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in this problem larger numberbe x and smaller number be y
difference between the numbers x-y=3
x-y=3 add y on both sides
x-y+y=3+y
x=3+y
square of the smaller number=4 times the larger number
y^2=4*x
y^2=4(3+y)
y^2=12+4y
subtract 12 and 4y on both sides
y^2-4y-12=12+4y-12-4y
y^2-4y-12=0
on adding -6 and +2 we get -4 and multiplying -6and 2 we get -12
y^2 -4y-12=0 can be written as
y^2-6y+2y-12=0
y(y-6)+2(y-6)=0 here 1st two terms y is common term and last two terms 2 is common term
now y0--6 is common term
(y-6)(y+2)=0 here equating these two terms with zero we get y value
y-6=0
adding 6 on both sides
y-6+6=0+6
y=6
y+2=0
subtract 2 on both sides
y+2-2=0-2
y=-2
here we take only positive values
y=6 only
now x=3+y=3+6=9
x=9 and y=6
check this problem
difference of these two numbers=9-6=3

two positive numbers differ by 5 means
here positive numbers are x and y
x-y=5
their sum is x+y
square of their sum=(x+y)^2
(x+y)^2=169
here 169 can be written as 13^2
(x+y)^2=(13)^2
here powers equal then bases equal
x+y=13 and x-y=5
solving these two equations we get x and y values
x+y=13
x-y=5
-------
2x=18
dividing both sides by 2
2x/2=18/2
x=9
x+y=13
9+y=13
subtract 9 on both sides
9+y-9=13-9
y=4
x=9 and y=4
check:
x+y=13
9+4=13

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in a triangle perimeter= sum of the 3 sides
perimeter=1st side+2nd side+3rd side
19x^2+12x+35cm=(7x^2+5x+9)+(8x^2-6x+12)+3rd side
here do addition carefully add only like terms means x^2 terms and add xterms and constant terms
19x^2+12x+35=(7x^2+8x^2)+(5x-6x)+(9+12)+3rd side
19x^2+12x+35=15x^2-x+21+3rd side
we subtract 15x^2 on both sides
19x^2-15x2+12x+35=(15x^2-15x^2)-x+21+3rd side
4x^2+12x+35=-x+21+3rd side
add x on both sides
4x^2+12x+x+35=(-x+x)+21+3rd side
4x^2+13x+35=21+ 3rd side
subtract 21 on both sides
4x^2+13x+35-21=(21-21)+ 3rd side
4x^2+13x+14=3rd side

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sol
the ratio of 3x-7y+4 to 4x+3y+12 is 1/3 means
(3x-7y+4)/(4x+3y+12)=1/3
on cross multiplication
3*(3x-7y+4)=1*(4x+3y+12)
9x-21y+12=4x+3y+12 for isolating x we subtract both sides by 4x,3y,12
9x-21y+12-4x-3y-12=4x+3y+12-4x-3y-12
5x-24y=0
again add 24y on both sides
5x-24y+24y=0+24y
5x=24y
dividing both sides by 5
5x/5=24y/5
x=24y/5
again dividing by y
x/y=24y/5y
x/y=24/5
the answer is d)

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sol:
cylinder high=10 inches
diameter=5 inches
radius=diameter/2=5/2=2.5 inches
volume of the cylinder= pi*r^2*h
v= 22/7*2.5*2.5* 10
v=3.14*2.5*2.5*10=196.43 cubic inches

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sol:
the equation of the line is of the form ax+by+c=0
then the slope= -(x coefficient/y coefficient)= -(a/b)
x-intercept= -(c/a)
y-intercept= -(c/b)
now convert the equation is of the form ax+by+c=0
given equation is
10x=3y+15
subtract 3y and 15 on both sides
10x-3y-15=3y+15-3y-15
10x-3y-15=0
now the equation of the form ax+by+c=0
here
a=10 ,b= -3, c= -15
slope=-(a/b)= -(10/-3)=-(-10/3)=10/3 (.e;-*-=+)
y-intercept=-(c/b)=-(-15/-3)= - (15/3)= -15/3 (i.e;-/-=+, -*+=-)
slope=10/3,y-intercept= -15/3

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sol:
slope of the line passing through the points (x1,y1) and (x2,y2) is
(y2-y1)/(x2-x1)
slope of the line passing through the points (7,-5) (-1,8)
here x1=7,y1=-5,x2=-1,y2=8
slope=(y2-y1)/(x2-x1)
slope=(8-(-5))/(-1-7)
slope=(8+5)/(-8)
slope=13/(-8)
slope= - 13/8

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sol:
in right angle triangle
(hypotenuse)^2=sum of the squares of the remaining two sides
this is Pythagorean Theorem
in this problem hypotenuse(c)=8cm and one side (a)=4cm
then we find the other side using above formula
c^2=a^2+b^2
8^2=4^2+b^2
64=16+b^2
subtract 16 on both sides
64-16=16-16+b^2
48=b^2
b= sqrt(48)
b=6.9 cm

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sol:
in x^5 the base=x and exponent is 5
in x^2 the base =x and exponent is 2
here bases are same so we can subtract the exponents
x^5/x^2= x^(5-2)=x^3

2nd method
in x^5 the base=x and exponent is 5 , x^5 means we can multiply the base 5 times
means
x^5=x*x*x*x*x
in x^2 the base=x and exponent is 2,x^2 means we can multiply the base 2 times
means
x^2=x*x
now
x^5/x^2=(x*x*x*x*x)/(x*x)
after cancellation we have x*x*x
x^5/x^2=x*x*x=x^3

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using this method solve these type of problems are easy
formula is used to solve these type of problems
amount A= p
here p= $75,000
rate of interest per year =6%, this is very important step convert rate of interest per half year for this we divide the rate of interest by 2
now rate of interest for half year=6%/2=3% ,now r=3%

now the time period=4 1/2 years=in 4 years we are having 8 half years ,in 4 1/2 years we are having 9 half years, now n=9
p=$75,000,r=3%,n=9
amount A=p
substitute p,r,n values in this formula we get amount compounded half-yearly
A=$75,000
A=$75,000
A=$75,000
A=$75,000*1.03*1.03*1.03*1.03*1.03*1.03*1.03*1.03*1.03
do this using calculator

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equation of the line=
5x+2y=1
the equation is of the form ax+by+c=0 then
the slope of the line= -(x coefficient)/(y coefficient)
step1:1st we convert given line in to ax+by+c=0 for this we subtract 1 on both sides
5x+2y-1=1-1
5x+2y-1=0
now the equation is of the form ax+by+c=0
slope= -(x coefficient)/(y coefficient)
here x coefficient=5,y coefficient=2
slope= -(5)/2

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perimeter=sum of the three sides of a triangle
here the sides are x,x+1 and x+2 means
1st side= x
2nd side=x+1
and 3rd side=x+2
perimeter=1st side+2nd side+3rd side
given perimeter=12
12=(x)+(x+1)+(x+2)
12=x+x+1+x+2
12=3x+3
isolate x subtract 3 on both sides
12-3=3x+3-3
9=3x
dividing both sides by 3
9/3=3x/3
3=x
1st side=3
2nd side=x+1=3+1=4
3rd side=x+2=3+2=5

check:
1st side+2nd side +3rd side=perimeter
perimeter=3+4+5=12 which is given in our problem

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let no of pounds of soybean meal be x pounds
and no of pounds of cornmeal be y pounds
x pounds of 18% protein and y pounds of 9% protein should be mixed we get 360lb of 10% protein
x+y=360lb -------(1)equation
y=360-x(i.e; for isolating y subtract x from both sides)
x * (18%)+ y*(9%) =360*(10%)
x*(18/100)+y*(9/100)=360*(10/100)
18x/100+9y/100=3600/100
multiplying every term by 100
(18x/100)*100+(9y/100)*100=(3600/100)*100
after removing 100 on both sides we have
18x+9y=3600------------(2) equation
substitute y value in 2 equation
18x+9(360-x)=3600
18x+9*360-9*x=3600
18x+3240-9x=3600
9x+3240=3600(subtracting 9x from 18x we get 9x)
isolate x on one side subtract 3240 on both sides
9x+3240-3240=3600-3240
9x=360
dividing both sides by 9 we get
9x/9 =360/9
x=40
substitute x value in y=360-x
y=360-x
y=360-40=320
no of pounds of soybean meal=40 pounds
no of pounds of cornmeal=320 pounds

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sol:
given function be g(x)=5x
a) g(-10)=
substitute x= -10 in the above function
g(-10)=5*(-10)=-50
g(-10)=-50
b)
g(6)
substitute x=6 in the above function
g(6)=5*6=30
g(6)=30
c)
g(32)
substitute x=32 in the above function
g(32)=5*32=160
g(32)=160

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sol:
equation the line parallel to the given point (4,-6) is 3x-7y=k
substitute x=4 and y=-6 in 3x-7y=k
3*4-7*(-6)=k
12+42=k
k=54
equation of the line parallel to the given line is
3x-7y=54

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sol:
let width be x
twice the width=2x
length of the room=5 less than twice the width=2x-5
perimeter of the rectangle=2*(length+width)=2*(2x-5+x)=2*(3x-5)
according to the problem perimeter of the rectangle=80 feet
2*(3x-5)=80
dividing both sides by 2
2*(3x-5)/2=80/2
3x-5=40
adding both sides by 5
3x-5+5=40+5
3x=45
dividing by 3 on both sides
3x/3=45/3
x=15
width=15 feet
and length=2x-5=(2*15)-5=30-5=25 feet
check this problem:
perimeter =2*(length+width)=2(25+15)=2*40=80 feet
which is given

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sol:
let the number be x
its reciprocal=1/x
7 times its reciprocal=7*1/x=7/x
according to the problem
sum of a number and 7 times its reciprocal=16/3
x+7/x=16/3
multiply both sides by x
x*x+7/x*x=16/3*x
x^2+7=16x/3
multiply both sides by 3
x^2*3+7*3=16x/3*3
3x^2+21=16x
transpose 16x from R.H.S to L.H.S
3x^2-16x+21=0
on multiplying two numbers we get 63 and adding or subtracting two numbers we get 16
those numbers are 7 and 9
7*9=63
and 7+9=16
3x^2-16x+21=0 can be written as
3x^2-9x-7x+21=0
3x(x-3)-7(x-3)=0
(3x-7)(x-3)=0
3x-7=0
3x=7
x=7/3
x-3=0
x=3
the number is 7/3 or 3

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sol:
equation of the line passing through (x1,y1) and slope m is
y-y1=m(x-x1)
in this problem the coordinates (x1,y1)=(2,4) means x1=2 and y1=4 and slope m=4
substitute these values in the above formula
y-4=4(x-2)
y-4=4x-8
transpose L.H.S part to R.H.S
4x-8-y+4=0
4x-y-4=0
which is the requried equation passing through (2,4) and slope 4

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sol:
equation of the line passing through the points (x1,y1),(x2,y2) is
y-y1/y2-y1=x-x1/x2-x1
equation of the line passing through the points (4,2)and (12,5) is
x1=4,y1=2,x2=12,y2=5
substitute the values in the above formula
y-2/5-2=x-4/12-4
y-2/3=x-4/8
on cross multiplication
8(y-2)=3(x-4)
8y-16=3x-12
transpose L.H.S part to R.H.S
3x-12-8y+16=0
3x-8y+4=0

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sol:
let speed of boat in still water =x mph
speed of current=6mph
speed of boat in downstream=speed of boat in still water+speed of current
speed of boat in down stream=x+6 mph
speed of boat in up stream=speed of boat in still water-speed of current
speed of boat in up stream=x-6
time taken to travel down stream=4 hours
time taken to travel up stream=5 hours
distance=speed*time
distance travel downstream=speed of boat downstream/time taken to travel down stream
distance travel down stream=(x+6)*4
distance travel upstream=speed of boat upstream/time taken to travel up stream
distance travel up stream=(x-6)*5
according to the problem
distance travel downstream=distance travel up stream
(x+6)*4=(x-6)*5
4x+24=5x-30
transpose 4x on R.H.S side and -30 on L.H.S
24+30=5x-4x
54=x
speed of boat in still water=54 mph

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sol:
let the angle be x
its complement angle=90-x
according to the problem
x=25+(90-x)
90-x transpose to L.H.S side
x-(90-x)=25
x-90+x=25
2x-90=25
adding 90 on both sides
2x-90+90=25+90
2x=115
dividing by 2 on both sides
2x/2=115/2
x=57.5
sol:
let the angle be x
supplement of angle=180-x
four times the measure of the angle=4x
according to the problem
supplement of angle=four times measure of angle-45
180-x=4x-45
transpose 45 on L.H.S side and x on R.H.S side
180+45=4x+x
225=5x
dividing both sides by 5
225/5=5x/5
45=x
supplement of angle=180-45=135

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sol:
the given equation is d=44t
if d=66
substitute d value in the above equation
66=44*t
dividing both sides by 44
66/44=44t/44
3/2=t
t=3/2

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sol:
let the angle be x
fourteen times the angle=14x
complement of the angle=90-x
according to the problem
complement of the angle= 14 times the angle
90-x=14x
adding x on both sides
90-x+x=14x+x
90=15x
dividing both sides by 15 we get
90/15=15x/15
6=x
the angle=6 degrees

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sol:
x^2+4x-77=0
product of two numbers we get 77
and adding or subtracting we get 4
those numbers are 11 and 7
their product=11*7=77
subtracting those numbers=11-7=4
above equation can be written as
x^2+11x-7x-77=0
here 4 terms are there ,in 1 and 2 terms x is common
and 3 and 4 terms 7 is common
x(x+11)-7(x+11)=0
now x+11 is common term
(x+11)(x-7)=0
x+11=0
x=-11
x-7=0
x=7
the roots are 7 and -11

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sol:
let Wang's age be x
Mary is twice Wnag's=2*x
kat is 2 years younger the wang=x-2
according to the problem
sum of their ages=46
x+2x+x-2=46
4x-2=46
adding 2 on both sides
4x-2+2=46+2
4x=48
dividing by 4 on both sides
4x/4=48/4
x=12
Wang's age=12 years
Mary age=2*Wang's age=2*12=24 years
Kat age=Wang's age-2=12-2=10 years
check:
sum of their ages=46
Wang's age+Mary age+Kat age=12+24+10=46
L.H.S=R.H.S

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sol:
in this problem 3 circles are given
let the radius of the 1 st circle be r
circumference of 1st circle=2*pi*r
and diameter of the 2nd circle=36cm
circumference of 2nd circle=pi*diameter of 2nd circle=pi*36=36pi
diameter of the 3rd circle=20cm
circumference of the 3rd circle=pi*20=20pi
according to the problem
circumference of 1st circle=circumference of 2nd circle +circumference of 3rd circle
2*pi*r=36pi+20pi
2*pi*r=56pi
dividing both sides by pi
2*pi*r/pi=56pi/pi
2r=56
dividing both sides by 2
2r/2=56/2
r=28cm

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sol:
y^3(y^3-y)
here y^3 can be written as y*y*y
substitute y value
y^3=y*y*y=3*3*3=27
y^3(y^3-y)
here substitute y^3 value and y value
=27(27-3)
=27(24)
=27*24=648

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sol:
Area of an equilateral triangle=squareroot(3)/4*a^2
inthe above formula a is side of an equilateral triangle
according to the question area of an equilateral triangle=57 cm^2
squareroot(3)/4*a^2=57
multiplying both sides by 4/root(3)
root(3)/4*4/root(3)*a^2=57*4/1.732 (root(3)=1.732)
after cancilation
a^2=228/1.732
a^2=131.63
a=root(131.63)
a=11.47
now the perimeter of equilateral triangle=root(3)/2*a
=1.732/2*(11.47)
=0.866*11.47
=9.93cm

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sol:
solving a Quadratic equation
step1: write the given Quadratic equation in standard form,ax^2+bx+c=0
step2: factorize the trinomial ax^2+bx+c and express it as the product of two linear factors
step3: put each linear factor equal to zero
thus,two values of x will be obtained,these are the roots of the given quadratic equation.
sol:
we have 2x^2+20= -13x
step1:given equation in standard form is 2x^2+20+13x=0
here -13x is transpose to L.H.S it convert into +13x
step2:
now we find the factors of this quadratic equation
2x^2+13x+20=0
now find two numbers with sum=13 and product=(2*20)=40
clearly,such numbers are 8 and 5,adding these numbers we get 8+5=13 and product we get 8*5=40
now 2x^2+13x+20=0 can be written as
2x^2+8x+5x+20=0
(2x^2+8x)+(5x+20)=0
2x(x+4)+5(x+4)=0
(x+4)(2x+5)=0
step3:
x+4=0 or 2x+5=0
x+4=0
x=-4 ,transpose +4 from L.H.S to R.H.S we get -4
2x+5=0
2x= -5
dividing by 2 on both sides
2x/2=-5/2
x= -5/2
x= -4 or -5/2
these are the roots of the quadratic equation

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sol:
speed=distance/time
according to the question
speed of upstream=distance/time=12/3=4km/hr
according to the question speed of boat in still water=9 km/hr
let speed of stream= x km/hr
now speed of upstream=speed of boat in still water-speed of stream
substitute above values in this formula ,we get
4km/hr=9 km/hr-x
transposing x from R.H.S to L.H.S and 4 from L.H.S to R.H.S
x=9 km/hr-4 km/hr= 5km/hr
speed of stream=5 km/hr
check:speed of upstream=4km/hr
speed of upstream=speed of boat in still water-speed of stream
=9 km/hr- 5 km/hr=4 km/hr
L.H.S= R.H.S (as requried)

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sol:
6x^2=5y
y=6x^2/5
now put x=1 then y=6*(1^2)/5=6/5=1.2 here(x,y)=(1,1.2)
now put x=2 then y=6*(2^2)/5=24/5=4.8 here (x,y)=(2,4.5)
now put x=3 then y=6*(3^2)/5=54/5=10.8 here(x,y)=(3,10.8)
now put x=0 then y=6*(0^2)/5=0/5=0 here (x,y)=(0,0)
now put x=-1 then y-=6*(-1^2)/5=6/5=1.2 here (x,y)=(-1,1.2)
now put x=-2 then y=6*(-2^2)/5=24/5=4.8 here (x,y)=(-2,4.8)
now put x=-3 then y=6*(-3^2)/5=54/5=10.8 here (x,y)=(-3,10.8)
the points are (1,1.2),(2,4.5),(3,10.8),(0,0),(-1,1.2),(-2,4.5),(-3,10.8)
using these points draw the graph

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sol
y^2=10x
dividing by 10 on both sides
y^2/10=10x/10
x=y^2/10
put y=1 then x=y^2/10=1^2/10=1/10=0.1
put y=2 then x=y^2/10=2^2/10=2*2/10=4/10=0.4
put y=3 then x=y^2/10=3^2/10=9/10=0.9
put y=0 then x=y^2/10=0/10=0
put y=-1 then x=y^2/10=(-1)^2/10=1/10=0.1
put y=-2 then x=y^2/10=(-2)^2/10=4/10=0.4
put y=-3 then x=y^2/10=(-3)^2/10=9/10=0.9
the coordinates (x,y) are (0.1,1),(0.4,2),(0.9,3),(0,0),(0.1,-1),(0.4,-2),(0.9,-3)
now draw the graph of the above coordinates using graph paper