Hi,
Please solve this question and explain the approach also.
1.If a,b,c,d are in continued proportion then a-d/b-c >= x .What is the value of x.
options are:
a 2
b 3
c 1
d 0
e 4.
Thanks in advance,
Hi , Plz solve this question
Find the 28383rd term of
1234567891011...
I forgot the options.Sorry
rahul005 SaysA person purchased a cupboard and a cot for RS 18,000.He sold the cupboard at a profit of 20% and the cot at a profit of 30%.If his total profit was 25.833%,find the cost price of the cupboard(ans. is RS 7500).............can u solve dis with the process of alligation??if yes,then pls show.
20% 25.833% 30%
-----------|-------------
|
n1 n2
Now n1+n2 = 18000
From alligation (n1+n2) correspond to (A2-A1)= (30-20)=10
and n1 corresponds to (A2-Aw)=(30-25.833)=4.167
therefore n1 = (A2-Aw)*(n1+n2)/(A2-A1)
= 4.167*18000/10
= 7500.6 ans
rahul005 SaysA vessel is full of a mixture of kerosene and petrol in which there is 18% kerosene.8 litres are drawn off and then the vessel is filled with petrol.If the kerosene is now 15%,how much does the vessel hold?(ans.48 litres)......pls solve it through the process of alligation.
Consider the case after the 8 litres of the mixture has been removed.
Since petrol is added we need to solve taking %age of petrol in mixture.
Now, the mixture is still 18% kerosene i.e. 82% petrol and let the qty of the mix be x.
after 8 ltrs of 100% petrol is added the mixture becomes 85% petrol.
A1= 82%, A2= 100%, Aw=85%
n1=x, n2=8
A1............. Aw...............A2
------------|-------------
n1...................................n2
By alligation n1 = n2*(A2-Aw)/(Aw-A1) = 8* 15/3 = 40
Capacity of the tank = n1+n2 = 40 +8 = 48 ans...
How did u calculate dude?
mightyrajat SaysHow did u calculate dude?
which step you are not able to understand..............
Hi , Plz solve this question
Find the 28383rd term of
1234567891011...
I forgot the options.Sorry
It is the series of numbers.
1 to 9 - 9 numbers of 1 digit each - i.e 9 terms
10 to 99 - 90 numbers of 2 digits each - i.e 180 terms
100 to 999 - 900 numbers of 3 digits each - i.e. 2700 terms
1000 to 9999 - 9000 numbers of 4 digits each - i.e. 36000 terms
Since we have to find 28383rd term It must be a part of 4 digit number. i.e. from 10000 onwards.
12345678910111213.....9989991000100110021003100410051006.......
Now upto 999 we have 2889 terms (9+180+2700)
So 28383rd term will be 28383-2889= 25494 th from 999.
In 4 digit number every new number starts from the 5th term.
100010011002............. i.e 1000 takes up 4 terms & 1001 starts from 5th term.
Now 25194 = 4*6373 + 2
ie our requd no. is the 2nd digit of the number 6373 i.e. 3.
Answer = 3
Please help me with the approach..
If x is the first term,y is the nth term and p is the product of n terms of a GP, then what is p^2 ?
a.xy^(n-1)
b.xy^n
c.xy^(1-n)
d.xy^(n/2)
the terms in brackets is the power.
Please help me with the approach..
If x is the first term,y is the nth term and p is the product of n terms of a GP, then what is p^2 ?
a.xy^(n-1)
b.xy^n
c.xy^(1-n)
d.xy^(n/2)
the terms in brackets is the power.
Take an example buddy....
1 2 4 8 16 is a GP
x=1, Let n=3 so y=4 p=8
8^2=64, which is 4^3, option 2 is correct.
Just to be sure,, 3 6 12 24
x=3 Let n=3, y=12 p=216
216^2=46656 and 36 ^3=46656
option 2
Please help me with the solutions for these questions on Progressions.
Thank you :)
1)The sum of 5 nos in AP is 30 and the sum of the squares is 220. Which is the 3rd term?
Ans:9
2) 4 Geometric means are inserted between 1/8 and 128 . Find the third Geometric mean.
Ans: 8
3) Two nos A and B are such that their GM is 20% lower than their AM.
Find the ratio betn nos.
Ans: 4:1
4) A no is divided into 4 parts that are in AP such that the ratio of product of 1st and 4th to the product of 2nd and 3rd is 2:3.
Find the largest part.
Ans: 8
5)The sum of an infinite GP whose common ratio is less than 1 is 32 and the sum of the first two terms is 24. What will be the 3rd term?
Ans: 4
6) What will be the value of x^(1/2) * x^(1/4) * x^(1/
... to infinity.
Ans:x
7) Determine the first term of GP, the sum of whose 1st term and 3rd term is 40 and the sum of 2nd and 4th term is 80
ans: 8 Find the 2nd term of AP if the sum of its first five even terms is equal to 15 and the sum of the first three terms is equal to -3.
Ans: -1
9) The sum of the first three terms of AP is 30 and the sum of the squares of the 1st and 2nd term is 116. Find the 7th term of the progression if its 5th term is known to be exactly divisible by 14.
ANs: 40
10) A is the sum of n terms of series 1+ 1/4 + 1/16 + ...
B is the sum of 2n terms of series 1+ 1/2 + 1/4 + ....
Find A/B.
Ans: 2/3
11) Find the sum of series: 1/(1*5) + 1/(5*9) + 1/(9*13) .... + 1/(221*225)
Ans: 56/225
12) If in a decreasing AP, sum of all its terms except for the first term is equal to -36. The sum of all its terms except for the last term is 0.
And the difference of the 10th and the 6th term is equal to -16.
Then what will be the first term of the series?
Ans: 16
13)The 1st and 3rd terms of an AP are equal, respectively, to the 1st and 3rd term of GP. And the 2nd term of AP exceeds the 2nd term of GP by 0.25
Calculate the sum of the first 5 terms of the AP is its 1st term is equal to 2.
Ans: 2.5 or 2.75
14) The sum of the squares of 5th and 11th term of an AP is 3 and the product of 2nd and 14th term is Q. Find the product of 1st and 15th term of AP.
ANs: (98Q+39)/90
15) The ration of harmonic mean of 2 nos to the Geometric mean is 12:13,
find the ratio of the nos.
Ans: 3/4 or 4/5
16)Find the sum of all whole number divisors of 720.
Ans:2624
17) If 3 positive real nos are in AP such that xyz=4, then what will be the minimum value of y?
Ans:2^(2/3)
1 The sum of first n terms of the AP is equal to half the sum of next n terms of the same progression. Find the ratio of sum of first 3n terms of the progression to the sum of its first n terms.
Ans: 8
19) Find the sum of all 3 digit whole nos less than 500 that leave a remainder of 2 when they are divided by 3.
Ans: 39767
20) If the first two terms of a HP are 2/5 and 12/13 resp, which of the folln terms is the largest term?
a)4th term b)5th c)6th d)7th
Ans: 4th
1)The sum of 5 nos in AP is 30 and the sum of the squares is 220. Which is the 3rd term?
correct answer is 6
explanation:
T5= n/2(2a+(n-1)d)
30 = 5/2(2a+4d)
30= 5a+10d
first term is a and common diff is d
thus, a^2 + (a+d)^2 + (a+2d)^2 + (a+3d)^2 + (a+4d)^2 = 220 (given)
After simplifying, 5a^2 + 20ad + 30d^2 = 220
5a^2 + 10ad + 10ad + 20d^2 + 10d^2 = 220
a(5a+10d) + 2d (5a + 10d ) + 10d^2 = 220
but 5a + 10d = 30
thus, 30a + 60d + 10d^2 = 220
6(5a + 10d) + 10d^2 = 220
180 + 10d^2 = 220
10d^2 = 40
d^2= 4
d = 2 and d = -2
consider d = 2
5a + 10d = 30 gives a = 2
thus series, 2, 4,6, 8 ,10
thus third term =6
4 Geometric means are inserted between 1/8 and 128 . Find the third Geometric mean.
lets say first tem a and last term b
a= 1/8 and b = 128
we need to find 4 terms ie 1/8, _ , _ , _ , _, 128
r = (b/a)^(1/n+1)
r = (128 / 1/^ 1/5
r = (2^10)^1/5
r = 4
thus, series 1/8, 1/2, 2 , 8, 32, 128
hence third inserted term is 8.
It is the series of numbers.
1 to 9 - 9 numbers of 1 digit each - i.e 9 terms
10 to 99 - 90 numbers of 2 digits each - i.e 180 terms
100 to 999 - 1000 numbers of 3 digits each - i.e. 3000 terms
1000 to 9999 - 10000 numbers of 4 digits each - i.e. 40000 terms
Since we have to find 28383rd term It must be a part of 4 digit number. i.e. from 10000 onwards.
12345678910111213.....9989991000100110021003100410051006.......
Now upto 999 we have 3189 terms (9+180+3000)
So 28383rd term will be 28383-3189= 25194 th from 999.
In 4 digit number every new number starts from the 5th term.
100010011002............. i.e 1000 takes up 4 terms & 1001 starts from 5th term.
Now 25194 = 4*6298 + 2
ie our requd no. is the 2nd digit of the number 6298 i.e. 2.
Answer = 2
I have a doubt regarding the bold part. How 100 to 999 is 1000 numbers? Shouldn't it be 900? Same about 1000 to 9999.
The answer may be 3.
Thanks.
I have a doubt regarding the bold part. How 100 to 999 is 1000 numbers? Shouldn't it be 900? Same about 1000 to 9999.
The answer may be 3.
Thanks.
You are right, the answer is indeed 3. I had solved this question about a year ago, here's the solution:
http://www.pagalguy.com/forum/quantitative-questions-and-answers/23813-quant-by-arun-sharma-167.html#post1636356
Two nos A and B are such that their GM is 20% lower than their AM.
Find the ratio betn nos.
X = a+b /2 and Y = root (ab)
x/y = 10/8 = 5/4
5/4 = a+b/2*root(ab)
10/4 = a+b/root(ab)
25/4 = (a+b)^2/ab
simplifying,
4a^2- 17ab - 4b^2 = 0
solving, a-4b = 0 or 4a-b = 0
thus a:b = 4:1 or 1:4
I have a doubt regarding the bold part. How 100 to 999 is 1000 numbers? Shouldn't it be 900? Same about 1000 to 9999.
The answer may be 3.
Thanks.
Thanks Buddy for pointing out the error................
q1. i think answer should be none of these....
q2. answer is 1180....
q3. answer is 40...
q4. answer is CBD
can u gv a detailed sol'n of the q2.mine is not matchng..
thanks in advance.
can u gv a detailed sol'n of the q2.mine is not matchng..
thanks in advance.
This was the question:
74. Three mangoes, four guavas and five watermelons cost Rs 750. Ten watermelons, six mangoes and 9 guavas cost Rs 1580. What is the cost of six mangoes, ten watermelons and 4 guavas?
a) 1280 b) 1180 c) 1080 d) cannot be determined
Given,
3m + 4g + 5w = 750 --------------------- (1)
6m + 9g + 10w = 1580 ------------------ (2)
To find,
6m + 4g + 10w ---------------------(x)
Multiplying (1) by 2,
6m + 8g + 10w = 1500 ------------------------ (3)
(2) - (3)
g = 80 ----------------- (4)
Thus, (x) = (3) - 4*(4)
6m + 4g + 10w = 1180
hello puys,i have some problems,need your help,all from number system.first LOD2 QUESTION62-M is a two digit number which has the property that:the product of factorial of its digit>sum of factorial of its digits.how many values of M exists? option A)56 B)64 C)63 D)none of these.........ans-(c)
sorry for writing each problem seperately,just to avoid confusion.........my second problem is LOD2 QUESTION66 the series of numbers(1,1/2,1/3.....1/1972)is taken.now two number taken from this series(the first two)say x,y.then the operation x+y+xy is performed to get consolidated number.the process is repeated.what will be the value of set after all the numbers are consolidated into one number. OPTIONS A)1970 B)1971 C)1972 D)none of these............................ans(c)
LOD2 QUESTION78 (1!)^3+(2!)^3+(3!)^3+.......+(1152!)^3 IS DIVIDED BY 1152? options a)125 b)225 c)325 d)205
LOD-2 NUMBER SYSTEM PROBLEM92 what is remainder when 2(8!)-21(6!) divides 14(7!)+14(13!)? options a)1 b)7! c)8! d)9!.....ans(b)................. LOD2 NUMBER SYSTEM PROBLEM 93 how many integer values of x and y are such that 4x+7y=3,when modulus of x
LOD3 PROBLEMS,NUMBER SYSTEM,QUESTION11-13:at a particular time in the twenty first century there were seven bowlers in the indian cricket team list of 16 players shortlisted to play for next world cup.staticians discovered that if you look at number of wckets taken by any of the 7bowlers of the current indian cricket team,the number of wickets taken by them has a strange property.the numbers are such that any team selection of 11 players(having 1 to 7 bowlers)by using the number of wickets taken by each bowlers and attaching coeffients of+1,0 or-1 to each value available and adding the resultant values,any number from 1 to 1093,both included could be formed.if we denote w1,w2,w3,w4,w5,w6 and w7 as the seven values in the ascending order what could be the answer to the following quesions:(Q47) find the value of w1+2w2+3w3+4w4+5w5+6w6....options a)2005 b)1995 c)1985 d)none of these.....................ans(a)..............(Q4find the index of the largest power of 3 contained in the product w1.w2.w3.w4.w5.w6.w7....option a)15 b)10 c)21 d)6..............ans(c).......................Q(49)if the sum of the seven coefficients is 0,find the smallest number that can be obtained....option a)-1607 b)-729 c)-1041 d)-1054.........ans(c)