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The denominator of a fraction is 4 less than the numerator. If the denominator is decreased by 2 and the numerator is increased by 1, then the numerator is eight times the denominator. Find the numerator of the fraction.
Let the fraction be x/y
According to the conditions y = x  4 x  y = 4 ...(1)
8*(y2) = (x+1) 8y16=x+1 x8y = 17...(2)
Subtracting (1) from (2),
we get x  8y  x + y
= 17  4 7y = 21 y
= 3 x = 4 + y
= 4+ 3 = 7
The numerator is 7
A father is four times as old as his son is. Five years back he was seven times as old as his son was then. What is the present age of the father?
Let the age of the son and father be x and y respectively. 5 years back, their ages were (x5) and (y5)
According to the given conditions
4x = y 4xy = 0 ...(1)
7(x5) = (y5) 7x  35
= y 5 7x y = 30 ...(2)
Subtracting (1) from (2),
we get 7x  y  4x + y = 30  0 3x = 30 x = 10 years
y = 4*10 = 40 years
The present age of the father is 40 years.
Let the number be x.
20% of x = 35% of 144
= 35*144/100
= 50.4 20% of x = 50.4 20x/100 = 50.4 x
= 50.4*100/20 = 252
The required number is 252
Find the positive value of x, for which AB = 5 units for points A(3,x/2) and B(1,x+1).
AB = 5
AB =
5=
Squaring both sides,
we get
25 = 16 + x^{2/4}+1+x x^{2/4} + x 8
=0 x^{2}+4x32=0 x^{2}+8x4x32
=0 x(x+8)4(x+8)=0 (x4)(x+8)=0
x = 4 [x^{2} = x*x]
When two lines are parallel, what is the difference of their slopes equal to?
When two lines are parallel, their slopes are equal. The differene of their slopes is equal to 0.
Find the distance of the midpoint of the line joining L(5,1) and M(1,1) from the point O(2,0).
Mid point of LM is given by
((51)/2, (11)/2)
= (4/2,0/2) =(2,0)
The distance of (2,0) from O(2,0) is equal to 0.
For what value of k is the line (k3)x(4k^{2})y+k^{2}7k+6=0 parallel to the x axis? [k^{2}=k*k]
For the line to be parallel to the x axis, its slope is 0 Slope =  coefficient of x/coefficient of y = (k3)/(4k^{2}) = 0
Hence, k3=0 k=3 For k=3,
the given line is parallel to the x axis.
Find the radius of the circle x^{2}+y^{2}12x+11=0 [x^{2}=x*x]
x^{2}+y^{2}12x+11=0
(x^{2}12x)+y^{2}+11=0
(x6)^{2}+(y+0)^{2} 36 +11 = 0
(x6)^{2}+(y+0)^{2} = 3611=25
(x6)^{2}+(y+0)^{2} = 5^{2}
The radius of the circle is 5 units.
The sum of three numbers is 132. The first number is twice the second and the third number is onethird of the first. Find the first number
Let the second number be 3x.
The first number will be 2*3x = 6x
The third number will be
1/3*6x = 2x 6x+3x+2x = 132
11x=132 x
= 132/11
= 12
The first number will be 6x = 6*12 = 72
Find the value of [27^{(2/3)}]/[64^{(4/3)}]. [27^{2}=27*27]
[27^{(2/3)}]/[64^{(4/3)}] =
[(3^{3})^{(2/3)}]/[(4^{3})^{(4/3)}]
= [3^{(3*2/3)}]*[4^{(3*4/3)}]
= (3^{2})*(4^{4}) = 9*256 = 2304
Threefourth of a number is 150 greater than threefourteenth of the number. Find the number.
Let the number be x.
According to the conditions,
3x/4 150
= 3x/14 x/4  50
= x/14 x*14  50*4*14
= x*4 14x  2800
= 4x 10x = 2800 x
= 280
If a/(a+b) = 17/23, then fill in the blank (a+b)/(ab) = /11
a/(a+b) = 17/23
Hence,
when a+b = 23,
a = 17
Thus,
b = 23
 a = 23  17
= 6 (a+b)/(ab)
= (17+6)/(176) = 23/11
Hence, the number to be filled in the blank is 23.
5 men can complete a work in 2 days, 4 women can complete it in 3 days and 5 children can complete it in 3 days. In how many days can 1 man, 1 woman and 1 child complete it working together?
Work done by one man in 2 days = 1/5
Work done by one man in one day = 1/(5*2) = 1/10
Work done by one woman in one day = 1/(4*3) = 1/12
Work done by one child in one day = 1/(5*3) = 1/15
Work done by one man, one child and one woman in one day
= 1/10+1/12+1/15 = (6+5+4)/(5*2*2*3 )
=15/(15*4) = 1/4
They will take 4 days to complete the work.
In a kilometer race A can give B a start of 50 meters and B can give C a start of 40 meters. A start of how many meters can A give C in a 2 km race?
When A completes 1000 meters, B covers 950 meters.
When B completes 1000 meters, C covers 960 meters.
Distance covered by A/ distance covered by B = 1000/950
Distance covered by B/distance covered by C = 1000/960
Distance covered by A = 1000/950 * distance covered by B = 1000/960 * 1000/950*distance covered by C Distance covered by C,
when A covers 2 km =
(950*960)/(1000*1000)*2000
= 1824 metres
A gives C a start of
20001824 = 176 meters
On what sum is the difference in compound interest and simple interest for 3 years at 5% per annum Rs.61?
Let CI and SI be the compound interest and simple interest.
Let P, R, T be the principle, rate and time.
SI = P*R*T/100 = P*3*5/100
= 15P/100 CI = P[(1+R/100)^{T1}]
= P[(1+5/100)^{31}] CISI
= 61 P[(21/20)^{31}]  15P/100 = 61 P[9261/8000 1 3/20]
= 61 P[(926180001200)/8000
= 61 P = 61*8000/(61)
= 8000
The sum invested was Rs.8000 [100^{T}=100*100*...T times]
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