Test: Numeric Entry- 3 - GRE MCQ

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

(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

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]

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

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

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]

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]

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.

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.

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]

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

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.

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]

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%.

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.