying the given expressions, first brackets are to be removed in the order: ( ), { }, [ ]\n\nStep2: The operations are to be performed strictly in the order: Division, Multiplication, Addition and Subtraction\n\n## Mensuration\n\n![](https://github.com/Psingh12354/Aptitude/blob/main/Men1.jpg)\n\n![](https://github.com/Psingh12354/Aptitude/blob/main/Men2.jpg)\n\n\n## Alligation or Mixture\n\n**Important formula : **\n\n- final mixture = initial misture(1-replaced amount/total amount)^n\n\n## Series\n\n**Important Formula**\n\n- Sum of Square of n natural number= n(n+1)(2n+1)/2\n\n- Sum of n natural number=n(n+1)/2\n\n- Sum in ap= nx(a+l)/2\n\n## Average\n\n- Avg of first n odd number = n\n\n- Avg of first n even number = n+1\n\n- Avg of n naturat number = (n+1)/2\n\n- Avg of number * total number = Sum\n---\n\n## Formulas and Tricks [](https://github.com/Psingh12354/Aptitude/blob/main/Formulas.jpeg)\n\n\u003ch2 align=center\u003e\u003cb\u003e\u003ci\u003eImportant Links\u003c/i\u003e\u003c/b\u003e\u003c/h2\u003e\n\n- [Population Based Question](https://www.math-shortcut-tricks.com/percentage-calculation-on-population-shortcut-tricks/)\n\n- [Ratio And Proportion](https://www.indiabix.com/aptitude/ratio-and-proportion/)\n\n- [Code/Decode](https://www.hitbullseye.com/Coding-Decoding-Questions.php)\n\n- [Log](https://www.indiabix.com/aptitude/logarithm/)\n\n- [Amcat Verbal Question](https://www.faceprep.in/amcat-exam/amcat-verbal-questions-previously-asked-questions/)\n\n- [Negative Remainder](https://bodheeprep.com/remainder-theorem-concepts-tricks-questions-for-cat-exam)\n\n- [Eligibility test](https://www.tutorialspoint.com/reasoning/reasoning_eligibility_test_examples.htm)\n\n---\n\n---\n\n\u003ch2 align=center\u003e\u003cb\u003e\u003ci\u003ePractice Questions\u003c/i\u003e\u003c/b\u003e\u003c/h2\u003e\n\n---\n\n### 1) What is the LCM of 147/64 and 30/44?\n\n1) 735/2\n\n2) 735/704\n\n3) 3/704\n\n4) 3/735\n\n\n\nSolution\n\n```\nOption A\n\nFind the LCM of the numerators.\n\nLCM (147, 30) = 1470\n\nStep 2: Find the HCF of denominators.\n\nHCF (64, 44) = 4\n\nSo, the LCM of 147/64 and 30/44 is (LCM of Numerators) / (HCF of Denominators) = 1470 / 4 = 735/2\n```\n\n### 2) A man rows a boat at a speed of 15 mph in still water. Find the speed of the river if it takes her 4 hours 30 minutes to row a boat to a place 30 miles away and return.\n\nA) 5 mph\n\nB) 10 mph\n\nC) 12 mph\n\nD) 20 mph\n\n\n\nSolution\n\n```\nOption C\n\nLet the speed of the river=x mph, then\n\nTime taken row 30 miles upstream and 30 miles downstream = 30/(15-x) + 30/(15+x) = 9/2\n\n= 10/(15-x) + 10/(15+x) = 3/2\n\n= 2[10(15+x) + 10(15-x)] = 3(15-x)²\n\n= 300 + 20x + 300 – 20x = 675 -3x²\n\nx² = 25 or x=5\n```\n\n### 3) Working 5 hours a day, A can Complete a work in 8 days and working 6 hours a day, B can complete the same work in 10 days. Working 8 hours a day, they can jointly complete the work in how many days?\n\nA) 3 days\n\nB) 4 days\n\nC) 4.5 days\n\nD) 5.4 days\n\n\n\nSolution\n\n```\noption A\n\nWorking 5 hours a day, A can complete the work in 8 days = 5*8 = 40 hours\n\nWorking 6 hours a day, B can complete the work in 10 days = 6*10 = 60 hours\n\n(A+B)’s 1 hour’s work = (1/40+1/60)\n\n=(3+2)/120\n\n= 1/24\n\nHence, A and B can complete the work in 24 hours which is equal to 3 days.\n```\n\n### 4) A mixture of 40 liters of milk and water contains 10% water. How much water should be added to this so that water may be 20% in the new mixture?\n\nA) 6.5 liters\n\nB) 5 liters\n\nC) 4 liters\n\nD) 7.5 liters\n\n\n\nSolution\n```\nOption B\n\nA mixture of 40 liters of milk = 36 liters of Milk and 4 liters of water = 90:10 ratio\n\nNow the new mixture should be in the ratio of 80:20\n\nHence 80% is equivalent to 36 liters (no addition of milk is done)\n\n100% is (36/80)*100 = 45 liters of milk is present in the new mixture\n\nThus water shall be added= (45 – 36 – 4) = 5 liters of water\n\n**or**\n\n```\n![](https://github.com/Psingh12354/Apptitude/blob/main/a1.jpg)\n\n\n### 5) Four different electronic devices make a beep after every 30 minutes, 1 hour, 3/2 hour and 1 hour 45 minutes respectively. All the devices beeped together at 12 noon. They will again beep together at:\n\nA) 12 midnight\n\nB) 3 a.m\n\nC) 6 a.m\n\nD) 9 a.m\n\n\n\nSolution\n\n```\nOption D\n\nFour different devices beep after every 30 mins, 60 mins, 90 mins and 105 mins.\n\nLCM of 30,60,90 and 105 is 1260.\n\nWhich means the devices beep together after every 1260 mins = 1260/60 = 21 hours\n\nHence 12 noon + 21 hours = 9 a.m\n```\n\n### 6) In a 100 m race, A can beat B by 25 m and B can beat C by 4 m. In the same race, A can beat C by?\n\nA) 21 m\n\nB) 26 m\n\nC) 28 m\n\nD) 29 m\n\n\n\nSolution\n\n```\nOption C\n\nWhen A travels 100 m, B travels 75 m. Hence A:B = 100:75\n\nWhen B travels 100 m, C travels 96 m. Hence B:C = 100:96\n\nWhen B Travels 75 m, C travels (96 x 75)/100 = 72 m\n\nHence B:C = 75:72.\n\nTherefore, A:B:C = 100:75:72.\n\nSo, when A Travels 100 m, C travels 72 m.\n\nTherefore, A beat C by 28 m\n```\n\n### 7) In an examination, 70% of students passed in physics, 65% in chemistry, 27% failed in both subjects. The percentage of students who passed is?\n\nA) 62\n\nB) 68\n\nC) 66\n\nD) 69\n\n\n\nSolution\n\n```\nOption A\n\n70% students passed in physics = 30% failed in Physics.\n\n65% students passed in Chemistry = 35% failed in Chemistry\n\nPercentage of students failed in both subject = 27%.\n\nPercentage of students failed = 35 + 30 – 27\n\n= 38%.\n\nPercentage of students passed = 100 – 38% = 62%\n```\n\n### 8) If 15 oxen or 20 cows can eat the grass of a field in 80 days, then in how many days will 6 oxen and 2 cows eat the same grass?\n\nA) 40\n\nB) 60\n\nC)100\n\nD) 160\n\n\n\nSolution\n\n```\nOption D\n\n15 oxen take 80 days so, 6 oxen take x days\n\nx = 15*80/6 = 200 days\n\n20 oxen also take 80 day. So, 2 cows take y days\n\ny = 20*80/2 = 800 days\n\nTogether work will be done in 800*200/(800+200) = 160 days\n```\n\n### 9) Simplify {(3 * 2.333 + 2)/3} / (1/10 of 100 + 4.8181)\n\nA) 297/10377\n\nB) 188/121\n\nC) 21/34\n\nD) 33/163\n\n\n\nSolution\n\n\n![](https://github.com/Psingh12354/Apptitude/blob/main/a2.jpg)\n\n```\nSolution: Option D\n```\n\n### 10) In how many ways can 6 lottery tickets be distributed among 4 different people if all of the four different people can get any number of tickets?\n\nA) 6C4\n\nB) 6P4\n\nC) 4^6\n\nD) 6^4\n\nSolution \n```\nC is correct because all 4 can get 4 ticket one by one\n```\n\n### 11) A salesman has the liberty to sell a hair dryer in his store at a price between Rs. 300 and Rs. 700. Profit earned by selling the hair dryer for Rs. 650 is twice the loss incurred when it is sold for Rs. 350. What is the cost price of the hair dryer?\n\nA) 550\n\nB) 450\n\nC) 350\n\nD) 150\n\n\n\nSolution\n\n```\nOption B\n\nGoing through the options,\n\nTaking Cost Price as Rs 450.\n\nProfit = 650 – 450 = 200\n\nLoss = 350 – 450 = 100\n\nClearly profit is twice the loss incurred.\n\nHence, Rs 450 is the correct option.\n```\n\n### 12) Ronald and Elan are working on an assignment. Ronald takes 6 hours to type 32 pages on a computer, while Elan takes 5 hours to type 40 pages. How much time will they take, working together on two different computers to type an assignment of 110 pages?\n\nA) 7 hours 30 minutes\n\nB) 8 hours\n\nC) 8 hours 15 minutes\n\nD) 8 hours 25 minutes\n\n\n\nSolution\n\n```\nOption C\n\nIn 1hr, Ronald types = 32/6 pages and Elan types = 40/5 pages\n\nIf they work together they will type = 32/6 + 40/5 = 40/3 pages in 1 hr\n\nTime needed  to complete the assignment is = (3 x 110)/40 = 33/4 = 8hrs 15mins\n\nHence, the time required is 8 hrs 15 mins.\n```\n\n### 13) A television manufacturing company has decided to increase the sale to beat the economic slowdown. It decides to reduce the price of television sets by 25% as a result of which the sales increased by 20%. What is the effect on the total revenue of the company?\n\nA) Decreased by 20%\n\nB) Increased by 20%\n\nC) Increased by 10%\n\nD) Decreased by 10%\n\n\n\nSolution\n\n```\n\nOption D\n\nLet the initial Price = Rs.100 and initial sales = 100\n\nSo, the initial revenue = Rs. 10000\n\nNow, the price is reduced to 25% which is equal to Rs.75 and Sales is increased by 20% which is equal to 120.\n\nNow new revenue = 120 x 75 = Rs. 9000\n\nChange in revenue = (10000 – 9000) = Rs.1000 decrease\n\n% decrease = (1000/10000) x 100 = 10%\n\nHence, the correct option is decrease of 10%.\n```\n\n### 14) Ravi has a bag full of 10 Nestle and 5 Cadbury chocolates. Out of these, he draws two chocolates. What is the probability that he would get at least one Nestle chocolate?\n\nA) 19/21\n\nB) 3/7\n\nC) 2/21\n\nD) 1/3\n\n\n\nSolution\n\n```\nOption A\n\nProbability of getting atleast one nestle chocolate = [(10C1 x 5C1) + 10C2] / 15C2\n\n[(10 x 5) + (10 x 9)/2] / [(15 x 14)/2] = 19/21.\n\nHence, the required probability is 19/21.\n```\n\n### 15) Anil and Ruhi started a business by investing Rs 2000 and Rs 2800 respectively. After 8 months, Anil added Rs 600 and Ruhi added Rs 400. At the same time Teena joined them with Rs 4200. Find the share of Teena if they get a profit of Rs 34,300 after a year.\n\nA) Rs 7490\n\nB) Rs 7350\n\nC) Rs 8250\n\nD) Rs 8530\n\nE) None of these\n\nSolution\n\n```\nOption B\n\nSolution:\nShare of Anil : Share of Ruhi : Share of Teena is\n2000×8 + 2600×4 : 2800×8 + 3200×4 : 4200×4\n33 : 44 : 21\nso share of Teena = 21/(33+44+21) × 34300 = Rs 7350\n```\n\n### 16) A sum of Rs 7000 is deposited in two schemes. One part is deposited in Scheme A which offers 8% rate of interest. Remaining part is invested in Scheme B which offers 10% rate of interest compounded annually. If interest obtained in scheme A after 4 years is Rs 226 more than the interest obtained in scheme B after 2 years, find the part deposited in scheme B.\n\nA) Rs 3200\n\nB) Rs 4500\n\nC) Rs 3800\n\nD) Rs 3500\n\nE) Rs 2800\n\nSolution: \n\n```\nOption C\n\nSolution:\nRs 3800\nSolution:\n(7000-x)*8*4/100 = x [ (1 + 10/100)2 – 1] + 226\n70*8*4 – 32x/100 = 21x/100 + 226\n2240 – 226 = 53x/100\n2014 = 53x/100\nSo, x = Rs 3800\n```\n\n### 17) A work which is completed by 20 men in 8 days can be completed by 25 women 12 days. 16 men and 10 women start doing the work. After 3 days, they leave. If the remaining work is to be completed in 6 days by x number of men, find x.\n\nA) 16\n\nB) 18\n\nC) 12\n\nD) 10\n\nE) 22\n\nSolution:\n\n```\nOption A\n\nSolution: \n20 men in 8 days so 16 men in 20 × 8/16 = 10 days and\n25 women in 12 days so 10 women in 25 × 12/10 = 30 days\nSo in 3 days, they complete (1/10 + 1/30) × 3 = 2/5\nSo remaining work = 1 – 2/5 = 3/5\n20 m 1 work in 8 days and x men 3/5 work in 6 days\nSo 20 × 8 × 3/5 = x × 6 × 1\nSo, x = 16 men\n```\n\n### 18) There are 140 tickets (numbered 1 to 140) in a bowl. Find the probability of choosing a ticket which bears multiple of either 3 or 7.\n\nA) 3/5\n\nB) 2/9\n\nC) 1/8\n\nD) 3/7\n\nE) None of these\n\nSolution: \n\n```\nOption D\n\nSolution\nNumber of multiples of 3 in 140 = 140/3 = 46\nNumber of multiples of 7 in 140 = 140/7 = 20\nNumber of multiples of 3×7= 21 in 140 = 140/21 = 6\nSo required probability = (46+20 – 6)/140 = 60/140 = 3/7\n```\n\n### 19) Introducing a man, a woman said, \"He is the only son of the mother of my mother.\" How is the woman related to the man?\n\nA.\tMother\n\nB.\tSister\n\nC.\tNiece\n\nD.\tMaternal aunt\n\nSolution\nC\n\n### 20) There are six persons A. B, C, D, E and F. C is the sister of F. B is the brother of E's husband. D is the father of A and grandfather of F. There are two fathers, three brothers and a mother in the group. Who is the mother ?\n\nA) A\t\n\nB) B\n\nC) C\t\n\nD) E \n\nSolution:\n\n```\nE\nSolution:\nD is father of A and grandfather of F. So, A is father of F. \n\nThus. D and A are the two fathers. C is the sister of F So. C is the daughter of A. \n\nSince there is only one mother, it is evident that E is the wife of A and hence the mother of C and F. \n\nSo, B is brother of A There are three brothers. So. F is the brother of C. \n\n Clearly, A is E's Husband.\n```\n \n ### 21) Pointing to a girl in the photograph, Amar said, \"Her mother's brother is the only son of my mother's father.\" How is the girl's mother related to Amar ? \n \nA) Mother\n\nB) Sister\n\nC) Aunt \t\n\nD) Grandmother\n\nSolution:\n\n```\nC\n\nSolution:\nOnly son of Amar's mother's father — Amar's maternal uncle. \n\nSo, the girl's maternal uncle is Arnar's maternal uncle. Thus, the girl's mother is Amar's aunt. \n```\n\n### 22) The total age of A and B is 12 years more than the total age of B and C. C is how many years younger than A ?\n\nA) 12\n\nB) 13\n\nC) 14\t\n\nD) 15\n\nSolution: \n\n```\nA\n\nSolution:\nA+B=B+C+12\nso\nA=12\n```\n\n### 23) The ratio of the present ages of P and Q is 3 : 4. Five years ago, the ratio of their ages was 5 : 7. Find their present ages.\n\nSolution: \n\n```\nP=(3*2*5)/1=30\nQ=(4*2*5)/1=40\n```\n\n### 24) The average age of a group of 10 students is 15 years. When 5 more students join the group, the average age increase by 1 year. The average age of the new students is?\n\nSolution: \n\n```\n(150+x)/15=16.\n\n=)150+x=240\n\n=x=90\n```\n\nAverage of their ages=sum/number=)90/5=18\n\n### 25) An accurate clock shows 8 o'clock in the morning. Through how may degrees will the hour hand rotate when the clock shows 2 o'clock in the afternoon?\n\nA. 154°\n\nB. 170°\n\nC. 160°\t\n\nD. 180°\n\nSolution: \n\n```\nWe know that angle traced by hour hand in 12 hrs = 360°\n\nFrom 8 to 2, there are 6 hours.\n\nAngle traced by the hour hand in 6 hours =  6×360/12= 180°\n```\n\n### 26) At 3:40, the hour hand and the minute hand of a clock form an angle of:\n\nA.\t120°\n\nB.\t125°\n\nC.\t130°\n\nD.\t135°\n\nSolution: \n\n```\nC\nSolution: \nAngle traced by hour hand in 12 hrs. = 360°.\n\nAngle traced by it in\t11/3\thrs =\t(360/12\tx\t11/3)°\t= 110°.\nAngle traced by minute hand in 60 min. = 360°.\n\nAngle traced by it in 40 min. =\t(360/60x40)°= 240°.\n Required angle (240 - 110)° = 130°.\n ```\n \n ### 27) A person crosses a 600 m long street in 5 minutes. What is his speed in km per hour?\n\nA.\t3.6\n\nB.\t7.2\n\nC.\t8.4\n\nD.\t10\n\nSolution: \n\n```\nB\nSolution: \n\nSpeed =600/(5 x 60)= 2 m/sec.\nConverting m/sec to km/hr (see important formulas section)\n   =\t(2 x18/5)km/hr= 7.2 km/hr.\n   ```\n   \n ### 28) A train of length 110 meters is running at a speed of 60 kmph. In what time, it will pass a man who is running at 6 kmph in the direction opposite to that in which the train is going?\nA. 10\n\nB. 8\n\nC. 6\n\nD. 4\n\nSolution:\n\n```\nC\n\nOpposite direction \nspeed=60+6=66km/h\ntime=distance/speed=110/66=5/3 km/h\nin m/s 5/3x18/5=6\n```\n\n### 29) Two trains of equal length, running in opposite directions, pass a pole in 18 and 12 seconds. The trains will cross each other in:\n\nA. 14.4 seconds\n\nB. 15.5 seconds\n\nC. 18.8 seconds\n\nD. 20.2 seconds\n\nSolution :\n\n```\nA\nSolution : \nLet length of each train be x meter.\n\nThen, speed of 1st train = x/18m/sec\n\nSpeed of 2nd train = x/12 m/sec\n\nNow,\n\nWhen both trains cross each other, time taken\n\n=2x/(x/18+x/12)=2x/(2x+3x)/36=(2x X 36)/5x=725=14.4seconds\n```\n\n### 30) What will come in place of question mark (?) in the following series?\n\n767 495 359 291 257 ?\n\nA. 230 \nB. 240\nC. 250\nD. 280\nE. 260\n\nSoluton : \n\n```\nB\n\nSolution: \n\n797-495=272\n495-359=136\nSo which means it is half of previous diffrences\n272/2=136\n291-257=32\n34/2=17\nSo subtract 17\n257-17=240\n\n```\n\n### 31)  What will come in place of question mark (?) in the following series?\n\n50 67 33 84 16 ?\n\nA. 101\nB. 109\nC. 107\nD. 103\nE. 201\n\nSolution: \n\n```\nA\n\nSolution: \n17 is the gap once increase and than decrease follow this order you will get the answer\n```\n\n### 32) In each of the following number series given, one particular number is wrong. Find out that wrong number in each series.\n\n### A) 2 3 10 38 172\n\n1. 192\n\n2. 10\n\n3. 38\n\n4. 2\n\n5. 3\n\nSolution: \n\n```\n3\n\nSolution: \nLogic is 2×1 + 1 = 3, 3 × 2 + 4 =10, 10 × 3 + 9 = 39, 39 × 4 + 16 = 172…. So in place of 38, it should be 39.\n```\n\n### 33) Find the greatest 6-digit number, which is a multiple of 12.\n\n1. 999980\n\n2. 999990\n\n3. 999984\n\n4. None of these\n\nSolution: \n\n```\nGreatest six-digit number is 999999. Divide this number by 12 and get remainder as 3. Since the remainder is 3, if you subtract 3 from the number, the remaining number will be a multiple of 12. So the greatest such number will be 999999 – 3 =999996.\n```\n\n### 34) What is the sum of all natural numbers between 100 and 200 which are multiples of 3?\n\n1. 5000\n\n2. 4950\n\n3. 4980\n\n4. 4900\n\n5. None of these\n\nSolution:\n\n```\nMultiples of 3 between 100 and 200 are 102, 105, 108,… ,198.\nHere, the first term = 102\nlast term = 198\nLet the number of Multiples of 3 between 100 and 200 = n\n\nW.K.T: Arithmetic Progression Formula:\nan = a1 + (n - 1)d\nWhere, an = last term = 198\na1 = first term = 102\nd = common difference = 105 - 102 = 3\n---\u003e 198 = 102 + (n - 1) * 3\n---\u003e 198 - 102 = (n - 1) * 3\n---\u003e 96 = (n - 1) * 3\n---\u003e (n - 1) = 96/3 = 32\n---\u003e n = 32 + 1\n---\u003e n = 33\n\nFormula:\nSum of n terms = Sn = (n/2) * (a + l)\nwhere n = number of elements = 33\na = first term = 102\nl = last term = 198\nThus, using the above formula, Sum of all natural numbers between 100 and 200 which are multiples of 3 = (33/2) * (102 + 198)\n= (33/2) * 300\n= 33 * 150\n= 4950\n```\n\n### 35) The sum of three numbers in an Arithmetic Progression is 45 and their product is 3000. What are the three numbers?\n\nA. 5, 15, 25\n\nB. 12, 15, 18\n\nC. 10, 15, 20\n\nD. -10, -15, -20\n\nSolution:\n\n```\nAssuming that the numbers are (a – d), a, (a + d) and their sum is 45, we get the middle number as 15. Now, the product (a – d) (a + d) = 200. Solving, we get d = 5. Therefore, the numbers are 10, 15 and 20.\n```\n\n### 36) if 1 is added to the denominator of a fraction, the fraction becomes 1/2. if 1 is added to the numerator of the fraction, the fraction becomes 1. find the fraction\n\nSolution: \n\n```\nx/(y+1)=1/2\n\nand\n\n(x+1)/y=1\n\n2x-y=1 ....eq (1)\n\nx=y-1 ....eq (2)\n\nby solving eq 1 and 2 we get\n\nx=2 and y=3\n```\n\n### 37) After the division of a number successively by 3, 4 and 7, the remainder obtained is 2, 1 and 4 respectively. What will be remainder if 84 divide the same number?\n\nA. 80\n\nB. 75\n\nC. 42\n\nD. 53\n\nAnswer: \n\n```\nOption D\n\nSolution: \nAs the Number gives a remainder of 4 when it is divided by 7, then the number must be in form of (7x + 4)\n\nThe same gives remainder 1 when it is divided 4, so the number must be in the form of {4 × (7x + 4) + 1}\n\nAlso, the number when divided by 3 gives remainder 2, thus number must be in form of [3 × {4 × (7x + 4) + 1} + 2]\n\nNow, On simplifying,\n[3 × {4 × (7x + 4) + 1} + 2]\n= 84x + 53\nWe get the final number 53 more than a multiple of 84 Hence, if the number is divided by 84,\nThe remainder will be 53\n```\n\n### 38) If the sum of two numbers is 55 and the H.C.F. and L.C.M. of these numbers are 5 and 120 respectively, then the sum of the reciprocals of the numbers is equal to:\n\nA) 55/601\n\nB) 601/55\n\nC) 11/120\n\nD) 120/11\n\nSolution: \n\n```\nThen, a + b = 55 and ab = 5 x 120 = 600.\nThe required sum =1a+1b = a+bab= 55600=11120\n```\n\n### 39) In a certain code language, ‘sea is deep’ means ‘213’, ‘sky is blue’ means ‘514’ and ‘sea looks blue’ means ‘264’\n\n### What number is the code for ‘sky’?\n\nSolution:\n```\n5\n```\n\n### 40) In the following questions, one term in the number series is wrong. Find the wrong term.\n\n2, 3, 5, 8, 13, 20, 34\n\nSolution: \n\n```\nfirst+second=third\nfollow this order you get **20** as a answer\n```\n\n### 41) In the following question, one term in the number series is wrong. Find the wrong term.\n\n196, 169, 144, 121, 100, 80\n\nSolution: \n\n```\n80\nFirst - Second=2\n```\n\n### 42) SCD, TEF, UGH, ____, WKL\n\nA.\tCMN\n\nB.\tUJI\n\nC.\tVIJ\t\n\nD.\tIJT\n\nSolution: \n\n```\nC\n```\n\n### 43) In a certain code 'MISSIONS' is written as 'MSIISNOS'. How is 'ONLINE' written in that code?\n\n1. OLNNIE\n\n2. ONILEN\n\n3. NOILEN\n\n4. LNOENI\n\n5. ONNLIE\n\nSolution: \n\n```\n1\n```\n\n### 44) The population of a city A which is 68000 decreases at the rate of 1200/year. Population of city B which is 42000, increases at the rate of 800 per year. Find in how many years, the population of cities A and B are equal?\n\na. 9 years\n\nb. 10 years\n\nc. 13 years\n\nd. 15 years\n\nSolution: \n\n```\nCorrect Option: (c)\n\nWe have to find the population of cities A and B after x years.\n\nStep 1: Population of city A = 68000, decreases at the rate of 1200/year\n68000 – 1200x\n\nStep 2: Population of city B = 42000, increases at the rate of 800/year\n42000 + 800x\n\nStep 3: Find after how many population of cities A and B are equal.\n\nPopulation of city A = Population of city B\n68000 – 1200x = 42000 + 800x\n68000 – 42000 = 1200x + 800x\n26000 = 2000x\nx = 13\n```\n\n\n### 45) A contractor pays Rs. 20 to a worker for each day and the worker forfeits Rs. 10 for each day if he is idle. At the end of 60 days, the worker gets Rs. 300. Find for how many days the worker was idle?\n\na. 28 days\n\nb. 30 days\n\nc. 34 days\n\nd. 40 days\n\nSolution:\n\n```\nCorrect Option: (b)\n\nStep 1: Number of days worked by the worker = 60 and he remained idle for x days. Therefore, number of days worked = (60 – x)\n\nStep 2: Each day he was getting paid Rs. 20. Therefore, the payment received for working days = (60 – x) 20\n\nStep 3: After subtracting the amount which he forfeited, he receives Rs. 300.\n\nTherefore,\n(60 – x) 20 – 10x = 300\n1200 – 20x – 10x = 300\n900 = 30x\nx = 30 days\n```\n\n### 46) In a farm, along with 50 hens, there were 45 goats and 8 horses and some farmers. If total number of feet be 224 more than number of heads, then find the number of farmers.\n\na. 11\n\nb. 15\n\nc. 16\n\nd. 18\n\nSolution: \n\n```\nCorrect Option: (b)\n\nLet’s the number of farmers be y.\n\nStep 1: Find number of heads\n= (50 hens + 45 goats + 8 horses + y farmers)\n= (103 + y)\n\nStep 2: Number of feet\n= [(Hens 2 × 50) + (45 × 4) + (8 × 4) + (y × 2)]\n= [100 + 180 + 32 + 2y]\n= 312 + 2y\n\nStep 3: Find number of farmers\n(312 + 2y) – (103 + y) = 224\n312 + 2y – 103 – y = 224\ny = 15\n```\n\n### 47) What is the difference in the place value of 5 in the numeral 754853?\n\nA) 49500\n\nB) 49950\n\nC) 45000\n\nD) 49940\n\nSolution: \n```\nThe Correct answer is (B)\n\nAnswer with explanation:\n\nThe digit 5 has two place values in the numeral, 5 * 105 = 50,000 and 5 * 101 = 50.\n\n∴Required difference = 50000 - 50 = 49950\n```\n\n### 48) What is the compound interest on Rs. 2500 for 2 years at rate of interest 4% per annum?\n\nA) Rs. 180\n\nB) Rs. 204\n\nC) Rs. 210\n\nD) Rs. 220\n\nSolution:\n```\nOption B\nCI=P(1+r/100)^t\n\nCI=2500*(1+4/100)^2\nCI=2704\nSo the diffrenece is 204\n```\n\n### 48) Sohan started a business with a capital of Rs. 80000. After 6 months Mohan joined as a partner by investing Rs. 65000. After one year they earned total profit Rs. 20000. What is share of Sohan in the profit?\n\nA) Rs. 5222.2\n\nB) Rs. 5777.7\n\nC) Rs. 6222.2\n\nD) Rs. 6777.7\n\nSolution: \n```\nOption B\n80000*12/65000*6=32/13\n113/32*20000=5777.7rs\n```\n\n### 49) If January 1, 1996, was Monday, what day of the week was January 1, 1997?\n\nA) Thursday\n\nB) Wednesday\n\nC) Friday\n\nD) Sunday\n\nSolution: \n```\nThe correct option is (B)\n\nExplanation:\n\nThe year 1996 is divisible by 4, so it is a leap year with 2 odd days.\n\nAs per the question, the first day of the year 1996 was Monday, so the first day of the year 1997 must be two days after Monday. So, it was Wednesday.\n```\n\n### 50) The speed of a boat in still water is 5km/hr. If the speed of the boat against the stream is 3 km/hr, what is the speed of the stream?\n\nA) 1.5 km/hr\n\nB) 2 km/hr\n\nC) 2.5 km/hr\n\nD) 1 km/hr\n\nSolution:\n```\nThe correct answer is B\n\nAnswer with explanation:\n\nLet the speed of stream = X km/hr\n\nSpeed of boat = 5 km/hr\n\nSpeed upstream = 3km/hr\n\nApply formula: Speed upstream = speed of boat - speed of stream\n\n∴ 3 = 5 - X\n\nX = 5 - 3 = 2 km/hr\n```\n\n### 51)  How many times the hands of a clock coincide in a day?\n\nA) 24\n\nB) 22\n\nC) 23\n\nD) 21\n\nSolution: \n```\nThe Correct answer is (B)\n\nExplanation:\n\nThe hands of a clock coincide only once between 11 O' clock and 1 O' clock, so in every 12 hours, the hands of a clock will coincide for 11 times.\n\n∴ In a day or 24 hours, the hands of a clock will coincide for 22 (11+11) times.\n```\n\n### 52) A pipe can fill a tank in 6 hours and another pipe can empty the tank in 12 hours. If both the pipes are opened at the same time,the tank can be filled in\n\nA) 10 hours\n\nB) 12 hours\n\nC) 14 hours\n\nD) 16 hours\n\nSolution:\n\n```\nOption B\nWater enter in 1 hr=1/6\nWater empty in 1 hr=1/12\nnet=1/6-1/12=1/12\nor 12hr\n```\n\n### 53) A running man crosses a bridge of length 500 meters in 4 minutes. At what speed he is running?\n\nA) 8.5 km/s\n\nB) 7.5 km/s\n\nC) 9.5 km/s\n\nD) 6.5 km/s\n\nSolution: \n\n```\nOption B\nsec=4*60=240s\nspeed=500/240=25/12m/s\nin km/s speed is =25/12*18/5=7.5km/s.\n```\n### 54) The surface area of a cube is 600 cm2. The length of its diagonal is\n\nSolution: \n```\nSurface area of cube=6a^2\n600=6*a^2\na=10\ndiagonal of cube=sqrt(3)*a\nans=sqrt(3)*10\n```\n\n### 55) \t\nThe H.C.F. of two numbers is 23 and the other two factors of their L.C.M. are 13 and 14. The larger of the two numbers is:\n\nA.\t276\n\nB.\t299\n\nC.\t322\n\nD.\t345\n\nSolution:\n```\nAnswer: Option C\n\nExplanation:\n\nClearly, the numbers are (23 x 13) and (23 x 14).\n\n Larger number = (23 x 14) = 322.\n ```\n \n ### 56) \t\nSix bells commence tolling together and toll at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together ?\n\nA.\t4\n\nB.\t10\n\nC.\t15\n\nD.\t16\n\nSolution:\n```\nAnswer: Option D\n\nExplanation:\n\nL.C.M. of 2, 4, 6, 8, 10, 12 is 120.\n\nSo, the bells will toll together after every 120 seconds(2 minutes).\n\nIn 30 minutes, they will toll together\t30\t+ 1 = 16 times.\n2\n```\n\n### 57) \t\nLet N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then sum of the digits in N is:\n\nA.\t4\n\nB.\t5\n\nC.\t6\n\nD.\t8\n\nSolution:\n```\nAnswer: Option A\n\nExplanation:\n\nN = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305)\n\n  = H.C.F. of 3360, 2240 and 5600 = 1120.\n\nSum of digits in N = ( 1 + 1 + 2 + 0 ) = 4\n```\n\n### 58) The greatest number of four digits which is divisible by 15, 25, 40 and 75 is:\n\nA.\t9000\n\nB.\t9400\n\nC.\t9600\n\nD.\t9800\n\nSolution\n```\nAnswer: Option C\n\nExplanation:\n\nGreatest number of 4-digits is 9999.\n\nL.C.M. of 15, 25, 40 and 75 is 600.\n\nOn dividing 9999 by 600, the remainder is 399.\n\n Required number (9999 - 399) = 9600.\n```\n\n### 59) If 15 men can reap the crops of a field in 28 days, in how many days will 5 men reap it?\n\nA) 50 days\n\nB) 60 days\n\nC) 84 days\n\nD) 9.333 days\n\nSolution: \n```\nAnswer: C\n\nExplanation:\n\nLet 5 men can reap a field in x days\nSo, put the same quantities on the same side.\nMen: Days\nNow, Men and Days are inversely proportional to each other. If we increase the number of men fewer days will be required to complete the work.\nInversely proportional means Apti Chain Rule\nApti Chain Rule\n\ni.e., 5: 15 = 28: x\nOr, x = (28*15)/ 5\nOr, x = 84 days\nHence, 5 men can reap a field in 84 days.\n```\n\n### 60)  Find the logarithm of 1/256 to the base 2√2.\n\nA) 16\n\nB) 13/5\n\nC) -16/3\n\nD) 12\n\nSolution: \n```\nAnswer: C\n\nExplanation:\n\nLet log2√2 [1/256] = x\n\nWe know that loga y = x is similar to ax = y\n\nSo, we can write it as [1/256] = (2√2) x\n\nOr, (2√2) x = [1/28]\n\nOr, [21 * 21/2]x = 1/28\n\nOr, 23x/2 = 2-8\n\nTherefore, 3x/2 = -8\n\nHence, x = (-8 * 2)/ 3 = -16/3\n```\n\n### 61) In what ways the letters of the word \"RUMOUR\" can be arranged?\n\nSolution: \n```\n(6!)/(2!)(2!)=180\n```\n\n### 62) Find the remainder when 1! + 2! + 3! +4! + 5! + ————– 1000! is divided by 8?\n\nSolution:\n```\nYou can see number less than 3! are not divisible by 8 so it decide your output\n(1!+2!+3!)=9\n9%8=1\n1 is the answer\n```\n\n### 63) 106 x 106 - 94 x 94 = ?\n\nA.\t2400\t\n\nB.\t2000\n\nC.\t1904\n\nD.\t1906\n\nE.\tNone of these\n\nSolution: \n```\n106 x 106 - 94 x 94\t= (106)2 - (94)2\n= (106 + 94)(106 - 94)    [Ref: (a2 - b2) = (a + b)(a - b)]\n= (200 x 12)\n= 2400.\n```\n\n### 64)  \t\nThe difference of two numbers is 1365. On dividing the larger number by the smaller, we get 6 as quotient and the 15 as remainder. What is the smaller number ?\n\nA.\t240\n\nB.\t270\t\n\nC.\t295\n\nD.\t360\n\nAnswer: Option B\n\nExplanation:\n```\nLet the smaller number be x. Then larger number = (x + 1365).\n\n x + 1365 = 6x + 15\n\n 5x = 1350\n\n x = 270\n\nSmaller number = 270\n```\n\n### 65) The sum of first 45 natural numbers is:\n\nA.\t1035\n\nB.\t1280\n\nC.\t2070\n\nD.\t2140\n\nAnswer: Option A\n\nExplanation:\n\n```\nLet Sn =(1 + 2 + 3 + ... + 45). This is an A.P. in which a =1, d =1, n = 45.\n\nSn =\tn\t[2a + (n - 1)d]\t=\t45\tx [2 x 1 + (45 - 1) x 1]\t=\t\t45\tx 46\t\t= (45 x 23)\n2\t2\t2\n= 45 x (20 + 3)\n\n= 45 x 20 + 45 x 3\n\n= 900 + 135\n\n= 1035.\n\nShorcut Method:\n\nSn =\tn(n + 1)\t=\t45(45 + 1)\t= 1035.\n2\t2\n```\n\n### 66) Ronald and Elan are working on an assignment. Ronald takes 6 hours to type 32 pages on a computer, while Elan takes 5 hours to type 40 pages. How much time will they take working together on two different computers to type an assignment of 110 pages ?\n\n[A]7 hours 30 minutes\n\n[B]8 hours\n\n[C]8 hours 15 minutes\n\n[D]8 hours 25 minutes\n\nSolution) \n```\nC)\nRonald 1 hr work = 32/6=16/3\nElan 1 hr work = 40/5=8\nShow both work in an hr=8+16/3=40/3\nShow for 110 pages it will take 110/(40/3) or (110 x 3)/40=33/4hr \nSince: convert it into hr 4*8=32 1 left in 1 hr 60 min 60/4=15min\n\nShow final answer is 8hr 15 mon\n```\n\n### 67) 4^x + 6^x = 9^x\n\nSolution: \n\n[link to solution](https://www.youtube.com/watch?v=6AwfRXKvGsM)\n\n```\n4^x/4^x + 6^x/4^x = 9^x/4^x\nNow,\n1 + (3/2)^x=(3/2)^(2x)\nConsider (3/2)^x=u\nThen,\n1 + u = u^2\nSimplifying this\n\n0 = u^2 -u -1\n\nBy solving we get\nu = (1 + sqrt(5))/2\n\nand this equal to\n(1 + sqrt(5))/2 = (3/2)^x\nTake log both side\n\nand you get 1.187 approx value.\n```\n\n### 68) The diameter of a wheel is 98 cm. The number of revolutions in which it will have to cover a distance of 1540 m is\n\n```\nSolution:\ncircumference of an wheel=πd\n=22/7×98\n22×14\n=308cm =1 revolution\ndistance covered\n1540×100=154000\nnow,154000÷308\n500 rotations\n```\n\n### 69) An error 2% in excess is made while measuring the side of a square. The percentage of error in the calculated area of the square is:\nA. 2%\n\nB. 2.02%\n\nC. 4%\n\nD. 4.04%\n\nE. None of these\n\nAnswer: Option D\n\n```\n100 cm is read as 102 cm.\n∴ A1 = (100 x 100) cm2 and A2 (102 x 102) cm2\n(A2 - A1) = [(102)2 - (100)2]\n              = (102 + 100) x (102 - 100)\n              = 404 cm2\n∴ Percentage error\n=(404100×100×100)%=4.04%\n```\n\n### 69) A copper wire when bent in the form of a square encloses an area of 484 sq cm if the same wire is bent in the form of a circle the area enclosed by it is\n\n```\nArea of the square=484cm \nside-22cm\nperimeter=22*4=88cm\ncircumfrence of cirlce is 2*pi*r=88\nr=14cm\narea=pi*r*r=616cm^2\n ```\n\n### 70) When two dice are rolled, what is the probability that the sum is either 5 or 8?\n\nSolution\n```\nP(A)=1/9\nP(B)=1/6\nP(C)=26/36=13/18\nApply GP\nyou get 2/5 ans\n```\n\n","project_url":"https://awesome.ecosyste.ms/api/v1/projects/github.com%2Fpsingh12354%2Faptitude","html_url":"https://awesome.ecosyste.ms/projects/github.com%2Fpsingh12354%2Faptitude","lists_url":"https://awesome.ecosyste.ms/api/v1/projects/github.com%2Fpsingh12354%2Faptitude/lists"}