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A sum of money becomes 8 times itself in 15 years when placed at compound interest. In how many years will it double itself?Correct
Let P, R and T be the principle, rate and time and A be the amount.
A = P(1+R/100)^{T}
8P=P(1+R/100)^{15}
8= (1+R/100)^{15}
2^{3}= [(1+R/100)^{5}]^{3}
2 = (1+R/100)^{5}
Hence, the amount doubles itself in 5 years.
[R^T=R*R*...*T times]
If (x^{3}+1/x^{3}) = 52, then find the value of (x+1/x). [x^{3}=x*x*x]
(x+1/x)^{3} = (x^{3}+1/x^{3}) + 3(x+1/x) (x+1/x)^{3}
= 52 + 3(x+1/x) Let (x+1/x)
= y We have y^{3}
=52+3y y^{3}  3y  52
= 0 Clearly
y = 4
Hence,
(x+1/x) = 4
If x+1/x = 2, then find the value of x^{3}+1/x^{3}. [x^{3}=x*x*x]
x+1/x = 2 (x+1/x)^{3}
= x^{3}+1/x^{3} + 3*x*1/x(x+1/x)
Substituting the value of (x+1/x),
we get 2^{3} = x^{3}+1/x^{3} + 3(2) x^{3}+1/x^{3}
= 8  6
= 2
Find the value of k for which the equations 4x+5y = 3 and kx + 15y = 9 have infinitely many solutions
For the equations to have infinitely many solutions,
4/k = 5/15 = 3/9
k = 4*15/5 = 4*3 = 12
The equations have infinitely many solutions for
k = 12
The difference between two numbers is 2 and the difference between their squares is 24. Find the larger number.
Let the smaller number be x and the greater number be y.
y  x = 2 ...(1)
y^{2}  x^{2} = 24 ...(2)
Substituting y = x+2 from equation (1) in equation (2) we get
(x+2)^{2} x^{2} = 24 x^{2}+4x+4 x^{2}
= 24 4x+4=24 4x = 244 = 20
x = 20/4 = 5
y = x+2 = 5+2 = 7
The larger number is 7.
[x^{2}=x*x]
Find the value of (bxay), if x/a+y/b = 2 and ax by = a^{2}b^{2}. [a^{2}=a*a]
x/a+y/b = 2 bx + ay = 2ab ...(1)
and
axby = a^{2}b^{2} ...(2)
Multiplying (1) by b and (2) by a, we get
(b^{2})x+aby + (a^{2})x aby
= 2ab^{2} + a^{3} ab^{2} (a^{2}+b^{2})x
= a(a^{2}+b^{2}) x = a Putting x = a in (1),
we get
ba+ay = 2ab
y = 2bb=b
Hence,
x = a and y = b bxay = baab
= 0
Find the first term of a GP whose second term is 2 and the sum to infinity is 8.
Let the first term of the GP be a and the common ratio be r
The second term is given by ar and the sum to infinity is given by
S=a/(1r) ar = 2 and a/(1r) = 8
r = 2/a and a = 8  8r
Hence, a = 8  8*2/a a^2 = 8a 16 a^28a+16=0 (a4)^2=0 a = 4
Hence,
the first term of the GP = a = 4 [a^2=a*a]
How many elements does the power set of A contain if A = {x,y}?
A = {x,y} The set A contains 2 elements.
The number of element in the power set of a set containing n elements is given by
2^n. Power set of A contains 2^2 = 4 elements.
[2^2=2*2]
The radius of the base of a solid cylinder is x cm and its height is 3 cm. It is recast into a cone of the same radius. Find the height of the cone in cm.
Volume of a cylinder = pi*r^2*h,
where pi = 22/7 and r and h are the radius and height of the cylinder respectively.
Volume of cone = 1/3*pi*r^2*h
where pi = 22/7, and r and h are the radius and height of the cone respectively.
Volume of the given cylinder = pi*x^2*3
= 3*pi*x^2
Volume of cone = 1/3*pi*x^2*h = pi*x^2*h/3
Since the two volumes are equal
3*pi*x^2 = pi*x^2*h/3 3=h/3 h = 3*3 = 9 cm
The height of the cone is 9 cm.
[pi=22/7, r^2=r*r]
Cards numbered 1 through 10 are placed in an urn. One ticket is drawn at random. In how many chances out of 20 shall the card have a prime number written on it?
There are 10 cards in the urn. Favourable numbers = 2, 3, 5, 7
Required probability
= 4/10 = 8/20
There are 8 chances out of 20 of drawing a prime number.
Find the value of a if (xa) is the g.c.d. of x^2x6 and x^2+3x18. [x^2=x*x]
Since (xa) is the g.c.d of the given polynomials, (xa) is a factor of both the polynomials.
Putting the value a in place of x in the two polynomials,
we get a^2a6=0
and
a^2+3a18=0 a^2a6
=a^2+3a18 a6
=3a18
3a+a =186
4a=12
a=3
The amount of money returned by Sam to Pam after two years was Rs.5500. How much money did Pam lend at 5% rate of simple interest?
Let P, R, T and SI be the principle, rate, time and simple interest.
Amount = SI+ P = P*R*T/100 + P 5500
= P*5*2/100 + P 5500
= (P+10P)/10 P = 5500*10/11
= 5000 Rs.
5000 were lent
In a twodigit number, the sum of the digits is 12. The smaller digit subtracted from the larger digit gives us 4. How many such numbers are possible?
Let the two digits be x and y such that x>y.
According to the conditions,
x+y = 12...(1)
x  y = 4...(2)
Adding (1) and (2), we get
x+y +x y = 12+4 2x = 16
x = 8 y
= 128 = 4
The possible numbers are 84 and 48.
Hence, two such numbers are possible.
One man can complete a work in 25 days and one woman can complete it in 10 days. In how many days can 5 men and 3 women complete the work?
One man can complete the work in 25 days.
Work completed by one man in one day = 1/25
One woman can complete the work in 10 days.
Work completed by one woman in one day
= 1/10
Work completed by 5 men and 3 women in one day
= 5*1/25+3*1/10
= 1/5+3/10 = (2+3)/10
= 5/10 =1/2
5 men and 3 women complete the work in 2 days.
P(n,4) = 20*P(n,2) n!/(n4)!
= 20*n!/(n2)! (n2)!
= 20*(n4)! (n2)(n3)(n4)!
= 20*(n4)! (n2)(n3)
=20 n^25n+620=0
n^25n14=0
n^27n+2n14=0
n(n7)+2(n7)=0
n=2,7
Since n>0,
we have n = 7
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