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# Test: Numeric Entry- 3

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

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*Answer can only contain numeric values
Test: Numeric Entry- 3 - Question 1

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

Detailed Solution for Test: Numeric Entry- 3 - Question 1

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
Test: Numeric Entry- 3 - Question 2

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

Detailed Solution for Test: Numeric Entry- 3 - Question 2

(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
Test: Numeric Entry- 3 - 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.

Detailed Solution for Test: Numeric Entry- 3 - Question 3

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
Test: Numeric Entry- 3 - 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?

Detailed Solution for Test: Numeric Entry- 3 - Question 4

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
Test: Numeric Entry- 3 - 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.

Detailed Solution for Test: Numeric Entry- 3 - Question 5

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
Test: Numeric Entry- 3 - Question 6

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

Detailed Solution for Test: Numeric Entry- 3 - Question 6

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
Test: Numeric Entry- 3 - 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]

Detailed Solution for Test: Numeric Entry- 3 - Question 7

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
Test: Numeric Entry- 3 - 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?

Detailed Solution for Test: Numeric Entry- 3 - Question 8

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
Test: Numeric Entry- 3 - 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.

Detailed Solution for Test: Numeric Entry- 3 - Question 9

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
Test: Numeric Entry- 3 - 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.

Detailed Solution for Test: Numeric Entry- 3 - Question 10

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
Test: Numeric Entry- 3 - 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.

Detailed Solution for Test: Numeric Entry- 3 - Question 11

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

*Answer can only contain numeric values
Test: Numeric Entry- 3 - Question 12

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

Detailed Solution for Test: Numeric Entry- 3 - Question 12

Since, the points lie on a circle, all the possible quadrilaterals shall be cyclic.
= 6!/(2!4!) = 6*5/2 = 15
Hence,
15 cyclic quadrilaterals can be drawn.

*Answer can only contain numeric values
Test: Numeric Entry- 3 - Question 13

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

Detailed Solution for Test: Numeric Entry- 3 - Question 13

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
Test: Numeric Entry- 3 - Question 14

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

Detailed Solution for Test: Numeric Entry- 3 - Question 14

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
Test: Numeric Entry- 3 - 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?

Detailed Solution for Test: Numeric Entry- 3 - Question 15

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.

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