M.Sc. Math and ICWAI
20 years home tuitions and coaching centres and 3 years online tuitions I can teach Math upto college level with a guaranteed cent per cent results. Expert in coaching for ICWAI, ACS, CA business Math, accounts, financial accounts, and management accounts
Home Tuition Only
Creates a free environment and makes student shed away all barriers in mind about the subject, motivates the student to do himself all problems, and guarantees high rank and aggregate. Assignment help, teaching a particular topic, etc.
350.00
--
--
--
--
--
--
--
Class 9 - 10 | Mathematics, Physics, All Boards, All Medium | INR 350.00 /hour |
Class 11 - 12 | Mathematics, Statistics, All Boards, All Medium | INR 350.00 /hour |
College Level | Costing, Business Mathematics, English, Mathematics | INR 600.00 /hour |
Test Preparation | MAT, ACS, MPSC | INR 600.00 /hour |
(t^2, 2t) and (u^2, 2u) would be coordinates of A and B.
(t^2, 2t) and (u^2, 2u) would be coordinates of A and B.
They can meet when they all come to starting point for the first time. Hence LCM of 8, 15 and 12 would be
120.
In 2 hours all three can meet
They can meet when they all come to starting point for the first time. Hence LCM of 8, 15 and 12 would be
120.
In 2 hours all three can meet
Compound interest = P(1.12)^4- P =2850 (given)
i.e. P=4969.32
Simple interest = P(48)/100 = P(0.48)
=2385.27
Compound interest = P(1.12)^4- P =2850 (given)
i.e. P=4969.32
Simple interest = P(48)/100 = P(0.48)
=2385.27
Together work done per day = 1/20+1/18 = 38/360
In 5 days work completed = 5(38/360) = 19/36:
work left = 1-19/36 = 17/36
Together work done per day = 1/20+1/18 = 38/360
In 5 days work completed = 5(38/360) = 19/36:
work left = 1-19/36 = 17/36
Upstream = x-y=28/7 and downstream = x+y =56/7 where x = speed of boat and y = speed of current
Subtract to get -2y = -4 or y = 2km/hou
Upstream = x-y=28/7 and downstream = x+y =56/7 where x = speed of boat and y = speed of current
Subtract to get -2y = -4 or y = 2km/hou
20S = 18C
Loss = C-S = C-20C/18 = -C/9
Percentate= 1/9(100) = 11.01% of cost or
10% of selling price
20S = 18C
Loss = C-S = C-20C/18 = -C/9
Percentate= 1/9(100) = 11.01% of cost or
10% of selling price
Interest for 4 years = PNR/100 = 700/-.
Amount to be repaid = 2500+700 = 3200/-
Interest for 4 years = PNR/100 = 700/-.
Amount to be repaid = 2500+700 = 3200/-
Interest for 4 years = PNR/100 = 700/-.
Amount to be repaid = 2500+700 = 3200/-
Interest for 4 years = PNR/100 = 700/-.
Amount to be repaid = 2500+700 = 3200/-
20S = 18C
Loss = C-S = C-20C/18 = -C/9
Percentate= 1/9(100) = 11.01% of cost or
10% of selling price
20S = 18C
Loss = C-S = C-20C/18 = -C/9
Percentate= 1/9(100) = 11.01% of cost or
10% of selling price
20S = 18C
Loss = C-S = C-20C/18 = -C/9
Percentate= 1/9(100) = 11.01% of cost or
10% of selling price
20S = 18C
Loss = C-S = C-20C/18 = -C/9
Percentate= 1/9(100) = 11.01% of cost or
10% of selling price
Train length = distance = 130 m.
Relative speed = x-4 where x = speed of train.
(x-4) 8/3600 = 130/1000:
80x-320 = 130(36)
x = 62.5km/hr
Train length = distance = 130 m.
Relative speed = x-4 where x = speed of train.
(x-4) 8/3600 = 130/1000:
80x-320 = 130(36)
x = 62.5km/hr
Work done by both per day = 1/20+1/30 = 1/12
i.e. for 12 days wages = 3840/-
For one day wages = 3840/12 = 320/-(Ans)
Work done by both per day = 1/20+1/30 = 1/12
i.e. for 12 days wages = 3840/-
For one day wages = 3840/12 = 320/-(Ans)
Let R be the radius of small and r the big cirlce.
Then pi(R^2+r^2 ) = 97 and R+r= 13 :
i.e. pi{(13-r)^2+r^2)}=97
169+2r^2-26r = 30.89
No real roots and hence no solution
Let R be the radius of small and r the big cirlce.
Then pi(R^2+r^2 ) = 97 and R+r= 13 :
i.e. pi{(13-r)^2+r^2)}=97
169+2r^2-26r = 30.89
No real roots and hence no solution
Since tangent makes 90 degrees with Radius, we have a right angled triangle AOP:
AP^2 = 61^2-11^2 = 50x72
AP = 5x12 = 60 cm.
Perimeter of quadrilateral= 2(AP+OA) = 142 cm.
Since tangent makes 90 degrees with Radius, we have a right angled triangle AOP:
AP^2 = 61^2-11^2 = 50x72
AP = 5x12 = 60 cm.
Perimeter of quadrilateral= 2(AP+OA) = 142 cm.
Let the circle have equation x^2+y^2+2gx+2fy+c=0: The points on the circle are (0,.0)(a,0) and (0,b)
Hence c =0: a^2+2ga =0: and b^2+2fb =0:
a =-2g and b =-2f
Or equation is x^2+y^2-ax-by =0
Let the circle have equation x^2+y^2+2gx+2fy+c=0: The points on the circle are (0,.0)(a,0) and (0,b)
Hence c =0: a^2+2ga =0: and b^2+2fb =0:
a =-2g and b =-2f
Or equation is x^2+y^2-ax-by =0
Leaf:Wood = 3:2 = 12: 8 and Leaf:Forest = 4:5 = 12:15
Hence ratio of Leaf:Wood:Forest = 12:8:15
Leaf share =12(1400)/35 = 480
Leaf:Wood = 3:2 = 12: 8 and Leaf:Forest = 4:5 = 12:15
Hence ratio of Leaf:Wood:Forest = 12:8:15
Leaf share =12(1400)/35 = 480
There are 151 girls and 10 boys.
Hence ratio of boys:girls= 10:151
There are 151 girls and 10 boys.
Hence ratio of boys:girls= 10:151
r1+r2 = 13 and pi (r1^2+r2^2) = 97
Solving we get r1 = 4 and r2 =9
Area of smaller circle =16pi
r1+r2 = 13 and pi (r1^2+r2^2) = 97
Solving we get r1 = 4 and r2 =9
Area of smaller circle =16pi
If Seva takes less number of days, then Seva is more efficient.
Their effeciency ratio is = 1/16 :1/20
= 5:4
If Seva takes less number of days, then Seva is more efficient.
Their effeciency ratio is = 1/16 :1/20
= 5:4
Total of all 15 persons = 15(29)
Total of two persons = 2(55)
Hence total of 13 persons = 15(29)-2(55)=325
Average of 13 persons = 25 years
Total of all 15 persons = 15(29)
Total of two persons = 2(55)
Hence total of 13 persons = 15(29)-2(55)=325
Average of 13 persons = 25 years
Let the number be x. Then 0.82x-0.28x = 540:
x = 1000
28% of 1000 = 280 (answer)
Let the number be x. Then 0.82x-0.28x = 540:
x = 1000
28% of 1000 = 280 (answer)
Difference between Rita and Sushant as per ratio is 7-5 = 2
2 is equivalent to 4000 means
Total 15 is equivalent to 30000
Tuhin and Sushant total = 8(30000)/15 = 16000
Difference between Rita and Sushant as per ratio is 7-5 = 2
2 is equivalent to 4000 means
Total 15 is equivalent to 30000
Tuhin and Sushant total = 8(30000)/15 = 16000
Imagine a graph with origin at starting point. East 10 km, means he reached (10,0). Then left 6 km means he reaches (10, 6). When he turns left he walks in western direction and hence new poisition is (4,6). Last again turns left means goes towards south and reaches (4,0)
Distance from origin = 4 kms.
Imagine a graph with origin at starting point. East 10 km, means he reached (10,0). Then left 6 km means he reaches (10, 6). When he turns left he walks in western direction and hence new poisition is (4,6). Last again turns left means goes towards south and reaches (4...
At present average of old family members = 27+5 = 32
Total age = 32(5) = 160 years
New person added total = 160+x
Average = 36 = (160+x)/6
Hence x = new person age = 56
At present average of old family members = 27+5 = 32
Total age = 32(5) = 160 years
New person added total = 160+x
Average = 36 = (160+x)/6
Hence x = new person age = 56
60% are males and literate males = 60(20)% = 12% of 1000= 120
Total literate = 250 and females literate = 130
Percentage = 130/1000 (100) = 13%
60% are males and literate males = 60(20)% = 12% of 1000= 120
Total literate = 250 and females literate = 130
Percentage = 130/1000 (100) = 13%
At present girls = 5(720)/12 = 300 and boys = 420
To make them equal 120 girls must be admitted new.
At present girls = 5(720)/12 = 300 and boys = 420
To make them equal 120 girls must be admitted new.
CP 12 pens is equal to SP 15 pens. and 12 <15.
There will be a loss. So % loss is given by ( 15-12/15)*100= 20. So loss= 20%
CP 12 pens is equal to SP 15 pens. and 12 <15.
There will be a loss. So % loss is given by ( 15-12/15)*100= 20. So loss= 20%
If total population is 500, then 300 males and 200 females
Of these, 60 males and 50 females are graduates.
Hence 110/500 i.e. 22% are graudates.
Non graduates are 78%
If total population is 500, then 300 males and 200 females
Of these, 60 males and 50 females are graduates.
Hence 110/500 i.e. 22% are graudates.
Non graduates are 78%
No of subsets of first set = 2^m and second set = 2^n
Difference = 56
2^m-2^n = 2^3 (7)
i.e. 2^(m-n)(2^n-1) = 2^3 (7)\
Comparing we get m-n = 3 and n = 3
m = 6 and n=3
No of subsets of first set = 2^m and second set = 2^n
Difference = 56
2^m-2^n = 2^3 (7)
i.e. 2^(m-n)(2^n-1) = 2^3 (7)\
Comparing we get m-n = 3 and n = 3
m = 6 and n=3
If length =l, then time for 162+l metres is 18 seconds and time for 120+l is 15 sec
Speed is the same. Hence 162+l/18 = 120+l/15
Cross multiply and simplify : 810+5l = 720+6l
l = 90 metres.
If length =l, then time for 162+l metres is 18 seconds and time for 120+l is 15 sec
Speed is the same. Hence 162+l/18 = 120+l/15
Cross multiply and simplify : 810+5l = 720+6l
l = 90 metres.
Time taken = distance/speed = 1.56 x 10^(-16)/3x10^8
= 0.52x10^-24
= 5.2 x 10^(-25) seconds
Time taken = distance/speed = 1.56 x 10^(-16)/3x10^8
= 0.52x10^-24
= 5.2 x 10^(-25) seconds
2% damaged = 700/35 = 20
Hence shipped = 100% = 20(50) = 1000 boxes.
2% damaged = 700/35 = 20
Hence shipped = 100% = 20(50) = 1000 boxes.
Water content = 25 (20%) = 5 kg.
If x kg is added then percent = (5+x)/(25+x) = 1/3
i.e. 15+3x = 25+x
x = 5 kg.
Water content = 25 (20%) = 5 kg.
If x kg is added then percent = (5+x)/(25+x) = 1/3
i.e. 15+3x = 25+x
x = 5 kg.
For this we have to find out the greatest common factor of 420 and 130
It will be 10.
For this we have to find out the greatest common factor of 420 and 130
It will be 10.
For crossing each other distance travelled = sum of lengths = 400+l (say)
Relative speed = 50+40 = 90km /hour = 90(5/18) = 25 m/sec
(400+l)/25 = 30
Or l = 750-400 = 350 m.
For crossing each other distance travelled = sum of lengths = 400+l (say)
Relative speed = 50+40 = 90km /hour = 90(5/18) = 25 m/sec
(400+l)/25 = 30
Or l = 750-400 = 350 m.
Let cost be 100.
Market price = 100+30 = 130
Discount = 20% = 26
Selling price = 104
Hence profit = 104-100 = 4%
Let cost be 100.
Market price = 100+30 = 130
Discount = 20% = 26
Selling price = 104
Hence profit = 104-100 = 4%
(1,1) lies on the curve. Hence f(1) = 1 = 1+b-b =1
The tangent has slope has f'(1) = 2x+b at (1,1)
= 2+b
Equation is y-1 = (b+2)(x-1)
y = (b+2)x-b-1
y intercept = -b-1 and x intercept is b+1/b+2
Area of triangle = 1/2 bh = -(b+1)^2/(b+2)/2 =2
b =-3 satisfies this equation.
-(b+1)^2(b+2) = 4
(1,1) lies on the curve. Hence f(1) = 1 = 1+b-b =1
The tangent has slope has f'(1) = 2x+b at (1,1)
= 2+b
Equation is y-1 = (b+2)(x-1)
y = (b+2)x-b-1
y intercept = -b-1 and x intercept is b+1/b+2
Area of triangle = 1...
Earth is in the form of sphere.
Volume = 4/3 pir^3
Density= Mass/vol = 5.975x10^24(3/4)(7/22) (1/6.37^3 x10^18)
= 5516.406 kg/m^3
Earth is in the form of sphere.
Volume = 4/3 pir^3
Density= Mass/vol = 5.975x10^24(3/4)(7/22) (1/6.37^3 x10^18)
= 5516.406 kg/m^3
Let the position of the particle be f(t) = x
Differentiate and find f'(t)
Velocity at t=3 is f'(3) and velocity at t =6 is f'(6)
Let the position of the particle be f(t) = x
Differentiate and find f'(t)
Velocity at t=3 is f'(3) and velocity at t =6 is f'(6)
Since time taken is the same let t be the time.
Angular velocity = 2pi/t hence same for both.
Since time taken is the same let t be the time.
Angular velocity = 2pi/t hence same for both.
The course length = 3 miles
Distance travelled = 2(3) = 6 miles
Average speed = distance/time = 6/(20+25) miles /minute
= 360/45 miles per hour
= 8 mph
The course length = 3 miles
Distance travelled = 2(3) = 6 miles
Average speed = distance/time = 6/(20+25) miles /minute
= 360/45 miles per hour
= 8 mph
Join AD, it would be perpendicular to BC cutting BC at E.
ED = 2r - altitude of triangle= a/sq rt 2 - sqrt 3 a/2
BD = DC = BE in 30 = 1/2 ( a/sq rt 2 - sqrt 3 a/2)
Join AD, it would be perpendicular to BC cutting BC at E.
ED = 2r - altitude of triangle= a/sq rt 2 - sqrt 3 a/2
BD = DC = BE in 30 = 1/2 ( a/sq rt 2 - sqrt 3 a/2)
Join OR Then OR = OS = OP = RQ
So in triangle ORQ , angle ORQ = 180-2y:
So angle ORS= OSR = 90-y
so SOR angle =180-4y
i.e. angles x+180-4y+y = 180
Or x= 3y
Join OR Then OR = OS = OP = RQ
So in triangle ORQ , angle ORQ = 180-2y:
So angle ORS= OSR = 90-y
so SOR angle =180-4y
i.e. angles x+180-4y+y = 180
Or x= 3y
Radius of incircle = Area / semi perimeter
= A/s
Area = sq rt (32)(12)(12)(8) = 192 cm^2
s = 32
Incircle radius =192/32 = 6 cm.
Radius of incircle = Area / semi perimeter
= A/s
Area = sq rt (32)(12)(12)(8) = 192 cm^2
s = 32
Incircle radius =192/32 = 6 cm.
For equilateral triangle circumradius = a/rt 3 where s is one side
Hence here side = R ( sqrt 3)
Option C
For equilateral triangle circumradius = a/rt 3 where s is one side
Hence here side = R ( sqrt 3)
Option C
Let d be the distance between two towers. and a between first tower and point of intersection, B second and point of intersection.
Then by similar triangles we get z/x = a/d and z/y = b/d
Adding z/x+z/y= (a+b)/x =1
Or z(x+y)/xy =1
z = xy/(x+y)
Let d be the distance between two towers. and a between first tower and point of intersection, B second and point of intersection.
Then by similar triangles we get z/x = a/d and z/y = b/d
Adding z/x+z/y= (a+b)/x =1
Or z(x+y)/xy =1
z = xy/(x+...
Let d be the distance between tower and chimney. Then tan 60 = h/d where h is height of tower.
Also tan 30 = (h-40)/d
Dividing 3 = h/(h-40) Or 3h-120 = h
h = 60 m.
Height is less than 60 hence meets pollution norms.
Let d be the distance between tower and chimney. Then tan 60 = h/d where h is height of tower.
Also tan 30 = (h-40)/d
Dividing 3 = h/(h-40) Or 3h-120 = h
h = 60 m.
Height is less than 60 hence meets pollution norms.
S7 = 7/2 (2a+6d) = 63 and S14 = 7(2a+13d)
S14 = S7+161 = 224
2a+13d =32
2a+6d = 18 From S7
7d =14 and d =2. a = 9-3d=3
28th term= 3+27(2) = 57
63+67d = 161
67d = 96
S7 = 7/2 (2a+6d) = 63 and S14 = 7(2a+13d)
S14 = S7+161 = 224
2a+13d =32
2a+6d = 18 From S7
7d =14 and d =2. a = 9-3d=3
28th term= 3+27(2) = 57
63+67d = 161
67d = 96
Area of rectangle = curved SA of cylinder
i.e. 880 = 2(3.14) (r)(40)
r = 7/2 = 3.5
Area of rectangle = curved SA of cylinder
i.e. 880 = 2(3.14) (r)(40)
r = 7/2 = 3.5
S 30 = 15[2a+29d] =30a + 435 d and
S20 = 10[2a+19d] and S10 = 5[2a+9d]
S20-s10 = 10a+ 145d
3(S20-S10) = 30a+435d
= S30
S 30 = 15[2a+29d] =30a + 435 d and
S20 = 10[2a+19d] and S10 = 5[2a+9d]
S20-s10 = 10a+ 145d
3(S20-S10) = 30a+435d
= S30
Since tangents have equal length we have XP = XQ and AO = AR and BQ = BR
XB+BR = XB + BQ = XQ and similarly XA+AR = XP and we already have XQ = XP
Hence proved.
Since tangents have equal length we have XP = XQ and AO = AR and BQ = BR
XB+BR = XB + BQ = XQ and similarly XA+AR = XP and we already have XQ = XP
Hence proved.
Use section formula for the 4 points as ratio 1:4, 2:3, 3:2, 4:1.
First point = (2, 6) and second point = (3, 5), III point = (4, 4) IV point = (5, 3)
Use section formula for the 4 points as ratio 1:4, 2:3, 3:2, 4:1.
First point = (2, 6) and second point = (3, 5), III point = (4, 4) IV point = (5, 3)
First triangle can be drawn using the given sides.
Second triangle is similar to the first since sides are proportional. Sides of the second triangle would be
3/5 times i.e. 3, 18/5, 21/5 cm. With these measurement drawn second triangle also.
First triangle can be drawn using the given sides.
Second triangle is similar to the first since sides are proportional. Sides of the second triangle would be
3/5 times i.e. 3, 18/5, 21/5 cm. With these measurement drawn second triangle also.
Surface area of the remaining solid = SA of 5 sides of cube + curved sA of hemisphere
= 7(7)(7)+2(22/7) (7/2) (7/2)
= 343+77
= 420 sq cm
Surface area of the remaining solid = SA of 5 sides of cube + curved sA of hemisphere
= 7(7)(7)+2(22/7) (7/2) (7/2)
= 343+77
= 420 sq cm
If h is the height of the light house, and d the distance between the first ship and light house, then
tan 30 = h/d and tan 45 = h/(100-d)
Divide to get
1/sq rt 3 = (100-d)/d
d = 100 rt 3 - rt 3 d
= 173.2 - 1.732 d
d = 173.2/2.732 = 63.4
h = 100-d = 36.6 m
If h is the height of the light house, and d the distance between the first ship and light house, then
tan 30 = h/d and tan 45 = h/(100-d)
Divide to get
1/sq rt 3 = (100-d)/d
d = 100 rt 3 - rt 3 d
= 173.2 - 1.732 d
d = 173.2/2....
Ratio between sides = square root of ratio between areas
= 3:1
Hence option d
Ratio between sides = square root of ratio between areas
= 3:1
Hence option d
Relative velocity = v1+v2 = 60+v2 kmph
Distance = l1+l2 = 180+270 m = 0.45 km.
0.45 = 10.8(60+v2)/3600
60+v2 = 150
v2= 90kmph
Relative velocity = v1+v2 = 60+v2 kmph
Distance = l1+l2 = 180+270 m = 0.45 km.
0.45 = 10.8(60+v2)/3600
60+v2 = 150
v2= 90kmph
Let speeds be 2x and 3x and distances 3y and 7y.
Then time taken = 3y/2x and 7y/3x
Ratio = 3/2:7/3 = 9:14
Option B
Let speeds be 2x and 3x and distances 3y and 7y.
Then time taken = 3y/2x and 7y/3x
Ratio = 3/2:7/3 = 9:14
Option B
We multiply the capital by number of months remained in the business.
For Sreekumar, 30000(6) +21000(6)= 306000
For Pete 30000(6)+18000(6) = 288000
For Ahmed 30000(6)+35000(6) = 390000
Hence ratio of profit share = 306:288:390
Ahmed share = 390(88560)/984
=35100
We multiply the capital by number of months remained in the business.
For Sreekumar, 30000(6) +21000(6)= 306000
For Pete 30000(6)+18000(6) = 288000
For Ahmed 30000(6)+35000(6) = 390000
Hence ratio of profit share = 306:288:390
Ahme...
Side of the square = 2(16+24)/4 = 20 cm.
Circumference of semi circle with radius 10 cm = 3.14(10) = 31.4 cm
Option b
Side of the square = 2(16+24)/4 = 20 cm.
Circumference of semi circle with radius 10 cm = 3.14(10) = 31.4 cm
Option b
v=u+at here v =0 a = -4 and u = 30
Hence 0 = 30-4t
t = 6.25 seconds
Option c
v=u+at here v =0 a = -4 and u = 30
Hence 0 = 30-4t
t = 6.25 seconds
Option c
V = pir^2 h
V'= pi (r^2 h'+2rr'h)
=pi(16(-4)+2(4)(3)(6))
= pi(80)
V = pir^2 h
V'= pi (r^2 h'+2rr'h)
=pi(16(-4)+2(4)(3)(6))
= pi(80)
Left side = 5^(m+3)
So m+3 =5 or m = 2
Left side = 5^(m+3)
So m+3 =5 or m = 2
Total people surveyed = 20(100/10) = 200
ii) The darkest shaded region music
iii) Classical = 100 semi classical =200 Dark shaded =
Total people surveyed = 20(100/10) = 200
ii) The darkest shaded region music
iii) Classical = 100 semi classical =200 Dark shaded =
Distance covered = 72/18 cm = 4cm.
Distance covered = 72/18 cm = 4cm.
In 5 hours distance travelled = 5(60)(14)/25 = 168 km.
In 5 hours distance travelled = 5(60)(14)/25 = 168 km.
Since there are two outcomes and n is large we can use Poisson distribution.
Mean for 2000 houses = 2
P(x=5) = 0.03608
Option A
Since there are two outcomes and n is large we can use Poisson distribution.
Mean for 2000 houses = 2
P(x=5) = 0.03608
Option A
Model ship and real ship are similar and hence dimensions are proportional.
Hence length = 28(9/12) = 21 cm
Model ship and real ship are similar and hence dimensions are proportional.
Hence length = 28(9/12) = 21 cm
Thickness of 5 books = 5x20 = 100 mm.
Thickness of paper sheets 5 = 5(0.016) = 0.08 mm.
Total thickness = 100.08 mm
Thickness of 5 books = 5x20 = 100 mm.
Thickness of paper sheets 5 = 5(0.016) = 0.08 mm.
Total thickness = 100.08 mm
5m 60 cm/10 m 50 cm = 3m 20 cm/x
i) x = Length of shadow cast = 320(105)/560 = 60 cm.
5m 60 cm/10 m 50 cm = 3m 20 cm/x
i) x = Length of shadow cast = 320(105)/560 = 60 cm.
i) Sugar crystls in 5 kg = 5/2 (9)(10^6) = 2.25(10^7)
ii) In 1.2 kg = 1.2/2 (9)(10^6) = 5.4(10^6)
i) Sugar crystls in 5 kg = 5/2 (9)(10^6) = 2.25(10^7)
ii) In 1.2 kg = 1.2/2 (9)(10^6) = 5.4(10^6)
Bottles that will be filled in 5 hours = 5/6(840) = 700.
Bottles that will be filled in 5 hours = 5/6(840) = 700.
Enlarged length = 20000(5)/50000= 2 cm.
Enlarged length = 20000(5)/50000= 2 cm.
Speeds = 2x and 3x and distances = 3y and 7y respectively.
Time taken = 3y/2x and 7y/3x
Ratios = 3/2: 7/3 = 9:14
Option B
Speeds = 2x and 3x and distances = 3y and 7y respectively.
Time taken = 3y/2x and 7y/3x
Ratios = 3/2: 7/3 = 9:14
Option B
i ) P neither green nor blue = P yellow =7/18
ii) P not blue= 1- P blue=1-6/18 = 2/3
i ) P neither green nor blue = P yellow =7/18
ii) P not blue= 1- P blue=1-6/18 = 2/3
1) P for an ace = 4/52 = 1/13
ii) P for not a red card = P for black card =26/.52 = 1.2
1) P for an ace = 4/52 = 1/13
ii) P for not a red card = P for black card =26/.52 = 1.2
The two triangles are similar with ratio of sides = 5:8
Hence area will have ratios = 25:64
So area of DOB= 64(60)/25 = 153.6 cm square
The two triangles are similar with ratio of sides = 5:8
Hence area will have ratios = 25:64
So area of DOB= 64(60)/25 = 153.6 cm square
PT * TQ = ST*TR
Also by Pythagorean theorem , TS = root of 6^2 -5^2 = rt 11
hence 4(5) = rt 11 (TR)
TR = 6.03 cm
Option c
PT * TQ = ST*TR
Also by Pythagorean theorem , TS = root of 6^2 -5^2 = rt 11
hence 4(5) = rt 11 (TR)
TR = 6.03 cm
Option c
Distance between vertices= Distance between (8,3) and (14,3)
= 6 units
Distance between vertices= Distance between (8,3) and (14,3)
= 6 units
8% is equivalent to 1000 units
Hence investment = 100% = 100(1000)/8 = 12500
For investing 12500 he has to buy for double the amount.
Hence 25000 is answer
Option C
8% is equivalent to 1000 units
Hence investment = 100% = 100(1000)/8 = 12500
For investing 12500 he has to buy for double the amount.
Hence 25000 is answer
Option C
Question is incomplete
Question is incomplete