my take:: see u can write the general term as 1/(n*n+4) now in this case 1/4(1/n-1/n+4) now applying values we get 1/4(1-1/5+1/5-1/9.......1/224-1/225) all the terms will cancel out except 1/4(1-1/225) so the answer is 56/225
1) how many kilograms of sugar worth rs.3.60/kg should b mixed with 8kg of sugar worth rs.4.20/kg such tht by selling d mixture at rs.4.40/kg there may b a gain of 10%
a) 6kg b) 3kg c) 2kg d) 5kg e) 4kg
2) a m ixture of 125 gallon of wine and water contains 20% water.how much water must b added to d mixture in order to increase the percentage of water to 25% of d new mixture?
a) 10gals b) 8.5gals c) 8gals d) 6.66gals e) 8.33gals
1) how many kilograms of sugar worth rs.3.60/kg should b mixed with 8kg of sugar worth rs.4.20/kg such tht by selling d mixture at rs.4.40/kg there may b a gain of 10%
a) 6kg b) 3kg c) 2kg d) 5kg e) 4kg
2) A mixture of 125 gallon of wine and water contains 20% water.how much water must b added to d mixture in order to increase the percentage of water to 25% of d new mixture?
a) 10gals b) 8.5gals c) 8gals d) 6.66gals e) 8.33gals
Q1. First of all you need to find the CP of the new mixture which can be got as Rs 4.4/1.1 = Rs 4. (10% profit). Now it becomes a ratio proportion problem where in x kg of 3.6 mix + 8 kg of 4.2 mix gives u (x+kg of Rs 4 mix. Solve this you get x= 4. Thats your answer.
Q2. 125 gall of mix has 100 gall of wine 25 gall of water. if you add x gall of water to this. Mix becomes 125+x gall . water is 25 +x gall.
(25+x)/(125+x) * 100% =25%
Solve x n you get x=8.33galls.
Let me know if the answers are correct n you r able to follow.
1) how many kilograms of sugar worth rs.3.60/kg should b mixed with 8kg of sugar worth rs.4.20/kg such tht by selling d mixture at rs.4.40/kg there may b a gain of 10%
a) 6kg b) 3kg c) 2kg d) 5kg e) 4kg
2) A mixture of 125 gallon of wine and water contains 20% water.how much water must b added to d mixture in order to increase the percentage of water to 25% of d new mixture?
a) 10gals b) 8.5gals c) 8gals d) 6.66gals e) 8.33gals
By mixture and alligation the answer is 4 kg
and for the second one just check in case of option 8.33 is satisfying
1) how many kilograms of sugar worth rs.3.60/kg should b mixed with 8kg of sugar worth rs.4.20/kg such tht by selling d mixture at rs.4.40/kg there may b a gain of 10%
a) 6kg b) 3kg c) 2kg d) 5kg e) 4kg
2) A mixture of 125 gallon of wine and water contains 20% water.how much water must b added to d mixture in order to increase the percentage of water to 25% of d new mixture?
a) 10gals b) 8.5gals c) 8gals d) 6.66gals e) 8.33gals
1) use the allegation formula to get the answer i.e qty cheap / qty dear = (CP of dear - Avg CP) / (Avg CP - CP of cheap)
the ans is e) 4kg
2) (25 + x) / (125 + x) = 1 / 4, the ans is e) 8.33gals
help me to understand Q no.47 of LOD III(number system). the question is very long(it's a passage type) so i m nt posting here. actually i m not able to get any conclusion frm that passage.
help me to understand Q no.47 of LOD III(number system). the question is very long(it's a passage type) so i m nt posting here. actually i m not able to get any conclusion frm that passage.
well, i chkd the ques..vil post it in DI thread ...
can sum1 explain how dis sum can b done... How many batting orders are possible for da indian cricket team if there is a squad of 15 to choose from such that sachin tendulkar is always chosen? a)1001. 11! b)364. 11! c)11! d) 15. 11! thnx in advance....
can sum1 explain how dis sum can b done... How many batting orders are possible for da indian cricket team if there is a squad of 15 to choose from such that sachin tendulkar is always chosen? a)1001. 11! b)364. 11! c)11! d) 15. 11! thnx in advance....
out of 15 the 1 is fixed so we need to choose the 10 players out of 14 available which can be done in 14C10 ways so the answer is
14C10*11! also they can be arranged in 11! ways Option 1 1001*11!
Q.67-Lod2-NS-> K is a 3-digit number such tat the ratio of number to the sum of its digits is least. what is the difference between hundreds and the tens digit of K. (a) 9 (b) 8 (c) 7 (d) none of these.
Q.68 -> In question 67, what can be said about the difference between tens and units digit. (a) 0 (b) 1 (c) 2 (d) none of these
Q69 -> For the above questions, for how many values of K wil the ratio be highest. (a) 9 (b) 8 (c) 7 (d) none of these
Q85 -> Find last two digits of the following number (201*202*203*204*246*247*248*249)^2 (a) 36 (b) 56 (c) 76 (d) 16
Q89 -> The super computer at Ram Mohan Roy Seminary takes an i/p of a no. N & a X where X is a factor of N. In a particular case N is equal to 83p796161q & X is equal to 11 where 0
A triangular number is defined as a number which has the property of being expressed as a sum of consecutive natural numbers starting with 1. How many triangular numbers less than 1000, have the property that they are the difference of squares of 2 consecutive natural numbers? (a)20 (b)21 (c)22 (d)23
Q93. -> How many integer values of x & y are there such that 4x+7y=3 while x (a)144 (b)141 (c)143 (d)142
Q95 -> How many 2-digit numbers less than or equal to 50, have product of the factorials of their digits less than or equal to sum of the factorials of their digits. (a) 17 (b) 16 (c) 15 (d) none of these
Also, questions 97 to 100 from LOD-2(Number systems)
Q.67-Lod2-NS-> K is a 3-digit number such tat the ratio of number to the sum of its digits is least. what is the difference between hundreds and the tens digit of K. (a) 9 (b) 8 (c) 7 (d) none of these.