[Q] Trigonometry Tricks - 4 [Maximum and Minimum Values]

Few points to remember:

1. sinθ and cosθ both have "1" as their maximum value and "-1" as their minimum value. Hence the values of sinx, sin2x, cosx, cos3x, etc. lie between -1 and 1.
2. For sin²x and cos²x
• Minimum value = 0
• Maximum value = 1
3. For sinxcosx
• Minimum value = -1/2
• Maximum value = 1/2
4. Minimum value of (sin θ cos θ)n = (-1/2)n

Some SSC CGL questions :

      Q. 1) What is the least value of 2sin²θ + 3cos²θ

       (A) 1            (B) 2             (C) 3            (D) 0

      Since 2 is less than 3
      Minimum value = 2

      Answer: (B)

       

     Hence the least value of 4sec²θ + 9cosec²θ = 13 + 12 = 25

     Answer: (C)

      Q. 4) The maximum of 3sinx - 4cosx is

       (A) -1               (B) 5                (C) 7               (D) 9

       Maximum value = √(3^2 + 4^2) = √25 = 5
       Answer: (B)

Sometimes they ask the minimum/maximum value of a function. In CGL, that function would always be quadratic (ax² + bx + c). Please note that a quadratic function can't have both maximum and minimum values. 

• If "a" is positive, then the quadratic function will only have a minimum value. The maximum value would be infinite.
• If "a" is negative, then the quadratic function will only have a maximum value. The minimum value would be infinite.

But the process to find both minimum and maximum values is same. Hence you shouldn't be worried about the words "maximum" or "minimum". When finding the minimum/maximum value of a function, we use the concept of "differentiation". Although differentiation is a wide topic in itself, but for CGL purpose, we only have to learn the basics.

Just remember following things:

1. Differentiation of axⁿ = a*n*x^n-1. Hence differentiation of 4x³ = 12x² and differentiation of 3x² = 6x
2. Differentiation of ax = a. Hence differentiation of 4x is 4.
3. Differentiation of any constant is zero. Hence differentiation of 5 is 0.
4. If you have to differentiate ax² + bx + c, just differentiate each of its term separately and add the result.
5. Process of finding the minimum value of a function :
1. Differentiate the function
2. What ever result you get, equate it with zero
3. Find the value of x
4. Put this value of x in the original function to get the minimum value.

Let us solve a CGL question:

Q. 5) Find the minimum value of (x - 2)(x - 9)
(A) -11/4             (B) 49/4             (C) 0            (D) -49/4

(x - 2)(x - 9) = x^2 - 11x + 18

Now we will differentiate (x^2 - 11x + 18)

Note that in this question, the value of a is positive, i.e., +1 and hence the question has asked you the "minimum" value. Had the value of "a" been negative (like -1), they would have asked you the "maximum" value.

Hence, differentiation of (x^2 - 11x + 18) = 2x - 11

Now 2x - 11 = 0

x = 11/2

Put x = 11/2 in (x^2 - 11x + 18) to get the minimum value

Minimum value of x^2 - 11x + 18 = (11/2)^2 - 11*(11/2) + 18 = -49/4

Answer: (D)

If you have any doubt in this article, please drop a comment...
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[Q] Trigonometry Tricks - 3

Sine and Cosine Rules

See the image above and mug the formulas thoroughly. Sin and cosine rules are important in trigonometry and can help you in solving some complex questions.

Q. 1)

The figure will look something like this -

In triangle ACD
sin∠CAD/3x = sin45/AD   ... (1)
In triangle ABD
sin∠BAD/x = sin60/AD    ...  (2)

Divide equation (2) by (1)

3* sin∠BAD/sin∠CAD = sin60/sin45

sin∠BAD/sin∠CAD = √6/6 = 1/√6

Answer: (C)

Similarly you can use cosine law to find any angle, if all the sides are given or to find a side, if the other two sides and an angle is given.

Q. 2)  In the below figure, ABC is right angled at B and AD = CD. If ∠ACB=30, find ∠ABD

          (A) 30                (B) 60                (C) 45                (D) 75

∠BAD = 180 - (90 + 30) = 60

In triangle ABD

sin∠ABD/AD = sin60/BD  [Since ∠BAD = 60]

In triangle BCD

sin∠CBD/CD = sin30/BD

Divide equation (2) by (1)

sin∠CBD/sin∠ABD = sin30/sin60     [AD = CD and hence they will cancel out]

Now, sin∠CBD = sin(90 - ∠ABD) = cos∠ABD  [Click here if you don't know how to convert sin into cos]

Hence, cos∠ABD/sin∠ABD = sin30/sin60

cot∠ABD = 1/√3

Hence ∠ABD = 60

Answer: (B)

Alternative Method

The figure given in this question is very important and at times it is embedded in some other figures. There is one short-cut  to calculate the angle.

Imagine the triangle in circumscribed in a circle

Now, D will be the centre of the triangle with diameter AC. We can say with surety that AC is the diameter of the circle because ∠ABC = 90, and we know angle in a semicircle is right angle. Moreover AD = CD, hence D is the midpoint of the diameter or the centre of the circle

Now you can see that AD, CD and BD are radii of the circle. Hence AD = CD = BD

∠BAD = 180 - (90 + 30) = 60

∠ABD = ∠BAD = 60 [Since AD = BD]

Answer: (B)

Now let us see a CGL question, in which the above figure was embedded.

Q. 3) G is the centroid of Triangle ABC, and AG=BC. Find angle BGC.

(A) 60              (B) 90               (C) 120                (D) 75

Let AG = 2x
Then BG = x (centroid divided the median in 2:1 ratio)
BC = AG = 2x
Let AG when extended cuts BC at D
Then D is the midpoint of BC  (as AD is the median)
BD = DC = x  [Since BC = 2x]
Now DG = BD = DC = x
That means D is the centre of a circle with diameter BC and one of the radius as DG.
Hence BGC = 90 (angle in a semi-circle)

Answer: (B)

Q. 4)

sec²x + tan²x = 5/3

We know, sec²x – tan²x = 1

Adding the above two equations

2sec²x = 8/3 or sec²x = 4/3

secx = 2/√3

That means x = 30

cos2x = cos60 = 1/2
Answer: (C)
Method 2:
sec²x = 4/3
That means, cos^2x = 3/4 and sin^2x = 1 - 3/4 = 1/4
cos2x = cos^2x - sin^2x = 3/4 - 1/4 = 1/2

Q. 5)

Multiply and Divide RHS by 2

(cosx - sinx)/(cosx + sinx) = (√3/2 - 1/2)/(√3/2 + 1/2)

Match RHS with LHS and you can easily see x = 30

Answer: (A)

Method 2

Cross multiply

(cosx - sinx)(√3 + 1) = (cosx + sinx)(√3 - 1)

Solve it and you will get, tanx = 1/√3

Hence x = 30

Q. 6)

Put θ = 45
2y*cos45 - x*sin45 = 0
2y = x  ... (1)
2x*sec45  - y*cosec45 = 3
2x - y = 3/√2
4y - y = 3/√2  [Put x = 2y]
y = 1/√2
Hence, x = 2y = √2
x^2 + 4y^2 = 2 + 4*(1/2) = 2 + 2 = 4
Answer: (C)


Some important values to mug:

1.  sin15 = (√3 - 1) /2√2
2.  cos15 = (√3 + 1) /2√2
3.  tan15 = 2 - √3
4.  cot15 = 2 + √3

These values are very important to solve some tricky trigonometric questions. Examples :

Q. 7) 

Although you can solve this question with the direct formula, which I discussed in Geometry Tricks - 1. But let us assume, you forget that formula. In such cases, the values of sin15 and cos15 will come handy.

Let the perpendicular and base of the triangle be P and B, respectively.

sin15 = Perpendicular/Hypotenuse = P/100

P = sin15*100
Similarly, B = cos15*100

Area = 1/2 * P * B = 1/2 * sin15 * 100 * cos15 * 100 =  (√3 - 1) /2√2 * (√3 + 1) /2√2 * 100 * 100/2

Area = 100*100/8 = 1250

Answer: (D)

Q. 8)

       A) 0               B) 1                 C) -1                 D) 2

Put x = 15
= cot15/(cot15 - cot45) + tan15/(tan15 - tan45)
= (2 + √3)/(1 + √3) + (2 - √3)/(1 - √3)
= 1
Answer: (B)


Some candidates are asking me the book which I refer to write these articles. But, I can assure you that all these tricks are authored by me and you won't find them in any book or coaching...

If you have any doubt in this article, please drop a comment...
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[Q] Algebra Tricks - 4

Please read Algebra Part - 1 and Algebra Part - 2, before reading this post.

Q. (1) 

                (A) 3                (B) 4                 (C) 6                   (D) 9

Although this equation is symmetrical, and hence we can assume x = y = z to solve it. But you should know one more thing about such equations. If the sum of certain number of terms is zero, you can assume each term to be zero. That means,

(4x - 3)/x = 0 or x = 3/4

(4y - 3)/y = 0 or y = 3/4

(4z - 3)/z = 0 or z = 3/4

So, 1/x + 1/y + 1/z = 4

Answer: (B)

Q. (2) If x^2 = y + z, y^2 = z + x, z^2 = x + y, then find the value of 

            

     (A) 1                (B) 2                 (C) 0                   (D) -1

Symmetrical equation, hence x = y = z

x^2 = x + x

x^2 = 2x

x = 2

Hence x = y = z = 2

Put in the expression

= 1/3 + 1/3 + 1/3

= 1

Answer: (A)

Q. (3)

This is a very famous question-type. In such questions we take everything on RHS to LHS and then try to make squares. You will get,

(x - 1)^2 + (y + 1)^2 + (z + 1)^2 = 0

And like I said before, if the sum of certain number of terms is zero, we can assume each term to be zero.

(x - 1)^2 = 0, (y + 1)^2 = 0, (z + 1)^2 = 0

Hence, x = 1, y = -1 and z = -1

Put these values in (2x - 3y + 4z)

= 2(1) -3(-1) + 4(-1)

= 1

Answer: (D)

Q. (4)

Take everything to LHS,

(x - 1)^2 + y^2 = 0

Hence, x = 1 and y = 0

Put these values in the expression

= (1)^3 + 0^5

= 1

Answer: (D) 

Q. (5)

               (A) 5/12                    (B) 12/5                     (C) 5/7                    (D) 7/5

By Componendo and Dividendo, whenever you see any equation written in the form

(m + n)/(m - n) = p

You can directly write m/n = (p + 1)/(p - 1)

In this question

m = √(3 + x), n = √(3 - x), p = 2

Hence, by Componendo-Dividendo

Squaring both sides

(3 + x)/(3 - x) = 9

Again apply componendo-dividendo

3/x = (9 + 1)/(9 - 1)

3/x = 5/4

x = 12/5

Answer: (B)

Q. (6) If x = 332, y = 333, z = 335, then the value of x^3 + y^3 + z^3 - 3xyz is

(A) 10000             (B) 7000              (C) 9000               (D) 8000

There is one more formula for a^3 + b^3 + c^3 - 3abc, apart from the one which you know

Hence,

x^3 + y^3 + z^3 - 3xyz = 1/2(332 + 333 + 335)[(332 - 333)^2 + (333 - 335)^2 + (335 - 332)^2]

= 1/2 (1000)[1 + 4 + 9]

= 7000

Answer: (B)

Q. 7)

(A) -1                 (B) 3abc                   (C) 1                      (D) 0

a + b + c = 0 is symmetrical

Whenever any symmetrical equation is equal to zero, and the expression whose value is asked, is also symmetrical( and the numerator of the terms is also 1), then the value of that expression will also be zero.

Answer: (D)

Q. (8)

(A) -2                (B) -1/2                   (C) 0                   (D) 1/2

Same logic

Value of the expression = 0

Answer: (C)

Q. (9)

                             (A) 9            (B) 0                (C) 8                (D)

Although a + b + c = 0, is a symmetrical equation, and the expression ((a+b)/c+ (b+c)/a+ (c+a)/b)(a/(b+c)+ b/(c+a)+c/(a+b)) is also symmetrical. But the numerator of the terms is not 1. Hence we can't say that the answer is zero.

We will solve it by assuming a=b=c. Hence

= (2 + 2 + 2)(1/2 + 1/2 + 1/2)

= 6 * 3/2

= 9

Answer: (A)

Q. (10)

ab + bc + ca = 0 is symmetrical

Hence value of the expression = 0

Answer: (B)

If you have any doubts in this article, please drop a comment...

As the exam date has been extended, I have updated the book-list for SSC, because now you have more time to practise :)

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[Q] Trigonometry Tricks - 2

Please read Part-1 before reading this article.

This article covers some concepts that are taught in classes 11th and 12th, and hence new for arts/commerce candidates.

Few things which the above figure represents :

• There are 4 quadrants (shown with I, II, III and IV).
• Quadrant I - 0° to 90°
• Quadrant II - 90° to 180°
• Quadrant III - 180° to 270°
• Quadrant IV - 270° to 360°


• In the first quadrant, all the trigonometric functions are positive. So the values of sin56, cos18, tan89, cot67, cosec33, etc. are positive. Note that I have taken the angles 56, 18, 89, 67, 33 and all of them are less than 90(hence belong to the first quadrant) 
• In the second quadrant, only sin and cosec are positive, and rest are negative. Hence sin91, sin135, cosec120, sin116, etc. are positive while cos135, cot120, tan95, etc, are negative. 
• In the third quadrant, only tan and cot are positive, and rest are negative. Hence tan200, tan198, cot255, etc. are positive while cos255, sin220, cosec265, etc, are negative. 
• In the fourth quadrant, only cos and sec are positive, and rest are negative. Hence cos300, cos350, sec290, sec285, etc. are positive while tan359, cot355, sin340, etc, are negative. 
• The mnemonic to remember which trigonometric function is positive in which quadrant is - All Students Take Calculus. "All" is the first word of the sentence and hence represents the first quadrant. All trigonometric functions are positive in the 1st quadrant. Second initial is "S" which represents sin (indicating sin/cosec are positive in 2nd quadrant). Third initial is "T" which represents tan (indicating tan/cot are positive in 3rd quadrant). Fourth initial is "C" which represents cos (indicating cos/sec are positive in 4th quadrant).
• Gist: 
• All trigonometric functions are positive in the 1st quadrant
• sin/cosec are positive in 2nd quadrant (sin and cosec are reciprocal of each other and hence their signs are same)
• tan/cot are positive in 3rd quadrant
• cos/sec are positive in 4th quadrant

Converting trigonometric functions :

sin(90 - A) = cosA, and hence sin65 = sin(90 - 25) = cos25
sin(90 + A) = cosA, and hence sin135 = sin(90 + 45) = cos45
cos(90 - A) = sinA, and hence cos85 = cos(90 - 5) = sin5
cos(90 + A) = -sinA, and hence cos135 = cos(90 + 45) = -sin45 = -(1/√2)
where A is any acute angle

Explanation

• sin(90 + A) refers to a value in the 2nd quadrant because A is an acute angle and hence (90 + A) would cover angles from 90 to 180 degrees (depending upon the value of A). 90 to 180 degrees is the range of 2nd quadrant. Now we have seen that in the second quadrant sin is positive. Hence sin(90 + A) = +cosA.
• cos(90 + A) = -sinA, because in the second quadrant cos is negative.
• (90 - A) represents the 1st quadrant and in the 1st quadrant, all the trigonometric functions are positive, hence:
• sin(90 - A) = +cosA and hence sin75 = sin(90 - 15) = cos15 (here A = 15) 
• cos(90 - A) = +sinA 
• tan(90 - A) = +cotA 
• cot(90 - A) = +tanA 
• sec(90 - A) = +cosecA 
• cosec(90 - A) = +secA
• Now instead of 90 degrees if we have 180 degrees, then the functions are not converted. E.g. 
• sin(180 - A) = sinA, and hence sin135 = sin(180 - 45) = sin45 
• cos(180 - A) = -cosA and hence cos165 = cos(180 - 15) = -cos15 
• tan(180 - A) = -tanA 
• cosec(180 - A) = cosecA 
• sec(180 - A) = -secA
• Note that in the above lines only sin and cosec are positive, because (180 - A) represents 2nd quadrant and in the second quadrant only sin and cosec are positive.

Caution: While converting, please keep in mind that we check the sign of the function which is about to get converted. So if you want to convert sinX into cosY, first check where does X lie (in which quadrant), and then check the sign of "sin" (not cos) in that quadrant. If sin is positive in that quadrant, write sinX = +cosY, else write sinX = -cosY.

Gist:

• sin is converted into cos
• tan is converted into cot
• sec is converted into cosec

You need only this much knowledge to solve SSC questions [You don't have to do PhD after all :)]
Now let us solve some CGL questions:

Q. 1)

We have to convert sin3A into cos.
3A is an acute angle and hence sin3A lies in the 1st quadrant.
sin3A = +cos(90 - 3A)  [sin is positive in 1st quadrant, hence we have written +cos(90 - 3A)]
Hence, cos(90 - 3A) = cos(A - 26)
90 - 3A = A - 26 [Equating cos]
4A = 116
or A = 29

Answer : (A)
Note: In this question we had to convert sin into cos, hence we checked the sign of sin.

Q. 2)

cos20 = cos(90 - 70) = +sin70 [cos20 lies in the 1st quadrant and cos is positive in the 1st quadrant, hence we have written +sin70]
Hence, sin5θ = sin70
5θ = 70 [Equating sin]
θ = 14
Answer : (D)
Note: In this question, don't write sin5θ = cos(90 - 5θ), because this formula is applicable only for acute angles and 5θ is not necessarily an acute angle

I hope this concept of quadrants and conversion is clear.

Moving on, I discussed the basic trick of Trigonometry (putting the value of theta) in Part-1. Since this trick is extremely important, in each article of trigonometry I will solve some questions by assuming the value of θ so that you imbibe that method well. For reference, I am attaching the values-

Let us take some more questions from CGL-

Q. 3)

In this question you cant take θ = 45, 90 because that will make 2cos^4θ - cos^2θ = 0, and we know that denominator can never be zero. So you are left with two values: 0, 30 or 60
Take θ = 0
sec0 = 1
sin0 = 0
cos0 = 1
Put the above values and you will get the value of the expression as 1.
Answer: (A)

Q. 4)

Put A = 30
The value of the expression = 1/2/(1 + √3/2) + 1/2(1 - √3/2) = (2 - √3) + (2 + √3) = 4
Now put A = 30 in all the 4 options
(A) 4              (B) 4/√3                  (C) 1                  (D) √3

Answer : (A)

Note : Don't put A = 45 in this question, because then options A and B will give the same output.

Q. 5)

Put θ = 45
a = 0, b = 0
Value of the expression = (0 + 4)(0 - 1)^2 = 4
Answer : (A)

Q. 6)

We had seen earlier that when we have to assume two angles, it is best to assume them as 30 and 60.
α = 30 and β = 60

Then a = √3 and b = 1/√3

Then sin²β = 3/4

Put a = √3 and b = 1/√3 in all the 4 options and check which one of the2m is giving 3/4 as the output

Answer : (C)

As always, in case you have any doubt/doubts in this article, please drop a comment. Print a PDF of this article if you want to.

Part - 3 to follow soon...

Keep reading :)

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SSC Tier-1 Postponed

Official Notice

The SSC Tier-1 exam has been postponed and this throws everyone in a state of dilemma. What should we do now? Would it be wise to study anything as the pattern is going to change?

Well! You can never be sure of the pattern, but one thing is clear - Quantitative Aptitude and English aren't going any where. So it would be wise to give general awareness and reasoning a break and focus your attention on English/Quantitative Aptitude for the time being. Tier-1 will most probably be held on 21st Aug and 28th August, so you have plenty of time to hone your english grammar/vocab and aptitude.

[Q] Speed, Time & Distance Tricks - 2

Please read Part-1 before reading this post. In this article I have covered questions related to Boat and Streams, Gun-shots and some other left over topics.

Q. 1)

Apply the direct formula
Average speed for the complete journey = 2XY/(X + Y) = 2*20*30/50 = 24 km/hr
Answer : (D)

        Q. 2)

Let the total distance be 100 km

Average Speed = Total Distance/Total Time

Total time = 70/20 + 10/25 + 20/8 = 3.5 + 0.4 + 2.5 = 6.4

Average Speed = 100/6.4 = 15.625 m.p.h

Q. 3)

This is again a very frequently asked question. Let the distance of his school be X km.

(Time taken to reach the school at 3 km/hr) - (Time taken to reach the school at 4 km/hr) = (10 + 10) minutes or 1/3 hours

X/3 - X/4 = 1/3

Hence X = 4 km

Answer : (B)

Direct Formula

Distance = S1*S2/(S1 - S2) * Time difference

S1 = 4 km/hr, S2 = 3 km/hr, Time Difference = 10 - (-10) = 20 minutes or 1/3 hours

Distance = 4*3/(4-3) * 1/3 = 4 km

Note: In the above formula, while calculating the time difference, "late" time is written with negative sign.

Q. 4)

Method 1

Let time taken by second runner  = t. So time taken by first runner = t + 32/60 = t + 8/15

Since distance is constant, hence speed and time are inversely proportional

S2/S1 = T1/T2

16/15 = (t + 8/15)/t

16/15 = 1 + 8/15t

1/15 = 8/15t

t = 8 hours

So second runner takes 8 hours to cover the distance with a speed of 16 km/hr

Hence distance = 8*16 = 128 km

Method 2

Let the distance be X km. Then,

X/15 - X/16 = 32/60

Solve for X

X = 128 km

Answer: (A)

Method 3 (Direct formula)-

In such questions you can use the same formula you used for Q. (3)

Distance =  S1*S2/(S1 - S2) * Time difference

Distance = 16*15/(16 - 15) * (32/60) = 16 * 15 * 32/60 = 128 km

      Q. 5)

A man rows down a river 15 km in 3 hrs.

Hence, Downstream Speed(v) = 15/3 = 5 km/hr

Similarly, Upstream Speed(u) = 15/7.5 = 2 km/hr

v = Rate in still water + Rate of stream

u = Rate in still water - Rate of stream

Add the above 2 equations-

Rate in still water = (v + u)/2 = (5 + 2)/2 = 3.5 km/hr

Answer: (C)

Q. 6)

We have in the above question-

Speed of the current = (v - u)/2

u = 36/6 = 6 km/hr

v = 48/6 = 8 km/hr

Speed of the current = (8 - 6)/2 = 1 km/hr

Answer: (D)

Q. 7)

Let the distance be X km. Let he takes 't' time downstream, then he will take '2t' time upstream.

Downstream speed(v) = X/t

Upstream speed(u) = X/2t

Speed of the boat in still water/Speed of the current = (v + u)/(v - u) = (X/t + X/2t)/(X/t - X/2t)

= 3/2 : 1/2

= 3 : 1

Answer : (B)

Direct formula-

So, Speed of the boat in still water/Speed of the current = (2t + t)/(2t - t) = 3 : 1

Q. 8)

Given,

24/u + 28/v = 6     or    12/u + 14/v = 3    ... (1)

30/u + 21/v = 6.5                                       ... (2)

The best way to solve (1) and (2) is by eliminating a variable.

Multiply equation (1) by 3

36/u + 42/v = 9      ... (3)

Multiply equation (2) by 2

60/u + 42/v = 13    ... (4)

Subtract equation (3) from (4)

24/u = 4

u = 6 km/hr

Put u = 6 in equation (1)

v = 14 km/hr

Speed of the boat in still water = (u + v)/2 = (6 + 14)/2 = 10 km/hr

Answer: (D)

Q. 9) Two guns were fired from the same place at an interval of 13 minutes but a person in a train approaching the place hears the second shot 12 mins 30 seconds after the first. Find the speed of the train(approx) supposing that sound travels at 330 m/s.

A. 40             B. 47               C. 55                 D. 60

Distance travelled by sound in 30 sec = Distance travelled by train in 12 min 30 sec

Let the speed of the train be X m/sec

Distance travelled by sound in 30 sec = 330*30 metres

Distance travelled by train in 12 min 30 sec (750 sec) = X*750

330*30 = X*750

X = 13.2 m/sec = 13.2 * 18/5 km/hr = 47.52 km/hr

Answer: 47 km/hr

Explanation

When you hear the gun shot, that means the sound has travelled to your ears.

First consider a simple scenario when the train is not moving. When the two shots are fired from A, a person sitting in the train will hear them at an interval of 13 minutes only. The sound travels the distance from A to B.

Now let us consider the scenario when the train is moving from B to A. When the first shot is fired, the sound will travel from A to B and the person sitting inside the train will hear it instantly. Now when the second shot is fired after 13 minutes, the sound would not have to travel from A to B, because the person sitting inside the train is not at B any more. He has moved from position B to X. Hence the sound only needs to travel from A to X. 

Hence in this case, the person is hearing the shot after 12 minutes 30 seconds. Instead of travelling for 13 minutes (from A to B), now the sound is travelling only for 12 min 30 sec (from A to X). Hence we can say,

AB = Distance travelled by sound in 13 minutes

AX = Distance travelled by sound in 12 minutes 30 seconds

XB = Distance travelled by sound in 30 seconds              ... (1)

After 12 minutes 30 seconds, the sound moves from A to X and also the train moves from B to X.

BX = Distance travelled by train in 12 minutes 30 seconds              ...  (2)

Hence from (1) and (2) we can say-

Distance travelled by sound in 30 sec = Distance travelled by train in 12 min 30 sec

Q. 10) Two guns were fired from the same place at an interval of 10 minutes and 30 seconds, but a person in a train approaching the place hears second shot 10 minutes after the first. The speed of train (in km/hr), supposing that sound travels at 330m/s is:

A. 19.8             B. 58.6               C. 59.4                 D. 111.8

Distance travelled by sound in 30 sec = Distance travelled by train in 10 minutes (600 sec)

330*30 = X*600

X = 16.5 m/sec or 59.4 km/hr

Answer: (C)

I have covered almost all the type of questions that are asked by SSC from this topic. If you have any doubt in STD, please drop comment.

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