When Ravi's brother was born, let Ravi's father's age = x years and mother's age = y years.
Then, sister's age = (x - 28) years. So, x - 28 = 4 or x = 32.
Ravi's age = (y - 26) years. Age of Ravi's brother = (y - 26 + 3) years = (y - 23) years.
Now, when Ravi's brother was born, his age = 0 i.e. y - 23 = 0 or y = 23.
Clearly, we have :
A = B - 3 ...(i)
D + 5 = E ...(ii)
A+C = 2E ...(iii)
B + D = A+C = 2E ...(iv)
A+B + C + D + E=150 ...(v)
From (iii), (iv) and (v), we get: 5E = 150 or E = 30.
Putting E = 30 in (ii), we get: D = 25.
Putting E = 30 and D = 25 in (iv), we get: B = 35.
Putting B = 35 in (i), we get: A = 32.
Putting A = 32 and E = 30 in (iii), we get: C = 28.
Let the numbers be 3x, 3x + 3 and 3x + 6.
Then,
3x + (3x + 3) + (3x + 6) = 72
9x = 63
x = 7
Largest number = 3x + 6 = 27.
=> Second largest number = 27 - 3 = 24
Since one of the numbers on the dial of a telephone is zero, so the product of all the numbers on it is 0.
Let the number of boys and girls participating in sports be 3x and 2x respectively.
Then, 3x = 15 or x = 5.
So, number of girls participating in sports = 2x = 10.
Number of students not participating in sports = 60 - (15 + 10) = 35.
Let number of boys not participating in sports be y.
Then, number of girls not participating in sports = (35 -y).
Therefore (35 - y) = y + 5 <=> 2y<=> 30<=> y = 15.
So, number of girls not participating in sports = (35 - 15) = 20.
Hence, total number of girls in the class = (10 + 20) = 30.
Let R, G and B represent the number of balls in red, green and blue boxes respectively.
Then,
R + G + B = 108 ...(i),
G + R = 2B ...(ii)
B = 2R ...(iii)
From (ii) and (iii), we have G + R = 2x 2R = 4R or G = 3R.
Putting G = 3R and B = 2R in (i), we get:
R + 3R + 2R = 108 => 6R = 108 => R = 18.
Therefore Number of balls in green box = G = 3R = (3 x 18) = 54.
We can get this by any of two explanations.
Answer 1: 41
7 = 3 + 4 = 34
Similarly, 5 = 4 + 1 = 41
Answer 2 : 41
7 - 3 = 4 => 34
Similarly, 5 - 4 = 1 => 41
let 'b' be the Length of bridge from cow to the near end of the bridge and 'a' be the distance of the train from the bridge.
'x' be speed of cow => '4x' speed of train
Then the total length of the bridge 2b + 10.
(a-2)/4x = b/x
=> a-2 = 4b........(1)
Now if it had run in opposite direction
(a+2b+10-2)/4x = (b+10-2)/x
=> a - 2b = 24......(2)
Solving (1) and (2)
b = 11 ,
Therefore length of the bridge is 2 x 11 + 10 = 32mts.
Let the total number of workers be x. Then,
Number of women = x/3 and number of men = x - x/3 = 2x/3
Number of women having children = x/18
Number of men having children = x/3
Number of workers having children = (x/18) + (x/3) = 7x/18
Therefore, workers having no children = x - 7x/18 = 11/18 of all workers.
Clearly, the required number would be such that it leaves a remainder of 1 when divided by 3, 4, 5, or 6 and no remainder when divided by 7. Thus, the number must be of the form (L.C.M of 3, 4, 5, 6) x + 1 i.e., (60x + 1 ) and a multiple of 7. Clearly, for x = 5, the number is a multiple of 7. So the number is 301.
Candidates passed in atleast four subjects
= (Candidates passed in 4 subjects) + (Candidates Passed in all 5 subjects)
= (Candidates failed in only 1 subject ) + ( Candidates passed in all subjects)
= (78 + 275 + 149 + 147 + 221) + 5685 = 870 + 5685 = 6555
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