Surbhih Saysthis is d question in arun sharma... it means dis is a wrong question??
Are you sure that you've copied the exact question and posted here?
If so, then it seems to be ambiguous.
Are you sure that you've copied the exact question and posted here?
If so, then it seems to be ambiguous.
Hi,
1) There are 576 boys and 448 girls in a school that are to be divided into equal sections of either boys or girls alone. Find the total number of sections thus formed
a)24
b)32
c)16
d)20
The approach that I followed in solving the question is finding the HCF of both of these numbers which comes out to be 8. Now as we have to find the total number of sections I added 8+8 and the answer comes out to be 16 which is correct in this case. I just wanted to know that whether I am correct in my approach or doing any wrong. Any better way is appreciable.
2) A milkman has three different quantities of milk . 403 gallons of 1st quantity, 465 gallons of 2nd quantity and 496 gallons of 3rd quality. Find the least possible number of bottle of equal size in which different milk of different qualities can be filled without mixing?
a)34
b)46
c)26
d)44
e)45
I was trying to employ the same approach that I tries in the above question. The HCF was coming out to be 31 . But adding them 3 times gives me 93 which is not the option in this case.
Kindly Help
Thanks
Hi,
1) There are 576 boys and 448 girls in a school that are to be divided into equal sections of either boys or girls alone. Find the total number of sections thus formed
a)24
b)32
c)16
d)20
The approach that I followed in solving the question is finding the HCF of both of these numbers which comes out to be 8. Now as we have to find the total number of sections I added 8+8 and the answer comes out to be 16 which is correct in this case. I just wanted to know that whether I am correct in my approach or doing any wrong. Any better way is appreciable.
2) A milkman has three different quantities of milk . 403 gallons of 1st quantity, 465 gallons of 2nd quantity and 496 gallons of 3rd quality. Find the least possible number of bottle of equal size in which different milk of different qualities can be filled without mixing?
a)34
b)46
c)26
d)44
e)45
I was trying to employ the same approach that I tries in the above question. The HCF was coming out to be 31 . But adding them 3 times gives me 93 which is not the option in this case.
Kindly Help
Thanks
for the second ques...
403 = 31 x 13
465 = 31 x 15
496 = 31 x 16
so we can have (13 + 15 + 16)= 44 bottles of say 31 gallons.
Hi,
1) There are 576 boys and 448 girls in a school that are to be divided into equal sections of either boys or girls alone. Find the total number of sections thus formed
a)24
b)32
c)16
d)20
The approach that I followed in solving the question is finding the HCF of both of these numbers which comes out to be 8. Now as we have to find the total number of sections I added 8+8 and the answer comes out to be 16 which is correct in this case. I just wanted to know that whether I am correct in my approach or doing any wrong. Any better way is appreciable.
You have mistake in calculation as well as approach.
HCF(576,448 ) = 64
576 = 64*9
448 = 64*7
Now, the section strength will be 64
there will be 9 sections of boys and 7 sections of girls ... total 16
PS: I think the question is either missing a crucial point in the statement or you have missed to type it.
The question should Minimum number of sections formed
Only then I will take HCF of 576 and 448 to determine the maximum students that can come in a section.
hai
please solve these questions .......
percentages(lod1)
Arjit sharma generally wears his fathers coat.his cousin poked him one day that he was wearing a coat of length more than his height by 15%.if the length of arjits fathers coat is 120cm then find the actual length of his coat
50% of a% 0f b is 75% of b% of c.which of the following is c
1.5a , 0.667a , 0.5a , 1.25a , 1.66a
At an election the candidate who got 56% of the votes cast won by 144 votes. find the total no; of voters on the voting list if 80% people cast their vote and there were no invalid votes
For the number 7200 find.
1 The sum and number of all factors
2. The sum and number of even factors
3. The sum and number of odd factors
4. The sum and number of factors divisible by 25
5. The sum and number of factors divisible by 40
6. The sum and number of factors divisible by 150
7. The sum and number of factors not divisible by 75
8. The sum and number of factors not divisible by 24
hai
please solve these questions .......
percentages(lod1)
Arjit sharma generally wears his fathers coat.his cousin poked him one day that he was wearing a coat of length more than his height by 15%.if the length of arjits fathers coat is 120cm then find the actual length of his coat
50% of a% 0f b is 75% of b% of c.which of the following is c
1.5a , 0.667a , 0.5a , 1.25a , 1.66a
At an election the candidate who got 56% of the votes cast won by 144 votes. find the total no; of voters on the voting list if 80% people cast their vote and there were no invalid votes
For question 1:
If there is an increase of N/D over a particular no. 'x' (where N/D is a fraction of x) to get another number 'y', the similar decrease from 'y' to get 'x', when represented as a fraction of 'y' would be N/(N+D)
in this case the increase is 15% = 3/20
thus the decrease would be 3/(20+3) = 3/23
length of his father's coat = 120 cm
length of his coat = 120 * (1 - 3/23) cm = 104 cm (approx)
For question 2:
50% of a% 0f b is 75% of b% of c
or, (50/100)*(a/100)*b = (75/100)*(b/100)*c
or, ab/200 = 3bc/400
or, 2a/3 = c
or, c = .667 a
For q 3:
assuming there 2 candidates in the election;
the candidate who won got 56% of the votes cast, thus the candidate who lost got 44% of the votes cast
the difference is
12% of the votes cast = 144
votes cast = 144/12% = 144*100/12 = 1200
as 80% of the total voters cast their votes
80% of the total voters = total votes cast = 1200
total voters = 1200/80% = 1500
For the number 7200 find.
1 The sum and number of all factors
2. The sum and number of even factors
3. The sum and number of odd factors
4. The sum and number of factors divisible by 25
5. The sum and number of factors divisible by 40
6. The sum and number of factors divisible by 150
7. The sum and number of factors not divisible by 75
8. The sum and number of factors not divisible by 24
7200 = 2^5 * 3^2 * 5^2
1.
no. of factors = (5+1)(2+1)(2+1) = 54 -------------------------(f1)
sum of factors = (2^6 - 1)(3^3 - 1)(5^3 - 1)/(2-1)(3-1)(5-1)-----(s1)
3.
no. of odd factors = factors of 3^2*5^2 = (2+1)(2+1) = 9 -----------(f3)
sum of factors ... use the same process -----------------------(s3)
2.
no. of even factors = (f1) - (f3)
sum of even factors = (s1)-(s3)
4.
7200 = 25 * 2^5*3^2 ... 25 should be in each factor ...
no. of factors div by 25 = no of factors of 2^5 * 3^2
work on similar lines for all the rest ...
let the number be = l
let 2 other natural numbers be m and n , such that
m^2= 100+l
n^2= 169+l
n^2-m^2=69=3.(23)
(n-m)(n+m)=3.(23)
clearly, n-m=3 and n+m=23
solving v get n=13 and m=10, from which l =0, which is not a natural number. So, according to me, there is no natural number which satisfies the given conditions.If i hav gone wrong smwhere plz point it out to me.
you got this completely wrong. This is old post but i am posting right answer for future aspirants.
m^2=100+x
n^2=169+x
n^2-m^2=69
(m+n)(m-n)=69
100 n 169 both r sq. no. doesn't mean x is also a sq. no.
diff. between any two consecutive sq. no is always an odd no that too follow consecutive pattern. 1 4 9 16 25 36 49 64 ..... now find diff. between them. u will get 3 5 7 9 11 13 15....
so there must be two consecutive sq no with diff. of 69
so as m2 n n2 are conse. sq.no. so m ,n must be cose. natural nos
so m-n=1
so m+n=69 now find two consecutive natural no whose sum is 69
69/2=34.5
so m=35 n=34
Puys ,
Please help !
Abdul goes to the market to buy bananas. If he can bargain and reduce the price per dozen by Rs 2 , he can buy 3 dozen bananas instead of 2 dozen with the money he has . How much money does he have ?
a)Rs 6
b)Rs 12
c)Rs 18
d)Rs 24
I got the solution using options :grin:, but can anyone provide a different approach ?
Puys ,
Please help !
Abdul goes to the market to buy bananas. If he can bargain and reduce the price per dozen by Rs 2 , he can buy 3 dozen bananas instead of 2 dozen with the money he has . How much money does he have ?
a)Rs 6
b)Rs 12
c)Rs 18
d)Rs 24
I got the solution using options :grin:, but can anyone provide a different approach ?
original price = x per dozen
bargain price = x-2 per dozen
2x = 3x-6
x = 6
so , he have 12 rs
2) 544=2^5*17^1.....num of divisors will be (5+1)*(1+1)=12-2=10
5+1=6(the power of 2 is 5 so according to the formula of getting number of divosrs or factors we have to add 1 and same we will do in case of power of 17 ) now we have subtract 2 factos(1 and 544)thn we get totoal divisors 12-2=10 for 544
3) again 12-2(1 and 2 and less thn 3)=10
4)sum of divosrs...2^5*17^1.....(2^0+2^1+2^2+2^3+2^4+2^5)*(17^0+17^1)=1134-545=589
5) sum of odd devisors= 2^0*=18
6) sum of even divsors of 4096=2^12
(2^1+2^2...............2^12)=8190...we wont add 2^0 as we are taking sum of even factors :)
plz someone ans q1 :|
1080 means 2^3 * 3^3 * 5
therefore the number of divisors of 1080 is (3+1)*(3+1)*(1+1)= 32 ;
Now , from 2^3 * 3^3 * 5 , we get the perfect squares as 2^2 , 3^2 , and (2*3)^2 , these are possible perfect squares . so excluding them all , we will get 32-3 = 29 divisors . So 29 is the correct answer ..
this lod II question is from the chapter averages
There were 42 students in a hostel. Due to the admission of 13 new students, the expenses of the mess increase by Rs 31 per day while the average expenditure per head diminished by Rs 3. What was the original expenditure of the mess?
a) Rs 633.23
b) 583.3
c) 623.3
d) 632
e) none of these
plz help
my approach was...
42x = M
65(x-3) = m+31
this lod II question is from the chapter averages
There were 42 students in a hostel. Due to the admission of 13 new students, the expenses of the mess increase by Rs 31 per day while the average expenditure per head diminished by Rs 3. What was the original expenditure of the mess?
a) Rs 633.23
b) 583.3
c) 623.3
d) 632
e) none of these
plz help
my approach was...
42x = M
65(x-3) = m+31
Let avg. expense for 1 day= a
Total cost= 42a
After admission of 13 students, cost= 42a+31
Avg. expense = a-3
Total expense for 55 students= 55(a-3)
=> 42a+31 = 55(a-3)
Whatever 'a' will come, multiply it by 42, you will get the answer
1. Three math classes X,Y,Z take an algebra test,
average score of class X is 83,
average score of class Y is 76,
average score of class Z is 85,
what is average score of classes X,Y,Z?
2. The total expenses of of a boarding house are patly fixed
and partly varying linerly with the number of boarders. average exlense per boarder is Rs. 700 when there are 25 boarders and Rs. 600 when there are 50 boarders. What is the average expense per boarder when there
are 100 boarder?
1. Three math classes X,Y,Z take an algebra test,
average score of class X is 83,
average score of class Y is 76,
average score of class Z is 85,
what is average score of classes X,Y,Z?
Depends on the number of students in each class X,Y and Z. So, cant be determined
2. The total expenses of of a boarding house are patly fixed
and partly varying linerly with the number of boarders. average exlense per boarder is Rs. 700 when there are 25 boarders and Rs. 600 when there are 50 boarders. What is the average expense per boarder when there
are 100 boarder?
Let X be fixed expense and Y linear expense/head
so,
X + 25Y = 25*700 and
X + 50Y = 50*600
=> 25Y = 25*500
=> Y = 500
=> X = 5000
So, expenses for 100 people per head = (5000 + 100 * 500) / 100 = 550
:D kool !! wud love to race :)
but on a serious note trains dont run at tat speed and i think v ran away at gr8 speeds from the correct sol.
why did u divide...330*23 by 30?
Depends on the number of students in each class X,Y and Z. So, cant be determined
Let X be fixed expense and Y linear expense/head
so,
X + 25Y = 25*700 and
X + 50Y = 50*600
=> 25Y = 25*500
=> Y = 500
=> X = 5000
So, expenses for 100 people per head = (5000 + 100 * 500) / 100 = 550
in ist question..
why average score can't be average of X Y Z class?
Depends on the number of students in each class X,Y and Z. So, cant be determined
Let X be fixed expense and Y linear expense/head
so,
X + 25Y = 25*700 and
X + 50Y = 50*600
=> 25Y = 25*500
=> Y = 500
=> X = 5000
So, expenses for 100 people per head = (5000 + 100 * 500) / 100 = 550
in ist question..
why average score can't be average of X Y Z class?
in ist question..
why average score can't be average of X Y Z class?
To understand it better lets take the number of students in X as 10 Y as 20 and Z as 30. Try and calculate the average.
Then again try using X as 20 Y as 30 and Z as 10.
You'll get the answer to your question.