1. A sum of money placed at CI doubles itself in 3 years.In how many years will it amount to 8 times itself?2.If the difference between CI and SI on a certain sum of money for 3 years at 2% p.a is rs 604,what is the sum?
Brass is an alloy of copper and zinc. Bronze is an alloy containing 80% copper , 4% zinc & 16% tin. A fused mass of brass and bronze is found to contain 74% copper, 16% zinc & 10% tin.The ratio of copper and zinc in brass isa.64% & 36% b. 33% & 67% c.35% & 65% d. None of these.Please help
ans:- a)
let mixture of brass(Bs)+ bronze(Bz) be 100gm
so cu=74gm, zn=16gm, tin=10gm.
bt since brass hs no tin, therefore dis whole tin hs come frm bronze. so 16% of Bronze= 10gm
Brass is an alloy of copper and zinc. Bronze is an alloy containing 80% copper , 4% zinc & 16% tin. A fused mass of brass and bronze is found to contain 74% copper, 16% zinc & 10% tin.The ratio of copper and zinc in brass isa.64% & 36% b. 33% & 67% c.35% & 65% d. None of these.Please help
Use ALLIGATION to determine the ratio of brass to bronze in the fused mass. Brass = 0% tin. Bronze = 16% tin. Fused mass = 10% tin. Step 1: Plot the 3 percentages on a number line, with the percentages in brass and bronze (0% and 16%) on the ends and the percentage in the fused mass (10%) in the middle. Brass 0%----------------10%-------------------16% Bronze Step 2: Calculate the distances between the percentages. Brass 0%--------10------10%-------6-----------16% Bronze The ratio of the distances = 10:6 = 5x : 3x. The ratio of the distances must be the same for COPPER. Percentage of copper in the fused mass = 74%. Percentage of copper in bronze = 80%. Let the percentage of copper in brass = C. Step 3: Plot the percentages for COPPER on the number line. Brass C%--------5x------74%-------3x-----------80% Bronze The number line above implies the following: 3x = 80-74 3x = 6 x=2. Since 5x = 5*2 = 10, the distance between C and 74 = 10. Thus, C = 74-10 = 64. Thus, brass is 64% copper and 36% zinc.
If the number A is even,which of the following will true?
(A)3A will always be divisible by 6
(b) 3A+5 will always be divisible by 11.
(c) (ASQUARE+3)/4 will be divisible by 7
(d)All of these
please help in it,option 2 and 3 will be incorrect...,in the book it is written as option 1 as correct but if we consider the value of A as 4,2...then it fall incorrect as it will not be divisible by 2 or 3....
Please solve the following question : LOD 2 ,Number SystemsQ 79.A set S is formed by including some of the First one thousand naural numbers.S contains the max no. of numbers such that they satisfy the following questions.1.No number of the set S is prime.2.When the numbers of the set S are selected two at a time , we always see co prime numbers.what is the number of elements in the set S?a)11 b)12 c)13 d)7
I saw this question somewhere on here and this was the solution given:
"Since all are composite numbers and all are co-prime to each other, we can say that all the numbers are product of prime numbers.
Also, 32 is greater than 1000, both the prime numbers in the product can not be greater than 31.
So, basically we just have to count number of prime numbers less than or equal to 31, i.e., 11 in numbers.
Also, we can have 1, as its neither prime nor composite.
=> We can have 12 number of elements "
Could someone please explain how the number of primes within 31 is equal to number of elements in set S? I think I've missed something. :(
@Sybar - co-prime means the numbers do not have common factors except of course 1.....the only way the condition in the question could be true is when each number in the set is prime or the second power of prime....the question says they can not be prime....so they have to be second power of the prime.....having said that, 32 square > 1000, so each number of the set has to be 31 square or less....
@UnCharismatic what do u mean!!.....3*A is always divisible by 3 as long as A is not a fraction....nd the question says that A is even (divisible by 2).....so 3A is divisible by 6...did i miss something here???
Production Pattern for the no.of units (in cubic feet) per day: days No.of units 1 150 2 180 3 120 4 250 5 160 6 120 7 150
For a truck that can carry 2000 cubic feet,hiring cost per day is 1000.Storing cost per cubic feet is Rs 5 per day.Any residual material left at the end of the 7th day has to be transferred.
Q1. If all units are to be sent to market,then on which days should the trucks be hired to minimize cost :
A. 2,4,6,7 B.7 C.2,4,5,7 D.None
Correct : A
Q2.If the storage cost is reduced to Rs 0.9 per cubic feet per day,then on which day/days should the truck be hired. A.4 B.7 C.4 and 7 D.None
"A mock test is taken at AMS learning systems .The test paper comprises of questions in three levels of difficulty --LOD1,LOD2and LOD3The following gives the details of the positive and negative marks attached to each question type : difficulty level________positive marks __________negative marks LOD1 _____________ 4_____________________2_____________ LOD2_______________3_____________________1.5___________ LOD3______________2______________________1_____________
The test had 200 questions with 80 on LOD 1 and 60 each on LOD2 and LOD3
Q1.If a student has solved 100 questions exactly and scored 120 marks , the maximum number of incorrect questions that he/she might have marked is : a 44 b 56 c 60 d none of these
Q2:-If Amit attempted the least number of questions and got a total of 130 marks , and if it is known that he attempted at least one of every type , then the number of questions must have attempted is :a 34 b 35 c 60 d none of these
Q3:-in the above question what is the least number of question he might have got incorrect ?a 0 b 1 c 2
Solution 1)For maximum number of incorrect questions, he should solve very less number of correct questions. In other words, he should have scored all his positive marks from the 4 mark questions (LOD1) and should have got all the negatives from the -1 mark (LOD3).
Let 'x' be the number of incorrect questions that he has solved and hence (100-x) would be the number of correct questions that he would have solved.
Thus, 4(100-x) -1(x) = 120
Solving, we get 400 - 5x = 120
or x = 56."
That's another question I found on the locked thread. I have a lot of problem with all three questions but maybe if I understand the first one I'll get the others.
In other words, he should have scored all his positive marks from the 4 mark questions (LOD1) and should have got all the negatives from the -1 mark (LOD3) Why? Why not all the positive marks from LOD 2 and all the negatives from some other LOD instead? Also the number of questions in LOD1 is different from LOD2 and LOD: 80, 60 and 60. Doesn't that affect that logic?
@Sybar no probs bro...the questions u have been posting are probably the 'toughest' (800 level) that you coud get on the test....I found the real test not so difficult really....keep practicing!!
Lets look at the new Q1....lets say the student marked the max number of incorrect questions to lead him to the 120 marks that he got and the marks he got deducted is m.....so if u look at your LOD1, 2 and 3....if he lost m marks and all of them were LOD2, he would have actually answered m/1.5 incorrectly, and if those were all LOD 1, then m/2 incorrectly whereas if all incorrect answers were LOD1, then he must have answered m/1 incorrectly...so which of these gives you the max number of incorrect answers??? of course it has to be m/1...... Also, to max out the number of incorrect answers, the number of correct answers will have to be a minimum or equal to total marks obtained for correct answers/ 4.......that is how u get to the equation in the solution....
4*(100-m)-1*m; i.e., max marks for 1 correct answer*total correct answers - minimum marks u get for 1 incorrect * total incorrect....got it??
Q2- Amit got 130 and answered the least number of questions. Now let us see how we get at 130 with the least number of questions. 120 is divisible by 4 (we use the highest possible markes 4 because that will translate to the least number of questions answered), so lets say he answered 120/4= 30.....now we have 10 more marks to adjust....3 more LOD1 correct is 132 and lets say one LOD1 incorrect, so that becaomes 130, therefore 34 questions....or else lets say 120+2 LOD2 correct + 2 LOD3 correct= 130, therefore 34 questions again....do not think further since 34 is one of the answers to the question u have, infact it is the lowest answer that you have....so the answer is A
Q3-In the solution to Q2 that I stated above if you refer 'or else lets say 120+2 LOD2 correct + 2 LOD3 correct= 130, therefore 34 questions again', then u can see that the smallest incorrect answers coud be 0. Therefore 0 is the answer.
Someone please help!