Test: Numeric Entry- 4 - GRE MCQ

Area of the cardboard = length*breath
= 21*12 = 252 sq.cm.
Diagonal of a square of side x cm is equal to x*sqrt(2) cm.
Hence,
side of square = 3 cm
Area of square = side*side = 3*3 = 9 sq.cm.
Area of the remaining cardboard
= 252-9
= 243 sq.cm.

Let the original height be h and radius of the base be r.
Original volume will be = (1/3)*pi*r²*h
New height will be 3h New radius will be 2r New volume will be
= (1/3)*pi*(2r)²*(3h)
= (1/3)*pi*r²*h = (1/3)*pi*r^2*h*12
New volume = 12*original volume
Hence, the volume increases 12 times.
[pi=22/7, r²=r*r]

Step 1: Calculate the volume of water needed to raise the water level in the reservoir.

The dimensions of the reservoir are:
- Length (L) = 120 m
- Width (W) = 75 m
- Height (H) = 2.4 m (the height the water needs to rise)

The volume (V) of the water needed can be calculated using the formula for the volume of a cuboid:
V=L×W×H

Substituting the values:
V=120m×75m×2.4m
V=21600m³

Step 2: Determine the volume of water flowing through the pipe.

The dimensions of the square pipe are:
- Width = 20 cm = 0.2 m (since 1 m = 100 cm)
- Height = 20 cm = 0.2 m

The volume of the pipe can be calculated using the formula for the volume of a cuboid:
Volume of pipe=Area of cross-section×Length

The area of the cross-section (A) of the square pipe is:
A=Width×Height=0.2m×0.2m=0.04m²

Let x be the speed of water flow in meters per hour. The length of water that flows through the pipe in 18 hours is:
Length=18hours×xm/h=18xm

Thus, the volume of water flowing through the pipe in 18 hours is:
Volume of pipe=A×Length=0.04m²×18xm
Volume of pipe=0.72xm³

Step 3: Set the volumes equal to each other.

Since the volume of water flowing into the reservoir through the pipe must equal the volume of water needed to raise the water level, we can set up the equation:
0.72x=21600

Step 4: Solve for x.

Now, we will solve for x:
x=216000 / 0.72
x=30000m/h

P(x-1,3) : P(x,4)
= 1: 9 (x-1)!/(x-1-3)! : x!/(x-4)!
= 1:9 (x-1)!/(x-4)! : x(x-1)!/(x-4)!
= 1:9 1:x = 1:9
x = 9

P(10,r) = 30240 10!/(10-r)! = 30240
10!/(10-r)! = 3024*10 10!/(10-r)!
= 336*9*10 10!/(10-r)!
= 42*8*9*10 10!/(10-r)!
= 6*7*8*9*10 10!/(10-r)!
= 10!/5! 10!/(10-r)!
= 10!/(10-5)!
Hence,
r = 5

The first term of the AP, a, is 1 and the common difference, d, is 3.
The sum of n terms is given by
Sn = (n/2)[2a+(n-1)d]
Sum of 20 terms
= (20/2)[2*1+(20-1)*3]
= 10(2+19*3)
= 10(59)
= 590

Let CP and SP be cost price and selling price respectively.
Let the CP of the first item be Rs.x and that of the second item be Rs.(410-x)
Selling price of the first item SP = CP(100-loss%)/100 = x(100-20)/100 = 8x/10
Selling price of the second item SP = CP(100+gain%)/100
= (410-x)(100+25)/100
= (410-x)(125/100)
Since both the items were sold at the same price,
we have 8x/10 = (410-x)(125/100) 80x
= 410*125 - 125x 80x + 125x
= 51250 205x=51250 x = 250
The second item would cost 410-250 = 160
The cheaper item costs Rs.160

Let the unknown length be x m. Area of trapezium = (1/2)*sum of parallel sides*distance between parallel sides
2500 = (1/2)*(70+x)*40 70+x
= 2500*2/40 70+x
= 125 x = 125-70 = 55
The other side of the trapezium is 55 m long.

x²+3x+2=(x+1)(x+2) (x+1) and (x+2) should be the factors of x⁴+ax²+bx+5
Putting x = -1 and x = -2,
we get
x⁴+ax²+bx+5 = (-1)⁴+a(-1)²+b(-1)+5=0
1+a-b+5=0 a-b = -6 ...(1)
x⁴+ax²+bx+5 = (-2)⁴+a(-2)²+b(-2) + 5=0 16 + 4a -2b+5=0 4a-2b = -21...(2)
Multiplying (1) by 4 and subtracting (2) from it,
we get
4a - 4b -4a +2b = -24 + 21 -2b = -3
b = 3/2 a = b-6 = 3/2-6
= (3-12)/2 = -9/2 b -a
= 3/2+9/2
= 12/2
= 6

x+1/x = 5 (x+1/x)³
= 5³ x^³+1/x³+3(x+1/x)
= 125 x³+1/x³
= 125 -3*5 =125-15
= 110 x³+1/x³
= 110

Let P, R and T be the principle, rate and time for the simple interest SI
SI = P*R*T/100
Since the SI is the same for both of them,
we have
1800*3*T/100
= 2250*3*4/100
T = 2250*3*4/(1800*3) = 5
The money was lent for 5 years

Let the centre of the circle be the point O and let the point at which the tangent drawn from point P meets the circle be T.
PTO will be a right triangle right angled at T.
PT = 63 cm and PO = 65 cm.
Applying Pythagoras theorem to the triangle PTO and squaring both the sides,
we get PO² = PT²+TO² 65²=63²+TO² 4225 = 3969 + TO² TO²
= 4225-3969 = 256
TO = 16 cm
The radius of the circle is 16 cm.
[PO²=PO*PO]

Volume of cylinder = π*r²*h,
where r is the radius and h is the height. Internal radius = 20 cm and external radius = 29 cm
Volume of metal used
= π*(29)²*h - pi(20)²*h
= π*h[841-400]
= π*h*441
Volume of solid cylinder
= π*h*441
= π*h*R²,
where R is the radius of the solid cylinder
R² = 441 R = sqrt(441) = 21
The radius of the solid cylinder is 21 cm.
[π=22/7, r²=r*r]

Let MP be the marked price, SP be the selling price and CP be the cost price.
MP = 100*SP/(100-discount%)
= 100*1134/(100-19) = 1400
If the books were sold at the primted price the selling price would be Rs1400 CP = 100*SP/(100+profit%)
= 100*1400/(100+40)
= 100*1400/140 =1000
The cost price of each book is Rs.1000

Volume of the cylinder =π*r²*h,
where r is the radius and h is the height of the cylinder.
Radius of the cylinder = 6/2 = 3cm
Volume = π*3²*4 = pi*36
Volume of sphere = 4/3*π*r³,
where r is the radius π*36 = 4/3*π*r³ 36*3/4 = r³
r³ = 27
r = 3cm
The radius of the sphere is 3 cm.
[π=22/7, r²=r*r]