Let 'x' be the number of new Rs.2000 notes.
Given for one new 2000 note there are three old 500 notes.
i.e, 3x
Given 10 more new Rs.2000 notes are added to the collection
and the ratio of new Rs.2000 notes to old Rs.500 notes
Therefore, number of new Rs.2000 notes = x+10 = 30.
old Rs.500 notes = 3x = 60.
Thus, the total number of notes in the collection = 30 + 60 = 90.
If Vijay gives 'x' marbles to Ajay then Vijay and Ajay would have V - x and A + x marbles.
V - x = A + x --- (1)
If Ajay gives 2x marbles to Vijay then Ajay and Vijay would have A - 2x and V + 2x marbles.
V + 2x - (A - 2x) = 30 => V - A + 4x = 30 --- (2)
From (1) we have V - A = 2x
Substituting V - A = 2x in (2)
6x = 30 => x = 5.
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
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.
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.
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.
3T+12C = 48,250----(1)
21T+36C = ?
multiply eq (1) with 3
i.e,, 3(7T+12C) = 48,250*3
21T+36C = 1,44,750/-
Let number of bangles kushi purchase of each type = x
From the given data,
34x + 42x + 6x = 1722
x = 1722/82 = 21.
Let the number of apples in each tree of the 5 farmers be a, b, c, d,e respectively. Therefore total no of apples are 7a, 9b, 11c, 13d and 14e respectively.
Given,
7a+1 = 9b+3 = 11c-1 = 13d+3 = 14e-6 = x
9b+3 = 13d+3
===> 9b = 13d
so take b = 13 and d = 9
9b+3 = 9*13+3 = 120
ie, x=120
substituting
7a+1 = 120
7a = 119
a = 17
11c-1 = 120
11c = 121
c=11
14e-6 = 120
14e = 126
e = 9
Yields per tree in the orchards of d 3rd & 4th farmers= 11,9
Assume the family has X girls and Y boys, according to the question,
we can have 2 equations:
X-1=Y, 2(Y-2)=X.
Solve the equations, we get X=4, Y=3.
There were all parrots but six means that six birds were not parrots but only pigeons and koels.
Similarly, number of parrots + number of koels = 6 and number of parrots + number of pigeons = 6.
This is possible when there are 3 parrots, 3 pigeons and 3 koels i.e. 9 birds in all.
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