Test: Numeric Entry- 5 - GRE MCQ

Let the seventh observation be x. Average = sum of observations/number of observations Sum of observations
= 12*6 = 72
The new average is 12-1 = 11
Sum of observations = 72 + x
Number of obervations
= 6+1 = 7 11 = (72+x)/7 77
= 72+x x = 77 - 72 = 5
The seventh observation is 5

Let the number of sides be x.
The number of diagonals is given by
x(x-3)/2 x(x-3)/2
= 44 x^2-3x-88=0
x^2 -11x+8x-88=0
x(x-11)+8(x-11)=0
x=-8,
11 Since x cannot be negative,
x = 11
The polygon has 11 sides.

Let the number of apples bought be x.
Price per apple = 110/x
Had he bought 1 more, the number of apples would have been x+1 Price per apple would have been 110/(x+1)
The apples would have costed Re.1 lesser.
Hence, we have
110/(x+1) = 110/x-1
110/(x+1) = (110-x)/x
110x = (110-x)(x+1)
110x = 110x -x^2+110-x
x^2+x-110=0
x^2+11x-10x-110=0
x(x+11)-10(x+11)=0
(x-10)(x+11)=0
x=10,-11
Rejecting negative value,
we get x = 10
The number of apples bought was 10.

The longest possible stick that can be fitted shall be along the diagonal of the cuboid.
Diagonal of a cuboid with length, breadth and height as l, b and h is given by
sqrt(l²+b²+h²)
= sqrt(12²+5²+84²)
= sqrt(144+25+7056)
= sqrt(7225) = 85
The length of the stick should be 85 cm.
[l²=l*l]

Let the original side of the square be x units.
Area of square = side*side = x² sq.units
Length of the rectangle formed = x + 30% of x = x + 30x/100 = 130x/100
Breadth of the rectangle formed = y + 20% of y = y + 20y/100 = 120y/100
Area of the rectangle
= length*breadth
= 130x/100*120x/100
= 156x²/100
Percentage increasse in area
= (area of rectangle-area of square)/ area of square *100
= (156x²/100 - x²)/x²*100
= (156-100)/100*100
= 56% [x²=x*x]

Let the two numbers be x and y such that
x>y x²+y² = (37/6)xy...(1)
x + y = 35...(2)
We know that
(x+y)² = x²+y²+2xy
Substituting from (1) and (2),
we get 35² = (37/6)xy + 2xy
1225 = (37xy+12xy)/6
49xy = 1225*6 xy = 1225*6/49 xy = 150 Putting x = 150/y in (2),
we get
150/y + y = 35
150 + y² = 35y y² - 35y +150=0 y² - 30y - 5y +150=0 y(y-30) -5(y-30) = 0 (y-5)(y-30) = 0 y = 30,5
The two numbers are 30 and 5 x - y = 30 - 5 = 25 [x²=x*x]

Let the SP be the selling price and CP be the cost price of the item.
Let MP be the marked price.
For the first discount, the MP was Rs.65 SP = (100-discount%)*MP/100
= (100-10)*65/100
= 90*65/100
= 58.50
For the second discount, the SP will be Rs. 56.16 and the MP will be Rs.58.50 56.16 = (100-discount%)*58.50/100 100-Discount%= 96 Discount% = 4%

Let the speed of the water in the stream be x km/hr.
Let the speed of the man in still water be y km/hr When he goes upstream, the relative speed is (y-x) km/hr
When he goes downstream, the relative speed is (y+x) km/hr
According to the given conditions, Distance = Speed * Time
12 = (y-x)*5...(1)
22 = (y+x)*5...(2)
Subtracting (1) from (2),
we get
22 - 12 = 5y + 5x -5y +5x 10 = 10x x = 1km/hr
Speed of the water is 1 km/hr

Let the two numbers be 6x and 13x.
LCM of 6x and 13x = 6*13*x = 78x
78x = 312
x = 312/78 = 4
The two numbers are 6*4 and 13*4 i.e.
24 and 52 The sum of the numbers is
24+52 = 76

198 Explanation:
(x/y+y/x) = 6 (x/y+y/x)³
= 6³ (x/y)³ + (y/x)³ +3(x/y*y/x)(x/y+y/x)
= 216 (x/y)³+(y/x)³
= 216 - 3*6
= 216 - 18
= 198

a²+b²
=5ab a²/ab+b²/ab
=5ab/ab a/b+b/a
=5 (a/b+b/a)³
=5³ (a/b)³+(b/a)³+3(a/b*b/a)(a/b+b/a)
= 125 (a/b)³+(b/a)³
=125-3*5 a³/b³+b³/a³
=125-15
=110

Let CD be the altitude of the triangle with D lying on AB, which is the base.
Side of square = perimeter/4
The altitude of the triangle will be = 140/4 = 35cm.
Area of triangle = 1/2*base*altitude 1/2*base*35 = 420
base = 2*420/35 = 24cm
In triangle ADC, AD will be perpendicular to CD AD = 24/2=12 and CD = 35 AC = sqrt(AD²+DC²)
= sqrt(12²+35²)
= sqrt(144+1225)
= sqrt(1369)
= 37 cm
Each of the equal sides in 37 cm long.
[x²=x*x]

The diagonals of a rhombus bisect each other at right angles.
Hence, we get four right angled triangles with two sides 72/2 and 30/2 cm long Hypotenuse of the triangle
= sqrt(36²+15²)
= sqrt(1296+225)
= sqrt(1521)
= 39 cm
Each side of the rhombus is 39 cm long.
[x²=x*x]

Let the number of hens be x and the number of cows be y.
Accordng to the conditions
x+y=48...(1)
2x+4y=140...(2)
Multiply (2) by 2 and subtract from
(2) 2x+4y-2x-2y=140-96
2y=44
y = 44/2=22
There are 22 cows in the group.