# Numeric Entry Practice Test 2

**Question 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 Answer: 5 years**

**Correct Answer: 5 years.**

**Explanation:**

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]

**Question 2**

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

[x^{3}=x*x*x]

**Correct Answer: 4**

**Explanation:**

(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

**Question 3**

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

[x^{3}=x*x*x]

**Correct Answer: 2**

**Explanation:**

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

**Question 4**

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

**Correct Answer: 12**

**Explanation:**

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

**Question 5**

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

**Correct Answer: 7**

**Explanation:**

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]

**Question 6**

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

[a^{2}=a*a]

**Correct Answer: 0**

**Explanation:**

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

**Question 7**

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

**Correct Answer: 4**

**Explanation:**

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]

**Question 8**

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

**Correct Answer: 4**

**Explanation:**

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]

**Question 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.**

**Correct Answer: 9 cm**

**Explanation:**

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]

**Question 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? **

**Correct Answer: 8**

**Explanation:**

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.

**Question 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]

**Correct Answer: 3**

**Explanation:**

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

**Question 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?**

**Correct Answer: Rs.5000**

**Explanation:**

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

**Question 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?**

**Correct Answer: 2**

**Explanation:**

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.

**Question 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?**

**Correct Answer: 2**

**Explanation:**

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.

**Question 15**

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

**Correct Answer: 7**

**Explanation:**

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