a)9:1 b)2:1 c)4:3 d)5:4
Q) There are four baskets, W, X, Y and Z, each containing a particular variety of oranges. The number of oranges in any basket is unknown. The average weight of the oranges in the four baskets combined is 120 gm. The average weights of the oranges in the baskets W, X, Y and Z, individually, are 90 gm, 100 gm, 144 gm and 160 gm respectively. If the average weight of the oranges in W and X put together is 96 gm and the average weight of the oranges in X and Y put together is 120 gm, what is the ratio of the number of oranges in basket W to that in basket Z?
@ron123 said:Q) There are four baskets, W, X, Y and Z, each containing a particular variety of oranges. The number of oranges in any basket is unknown. The average weight of the oranges in the four baskets combined is 120 gm. The average weights of the oranges in the baskets W, X, Y and Z, individually, are 90 gm, 100 gm, 144 gm and 160 gm respectively. If the average weight of the oranges in W and X put together is 96 gm and the average weight of the oranges in X and Y put together is 120 gm, what is the ratio of the number of oranges in basket W to that in basket Z?a)9:1 b)2:1 c)4:3 d)5:4
4:3?
I am getting 4:6:5:3 for the 4 baskets...
regards
scrabbler
I am getting 4:6:5:3 for the 4 baskets...
regards
scrabbler
@ron123 said:Q) There are four baskets, W, X, Y and Z, each containing a particular variety of oranges. The number of oranges in any basket is unknown. The average weight of the oranges in the four baskets combined is 120 gm. The average weights of the oranges in the baskets W, X, Y and Z, individually, are 90 gm, 100 gm, 144 gm and 160 gm respectively. If the average weight of the oranges in W and X put together is 96 gm and the average weight of the oranges in X and Y put together is 120 gm, what is the ratio of the number of oranges in basket W to that in basket Z?a)9:1 b)2:1 c)4:3 d)5:4
4:3?
@ron123 said:Q) There are four baskets, W, X, Y and Z, each containing a particular variety of oranges. The number of oranges in any basket is unknown. The average weight of the oranges in the four baskets combined is 120 gm. The average weights of the oranges in the baskets W, X, Y and Z, individually, are 90 gm, 100 gm, 144 gm and 160 gm respectively. If the average weight of the oranges in W and X put together is 96 gm and the average weight of the oranges in X and Y put together is 120 gm, what is the ratio of the number of oranges in basket W to that in basket Z?a)9:1 b)2:1 c)4:3 d)5:4
W - a
X - b
Y - c
Z - d
120(a+b+c+d)=90a+100b+144c+160d
96(a+b)=90a+100b
==> 3a=2b
120(b+c)=100b+144c
==> 5b=6c
120a+180a+150a+120d=90a+150a+180a+160d
==> 30a=40d
a:d => 4:3
c) 4:3
X - b
Y - c
Z - d
120(a+b+c+d)=90a+100b+144c+160d
96(a+b)=90a+100b
==> 3a=2b
120(b+c)=100b+144c
==> 5b=6c
120a+180a+150a+120d=90a+150a+180a+160d
==> 30a=40d
a:d => 4:3
c) 4:3
@ron123 said:Q) There are four baskets, W, X, Y and Z, each containing a particular variety of oranges. The number of oranges in any basket is unknown. The average weight of the oranges in the four baskets combined is 120 gm. The average weights of the oranges in the baskets W, X, Y and Z, individually, are 90 gm, 100 gm, 144 gm and 160 gm respectively. If the average weight of the oranges in W and X put together is 96 gm and the average weight of the oranges in X and Y put together is 120 gm, what is the ratio of the number of oranges in basket W to that in basket Z?a)9:1 b)2:1 c)4:3 d)5:4
Let no of oranges be w,x,y,z In W,X,Y,Z
Weight W = 90w, X=100x, Y=144y, Z=160z
90w+100x+144y+160z=120(w+x+y+z)
24y+40z=30w+20x---1
90w+100x=96(w+x)
4x=6w;x=(3/2)w---2
100x+144y=120(x+y)
y=(10/24)x=(30/24)w----3
Put 3 and 2 in 1
24*(30/24)w+40z=30w+20*(3/2)w
30w+40z=30w+30w
30w=40z
w/z=4:3
Weight W = 90w, X=100x, Y=144y, Z=160z
90w+100x+144y+160z=120(w+x+y+z)
24y+40z=30w+20x---1
90w+100x=96(w+x)
4x=6w;x=(3/2)w---2
100x+144y=120(x+y)
y=(10/24)x=(30/24)w----3
Put 3 and 2 in 1
24*(30/24)w+40z=30w+20*(3/2)w
30w+40z=30w+30w
30w=40z
w/z=4:3
Good night friends. :)
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Annie, Benny and Cathy houses forms an equilateral triangle boundary of side 1 unit, with a square grass field of largest possible area within that boundary. There is a circular fountain such that it covers (pie/8)th area of the field and the center of the field and fountain coincide.
Four points P1, P2, P3, P4are chosen at the floor of the fountain such that they are at NE, NW, SE, SW wrt to the center. There are a total of 12 ants, 3 each of same category P(i) at each point P(i). Ants started to move towards the boundary by using the shortest possible path.
Find the summation of total length covered by each ant. If no two ants of same category should be there at any of the boundary.
Consider house to be a point object and field to be flat.
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Annie, Benny and Cathy houses forms an equilateral triangle boundary of side 1 unit, with a square grass field of largest possible area within that boundary. There is a circular fountain such that it covers (pie/8)th area of the field and the center of the field and fountain coincide.
Four points P1, P2, P3, P4are chosen at the floor of the fountain such that they are at NE, NW, SE, SW wrt to the center. There are a total of 12 ants, 3 each of same category P(i) at each point P(i). Ants started to move towards the boundary by using the shortest possible path.
Find the summation of total length covered by each ant. If no two ants of same category should be there at any of the boundary.
Consider house to be a point object and field to be flat.
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@Torque024 said:Good night friends. ==================================================Annie, Benny and Cathy houses forms an equilateral triangle boundary of side 1 unit, with a square grass field of largest possible area within that boundary. There is a circular fountain such that it covers (pie/8)th area of the field and the center of the field and fountain coincide.Four points P1, P2, P3, P4are chosen at the floor of the fountain such that they are at NE, NW, SE, SW wrt to the center. There are a total of 12 ants, 3 each of same category P(i) at each point P(i). Ants started to move towards the boundary by using the shortest possible path.Find the summation of total length covered by each ant. If no two ants of same category should be there at any of the boundary.Consider house to be a point object and field to be flat. ==================================================
2rt3?
regards
scrabbler
regards
scrabbler
Q. Eight letters are to be placed in eight addressed envelopes. If the letters are placed at random into the envelopes, the probability of the event where none of the eight letters is placed into its corresponding envelope is _______
A. (1/2!)-(1/3!)+(1/4!)-(1/5!)+(1/6!)-(1/7!)+(1/8!)
Now please explain how to solve the problem..
@aakhrijang said:Q. Eight letters are to be placed in eight addressed envelopes. If the letters are placed at random into the envelopes, the probability of the event where none of the eight letters is placed into its corresponding envelope is _______A. (1/2!)-(1/3!)+(1/4!)-(1/5!)+(1/6!)-(1/7!)+(1/8!)Now please explain how to solve the problem..
Simple derangement concept..
@ron123 said:Q) There are four baskets, W, X, Y and Z, each containing a particular variety of oranges. The number of oranges in any basket is unknown. The average weight of the oranges in the four baskets combined is 120 gm. The average weights of the oranges in the baskets W, X, Y and Z, individually, are 90 gm, 100 gm, 144 gm and 160 gm respectively. If the average weight of the oranges in W and X put together is 96 gm and the average weight of the oranges in X and Y put together is 120 gm, what is the ratio of the number of oranges in basket W to that in basket Z?a)9:1 b)2:1 c)4:3 d)5:4
W X
90 100
96
4 6
X Y
100 144
120
24 20
6 5
W:X:Y
4:6:5
4x*90+6x*100+5x*144+16z=(15x+z)*120
z=3x
W:X:Y:Z
4:6:5:3
A bag has coins in the denominations of 50 p , 25 p , 20 p in the ration 4:2:1. If the total value of the coins is Rs 54, find the number of 25 p coins in the bag.
@gupanki2 said:A bag has coins in the denominations of 50 p , 25 p , 20 p in the ration 4:2:1. If the total value of the coins is Rs 54, find the number of 25 p coins in the bag.
Let the coins be 40x, 20x, 10x
So, 40x*(1/2) + 20x*(1/4) + 10x*(1/5) = 54
=> 20x + 5x +2x =54
x = 2
Number of 25p coins = 20x = 40
With the use of three different weights, namely, 1gm, 3gm and 9gm, how many objects of different weights can be weighed, if the objects to be weighed and the given weights may be placed in either pan of the scale?
@jain4444 said:With the use of three different weights, namely, 1gm, 3gm and 9gm, how many objects of different weights can be weighed, if the objects to be weighed and the given weights may be placed in either pan of the scale?
Weights are 3^0, 3^1, 3^2
Every weight upto 13gm (1 + 3 + 9) -> 13 ?
@jain4444 said:With the use of three different weights, namely, 1gm, 3gm and 9gm, how many objects of different weights can be weighed, if the objects to be weighed and the given weights may be placed in either pan of the scale?
13?
@jain4444 said:With the use of three different weights, namely, 1gm, 3gm and 9gm, how many objects of different weights can be weighed, if the objects to be weighed and the given weights may be placed in either pan of the scale?
13 bro!
does anyone has pdf unlocking softwre..
On a particular day in a cafeteria, out of 30 persons, 20 ordered tea, 10 ordered coffee and 14 ordered ice tea. If no person ordered all the three drinks and 8 persons ordered none of the three drinks, then how many persons ordered both ice tea and coffee?
3
4
2
1
3
4
2
1
@Cat.Aspirant123 As none of them ordered all 3 drinks, it lefts us with
30-8-20=2 peoples who ordered both coffee and ice tea