# Numeric Entry Practice Test 3

**Question 1**

**Find r when C(n,r) = 35 and P(n,r) = 840**

**Correct Answer: 4**

**Explanation:**

From the given conditions, we get

C(n,r) = n!/[r!(n-r)!] = 35...(1)

P(n,r) = n!/(n-r)! = 840...(2)

Dividing (2) by (1), we get

n!/(n-r)!/n!/[r!(n-r)!] = 840/35

r! = 24

r! = 1*2*3*4 = 24

r = 4

**Question 2**

**If (n+1)! = 30*(n-1)!, then find n.**

**Correct Answer: 5**

**Explanation:**

(n+1)! = 30*(n-1)!

(n+1)*n*(n-1)! = 30(n-1)!

(n+1)*n = 30

n^{2} + n - 30 = 0

n^{2} +6n-5n - 30 =0

n(n+6) - 5(n+6) = 0

n = -6, 5

Since n cannot be negative, n = 5

**Question 3**

**Find x if (0,0), (3, sqrt(3)) and (x, 2*sqrt(3)) are the vertices of an equilateral triangle in the first quadrant.**

**Correct Answer: 0**

**Explanation:**

Let the vertices (0,0), (3, sqrt(3)) and (x, 2*sqrt(3)) be A, B and C respectively.

Since the triangle is equilateral, we have

AB = BC = CA

AB^{2} = BC^{2} = CA^{2}

(3-0)^{2}+(sqrt(3)-0)^{2} = (3-x)^{2} + (2*sqrt(3) - sqrt(3))^{2} = (x-0)^{2} + (2*sqrt(3)-0)^{2}

9 + 3 = 9 - 2x + x^{2} +3 = x^{2} + 12

Hence, x^{2}+12 = 12

x = 0

[AB^{2}=AB*AB]

**Question 4**

**C can complete the work in 18 days if he works alone. B takes 3 days lesser than C and A takes 5 days lesser than B to complete the work.
On which day will they complete the work if C works for the initial two days only?**

**Correct Answer: 6**

**Explanation:**

C takes 18 days to complete the work.

B takes 3 days lesser. Hence, B takes 18-3 = 15 days.

A takes 5 days lesser than B. Hence, A takes 15-5 = 10 days.

Work done by A in one day = 1/10

Work done by B in one day = 1/15

Work done by C in one day = 1/18

Work done by the three of them in two days = 2(1/18+1/10+1/15)

= 2(5+9+6)/90

= 2*20/90 = 4/9

Remaining work = 1-4/9 = 5/9

Work done by A and B in one day = 1/10+1/15

= (3+2)/30

=1/6

Time taken by A and B to complete the remaining work = 5/9*6/1

= 10/3 = 3.33

The work in completed on 2+3.33 = 6th day

**Question 5**

**If A and B are two mutually exclusive events and P(A) = 2/3 and P(B) = 4/9, then find the probability of the occurrance of A and B together.**

**Correct Answer: 0**

**Explanation:**

Since A and B are mutually exclusive events, they do not occur together at all.

Hence, the required probability is zero.

**Question 6**

**Find the diagonal of a cuboid whose volume is 144 cc and the dimensions of its base are 12cm and 4cm.**

**Correct Answer: 13cm**

**Explanation:**

The length (l) and breadth (b) of the base are 12 cm and 4 cm respectively.

Height (h) of the cuboid = volume/(l*b)

= 144/(12*4)

= 3 cm

Diagonal of the cuboid = sqrt(l^{2}+b^{2}+h^{2})

= sqrt(12^{2}+4^{2}+3^{2})

= sqrt(144+16+9)

= sqrt(169)

= 13 cm

The diagonal of the cuboid is 13 cm.

[l^{2}=l*l]

**Question 7**

**Find the value of k for which both the equations 12x ^{2}+4kx+3=0 and (k+1)x^{2}-2(k-1)x+1=0 have equal real roots.**

[x^{2}=x*x]

**Correct Answer: 3**

**Explanation:**

Since the two equations have equal real roots, their discriminants are 0.

For 12x^{2}+4kx+3=0,

D = 0

D^{2} = 0

b^{2}-4ac=0

(4k)^{2} - 4*12*3 = 0

16k^{2} - 144=0

k^{2} = 144/16 = 9

k = -3, 3

For (k+1)x^{2}-2(k-1)x+1=0

b^{2}-4ac=0

[2*(k-1)]^{2}-4*(k+1)*1=0

4(k^{2}-2k+1)-4(k+1)=0

4[k^{2}-2k+1-k-1]=0

k^{2} - 3k = 0

k(k-3)=0

k=0,3

Hence, we have k = 3

[k^{2}=k*k]

**Question 8**

**What percent is the first number of the second if the first number is 120% of the third and the second number is 150% of the third?**

**Correct Answer: 80%**

**Explanation:**

Let the first, second and the third numbers be x, y and z respectively.

x = 120% of z

= 120/100*z

y = 150% of z

= 150/100*z

x/y*100 = (120z/100)/(150z/100)*100

= 80%

x is 80% of y.

**Question 9**

**The difference between two numbers is 2. Each number is less than 14 and their sum is greater than 22. Find the greater number of the two.**

**Correct Answer: 13**

**Explanation:**

Let the two numbers be x and y such that x < y.

y – x = 2, x < 14, y < 14, x + y > 22

y = x + 2, x < 14, x + 2 < 14, x + x + 2 > 22

x < 14, x < 12, 2x + 2 > 22

x < 12, x +1 > 11

x < 12, x > 10

Hence, x can be 11 and the corresponding value of y is 13.

The two numbers are 11 and 13.

**Question 10**

**The sum of the squares of the two positive numbers is 68 and the square of their difference is 36. Find the sum of the numbers.**

**Correct Answer: 10**

**Explanation:**

Let the numbers be x and y, x>y

x^{2}+y^{2}=68 and (x-y)^{2}=36

(x-y)^{2} = x^{2}+y^{2}-2xy

36 = 68-2xy

2xy=68-36=32

xy=32/2=16

(x+y)^{2}=x^{2}+y^{2}+2xy

= 68+2*16

=68+32=100

x+y = sqrt(100) = 10

[x^{2}=x*x]

**Question 11**

**The difference between the time when the lightening was seen and the time when the thunder was heard is 10 seconds. Sound covers a distance 330 meters in one second. Find the distance of the thundercloud from the point of observation in meters.**

**Correct Answer: 3300 meters**

**Explanation:**

Distance = speed*time

= 330*10

= 3300 meters

**Question 12**

**Six points lie on a circle. How many cyclic quadrilaterals can be formed by joining the points?**

**Correct Answer: 15**

**Explanation:**

Since, the points lie on a circle, all the possible quadrilaterals shall be cyclic.

Number of quadrilateals = C(6,4)

= 6!/(2!4!)

= 6*5/2

= 15

Hence, 15 cyclic quadrilaterals can be drawn.

**Question 13**

**When a number 'a' is increased by 17, it equals 60 times its reciprocal. How many values of 'a' are possible?**

**Correct Answer: 2**

**Explanation:**

According to the conditions, we have

a+17=60*1/a

a^{2}+17a-60=0

a^{2} +20a-3a-60=0

a(a+20)-3(a+20)=0

(a-3)(a+20)=0

a=3, -20

When a = 3,

a+17 = 3+17 = 20 = 60*1/3

When a = -20,

a+17 = -20+17 = -3 = 60*(-1/20)

Hence, the two values of 'a' are valid. There are two possible values of 'a'.

[a^{2}=a*a]

**Question 14**

**The tax on a commodity decreases by 10% and the consumption increases by 20%. What is the percentage increase in the revenue?**

**Correct Answer: 8%**

**Explanation:**

Revenue = Tax*Consumption

Let the original tax be x and consumption be y.

According to the given conditions, the new tax and consumption will be

(x-10x/100) = 90x/100 and (y+20y/100) = 120y/100

The original revenue would be xy and the new revenue would be (90x/100)*(120y/100)

Perentage change in revenue = (new revenue-old revenue)/old revenue * 100

= (90x/100*120x/100-xy)/xy*100

= (1.08xy-xy)/xy*100

= 0.08*100

= 8%

The increase in revenue will be 8%.

**Question 15**

**A shopkeeper bought flowers at the rate of 8 for Rs.34. He sold them at 12 flowers for Rs.57 and gained Rs. 900. How many flowers were there?**

**Correct Answer: 1800**

**Explanation:**

Let x be the number of flowers bought and then sold by the shopkeeper.

Let C.P. and S.P. be the cost price and selling price.

Total C.P. = 34/8*x = 17x/4

Total S.P. = 57/12*x = 57x/12

Gain = Total SP - Total CP

900 = 57x/12 - 17x/4

(57x-51x)/12 = x/2 = 900

x = 900*2=1800

Number of flowers = 1800.