In a 500m race, the ratio of speed of two runners vinay and shyam is 3 : 4. If vinay has a start f 140m then vinay wins by

a. 15m
b. 20m
c. 25m
d. 30m

option b= 20m

In a 500m race, the ratio of speed of two runners vinay and shyam is 3 : 4. If vinay has a start f 140m then vinay wins by

a. 15m
b. 20m
c. 25m
d. 30m

my take option b... take speeds as 3m/s and 4m/s vinay travels 360m and completes 500m and shyam will travel 4*360/3 => 480m so it's b)

RanaG Says

take sides as 3 , 4 and 5 ==> option d satisfies...

If the sum of the squares of the sides of a triangle is denoted by A and its perimeter by P, then which of the following is true about A^2/P?

1] A^2/P<2
2] c<=A^2/P<=4
3] 2<=A^2/P<=3
4] 4<=A^2/P

4th option hai kya...take sides as 3,4,5 and checked

OA answer is option 3.....
Mani bhai pls see

I guess printing mistake. Even i got the same ans but just to make sure i am nt making a mistake, i posted it......

If the sum of the squares of the sides of a triangle is denoted by A and its perimeter by P, then which of the following is true about A^2/P?

1] A^2/P<2
2] c<=A^2/P<=4
3] 2<=A^2/P<=3
4] 4<=A^2/P

take sides as 3 , 4 and 5 ==> option d satisfies...

In a 500m race, the ratio of speed of two runners vinay and shyam is 3 : 4. If vinay has a start f 140m then vinay wins by

a. 15m
b. 20m
c. 25m
d. 30m

If the sum of the squares of the sides of a triangle is denoted by A and its perimeter by P, then which of the following is true about A^2/P?

1] A^2/P<2
2] c<=A^2/P<=4
3] 2<=A^2/P<=3
4] 4<=A^2/P

4th option hai kya...take sides as 3,4,5 and checked

When the curve y=x^2+4x-5 and y=1/(1+x^2) are drawn in the x-y plane, how many times do they intersect for values of x>0?

1] Never
2] Once
3] Twice
4] More than twice

P.S: _________/\________ Mani bhai

y(x^2 + 1) = 1,and x is always positive which means the curve would be symmetric

so,it's 2 and the picture would be somewhat like this:

even if we tk the solns for x<0 we get only 2 intersecting pts.

option d)

I guess printing mistake. Even i got the same ans but just to make sure i am nt making a mistake, i posted it......

If the sum of the squares of the sides of a triangle is denoted by A and its perimeter by P, then which of the following is true about A^2/P?

1] A^2/P<2
2] c<=A^2/P<=4
3] 2<=A^2/P<=3
4] 4<=A^2/P

Sanket Yagnik Says

Thanx for solving bcoz OA is given as option d

even if we tk the solns for x<0 we get only 2 intersecting pts.

option d)

2)once

draw the graphs and we can have only 1 intersecting pt

Thanx for solving bcoz OA is given as option d

When the curve y=x^2+4x-5 and y=1/(1+x^2) are drawn in the x-y plane, how many times do they intersect for values of x>0?

1] Never
2] Once
3] Twice
4] More than twice

P.S: _________/\________ Mani bhai

2)once

draw the graphs and we can have only 1 intersecting pt

Puys pls help me solve this simple one...

A motorboat went downstream for 28km and immediately returned. It took the boat twice as long to make the return trip. If the speed of the river flow were twice as high, the trip downstream and back would take 672 mins. Find the speed of the the boat in still water and the speed of the river flow.

a. 9km/h, 3km/h
b. 9km/h, 6km/h
c. 8km/h, 2km/h
d. 12km/h, 3km/h.


pls explain the solution frens....

Let speed of boat be v and speed of stream be u:

Downstream 28/(v+u) = Time (t, say)

and

Upstream 28/(v-u)= 2t



so we have from above eqs. v=3u


Now,



when river speed is twice: 2u

then:

{28/(v+2u)} + {28/(v-2u)} = 672/60 (for converting min to hr as answer is in km/hr)


Solve above eq with the help of v=3u and u will get ans as : 9 and 3 km/hr


OPT A

Puys pls help me solve this simple one...

A motorboat went downstream for 28km and immediately returned. It took the boat twice as long to make the return trip. If the speed of the river flow were twice as high, the trip downstream and back would take 672 mins. Find the speed of the the boat in still water and the speed of the river flow.

a. 9km/h, 3km/h
b. 9km/h, 6km/h
c. 8km/h, 2km/h
d. 12km/h, 3km/h.


pls explain the solution frens....

U can also check by option
Let v be the speed of boat and u be the speed of river
it is given that v+u=2(v-u)
=> v=3u
only option a satisfies the above condition

When the curve y=x^2+4x-5 and y=1/(1+x^2) are drawn in the x-y plane, how many times do they intersect for values of x>0?

1] Never
2] Once
3] Twice
4] More than twice

P.S: _________/\________ Mani bhai

Puys pls help me solve this simple one...

A motorboat went downstream for 28km and immediately returned. It took the boat twice as long to make the return trip. If the speed of the river flow were twice as high, the trip downstream and back would take 672 mins. Find the speed of the the boat in still water and the speed of the river flow.

a. 9km/h, 3km/h
b. 9km/h, 6km/h
c. 8km/h, 2km/h
d. 12km/h, 3km/h.


pls explain the solution frens....

It's given that Sb*t + Ss*t = 28 ->1

2Sb*t - 2Ss*t = 28 where Sb and Ss are speeds of boat in still water and the stream

Solving these two, u get Sb*t = 21 and Ss*t = 7

here ratio of speed of boat and stream is 3:1

Only Option a fits here...

Puys pls help me solve this simple one...

A motorboat went downstream for 28km and immediately returned. It took the boat twice as long to make the return trip. If the speed of the river flow were twice as high, the trip downstream and back would take 672 mins. Find the speed of the the boat in still water and the speed of the river flow.

a. 9km/h, 3km/h
b. 9km/h, 6km/h
c. 8km/h, 2km/h
d. 12km/h, 3km/h.


pls explain the solution frens....

The two adjacent sides of a quadrilateral are 6 cm and 8 cm long. What is the maximum possible area of the quadrilateral (in sq cm) if it is inscribed in a circle of radius 5 cm?
a: 45
b:48
c: 49
d: 50
please tell me how to find ths.

yeah 49 wil be the maxm area posible

The two adjacent sides of a quadrilateral are 6 cm and 8 cm long. What is the maximum possible area of the quadrilateral (in sq cm) if it is inscribed in a circle of radius 5 cm?
a: 45
b:48
c: 49
d: 50
please tell me how to find ths.

it should be 49

_/(6^2+8^2)=10

diagonal would be 10 which is diameter of the circle

so area of that triangle=1/2*6*8=24

now another triangle would have one side 10 (diagonal)

other two sides would be equal for maximum area= 5_/2,5_/2

area of this triangle=1/2*5_/2*5_/2=25

total area=24+25=49

The two adjacent sides of a quadrilateral are 6 cm and 8 cm long. What is the maximum possible area of the quadrilateral (in sq cm) if it is inscribed in a circle of radius 5 cm?
a: 45
b:48
c: 49
d: 50
please tell me how to find ths.

solve the problem in the attachment
detailing on it

11/21

6:5
(a,p)=alcohol,pepsi
vessel A(a,p)------------------Vessel B (a,p)
1,0---------------------------0,1
1/2-------------------------1/2,1
1/2+1/4,1/2-----------------1/4,1/2
1/4+1/8,1/4-----------------1/4+1/4+1/8,1/2+1/4
1/4+1/8+1/4+1/16,1/4+1/4+1/8----1/4+1/16,1/4+1/8
11/16,5/8------------5/16,3/8

fraction of alcohol(vessel A)= 11/21
p:a vessel 2=6:5

i think what he means by repeating the process is transferring from 1st vessel to 2nd and then from 2nd to 1st. So total iteration would be 4 and in that case answer would be 11/21 for Q-1

PS: ____/|\________ to all quant lovers

solve the problem in the attachment
detailing on it

T....=..A.+.P........T.=..A.+.P
100.=100+0........100=.0.+.100
50...=.50+0........150=.50.+100
125.=.75+50........75=.25.+.50
62.5=37.5+25...37.5=62.5+ 75

Ans:1
37.5/62.5
15/25
3/5
d. N.O.T.

Ans:2
75/62.5
3/2.5
6:5
Option(a)

Should've chosen a higher base number instead of 100, maybe 200.

..............................First Vessel..................................Second Vessel
...............................A = 200......................................P = 200
1st iteration:.............A = 100......................................P = 200, A = 100
2nd iteration:............A = 150, P = 100.........................P = 100, A = 50
3rd iteration:.............A = 75, P = 50............................P = 150, A = 125

1.) Ratio of A/(A+P) = 75/125 = 3/5 ......None of these

2.) Ratio of P/A = 150/125 = 6/5 .........Option a.)

ans is 14 > c
15 > a

solve the problem in the attachment
detailing on it

let initially there be 100l of alcohol and 100l of pepsi

after 1st operation
first solution second solution
alcohol: pepsi alcohol
50l 100l 50l

second operation
first solution second solution
al pep pep al
75l 50l 50l 25l

finally
al pep al pep
37.5 25 62.5 75

1) fraction of alcohol in first container=37.5/62.5=60%
2)pepsi:alcohol=6:5

p.s :good morning varun ji and other puys

solve the problem in the attachment
detailing on it

is it 1/D times in the question?- when he mixes for 2nd time?

solve the problem in the attachment
detailing on it

..............................First Vessel..................................Second Vessel
...............................A = 200......................................P = 200
1st iteration:.............A = 100......................................P = 200, A = 100
2nd iteration:............A = 150, P = 100.........................P = 100, A = 50
3rd iteration:.............A = 75, P = 50............................P = 150, A = 125

1.) Ratio of A/(A+P) = 75/125 = 3/5 ......None of these

2.) Ratio of P/A = 150/125 = 6/5 .........Option a.)

solve the problem in the attachment
detailing on it

devanshu97 Says

A and B pick up a ball at random from a bag containing M red and N yellow colored balls one after the other replacing the ball everytime till one of them gets red ball. The first one to get a red ball is declared the winner. If A begins the game and odds in favor of his winning the game are 3 to 2, then find the ratio of M:N

Any1..try this

Probability = M/(M + N) + NM/(M + N) + .....
= {M/(M + N)}/{(M + 2MN)/(M + N)}
= (M + N)/(M + 2N) = 3/5

=> M/N = 3/2

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