Q. 1

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

• EXPLAINATION:
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]

Q. 2

If (x^3+1/x^3) = 52, then find the value of (x+1/x).

[x^3=x*x*x]

• EXPLAINATION:
(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

Q. 3

If x+1/x = 2, then find the value of x^3+1/x^3.

[x^3=x*x*x]

• EXPLAINATION:
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

Q. 4

Find the value of k for which the equations 4x+5y = 3 and kx + 15y = 9 have infinitely many solutions

• EXPLAINATION:
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

Q. 5

The difference between two numbers is 2 and the difference between their squares is 24. Find the larger number.

• EXPLAINATION:
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 = 24-4 = 20
x = 20/4 = 5
y = x+2 = 5+2 = 7
The larger number is 7.

[x^2=x*x]

Q. 6

Find the value of (bx-ay), if x/a+y/b = 2 and ax -by = a^2-b^2.

[a^2=a*a]

• EXPLAINATION:
x/a+y/b = 2
bx + ay = 2ab …(1)
and ax-by = 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 = 2b-b=b
Hence, x = a and y = b
bx-ay = ba-ab = 0

Q. 7

Find the first term of a GP whose second term is 2 and the sum to infinity is 8.

• EXPLAINATION:
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/(1-r)
ar = 2 and a/(1-r) = 8
r = 2/a and a = 8 – 8r
Hence, a = 8 – 8*2/a
a^2 = 8a -16
a^2-8a+16=0
(a-4)^2=0
a = 4
Hence, the first term of the GP = a = 4

[a^2=a*a]

Q. 8

How many elements does the power set of A contain if A = {x,y}?

• EXPLAINATION:
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]

Q. 9

The radius of the base of a solid cylinder is x cm and its height is 3 cm. It is re-cast into a cone of the same radius. Find the height of the cone in cm.

• EXPLAINATION: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]

Q. 10

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?

• EXPLAINATION:
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.

Q. 11

Find the value of a if (x-a) is the g.c.d. of x^2-x-6 and x^2+3x-18.

[x^2=x*x]

• EXPLAINATION:
Since (x-a) is the g.c.d of the given polynomials, (x-a) is a factor of both the polynomials.
Putting the value a in place of x in the two polynomials, we get
a^2-a-6=0 and a^2+3a-18=0
a^2-a-6=a^2+3a-18
-a-6=3a-18
3a+a=18-6
4a=12
a=3

Q. 12

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?

• EXPLAINATION:
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

Q. 13

In a two-digit 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?

• EXPLAINATION:
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 = 12-8 = 4
The possible numbers are 84 and 48.
Hence, two such numbers are possible.

Q. 14

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?

• EXPLAINATION:
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.

Q. 15

Find n when n>0 and P(n,4) = 20*P(n,2).

• EXPLAINATION:
P(n,4) = 20*P(n,2)
n!/(n-4)! = 20*n!/(n-2)!
(n-2)! = 20*(n-4)!
(n-2)(n-3)(n-4)! = 20*(n-4)!
(n-2)(n-3)=20
n^2-5n+6-20=0
n^2-5n-14=0
n^2-7n+2n-14=0
n(n-7)+2(n-7)=0
n=-2,7
Since n>0, we have n = 7

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