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Mathematics: Post your doubts here!

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i have doubt in 0580/41/m/j/12 in maths past paper question no 6 plz some one can kindly help me fast
 
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i have doubt in 0580/41/m/j/12 in maths past paper question no 6 plz some one can kindly help me fast
1) firstly form an equation (2x-1) (4x-7) = 1
2) simplify by removing the brackets.
for the second question use the quadratic formula and calculate the height of the parallelogram
 
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As per the question number Q-2 B. For the number of people of families of another n students. As you already know:

mean = summation of observations/ no of observations

As given, the mean of these number of people of families of another n students is 3.
Hence substituting in the formula of mean:

3 = no of people in families of another n students/n

Hence, no of people in families of another n students = 3n

Mean number of all(40 + 3n) students = total number of families/ total number of students
= 190+3n/ 40+n




We hope it helped you.
Feel free to come again.
ciemathematics.com team.
 
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Q. In How Many Years a sum of Rs 3000 would amount Rs 6130.43 at 6% compounded quarterly ?
Q. Find the amount of Rs 250 invested at the end of each of 5 successive years at 6% interest compounded annually ?

plz help me as soon as possible with all steps :(
 
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Q. In How Many Years a sum of Rs 3000 would amount Rs 6130.43 at 6% compounded quarterly ?
Q. Find the amount of Rs 250 invested at the end of each of 5 successive years at 6% interest compounded annually ?

plz help me as soon as possible with all steps :(

Ans.1. A=P(1+R/100)^T (where A is the final amount, P is the principal amount, R is the rate and T is the time)
6130.43 = 3000(1.06)^T
6130.43/3000 = 1.06^T
(6130.43/3000)^1/12 = 1.06
T=12
as the interest is compounded quarterly, 12/4 = 3 years

Ans.2. A = 250(1.06)^5
=Rs.334.6
 
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Ans.1. A=P(1+R/100)^T (where A is the final amount, P is the principal amount, R is the rate and T is the time)
6130.43 = 3000(1.06)^T
6130.43/3000 = 1.06^T
(6130.43/3000)^1/12 = 1.06
T=12
as the interest is compounded quarterly, 12/4 = 3 years

Ans.2. A = 250(1.06)^5
=Rs.334.6

thank you :)

My Next Questions :)

Q. If two linear equations in two unknowns have no common solution, these equations are called ?
A. Consistent B. Inconsistent C.Identical D. Independent

Q. Solve x- 12x + 36 = 0
By Factorization Method
By Quadratic Formula

Q. The Sum Of three Consecutive intergers is 54. What are the intergers?
 
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thank you :)

My Next Questions :)

Q. If two linear equations in two unknowns have no common solution, these equations are called ?
A. Consistent B. Inconsistent C.Identical D. Independent

Q. Solve x- 12x + 36 = 0
By Factorization Method
By Quadratic Formula

Q. The Sum Of three Consecutive intergers is 54. What are the intergers?

1)the factorisation method is simple
take the middle value of (x) as the sum (-12)
multiply the first value of (x2) with the last no. (36)
thus we get the sum as -12 and the product as 36
now find two no.s which when added hav a sum of -12 and a product of 36
they are -6, -6 and since the coefficient of (x2) so we can directly write it as (x-6) (x-6)
 
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2) let the first no. be (x) so the next two consective no.s are x+1 and x+2
s0 x+x+1+x+2=54
so x=3

x+x+1+x+2=54
so x=17
the integers are 17, 18 and 19
sweetiepie
the quadratic formula for ax^2+bx+c=0 is[1]s
NumberedEquation2.gif

For ax2 + bx + c = 0, the value of x is give
so -(-12)+-root(12)^2(-4*1*36)/2*1
12+-root(144-144)/2
12+-root(0)/2
12/2=6
so x=6
 
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thank you :)

My Next Questions :)

Q. If two linear equations in two unknowns have no common solution, these equations are called ?
A. Consistent B. Inconsistent C.Identical D. Independent

Q. Solve x- 12x + 36 = 0
By Factorization Method
By Quadratic Formula

Q. The Sum Of three Consecutive intergers is 54. What are the intergers?
for the first question, i think inconsistent

source:
http://www.google.com/url?sa=t&rct=j&q=independent equations&source=web&cd=2&cad=rja&ved=0CDYQFjAB&url=http://en.wikipedia.org/wiki/System_of_linear_equations&ei=zATXUNK5D8uS0QWx8ICwAg&usg=AFQjCNGA-88TP0T5tyKqCkIlcGH_k1uSOA
 
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10ai) 8(8+1)/2 ; 36
ii) 400(400+1)/2 ; 80200
bi) for 1+2+3+4+5+6..................+n=n(n+1)/2
so cross multiply it by 2 so it comes to 2n(n+1)
bii) 200(200+1)/2
biii) so subtract 80200- 40200
c) 2n(2n+1)/2
 
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