Aptitude & reasoning – Arunachal Pradesh PCS Free Notes

TIME & DISTANCE

June 4, 2017October 19, 2024 by PSC Notes

In this module we will deal with basic concepts of time and distance, speed, average speed, conversion from km/h to m/s and vice versa. This chapter will form the basis of further concept of relative speed which is used in train and boat problems.

Important Formulas

Speed=Distance/Time
Distance=Speed×Time
Time=Distance/Speed
To convert Kilometers per Hour(km/hr) to Meters per Second(m/s)
x km/hr=(x×5)/18m/s
To convert Meters per Second(m/s) to Kilometers per Hour(km/hr)
x m/s=(x×18)/5 km/hr
If a car covers a certain distance at x kmph and an equal distance at y kmph, the average speed of the whole journey = 2xy/(x+y) kmph
Speed and time are inversely proportional (when distance is constant) ⇒Speed ∝1/Time (when distance is constant)
If the ratio of the speeds of A and B is a : b, then the ratio of the times taken by them to cover the same distance is 1/a:1/b or b : a

Solved Examples

Level 1

1.A person crosses a 600 m long street in 5 minutes. What is his speed in km per hour?
A. 8.2	B. 4.2
C. 6.1	D. 7.2

Answer : Option D

Explanation :

Distance = 600 meter

time = 5 minutes = 5 x 60 seconds = 300 seconds

Speed = distance/time=600/300=2m/s=(2×18)/5 km/hr=36/5 km/hr=7.2 km/hr

2.Two boys starts from the same place walking at the rate of 5 kmph and 5.5 kmph respectively in the same direction. What time will they take to be 8.5 km apart?
A. 17 hr	B. 14 hr
C. 12 hr	D. 19 hr

Answer : Option A

Explanation :

Relative speed = 5.5 – 5 = .5 kmph (because they walk in the same direction)

distance = 8.5 km

Time = distance/speed=8.5/.5=17 hr.

3.Walking 6/7^th of his usual speed, a man is 12 minutes too late. What is the usual time taken by him to cover that distance?
A. 1 hr 42 min	B. 1 hr
C. 2 hr	D. 1 hr 12 min

Answer : Option D

Explanation :

New speed = 6/7 of usual speed
Speed and time are inversely proportional.
Hence new time = 7/6 of usual time
Hence, 7/6 of usual time – usual time = 12 minutes
=>1/6 of usual time = 12 minutes => usual time = 12 x 6 = 72 minutes = 1 hour 12 minutes

4.A man goes to his office from his house at a speed of 3 km/hr and returns at a speed of 2 km/hr. If he takes 5 hours in going and coming, what is the distance between his house and office?
A. 3 km	B. 4 km
C. 5 km	D. 6 km

Answer : Option D

Explanation :

If a car covers a certain distance at x kmph and an equal distance at y kmph,the average speed of the whole journey = 2xy/(x+y) kmph

Hence, average speed = (2×3×2)/(2+3)=12/5 km/hr .

Total time taken = 5 hours

⇒Distance travelled=(12/5)×5=12 km

⇒Distance between his house and office =12/2=6 km

5.If a person walks at 14 km/hr instead of 10 km/hr, he would have walked 20 km more. What is the actual distance travelled by him?
A. 80 km	B. 70 km
C. 60 km	D. 50 km

Answer : Option D

Explanation :

Assume that the person would have covered x km if travelled at 10 km/hr

⇒Speed = Distance/Time=x/10….. (Equation1)

Give that the person would have covered (x + 20) km if travelled at 14 km/hr
⇒Speed = Distance/Time=(x+20)/14….. (Equation2)

From Equations 1 and 2,
X/10=(x+20)/14⇒14x=10x+200⇒4x=200⇒x=200/4=50

6.A car travels at an average of 50 miles per hour for 212 hours and then travels at a speed of 70 miles per hour for 112 hours. How far did the car travel in the entire 4 hours?
A. 210 miles	B. 230 miles
C. 250 miles	D. 260 miles

Answer : Option B

Explanation :

Speed1 = 50 miles/hour

Time1 = 2*(1/2) hour=5/2 hour

⇒Distance1 = Speed1 × Time1 = (50×5)/2=25×5=125 miles

⇒Speed2 = 70 miles/hour

Time2 = 1*1/2 hour=3/2 hour

Distance2 = Speed2 × Time2 = 70×3/2=35×3=105 miles

Total Distance = Distance1 + Distance2 =125+105=230 miles

7.Sound is said to travel in air at about 1100 feet per second. A man hears the axe striking the tree, ¹¹/₅ seconds after he sees it strike the tree. How far is the man from the wood chopper?
A. 1800 ft	B. 2810 ft
C. 3020 ft	D. 2420 ft

Answer : Option D

Explanation :

Speed of the sound = 1100 ft/s ⇒Time = ¹¹/₅ second

Distance = Speed × Time = 1100 ×11/5=220×11=2420 ft

8.A man walking at the rate of 5 km/hr crosses a bridge in 15 minutes. What is the length of the bridge (in meters)?
A. 1250	B. 1280
C. 1320	D. 1340

Answer : Option A

Explanation :

Speed = 5 km/hr

Time = 15 minutes = 1/4 hour

Length of the bridge = Distance Travelled by the man

= Speed × Time = 5×1/4 km

=5×1/4×1000 metre=1250 metre

Level 2

1.A man takes 5 hours 45 min in walking to a certain place and riding back. He would have gained 2 hours by riding both ways. The time he would take to walk both ways is
A. 11 hrs	B. 8 hrs 45 min
C. 7 hrs 45 min	D. 9 hts 20 min

Answer : Option C

Explanation :

Given that time taken for riding both ways will be 2 hours lesser than the time needed for waking one way and riding back From this, we can understand that time needed for riding one way = time needed for waking one way – 2 hours Given that time taken in walking one way and riding back = 5 hours 45 min Hence The time he would take to walk both ways = 5 hours 45 min + 2 hours = 7 hours 45 min In fact, you can do all these calculations mentally and save a lot of time which will be a real benefit for you. 2.A man complete a journey in 10 hours. He travels first half of the journey at the rate of 21 km/hr and second half at the rate of 24 km/hr. Find the total journey in km.
A. 121 km	B. 242 km
C. 224 km	D. 112 km

Answer : Option C

Explanation :

distance = speed x time
Let time taken to travel the first half = x hr
then time taken to travel the second half = (10 – x) hr
Distance covered in the first half = 21x
Distance covered in the second half = 24(10 – x)
But distance covered in the first half = Distance covered in the second half
=> 21x = 24(10 – x) => 21x = 240 – 24x => 45x = 240 => 9x = 48 => 3x = 16⇒x=16/3

Hence Distance covered in the first half = 21x=21×16/3=7×16=112 km. Total distance = 2×112=224 km

3.A car traveling with 5/7 of its actual speed covers 42 km in 1 hr 40 min 48 sec. What is the actual speed of the car?
A. 30 km/hr	B. 35 km/hr
C. 25 km/hr	D. 40 km/hr

Answer : Option B

Explanation :

Time = 1 hr 40 min 48 sec = 1hr +40/60hr+48/3600hr=1+2/3+1/75=126/75hr

Distance = 42 kmSpeed=distance/time=42(126/75) =42×75/126

⇒5/7 of the actual speed = 42×75/126

⇒actual speed = 42×75/126×7/5=42×15/18=7×15/3=7×5=35 km/hr

4.A man covered a certain distance at some speed. If he had moved 3 kmph faster, he would have taken 40 minutes less. If he had moved 2 kmph slower, he would have taken 40 minutes more. What is the the distance in km?
A. 36	B. 38
C. 40	D. 42

Answer : Option C

Explanation :

Let the distance be x km , the speed in which he moved = v kmph

Time taken when moving at normal speed – time taken when moving 3 kmph faster = 40 minutes

⇒x/v−x/(v+3)=40/60⇒x[1/v−1/(v+3)]=2/3⇒x[(v+3−v)/v(v+3)]=2/3

⇒2v(v+3)=9x…………….(Equation1)

Time taken when moving 2 kmph slower – Time taken when moving at normal speed = 40 minutes
⇒x/(v−2)−x/v=40/60⇒x[1/(v−2)−1/v]=2/3

⇒x[(v−v+2)/v(v−2)]=2/3⇒x[2/v(v−2)]=2/3

⇒x[1/v(v−2)]=1/3⇒v(v−2)=3x…………….(Equation2)

Equation1/Equation2

⇒2(v+3)/(v−2)=3⇒2v+6=3v−6⇒v=12

Substituting this value of v inEquation1⇒2×12×15=9x

=>x= (2×12×15)/9= (2×4×15)/3=2×4×5=40. Hence distance = 40 km

5.In covering a distance of 30 km, Arun takes 2 hours more than Anil. If Arun doubles his speed, then he would take 1 hour less than Anil. What is Arun’s speed?
A. 8 kmph	B. 5 kmph
C. 4 kmph	D. 7 kmph

Answer : Option B

Explanation :

Let the speed of Arun = x kmph and the speed of Anil = y kmph
distance = 30 km

We know that distance/speed=time. Hence, 30/x−30/y=2………..(Equation1)

30/y−30/2x=1………..(Equation2)

Equation1 + Equation2⇒30/x−30/2x=3⇒30/2x=3⇒15/x=3⇒5/x=1⇒x=5. Hence Arun’s speed = 5 kmph

6.A car travels first 160 km at 64 km/hr and the next 160 km at 80 km/hr. What is the average speed for the first 320 km of the tour?
A. 70.24 km/hr	B. 74. 24 km/hr
C. 71.11 km/hr	D. 72.21 km/hr

Answer : Option C

Explanation :

If a car covers a certain distance at x kmph and an equal distance at y kmph,the average speed of the whole journey = 2xy/(x+y) kmph.

By using the same formula, we can find out the average speed quickly average speed = (2×64×80)/(64+80)=(2×64×80)/144⇒ (2×32×40)/36= (2×32×10)/9⇒ (64×10)/9=71.11 kmph

7.A man rides his bicycle 10 km at an average speed of 12 km/hr and again travels 12 km at an average speed of 10 km/hr. What is his average speed for the entire trip approximately?
A. 11.2 kmph	B. 10 kmph
C. 10.2 kmph	D. 10.8 kmph

Answer : Option D

Explanation :

Total distance travelled = 10 + 12 = 22 km

Time taken to travel 10 km at an average speed of 12 km/hr = distance/speed=10/12 hr

Time taken to travel 12 km at an average speed of 10 km/hr = distance/speed=12/10 hr

Total time taken =10/12+12/10 hr

Average speed = distance/time=22/(10/12+12/10)=(22×120)/{(10×10)+(12×12)}

(22×120)/244=(11×120)/122=(11×60)/61=660/61≈10.8 kmph

8.An airplane covers a certain distance at a speed of 240 kmph in 5 hours. To cover the same distance in 123 hours, it must travel at a speed of:
A. 660 km/hr	B. 680 km/hr
C. 700 km/hr	D. 720 km/hr

Answer : Option D

Explanation :

Speed and time are inversely proportional ⇒Speed ∝ 1/Time (when distance is constant)

Here distance is constant and Speed and time are inversely proportional

Speed ∝ 1/Time⇒Speed1/Speed2=Time2/Time1

⇒240/Speed2=(1*2/3)5⇒240/Speed2=(5/3)/5⇒240/Speed2=1/3⇒Speed2=240×3=720 km/hr

9.A train can travel 50% faster than a car. Both start from point A at the same time and reach point B 75 kms away from A at the same time. On the way, however, the train lost about 12.5 minutes while stopping at the stations. What is the speed of the car?
A. 80 kmph	B. 102 kmph
C. 120 kmph	D. 140 kmph

Answer : Option C

Explanation :

Let speed of the car = x kmph

Then speed of the train = x *(100+50)/100=150 x /100=3 x /2 kmph

Time taken by the car to travel from A to B=75/x hours

Time taken by the train to travel from A to B=75/(3 x /2)+12.5/60 hours

Since both start from A at the same time and reach point B at the same time

75/x=75/(3 x /2)+12.5/60⇒25/x=12.5/60⇒x=(25×60)/12.5=2×60=120

TIME AND WORK

In these problems the number of persons, quantity of work done and time taken are important factors. Also time taken by a person depends on the efficiency of that person which comes into picture when different people do the work such as women, children do the work alongside the men. The problems related to time and work can be solved by two major approaches – ratio & proportions and unitary method. Let us proceed to find some formulae related to these questions.

Important Formulas – Time and Work

If A can do a piece of work in n days, work done by A in 1 day = 1/n

If A does 1/n work in a day, A can finish the work in n days

If M1 men can do W1 work in D1 days working H1 hours per day and M2 men can do W2 work in D2 days working H2 hours per day (where all men work at the same rate), then

M1 D1 H1 / W1 = M2 D2 H2 / W2

If A can do a piece of work in p days and B can do the same in q days, A and B together can finish it in pq / (p+q) days

If A is thrice as good as B in work, then

Ratio of work done by A and B = 3:1

Ratio of time taken to finish a work by A and B = 1: 3

SOLVED EXAMPLES

Level 1

1.P is able to do a piece of work in 15 days and Q can do the same work in 20 days. If they can work together for 4 days, what is the fraction of work left?
A. 8/15	B. 7/15
C. 11/15	D. 2/11

Answer : Option A

Explanation :

Amount of work P can do in 1 day = 1/15

Amount of work Q can do in 1 day = 1/20

Amount of work P and Q can do in 1 day = 1/15 + 1/20 = 7/60

Amount of work P and Q can together do in 4 days = 4 × (7/60) = 7/15

Fraction of work left = 1 – 7/15= 8/15

2.A can do a piece of work in 4 hours . A and C together can do it in just 2 hours, while B and C together need 3 hours to finish the same work. B alone can complete the work in — hours.
A. 12 hours	B. 6 hours
C. 8 hours	D. 10 hours

Answer : Option A

Explanation :

Work done by A in 1 hour = 1/4

Work done by B and C in 1 hour = 1/3

Work done by A and C in 1 hour = 1/2

Work done by A,B and C in 1 hour = 1/4+1/3 = 7/12

Work done by B in 1 hour = 7/12 – 1/2 = 1/12

=> B alone can complete the work in 12 hours

3.A completes 80% of a work in 20 days. Then B also joins and A and B together finish the remaining work in 3 days. How long does it need for B if he alone completes the work?
A. 37 ½ days	B. 22 days
C. 31 days	D. 22 days

Answer : Option A

Explanation :

Work done by A in 20 days = 80/100 = 8/10 = 4/5

Work done by A in 1 day = (4/5) / 20 = 4/100 = 1/25 — (1)

Work done by A and B in 3 days = 20/100 = 1/5 (Because remaining 20% is done in 3 days by A and B)

Work done by A and B in 1 day = 1/15 —(2)

Work done by B in 1 day = 1/15 – 1/25 = 2/75

=> B can complete the work in 75/2 days = 37 ½ days

4.P can finish a work in 18 days. Q can finish the same work in 15 days. Q worked for 10 days and left the job. How many days does P alone need to finish the remaining work?
A. 8	B. 5
C. 4	D. 6

Answer : Option D

Explanation :

Work done by P in 1 day = 1/18

Work done by Q in 1 day = 1/15

Work done by Q in 10 days = 10/15 = 2/3

Remaining work = 1 – 2/3 = 1/3

Number of days in which P can finish the remaining work = (1/3) / (1/18) = 6

5.Anil and Suresh are working on a special assignment. Anil needs 6 hours to type 32 pages on a computer and Suresh needs 5 hours to type 40 pages. If both of them work together on two different computers, how much time is needed to type an assignment of 110 pages?
A. 7 hour 15 minutes	B. 7 hour 30 minutes
C. 8 hour 15 minutes	D. 8 hour 30 minutes

Answer : Option C

Explanation :

Pages typed by Anil in 1 hour = 32/6 = 16/3

Pages typed by Suresh in 1 hour = 40/5 = 8

Pages typed by Anil and Suresh in 1 hour = 16/3 + 8 = 40/3

Time taken to type 110 pages when Anil and Suresh work together = 110 × 3 /40 = 33/4

= 8 ¼ hours = 8 hour 15 minutes

6.P works twice as fast as Q. If Q alone can complete a work in 12 days, P and Q can finish the work in — days
A. 1	B. 2
C. 3	D. 4

Answer : Option D

Explanation :

Work done by Q in 1 day = 1/12

Work done by P in 1 day = 2 × (1/12) = 1/6

Work done by P and Q in 1 day = 1/12 + 1/6 = ¼

=> P and Q can finish the work in 4 days

7.A work can be finished in 16 days by twenty women. The same work can be finished in fifteen days by sixteen men. The ratio between the capacity of a man and a woman is
A. 1:3	B. 4:3
C. 2:3	D. 2:1

Answer : Option B

Explanation :

Work done by 20 women in 1 day = 1/16

Work done by 1 woman in 1 day = 1/(16×20)

Work done by 16 men in 1 day = 1/15

Work done by 1 man in 1 day = 1/(15×16)

8.P,Q and R together earn Rs.1620 in 9 days. P and R can earn Rs.600 in 5 days. Q and R in 7 days can earn Rs.910. How much amount does R can earn per day?

A. Rs.40

B. Rs.70

C. Rs.90

D. Rs.100

Answer : Option B

Explanation :

Amount Earned by P,Q and R in 1 day = 1620/9 = 180 —(1)

Amount Earned by P and R in 1 day = 600/5 = 120 —(2)

Amount Earned by Q and R in 1 day = 910/7 = 130 —(3)

(2)+(3)-(1) => Amount Earned by P , Q and 2R in 1 day

– Amount Earned by P,Q and R in 1 day = 120+130-180 = 70

=>Amount Earned by R in 1 day = 70
Ratio of the capacity of a man and woman =1/(15×16) : 1/(16×20) = 1/15 : 1/20

= 1/3 :1/4 = 4:3

Level 2

1.P, Q and R can do a work in 20, 30 and 60 days respectively. How many days does it need to complete the work if P does the work and he is assisted by Q and R on every third day?
A. 10 days	B. 14 days
C. 15 days	D. 9 days

Answer : Option C

Explanation :

Amount of work P can do in 1 day = 1/20

Amount of work Q can do in 1 day = 1/30

Amount of work R can do in 1 day = 1/60

P is working alone and every third day Q and R is helping him

Work completed in every three days = 2 × (1/20) + (1/20 + 1/30 + 1/60) = 1/5

So work completed in 15 days = 5 × 1/5 = 1

Ie, the work will be done in 15 days

2.A is thrice as good as B in work. A is able to finish a job in 60 days less than B. They can finish the work in – days if they work together.
A. 18 days	B. 22 ½ days
C. 24 days	D. 26 days

Answer : Option B

Explanation :

If A completes a work in 1 day, B completes the same work in 3 days

Hence, if the difference is 2 days, B can complete the work in 3 days

=> if the difference is 60 days, B can complete the work in 90 days

=> Amount of work B can do in 1 day= 1/90

Amount of work A can do in 1 day = 3 × (1/90) = 1/30

Amount of work A and B can together do in 1 day = 1/90 + 1/30 = 4/90 = 2/45

=> A and B together can do the work in 45/2 days = 22 ½ days

3.P can do a work in the same time in which Q and R together can do it. If P and Q work together, the work can be completed in 10 days. R alone needs 50 days to complete the same work. then Q alone can do it in
A. 30 days	B. 25 days
C. 20 days	D. 15 days

Answer : Option B

Explanation :

Work done by P and Q in 1 day = 1/10

Work done by R in 1 day = 1/50

Work done by P, Q and R in 1 day = 1/10 + 1/50 = 6/50

But Work done by P in 1 day = Work done by Q and R in 1 day . Hence the above equation can be written as Work done by P in 1 day × 2 = 6/50

=> Work done by P in 1 day = 3/50

=> Work done by Q and R in 1 day = 3/50

Hence work done by Q in 1 day = 3/50 – 1/50 = 2/50 = 1/25

So Q alone can do the work in 25 days

4.6 men and 8 women can complete a work in 10 days. 26 men and 48 women can finish the same work in 2 days. 15 men and 20 women can do the same work in – days.

A. 4 days

B. 6 days

C. 2 days

D. 8 days

Answer : Option A

Explanation :

Let work done by 1 man in 1 day = m and work done by 1 woman in 1 day = b

Work done by 6 men and 8 women in 1 day = 1/10

=> 6m + 8b = 1/10

=> 60m + 80b = 1 — (1)

Work done by 26 men and 48 women in 1 day = 1/2

=> 26m + 48b = ½

=> 52m + 96b = 1— (2)

Solving equation 1 and equation 2. We get m = 1/100 and b = 1/200

Work done by 15 men and 20 women in 1 day

= 15/100 + 20/200 =1/4

=> Time taken by 15 men and 20 women in doing the work = 4 days

5.Machine P can print one lakh books in 8 hours. Machine Q can print the same number of books in 10 hours while machine R can print the same in 12 hours. All the machines started printing at 9 A.M. Machine P is stopped at 11 A.M. and the remaining two machines complete work. Approximately at what time will the printing of one lakh books be completed?
A. 3 pm	B. 2 pm
C. 1:00 pm	D. 11 am

Answer : Option C

Explanation :

Work done by P in 1 hour = 1/8

Work done by Q in 1 hour = 1/10

Work done by R in 1 hour = 1/12

Work done by P,Q and R in 1 hour = 1/8 + 1/10 + 1/12 = 37/120

Work done by Q and R in 1 hour = 1/10 + 1/12 = 22/120 = 11/60

From 9 am to 11 am, all the machines were operating.

Ie, they all operated for 2 hours and work completed = 2 × (37/120) = 37/60.

6.A can complete a work in 12 days with a working of 8 hours per day. B can complete the same work in 8 days when working 10 hours a day. If A and B work together, working 8 hours a day, the work can be completed in — days.
A. 5 ⁵⁄₁₁	B. 4 ⁵⁄₁₁
C. 6 ⁴⁄₁₁	D. 6 ⁵⁄₁₁

Answer : Option A

Explanation :

A can complete the work in 12 days working 8 hours a day

=> Number of hours A can complete the work = 12×8 = 96 hours

=> Work done by A in 1 hour = 1/96

B can complete the work in 8 days working 10 hours a day

=> Number of hours B can complete the work = 8×10 = 80 hours => Work done by B in 1 hour = 1/80

Work done by A and B in 1 hour = 1/96 + 1/80 = 11/480 => A and B can complete the work in 480/11 hours. A and B works 8 hours a day.

Hence total days to complete the work with A and B working together = (480/11)/ (8) = 60/11 days = 5 ⁵⁄₁₁ days

Pending work = 1- 37/60 = 23/60

Hours taken by Q an R to complete the pending work = (23/60) / (11/60) = 23/11

which is approximately equal to 2. Hence the work will be completed approximately 2 hours after 11 am ; ie around 1 pm

7.If daily wages of a man is double to that of a woman, how many men should work for 25 days to earn Rs.14400? Given that wages for 40 women for 30 days are Rs.21600.
A. 12	B. 14
C. 16	D. 18

Answer : Option C

Explanation :

Wages of 1 woman for 1 day = 21600/(40×30)

Wages of 1 man for 1 day = (21600×2)/(40×30)

Wages of 1 man for 25 days = (21600×2×25)/(40×30)

Number of men = 14400/(21600×2×25)/(40×30)=144/(216×50)/40×30)=144/9=16

8.There is a group of persons each of whom can complete a piece of work in 16 days, when they are working individually. On the first day one person works, on the second day another person joins him, on the third day one more person joins them and this process continues till the work is completed. How many days are needed to complete the work?
A. 3 ¹⁄₄ days	B. 4 ¹⁄₃ days
C. 5 ¹⁄₆ days	D. 6 ¹⁄₅ days

Answer : Option C

Explanation :

Work completed in 1st day = 1/16

Work completed in 2nd day = (1/16) + (1/16) = 2/16

Work completed in 3rd day = (1/16) + (1/16) + (1/16) = 3/16

An easy way to attack such problems is from the choices. You can see the choices are

very close to each other. So just see one by one.

For instance, The first choice given in 3 ¹⁄₄

The work done in 3 days = 1/16 + 2/16 + 3/16 = (1+2+3)/16 = 6/16

The work done in 4 days = (1+2+3+4)/16 = 10/16

The work done in 5 days = (1+2+3+4+5)/16 = 15/16, almost close, isn’t it?

The work done in 6 days = (1+2+3+4+5+6)/16 > 1

Hence the answer is less than 6, but greater than 5. Hence the answer is 5 ¹⁄₆ days.

(Just for your reference, work done in 5 days = 15/16)

Pending work in 6th day = 1 – 15/16 = 1/16.

In 6th day, 6 people are working and work done = 6/16.

To complete the work 1/16, time required = (1/16) / (6/16) = 1/6 days.

Hence total time required = 5 + 1/6 = 5 ¹⁄₆ days

ANALOGY LEVEL 1

June 4, 2017October 18, 2024 by PSC Notes

When you draw an analogy between two things we compare them for the purpose of explanation. If a scientist says that earth’s forest functions as human lungs then we instantly draw an explanation that both lungs and trees intake important elements from air. As far as SSC exam is concerned this is one of the trickiest section. We intend to comprehend it by solving as many different types of questions as is asked in papers. Presently in the level 1 exam a good aspirant must be able to solve at least 7 out of 10 questions given.

In each of the following questions, select the related word/number from the given alternative:

Flow : River :: Stagnant : ? A. Rain B. Stream C. Pool D. Canal

Ornithologist : Bird :: Archaeologist : ? A.Islands B. Mediators C. Archaeology D.Aquatic

Peacock : India :: Bear : ? A.Australia B. America C. Russia D. England

Given set: (3,7,15) A. 2,6,10 B. 4,8,18 C. 5,9,17 D. 7,12,19

Given set: (63,49,35) A. 81,63,45 B. 64,40,28 C. 72,40,24 D. 72,48,24

3 : 243 :: 5 : ? A. 405 B. 465 C. 3125 D. 546

5 : 36 :: 6 : ? A. 48 B. 50 C. 49 D. 56

TALE : LATE :: ? : CAFE A. FACE B. CAEF C. CAFA D. FEAC

AZBY : DWEV :: HSIR : ? A. JQKO B. KPOL C. KPLO D. KOLP

DE : 45 :: BC : ? A. 34 B. 23 C. 56 D. 43

SOLUTION TO ANALOGY LEVEL 1

Answer: Option C

Explanation: As Water of a River flows similarly water of Pool is Stagnant. Answer & Explanation

Answer: Option C

Explanation: As Ornithologist is a specialist of Birds similarly Archaeologist is a specialist of Archaeology.

Answer: Option C

Explanation: As Peacock is the national bird of India, similarly Bear is the national animal of Russia.

Ans. Option C

Explanation: 1st number+4 = 2nd number

2nd number+8= 3rd number

Ans. Option A

Explanation: 63= 7*9

49= 7*7

35= 7*5

Ans. Option C

Explanation: 3^5=243

5^5=3125

Ans. Option A

Explanation: 13^2+13=182

18^2+18=342

19^2+19=380

Ans. Option A

Explanation: The first and the third letters has been interchanged

Ans. Option C

Explanation: Pairs of opposite letter

A&Z, B&Y, similarly H&S, I&R, K&P, L&O

Ans. Option B

Explanation: D=4, E=5, similarly B=2, C=3

Important Formulas – Mixtures and Alligations

June 4, 2017October 17, 2024 by PSC Notes

Alligation
It is the rule which enables us to find the ratio in which two or more ingredients at the given price must be mixed to produce a mixture of a specified price.
Mean Price
Mean price is the cost price of a unit quantity of the mixture

Suppose a container contains x of liquid from which y units are taken out and replaced by water.After n operations, the quantity of pure liquid = [x(1−y/x)n]

Rule of Alligation
If two ingredients are mixed, then

(Quantity of cheaper/Quantity of dearer)=(P. of dearer – Mean Price)/(Mean price – C.P. of cheaper)

Cost Price(CP) of a unit quantity of cheaper (c)	Cost Price(CP) of a unit quantity of dearer (d)


Mean Price
(m)

(d – m)	(m – c)

=> (Cheaper quantity) : (Dearer quantity) = (d – m) : (m – c)

Solved Examples

Level 1

1.A container contains 40 litres of milk. From this container 4 litres of milk was taken out and replaced by water. This process was repeated further two times. How much milk is now contained by the container?
A. 26 litres	B. 29.16 litres
C. 28 litres	D. 28.2 litres

Answer : Option B

Explanation :

Assume that a container contains x of liquid from which y units are taken out and replaced

by water. After n operations, the quantity of pure liquid

=[x(1−y/x)n]

Hence milk now contained by the container = 40(1−4/40)3=40(1−1/10)3

40×9/10×9/10×9/10=(4×9×9×9)/100=29.16

2.A jar full of whiskey contains 40% alcohol. A part of this whisky is replaced by another containing 19% alcohols and now the percentage of alcohol was found to be 26%. The quantity of whisky replaced is
A. 43	B. 34
C. 32	D. 23

Answer : Option D

Explanation :

Concentration of alcohol in 1st Jar = 40%

Concentration of alcohol in 2nd Jar = 19%

After the mixing, Concentration of alcohol in the mixture = 26%

By the rule of alligation,

Concentration of alcohol in 1st Jar	Concentration of alcohol in 2nd Jar
40%	19%

Mean
26%

7	14

Hence ratio of 1st and 2nd quantities = 7 : 14 = 1 : 2

3.In what ratio should rice at Rs.9.30 per Kg be mixed with rice at Rs. 10.80 per Kg so that the mixture be worth Rs.10 per Kg ?
A. 7 : 8	B. 8 : 7
C. 6 : 7	D. 7 ; 6

Answer : Option B

Explanation :

By the rule of alligation, we have

Cost of 1 kg rice of 1st kind	Cost of 1 kg rice of 2nd kind
9.3	10.80

Mean Price
10

10.8-10 = .8	10 – 9.3 = .7

Required ratio = .8 : .7 = 8 : 7.

4.In what ratio must water be mixed with milk costing Rs.12 per litre in order to get a mixture worth of Rs.8 per litre?
A. 1 : 3	B. 2 : 2
C. 1 : 2	D. 3 : 1

Answer : Option C

Explanation :

By the rule of alligation, we have

Cost Price of 1 litre of water	Cost Price of 1 litre of milk
0	12

Mean Price
8

12-8=4	8-0=8

Required Ratio = 4 : 8 = 1 : 2

5.In 1 kg mixture of iron and manganese 20% of manganese. How much iron should be added so that the proportion of manganese becomes 10%
A. 1.5 Kg	B. 2 Kg
C. .5 Kg	D. 1 Kg

Answer : Option D

Explanation :

By the rule of alligation, we have

Percentage concentration of manganese in the mixture : 20	Percentage concentration of manganese in pure iron : 0


Percentage concentration of manganese in the final mixture
10

10 – 0 = 10	20 – 10 = 10

=> Quantity of the mixture : Quantity of iron = 10 : 10 = 1 : 1

Given that Quantity of the mixture = 1 Kg

Hence Quantity of iron to be added = 1 Kg

6.A trader has 1600Kg of sugar. He sells a part at 8% profit and the rest at 12% profit. If he gains 11% on the whole , find the quantity sold at 12%.
A. 1200 Kg	B. 1400 Kg
C. 1600 Kg	D. 800 Kg

Answer : Option A

Explanation :

By the rule of alligation, we have

% Profit by selling part1	% Profit by selling part2
8	12

Net % Profit
11

12 – 11 = 1	11 – 8 = 3

=>Quantity of part1 : Quantity of part2 = 1 : 3

Given that total quantity = 1600 Kg

Hence quantity of part2 (quantity sold at 12%) = 1600×3/4=1200

7.How many litres of water must be added to 16 liters of milk and water contains 10% water to make it 20% water in it
A. 4 litre	B. 2 litre
C. 1 litre	D. 3 litre

Answer : Option B

Explanation :

By the rule of alligation, we have

% Concentration of water in pure water : 100	% Concentration of water in the given mixture : 10


Mean % concentration
20

20 – 10 = 10	100 – 20 = 80

=> Quantity of water : Quantity of the mixture = 10 : 80 = 1 : 8

Here Quantity of the mixture = 16 litres

=> Quantity of water : 16 = 1 : 8

Quantity of water = 16×1/8=2 litre

8.A merchant has 1000 kg of sugar part of which he sells at 8% profit and the rest at 18% profit. He gains 14% on the whole. The Quantity sold at 18% profit is
A. 300	B. 400
C. 600	D. 500

Answer : Option C

Explanation :

By the rule of alligation,we have

Profit% by selling 1st part	Profit% by selling 2nd part
8	18

Net % profit
14

18-14=4	14-8=6

=> Quantity of part1 : Quantity of part2 = 4 : 6 = 2 : 3

Total quantity is given as 1000Kg

So Quantity of part2 (Quantity sold at 18% profit) = 1000×3/5=600Kg

Level 2

1.Tea worth Rs. 126 per kg and Rs. 135 per kg are mixed with a third variety of tea in the ratio 1 : 1 : 2. If the mixture is worth Rs. 153 per kg, what is the price of the third variety per kg ?
A. Rs.182.50	B. Rs.170.5
C. Rs.175.50	D. Rs.180

Answer : Option C

Explanation :

Tea worth Rs. 126 per kg and Rs. 135 per kg are mixed in the ratio 1 : 1

So their average price = (126+135)2=130.5

Hence let’s consider that the mixture is formed by mixing two varieties of tea.

one at Rs. 130.50 per kg and the other at Rs. x per kg in the ratio 2 : 2, i.e.,

1 : 1. Now let’s find out x.

By the rule of alligation, we can write as

Cost of 1 kg of 1st kind of tea	Cost of 1 kg of 2nd kind of tea
130.50	x

Mean Price
153

(x – 153)	22.50

(x – 153) : 22.5 = 1 : 1

=>x – 153 = 22.50

=> x = 153 + 22.5 = 175.5

2.A milk vendor has 2 cans of milk. The first contains 25% water and the rest milk. The second contains 50% water. How much milk should he mix from each of the containers so as to get 12 litres of milk such that the ratio of water to milk is 3 : 5?
A. 5litres, 7 litres	B. 7litres, 4 litres
C. 6litres, 6 litres	D. 4litres, 8 litres

Answer : Option C

Explanation :

Let cost of 1 litre milk be Rs.1

Milk in 1 litre mix in 1st can = 3/4 litre

Cost Price(CP) of 1 litre mix in 1st can = Rs.3/4

Milk in 1 litre mix in 2nd can = 1/2 litre

Cost Price(CP) of 1 litre mix in 2nd can = Rs.1/2

Milk in 1 litre of the final mix=5/8

Cost Price(CP) of 1 litre final mix = Rs.5/8

=> Mean price = 5/8
By the rule of alligation, we can write as

CP of 1 litre mix in 2nd can	CP of 1 litre mix in 1st can
1/2	3/4

Mean Price
5/8

3/4 – 5/8 = 1/8	5/8 – 1/2 = 1/8

=> mix in 2nd can :mix in 1st can = 1/8 : 1/8 = 1:1

ie, from each can, 12×12=6 litre should be taken

3.Two vessels A and B contain spirit and water in the ratio 5 : 2 and 7 : 6 respectively. Find the ratio in which these mixture be mixed to obtain a new mixture in vessel C containing spirit and water in the ration 8 : 5 ?
A. 3: 4	B. 4 : 3
C. 9 : 7	D. 7 : 9

Answer : Option D

Explanation :

Let Cost Price(CP) of 1 litre spirit be Rs.1

Quantity of spirit in 1 litre mixture from vessel A = 5/7

Cost Price(CP) of 1 litre mixture from vessel A = Rs. 5/7

Quantity of spirit in 1 litre mixture from vessel B = 7/13

Cost Price(CP) of 1 litre mixture from vessel B = Rs. 7/13

Quantity of spirit to be obtained in 1 litre mixture from vessel C = 8/13

Cost Price(CP) of 1 litre mixture from vessel C = Rs. 8/13 = Mean Price

By the rule of alligation, we can write as

CP of 1 litre mixture from vessel A	CP of 1 litre mixture from vessel B
5/7	7/13

Mean Price
8/13

8/13 – 7/13 = 1/13	5/7 – 8/13 = 9/91

=> Mixture from Vessel A : Mixture from Vessel B = 1/13 : 9/91 = 7 : 9 = Required Ratio

4.How many kilograms of sugar costing Rs. 9 per kg must be mixed with 27 kg of sugar costing Rs. 7 per Kg so that there may be a gain of 10 % by selling the mixture at Rs. 9.24 per Kg ?
A. 60 Kg	B. 63 kg
C. 58 Kg	D. 56 Kg

Answer : Option B

Explanation :

Selling Price(SP) of 1 Kg mixture= Rs. 9.24

Profit = 10%

Cost Price(CP) of 1 Kg mixture= 100×SP /(100+Profit%) =100×9.24/(100+10)

=100×9.24/110=92.4/11=Rs.8.4

By the rule of alligation, we have

CP of 1 kg sugar of 1st kind	CP of 1 kg sugar of 2nd kind
Rs. 9	Rs. 7

Mean Price
Rs.8.4

8.4 – 7 = 1.4	9 – 8.4 = .6

ie, to get a cost price of 8.4, the sugars of kind1 and kind2 should be mixed in the

ratio 1.4 : .6 = 14 : 6 = 7 : 3

Let x Kg of kind1 sugar is mixed with 27 kg of kind2 sugar

then x : 27 = 7 : 3

⇒x/27=7/3

⇒x=(27×7)/3=63

5.A container contains a mixture of two liquids P and Q in the ratio 7 : 5. When 9 litres of mixture are drawn off and the container is filled with Q, the ratio of P and Q becomes 7 : 9. How many litres of liquid P was contained in the container initially?
A. 23	B. 21
C. 19	D. 17

Answer : Option B

Explanation :

Let’s initial quantity of P in the container be 7x

and initial quantity of Q in the container be 5x

Now 9 litres of mixture are drawn off from the container

Quantity of P in 9 litres of the mixtures drawn off = 9×7/12=63/12=21/ 4

Quantity of Q in 9 litres of the mixtures drawn off = 9×5/12=45/12=1/54

HenceQuantity of P remains in the mixtures after 9 litres is drawn off =7x−21/4

Quantity of Q remains in the mixtures after 9 litres is drawn off =5x−15/4

Since the container is filled with Q after 9 litres of mixture is drawn off,

Quantity of Q in the mixtures=5x−15/4+9=5x+21/4

Given that the ratio of P and Q becomes 7 : 9

⇒(7x−21/4):(5x+21/4)=7:9

⇒(7x−21/4)(5x+21/4)=7/9

63x−(9×21/4)=35x+(7×2/14)

28x=(16×21/4)

x=(16×21)/(4×28)

litres of P contained in the container initially = 7x=(7×16×21)/(4×28)=(16×21)/(4×4)=21

6.A dishonest milkman sells his milk at cost price but he mixes it with water and thereby gains 25%. What is the percentage of water in the mixture?
A. 25%	B. 20%
C. 22%	D. 24%

Answer : Option B

Explanation :

Let CP of 1 litre milk = Rs.1

Given that SP of 1 litre mixture = CP of 1 Litre milk = Rs.1

Given than Gain = 25%

Hence CP of 1 litre mixture = 100/(100+Gain%)×SP

=100(100+25)×1

=100/125=4/5

By the rule of alligation, we have

CP of 1 litre milk	CP of 1 litre water
1	0

CP of 1 litre mixture
4/5

4/5 – 0 = 4/5	1- 4/5 = 1/5

=> Quantity of milk : Quantity of water = 4/5 : 1/5 = 4 : 1

Hence percentage of water in the mixture = 1/5×100=20%

SURDS

June 4, 2017October 16, 2024 by PSC Notes

Surds

A surd is a square root which cannot be reduced to a rational number.

For example, is not a surd.

However is a surd.

If you use a calculator, you will see that and we will need to round the answer correct to a few decimal places. This makes it less accurate.

If it is left as , then the answer has not been rounded, which keeps it exact.

Here are some general rules when simplifying expressions involving surds.

a^mx aⁿ = a^m^+ n

a^m	= a^m^– n
aⁿ	= a^m^– n

(a^m)ⁿ= a^mn

(ab)ⁿ= aⁿbⁿ

a	n	=	aⁿ
b	n	=	bⁿ

a⁰= 1

Questions

Level-I

(17)^3.5 x (17)^? = 17⁸

A.	2.29
B.	2.75
C.	4.25
D.	4.5

If		a		x – 1	=		b		x – 3	, then the value of x is:
		b					a

Given that 10^0.48 = x, 10^0.70 = y and x^z = y², then the value of z is close to:

A.	1.45
B.	1.88
C.	2.9
D.	3.7

If 5^a = 3125, then the value of 5^{(a – 3)} is:

A.	25
B.	125
C.	625
D.	1625

If 3^(x – y) = 27 and 3^(x + y) = 243, then x is equal to:

A.	0
B.	2
C.	4
D.	6

.6.

(256)^0.16 x (256)^0.09 = ?

A.	4
B.	16
C.	64
D.	256.25

The value of [(10)¹⁵⁰ ÷ (10)¹⁴⁶]

A.	1000
B.	10000
C.	100000
D.	10⁶

1	+	1	+	1	= ?
1 + x^(b – a) + x^(c – a)		1 + x^(a – b) + x^(c – b)		1 + x^(b – c) + x^(a – c)

A.	0
B.	1
C.	x^a^{– b – c}
D.	None of these

(25)^7.5 x (5)^2.5 ÷ (125)^1.5 = 5^?

A.	8.5
B.	13
C.	16
D.	17.5
E.	None of these

10.

(0.04)^-1.5 = ?

A.	25
B.	125
C.	250
D.	625

Level-II

11.

(243)ⁿ^/5 x 3^{2n + 1}	= ?
9ⁿ x 3ⁿ^{– 1}

A.	1
B.	2
C.	9
D.	3ⁿ

12.

1	+	1	= ?
1 + a^(n – m)		1 + a^(m – n)

a^m^+ n

13.

If m and n are whole numbers such that mⁿ = 121, the value of (m – 1)ⁿ^{+ 1} is:

A.	1
B.	10
C.	121
D.	1000

14.

x^b

(b + c – a)

x^c

(c + a – b)

x^a

^{(a + b – c)}

= ?

x^c

x^a

x^b

A.	x^abc
B.	1
C.	x^ab^+ bc + ca
D.	x^a^+ b + c

If 5√5 * 5³÷ 5^-3/2= 5^a+2 , the value of a is:
A. 4
B. 5
C. 6
D. 8

16.(132)⁷ ×(132)^? =(132)^11.5.

A. 3
B. 3.5
C. 4
D. 4.5

17. (ab)x−2=(ba)x−7. What is the value of x ?

A. 3
B. 4
C. 3.5
D. 4.5

18. (0.04)^-2.5 = ?

A. 125
B. 25
C. 3125
D. 625

Answers

Level-I

Answer:1 Option D

Explanation:

Let (17)^3.5 x (17)^x = 17⁸.

Then, (17)^{3.5 + x} = 17⁸.

3.5 + x = 8

x = (8 – 3.5)

x = 4.5

Answer:2 Option C

Explanation:

Given	a		x – 1	=		b		x – 3
	b					a

x – 1

-(x – 3)

(3 – x)

x – 1 = 3 – x

2x = 4

x = 2.

Answer:3 Option C

Explanation:

x^z = y² 10^(0.48z) = 10^{(2 x 0.70)} = 10^1.40

0.48z = 1.40

z =	140	=	35	= 2.9 (approx.)
	48		12

Answer:4 Option A

Explanation:

5^a = 3125 5^a = 5⁵

a = 5.

5^{(a – 3)} = 5^{(5 – 3)} = 5² = 25.

Answer:5 Option C

Explanation:

3^x^– y = 27 = 3³ x – y = 3 ….(i)

3^x^+ y = 243 = 3⁵ x + y = 5 ….(ii)

On solving (i) and (ii), we get x = 4

Answer:6 Option A

Explanation:

(256)^0.16 x (256)^0.09 = (256)^{(0.16 + 0.09)}

= (256)^0.25

= (256)^(25/100)

= (256)^(1/4)

= (4⁴)^(1/4)

= 4^4(1/4)

= 4¹

= 4

Answer:7 Option B

Explanation:

(10)¹⁵⁰ ÷ (10)¹⁴⁶ =	10¹⁵⁰
	10¹⁴⁶

= 10^{150 – 146}

= 10⁴

= 10000.

Answer:8 Option B

Explanation:

Given Exp. =

	1 +	x^b	+	x^c
		x^a		x^a

	1 +	x^a	+	x^c
		x^b		x^b

	1 +	x^b	+	x^a
		x^c		x^c

=	x^a	+	x^b	+	x^c
	(x^a + x^b + x^c)		(x^a + x^b + x^c)		(x^a + x^b + x^c)

=	(x^a + x^b + x^c)
	(x^a + x^b + x^c)

= 1.

Answer:9 Option B

Explanation:

Let (25)^7.5 x (5)^2.5 ÷ (125)^1.5 = 5^x.

Then,	(5²)^7.5 x (5)^2.5	= 5^x
	(5³)^1.5

	5^{(2 x 7.5)} x 5^2.5	= 5x
	5^{(3 x 1.5)}

	5¹⁵ x 5^2.5	= 5^x
	5^4.5

5^x = 5^{(15 + 2.5 – 4.5)}

5^x = 5¹³

x = 13.

Answer:10 Option B

Explanation:

(0.04)^-1.5 =		4		-1.5
		100

=		1		-(3/2)
		25

= (25)^(3/2)

= (5²)^(3/2)

= (5)^{2 x (3/2)}

= 5³

= 125.

Level-II

Answer:11 Option C

Explanation:

Given Expression

=	(243)^(n/5) x 3^{2n + 1}
	9ⁿ x 3ⁿ^{– 1}

=	(3⁵)^(n/5) x 3^{2n + 1}
	(3²)ⁿ x 3ⁿ^{– 1}

=	(3^{5 x (n/5)} x 3^{2n + 1})
	(3²ⁿ x 3ⁿ^{– 1})

=	3ⁿ x 3^{2n + 1}
	3²ⁿ x 3ⁿ^{– 1}

=	3^{(n + 2n + 1)}
	3^{(2n + n – 1})

=	3^{3n + 1}
	3^{3n – 1}

= 3^{(3n + 1 – 3n + 1)} = 3² = 9.

Answer:12 Option C

Explanation:

	1 +	aⁿ
		a^m

	1 +	a^m
		aⁿ

1 + a^(n – m)

1 + a^(m – n)

=	a^m	+	aⁿ
	(a^m + aⁿ)		(a^m + aⁿ)

=	(a^m + aⁿ)
	(a^m + aⁿ)

= 1.

Answer:13 Option D

Explanation:

We know that 11² = 121.

Putting m = 11 and n = 2, we get:

(m – 1)ⁿ^{+ 1} = (11 – 1)^{(2 + 1)} = 10³ = 1000.

Answer:14 Option B

Explanation:

Given Exp.

= x^{(b – c)(b + c – a)} . x^{(c – a)(c +a – b)} . x^{(a – b)(a + b – c)}

= x^{(b – c)(b + c) – a(b – c)} . x^{(c – a)(c + a) – b(c – a)} . x^{(a – b)(a + b) – c(a – b)}

= x^{(b2 – c2 + c2 – a2 + a2 – b2)} . x^{–a(b – c) – b(c – a) – c(a – b)}

= (x⁰ x x⁰)

= (1 x 1) = 1.

Answer:15 option C

Answer:16

Explanation

am.an=am+n

(132)⁷ × (132)^x = (132)^11.5

=> 7 + x = 11.5

=> x = 11.5 – 7 = 4.5

Answer:17

Explanation:

an=1a−n

(ab)x−2=(ba)x−7⇒(ab)x−2=(ab)−(x−7)⇒x−2=−(x−7)⇒x−2=−x+7⇒x−2=−x+7⇒2x=9⇒x=92=4.5

Answer:18

Explanation:

a−n=1/an

(0.04)−2.5=(1/.04)2.5=(100/4)2.5=(25)2.5=(52)2.5=(52)(5/2)=55=3125

MATHEMATICS AND QUATITUATIVE APTITUDE – SIMPLE INTEREST

June 4, 2017October 13, 2024 by PSC Notes

Introduction

Money is not free and it costs to borrow the money. Normally, the borrower has to pay an extra amount in addition to the amount he had borrowed. i.e, to repay the loan, the borrower has to pay the sum borrowed and the interest.

Lender and Borrower

The person giving the money is called the lender and the person taking the money is the borrower.

Principal (sum)

Principal (or the sum) is the money borrowed or lent out for a certain period. It is denoted by P.

Interest

Interest is the extra money paid by the borrower to the owner (lender) as a form of compensation for the use of the money borrowed.

Simple Interest (SI)

If the interest on a sum borrowed for certain period is calculated uniformly, it is called simple interest(SI).

Amount (A)

The total of the sum borrowed and the interest is called the amount and is denoted by A

The statement “rate of interest 10% per annum” means that the interest for one year on a sum of Rs.100 is Rs.10. If not stated explicitly, rate of interest is assumed to be for one year.

Let Principal = P, Rate = R% per annum and Time = T years. Then
Simple Interest, SI = PRT/100

From the above formula , we can derive the followings
P=100×SI/RT

R=100×SI/PT

T=100×SI/PR

Some Formulae

If a sum of money becomes n times in T years at simple interest, then the rate of interest per annum can be given be R = 100(n−1)/T %
The annual instalment which will discharge a debt of D due in T years at R% simple interest per annum =100D/ (100T+RT(T-1)/2)
If an amount P₁is lent out at simple interest of R₁% per annum and another amount P₂ at simple interest rate of R₂% per annum, then the rate of interest for the whole sum can be given by
R=(P1R1+P2R2)/ (P1+P2)
If a certain sum of money is lent out in n parts in such a manner that equal sum of money is obtained at simple interest on each part where interest rates are R₁, R₂, … , R_nrespectively and time periods are T₁, T₂, … , T_n respectively, then the ratio in which the sum will be divided in n parts can be given by (1/R1T1):(1/R2T2):⋯(1/RnTn)
If a certain sum of money P lent out for a certain time T amounts to P₁at R₁% per annum and to P₂at R₂% per annum, then P = (P2R1−P1R2)/ (R1−R2) and T = (P1−P2) ×100 years / (P2R1−P1R2)

SOLVED EXAMPLES

LEVEL 1

1. Arun took a loan of Rs. 1400 with simple interest for as many years as the rate of interest. If he paid Rs.686 as interest at the end of the loan period, what was the rate of interest?
A. 8%	B. 6%
C. 4%	D. 7%

Ans. Let rate = R%

Then, Time, T = R years

P = Rs.1400

SI = Rs.686

SI= PRT/100⇒686 = 1400 × R × R/100⇒686=14 Rx R ⇒49=Rx R ⇒R=7

i.e.,Rate of Interest was 7%. (D)

2. How much time will it take for an amount of Rs. 900 to yield Rs. 81 as interest at 4.5% per annum of simple interest?
A. 2 years	B. 3 years
C. 1 year	D. 4 years

Ans. P = Rs.900

SI = Rs.81

T = ?

R = 4.5%

T= 100×SI/PR = 100×81/(900×4.5) = 2 years (A)

3. A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years. The sum is :
A. Rs. 700	B. Rs. 690
C. Rs. 650	D. Rs. 698

Ans. Simple Interest (SI) for 1 year = 854-815 = 39

Simple Interest (SI) for 3 years = 39 × 3 = 117

Principal = 815 – 117 = Rs.698 (D)

4. A sum fetched a total simple interest of Rs. 929.20 at the rate of 8 p.a. in 5 years. What is the sum?
A. Rs. 2323	B. Rs. 1223
C. Rs. 2563	D. Rs. 2353

Ans. SI = Rs.929.20

P = ?

T = 5 years

R = 8%

P = 100×SI/RT=100×929.20/(8×5) = Rs.2323 (A)

5. What will be the ratio of simple interest earned by certain amount at the same rate of interest for 5 years and that for 15 years?
A. 3 : 2	B. 1 : 3
C. 2 : 3	D. 3 : 1

Solution 1
Let Principal = P

Rate of Interest = R%

Required Ratio = (PR×5/100)/ (PR×15/100) =1:3 (B)
Solution 2

Simple Interest = PRT100

Here Principal(P) and Rate of Interest (R) are constants

Hence, Simple Interest ∝ T

Required Ratio = Simple Interest for 5 years Simple Interest for 15 years=T1T2=515=13=1:3 (B)

6. A sum of money amounts to Rs.9800 after 5 years and Rs.12005 after 8 years at the same rate of simple interest. The rate of interest per annum is
A. 15%	B. 12%
C. 8%	D. 5%

Ans. Simple Interest for 3 years = (Rs.12005 – Rs.9800) = Rs.2205

Simple Interest for 5 years = 22053×5=Rs.3675

Principal (P) = (Rs.9800 – Rs.3675) = Rs.6125

R = 100×SI/PT=100×3675/(6125×5) =12% (B)

7. A lent Rs. 5000 to B for 2 years and Rs. 3000 to C for 4 years on simple interest at the same rate of interest and received Rs. 2200 in all from both of them as interest. The rate of interest per annum is:
A. 5%	B. 10%
C. 7%	D. 8%

Ans. Let the rate of interest per annum be R%

Simple Interest for Rs. 5000 for 2 years at rate R% per annum +Simple Interest for Rs. 3000 for 4 years at rate R% per annum = Rs.2200

⇒5000×R×2/100+3000×R×4/100=2200

⇒100R + 120R=2200⇒220R=2200⇒R=10

i.e, Rate = 10%. (B)

8. In how many years, Rs. 150 will produce the same interest at 6% as Rs. 800 produce in 2 years at 4½% ?
A. 4 years	B. 6 years
C. 8 years	D. 9 years

Ans. Let Simple Interest for Rs.150 at 6% for n years = Simple Interest for Rs.800 at 4½ % for 2 years

150×6×n/100=800×4.5×2/100

150×6×n=800×4.5×2

n=8 years (C)

LEVEL 2

1. Mr. Thomas invested an amount of Rs. 13,900 divided in two different schemes A and B at the simple interest rate of 14% p.a. and 11% p.a. respectively. If the total amount of simple interest earned in 2 years be Rs. 3508, what was the amount invested in Scheme B?
A. Rs. 6400	B. Rs. 7200
C. Rs. 6500	D. Rs. 7500

Ans. Let the investment in scheme A be Rs.x

and the investment in scheme B be Rs. (13900 – x)

We know that SI = PRT/100

Simple Interest for Rs.x in 2 years at 14% p.a. = x×14×2100=28x100Simple Interest for Rs.(13900 – x) in 2 years at 11% p.a. = (13900−x)×11×2/100 =22(13900−x)/100

Total interest =Rs.3508

Thus, 28x/100+22(13900−x)/100 = 3508

28x+305800−22x=350800

6x = 45000

x=45000/6=7500

Investment in scheme B = 13900 – 7500 = Rs.6400 (A)

2. A certain sum in invested for T years. It amounts to Rs. 400 at 10% per annum. But when invested at 4% per annum, it amounts to Rs. 200. Find the time (T).
A. 45 years	B. 60 years
C. 40 years	D. 50 Years

Solution 1
Let the principal = Rs.x

and time = y years

Principal,x amounts to Rs.400 at 10% per annum in y years

Simple Interest = (400-x)

Simple Interest = PRT/100

⇒ (400−x) = x×10×y/100

⇒ (400−x) = xy/10— (equation 1)

Principal,x amounts to Rs.200 at 4% per annum in y years

Simple Interest = (200-x)

Simple Interest = PRT/100

⇒ (200−x) = x×4×y/100

⇒ (200−x) = xy/25— (equation 2)

(equation 1)/(equation2)

⇒(400−x) / (200−x) = (xy/10)/(xy/25)

⇒ (400−x)/ (200−x) =25/10

⇒ (400−x)/ (200−x) =52

⇒800−2x = 1000−5x

⇒200=3x

⇒x =200/3 Substituting this value of x in Equation 1, we get,

(400−200/3) = (200y/3)/10

⇒ (400−200/3) = 20y/3

⇒1200−200=20y

⇒1000=20y

y=1000/20=50 years (D)

Solution 2
If a certain sum of money P lent out for a certain time T amounts to P₁ at R₁% per annum and to P₂ at R₂% per annum, then

P = (P2R1−P1R2)/ (R1−R2)

T = (P1−P2)x 100 years/(P2R1−P1R2)

R₁ = 10%, R₂ = 4%

P₁ = 400, P₂ = 200

T = (P1−P2)x 100 / (P2R1−P1R2) = (400−200)x 100 / (200×10−400×4)

=200 x 100/ (2000−1600) =200 ×100/400 = 12×100=50 years (D)

3. Mr. Mani invested an amount of Rs. 12000 at the simple interest rate of 10% per annum and another amount at the simple interest rate of 20% per annum. The total interest earned at the end of one year on the total amount invested became 14% per annum. Find the total amount invested.
A. Rs. 25000	B. Rs. 15000
C. Rs. 10000	D. Rs. 20000

Ans. If an amount P₁ is lent out at simple interest of R₁% per annum and another amount P₂ at simple interest rate of R₂% per annum, then the rate of interest for the whole sum can be given by

R= (P1R1+P2R2)/(P1+P2)

P₁ = Rs. 12000, R₁ = 10%

P₂ =? R₂ = 20%

R = 14%

14 = (12000×10+P2×20)/ (12000+P2)

12000×14+14P2 =120000+20P2

6P2=14×12000−120000=48000

⇒P2=8000

Total amount invested = (P₁ + P₂) = (12000 + 8000) = Rs. 20000 (D)

4. A sum of money is lent at S.I. for 6 years. If the same amount is paid at 4% higher, Arun would have got Rs. 120 more. Find the principal
A. Rs. 200	B. Rs. 600
C. Rs. 400	D. Rs. 500

Ans. This means, simple interest at 4% for that principal is Rs.120

P=100×SI/ RT=100×120/ (4×6) =100×30/6 = 100×5 = 500 (D)

5. The simple interest on Rs. 1820 from March 9, 2003 to May 21, 2003 at 7 ¹⁄₂% rate is
A. Rs. 27.30	B. Rs. 22.50
C. Rs. 28.80	D. Rs. 29

Ans. Time, T = (22 + 30 + 21) days = 73 days = 73/365 year=1/5 year

Rate, R = 7.5%=15/2%

SI = PRT/100 = 1820× (15/2) × (1/5)/100 = 1820 × (3/2)/100 = 910 × 3/100

= 2730/100 = 27.30 (A)

6. A sum of Rs. 7700 is to be divided among three brothers Vikas, Vijay and Viraj in such a way that simple interest on each part at 5% per annum after 1, 2 and 3 years respectively remains equal. The Share of Vikas is more than that of Viraj by
A. Rs.1200	B. Rs.1400
C. Rs.2200	D. Rs.2800

Ans. If a certain sum of money is lent out in n parts in such a manner that equal sum of money is obtained at simple interest on each part where interest rates are R₁, R₂, … , R_n respectively and time periods are T₁, T₂, … , T_n respectively, then the ratio in which the sum will be divided in n parts can be given by

1/R1T1:1/R2T2:⋯1/RnTn

T₁ = 1 , T₂ = 2, T₃ = 3

R₁ = 5 , R₂ = 5, R₃ = 5

Share of Vikas : Share of Vijay : Share of Viraj

= (1/5×1) : (1/5×2) : (1/5×3) = 1/1:1/2:1/3 = 6:3:2

Total amount is Rs. 7700

Share of Vikas = 7700×6/11=700×6 = 4200

Share of Viraj = 7700×2/11=700×2=1400

Share of Vikas is greater than Share of Viraj by (4200 – 1400) = Rs. 2800 (D)

7. David invested certain amount in three different schemes A, B and C with the rate of interest 10% p.a., 12% p.a. and 15% p.a. respectively. If the total interest accrued in one year was Rs. 3200 and the amount invested in Scheme C was 150% of the amount invested in Scheme A and 240% of the amount invested in Scheme B, what was the amount invested in Scheme B?
A. Rs.5000	B. Rs.2000
C. Rs.6000	D. Rs.3000

Ans. Let x, y and x be his investments in A, B and C respectively. Then

Then, Interest on x at 10% for 1 year

+ Interest on y at 12% for 1 year

+ Interest on z at 15% for 1 year

= 3200

x×10×1/100+y×12×1/100+z×15×1/100=3200

⇒10x+12y+15z=320000−−−(1)

Amount invested in Scheme C was 240% of the amount invested in Scheme B

=>z=240y/100 = 60y/25=12y/5−−−(2)

Amount invested in Scheme C was 150% of the amount invested in Scheme A

=>z=150x/100=3x/2

=>x=2z/3=2/3×12y/5=8y/5−−−(3)

From(1),(2) and (3),

10x + 12y + 15z = 320000

10(8y/5)+12y+15(12y/5)=320000

16y+12y+36y=320000

64y=320000

y=320000/64=10000/2=5000

i.e.,Amount invested in Scheme B = Rs.5000 (A)

DISCOUNT

June 4, 2017October 13, 2024 by PSC Notes

Discount

The discount is referred to the reduction in the price of some commodity or service. It may anywhere appear in the distribution channel in the form of modifications in marked price (printed on the item) or in retail price (set by retailer usually by pasting a sticker on the item) or in list price (quoted for the buyer). The discount is provided for the purpose of increasing sales, to clear out old stock, to encourage distributors, to reward potential customer etc. In short, the discount can serve as a way to attract customers for a particular item or service.

In math, discount is one of the easiest way to raise the customers of particular product. Discounts are a significant element of your online merchandising plan. You build discounts so that you can force sales on items or collection of products to your customers who convene particular conditions. In math, the discount problems can be solved by using discount formula.

The “discount rate” means the interest rate. Discount rate is based on the simple interest rate. To calculate simple interest rate, just find out the interest rate for one period (multiply by amount, interest rate, period) but calculate the discount rate, just multiply by the amount and an interest rate. This is called the define discount rate.

To calculate the discount rate, just multiply the amount by an interest rate. By using the Formula Discount rate DR = pr (p = principal amount, r = interest rate).

What is Discount Rate?

Discount rate is one of the simple ways to increase the customers of particular product. Discounts are a important element of your online merchandising strategy. You make discounts so that you can force sales on products or collection of products to your customers who meet certain particular conditions.

The formula used to calculate the discount is discount = marked price – selling price.

Here,

Selling price is what you actually pay for the item.

Marked price is the normal price of the item without a discount.

Discount is either a dollar rate or a percentage of the marked cost.

Discount Rate Definition

Discount Rate is the cost of the total amount generally less than its original value is called . In other words, a total bill will generally sell at a discount, and the discount rate is annualized percentage of this discount, that is percentage is adjusted to give an annual percentage.

Discount Rate Formula

Formula of the Discount Rate is:

Discount rate DR = pr

where,

p = principal amount
r = interest rate

Questions:

Level-I

1: Ricky purchase the dress. That dress rate was Rs1000 at 10% discount . Find discount rate? And then ricky how many dollars give to cashier?

2: Kalvin purchased land for 50000 dollars at 20% in 2000th year. Then 2004th year that land sales 3000 dollars. How many dollars he loss?

The marked price of a ceiling fan is $ 1250 and the shopkeeper allows a discount of 6% on it. Find the selling price of the fan.
A trader marks his goods at 40% above the cost price and allows a discount of 25%. What is his gain percent?
A dealer purchased a washing machine for $ 7660. He allows a discount of 12% on its marked price and still gains 10%. Find the marked price of the machine.
How much per cent above the cost price should a shopkeeper mark his goods so that after allowing a discount of 25% on the marked price, he gains 20%?
Find the single discount equivalent to two successive discounts of 20% and 10%.
A merchant who marked his goods up by 50% subsequently offered a discount of 20% on the marked price. What is the percentage profit that the merchant make after offering the discount?

Applied to a bill for Rs. 1,00,000 the difference between a discount of 40% and two successive discounts of 36% and 4% is:
On a 20% discount sale, an article costs Rs. 596. What was the original price of the article?

Level-II:

A discount of 15% on one article is the same as discount of 20% on a second article. The costs of the
A discount of 2 ½% is given to the customer on marked price of an article. A man bought the article for Rs. 39. The marked price of article is:
Printed price of an article is Rs. 900 but the retailer gets a discount of 40%. He sells the article for Rs. 900. Retailer’s gain percent is:
The marked price of a watch was Rs. 720. A man bought the same watch for Rs. 550.80, after getting two successive discounts. If the first discount was 10%, what was the second discount rate?
A shopkeeper marks his goods 20% above cost price, but allows 30% discount for cash. His net loss is:
A retailer buys 40 pens at the marked price of 36 pens from a wholesaler. If he sells these pens giving a discount of 1%, what is the profit percent?
A pizzeria has a coupon that reads, “Getoff a $9.00 cheese pizza.” What is the discount? What is the sale price of the cheese pizza?

18.In a video store, a DVD that sells for $15 is marked, “10% off.” What is the sale price of the DVD?

Answers:

Level-I:

Solution:1
Here,

Principal amount p = 1000 rs

Interest rate r = 10%

Discount rate DR = pr

DR = 1000*

= 100

The discount amount for the dress is 100.

Discount rate DR = 100.

Dress rate = principal amount – discount rate

= 1000 – 100

=900

Ricky gives 900 rs to cashier

Solution:2
Principal amount p = 50000 dollars

Interest rate r = 20%

Discount rate DR = pr

DR = 50000 x 2010020100 in 2000th year

= 10000

Discount rate DR = 1000 dollars in 2000th year.

The discount amount is 10000 dollars.

Discount rate DR = 50000*30/100 in 2004th year

Discount rate =15000 dollars.

The discount amount is 15000 dollars.

Loss Discount rate in 2004th year – Discount rate in 2000th year

=15000 dollars – 10000 dollars

=5000 dollars

Kalvin 5000 dollars losses in that land.

Solution:3

Marked price = $ 1250 and discount = 6%.

Discount = 6% of Marked Price

= (6% of $ 1250)

= $ {1250 × (6/100)}

= $ 75

Selling price = (Marked Price) – (discount)

= $ (1250 – 75)

= $ 1175.

Hence, the selling price of the fan is $ 1175.

Solution:4

Let the cost price be $ 100.

Then, marked price = $ 140.

Discount = 25% of Marked Price

= (25% of $ 140)

= $ {140 × (25/100)

= $ 35.

Selling price = (marked price) – (discount)

= $ (140 – 35)

= $ 105.

Gain% = (105 – 100) % = 5%.

Hence, the trader gains 5%.

Solution:5

Cost price of the machine = $ 7660, Gain% = 10%.

Therefore, selling price = [{(100 + gain%)/100} × CP]

= $ [{(100 + 10)/100} × 7660]

= $ [(110/100) × 7660]

= $ 8426.

Let the marked price be $ x.

Then, the discount = 12% of $x

= $ {x × (12/100)}

= $ 3x/25

Therefore, SP = (Marked Price) – (discount)

= $ (x – 3x/25)

= $ 22x/25.

But, the SP = $ 8426.

Therefore, 22x/25 = 8426

⇒ x = (8426 × 25/22)

⇒ x = 9575.

Hence, the marked price of the washing machine is $ 9575

Solution:6

Let the cost price be $ 100.

Gain required = 20%.

Therefore, selling price = $ 120.

Let the marked price be $x.

Then, discount = 25% of $x

= $ (x × 25/100)

= $ x/4

Therefore, selling price = (Marked Price) – (discount)

= $ {x – (x/4)

= $ 3x/4

Therefore, 3x/4 = 120

⇔ x = {120 × (4/3)} = 160

Therefore, marked price = $ 160.

Hence, the marked price is 60% above cost price.

Solution:7

Let the marked price of an article be $ 100.

Then, first discount on it = $ 20.

Price after first discount = $ (100 – 20) = $ 80.

Second discount on it = 10% of $ 80

= $ {80 × (10/100)} = $ 8.

Price after second discount = $ (80 – 8) = $ 72.
Net selling price = $ 72.

Single discount equivalent to given successive discounts = (100 – 72)% = 28%

Solution:8 The easiest way to solve these kinds of problems is to assume a value for the merchant’s cost price.
To make calculations easy, it is best to assume the cost price to be $100.

The merchant marks his goods up by 50%.
Therefore, his marked price (quoted price) = cost price + mark up.
Marked price = $100 + 50% of $100 = 100 + 50 = $150.

The merchant offers a discount of 20% on his marked price.
Discount offered = 20% of 150 = $30.

Therefore, he finally sold his goods for $150 – $30 = $ 120.
We assumed his cost to be $100 and he sold it finally for $120.

Therefore, his profit = $20 on his cost of $ 100.
Hence, his % profit = profit/cost price * 100 = 20/100*100 = 20%.

Solution:9 40% of Rs. 1,00,000 = Rs. 40,000
36% of 1,00,000 = 36000
4% of 36,000 = Rs. 2,560.
Therefore, two successive discounts on Rs. 1,00,000 = 36,000 + 2560 = Rs. 38,560.
Difference between a discount of 40% and two successive discounts of 36% and 4%
= 40,000 – 38,560
= Rs. 1,440

Solution:10 If the selling price of the article is S, then
S – 20% of S = 596
S – S/5 = 596
4S/5 = 596
⇒ S = 596 x 5/4
⇒ S = 745

Level-II

Solution:11Let the prices of two articles be X and Y
From the question 15X/100 = 20Y/100
X/Y = 20/15
Thus the ratio of prices of two articles is 4 : 3
Any two amounts in the ratio 4 : 3 will satisfy the condition.
In the above instance, Rs. 80 and Rs. 60 is the answer.

Solution:12 Formula for Marked Price = 100 x SP/(100 – d%) = 100 x 39/(100 – 2.5%)
= 3900 / 97.5
= Rs. 40.
Marked Price of Article is Rs. 40.

Solution:13 Retailer gets a discount of 40% means he buys it at 60% of the price
60% x 900 = Rs. 540
Profit on selling it at Rs. 900 = 900 – 540 = Rs. 360.
Profit % = (Profit / C.P) x 100 = (360 / 540) x 100 = 66²/₃
Retailer’s Gain percent is 66²/₃

Solution:1410% discount on 720 = Rs. 72
Cost after 1st discount = 720 – 72 = Rs. 648.
Cost after 2nd discount = Rs. 550.80
Therefore 2nd discount = 648 – 550.80 = Rs. 97.20
Discount % = (97.2 x 100)/648 = 15%
Second discount rate = 15%.

Solution:15 Let the cost price be Rs. 100.
M.P. (which is 20% above C.P.) = Rs. 120.
30% discount on Rs. 120 = Rs. 36.
Selling Price = Rs. 120 – 36 = Rs. 84
Cost Price = Rs 100 and Selling Price = Rs 84 {since CP > SP, it is a loss}
Loss% = (16/100) x 100 = 16%.
His net loss percent is 16%.

Solution:16 Assuming the M.P. of each pen to be Rs. 10, the M.P. of 36 pens = Rs. 360
Cost price of 40 pens = Rs. 360 (from the question)
Cost price of each pen = 360/40 = Rs. 9
Selling Price of each pen at a discount of 1% on a marked price of Rs. 10 = 99% x 10 = Rs. 9.90
Profit = 9.90 – 9.00 = Rs. 0.90
Profit % = (0.90/9.00) x 100 = 10%
Profit % = 10%.

Solution:17 The discount is $3.00 and the sale price is $6.00

Solution:18 The rate is 10%. Thus, the customer is paying 90% for the DVD

The sale price is: 0.90 x $15.00 = $13.50

The sale price is $13.50.

BOAT AND STREAM

June 4, 2017October 11, 2024 by PSC Notes

Boat and stream problems is a sub-set of time, speed and distance type questions where in relative speed takes the foremost role. We always find several questions related to the above concept in SSC common graduate level exam as well as in bank PO exam. Upon listing the brief theory of the issue below we move to the various kinds of problems asked in the competitive examination.

Important Formulas – Boats and Streams

Downstream
In running/moving water, the direction along the stream is called downstream.
Upstream
In running/moving water, the direction against the stream is called upstream.

Let the speed of a boat in still water be u km/hr and the speed of the stream be v km/hr, then
Speed downstream = (u+v) km/hr
Speed upstream = (u – v) km/hr

Let the speed downstream be a km/hr and the speed upstream be b km/hr, then
Speed in still water =1/2*(a+b)km/hr
Rate of stream = 1/2*(a−b) km/hr

Some more short-cut methods

Assume that a man can row at the speed of x km/hr in still water and he rows the same distance up and down in a stream which flows at a rate of y km/hr. Then his average speed throughout the journey
= (Speed downstream × Speed upstream)/Speed in still water=((x+y)(x−y))/xkm/hr

Let the speed of a man in still water be x km/hr and the speed of a stream be y km/hr. If he takes t hours more in upstream than to go downstream for the same distance, the distance
=((x* x –y* y)*t)/2ykm

A man rows a certain distance downstream in t₁ hours and returns the same distance upstream in t₂ If the speed of the stream is y km/hr, then the speed of the man in still water
=y((t2+t1) / (t2−t1)) km/hr

A man can row a boat in still water at x km/hr. In a stream flowing at y km/hr, if it takes him t hours to row a place and come back, then the distance between the two places
=t((x* x –y* y))/2xkm

A man takes n times as long to row upstream as to row downstream the river. If the speed of the man is x km/hr and the speed of the stream is y km/hr, then
x=y*((n+1)/(n−1))

Solved Examples

Level 1

1. A man’s speed with the current is 15 km/hr and the speed of the current is 2.5 km/hr. The man’s speed against the current is:
A. 8.5 km/hr	B. 10 km/hr.
C. 12.5 km/hr	D. 9 km/hr

Answer : Option B

Explanation :

Man’s speed with the current = 15 km/hr

=>speed of the man + speed of the current = 15 km/hr

speed of the current is 2.5 km/hr

Hence, speed of the man = 15 – 2.5 = 12.5 km/hr

man’s speed against the current = speed of the man – speed of the current

= 12.5 – 2.5 = 10 km/hr

2. In one hour, a boat goes 14 km/hr along the stream and 8 km/hr against the stream. The speed of the boat in still water (in km/hr) is:
A. 12 km/hr	B. 11 km/hr
C. 10 km/hr	D. 8 km/hr

Answer : Option B

Explanation :

Let the speed downstream be a km/hr and the speed upstream be b km/hr, then

Speed in still water =1/2(a+b) km/hr and Rate of stream =1/2(a−b) km/hr
Speed in still water = 1/2(14+8) kmph = 11 kmph.

3. A boatman goes 2 km against the current of the stream in 2 hour and goes 1 km along the current in 20 minutes. How long will it take to go 5 km in stationary water?
A. 2 hr 30 min	B. 2 hr
C. 4 hr	D. 1 hr 15 min

Answer : Option A

Explanation :

Speed upstream = 2/2=1 km/hr

Speed downstream = 1/(20/60)=3 km/hr

Speed in still water = 1/2(3+1)=2 km/hr

Time taken to travel 5 km in still water = 5/2= 2 hour 30 minutes

4. Speed of a boat in standing water is 14 kmph and the speed of the stream is 1.2 kmph. A man rows to a place at a distance of 4864 km and comes back to the starting point. The total time taken by him is:
A. 700 hours	B. 350 hours
C. 1400 hours	D. 1010 hours

Answer : Option A

Explanation :

Speed downstream = (14 + 1.2) = 15.2 kmph

Speed upstream = (14 – 1.2) = 12.8 kmph

Total time taken = 4864/15.2+4864/12.8 = 320 + 380 = 700 hours

5. The speed of a boat in still water in 22 km/hr and the rate of current is 4 km/hr. The distance travelled downstream in 24 minutes is:
A. 9.4 km	B. 10.2 km
C. 10.4 km	D. 9.2 km

Answer : Option C

Explanation :

Speed downstream = (22 + 4) = 26 kmph

Time = 24 minutes = 24/60 hour = 2/5 hour

distance travelled = Time × speed = (2/5)×26 = 10.4 km

6. A boat covers a certain distance downstream in 1 hour, while it comes back in 1¹⁄₂ hours. If the speed of the stream be 3 kmph, what is the speed of the boat in still water?
A. 14 kmph	B. 15 kmph
C. 13 kmph	D. 12 kmph

Answer : Option B

Explanation :

Let the speed of the boat in still water = x kmph

Given that speed of the stream = 3 kmph

Speed downstream = (x+3) kmph

Speed upstream = (x-3) kmph

He travels a certain distance downstream in 1 hour and come back in 1¹⁄₂ hour.

ie, distance travelled downstream in 1 hour = distance travelled upstream in 1¹⁄₂ hour

since distance = speed × time, we have

(x+3)×1=(x−3)*3/2

=> 2(x + 3) = 3(x-3)

=> 2x + 6 = 3x – 9

=> x = 6+9 = 15 kmph

7. A boat can travel with a speed of 22 km/hr in still water. If the speed of the stream is 5 km/hr, find the time taken by the boat to go 54 km downstream
A. 5 hours	B. 4 hours
C. 3 hours	D. 2 hours

Answer : Option D

Explanation :

Speed of the boat in still water = 22 km/hr

speed of the stream = 5 km/hr

Speed downstream = (22+5) = 27 km/hr

Distance travelled downstream = 54 km

Time taken = distance/speed=54/27 = 2 hours

8. A boat running downstream covers a distance of 22 km in 4 hours while for covering the same distance upstream, it takes 5 hours. What is the speed of the boat in still water?
A. 5 kmph	B. 4.95 kmph
C. 4.75 kmph	D. 4.65

Answer : Option B

Explanation :

Speed downstream = 22/4 = 5.5 kmph

Speed upstream = 22/5 = 4.4 kmph

Speed of the boat in still water = (½) x (5.5+4.42) = 4.95 kmph

9. A man takes twice as long to row a distance against the stream as to row the same distance in favor of the stream. The ratio of the speed of the boat (in still water) and the stream is:
A. 3 : 1	B. 1 : 3
C. 1 : 2	D. 2 : 1

Answer : Option A

Explanation :

Let speed upstream = x

Then, speed downstream = 2x

Speed in still water = (2x+x)2=3x/2

Speed of the stream = (2x−x)2=x/2

Speed of boat in still water: Speed of the stream = 3x/2:x/2 = 3 : 1

Level 2

1. A motorboat, whose speed in 15 km/hr in still water goes 30 km downstream and comes back in a total of 4 hours 30 minutes. The speed of the stream (in km/hr) is:
A. 10	B. 6
C. 5	D. 4

Answer : Option C

Explanation :

Speed of the motor boat = 15 km/hr

Let speed of the stream = v

Speed downstream = (15+v) km/hr

Speed upstream = (15-v) km/hr

Time taken downstream = 30/(15+v)

Time taken upstream = 30/(15−v)

total time = 30/(15+v)+30/(15−v)

It is given that total time is 4 hours 30 minutes = 4.5 hour = 9/2 hour

i.e., 30/(15+v)+30/(15−v)=9/2

⇒1(15+v)+1(15−v)=(9/2)×30=3/20

⇒(15−v+15+v)/(15+v)(15−v)=3/20

⇒30/(15*15−v*v)=3/20

⇒30/(225−v*v)=3/20

⇒10/(225−v* v)=1/20

⇒225−v* v =200

⇒v* v =225−200=25

⇒v=5 km/hr

2. A man rows to a place 48 km distant and come back in 14 hours. He finds that he can row 4 km with the stream in the same time as 3 km against the stream. The rate of the stream is:
A. 1 km/hr.	B. 2 km/hr.
C. 1.5 km/hr.	D. 2.5 km/hr.

Answer : Option A

Explanation :

Assume that he moves 4 km downstream in x hours

Then, speed downstream = distance/time=4/x km/hr

Given that he can row 4 km with the stream in the same time as 3 km against the stream

i.e., speed upstream = 3/4of speed downstream=> speed upstream = 3/x km/hr

He rows to a place 48 km distant and come back in 14 hours

=>48/(4/x)+48/(3/x)=14

==>12x+16x=14

=>6x+8x=7

=>14x=7

=>x=1/2

Hence, speed downstream = 4/x=4/(1/2) = 8 km/hr

speed upstream = 3/x=3/(1/2) = 6 km/hr

Now we can use the below formula to find the rate of the stream

Let the speed downstream be a km/hr and the speed upstream be b km/hr, then

Speed in still water =1/2*(a+b) km/hr

Rate of stream =12*(a−b) km/hr
Hence, rate of the stream = ½*(8−6)=1 km/hr

3. A boat running upstream takes 8 hours 48 minutes to cover a certain distance, while it takes 4 hours to cover the same distance running downstream. What is the ratio between the speed of the boat and speed of the water current respectively?

A. 5 : 6

B. 6 : 5

C. 8 : 3

D. 3 : 8

Answer : Option C

Explanation :

Let the rate upstream of the boat = x kmph

and the rate downstream of the boat = y kmph

Distance travelled upstream in 8 hrs 48 min = Distance travelled downstream in 4 hrs.

Since distance = speed × time, we have

x×(8*4/5)=y×4

x×(44/5)=y×4

x×(11/5)=y— (equation 1)

Now consider the formula given below

Let the speed downstream be a km/hr and the speed upstream be b km/hr, then

Speed in still water =1/2(a+b) km/hr

Rate of stream =1/2(a−b) km/hr
Hence, speed of the boat = (y+x)/2

speed of the water = (y−x)/2

Required Ratio = (y+x)/2:(y−x)/2=(y+x):(y−x)=(11x/5+x):(11x/5−x)

(Substituted the value of y from equation 1)

=(11x+5x):(11x−5x)=16x:6x=8:3

4. A man can row at 5 kmph in still water. If the velocity of current is 1 kmph and it takes him 1 hour to row to a place and come back, how far is the place?
A. 3.2 km	B. 3 km
C. 2.4 km	D. 3.6 km

Answer : Option C

Explanation :

Speed in still water = 5 kmph
Speed of the current = 1 kmph

Speed downstream = (5+1) = 6 kmph
Speed upstream = (5-1) = 4 kmph

Let the requited distance be x km

Total time taken = 1 hour

=>x/6+x/4=1

=> 2x + 3x = 12

=> 5x = 12

=> x = 2.4 km

5. A man can row three-quarters of a kilometer against the stream in 11¹⁄₄ minutes and down the stream in 7¹⁄₂minutes. The speed (in km/hr) of the man in still water is:
A. 4 kmph	B. 5 kmph
C. 6 kmph	D. 8 kmph

Answer : Option B

Explanation :

Distance = 3/4 km

Time taken to travel upstream = 11¹⁄₄ minutes

= 45/4 minutes = 45/(4×60) hours = 3/16 hours

Speed upstream = Distance/Time= (3/4)/ (3/16) = 4 km/hr

Time taken to travel downstream = 7¹⁄₂minutes = 15/2 minutes = 15/2×60 hours = 1/8 hours

Speed downstream = Distance/Time= (3/4)/ (1/8) = 6 km/hr

Rate in still water = (6+4)/2=10/2=5 kmph

6. A boat takes 90 minutes less to travel 36 miles downstream than to travel the same distance upstream. If the speed of the boat in still water is 10 mph, the speed of the stream is:
A. 4 mph	B. 2.5 mph
C. 3 mph	D. 2 mph

Answer : Option D

Explanation :

Speed of the boat in still water = 10 mph

Let speed of the stream be x mph

Then, speed downstream = (10+x) mph

speed upstream = (10-x) mph

Time taken to travel 36 miles upstream – Time taken to travel 36 miles downstream= 90/60 hours

=>36/(10−x)−36/(10+x)=3/2=>12/(10−x)−12/(10+x)=1/2=>24(10+x)−24(10−x)=(10+x)(10−x)

=>240+24x−240+24x=(100−x* x)=>48x=100− (x* x)=> x* x +48x−100=0

=>(x+50)(x−2)=0=>x = -50 or 2; Since x cannot be negative, x = 2 mph

7. At his usual rowing rate, Rahul can travel 12 miles downstream in a certain river in 6 hours less than it takes him to travel the same distance upstream. But if he could double his usual rowing rate for his 24-mile round trip, the downstream 12 miles would then take only one hour less than the upstream 12 miles. What is the speed of the current in miles per hour?
A. 2*1/3 mph	B. 1*1/3 mph
C. 1*2/3 mph	D. 2*2/3 mph

Answer : Option D

Explanation :

Let the speed of Rahul in still water be x mph
and the speed of the current be y mph

Then, Speed upstream = (x – y) mph
Speed downstream = (x + y) mph

Distance = 12 miles

Time taken to travel upstream – Time taken to travel downstream = 6 hours

⇒12/(x−y)−12/(x+y)=6

⇒12(x+y)−12(x−y)=6(x*x−y*y)

⇒24y=6(x*x−y*y)

⇒4y= x*x−y*y

⇒x * x =(y* y +4y)⋯(Equation 1)

Now he doubles his speed. i.e., his new speed = 2x

Now, Speed upstream = (2x – y) mph

Speed downstream = (2x + y) mph

In this case, Time taken to travel upstream – Time taken to travel downstream = 1 hour

⇒12/(2x−y)−12/(2x+y)=1

⇒12(2x+y)−12(2x−y)=4*x* x –y* y

⇒24y=4*x* x –y* y

⇒4*x* x = y* y +24y⋯(Equation 2)

(Equation 1 × 4)⇒4x* x =4(y* y +4y)⋯(Equation 3)

(From Equation 2 and 3, we have)

y* y +24y=4(y* y +4y)⇒y* y +24y=4y* y +16y⇒3y* y =8y⇒3y=8

y=8/3 mphi.e., speed of the current = 8/3 mph=2*2/3 mph

CLASSIFICATION LEVEL 1

June 4, 2017October 11, 2024 by PSC Notes

Classification involves putting things into a class or group according to particular characteristics so it’s easier to make sense of them, whether you’re organizing your shoes, your stock portfolio, or a group of invertebrates. From all competitive examination classification is one of the most important topics, this pattern come with lot of questions minimum they asking the 4 to 5 question from the classification topic. In the SSC CGL or SSC constable GD examination having the same topics from the reasoning section but the standard of the topic will be different, so most of the candidates preference for this topic to get the best score in the written examination.

Directions: Find the odd one out

A. Square B. Circle C. Rectangle D. Triangle

A. Cotton B. Terene C. Silk D. Wool

A. Light B. Wave C. Heat D. Sound

A. 81 : 243 B. 16 :64 C. 64 : 192 D. 25 : 75

A. 64 : 8 B. 80 : 9 C. 7 : 49 D. 36 : 6

A. 26 : 62 B. 36 : 63 C. 46 : 64 D. 56 : 18

A. ABZY B. BCYX C. CDVW D. DEVU

A. ACE B. FHJ C. KLM D. SUW

Find the wrong number in the series

441, 484, 529, 566, 625

484 B. 529 C. 625 D. 566

Find the wrong number in the series

232, 343, 454, 564, 676

676 B. 454 C. 343 D. 564

SOLUTION TO CLASSIFICATION LEVEL 1

B. Except circle, all others are geometrical figures consisting straight lines.

B. Except terene, all others are natural fibres.

B. Except wave, all others are different form of energy.

B. 81*3=243

64*3=192

25*3=75

But 16*4=64

D. Except D, in each pair one number is square root of the other.

D. Except D, in each pair the position of digits has been interchanged.

C. A+1=B & Z-1=Y

B+1=C & Y-1=X

D+1=E & V-1=U

But C+1=D & V+1=W

C. A+2=C & C+2=E

F+2=H & H+2=J

But K+1=L & L+1=M

D. 21^2=441

22^2=484

23^2=529

25^2=625

But (23.79)^2=566

D. 232+111=343

343+111=454

454+111=565 (but given 564)

MENSURATION

June 4, 2017October 10, 2024 by PSC Notes

Mensuration is the branch of mathematics which deals with the study of different geometrical shapes, their areas and Volume. In the broadest sense, it is all about the process of measurement. It is based on the use of algebraic equations and geometric calculations to provide measurement data regarding the width, depth and volume of a given object or group of objects

Pythagorean Theorem (Pythagoras’ theorem)

In a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides

c² = a² + b² where c is the length of the hypotenuse and a and b are the lengths of the other two sides

Pi is a mathematical constant which is the ratio of a circle’s circumference to its diameter. It is denoted by π

π≈3.14≈227

Geometric Shapes and solids and Important Formulas

Geometric Shapes

Description

Formulas

Rectangle

l = Length

b = Breadth

d= Length of diagonal

Area = lb

Perimeter = 2(l + b)

d = √l2+b2

Square

a = Length of a side

d= Length of diagonal

Area= a*a=1/2*d*d

Perimeter = 4a

d = 2√a

Parallelogram

b and c are sides

b = base

h = height

Area = bh

Perimeter = 2(b + c)

Rhombus

a = length of each side

b = base

h = height

d1, d2 are the diagonal

Area = bh(Formula 1)

Area = ½*d1*d2 (Formula 2 )

Perimeter = 4a

Triangle

a , b and c are sides

b = base

h = height

Area = ½*b*h (Formula 1) Area(Formula 2) = √S(S−a)(S−b)(S−c where S is the semiperimeter
S =(a+b+c)/2 (Formula 2 for area -Heron’s formula) Perimeter = a + b + c

Radius of incircle of a triangle of area A =AS
where S is the semiperimeter
=(a+b+c)/2

Equilateral Triangle

a = side

Area = (√3/4)*a*a Perimeter = 3a

Radius of incircle of an equilateral triangle of side a = a/2*√3

Radius of circumcircle of an equilateral triangle
of side a = a/√3

Base a is parallel to base b

Trapezium(Trapezoid in American English)

h = height

Area = 12(a+b)h

Circle

r = radius

d = diameter

d = 2r

Area = πr2 = 14πd2

Circumference = 2πr = πd

Sector of Circle

r = radius

θ = central angle

Area = (θ/360) *π*r*r
Arc Length, s = (θ/180)* π*r

In the radian system for angular measurement,
2π radians = 360°
=> 1 radian = 180°π
=> 1° = π180 radians
Hence,
Angle in Degrees
= Angle in Radians × 180°π
Angle in Radians
= Angle in Degrees × π180°

Ellipse

Major axis length = 2a

Minor axis length = 2b

Area = πab

Perimeter ≈

Rectangular Solid

l = length

w = width

h = height

Total Surface Area
= 2lw + 2wh + 2hl
= 2(lw + wh + hl)

Volume = lwh

Cube

s = edge

Total Surface Area = 6s²

Volume = s³

Right Circular Cylinder

h = height

r = radius of base

Lateral Surface Area
= (2 π r)h

Total Surface Area
= (2 π r)h + 2 (π r²)

Volume = (π r²)h

Pyramid

h = height

B = area of the base

Total Surface Area = B + Sum of the areas of the triangular sides

Volume = 1/3*B*h

Right Circular Cone

h = height

r = radius of base

Lateral Surface Area=πrs
where s is the slant height =√r*r+h*h
Total Surface Area
=πrs+πr2

Sphere

r = radius

d = diameter

d = 2r

Surface Area =4πr*r=πd*d

Volume =4/3πr*r*r=16πd*d*d

Important properties of Geometric Shapes
1. Properties of Triangle
  1. Sum of the angles of a triangle = 180°
  2. Sum of any two sides of a triangle is greater than the third side.

The line joining the midpoint of a side of a triangle to the positive vertex is called the median

The median of a triangle divides the triangle into two triangles with equal areas
Centroid is the point where the three medians of a triangle meet.
Centroid divides each median into segments with a 2:1 ratio

Area of a triangle formed by joining the midpoints of the sides of a given triangle is one-fourth of the area of the given triangle.
An equilateral triangle is a triangle in which all three sides are equal

In an equilateral triangle, all three internal angles are congruent to each other
In an equilateral triangle, all three internal angles are each 60°
An isosceles triangle is a triangle with (at least) two equal sides

In isosceles triangle, altitude from vertex bisects the base.

Properties of Quadrilaterals
Rectangle
1. The diagonals of a rectangle are equal and bisect each other
2. opposite sides of a rectangle are parallel

opposite sides of a rectangle are congruent

opposite angles of a rectangle are congruent
All four angles of a rectangle are right angles
The diagonals of a rectangle are congruent
Square

All four sides of a square are congruent
Opposite sides of a square are parallel

The diagonals of a square are equal
The diagonals of a square bisect each other at right angles
All angles of a square are 90 degrees.

A square is a special kind of rectangle where all the sides have equal length

Parallelogram

The opposite sides of a parallelogram are equal in length.
The opposite angles of a parallelogram are congruent (equal measure).

The diagonals of a parallelogram bisect each other.

Each diagonal of a parallelogram divides it into two triangles of the same area

Rhombus

All the sides of a rhombus are congruent
Opposite sides of a rhombus are parallel.
The diagonals of a rhombus bisect each other at right angles

Opposite internal angles of a rhombus are congruent (equal in size)

Any two consecutive internal angles of a rhombus are supplementary; i.e. the sum of their angles = 180° (equal in size)
If each angle of a rhombus is 90°, it is a square

Other properties of quadrilaterals

The sum of the interior angles of a quadrilateral is 360 degrees
If a square and a rhombus lie on the same base, area of the square will be greater than area of the rhombus (In the special case when each angle of the rhombus is 90°, rhombus is also a square and therefore areas will be equal)
A parallelogram and a rectangle on the same base and between the same parallels are equal in area.
Of all the parallelogram of given sides, the parallelogram which is a rectangle has the greatest area.
Each diagonal of a parallelogram divides it into two triangles of the same area
A square is a rhombus and a rectangle.

Sum of Interior Angles of a polygon
1. The sum of the interior angles of a polygon = 180(n – 2) degrees where n = number of sides Example 1 : Number of sides of a triangle = 3. Hence, sum of the interior angles of a triangle = 180(3 – 2) = 180 × 1 = 180 ° Example 2 : Number of sides of a quadrilateral = 4. Hence, sum of the interior angles of any quadrilateral = 180(4 – 2) = 180 × 2 = 360.

Solved Examples

Level 1

An error 2% in excess is made while measuring the side of a square. What is the percentage of error in the calculated area of the square?
4.04 %
2.02 %
4 %
2 %

Answer : Option A

Explanation :

Error = 2% while measuring the side of a square.

Let the correct value of the side of the square = 100
Then the measured value = (100×(100+2))/100=102 (∵ error 2% in excess)

Correct Value of the area of the square = 100 × 100 = 10000
Calculated Value of the area of the square = 102 × 102 = 10404

Error = 10404 – 10000 = 404
Percentage Error = (Error/Actual Value)×100=(404/10000)×100=4.04%

A towel, when bleached, lost 20% of its length and 10% of its breadth. What is the percentage of decrease in area?
30 %
28 %
32 %
26 %

Answer : Option B

Explanation :
Let original length = 100 and original breadth = 100
Then original area = 100 × 100 = 10000

Lost 20% of length
=> New length =( Original length × (100−20))/100
=(100×80)/100=80

Lost 10% of breadth
=> New breadth= (Original breadth × (100−10))/100
=(100×90)/100=90

New area = 80 × 90 = 7200

Decrease in area
= Original Area – New Area
= 10000 – 7200 = 2800

Percentage of decrease in area
=(Decrease in Area/Original Area)×100=(2800/10000)×100=28%

If the length of a rectangle is halved and its breadth is tripled, what is the percentage change in its area?
25 % Increase
25 % Decrease
50 % Decrease
50 % Increase

Answer : Option D

Explanation :
Let original length = 100 and original breadth = 100
Then original area = 100 × 100 = 10000

Length of the rectangle is halved
=> New length = (Original length)/2=100/2=50

breadth is tripled
=> New breadth= Original breadth × 3 = 100 × 3 = 300

New area = 50 × 300 = 15000

Increase in area = New Area – Original Area = 15000 – 10000= 5000
Percentage of Increase in area =( Increase in Area/OriginalArea)×100=(5000/10000)×100=50%

The area of a rectangle plot is 460 square metres. If the length is 15% more than the breadth, what is the breadth of the plot?
14 metres
20 metres
18 metres
12 metres

Answer : Option B

Explanation:

lb = 460 m² ——(Equation 1)

Let the breadth = b
Then length, l =( b×(100+15))/100=115b/100——(Equation 2)

From Equation 1 and Equation 2,
115b/100×b=460b2=46000/115=400⇒b=√400=20 m

If a square and a rhombus stand on the same base, then what is the ratio of the areas of the square and the rhombus?
equal to ½
equal to ¾
greater than 1
equal to 1

Answer : Option C

Explanation :

If a square and a rhombus lie on the same base, area of the square will be greater than area of the rhombus (In the special case when each angle of the rhombus is 90°, rhombus is also a square and therefore areas will be equal)

Hence greater than 1 is the more suitable choice from the given list

================================================================
Note : Proof

Consider a square and rhombus standing on the same base ‘a’. All the sides of a square are of equal length. Similarly all the sides of a rhombus are also of equal length. Since both the square and rhombus stands on the same base ‘a’,

Length of each side of the square = a
Length of each side of the rhombus = a

Area of the sqaure = a² …(1)

From the diagram, sin θ = h/a
=> h = a sin θ

Area of the rhombus = ah = a × a sin θ = a² sin θ …(2)

From (1) and (2)

Area of the square/Area of the rhombus= a² /a²sinθ=1/sinθ

Since 0° < θ < 90°, 0 < sin θ < 1. Therefore, area of the square is greater than that of rhombus, provided both stands on same base.

(Note that, when each angle of the rhombus is 90°, rhombus is also a square (can be considered as special case) and in that case, areas will be equal.

The breadth of a rectangular field is 60% of its length. If the perimeter of the field is 800 m, find out the area of the field.
37500 m²
30500 m²
32500 m²
40000 m²

Answer : Option A

Explanation :

Given that breadth of a rectangular field is 60% of its length
⇒b=(60/100)* l =(3/5)* l

perimeter of the field = 800 m
=> 2 (l + b) = 800
⇒2(l+(3/5)* l)=800⇒l+(3/5)* l =400⇒(8/5)* l =400⇒l/5=50⇒l=5×50=250 m

b = (3/5)* l =(3×250)/5=3×50=150 m

Area = lb = 250×150=37500 m2

What is the percentage increase in the area of a rectangle, if each of its sides is increased by 20%?
45%
44%
40%
42%

Answer : Option B

Explanation :
Let original length = 100 and original breadth = 100
Then original area = 100 × 100 = 10000

Increase in 20% of length.
=> New length = (Original length ×(100+20))/100=(100×120)/100=120

Increase in 20% of breadth
=> New breadth= (Original breadth × (100+20))/100=(100×120)/100=120

New area = 120 × 120 = 14400

Increase in area = New Area – Original Area = 14400 – 10000 = 4400
Percentage increase in area =( Increase in Area /OriginalArea)×100=(4400/10000)×100=44%

What is the least number of squares tiles required to pave the floor of a room 15 m 17 cm long and 9 m 2 cm broad?
814
802
836
900

Answer : Option A

Explanation :

l = 15 m 17 cm = 1517 cm
b = 9 m 2 cm = 902 cm
Area = 1517 × 902 cm²

Now we need to find out HCF(Highest Common Factor) of 1517 and 902.
Let’s find out the HCF using long division method for quicker results

902) 1517 (1

-902

—————–

615) 902 (1

————–

287) 615 (2

-574

—————–

41) 287 (7

-287

————

0
————

Hence, HCF of 1517 and 902 = 41

Hence, side length of largest square tile we can take = 41 cm
Area of each square tile = 41 × 41 cm²

Number of tiles required = (1517×902)/(41×41)=37×22=407×2=814

Level 2

A rectangular parking space is marked out by painting three of its sides. If the length of the unpainted side is 9 feet, and the sum of the lengths of the painted sides is 37 feet, find out the area of the parking space in square feet?
126 sq. ft.
64 sq. ft.
100 sq. ft.
102 sq. ft.

Answer : Option A

Explanation :

Let l = 9 ft.

Then l + 2b = 37
=> 2b = 37 – l = 37 – 9 = 28
=> b = 282 = 14 ft.

Area = lb = 9 × 14 = 126 sq. ft.

A large field of 700 hectares is divided into two parts. The difference of the areas of the two parts is one-fifth of the average of the two areas. What is the area of the smaller part in hectares?
400
365
385
315

Answer : Option D

Explanation :

Let the areas of the parts be x hectares and (700 – x) hectares.

Difference of the areas of the two parts = x – (700 – x) = 2x – 700

one-fifth of the average of the two areas = 15[x+(700−x)]2
=15×7002=3505=70

Given that difference of the areas of the two parts = one-fifth of the average of the two areas
=> 2x – 700 = 70
=> 2x = 770
⇒x=7702=385

Hence, area of smaller part = (700 – x) = (700 – 385) = 315 hectares.

The length of a rectangle is twice its breadth. If its length is decreased by 5 cm and breadth is increased by 5 cm, the area of the rectangle is increased by 75 sq.cm. What is the length of the rectangle?
18 cm
16 cm
40 cm
20 cm

Answer : Option C

Explanation :

Let breadth = x cm
Then length = 2x cm
Area = lb = x × 2x = 2x²

New length = (2x – 5)
New breadth = (x + 5)
New Area = lb = (2x – 5)(x + 5)

But given that new area = initial area + 75 sq.cm.
=> (2x – 5)(x + 5) = 2x² + 75
=> 2x² + 10x – 5x – 25 = 2x² + 75
=> 5x – 25 = 75
=> 5x = 75 + 25 = 100
=> x = 1005 = 20 cm

Length = 2x = 2 × 20 = 40cm

The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then what is the area of the park (in sq. m)?
142000
112800
142500
153600

Answer : Option D

Explanation :

l : b = 3 : 2 —-(Equation 1)

Perimeter of the rectangular park
= Distance travelled by the man at the speed of 12 km/hr in 8 minutes
= speed × time = 12×860 (∵ 8 minute = 860 hour)
= 85 km = 85 × 1000 m = 1600 m

Perimeter = 2(l + b)

=> 2(l + b) = 1600
=> l + b = 16002 = 800 m —-(Equation 2)

From (Equation 1) and (Equation 2)
l = 800 × 35 = 480 m
b = 800 × 25 = 320 m (Or b = 800 – 480 = 320m)

Area = lb = 480 × 320 = 153600 m²

It is decided to construct a 2 metre broad pathway around a rectangular plot on the inside. If the area of the plots is 96 sq.m. and the rate of construction is Rs. 50 per square metre., what will be the total cost of the construction?
Rs.3500
Rs. 4200
Insufficient Data
Rs. 4400

Answer : Option C

Explanation :
Let length and width of the rectangular plot be l and b respectively
Total area of the rectangular plot = 96 sq.m.
=> lb = 96

Width of the pathway = 2 m
Length of the remaining area in the plot = (l – 4)
breadth of the remaining area in the plot = (b – 4)
Area of the remaining area in the plot = (l – 4)(b – 4)

Area of the pathway
= Total area of the rectangular plot – remaining area in the plot
= 96 – [(l – 4)(b – 4)] = 96 – [lb – 4l – 4b + 16] = 96 – [96 – 4l – 4b + 16] = 96 – 96 + 4l + 4b – 16
= 4l + 4b – 16
= 4(l + b) – 16

We do not know the values of l and b and hence area of the pathway cannot be found out. So we cannot determine total cost of the construction.

A circle is inscribed in an equilateral triangle of side 24 cm, touching its sides. What is the area of the remaining portion of the triangle?
144√3−48π cm²
121√3−36π cm²
144√3−36π cm²
121√3−48π cm²

Answer : Option A

Explanation :
Area of an equilateral triangle = (3/√4)*a *a where a is length of one side of the equilateral triangle
Area of the equilateral Δ ABC = (3/√4)*a *a = (3/√4)*24*24=144√3 cm2⋯ (1)

Area of a triangle = 12bhwhere b is the base and h is the height of the triangle
Let r = radius of the inscribed circle. Then
Area of Δ ABC
= Area of Δ OBC + Area of Δ OCA + area of Δ OAB
= (½ × r × BC) + (½ × r × CA) + (½ × r × AB)
= ½ × r × (BC + CA + AB)
= ½ x r x (24 + 24 + 24)
= ½ x r x 72 = 36r cm² —-(2)

From (1) and (2),
144√3=36r⇒r=144√3/36=4√3−−−−(3)

Area of a circle = πr2 where = radius of the circle
From (3), the area of the inscribed circle = πr2=π(4√3)* (4√3)=48π⋯(4)

Hence, area of the remaining portion of the triangle
= Area of Δ ABC – Area of inscribed circle
144√3−48π cm2

What will be the length of the longest rod which can be placed in a box of 80 cm length, 40 cm breadth and 60 cm height?
√11600 cm
√14400 cm
√10000 cm
√12040 cm

Answer : Option A

Explanation :
The longest road which can fit into the box will have one end at A and other end at G (or any other similar diagonal).
Hence the length of the longest rod = AG

Initially let’s find out AC. Consider the right angled triangle ABC

AC² = AB² + BC² = 40² + 80² = 1600 + 6400 = 8000
⇒AC = √8000 cm

Consider the right angled triangle ACG

AG² = AC² + CG²
(√8000)²+60²=8000+3600=11600
=> AG = √11600 cm
=> Length of the longest rod = √11600cm

A rectangular plot measuring 90 metres by 50 metres needs to be enclosed by wire fencing such that poles of the fence will be kept 5 metres apart. How many poles will be needed?
30
44
56
60

Answer : Option C

Explanation :

Perimeter of a rectangle = 2(l + b)
where l is the length and b is the breadth of the rectangle

Length of the wire fencing = perimeter = 2(90 + 50) = 280 metres
Two poles will be kept 5 metres apart. Also remember that the poles will be placed along the perimeter of the rectangular plot, not in a single straight line which is very important.
Hence number of poles required = 280/5 = 56

PARTNERSHIP

June 4, 2017October 8, 2024 by PSC Notes

Partnership :

Partnership is an association of two or more parties, they put money for business.

Simple Partnership:

Simple partnership is one in which the capitals of the partners are invested for the same time. The profit or losses are divided among the partners in the ratio of their investments.

Compound Partnership:

Compound Partnership is one which the capitals of the partners are invested for different periods. In such cases equivalent capitals are calculated for a unit time by multiplying the capital with the number of units of time. The profits or losses are then divided in the ratio of these equivalent capitals. Tus the ratio of profits is directly proportional to both capital invested as time.

Working partner:

A partner who participates in the working and manages the business is called a Working Partner.

Sleeping Partner:

A partner who only invests capital but does not participate in the working of the business is called a Sleeping Partner.

Division of Profit and Loss:

1. Rule :When investment of all partners are for the same time, the loss or profit is distributed among partners in the ratio of investment.
Ex. Let P and Q invested Rs. a and b for one year in a business then share of profit and loss be ,

P’s share of profit : Q’s share profit = a : b

2.Rule : When investments are for different time period, then profit ratio is calculated as capital multiplied by length of investment

Ex. P’s share of profit : Q’s share profit = a* t1 : b* t2

Questions with solutions

Level-I

A, B and C enter into a partnership. They invest Rs. 40,000, Rs. 80,000 and Rs. 1,20,000 respectively. At the end of the first year, B withdrawns Rs. 40,000, while at the end of the second year, C withdraws Rs. 80,000. In what ratio will the profit be shared at the end of 3 years ?

Solution: A : B : C = (40,000 X 36) : (80,000 X 12 + 40,000 X 24) : (120,000 X 24 + 40,000 X 12) = 3: 4: 16

A, B, C enter into a partnership investing Rs. 35,000, Rs.45,000 and Rs.55,000 respectively. The respective shares of A, B, C in an annual profit of Rs.40,500 are ?

Solution : A : B : C = 35000 : 45000 : 55000 = 7 : 9 : 11.

A’s share = Rs (40500 x 7/27) = Rs. 10500

B’s share = Rs.(40500× 9/27) = Rs. 13500

C’s share = Rs.(40500×11/27)= Rs. 16500

In a business, Lucky invests Rs. 35,000 for 8 months and manju invests Rs 42,000 for 10 months. Out of a profit of Rs. 31,570. Manju’s share is 😕

Solution : lucky: Manju = (35000 X 8) : (42,000 X 10) = 2:3
Manju’s share = Rs.3/5×31570 = Rs. 18,942

Amar started a business investing Rs. 70,000. Ramki joined him after six months with an amount of Rs. 1,05,000 and Sagar joined them with Rs. 1.4 lakhs after another six months. The amount of profit earned should be distributed in what ratio among Aman, Rakhi and Sagar respectively, 3 years after Aman started the business ?

Solution: Amar : Ramki : Sagar =

(70000 X 36) : (105000 X 30) : (140000 X24) = 12 : 15 : 16.

5 . A begins a business with Rs 450 and is joined afterwards by B with Rs 300. After how many months does B join if the profits at the end of the year is divided in the ratio 2 : 1?

Solution.-.(B) Suppose B joins for x months.

Then, 450 ´12 = 2
300 ´ x 1

x =450× 6
300

x= 9 months

\B joins after (12 – 9) = 3 months.

Shekhar started a business investing Rs. 25,000 in 1999. In 2000, he invested an additional amount of Rs. 10,000 and Rajeev joined him with an amount of Rs. 35,000. In 2001, Shekhar invested another additional amount of Rs. 10,000 and Jatin joined them with an amount of Rs. 35,000. What will be Rajeev’s share in the profit of Rs. 1,50,000 earned at the end of 3 years from the start of the business in 1999?.

Solution : Shekhar : Rajeev : Jatin =

(25000 X 12 + 35000 X 12 + 45000 X 12) : (35000 X 24) :   (35000 X 12)
= 1260000   : 840000 : 420000 =   3 : 2 : 1.
Rajeev’s share   = Rs.(150000×26) =   Rs. 50000

A,B and C started a business with Rs.15000, Rs.25000 and Rs.35000 respectively. A was paid 10% of the total profit as a salary and the balance was divided in the ration of investment. If A’s share is Rs.4,200, then C’s share is: ?

Solution : A, B and C must divide their salaries in the ratio :

15,000 : 25,000:35,000 = 3:5:7
Assume total Profit = 100X.

then A share is 10% of 100X for managing business and 3/15 part of 90X for his investment (as the remaining profit is (100X – 10X = 90X)
So total A’s share = 10X + 315 × 90X = 4,200
⇒X = 150
Substituting X = 150 in 90X we get remaining profit for sharing. That is Rs.13,500
Now C’s share = 715×13,500 = Rs.6,300

Level-II

A and B invest in a business in the ratio 3 : 2. If 5% of the total profit goes to charity and A’s share is Rs. 855, the total profit is:

A.	Rs. 1425
B.	Rs. 1500
C.	Rs. 1537.50
D.	Rs. 1576

Answer:1 Option B

Explanation:

Let the total profit be Rs. 100.

After paying to charity, A’s share = Rs.		95 x	3		= Rs. 57.
			5

If A’s share is Rs. 57, total profit = Rs. 100.

If A’s share Rs. 855, total profit =		100	x 855		= 1500

A, B and C jointly thought of engaging themselves in a business venture. It was agreed that A would invest Rs. 6500 for 6 months, B, Rs. 8400 for 5 months and C, Rs. 10,000 for 3 months. A wants to be the working member for which, he was to receive 5% of the profits. The profit earned was Rs. 7400. Calculate the share of B in the profit.

A.	Rs. 1900
B.	Rs. 2660
C.	Rs. 2800
D.	Rs. 2840

Answer: 2 Option B

Explanation:

For managing, A received = 5% of Rs. 7400 = Rs. 370.

Balance = Rs. (7400 – 370) = Rs. 7030.

Ratio of their investments = (6500 x 6) : (8400 x 5) : (10000 x 3)

= 39000 : 42000 : 30000

= 13 : 14 : 10

B’s share = Rs.		7030 x	14		= Rs. 2660.
			37

3 .A, B and C enter into a partnership in the ratio : : . After 4 months, A increases his share 50%. If the total profit at the end of one year be Rs. 21,600, then B’s share in the profit is:

A.	Rs. 2100
B.	Rs. 2400
C.	Rs. 3600
D.	Rs. 4000

Answer:3 Option D

Explanation:

Ratio of initial investments =		7	:	4	:	6		= 105 : 40 : 36.
		2		3		5

Let the initial investments be 105x, 40x and 36x.

A : B : C =		105x x 4 +	150	x 105x x 8		: (40x x 12) : (36x x 12)
			100

= 1680x : 480x : 432x = 35 : 10 : 9.

Hence, B’s share = Rs.		21600 x	10		= Rs. 4000.
			54

A, B, C subscribe Rs. 50,000 for a business. A subscribes Rs. 4000 more than B and B Rs. 5000 more than C. Out of a total profit of Rs. 35,000, A receives:

A.	Rs. 8400
B.	Rs. 11,900
C.	Rs. 13,600
D.	Rs. 14,700

Answer:4 Option D

Explanation:

Let C = x.

Then, B = x + 5000 and A = x + 5000 + 4000 = x + 9000.

So, x + x + 5000 + x + 9000 = 50000

3x = 36000

x = 12000

A : B : C = 21000 : 17000 : 12000 = 21 : 17 : 12.

A’s share = Rs.		35000 x	21		= Rs. 14,700.
			50

Three partners shared the profit in a business in the ratio 5 : 7 : 8. They had partnered for 14 months, 8 months and 7 months respectively. What was the ratio of their investments?

A.	5 : 7 : 8
B.	20 : 49 : 64
C.	38 : 28 : 21
D.	None of these

Answer:5 Option B

Explanation:

Let their investments be Rs. x for 14 months, Rs. y for 8 months and Rs. z for 7 months respectively.

Then, 14x : 8y : 7z = 5 : 7 : 8.

Now,	14x	=	5	98x = 40y y =	49	x
	8y		7		20

And,	14x	=	5	112x = 35z z =	112	x =	16	x.
	7z		8		35		5

x : y : z = x :	49	x	:	16	x	= 20 : 49 : 64.
	20			5

A starts business with Rs. 3500 and after 5 months, B joins with A as his partner. After a year, the profit is divided in the ratio 2 : 3. What is B’s contribution in the capital?

A.	Rs. 7500
B.	Rs. 8000
C.	Rs. 8500
D.	Rs. 9000

Answer:6 Option D

Explanation:

Let B’s capital be Rs. x.

Then,		3500 x 12	=	2
		7x		3

14x = 126000

x = 9000.

A and B entered into partnership with capitals in the ratio 4 : 5. After 3 months, A withdrew of his capital and B withdrew of his capital. The gain at the end of 10 months was Rs. 760. A’s share in this profit is:

A.	Rs. 330
B.	Rs. 360
C.	Rs. 380
D.	Rs. 430

Answer:7 Option A

Explanation:

A : B =

4x x 3 +

4x –

x 4x

x 7

5x x 3 +

5x –

x 5x

x 7

= (12x + 21x) : (15x + 28x)

= 33x :43x

= 33 : 43.

A’s share = Rs.		760 x	33		= Rs. 330.
			76

A and B started a partnership business investing some amount in the ratio of 3 : 5. C joined then after six months with an amount equal to that of B. In what proportion should the profit at the end of one year be distributed among A, B and C?

A.	3 : 5 : 2
B.	3 : 5 : 5
C.	6 : 10 : 5
D.	Data inadequate

Answer:8 Option C

Explanation:

Let the initial investments of A and B be 3x and 5x.

A : B : C = (3x x 12) : (5x x 12) : (5x x 6) = 36 : 60 : 30 = 6 : 10 : 5.

A, B, C rent a pasture. A puts 10 oxen for 7 months, B puts 12 oxen for 5 months and C puts 15 oxen for 3 months for grazing. If the rent of the pasture is Rs. 175, how much must C pay as his share of rent?

A.	Rs. 45
B.	Rs. 50
C.	Rs. 55
D.	Rs. 60

Answer:9 Option A

Explanation:

A : B : C = (10 x 7) : (12 x 5) : (15 x 3) = 70 : 60 : 45 = 14 : 12 : 9.

C’s rent = Rs.		175 x	9		= Rs. 45.
			35

10.

A and B started a business in partnership investing Rs. 20,000 and Rs. 15,000 respectively. After six months, C joined them with Rs. 20,000. What will be B’s share in total profit of Rs. 25,000 earned at the end of 2 years from the starting of the business?

A.	Rs. 7500
B.	Rs. 9000
C.	Rs. 9500
D.	Rs. 10,000

Answer: 10 Option A

Explanation:

A : B : C = (20,000 x 24) : (15,000 x 24) : (20,000 x 18) = 4 : 3 : 3.

B’s share = Rs.		25000 x	3		= Rs. 7,500.
			10