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# Test: Numeric Entry- 3

## 15 Questions MCQ Test Section-wise Tests for GRE | Test: Numeric Entry- 3

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This mock test of Test: Numeric Entry- 3 for GRE helps you for every GRE entrance exam. This contains 15 Multiple Choice Questions for GRE Test: Numeric Entry- 3 (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Numeric Entry- 3 quiz give you a good mix of easy questions and tough questions. GRE students definitely take this Test: Numeric Entry- 3 exercise for a better result in the exam. You can find other Test: Numeric Entry- 3 extra questions, long questions & short questions for GRE on EduRev as well by searching above.
*Answer can only contain numeric values
QUESTION: 1

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

Solution:

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

*Answer can only contain numeric values
QUESTION: 2

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

Solution:

(n+1)! = 30*(n-1)! (n+1)*n*(n-1)!
= 30(n-1)! (n+1)*n
= 30 n2 + n - 30
= 0 n2 +6n-5n - 30
=0 n(n+6) - 5(n+6)
= 0 n = -6,
5 Since n cannot be negative,
n = 5

*Answer can only contain numeric values
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.

Solution:

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]

*Answer can only contain numeric values
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?

Solution:

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

*Answer can only contain numeric values
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.

Solution:

Since A and B are mutually exclusive events, they do not occur together at all.
Hence, the required probability is zero.

*Answer can only contain numeric values
QUESTION: 6

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

Solution:

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(l2+b2+h2)
= sqrt (122+42+32)
= sqrt (144+16+9)
= sqrt(169)
= 13 cm
The diagonal of the cuboid is 13 cm.
[l2=l*l]

*Answer can only contain numeric values
QUESTION: 7

Find the value of k for which both the equations 12x2+4kx+3=0 and (k+1)x2-2(k-1)x+1=0 have equal real roots. [x2=x*x]

Solution:

Since the two equations have equal real roots, their discriminants are 0.
For 12x2+4kx+3=0, D = 0
D^2 = 0 b2-4ac=0
(4k)2 - 4*12*3 = 0 16k2 - 144=0
k2 = 144/16 = 9
k = -3,
3 For (k+1)x2-2(k-1)x+1=0
b2-4ac=0 [2*(k-1)]2-4*(k+1)*1=0
4(k2-2k+1)-4(k+1)=0
4[k2-2k+1-k-1]=0
k2 - 3k = 0 k(k-3)=0 k=0,3
Hence, we have k = 3 [k2=k*k]

*Answer can only contain numeric values
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?

Solution:

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.

*Answer can only contain numeric values
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.

Solution:

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.

*Answer can only contain numeric values
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.

Solution:

Let the numbers be x and y,
x>y x2+y2=68 and (x-y)2
=36 (x-y)2
= x2+y2-2xy 36
= 68-2xy 2xy
=68-36=32 xy
=32/2=16 (x+y)2
=x2+y2+2xy = 68+2*16
=68+32=100 x+y
= sqrt(100)
= 10 [x2=x*x]

*Answer can only contain numeric values
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.

Solution:

Distance = speed*time
= 330*10
= 3300 meters

*Answer can only contain numeric values
QUESTION: 12

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

Solution:

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.

*Answer can only contain numeric values
QUESTION: 13

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

Solution:

According to the conditions,
we have a+17=60*1/a
a2+17a-60=0
a2 +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'.
[a2=a*a]

*Answer can only contain numeric values
QUESTION: 14

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

Solution:

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%.

*Answer can only contain numeric values
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?

Solution:

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.