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# Test: Numeric Entry- 4

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## 15 Questions MCQ Test Section-wise Tests for GRE | Test: Numeric Entry- 4

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*Answer can only contain numeric values
Test: Numeric Entry- 4 - Question 1

### From a rectangular cardboard of length 21 cm and breadth 12 cm, a square whose diagonal is of length 3*sqrt(2) cm is cut out. Find the area of the remaining cardboard.

Detailed Solution for Test: Numeric Entry- 4 - Question 1

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.

*Answer can only contain numeric values
Test: Numeric Entry- 4 - Question 2

### The height of a cone is trebled. The radius of the cone is doubled. By how many times will the volume of the cone increase?

Detailed Solution for Test: Numeric Entry- 4 - Question 2

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]

*Answer can only contain numeric values
Test: Numeric Entry- 4 - Question 3

### A rectangular reservoir is 120 m long and 75 m high. The cross-sectional area of the pipe through which water flows is 0.04 sq.m. and the water level in the reservoir rises 2.4 meters in 18 hours. Find the speed of water flowing into the reservoir.

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

Cross-sectional area of the pipe = 0.04 sq.m. Volume of water put into the reserviour in 18 hours
= 120*75*2.4
= 21600 cubic m
Volume of water put into the reserviour per hour = 21600/18 cubic m/hr
Required speed = Volume/area
= (21600/18)/0.04
= 30000 cubic m/hr

*Answer can only contain numeric values
Test: Numeric Entry- 4 - Question 4

P(x-1,3) : P(x,4) = 1: 9, find x.

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

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

*Answer can only contain numeric values
Test: Numeric Entry- 4 - Question 5

If P(10,r) = 30240, then find r.

Detailed Solution for Test: Numeric Entry- 4 - Question 5

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

*Answer can only contain numeric values
Test: Numeric Entry- 4 - Question 6

Find the sum of the first 20 terms of the AP 1, 4, 7, 10...

Detailed Solution for Test: Numeric Entry- 4 - Question 6

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

*Answer can only contain numeric values
Test: Numeric Entry- 4 - Question 7

Tim buys two items at different rates for a total of Rs. 410 and sells one at a loss of 20% and the other at a gain of 25%. If both the items were sold at the same price, then find the cost price of the cheaper item.

Detailed Solution for Test: Numeric Entry- 4 - Question 7

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

*Answer can only contain numeric values
Test: Numeric Entry- 4 - Question 8

Find the length of a parallel side of the trapezium if one of its parallel sides is 70 m long and its area is 2500 sq.m. The distance between the parallel sides is 40m.

Detailed Solution for Test: Numeric Entry- 4 - Question 8

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.

*Answer can only contain numeric values
Test: Numeric Entry- 4 - Question 9

Find (b-a) when x4+ax2+bx+5 is exactly divisible be x2+3x+2. [x^4=x*x*x*x]

Detailed Solution for Test: Numeric Entry- 4 - Question 9

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

*Answer can only contain numeric values
Test: Numeric Entry- 4 - Question 10

x+1/x = 5. Find x3+1/x3. [x3=x*x*x]

Detailed Solution for Test: Numeric Entry- 4 - Question 10

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

*Answer can only contain numeric values
Test: Numeric Entry- 4 - Question 11

Pam lent Rs. 1800 to Ritu. She also lent Rs 2250 to Tammy for four years. They both returned the same interest charged at the same rate of interest of 3% per annum. For how many years did Ritu borrow the money?

Detailed Solution for Test: Numeric Entry- 4 - Question 11

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

*Answer can only contain numeric values
Test: Numeric Entry- 4 - Question 12

The length of the tangent drawn from an exterior point P to the circle is 63 cm. The point P is at a distance of 65 cm from the centre of the circle. Find the radius of the circle.

Detailed Solution for Test: Numeric Entry- 4 - Question 12

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]

*Answer can only contain numeric values
Test: Numeric Entry- 4 - Question 13

A hollow cylinder of height 3m is melted to form a solid cylinder of the same height. The external and internal radii of the hollow cylinder are 29 cm and 20 cm respectively. Find the radius of the solid cylinder.

Detailed Solution for Test: Numeric Entry- 4 - Question 13

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]

*Answer can only contain numeric values
Test: Numeric Entry- 4 - Question 14

A shopkeeper sells each book for Rs 1134 after giving a discount of 19% on the marked price. Had he sold the books at the printed price, he would have earned a profit of 40%. Find the cost price of each book.

Detailed Solution for Test: Numeric Entry- 4 - Question 14

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

*Answer can only contain numeric values
Test: Numeric Entry- 4 - Question 15

A metallic cylinder was melted and a sphere was formed from the metal. The diameter of the cylinder is 6 cm and its length is 4 cm. Find the radius of the sphere.

Detailed Solution for Test: Numeric Entry- 4 - Question 15

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]

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