Q1) in an organisation,the daily average wages of 20 illiterate employees is decreased from Rs 25 to Rs 10. thus the average salary of all the literate and illiterate employees is decreased by Rs 10per day.the number of educated employees working in the organisation are:-
a) 15 b) 20
c)10 d) 25
Q2)
the average age of 8 persons in a committee is inctreased by 2 years when two men aged 35yrs and 45 yrs are substituted by two women.find the average age of two women.
a) 48 b) 45 c) 51 d) 42 e) 46
Q3)
The two women who are joining the group should have the weight equal to the total of the two men and in addition they should have the extra weight to increase the average of the group by 8*2=16 . Hence, the weight should be 35+45+16=96. Thus, the average will be 48.
I purchased a lottery ticket, but since i did not have rs 100 with me, I took a loan of rs 50 from a friend. The next week when the results were out i found that the first 5 digits tallied with the first prize but since the last digit was torn, I could not find out if i was the winner. When i contacted my friend, he told me that the six-digit number was such that its first 6 multiples had the same 6 digits in different orders. But for this piece of information and the loan he demanded half the prize money from me. What was the last digit?
I purchased a lottery ticket, but since i did not have rs 100 with me, I took a loan of rs 50 from a friend. The next week when the results were out i found that the first 5 digits tallied with the first prize but since the last digit was torn, I could not find out if i was the winner. When i contacted my friend, he told me that the six-digit number was such that its first 6 multiples had the same 6 digits in different orders. But for this piece of information and the loan he demanded half the prize money from me. What was the last digit?
1) 2 3) 5 2) 4 4) 7
is d ans 7 ??
my logic...since d 6th multiple of six digit no is six digit, 1st no is 1... i.e one of the first 6 multiples of 2/5/4/7 has to end with 1... only no 7 satisfies this condition ...hence it should be 7...
I purchased a lottery ticket, but since i did not have rs 100 with me, I took a loan of rs 50 from a friend. The next week when the results were out i found that the first 5 digits tallied with the first prize but since the last digit was torn, I could not find out if i was the winner. When i contacted my friend, he told me that the six-digit number was such that its first 6 multiples had the same 6 digits in different orders. But for this piece of information and the loan he demanded half the prize money from me. What was the last digit?
1) 2 3) 5 2) 4 4) 7
Answer 7
Logic
for the other options the last digit got for the multiples is repeating but for seven it comes as 7 4 1 8 5 2
my logic...since d 6th multiple of six digit no is six digit, 1st no is 1... i.e one of the first 6 multiples of 2/5/4/7 has to end with 1... only no 7 satisfies this condition ...hence it should be 7...
i may be completely wrong..pls correct me ??:
Hi...U r completelyright 😃 .....The ans is 7 and the number is 1,42,857 which is a 6 digit number occurring in a cyclic form.
pg 253 Qn 43 Arun Sharma In a box there are 5 red balls 4 blue balls and 3 green balls In how many ways can we select 4 balls out of it if all the balls in a box are identical? Puys the answer given is 495 but i dont know how to get to it. Plz help
pg 253 Qn 43 Arun Sharma In a box there are 5 red balls 4 blue balls and 3 green balls In how many ways can we select 4 balls out of it if all the balls in a box are identical? Puys the answer given is 495 but i dont know how to get to it. Plz help
is this a complete question? 4 balls can be selected from 12 balls in 12C4 ways= 495
Hey avinav,................ buddy...... wats the cut off of SIIB??????????????
Yaar plz watch the thread where you're posting.... There's a separate thread for SIIB as well.... Anyways, just answering your query now.... SIIB didn't declare any cutoff but should be around 75 for General category
pg 253 Qn 43 Arun Sharma In a box there are 5 red balls 4 blue balls and 3 green balls In how many ways can we select 4 balls out of it if all the balls in a box are identical? Puys the answer given is 495 but i dont know how to get to it. Plz help
i feel ans given by arun sharma is wrong... 12C4=495 the reason->
suppose we have 5 red balls as R1 R2 R3 R4 R5
in all the ways to select 4 balls-there will be also few ways where all 4 are red balls... so, there will be R1 R2 R3 r4 and R2 R3 R4 R5 among those cases but the problem is that all the balls are identical...that means we can not identify which 4 balls we have choosen.... or in other words, no of ways to select 4 balls where all 4 are red = 1
simply we have to distribute 4 in R B G R B G 0 4 0 0 3 1 0 2 2 0 1 3 -------------------->0 0 4 missing as max no of G balls =3 1 0 3 1 1 2 1 2 1 1 3 0 2 0 2 2 1 1 2 2 0 3 0 1 3 1 0 4 0 0
if suppose we take 3rd case-> 2B, 2 G balls-> no of ways to select 2 blue balls out of 4 identical blue balls and 2 green balls from 3 identical green balls =1 (as all balls are same)
that is total no of ways = 14
though i am not able to express myself clearly...but i am sure that this is correct(saving any calcln mistakes)
if he had said all balls are different- then ans would have 12C4 as 5 diffent red balls,4 different blue balls is equvalent to 9 different balls....then the colour won't matter...
I am not sure about the concept that the_hate has applied ..... i think its plain simple combinatin prolblem with answer being 12c 4 - some enlightened soul please throw light on any misunderstadings or flawed reasoning
I am not sure about the concept that the_hate has applied ..... i think its plain simple combinatin prolblem with answer being 12c 4 - some enlightened soul please throw light on any misunderstadings or flawed reasoning
the_hate has explained in the best possible way and I cannot do better than that. But I will try and change the words.Lets see if you agree to this. Assume the 5 red balls are numbered 1,2,3,4,5 and they are identical(numbers just for ur understanding). Now the way in which you can pick 4 balls out of the 5 is not 5C4 since all the balls are identical. So it doesnt make a difference if you pick 1,2,3,4 or 2,3,4,5. Both these combinations are considered the same. so the no of ways you can pick 4 balls is 1. Hope you are convinced by this.
i tried to search somethin for the same on net...found this... hope it helps-
(VI) Number of ways of selecting r things from n identical things is 1. Example: In how many ways 5 balls can be selected from 12 identical red balls? Ans. The balls are identical, total number of ways of selecting 5 balls = 1.
I am not sure about the concept that the_hate has applied ..... i think its plain simple combinatin prolblem with answer being 12c 4 - some enlightened soul please throw light on any misunderstadings or flawed reasoning
the_hate has explained in the best possible way and I cannot do better than that. But I will try and change the words.Lets see if you agree to this. Assume the 5 red balls are numbered 1,2,3,4,5 and they are identical(numbers just for ur understanding). Now the way in which you can pick 4 balls out of the 5 is not 5C4 since all the balls are identical. So it doesnt make a difference if you pick 1,2,3,4 or 2,3,4,5. Both these combinations are considered the same. so the no of ways you can pick 4 balls is 1. Hope you are convinced by this.
I purchased a lottery ticket, but since i did not have rs 100 with me, I took a loan of rs 50 from a friend. The next week when the results were out i found that the first 5 digits tallied with the first prize but since the last digit was torn, I could not find out if i was the winner. When i contacted my friend, he told me that the six-digit number was such that its first 6 multiples had the same 6 digits in different orders. But for this piece of information and the loan he demanded half the prize money from me. What was the last digit?
1) 2 3) 5 2) 4 4) 7
i did it by options...
suppose last digit is 2..then 1st multiples last digit = 2 2nd multiples last digit = 3 3rd multipes last digit = 4 4th multiples last digit = 5..... 5th multiples last digit = 0
Now as the number was such that its first 6 multiples had the same 6 digits in different orders.. so 0 cannot b one of the digits as if 0 is @first place the number becomes 6 digit number...
similarly for 5 2nd multiples last digit is 0 and for 4 5th multiples last digit is 0...
But we dont get zero in any multiple if last digit is 7...
can someone plz help me with this one (pg 254 Qn 8? Out of 4 men and 4 women a committee of 5 pple has to be created comprising a president a vice president and 3 secretaries. Find the number of ways of forming this committee such that there are at least 3 women in the committee and at least one woman holds the post of the president or the vice president. Puys the answer given is (4C4 x 4C1 x 5!/3!) +(4C2 x 4C3 x 5!/3!) - (4C2 x 4C3 x 2C2 x 2!) puys I understood what and why they hv done in the 1st & 2nd brackets but I don't understand wat hv they done in the 3rd bracket. Someone plz help !!
Hi.....First of all....This is not a question from Arun sharma...so kindly excuse me and direct me if need be. The question is;
A local club arranged a coconut race. An odd nuber of coconuts were kept at equal distances along a straight line. All the coconuts were to be piled around the centre coconut but only one coconut could be carried at a time. The one who can do it the fastest will be the winner. There was a distance of 10 m between any two adjacent coconuts. The contestants had to start from the first coconut from the either end. I calculated the total distance to be covered was 8.2 km. How many coconuts were there. 1) 25 3) 41 2) 37 4) 47
P.S: My concern is not the ans but the method. Can this be solved using simple concepts like average and stuff.....Have got the ans through a lengthy calculation. Was just wondering about the related concepts.
Let number of coconuts be n, let the person take from one end , then go to middle then the other end and so on. first time the distance traveled is (n-1)*10 for second n-2 *10 ..... now the total distance traveled = sum of 1 to n-1 *10
that is n(n-1)/2 =820 n(n-1)=1640=40*41 ans 3
Hi.....First of all....This is not a question from Arun sharma...so kindly excuse me and direct me if need be. The question is;
A local club arranged a coconut race. An odd nuber of coconuts were kept at equal distances along a straight line. All the coconuts were to be piled around the centre coconut but only one coconut could be carried at a time. The one who can do it the fastest will be the winner. There was a distance of 10 m between any two adjacent coconuts. The contestants had to start from the first coconut from the either end. I calculated the total distance to be covered was 8.2 km. How many coconuts were there. 1) 25 3) 41 2) 37 4) 47
P.S: My concern is not the ans but the method. Can this be solved using simple concepts like average and stuff.....Have got the ans through a lengthy calculation. Was just wondering about the related concepts.
can someone plz help me with this one (pg 254 Qn 8? Out of 4 men and 4 women a committee of 5 pple has to be created comprising a president a vice president and 3 secretaries. Find the number of ways of forming this committee such that there are at least 3 women in the committee and at least one woman holds the post of the president or the vice president. Puys the answer given is (4C4 x 4C1 x 5!/3!) +(4C2 x 4C3 x 5!/3!) - (4C2 x 4C3 x 2C2 x 2!) puys I understood what and why they hv done in the 1st & 2nd brackets but I dont understand wat hv they done in the 3rd bracket. Someone plz help !!
let me try to explain...
we have to select atleast 3 women and atleat 1 of then has to president or VP. We can select : either 4 women and 1 man OR 3 women and 2 men.
selection of these possibilities is given by bracket 1 and 2... now, for 4 W & 1 M, there is just a single man in the team...n he cannot simultaneously be president and VP. so atleast 1 women to be pres or vp is automatically satisfied..
now, consider 3 W & 2 M, there are 2 men on the team..and these 2 men could hold the position of president and vice president...which we do not want... so...selecting 3 W & 2 M , such that men hold position of president and VP can be done in (4C2 x 4C3 x 2C2 x 2!)..in all other ways of allocating 3 W and 2 M for various positions..atleast 1 women would be president or VP.
so ..total ways = all poss ways - ways you do not want... hence u subtract d 3rd bracket from sum of 1st 2...
explanation may not have been good...but i hope u got d point..
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