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# Problems on Numbers Questions

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In a group of donkeys and pigs, the numbers of legs are 16 more than twice the number of heads. The number of donkeys is

A6

B8

C11

D13

Explanation:

Let the number of donkeys be x and the number of pigs be y.
Then,
=> 4x + 2y = 2(x + y) = 16.
=> 2x + (2x + 2y) = (2x + 2y) + 16.
=> x = 8.

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A Student brought 4 1/3m of yellow ribbon, 61/6 m of red ribbons and 3 2/9m of blue ribbon to decorate his room. How many meters of ribbons didd he buy in all ?

A12 13/18m

B11 13/18 m

C10 13/18m

D13 13/18m

Explanation:

Length of yellow ribbon = 4 1/3 m = 13/3 m
Length of red ribbon = 6 1/6 m = 37/6 m
Length of blue ribbon = 3 2/9 m = 29/9 m
So, the total length of all 3 ribbons = (13/3 + 37/6 + 29/9) = (78+111+58)/18 = (247/18) = 13 13/18 m

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In a group of 15 workers, 5 are good at both Plumping and woodwork, 3 workers are good only at woodwork, and 3 are not good at either of them. How many of them are good at Plumping?

A7

B4

C5

D9

Explanation:

Here is no explanation for this answer

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In a particular library there are 5 computer clusters open for public use. Each Computer Clusters has 40 Tablets. In the 1st week, 75 tablets are used, while 155 tablets are used in the second week, 30 in the third week & 140 in 4th week. Approximately what fraction of the library's computer facilities are in use per month?

A1/3

B1

C1/2

DNone of these

Explanation:

Total number of tablets = 40*5 = 200

Fraction of library's computer facilities used,
1st week = 75/200 = 3/8
2nd week = 155/200 = 31/40
3rd week = 30/200 = 3/20
4th week = 140/200 = 7/10

Fraction of library's computer facilities used per month
= (3/8+31/40+3/20+7/10)/4
= 1/2

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If there are 1024*1280 pixels on a screen and each pixel can have around 16 million colors. Find the memory required for this?

A4 MB

B16 MB

C2 MB

DNone of these

Explanation:

Here is no explanation for this answer

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Ramesh, Ram, Kareem and Mohan collected coins of different countries. They collected 100 together. None collected less than 10. Each collected an even number. Each collected a different number. Based on these, we can say that the number of coins collected by the boy who collected the most could not have exceeded?

A54

B64

C58

D60

Explanation:

Here is no explanation for this answer

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A student is ranked 13th from right and 8th fromleft. How many students are there in totality?

A19

B16

C20

D21

Explanation:

Here is no explanation for this answer

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If three eighth of a number is 1257, the one fourth of the number will be:

A559

B670

C838

D926

Explanation:

Let number is x, then
(3/8)*x = 1257
=> x/8 = 419
=> x/4 = 838
=> (1/*x = 838

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In row of girls,there are 16 girls between priya and natashpriya is the thirty second girl from the left end of the row. If priya is nearer than natasha to the right end of the row, then how far away is natasha from the left end of the row?

A16

B14

C15

D13

Explanation:

Left end -... Natash... 16 gals .... Priya ...right end.
As priya is 32nd girl from left.
so, to get the position of Natash.
32 - {16+1=}17 = 15

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A triangle ABC is given, a line DE is parallel to base side and that cuts the triangle. The ratio of area of triangle to the area of trapezium.
Given DE/BC = 3/5.

A9/16

B5/2

C3/16

DNone of these

Explanation:

The line DE divides the triangle ABC into 2 parts - triangle ADE and trapezium DBCE.The question is to find the ratio of area(ADE) to area(DBCE).

Now triangle ADE is similar to triangle ABC.
=> DE/BC = AD/AB = AE/AC ....(1)
Drawing a perpendicular from A to BC meeting BC at G and meeting DE at F.
Now AF and AG are the heights of the triangles ADE and ABC resp.
Area of a triangle = 1/2(base*height).
Area of tri(ABC) = 1/2(BC*AG)
Area of trapezium(DBCE) = area(ABC)-area(ADE) = 1/2(BC*AG-DE*AF)
Ratio - area(ADE)/area(DBCE) = (DE*AF)/(BC*AG-DE*AF) = 1/[(BC*AG-DE*AF)/(DE*AF)]
(rearranging) = 1/[(BC/DE)*(AG/AF)-1]
Using (1) and (2),(BC/DE)*(AG/AF) = (BC/DE)^2
(using given DE/BC=3/5) => BC/DE=5/3 => (BC/DE)^2 = 25/9
The required ratio = 1/(25/9-1) =1/(16/9) = 9/16

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