q3. a sq. has a side of 40 cm. another sq. is formed by joining the mid-points of the sides of the given sq. and this process is repeated infinitely. find the perimetre of all the squares thus formed. a) 160(1 + sqrt2) b) 160(2 + sqrt2) c)160(2 - sqrt 2) d) 160(1-sqrt2) e) 160(3 - sqrt2) ans b
Side of Square 1 = 40 cm Side of Square 2 = 40/root2 cm Side of Square 3 = 40/root2 * root2 cm Side of Square 4 = 40/root2 * root2 * root2 cm...and so on...
sum of infinite AP = 40(2+root2)...multiply it with 4.....160(2+root2)
q2. the sum of the first n terms of the AP is equal to half the sum of the next n terms of he same progression. find the ratio of the sum of the first 3n terms of the progression to the sum of its 1st n terms a) 5 b)6 c)7 d)8 e)9 ans d
well the answer is 6....
sum of n terms = 1/2(sum of 2n terms - sum of n terms) 3 (sum of n terms) = sum of 2n terms from here we get 2a = (n+1)d sum of 3n terms:sum of n terms = 6...
thanx for the solutions, but some1 plz solve this question
q1. one side of the staircase is to be closed in by rectangular planks from the floor to each step. the width of each plank is 9 inches and their height are successively 6 inch, 12 inches, 18 inches and so on. there are 24 planks required in total. find the area in square feet. a)112.5 b)107 c)118.5 d) none of these ans a
thanx for the solutions, but some1 plz solve this question
q1. one side of the staircase is to be closed in by rectangular planks from the floor to each step. the width of each plank is 9 inches and their height are successively 6 inch, 12 inches, 18 inches and so on. there are 24 planks required in total. find the area in square feet. a)112.5 b)107 c)118.5 d) none of these ans a
We need to find the total area. Height of first plank is 6 inch and the height is increasing by 6 inches for each of the next plank. Hence the height of last plank (24th plank) is = 6 + (24-1) * 6 =144 inch Now the total area is = 9 * ( 6+ 12+ 18+ .... + 144) = 9 * 24 * ( 6 +144) /2 = 9*1800 sq inches = 16200 sq inches = 16200/144 sq ft = 112.5 sq ft.
Please solve this question by using alligations only:
A mixture of 125 gallons of wine and water contains 20% water. How much water must be added to the mixture in order to increase the %of water to 25% in the new mixture?
Please solve this question by using alligations only:
A mixture of 125 gallons of wine and water contains 20% water. How much water must be added to the mixture in order to increase the %of water to 25% in the new mixture?
10 gals 8.33 gals 8 gals 6.66 gals
Please give detailed explanation.
Thanks.:splat:
Answer is 8.33 gallons. Since 125 gallon of mixture contains 20% water, thus the mixture is of the form 25:100 of water:wine. Now, if we add x amount of water to the mixture, the ratio of water to the mixture would become 25%.
Suppose speed of Ravi is v and he walked for time t mins. speed of Ravi's wife = 5v. from first condition ravi saved 12 mins by walking. it means that if he has covered distance d then his wife would have covered 2d (going from the meeting point to the station and coming back) in 12 mins. hence (2 * v * t)/ 5v = 12 mins => t = 30 mins. It means ravi has walked for 30 mins. this distance can be covered by his wife in car in 30/5 = 6 mins. hence if his wife was at station when ravi arrived, they could have saved 30-6 = 24 mins and he could have reached by 5:36 o'clock. what is the write answer.
1) A set of 15 different words is given. In how many ways it is possible to choose the subset of noy more than 5 words? 1)4944 2) 4^15 3) 15^4 4) 4943
2)There are 10 subjects in the school day but 6th standard students have only 5 periods in a day. In how many ways can we form the timetable for the day for 6th standard? 1)5^10 2) 10^5 3) 252 4) 30240
1) A set of 15 different words is given. In how many ways it is possible to choose the subset of noy more than 5 words? 1)4944 2) 4^15 3) 15^4 4) 4943
2)There are 10 subjects in the school day but 6th standard students have only 5 periods in a day. In how many ways can we form the timetable for the day for 6th standard? 1)5^10 2) 10^5 3) 252 4) 30240
1) a set of 15 different words is given. In how many ways it is possible to choose the subset of noy more than 5 words? 1)4944 2) 4^15 3) 15^4 4) 4943
2)there are 10 subjects in the school day but 6th standard students have only 5 periods in a day. In how many ways can we form the timetable for the day for 6th standard? 1)5^10 2) 10^5 3) 252 4) 30240
1. Find 28383rd term in series 1234567891011.. a.3 b.4 c.9 d.7
Please take the pains of going through previous pages before asking a question. This one in particular, has been asked over and over again, and answered as well. Just a few posts in the last month or so in this very thread:
Hi can anybody pleas help me with these sums ?? LOD II Q 66 The series of niumbers ( 1, 1/2,1/2,1/4 .....1/1972) is taken and now two numbers are taken from the series ( the fisrst two say x,y ). Then the operation x+y+x.y is performed to get a cosolidated no. . The process is repeated > What will be the value of the set after all the no.s are consolidated into 1 no. a 1970 b 1971 c 1972 d NONE
I think the x+y term is to be dealt by sum of a HP but i dunno dat and for the x.y term i believe the number will be so small that we can approximate it to zero
Q 67 K is a three digit no. such that the ratio of the no. to the sum of the digits is least ? What's the diff. between the hundreds and tens digit of K
a 9 b 8 c 7 d NONE
shouldn't the ans. be 999/27 = 37 ??
Q 78 --- what's the remainder when (1!)^3+(2!)^3 + (3!)^3 ......(1152!)^3 is divided by 1152 ???
P.S. also tell me the trick for finding the last two digits when two or more no.s are multiplied e.g Q 80 101*102*103*197*198*199 has las two digits ?? a 54 b 64 c 74 d 84
Shudn't it be 64?? Please answer thes doubts of mine
Hi can anybody pleas help me with these sums ?? LOD II Q 66 The series of niumbers ( 1, 1/2,1/2,1/4 .....1/1972) is taken and now two numbers are taken from the series ( the fisrst two say x,y ). Then the operation x+y+x.y is performed to get a cosolidated no. . The process is repeated > What will be the value of the set after all the no.s are consolidated into 1 no. a 1970 b 1971 c 1972 d NONE
I think the x+y term is to be dealt by sum of a HP but i dunno dat and for the x.y term i believe the number will be so small that we can approximate it to zero
Q 67 K is a three digit no. such that the ratio of the no. to the sum of the digits is least ? What's the diff. between the hundreds and tens digit of K
a 9 b 8 c 7 d NONE
shouldn't the ans. be 999/27 = 37 ??
Q 78 --- what's the remainder when (1!)^3+(2!)^3 + (3!)^3 ......(1152!)^3 is divided by 1152 ???
P.S. also tell me the trick for finding the last two digits when two or more no.s are multiplied e.g Q 80 101*102*103*197*198*199 has las two digits ?? a 54 b 64 c 74 d 84
Shudn't it be 64?? Please answer thes doubts of mine
Q66-If we take first two nos like 1 and 1/2 then x+y+x.y= 2 Similarly if we take next no 1/3 then it will be 2+1/3+2.1/3 = 3 Similarly the sum of next two nos is 4 which means the sum of last terms will be 1972
Q67-In this case the ratio will be minimum at 199 as ratio-199/19 = 10.4 which means the diff between hundred and ten digit will be 8
Q78-For this que i don't know the method of solving it but when i have taken first three terms then there sum comes out to be 225
Q80-For this que i also calculated the ans to be 64.
I think these Solutions might be of some use for you..:-P
Q 66 The series of niumbers ( 1, 1/2,1/2,1/4 .....1/1972) is taken and now two numbers are taken from the series ( the fisrst two say x,y ). Then the operation x+y+x.y is performed to get a cosolidated no. . The process is repeated > What will be the value of the set after all the no.s are consolidated into 1 no. a 1970 b 1971 c 1972 d NONE Q 67 K is a three digit no. such that the ratio of the no. to the sum of the digits is least ? What's the diff. between the hundreds and tens digit of K
a 9 b 8 c 7 d NONE
Q 80 101*102*103*197*198*199 has las two digits ?? a 54 b 64 c 74 d 84
Q66. a+b+ab take 1,1/2 1+1/2 + 1/2 = 2 now 2,1/3 2+1/3+2/3 = 3..and so on 1971 + 1/1972 + 1971/1972 = 1972
Q67. For Number To be least 199/19 = 10.4...which gives..difference = 8
In order to maintain the price line, a trader allows a discount of 10% on the marked price of goods in his shop. However, he still makes a gross profit of 17% on the cost price. Find the profit percent he would have made on the selling price had he sold at the marked price. a) 23.07 b) 30 c) 21.21 d) 25
:splat: