Tuesday, 26 April 2016

[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 :)

Keep reading :)

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Thursday, 21 April 2016

[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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Monday, 18 April 2016

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.



Sunday, 17 April 2016

[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.
Keep Reading :)

To buy the super-hit SSC Hack-book, follow the below link-

Friday, 15 April 2016

[Q] Time and Work Tricks - 3

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


Here W  is the work. For e.g., if 5 men are cutting 10 trees in 2 days, working 4 hours per day. Then,
M = 5, D = 2, H = 4 and W = 10.

Q. 1)
H1 = 6
D1 = 18
D2 = 12
H2 = ?
We know, H1 * D1 = H2 * D2
6 * 18 = H2 * 12
H2 = 9 hours
Answer : (B)


Q. 2)

M1 = 15
D1 = 20
H1 = 8
M2 = 20
D2 = 12
H2 = ?
We know, M1 * D1 * H1 = M2 * D2 * H2
15*20*8 = 20*12*H2
H2 = 10 hours
Answer : (B)


Q. 3)

In this question, Taps = Men
Number of taps required = 20*9/15 = 12
Answer : (B)


Q. 4)

Let there be X number of men.
X men can finish a piece of work in 100 days. Hence total work = 100X
If there were (X - 10) men, it would have taken 110 days to finish the work. Total work in this case = 110(X - 10)
Total work remains the same. Hence,
100X = 110(X - 10)
X = 110
Answer : (D)


          Q. 5)

Efficiency of Subhash = 50/10 = 5 per hour
Efficiency of Subhash and Prakash = 300/40 = 7.5 per hour
Efficiency of Prakash = (Efficiency of Subhash and Prakash) - (Efficiency of Subhash) = 7.5 - 5 = 2.5
So Prakash can copy 2.5 pages per hour. To copy 30 pages, he would require 30/2.5 or 12 hours.
Answer : (D)


                                        Q. 6)

40 men can finish a work in 60 days. Hence, total work = 40*60 = 2400
Let the 10 men left after X days.
For X days, all the 40 men worked. Total work performed = 40X
Now when 10 men quit, only 30 men were left to do the work and they took (70 - X) more days to finish it.
Total work done by 30 men = 30*(70 - X)
Now, 40X + 30*(70 - X) = 2400
X = 30 days
Answer : (C)


Q. 7)

Let the total work = 360 units
Efficiency of A = 360/45 = 8 units
Efficiency of B = 360/40 = 9 units
Efficiency of A + B = 17 units
Let A left after X days.
For X days, both A and B worked. Hence work performed = 17X
B worked for 23 days. Hence work performed by B = 23*9 = 207 units
Now, 17X + 207 = 360
X = 9 days
Answer : (D)

Q. 8)

B and C together do 8/23 of the work, hence A does (1 - 8/23) or 15/23 of the work.
A should be paid = 15*5290/23 = Rs. 3450
Answer : (D)
Note : In this question, they have asked the wages of A. Had they asked the wages of B, firstly you would have calculated the work performed by B with the formula-
Work done by B = (Portion of work done by A and B) + (Portion of work done by B and C) - 1
Work done by B = 19/23 + 8/23 - 1 = 4/23
Wages of B = 4*5290/23 = Rs. 920

Q. 9)


This is a very famous question. A company employed 200 workers to complete a certain work in 150 days. Here the total work is not 200*150 because 200 workers and 150 days was only a plan. In reality, only 1/4th of the work has been done in 50 days. So if they go with the same pace, 200 workers will take 200 days to complete the work.
So total work = 200*200 units
200 workers have worked for 50 days. Hence they have finished 200*50 units of work.
Remaining work = 200*200 - 200*50 = 200*150
Let the number of additional workers required = X.
Now (200+X) workers will work for 100 days to finish the work as per the schedule.
Work they need to perform = (200 + X)*100
Now, (200 + X)*100 = 200*150
X = 100
Answer : (C)


Q. 10) A contractor undertook to finish a certain work in 124 days and employed 120 men.After 64 days,he found that he had already done 2/3 of work. How many men can be discharged now so that the work may finish in time?
A) 56             B) 44               C) 50               D) 60

120 workers finish 2/3 of the work in 64 days. So to complete the whole work, workers will take 64*3/2 or 96 days.
Total work to be performed = 120*96
Now the workers have already finished 2/3 of the work and only 1/3 work has to be performed.
Remaining work = 120*96/3 = 120*32
Let the contractor discharges X men. Remaining workers = 120-X. These workers will continue the work for (124-64) or 60 days. Hence,
120*32 = (120-X)*60
X = 56
Answer : (A)
Method 2
M1 = 120, D1 = 64, W1 = 2/3
M2 = 120 – x, D2 = 60, W2 = 1/3
(M1*D1)/W1 = (M2*D2)/W2
120*64*3/2 = (120 - x)*60*3
x = 56
Q. 11)

Let the second pipe fills the pool in X hours. Then first pipe takes (X+5) hours and the third pipe takes (X-4) hours to fill the pool. Now, 1st and 2nd pipe together take the same time to the fill the pool as the 3rd pipe alone. Hence,
1/(X+5) + 1/(X) = 1/(X - 4)
Solve this quadratic equation and you will get X = 10 hours
That means second pipe takes 10 hours to fill the pool while the third pipe takes 6 hours. Together they will take 10*6/(10 + 6) hours to fill the pool.
Answer: (B)


I have covered almost all the possible questions from Time and Work that are asked by SSC. If you have any doubt in this topic, please drop a comment.
Keep Reading :)

To buy the super-hit SSC Hack-Book, follow the below link-

Thursday, 14 April 2016

[Q] Time and Work Tricks - 2

Please read Part-1, before going through this post. In this article, I have solved questions where some workers quit in the middle, as well as some questions on efficiencies.


Note: In the complete Time and Work series, Efficiency would mean "Work Done in 1 day", and efficiency has been denoted by small letters, e.g. "a" means "Efficiency of A".


Q. 1)

If Pratibha finishes the work in X days, then Sonia will take 3X days to finish the same work
Given 3X – X = 60
Or X = 30
Pratibha takes 30 days and Sonia takes 90 days
Answer: (A)


Q. 2)


Let the total work be 24 units.
Efficiency of Sunil = 24/4 = 6 units (Since Sunil takes 4 days to complete the work)
Efficiency of Ramesh = 6 * 1.5 = 9 units (Since Ramesh is 1.5 times efficient as Sunil)
Efficiency of Dinesh = 24/6 = 4 units ((Since Sunil takes 6 days to complete the work))
Efficiency of (Sunil + Ramesh + Dinesh) = 6 + 9 + 4 = 19 units
Time required to finish the complete work = 24/19 days
Answer: (D)

Q. 3)


Let the total work be 15 units. Efficiency of A = a and Efficiency of B = b
A and B complete the work in 5 days.
Hence efficiency of A and B = 15/5 = 3 units
So, a + b = 3 … (1)
New efficiency of A = 2a
New efficiency of B = b/3
With new efficiency the work was completed in 3 days.
So, 2a + b/3 = 15/3 = 5 … (2)
Solve (1) and (2), you will get a = 12/5 = 2.4 units
So A will complete 15 units work in 15/2.4 or 25/4 days
Answer: (B)

Q. 4)

Let the total work be 24 units
Given, 3*Efficiency of A = Efficiency of B + Efficiency of C
3a = b + c
A, B and C compete the work in 24 days.
Hence, a + b + c = 24/24 = 1 or 4a = 1 [Put b + c = 3a]
a = 1/4 = 0.25 unit
A completes 0.25 unit work in 1 day. So to complete 24 units of work, he will take 24/0.25 = 96 days
Answer: (B)



Q. 5)


Let the total work be 7 units. Since they all complete the work in 7 days, so their total efficiency = 7/7 = 1 unit
Let efficiency of boy = x
Then efficiency of women = 2x
Efficiency of man = 4x
x + 2x + 4x = 1
7x = 1 or x = 1/7
The boy completes 1/7 work in 1 day, so to complete 7 units of work, he will take 49 days
Answer: (A)


Q. 6)

A does 1/2 as much work as B in 3/4 of the time. Hence A will do (1/2 + 1/2) or complete work in (3/4 + 3/4) or 1.5 times more time than B.
A = 1.5B  (where A = no. of days taken by A to finish the work and B = no. of days taken by B to finish the work)
Also A*B/(A+B) = 18
Put A = 1.5B in the above equation and solve
B = 30 days
Answer: (B)

                                       Q. 7)


Let the total work = 60 units
Efficiency of A = 60/20 = 3 units
Efficiency of B = 60/30 = 2 units
Efficiency of (A + B) = 5 units
Work done by A and B in 7 days = 5*7 = 35 units
Work left = 60 – 35 = 25 units
C completes 25 units of work in 10 days. Hence he will complete 60 units of work in 10* 60/25 = 24 days
Answer: (C)


Q. 8)

Let total work be 120 units.
Efficiency of A = 120/6 = 20 units
Efficiency of B = 120/12 = 10 units
Efficiency of C = 120/15 = 8 units
Work left = 7/8 * 120 = 105 units
Efficiency of A + B = 30 units
Hence time taken by A and B to complete 105 units of work = 105/30 = 3.5
Answer: (C)

Q. 9)


Let the total work = 80 units
Efficiency of (A + B + C) = 80/40 = 2 units
Work done by (A + B + C) in 16 days = 16 * 2 = 32 units
Remaining work = 80 – 32 = 48 units
B and C complete the remaining work (48 units) in 40 days.
Efficiency of B + C = 48/40 = 1.2 units
Efficiency of A = Efficiency of (A + B + C) - Efficiency of (B + C) = 2 – 1.2 = 0.8 unit
Time taken by A to complete the whole work = 80/0.8 = 100 days
Answer: (C)

Q. 10)

Let the total work = 360 units
Efficiency of A = 360/45 = 8 units
Efficiency of B = 360/40 = 9 units
Efficiency of A + B = 17 units
Let A left after x days, that means A and B worked together for x days. Total work done by A and B together = 17x
Then the remaining work is finished by B in 23 days. Hence work done by B alone = 23 * 9 = 207 units
So, 17x + 207 = 360
Or x = 9 days
Answer: (D)

Q. 11) 

This question appeared in SSC Tier-2 2015, and stumped many candidates. Although there is nothing tricky about it. 
Let the total work be 60 units.
p + q = 60/6 = 10
q + r = 60*7/60 = 7
Given, Total work done = 3 days work of P + 6 days work of Q and R
60 = 3*p + 6*(7)
p = 6
Hence time taken by P to complete the work = 60/6 = 10 days
p + q = 10, hence q = 4
q + r = 7, hence r = 3
Hence time taken by R to complete the work = 60/3 = 20 days
Difference = 20 - 10 = 10 days
Answer : (C)

Q. 12)  4 Men and 6 Women working together can complete the work in 10 days. 3 men and 7 women working together will complete the same work in 8 days. In how many days 10 women will complete this work?

One day work for a man = 1/m
One day work for a woman = 1/w
In one day, 4 men and 6 women will do 1/10 of the work. Hence,
4/m + 6/w = 1/10    ...  (i)
Similarly,
3/m + 7/w = 1/8      ...  (ii)
Multiply equation (i) with 3 and equation (ii) with 4
12/m + 18/w = 3/10
12/m + 28/w = 1/2
Subtract the equations
10/w = 1/5
So 10 women will complete the work in 5 days
Answer: (5)


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