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# Miscellaneous

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A motocycle has brand new tyres on both wheels. A Tyre is considered worn if its has run 15000 km on the rear wheel or 25000 km on the front wheel. What is the maximum possible number of kilometers that the motorcycle can run until the tyres become if the front and rear tyres are interchanged at the appropriate time?

ans: 18750

front can do work in =25000

rear can do work in=15000

together they will do max work in= x

1/x=(1/250000)+(1/15000)

x=75000/8=9375

this is the time or the km after which we can change them so they have travelled 9375 and when they are changed they can travel 9375 more

so total distance travelled =2*9375=18750 (answer)

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The number of positive integral solutions to the equation 4x + 5y + 2z = 111

4x+2y = 111-5z

Put z=1
2x+y=53
Possible values for y=1,3,...51
26 Solutions.

Now z=2. This case won't be possible.

Next, take z=3
2x+y=48
y=2,4,....,46
= 23 solution

It will make a series of: 26+23+21+18+16+13+11+8+6+3+1=146

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How many real solutions does the equation given below have?

(2x2 – 5)x^2 – 3x = 1

This post was modified 6 years ago by Utkarsh Garg

4 ?

L.H.S will only be equal to 1 when either base is 1 or power is 0

Base =1

2x^2= 6

x= +√3,-√3 .{2 values}

Power =0

x^2-3x=0

x(x-3x)=0

x=0,3 {2 values }

One more case will be there when base is -1 and power is even but we have to check it

Base=-1

2x^2= 4

x=+√2,-√2

Putting in power we don't get a rational no. Hence this will not be the case

Total values = 4

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If x,y and z are positive real numbers such that none of them is equal to 1. Further logxyz + logyz =5 and logzxy + logyx = 3. Find all possible real numbers a such that yz = xa

1,3/2

3,3/2

4,4/3

2,9/2

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An ice cream is made by mixing Elaichi and Pista flavoured creams. The 55-gallon container is mistakenly filled in such a way that it has 6% Eliachi flavoured cream. How many gallons must be removed and then replaced with a mixture containing 50% Eliachi cream so that the resulting mixture has 10% Eliachi cream?

2.5

5

In 55 gallon we have 6% = 3.3 gallon elaichi

Now let x gallon is taken out then elaichi taken out will be 6x/100 and replaced with same quantity which contains (1/2)x elaichi and then it becomes 10 %

{3.3- (6x/100) +x/2 }55 = 10/100

Solving this we get x= 5 gallon

That's divided by 55 in equation

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In trapezoid JKLM, JK is parallel to LM, angel J is a right angle and JK=4, JM=17,LM=12, and O lies on JM such that  angle JOK=1/2 of angle LOM. Find the ratio JO: OM.

8:9

PFA the solution

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Vinayak bought a ticket for the grand finale of the ‘Bull fighting’ challenge in Valencia, Spain. Unfortunately, he had to change his plans and decided to sell his ticket. He expected a lot of demand for the ticket but had to sell it for 1/2 of what he had initially quoted. This reduced his profits by 60%. His profit margin, in %, must have been

25

75

66.66

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In triangle ABC, right angled at B with integral lengths, secA – tanA = ½ where A is the angle opposite to side BC. Which of the following can be the inradius of this triangle?

1/2

3/4

1/4

1

secA - tanA = 1/2

h/b - p/b = 1/2

(h - p )/b = 1/2 ... (1)

We know , r = ( p + b - h)/2 ...  (2) [ in a right angled triangle. ]

from (1) and (2)
r = b/4.

smallest possible value of b is 4 ( as it forms a pythagorean triplet )

Hence r = 1.

SecA - tanA = 1/2 .......(1)

Since sec^2 (A) -tan^2 A= 1

Sec A - tan A = 2 .........(2)

2sec A = 5/2

Cos A = 4/5

Triangle is integral with sides in ratio 3,4,5

putting value we get 1

Alternatively :

If all sides of a right angled triangle are natural numbers then inradius is always a natural number .

r = ( sum of legs - hypotenuse )/2

Hence option (D).

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The coefficient of x6 in the expansion of (x2-6x+3).(1+x)7 is:

49

70

-63

-70

?

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0.01

1

10

100

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Raju having some coins want to distribute to his 5 son , 5 daughter and driver in a manner that , he gave fist coin to driver and 1/5 of remaining to first son he again gave one to driver and 1/5 to 2nd son and so on.... at last he equally distributed all the coins to 5 daughters. how many coins raju initially has?
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Given a, b and c are positive real numbers and log

PFA the solution

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Sir ,Kindly share the solution to this problem -

40 rs. and 10 paise?

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Discriminant of a second degree polynomial with integer coefficients cannot be:
A. 43
B. 33
C. 68
D. 25

discriminant cannot be a prime no, hence option A

why can't it be a prime number?

Discriminant of a second degree polynomial ax2 + bx + c ( say ) is b2 + 4ac

now square of an integer is either of the form 4k or 4k+1 so b2+4ac will never be of the form of 4k + 3.

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Given a, b and c are positive real numbers and log

we know ,x3 + y3 + z3 - 3xyz = (x + y + z)(x2 + y2 + z2 - xy - yz - zx)

for given values

(log

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