Could someone explain this ?
The series is of the form a1, a2, a3, ..... an.
The next term in the series is 2 more than the previous term. If a5-a2 = 12, then what is the first term ?
I got 4 but it's not the correct answer. Could someone explain ?
Thanks.
Could someone explain this ?
The series is of the form a1, a2, a3, ..... an.
The next term in the series is 2 more than the previous term. If a5-a2 = 12, then what is the first term ?
I got 4 but it's not the correct answer. Could someone explain ?
Thanks.
are you sure the question is correctly quoted, becoz as it says that "The next term in the series is 2 more than the previous term" , it means it is an AP with common difference 2.
on the contrary a5-a2=12 gives common difference = 4.
Its a contradiction... correct me if i m wrong.
guys pls tell me a short method to solve this one....
Time and distance LOD I 41
A & B run race of 2000m. First A gives B a strt of 200m and beats him by 30s. Nxt A gives B a strt of 3 min and is beaten by 1000m find the time in min which A & B run seperately
Could someone tell me the answer to this questions ?
If n is an even number, which one the following has to be odd?
1) 3n
2) 3xsquare/2 +1
3) 5(n)+4
4) 3x/2 + 1
The answer is the second choice. But, I think choice number 4 is also an option . Please explain.
Hey sarah,
Well the 2nd option will always be odd as the square of an even number will always be divisible by 2 and odd*even = even...and even + 1 will always be odd..
But in case 4 the number when divided by 2 might be odd or even..like if u take n as 6 ..it will give u 3*3 + 1 which is even....while if you take n= 4 you will get 3*2 + 1 which is odd.....
Thus option 2 will always be odd
dude, i think the answer in the book is correct.
2^2 divided by 9 gives remainder 2^2
remainder of 22^2 divided by 9 => 4^2 divided by 9 , ( 22 by 9 gives remainder 4)
remainder of 222^2 divided by 9 => 6^2 divided by 9 , ( 222 by 9 gives remainder 6)
remainder of 2222^2 divided by 9 => 8^2 divided by 9 , ( 2222 by 9 gives remainder
similarly doing, for ( 22....49times)^2 by 9 => (2*49)^2
so adding these, we get
= remainder on dividing (2^2 + 4^2 + 6^2 +..........+ 98^2) by 9
= remainder on dividing 2^2 ( 1^2 + 2^2 + 3^2 +.......+ 49^2) by 9
= remainder on dividing 2^2 * (using the sum of squares formula)
= remainder on dividing (4 * 49 * 25 * 33) by 9
= remainder on dividing (4 *4 * 7 * 6) by 9
=remainder on dividing (7*7*6) by 9
= remainder on dividing 24 by 9
= 6
Thanks for correcting.
Probably I had done some calculation mistakes
Could someone explain this ?
The series is of the form a1, a2, a3, ..... an.
The next term in the series is 2 more than the previous term. If a5-a2 = 12, then what is the first term ?
I got 4 but it's not the correct answer. Could someone explain ?
Thanks.
I think their is some flaw in this Ques.....what akhil is saying is right.....
Machine A can produce 80 nuts in 1 hour and machine B can produce 120 nuts in 1 hour. If both the machines work together and A is working for 8 hors and they producce 960 nuts, then how many hours would mahine B have been working along with machine A ?
Can someone explain this problem ?
nha0801 Saysplz send me the di sets at
Please remove your personal email id from your post as it's against the PG rules.
guys pls tell me a short method to solve this one....
Time and distance LOD I 41
A & B run race of 2000m. First A gives B a strt of 200m and beats him by 30s. Nxt A gives B a strt of 3 min and is beaten by 1000mA find the time in min which A & B run seperately
Assume
Sa is speed of A
Sb is speed of B
Condition 1: A gives B a start of 200m and beats by 30s
Say t1 is time taken by A to complete race
Sa=2000 / t1-------------(1)
B takes 30s more than A
Sb=1800 / (t1+30)--------(2)
Condition 2: A gives B a start of 3 min and is beaten by 1000m
Let time taken to finish race by B is t2
t2=2000/Sb---------------(3)
As A starts 3 min(180s) later and manages to complete only 1000m
t2=(1000/Sa)+180---------(4)
Start Solving:
1. Equating eq (3) and (4)
2000/Sb=(1000/Sa)+180
2. Substitute for Sb and Sa in eq (3) and (4) from (1) and (2) respectively
We get,
(t1/2)+180=(t1+30)*(10/9)=t2
which gives t1=240 s
Substituting value of t1 in equation
t2=(t1/2)+180 which comes above in step 1
t2=300
Hence Solved Sorry I don't have the book so can't verify answers. verify if found correct
are you sure the question is correctly quoted, becoz as it says that "The next term in the series is 2 more than the previous term" , it means it is an AP with common difference 2.
on the contrary a5-a2=12 gives common difference = 4.
Its a contradiction... correct me if i m wrong.
@ akhil Yes dude, you got it right.
@ sarah we will appreciate if you can post the question exactly as it is printed (or misprinted which i doubt)
Machine A can produce 80 nuts in 1 hour and machine B can produce 120 nuts in 1 hour. If both the machines work together and A is working for 8 hors and they producce 960 nuts, then how many hours would mahine B have been working along with machine A ?
Can someone explain this problem ?
Possible Explanation
Machine A produces 80 nuts/hr
So in 8 hrs it produces 8*80 = 640 nuts
Total nuts produced = 960
Nuts produced by B = 960- Nuts produced by A
= 320
As machine B produces 120 nuts/hr
Working time of B = 320/120
= 2 hrs 45 mins.
Hence Solved Please verify the answer and let me know too, Sarah.
arun sharma is driving me crazy
puys help me solve this acording to allegation method in arun sharma(general also acceptable if not alleagation..but preference is more for allegation )
400 studs took a exam in lucknow.60% of the boys and 80% of the girls cleared.if the total percentage of studs qualifying is 65%, how many girls appeard for exam??
page number 123 6q
plz on the same page help me with 7 & 8
arun sharma is driving me crazy
puys help me solve this acording to allegation method in arun sharma(general also acceptable if not alleagation..but preference is more for allegation )
400 studs took a exam in lucknow.60% of the boys and 80% of the girls cleared.if the total percentage of studs qualifying is 65%, how many girls appeard for exam??
page number 123 6q
plz on the same page help me with 7 & 8
Assume:
Total number of girls appeared is G
Then total number of boys is 400-G
Condition 1: 60% of the boys and 80% of the girls cleared
Total number of cleared girls = 0.8G
And total number of cleared boys = 0.6(400-G)
From above, total number of students cleared = 0.8G+0.6(400-G) -------(1)
Condition 2: total percentage of studs qualifying is 65%
Total number of students cleared is 0.65*400---------------------------(2)
Start solving
From eq (1) and (2)
0.8G+240-0.6G=260
0.2G=20
G=100
So, total 100 girls appeared for the examination.
Hence Solved P.S. The examination in consideration for this question is definately not held at IIM L. Heard that they have worse sex ratio.
arun sharma is driving me crazy
puys help me solve this acording to allegation method in arun sharma(general also acceptable if not alleagation..but preference is more for allegation )
400 studs took a exam in lucknow.60% of the boys and 80% of the girls cleared.if the total percentage of studs qualifying is 65%, how many girls appeard for exam??
page number 123 6q
plz on the same page help me with 7 & 8
Hey mate, please post the questions 7 and 8 as some cheapos like me haven't bothered to buy the book.
@ mods : Sorry for multiple posts.
Hey mate, please post the questions 7 and 8 as some cheapos like me haven't bothered to buy the book.
@ mods : Sorry for multiple posts.
ANYTHING FOR PUYS AT ANY TIME :-P
bro the girls problem u solved was awesome but it ll take a lot of time ...if we use allegation method and detect A1,A2 n1 n2 Aw the problem can be solved in only one step bro....but am thankfull for u
here goes the 7th problem:D
Aman purchased cow and a calf for 1300.he sold the calf at a profit of 20% and the cow at a profit of 25%.In this way his total profit was 23 1/13%.find the cost price of the cow.
again this can be done on the same lines bro
125%Cow+ (1300-cow)120%=1600(SP)
..but can u teach me thru allegation method bro....
like which one is A1 A2 n1 n2 and Aw....thanks in advance ...
few more problems on alligations ...
page 124 Arun sharma
17.A dishonest grocer professes to sell pure butter at CP but he mixes it with adulterated fat and gains 25%.find the %age of adultr fat in the mixture which is freely available.
22.there are 2 mixtures of honey and water, the quantity of honey in them being 25% and 75% of the mixture.if 2 gallons of the first are mixed with the 3 gallons of the second what will be the ratio of the honey to water in the new mixture?????
few more problems on alligations ...
page 124 Arun sharma
17.A dishonest grocer professes to sell pure butter at CP but he mixes it with adulterated fat and gains 25%.find the %age of adultr fat in the mixture which is freely available.
let the cost price of 1 unit of pure butter be 'c', and let the amount of adulterated fat he mixes with one unit of pure butter be 'x' kg.
so, since he is only selling (1-x) units of pure butter when he claims to be selling 1 unit butter, his actual cost price is the cost price of (1-x) units of pure butter for every unit he sells.
so the actual cost price of every unit adulterated butter he sells= c(1-x)
but he sells it at the cost price of 1 unit pure butter i.e = c
so profit= S.P - C.P
=c - c(1-x)
% profit =/ c(1-x) * 100 = 25%
solving we get x=1/5
that is equal to 20%
hope that helps....
few more problems on alligations ...
page 124 Arun sharma
22.there are 2 mixtures of honey and water, the quantity of honey in them being 25% and 75% of the mixture.if 2 gallons of the first are mixed with the 3 gallons of the second what will be the ratio of the honey to water in the new mixture?????
I could have solved both the question in one post but for the sake of clarity decided to give each a separate post.
consider 2 gallons of first mix.
honey content = 0.5 gallons (each gallon of first mix contains .25 gallon honey)
water content = 1.5 gallons
consider 3 gallons of second mix.
honey content = 2.25 gallons (each gallon of first mix contains .75 gallon
honey)
water content = .75 gallon
adding these two mixtures we get
honey = 0.5 + 2.25 = 2.75
water = 1.5 + 0.75 = 2.25
honey / water = 2.75 / 2.25
= 11 / 9
hello frinds.