A. 453600

B. 128000

C. 478200

D. 635630

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**ANSWER: A. 453600Solution: This is a simple problem of permutation and combination.Let us consider the number of times each alphabet is being used:E: 2N: 1C: 2Y: 1L: 1O: 1P: 1A: 2I: 1D: 1Now, total number of consonants: 7=> Total number of vowels: 613 letter word ENCYCLOPAEDIA has 6 vowels and there are 6 even spaces=>6 vowels can be arranged in 6 spaces in **

A. 10/19

B. 9/38

C. 9/19

D. 5/38

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**ANSWER: C. 9/19 Solution: There are a total of 20 balls out of which 2 balls are to be randomly selected=> There are a total of 10 black and 10 white balls so one can either get 2 black balls or 2 white balls.=> Number of cases in which both balls are of same color: 10C2 (Black) + 10C2 (White)=> Total number of cases: **

A. 2:3:1

B. 1:3:2

C. 2:1:3

D. 1:2:3

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**ANSWER: C. 2:1:3 Solution: Volume of a sphere= (2/3) πr**

A. 0.3010

B. 0.4331

C. 3.3222

D. 0.6990

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**ANSWER: C. 3.3222 Solution: By logarithmic properties we know that:Log**

A. Rs 420

B. Rs 360

C. Rs 450

D. Rs 480

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**ANSWER: D. Rs 480 Solution: Given data:Time(t) = 2 yearsRate(R) = 12.5%Compound Interest = 510 RsLet the principal amount be XFormula for total amount is: = X(1+R/100) **

A. 1:6

B. 1:3

C. 6:1

D. 3:1

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**ANSWER: A. 1:6 Solution: To start off this question let us assume that cost price of 1 litre milk is Rs 1Now need to make a mixture and sell this mixture at 1 Rs per litre such that the total gain on mixture is 16.667%.Therefore, CP of 1 litre of mixture becomes (quantity of milk)/ (quantity of mixture containing 1 L milk)*(price of 1 litre milk).=>(100/(100+50/3))*1=>CP of 1 litre milk of mixture: Rs 5/6As price of any amount of water is zero, and as 1 litre milk costs Rs 1.5/6litre of mixture will comprise entirely of cost of milk which means,1 litre of mixture will contain 5/6th amount of milk.=>Water is added in the ratio of (1-5/6)=1/6**

A. 50 m

B. 150 m

C. 200m

D. 130m

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**ANSWER: B. 150 m Solution: Let us assume for this question that the length of train is x metres and it is assumed to be running at the speed of y m/sec.A pole is assumed as a point object.=>Time taken by the train to pass the pole= x/y=>15=x/y=>y=x/15Now, the train passes the platform which is 100 m in length in 25 seconds.=>x+100/y=25=>x+100/25=yEquating speed generated from both cases:x+100/25=x/15 Therefore x = 150 metres.**

A. 3

B. 6

C. 11

D. 12

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**ANSWER: B. 6 Solution: The question is based on efficiency to do work.Ram can finish a puzzle in 3 hours. Shyam can finish the puzzle in 2 hours.=>In one hour Ram can finish 1/3rd of a puzzle=>In one hour Shyam can finish half the puzzleA total of 15 candies are to be shared amongst both of themHence Ram’s share must be = ((Work done by Ram in 1 hour)/(Work done by Shyam in one hour)+(Work done by Ram in hour))*15=> 1/3/((1/2)+(1/3))*15=> (1/3/(5/6))*15=> 6/15*(15)=>6 candies**

A. Rs 12,420

B. Rs 18,040

C. Rs 18,942

D. Rs 12,628

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**ANSWER: D. Rs 12,628 Solution: For a given business profit is directly dependent upon the capital invested and the time of investment.=> Ratio of shares of Akash and Akshay becomes: (35,000*8)/(42,000*10) = 2/3=>% of profit belonging to Akash: 2/(3+2)*(31,570)=>Rs 12,768**

A. 1

B. 2

C. 4

D. 7

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**ANSWER: B. 2 Solution: It is asked in the question that we need to find the number which when subtracted by 14,17 and 34,42 respectively will give their remainders in same ratio.Mathematically this can be represented in this form: => (14-x)/(17-x) = (34-x)/(42-x)=> Applying Componando &Dividendo=> C/D = (C+D)/(C-D)=> (31-2x)/(-3) = (76-2x)/(-8)=> 248-16x = 228-6x =>20=10x=>2=x**

**RE: Aptitude model placement papers with solution - set 9 -Sandeep (06/13/16)**- The questions were really helpful.

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