1)A milk man sold his buffalo at Rs.720/- at some profit.Had he sold his buffalo at Rs.510/-, the quantum of the loss incurred wud have been double that of profit earned. What is the cost price?
a)600 b)625 c)675 d)none
2)A Shopkeeper buys an article for Rs.400 and marks it for sale at a price that gives him 80 % profit on his cost.He however gives a 15% discount on the Marked price to his customer. Calculate the actual profit % made by shopkeeper ?
a)62% b)64% c)53% d)54%
3)Mithilesh makes 750 articles at a cost of 60 paise per article.He fixes the seeling price such tha if only 600 articles are sold,he would have made a profit of 40% on the outlay.However 120 articles got spoilt and he was able to sell 630 articles at this price.Find his actual profit percent of total outlay assuming that he unsold articles are useless.
The sides AB,BC,CA of a triangle ABC have 3,4,5 interior pts resptively on them. The total number of triangles tat can be constructed by using these pts as vertices are a) 212 b)210 c)205 d)190
How many 7 digit nos are there having digit 3 three times and digit 0 four times? 1)15 2)3^3 * 4^4 3)30 4) None of these
please can any one show me how to go abt these sums... i wud appreciate if i cud get the full explanation. Thanx
The sides AB,BC,CA of a triangle ABC have 3,4,5 interior pts resptively on them. The total number of triangles tat can be constructed by using these pts as vertices are a) 212 b)210 c)205 d)190
How many 7 digit nos are there having digit 3 three times and digit 0 four times? 1)15 2)3^3 * 4^4 3)30 4) None of these
please can any one show me how to go abt these sums... i wud appreciate if i cud get the full explanation. Thanx
How many 7 digit nos are there having digit 3 three times and digit 0 four times? 1)15 2)3^3 * 4^4 3)30 4) None of these
here first digit compulsory 3 to make 7 digit no
for remain 6 digit 3 two times ans 0 four times
so 6c2*4c4=15
ans is 15
The sides AB,BC,CA of a triangle ABC have 3,4,5 interior pts resptively on them. The total number of triangles tat can be constructed by using these pts as vertices are a) 212 b)210 c)205 d)190
How many 7 digit nos are there having digit 3 three times and digit 0 four times? 1)15 2)3^3 * 4^4 3)30 4) None of these
here first digit compulsory 3 to make 7 digit no
for remain 6 digit 3 two times ans 0 four times
so 6c2*4c4=15
ans is 15
The sides AB,BC,CA of a triangle ABC have 3,4,5 interior pts resptively on them. The total number of triangles tat can be constructed by using these pts as vertices are a) 212 b)210 c)205 d)190
ans is 5*6*7=210
hey cud you explain the second sum abt the triangles?
How do we identify that n! is of form 4n, 4n+1, 4n+2 or 4n+3....
This is used in questions like :-
Find the unit digit of (1234)^123!
Got the doubt while reading basics of number system.
Help appreciated.
For this particular question, no need scratching ur head on 4n, 4n+1, 4n+2 and all that stuff. Just remember that 123! is always going to be even. Now the repetition cycle of 4 is 2. Consider the following explanation: 4^1=4 4^=16 4^3=64 4^4=256 For even power unit digit of anything having 4 in the end is 6. Hence (1234)^123! has a unit digit of 6.
How many 7 digit nos are there having digit 3 three times and digit 0 four times? 1)15 2)3^3 * 4^4 3)30 4) None of these
here first digit compulsory 3 to make 7 digit no
for remain 6 digit 3 two times ans 0 four times
so 6c2*4c4=15
ans is 15
The sides AB,BC,CA of a triangle ABC have 3,4,5 interior pts resptively on them. The total number of triangles tat can be constructed by using these pts as vertices are a) 212 b)210 c)205 d)190
ans is 5*6*7=210
The answer to the triangle problem should be 205 (in my honest opinion). We are having total 12 points. We cant consider points A, B and C as the vertices' for the problem. So number of triangle possible is 12c3. But a set of 5 points are collinear, 4 points are collinear and 3 points are collinear. So we have to deduct them. 12c3-3c3-4c3-5c3= 220-1-4-10=205.
As for the other problem required answer is 6!/(4!*2!)=15.
The sides AB,BC,CA of a triangle ABC have 3,4,5 interior pts resptively on them. The total number of triangles tat can be constructed by using these pts as vertices are a) 212 b)210 c)205 d)190
ans is 5*6*7=210
I think the answer is 205....
No of cases: 1) When each of the vertices lies on different line segment: 3C1 x 4C1 x 5C1 = 60 2) When two vertices lie on one segment and the third one either of the other two: 3C2 x 9C1 + 4C2 x 8C1 + 5C2 x 7C1 = 27 + 48 + 70 = 145
No other cases possible. Hence the total possible cases = 60 + 145 = 205
I have one question from LOD II of Number Systems from Arun Sharma. The Question: Find the 28838rd term of the series:12345678910111213141516... The answer choices are a) 3 b) 4 c) 9 d) 7
The answer given is (c) 9 I have calculated this, but my answer is coming as (a) 3. Below is the explaination from my end From 1 to 9 Nos 9 From 10 to 99 Nos 180(90*2) From 100 to 999 Nos 2700(900*3) From 1000 to 6372 Nos 25492(6373*4)
Total : 28836
Hence from next number 6373 2nd number is 3 which makes a total of 28838. So my answer is (a) 3.
But as you know the answer given is 9.
Is there any mistake I have made? What is the correct way of calculating this if I am wrong somewhere? Please tell me.
I have one question from LOD II of Number Systems from Arun Sharma. The Question: Find the 28838rd term of the series:12345678910111213141516... The answer choices are a) 3 b) 4 c) 9 d) 7
The answer given is (c) 9 I have calculated this, but my answer is coming as (a) 3. Below is the explaination from my end From 1 to 9 Nos 9 From 10 to 99 Nos 180(90*2) From 100 to 999 Nos 2700(900*3) From 1000 to 6372 Nos 25492(6373*4)
Total : 28836
Hence from next number 6373 2nd number is 3 which makes a total of 28838. So my answer is (a) 3.
But as you know the answer given is 9.
Is there any mistake I have made? What is the correct way of calculating this if I am wrong somewhere? Please tell me.
well u hv made an obvious calculation mistake. It doesn't total up to 28836 as u hv done. It's actually coming 28381. u can recheck tht urself. And i too am not getting 9 as answer. I'm getting 7 as answer.
The question is : If you form a subset of integers choosen from between 1 to 3000, such that no two integers add up to a multiple of 9, what can be the maximum number of elements in the subset?
Answers: a) 1668 b)1332 c)1333 d)1334
I don't know any way to solve this problem in short time. Even I have not able to get the answer. The answer given is c)1333.
Hi Guys, Question : A set S is formed by including some of the First one thousand natural numbers. S contains the maximum number of numbers such that they satisfy the following conditions: 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?
My answer is If When the numbers of the set S are selected two at a time, we always see co prime numbers then it must be square of prime numbers like 2^2 = 4, 3^2 = 9, 25, 49, 121, 169, 289, 361, 429, 841, 961 = 31^2. which makes total of 11 numbers.
But the answer is 12.
Is 1 can be considered as one of these numbers(but 1 cannot be considered as a prime).
The question is : If you form a subset of integers choosen from between 1 to 3000, such that no two integers add up to a multiple of 9, what can be the maximum number of elements in the subset?
Answers: a) 1668 b)1332 c)1333 d)1334
I don't know any way to solve this problem in short time. Even I have not able to get the answer. The answer given is c)1333.
Can you please tell me how to do that?
A bit difficult to make understand. But i'll try my best. Divide all the nos from 1 to 3000 in following groups: 9x (1 number) 9x+1 (334 numbers) 9x+2 (334 numbers) 9x+3 (334 numbers) 9x+4 (332 numbers) Hence total numbers=1335 Another one of those where options are wrong. Can elaborate further if it's still not clear.
Hi Guys, Question : A set S is formed by including some of the First one thousand natural numbers. S contains the maximum number of numbers such that they satisfy the following conditions: 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?
My answer is If When the numbers of the set S are selected two at a time, we always see co prime numbers then it must be square of prime numbers like 2^2 = 4, 3^2 = 9, 25, 49, 121, 169, 289, 361, 429, 841, 961 = 31^2. which makes total of 11 numbers.
But the answer is 12.
Is 1 can be considered as one of these numbers(but 1 cannot be considered as a prime).
Any help?:shock:
1 is included,i think,as the definition of coprime numbers is those numbers who have just 1 as the common factor,which is true.so,1 has to be included,i think 😃
Can anyone provide the solution and the approach to this problem?
In an Examination the maximum marks for each of the three papaers is 50 each.The maximum marks for the fourth paper is 100.Find the number of ways with which a student can score 60% marks in aggregate
Can anyone provide the solution and the approach to this problem?
In an Examination the maximum marks for each of the three papaers is 50 each.The maximum marks for the fourth paper is 100.Find the number of ways with which a student can score 60% marks in aggregate
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