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Maths Shortcut Tricks (Calculate the Day of the week of any Given Date)

How to Calculate the Day of the week of any given date in no time. It’s a calendar calculation technique that I learned from a book. By Practicing it allows you to take a date, like 15 August 1947, and mentally calculate which day of the week it fell on.

Note: “Mod” means to divide the number and keep only the remainder. For Example, if the Total of all codes is 58 mod 7, then divide 58/7 and keep only the reminder i.e. 2

The Year Code
To calculate the Year Code, use this formula:

YY is the last two digits of the year. For the year 1947, it’s 47.
First, divide YY by 4 and discard the remainder: 47 div 4 = 11.
Then add 11 back into the YY number, which is 47 in this case, resulting in 58.
The next step is: 58 mod 7. We’ve divided 58/7 and left with a remainder of 2. That is the Year Code for 1947.
Hold that in memory while you calculate the items below.

The Month Code
This is easy — just memorize the number 033614625035:

Memorize it as (033)(614)(525)(035)

The Century Code
Just remember the number 4206420:
(420)6(420)

Leap Year Code

Note : The other thing to take into account is whether you are dealing with a leap year. If the date is from January or February of a leap year, you have to subtract one from your total before the final step and for other months except Jan and Feb.

If you can divide a Gregorian year by 4, it’s a leap year, unless it’s divisible by 100. But it is a leap year if it’s divisible by 400.
1992 is a leap year because you can divide it by four.
1900 is not a leap year because you can divide it by 100.
2000 is a leap year because you can divide it by 400. Calculating the Day

Back to the original formula:
(Year Code + Month Code + Century Code + Date Number – Leap Year Code) mod 7
For 15 August 1947, here are the results:
Year Code: 2
Month Code: 2
Century Code: 0
Date Number: 15 (the 15th of the month)
Leap Year Code: 0
So:

(2 + 2 + 15) mod 7 = 19 mod 7 = 5

Match the resulting number in the list below, and you’ll have the day of the week:

0 = Sunday
1 = Monday
2 = Tuesday
3 = Wednesday
4 = Thursday
5 = Friday
6 = Saturday

15 August 1947 was on a Friday.

Leap Year Example :

Lets calculate the Day for 18 July 1996
Year Code : 1 (96/4=24+96=120, 120 Mod 7 = 1)
Month Code : 6
Century Code: 0
Date Number : 18
Leap Year Code : 0 (No minus because it doesn’t fall in Jan and Feb)
So:
(1 + 6 + 18) mod 7 = 25 mod 7 = 4

18 July 1996 was on a Thursday

Another Leap Year Example :

1 January 2000: Start with ’00, leaving a Year Code of zero.
January has a Month Code of zero.
The Day Number is 1.
The Century Code for dates in the 2000s is 6.
2000 is a leap year since it can be divided by 400, and the date is in a January or February, so subtract 1 from the total in the final step.
Answer: 0 + 0 + 1 + 6 – 1 = 6.

1 January 2000 was a Saturday

Another Example
07 November 2017
Year Code : (17/4=leaving reminder we get 4) (4+17 Mod 7 = 0) because reminder is zero
Month Code : 3 (November)
Century Code : 6 (year falls in 2000 century)
Leap Year : 0 (No 2017 is not a leap year)
So.
(3 + 6 = 9 Mod 7 = 2)

The day is Tuesday!

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Maths Shortcut Tricks (Age Problems)

Maths Shortcut methods and tricks for Age-Related Problems was given below. Candidates those who are preparing for competitive exams in future may use this shortcuts trick to save time for other questions.

Question 1. If Sum of the ages of Father and Son is 70 years and the ratio of their ages is 3:2. What is the age of the Son?

Normal Method

Classification: If you use the above method, it will take minimum 35 to 40 seconds to get the answer. And by using the below shortcuts we can get the answer within 5 seconds.

Shortcut Method

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Step 1 : Bring down 2 = 2 Step 2 : Add 2 + 1 = 3 Step 3 : Add 2 + 1 + 4 = 7 Step 4 : Omit or cut 2 and add 1 + 4 + 3 = 8 Step 5 : Omit 1 and add 4 + 3 + 1 = 8 Step 6 : Omit 4 and 3 + 1 = 4 Step 7 : Omit 3 bring down 1 = 1 ∴ 21431 x 111 = 2378841

76859 x 111 = ?

If we get double digits in the process of addition, then apply balancing rule.
∴ 76859 x 111 = 8531349

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Maths : Problems on Average (Solved in Steps)

1. The average monthly income of A and B is Rs. 5050. The average monthly income of B and C is Rs. 6250 and the average monthly income of A and C is Rs.5200. What is the monthly income of A?

A. 2000

B. 3000

C. 4000

D. 5000

Explanation :

Let the monthly income of A = a
monthly income of B = b
monthly income of C = a
a +b=2×5050————(Equation1)
b + c=2×6250————(Equation2)
a +c=2×5200————(Equation3)
(Equation 1) + (Equation 3) – (Equation 2)
=> ( a +b +a +c) “ (b +c)
=(2×5050)+(2×5200)”(2×6250)
=> 2a = 2(5050 + 5200 – 6250)
=> a = 4000
=> Monthly income of A = 4000

2. In the first 10 overs of a cricket game, the run rate was only 3.2. What should be the run rate in the remaining 40 overs to reach the target of 282 runs?

A. 6.25,

B. 5.5,

C. 7.4 ,

D. 5

Explanation :

Runs scored in the first 10 overs = 10×3.2=32
Total runs = 282
remaining runs to be scored = 282 – 32 = 250
remaining overs = 40
Run rate needed = 250/40=6.25

3. A grocer has a sale of Rs. 6435, Rs. 6927, Rs. 6855, Rs. 7230 and Rs. 6562 for 5 consecutive months. How much sale must he have in the sixth month so that he gets an average sale of Rs.6500?

A. 4800

B. 4991

C. 5004

D. 5000

Explanation :

Let the sale in the sixth month = x
Then( 6435+6927+6855+7230+6562+x)/6=6500
= > 6435 + 6927 + 6855+ 7230 + 6562 + x
= 6 × 6500 = 39000
= > 34009 + x = 39000
= > x = 39000 – 34009 = 4991

4. The average of 20 numbers is zero. Of them, How many of them may be greater than zero, at the most?

A. 1

B. 20

C. 0

D. 19

Explanation :

Average of 20 numbers = 0
=>Sum of 20 numbers/20=0
=> Sum of 20 numbers = 0
Hence at the most, there can be 19 positive numbers.
(Such that if the sum of these 19 positive numbers is x, 20th number will be -x)

5. The captain of a cricket team of 11 members is 26 years old and the wicketkeeper is 3 years older. If the ages of these two are excluded, the average age of the remaining players is one year less than the average age of the whole team. Find out the average age of the team.

A. 23 years

B. 20 years

C. 24 years

D. 21 years

Explanation :

Number of members in the team = 11
Let the average age of the team = x
=>(Sum of the ages of all the 11 members of the team)/11=x
=> Sum of the ages of all the 11 members of the team = 11x
Age of the captain = 26
Age of the wicket keeper = 26 + 3 = 29
Sum of the ages of 9 members of the team excluding captain and wicket keeper = 11x – 26 – 29 = 11x – 55
Average age of 9 members of the team excluding captain and wicket keeper =(11x”55)/9
Given that( 11x”55)/9=(x”1)
=> 11x – 55 = 9(x – 1)
=> 11x – 55 = 9x – 9
=> 2x = 46
=>x=46/2=23 years

6. In Kiran’s opinion, his weight is greater than 65 kg but less than 72 kg. His brother does not agree with Kiran and he thinks that Kiran’s weight is greater than 60 kg but less than 70 kg. His mother’s view is that his weight cannot be greater than 68 kg. If all are them are correct in their estimation, what is the average of different probable weights of Kiran?

A. 70 kg

B. 69 kg

C. 61 kg

D. 67 kg

Explanation :

Let Kiran’s weight = x. Then
According to Kiran, 65 < x < 72 ——————(equation 1)
According to brother, 60 < x < 70 ——————(equation 2)
According to mother, xd”68 ——————(equation3)
Given that equation 1, equation 2 and equation 3 are correct. By combining these equations, we can write as 65<xd”68
That is x = 66 or 67 or 68
the average of different probable weights of Kiran = (66+67+68)/3=67

7. The average weight of 16 boys in a class is 50.25 kg and that of the remaining 8 boys is 45.15 kg. Find the average weights of all the boys in the class.

A. 48.55

B. 42.25

C. 50

D. 51.25

Explanation :

Average weight of 16 boys = 50.25
Total Weight of 16 boys = 50.25×16
Average weight of remaining 8 boys = 45.15
Total Weight of remaining 8 boys = 45.15×8
Total weight of all boys in the class = (50.25×16)+(45.15×8)
Total boys = 16+8=24
Average weight of all the boys = (50.25×16)+(45.15×8)/24
= (50.25×2)+(45.15×1)/3
= (16.75×2) + 15.05
=33.5+15.05=48.55

8. If the average marks of three batches of 55, 60 and 45 students respectively is 50, 55, 60, what is the average marks of all the students?

A. 53.23

B. 54.68

C. 51.33

D. 50

Explanation :

Average marks of batch1 = 50
Students in batch1 = 55
Total marks of batch1=55×50
Average marks of batch2 = 55
Students in batch2 = 60
Total marks of batch2=60×55
Average marks of batch3 = 60
Students in batch3 = 45
Total marks of batch3=45×60
Total students = 55 + 60 + 45 = 160
Average marks of all the students
= {(55×50)+(60×55)+(45×60)}/160
= (275+330+270)/16
= 875/16=54.68

9. A library has an average of 510 visitors on Sundays and 240 on other days. What is the average number of visitors per day in a month of 30 days beginning with a Sunday?

A. 290

B. 304

C. 285

D. 270

Explanation :

in a month of 30 days beginning with a Sunday, there will be 4 complete weeks and another two days which will be Sunday and Monday Hence there will be 5 Sundays and 25 other days in a month of 30 days beginning with a Sunday
Average visitors on Sundays = 510
Total visitors of 5 Sundays = 510×5
Average visitors on other days = 240
Total visitors of other 25 days = 240×25
Total visitors = (510×5)+(240×25)
Total days= 30
Average number of visitors per day ={ (510×5) + (240×25)}/30
= { (51×5)+(24×25) }/3
= (17×5) + (8×25)
=85+200=285

10. A student’s mark was wrongly entered as 83 instead of 63. Due to that, the average marks for the class got increased by half 1/2. What is the number of students in the class?

A. 45

B. 40

C. 35

D. 30

Explanation :

Let the total number of students = x
The average marks increased by 1/2 due to an increase of 83-63=20 marks.
But total increase in the marks = 1/2×x=x/2
Hence we can write asx/2=20•Ëx=20×2=40

11. A family consists of two grandparents, two parents and three grandchildren. The average age of the grandparents is 67 years, that of the parents is 35 years and that of the grandchildren is 6 years. The average age of the family is

A. 32.27 years

B. 31.71 years

C. 28.17 years

D. 30.57 years

Explanation :

Total age of the grandparents = 67×2
Total age of the parents = 35×2
Total age of the grandchildren = 6×3
Average age of the family ={(67×2)+(35×2)+(6×3)}/7=(134+70+18)/ 7=222/7=31.71

12. The average weight of A, B and C is 45 kg. If the average weight of A and B be 40 kg and that of B and C be 43 kg, what is the weight of B?

A. 31 kg

B. 2812 kg

C. 32 kg

D. 3012 kg

Explanation :

Let the weight of A, B and C are a,b and c respectively.
Average weight of A,B and C = a+b+c=45×3=135—equation(1)
average weight of A and B = 40 a+b=40×2=80—equation(2)
average weight of B and C = 43 b+c=43×2=86 —equation (3)
equation(2) +equation(3)- equation(1)
=> a+ b+ b +c “(a +b +c)=80+86”135
=>b=80+86″135=166″135=31
=> weight of B = 31

13. The average age of husband, wife and their child 3 years ago was 27 years and that of wife and the child 5 years ago was 20 years. What is the present age of the husband?

A. 40

B. 32

C. 28

D. 30

Explanation :

let the present age of the husband = h
present age of the wife = w
present age of the child = c
3 years ago, average age of husband, wife andtheir child = 27
= >Sum of age of husband, wife and their child before 3 years = 3×27=81
= > (h-3) + (w-3) + (c-3) = 81
= > h + w + c = 81+9 = 90 ————————— equation(1)
5 years ago, average age of wife and child = 20
= >Sum of age of wife and child before 5 years = 2×20=40
= > (w-5) + (c-5) = 40
= > w + c = 40+10 = 50 —————————equation(2)
Substituting equation(2) in equation(1)
= > h + 50 = 90
= > h = 90 – 50 = 40
= >present age of the husband = 40

14. The average weight of 8 person’s increases by 2.5 kg when a new person comes in place of one of them weighing 65 kg. What is the weight of the new person?

A. 70

B. 75

C. 85

D 80

Explanation :

Total increase in weight = 8×2.5=20
If x is the weight of the new person, total increase in weight = x – 65
=> 20 = x – 65
=> x = 20 + 65 = 85

15. A batsman makes a score of 87 runs in the 17th inning and thus increases his averages by 3. What is his average after 17th inning?

A. 39

B. 35

C. 42

D. 40.5

Explanation :

Let the average after 17 innings = x
Total runs scored in 17 innings = 17x
then average after 16 innings = (x-3)
Total runs scored in 16 innings = 16(x-3)
We know that Total runs scored in 16 innings + 87
= Total runs scored in 17 innings
= > 16(x-3) + 87 = 17x
= > 16x – 48 + 87 = 17x
= > x = 39

16. There are two divisions A and B of a class, consisting of 36 and 44 students respectively. If the average weight of divisions A is 40 kg and that of division b is 35 kg. What is the average weight of the whole class?

A. 38.25

B. 37.25

C. 38.5

D. 37

Explanation:

Total weight of students in division A = 36×40
Total weight of students in division B = 44×35
Total students = 36+44=80
Average weight of the whole class= {(36×40)+(44×35)}/80
= {(9×40)+(11×35)}/20
= {(9×8)+(11×7)}/4
= (72+77)/4
= 149/4=37.25

17. A student needed to find the arithmetic mean of the numbers 3, 11, 7, 9, 15, 13, 8, 19, 17, 21, 14 and x. He found the mean to be 12. What is the value of x?

A. 12

B. 5

C. 7

D. 9

Explanation :

(3+11+7+9+15+13+8+19+17+21+14+x)/12=12
= > (137+x)/12=12
= > 137 + x = 144
= > x = 144 – 137 = 7

18. Arun obtained 76, 65, 82, 67 and 85 marks (out in 100) in English, Mathematics, Chemistry, Biology and Physics. What is his average mark?

A. 53

B. 54

C. 72

D. 75

Explanation :

Average mark = (76+65+82+67+85)/5=375#5=75

19. The distance between two stations A and B is 778 km. A train covers the journey from A to B at 84 km per hour and returns back to A with a uniform speed of 56km per hour. Find the average speed of the train during the whole journey?

A. 69.0 km /hr

B. 69.2 km /hr

C. 67.2 km /hr

D. 67.0 km /hr

Explanation :

If a car covers a certain distance at x kmph and an equal distance at y kmph.
Then, 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 = 2×84×56# (84+56) = (2×84×56)#140==67.2

20. The average age of 36 students in a group is 14 years. When teacher’s age is included in it, the average increases by one. Find out the teacher’s
age in years?

A. 51 years,

B. 49 years

C. 53 years

D. 50 years

Explanation :

the average age of 36 students in a group is 14
= > Sum of the ages of 36 students = 36×14
When teacher’s age is included in it, the average increases by one => average = 15
= > Sum of the ages of 36 students and the teacher =37×15
Hence teachers age = 37×15″36×14=37×15″14(37″1)=37×15″37×14+14=37(15″14)+14=37+14=51

21. The average of five numbers id 27. If one number is excluded, the average becomes 25. What is the excluded number?

A. 30,

B. 40,

C. 32.5,

D. 35

Explanation :

Sum of 5 numbers = 5×27Sum of 4 numbers after excluding one number = 4×25Excluded number = 5×27″4×25=135″100=35

22. The average of six numbers is x and the average of three of these is y. If the average of the remaining three is z, then

A. None of these

B. x = y + z

C. 2x = y + z

D. x = 2y + 2z

Explanation :

Average of 6 numbers = x
=> Sum of 6 numbers = 6x
Average of the 3 numbers = y
=> Sum of these 3 numbers = 3y
Average of the remaining 3 numbers = z
=> Sum of the remaining 3 numbers = 3z
Now we know that 6x = 3y + 3z
=> 2x = y + z

23. The batting average for 40 innings of a cricket player is 50 runs. His highest score exceeds his lowest score by 172 runs. If these two innings are excluded, the average of the remaining 38 innings is 48 runs. Find out the highest score of the player.

A. 150

B. 174

C. 180

D. 166

Explanation :

Total runs scored by the player in 40 innings = 40×50
Total runs scored by the player in 38 innings after excluding two innings= 38×48
Sum of the scores of the excluded innings = 40×50″38×48=2000″1824=176
Given that the scores of the excluded innings differ by 172. Hence let’s take the highest score as x + 172 and lowest score as x
Now x + 172 + x = 176
= > 2x = 4
= >x=4# 2=2
highest score as x + 172 = 2 + 172 = 174

24. The average score of a cricketer for ten matches is 38.9 runs. If the average for the first six matches is 42, what is the average for the last four matches?

A. 34.25

B. 36.4

C. 40.2

D. 32.25

Explanation :

Total runs scored in 10 matches = 10×38.9
Total runs scored in first 6 matches = 6×42
Total runs scored in the last 4 matches = 10×38.9″6×42
Average of the runs scored in the last 4 matches
= (10×38.9″6×42)#4
= (389″252)#4
= 137#4=34.25

25. Suresh drives his car to a place 150 km away at an average speed of 50 km/hr and returns at 30 km/hr. What is his average speed for the whole journey?

A. 32.5 km/hr.

B. 35 km/hr

C. 37.5 km/hr

D. 40 km/hr

Explanation :

If a car covers a certain distance at x kmph and an equal distance at y kmph. Then, 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 = ( 2×50×30)#(50+30)
= (2×50×30)#80
= (2×50×3) # 8
= (50×3)#4
=(25×3)# 2=75#2=37.5

26. The average age of a husband and his wife was 23 years at the time of their marriage. After five years they have a one-year-old child. What is the average age of the family?

A. 21 years

B. 20 years

C. 18 years

D. 19 years

Explanation :

Total of the age of husband and wife = 2×23=46
Total of the age of husband and wife after 5 years
+ Age of the 1-year-old child =46+5+5+1=57
The average age of the family=57#3=19

27. In an examination, a student’s average marks were 63. If he had obtained 20 more marks for his Geography and 2 more marks for his history, his average would have been 65. How many subjects were there in the examination?

A. 12

B. 11

C. 13

D. 14

Explanation :

Let the number of subjects = x
Then, Total marks he scored for all subjects = 63x
If he had obtained 20 more marks for his Geography and 2 more marks for his history, his average would have been 65
=> Total marks he would have scored for all subjects = 65x
Now we can form the equation as 65x – 63x = the
additional marks of the student = 20 + 2 = 22
=> 2x = 22
=> x= 22#2
=11

28. The average salary of all the workers in a workshop is Rs.8000. The average salary of 7 technicians is Rs.12000 and the average salary of the rest is Rs.6000. How many workers are there in the workshop?

A. 21

B. 22

C. 23

D.24

Explanation:

Let the number of workers = x
Given that average
salary of all the workers = Rs.8000
then total salary of all workers = 8000x
Given that average salary of 7 technicians is Rs.12000
=>
Total salary of 7 technicians = 7×12000=84000
Number of the rest of the employees = (x – 7)
Average salary of the rest of the employees = Rs.6000
Total salary of the rest of the employees = (x – 7)(6000) 8000x = 84000 + (x – 7)(6000)
= > 8x = 84 + (x – 7)(6)
= > 8x = 84 + 6x – 42
= > 2x = 42
= >x=21

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Maths : Profit & Loss (Solved in Steps)

1. Aravind buys an old scooter for Rs. 4700 and spends Rs. 800 on its repairs. If he sells the scooter for Rs. 5800, what is his gain percent?

A. 12%

B. 10%

C. 44.7%

D. 5.45 %

Explanation :

Cost Price (CP) = 4700 + 800 = Rs. 5500
Selling Price (SP) = Rs. 5800.
Gain = (SP) – (CP) = 5800 – 5500= Rs. 300.
Gain%
= (GainCP)×100
= (300÷5500)×100
= 300÷55
= 60÷11
= 5.45

2. The cost price of 20 articles is the same as the selling price of x articles. If the profit is 25%, find out the value of x

A. 15

B. 25

C. 18

D. 16

Explanation :

Let the Cost Price (CP) of one article = 1
=> CP of x articles = x ——————————( 1)
CP of 20 articles = 20
Given that cost price of 20 articles is the same as the selling price of x articles
=> Selling price (SP) of x articles = 20————–( 2)
Given that Profit = 25%
(SP-CP)÷CP
=25÷100=1/4————( 3)
Substituting equations 1 and 2 in equation 3,
⇒(20-x) ÷x = 1/4
⇒80-4x=x
⇒ 5x=80
⇒ x=80÷5
=16

3. If the selling price is doubled, the profit triples. What is the profit percent?

A. 100

B. 105.33

C. 66.66

D. 120

Explanation :

Let the CP = x , SP = y
profit = SP –CP = y-x
If SP is doubled, SP = 2y
Now Profit = SP –CP = 2y – x
Given that If selling price is doubled, the profit triples
⇒2y–x=3(y-x)
⇒2y–x=3y-3x
⇒y=2x
⇒Profit%=(Profit÷CP)×100
=((y-x)÷x)×100
=((2x–x)÷x)×100
=(x÷x)×100=100%

4. In a shop, the profit is 320% of the cost. If the cost increases by 25% but the selling price remains constant, find out approximately what percentage of the selling price is the profit?

A. 250%

B. 100%

C. 70%

D. 30%

Explanation :

Let the CP = 100
Profit=(320÷100)×100
=320
SP=CP+Profit=100+320=420
If the cost increases by 25%,
New CP=(125÷100)×100=125
Selling Price is constant, hence New SP=420
Profit=SP–CP=420–125=295
Required Percentage =(295÷420)×100
=2950÷42
=70(approximately)

5. Sebastian bought bananas at 6 for a rupee. How many for a rupee must he sell to gain 20%?

A. 3

B. 4

C. 5

D. 6

Explanation :

CP of 6 bananas = 1
Gain = 20%
SP of 6 bananas=1×(120÷100)
=12÷10
⇒The number of bananas he should sell for Rs.1 for
a gain of 20%
=(10÷12)×6
=10÷2=5

6. An exporter expects a gain of 22.5% on his cost price. If in a week, his sale was of Rs. 392, what was his profit?

A. Insufficient Data

B. Rs. 80

C. Rs. 90

D. Rs. 72

Explanation :

SP=392
Gain=22.5%
CP=(100÷(100+Gain%))×SP
=(100÷(100+22.5))×392
=(100÷122.5)×392
=(1000÷1225)×392
=(40÷49)×392
=(40÷7)×56
=40×8
=320
Profit=SP-CP=392-320=7

7. The percentage profit earned by selling an item for Rs. 1920 is equal to the percentage loss incurred by selling the same item for Rs. 1280. At what price should the item be sold to make 25% profit?

A. Insufficient Data

B. Rs. 3000

C. Rs. 2000

D. Rs. 2200

Explanation :

Let CP = x
Percentage profit earned by selling an item for Rs. 1920
= ( (SP-CP)÷CP)×100
= ((1920-x)÷x)×100
Percentage loss incurred by selling the same item for Rs. 1280
=((SP-CP)÷CP)×100= ((x-1280)÷x)×100
Given that Percentage profit earned by selling an item for Rs. 1920=Percentage loss incurred by selling the same item for Rs. 1280
⇒((1920-x)÷x)×100=((x-1280)÷x)×100
⇒(1920-x)÷x=(x-1280)÷x
⇒ 1920–x = x–1280
⇒2x=1920+1280=3200
⇒ x=3200÷2=1600
Required Selling Price=CP×(125÷100)
=1600×(125÷100)=1600×(5÷4)=400×5=2000

8. A man buys a scooter for Rs. 1400 and sells it at a loss of 15%. What is the selling price of the scooter?

A. Rs. 1240

B. Rs. 1190

C. Rs. 1090

D. Rs. 1130

Explanation :

SP=1400×(85÷100)=14×85=1190

9. Murali purchased 20 dozens of toys at the rate of Rs. 375 per dozen. He sold each one of them at the rate of Rs. 33. Find out his profit percentage?

A. 3.5

B. 5.6

C. 4.1

D. 3.4

Explanation :

CP of 1 toy=(375÷12)
SP of 1 toy=33
Profit=SP-CP
=33-(375÷12)
Profit%=(Profit÷CP)×100
={(33-(375÷12)) ÷(375÷12)}×100
=(33×(12÷375)-1)×100
=(33×(4÷125)-1)×100
=(7÷125)×100
=(7÷5)×4
=28÷5
=5.6%

10. Some items were bought at 6 items for Rs. 5 and sold at 5 items for Rs. 6. What is the gain percentage?

A. 44%

B. 3313

C. 3123

D. 30%

Explanation :

CP of 1 item=5÷6
SP of 1 item=6÷5
Gain%=((SP-CP)÷CP)×100
={((6÷5)-(5÷6)) ÷(5÷6)}×100
={(36÷25-1)}×100
=(11÷25)×100
=11×4=44

11. 100 oranges were bought at the rate of Rs. 350 and sold at the rate of Rs. 48 per dozen. What is the percentage of profit or loss?

A. 11.29% Loss

B. 11.14% Profit

C. 14.29% Profit

D. 14.29% Loss

Explanation :

CP of 100 oranges=350
⇒CP of 1 orange=350÷100=3.5
SP of 12 oranges=48
⇒SP of 1orange=48÷12=4
Profit%={(SP-CP)÷CP}×100
={(4-3.5)÷3.5}×100= (.5÷3.5)×100=(1÷7)×100=100÷7=14.29

12. On selling 17 balls at Rs. 720, there is a loss equal to the cost price of 5 balls. What is the cost price of a ball?

A. Rs. 43

B. Rs. 60

C. Rs. 55

D. Rs. 34

Explanation :

Loss = (CP of 17 balls) – (SP of 17 balls) = (CP of 17 balls) – 720
Given that Loss = (CP of 5 balls)
=> (CP of 17 balls) – 720 = (CP of 5 balls) => (CP of 17 balls) – (CP of 5 balls) = 720
=> CP of 12 balls = 720
=> CP of 1 ball=720÷12=60

13. When an item is sold for Rs. 18,700, the owner loses 15%. At what price should that plot be sold to get a gain of 15%?

A. Rs. 25100

B. Rs. 24200

C. Rs. 25300

D. Rs. 21200

Explanation :

When an item is sold for Rs. 18,700, the owner loses 15%⇒SP = 18700 and Loss=15%
⇒CP={100÷(100 -Loss%)}×SP
={100÷ (100″15)}×18700={100÷85}×18700
To get a gain of 15%, SP ={(100+Gain%)÷100}×CP
={(100+15)÷100}×CP
=(115÷100)×CP
={(115÷100)×(100÷85)}×18700=(115÷85)×18700=(23÷17)×18700
=23×1100=25300

14. A trader mixes 26 kg of rice at Rs. 20 per kg with 30 kg of rice of other variety at Rs. 36 per kg and sells the mixture at Rs. 30 per kg. What is his profit percentage?

A. 6%

B. 5%

C. 4%

D. 7%

Explanation :

CP of 1st variety rice=20
CP of 2nd variety rice=36
CP of the 56 kg rice mixture=(26×20+30×36)=520+1080=1600
SP of the 1 kg rice mixture=30
SP of the 56 kg rice mixture=30×56=1680
Gain=SP-CP=1680-1600=80
Gain%=(Gain÷CP)×100=(80÷1600)×100=5%

15. If a material is sold for Rs.34.80, there is a loss of 25%. Find out the cost price of the material?

A. Rs.46.40

B. Rs.44

C. Rs.42

D. Rs.47.20

Explanation :

SP=34.80
Loss=25%
CP={100÷ (100-Loss%)}×SP
=[100÷ (100-25)]×34.80=(100÷75)×34.80
=[4×34.80]÷3
=4×11.60=46.40

16. By selling an item for Rs.15, a trader loses one-sixteenth of what it costs him. The cost price of the item is?

A. Rs.14

B. Rs.15

C. Rs.16

D. Rs.17

Explanation :

SP=15
Loss=CP÷16
Loss=CP-SP=CP-15
⇒(CP÷16)=CP-15
⇒(15 CP÷16)=15
⇒(CP÷16)=1
⇒CP=16

17. A fruit seller sells apples at the rate of Rs.9 per kg and thereby loses 20%. At what price per kg, he should have sold them to make a profit of 5%?

A. 11.32

B. 11

C. 12

D. 11.81

Explanation :

SP=9
Loss=20%
CP={100÷ (100-Loss%)}×SP={100÷ (10020)}×9={100÷80}×9=(5÷4)×9
To make a profit of 5%,
SP={(100+Gain%)÷100}×CP=[(100+5)
÷100]×CP
=(105÷100)×((5÷4)×9)=
=((21÷20)×(5÷4)×9)= 11.81

18. A trader gives 12% additional discount on the discounted price, after giving an initial discount of 20% on the labelled price of an item. The final sale price of the item is Rs.704. Find out the labelled price?

A. 1000

B. 2000

C. 1200

D. 920

Explanation :

Let the labelled price=x
SP=704
Initial Discount=20%
Price after initial discount=x×(80÷100)
⇒x×(80÷100)×(88÷100)=704
⇒x×(4÷5)×(22÷25)=704
⇒x=704×(25÷22)×(5÷4)= 1000

19. John purchased a machine for Rs. 80,000. After spending Rs.5000 on repair and Rs.1000 on transport he sold it with 25% profit. What price did he sell the machine?

A. Rs.107000

B. Rs.107500

C. Rs.108500

D. None of these

Explanation :

CP=80,000+5000+1000=86000
Profit=25%
SP=[(100+Gain%)÷100]×CP
=[(100+25)÷100]×86000
=(125÷100)×86000
=(5÷4)×86000=5×21500=107500

20. A shopkeeper sells his goods at cost price but uses a weight of 800 gm instead of kilogram weight. What is his profit percentage?

A. 18%

B. 40%

C. 25%

D. 20%

Explanation :

If a trader professes to sell his goods at cost price but uses false weights, then
Gain%=[(Error÷ (True Value-Error))×100]%
So here profit percentage=[(200÷ (1000200))×100]%
=[(200÷800)×100]%=25%

21. An item was sold for Rs.27.50 with a profit of 10%. If it was sold for Rs.25.75, what would have been the percentage of profit or loss?

A. 3%

B. 2%

C. 4%

D. 5%

Explanation :

SP = 27.5
Profit = 10%
CP=[100÷ (100+Profit%)]×SP
=[100÷(100+10)]×27.5
=2750÷110=25
If the item was sold for Rs.25.75,
Gain%=[(25.75-25) ÷25]×100=[0.75÷25]×100=3%

22. Prasanth bought a car and paid 10 % less than the original price. He sold it with 30% profit on the price he had paid. What percentage of profit did he earn the original price?

A. 17%

B. 16%

C. 18%

D. 14%

Explanation :

Let the original price=100
Then the price at which he purchased (CP) =90% of
100=90
Profit=30%
SP=((100+Profit%)÷100)×CP
=[(100+30)÷100]×90
=[130÷100]×90=13×9=117
Required %=[(117-100) ÷100]×100=17%

23. If a seller reduces the selling price of an item from Rs.400 to Rs.380, his loss increases by 2%. What is the cost price of the item?

A. 1000

B. 800

C. 1200

D. 1100

Explanation :

Initial Loss% = {[ CP-400] ÷CP}×100
If the SP is reduced from 400 to 380, Loss% = [ (CP380)÷CP]×100
It is given that If the SP is reduced from 400 to 380,
Loss% increases by 2
⇒[(CP-380)÷CP]×100- {[ CP-400] ÷CP}×100=2
⇒(CP-380)-(CP-400)=2×[CP÷100]
⇒20=2×[CP÷100]
⇒CP=(20×100)÷2=1000

24. If the selling price of an article is Rs. 250, profit percentage is 25%. Find the ratio of the cost price and the selling price

A. 5: 3

B. 3 : 5

C. 4 : 5

D. 5 : 4

Explanation :

SP = 250
Profit = 25%
CP=[100÷ (100+Profit%)]×SP
=[100÷ (100+25)]×250==[100÷125]×250=200
Required Ratio = 200 : 250 = 4:5

25. Sunil purchases two books at Rs.300 each. He sold one book 10% gain and other at 10% loss. What is the total loss or gain in percentage?

A. 10% gain

B. 1% loss

C. No loss or no gain

D. 1% gain

Explanation :

CP of two books = 300 + 300 = 600
SP of two books
= 300×(110÷100)+300×(90÷100)=(300÷100)×(110+90)=3×200=600
Total SP = Total CP
Hence No loss or no gain

Categories

Maths : Problems on Train (Solved in Steps)

1. A train is running at a speed of 40 km/hr and it crosses a post in 18 seconds. What is the length of the train?

A. 190 metres

B. 160 metres

C. 200 metres

D. 120 metres

Explanation :

Speed of the train = 40 km/hr
= 40000/3600 m/s
= 400/36 m/s
Time taken to cross = 18 s
Distance Covered = speed× time
= (400/36)× 18
= 200 m
Distance covered is equal to the length of the train = 200 m

2. A train ,130 meters long travels at a speed of 45 km/hr crosses a bridge in 30 seconds. The length of the bridge is

A. 270 m

B. 245 m

C. 235 m

D. 220 m

Explanation :

Assume the length of the bridge = x meter
Total distance covered = 130+x meter
total time taken = 30s
speed = Total distance covered /total time taken = (130+x)/30 m/s
=> 45 × (10/36) = (130+x)/30
=> 45 × 10 × 30 /36 = 130+x
=> 45 × 10 × 10 / 12 = 130+x
=> 15 × 10 × 10 / 4 = 130+x
=> 15 × 25 = 130+x = 375
=> x = 375-130 =245

3. A train has a length of 150 meters . it is passing a man who is moving at 2 km/hr in the same direction of thetrain, in 3 seconds. Find out the speed of the train.

A. 182 km/hr

B. 180 km/hr

C. 152 km/hr

D. 169 km/hr

Explanation :

Length of the train, l = 150m
Speed of the man = 2 km/hr
Relative speed = total distance/time = (150/3) m/s
= (150/3) × (18/5) = 180 km/hr
Relative Speed = Speed of train – Speed of man (As both are moving in the same direction)
=> 180 = Speed of train – 2
=> Speed of train = 180 + 2 = 182 km/hr

4. A train having a length of 240 m passes a post in 24 seconds. How long will it take to pass a platform having a length of 650 m?

A. 120 sec

B. 99 s

C. 89 s

D. 80 s

Explanation :

speed of the train = 240/24 = 10 m/s
time taken to pass a platform having a length of 650 m
= (240+650)/10 = 89 seconds

5. A train 360 m long runs with a speed of 45 km/hr. What time will it take to pass a platform of 140 m long?

A. 38 sec

B. 35 s

C. 44 sec

D. 40 s

Explanation :

Speed = 45 km/hr = 45×(10/36) m/s
= 150/12 = 50/4 = 25/2 m/s
Total distance = length of the train + length of the platform
= 360 + 140 = 500 meter
Time taken to cross the platform = 500/(25/2) = 500×2/25 = 40 seconds

6. Two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively . If they cross each other in 23 seconds, what is the ratio of their speeds?

A. Insufficient data

B. 3 : 1

C. 1 : 3

D. 3 : 2

Explanation :

Let the speed of the trains be x and y respectively length of train1 = 27x
length of train2 = 17y
Relative speed= x+ y
Time taken to cross each other = 23 s
=> (27x + 17 y)/(x+y) = 23
=> (27x + 17 y) = 23(x+y)
=> 4x = 6y
=> x/y = 6/4 = 3/2
ratio of their speeds=3:2

7. Two trains of equal length are running on parallel lines in the same direction at 46 km/hr and 36 km/hr. If the faster train passes the slower train in 36 seconds, what is the length of each train?

A. 88

B. 70

C. 62

D. 50

Explanation :

Assume the length of each train = x
Total distance covered for overtaking the slower train =
x+x = 2x
Relative speed = 46-36 = 10km/hr = (10×10)/36 = 100/36 m/s
Time = 36 seconds
2x/ (100/36) = 36
=> (2x × 36 )/100 = 36
=> x = 50 meter

8. Two trains having length of 140 m and 160 m long run at the speed of 60 km/hr and 40 km/hr respectively in opposite directions (on parallel tracks). The time which they take to cross each other, is

A. 10.8 s

B. 12 s

C. 9.8 s

D. 8 s

Explanation :

Distance = 140+160 = 300 m
Relative speed = 60+40 = 100 km/hr =
(100×10)/36 m/s
Time = distance/speed = 300 / (100×10)/36 = 300×36 /1000 = 3×36/10 = 10.8 s

9. Two trains are moving in opposite directions with speed of 60 km/hr and 90 km/hr respectively. Their lengths are 1.10 km and 0.9 km respectively. The slower train cross the faster train in — seconds

A. 56

B. 48

C. 47

D. 26

Explanation :

Relative speed = 60+90 = 150 km/hr (Since both trains are moving in opposite directions)
Total distance = 1.1+.9 = 2km
Time = 2/150 hr = 1//75 hr = 3600/75 seconds = 1200/25 = 240/5 = 48 seconds

10. A train passes a platform in 36 seconds. The same train passes a man standing on the platform in 20 seconds. If the speed of the train is 54 km/hr, The length of the platform is

A. None of these

B. 280 meter

C. 240 meter

D. 200 meter

Explanation :

Speed of the train = 54 km/hr = (54×10)/36 m/s = 15 m/s
Length of the train = speed × time taken to cross the man = 15×20 = 300 m
Let the length of the platform = L
Time taken to cross the platform = (300+L)/15
=> (300+L)/15 = 36
=> 300+L = 15×36 = 540
=> L = 540-300 = 240 meter

11. A train moves past a post and a platform 264 m long in 8 seconds and 20 seconds respectively. What is the speed of the train?

A. 79.2 km/hr

B. 69 km/hr

C. 74 km/hr

D. 61 km/hr

Explanation :

Let x is the length of the train and v is the speed
Time taken to move the post = 8 s
=> x/v = 8
=> x = 8v — (1)
Time taken to cross the platform 264 m long = 20 s
(x+264)/v = 20
=> x + 264 = 20v —(2)
Substituting equation 1 in equation 2, we get
8v +264 = 20v
=> v = 264/12 = 22 m/s
= 22×36/10 km/hr = 79.2 km/hr

12. Two trains having equal lengths, take 10 seconds and 15 seconds respectively to cross a post. If the length of each train is 120 meters, in what time (in seconds) will they cross each other when traveling in opposite direction?

A. 10

B. 25

C. 12

D. 20

Explanation :

speed of train1 = 120/10 = 12 m/s
speed of train2 = 120/15 = 8 m/s
if they travel in opposite direction, relative speed = 12+8 = 20 m/s
distance covered = 120+120 = 240 m
time = distance/speed = 240/20 = 12 s

13. A train having a length of 1/4 mile, is travelling at a speed of 75 mph. It enters a tunnel 3 ½ miles long. How long does it take the train to pass through the tunnel from the moment the front enters to the moment the
rear emerges?

A. 3 min

B. 4.2 min

C. 3.4 min

D. 5.5 min

Explanation :

Total distance = 3 ½ + ¼ = 7/2 + ¼ = 15/4 miles
Speed = 75 mph
Time = distance/speed = (15/4) / 75 hr = 1/20 hr = 60/20 minutes = 3 minutes

14. A train runs at the speed of 72 kmph and crosses a 250 m long platform in 26 seconds. What is the length of the train?

A. 270 m

B. 210 m

C. 340 m

D. 130 m

Explanation :

Speed= 72 kmph = 72×10/36 = 20 m/s
Distance covered = 250+ x where x is the length of the train
Time = 26 s
(250+x)/26 = 20
250+x = 26×20 = 520 mx = 520-250 = 270 m

15. A train overtakes two persons who are walking in the same direction to that of the train at 2 kmph and 4 kmph and passes them completely in 9 and 10 seconds respectively. What is the length of the train?

A. 62 m

B. 54 m

C. 50 m

D. 55 m

Explanation :

Let x is the length of the train in meter and v is its speed in kmph
x/9 = ( v-2)(10/36) —(1)
x/10 =( v-4) (10/36) — (2)
Dividing equation 1 with equation 2
10/9 = (v-2)/(v-4)
=> 10v – 40 = 9v – 18
=> v = 22
Substituting in equation 1, x/9 = 200/36 => x = 9×200/36 = 50 m

16. A train is travelling at 48 kmph . It crosses another train having half of its length, travelling in opposite direction at 42 kmph, in 12 seconds. It also passes a railway platform in 45 seconds. What is the length of the platform?

A. 500 m

B. 360 m

C. 480 m

D. 400 m

Explanation :

Speed of train1 = 48 kmph
Let the length of train1 = 2x meter
Speed of train2 = 42 kmph
Length of train 2 = x meter (because it is half of train1’s length)
Distance = 2x + x = 3x
Relative speed= 48+42 = 90 kmph = 90×10/36 m/s = 25 m/s
Time = 12 s
Distance/time = speed => 3x/12 = 25
=> x = 25×12/3 = 100 meter
Length of the first train = 2x = 200 meter
Time taken to cross the platform= 45 s
Speed of train1 = 48 kmph = 480/36 = 40/3 m/s
Distance = 200 + y where y is the length of the platform
=> 200 + y = 45×40/3 = 600
=> y = 400 meter

17. A train having a length of 270 meters is running at the speed of 120 kmph . It crosses another train running in opposite direction at the speed of 80 kmph in 9 seconds. What is the length of the other train?

A. 320 m

B. 190 m

C. 210 m

D. 230 m

Explanation :

Relative speed = 120+80 = 200 kmph = 200×10/36 m/s
= 500/9 m/s
time = 9s
Total distance covered = 270 + x ,where x is the length of other train
(270+x)/9 = 500/9
=> 270+x = 500
=> x = 500-270 = 230 meter

18. Two trains, each 100 m long are moving in opposite directions. They cross each other in 8 seconds. If one is moving twice as fast the other, the speed of the faster train is

A. 75 km/hr

B. 60 km/hr

C. 35 km/hr

D. 70 km/hr

Explanation :

Total distance covered = 100+100 = 200 m
Time = 8 s
let speed of slower train is v . Then the speed of the faster train is 2v
(Since one is moving twice as fast the other)
Relative speed = v + 2v = 3v
3v = 200/8 m/s = 25 m/s
=> v = 25/3 m/s
Speed of the faster train = 2v = 50/3 m/s = (50/3)×(36/
10) km/hr = 5×36/3 = 5×12 = 60 km/hr

19. Two stations P and Q are 110 km apart on a straight track. One train starts from P at 7 a.m. and travels towards Q at 20 kmph. Another train starts from Q at 8 a.m. and travels towards P at a speed of 25 kmph. At what time will they meet?

A. 10.30 a.m

B. 10 a.m

C. 9.10 a.m.

D. 11 a.m

Explanation :

Assume both trains meet after x hours after 7 am
Distance covered by train starting from P in x hours = 20x km
Distance covered by train starting from Q in (x-1) hours = 25(x-1)
Total distance = 110
=> 20x + 25(x-1) = 110
=> 45x = 135
=> x= 3
Means, they meet after 3 hours after 7 am, ie, they meet at 10 am

20. A train overtakes two persons walking along a railway track. The first person walks at 4.5 km/hr and the other walks at 5.4 km/hr. The train needs 8.4 and 8.5 seconds respectively to overtake them. What is the speed of the train if both the persons are walking in the same direction as the train?

A. 81 km/hr

B. 88 km/hr

C. 62 km/hr

D. 46 km/hr

Explanation :

Let x is the length of the train in meter and y is its speed in kmph
x/8.4 = (y-4.5)(10/36) —(1)
x/8.5 = (y-5.4)(10/36) —(2)
Dividing 1 by 2
8.5/8.4 = (y-4.5)/ (y-5.4)
=> 8.4y – 8.4 × 4.5 = 8.5y – 8.5×5.4
.1y = 8.5×5.4 – 8.4×4.5
=> .1y = 45.9-37.8 = 8.1
=> y = 81 km/hr

21. A train , having a length of 110 meter is running at a speed of 60 kmph. In what time, it will pass a man who is running at 6 kmph in the direction opposite to that of the train

A. 10 sec

B. 8 sec

C. 6 sec

D. 4 sec

Explanation :

Distance = 110 m
Relative speed = 60+6 = 66 kmph (Since the train and the man are in moving in opposite direction)
= 66×10/36 mps = 110/6 mps
Time = distance/speed = 110

22. A 300-metre long train crosses a platform in 39 seconds while it crosses a post in 18 seconds. What is the length of the platform?

A. 150 m

B. 350 m

C. 420 m

D. 600 m

Explanation :

Length of the train
= distance covered in crossing the post
= speed × time
= speed × 18
ie,300= speed × 18
Speed of the train = 300/18 m/s = 50/3 m/s
Time taken to cross the platform = 39 s
(300+x)/(50/3) = 39 s where x is the length of the platform
300+x = (39 × 50) / 3 = 650 meter
x = 650-300 = 350 meter

23. A train crosses a post in 15 seconds and a platform 100 m long in 25 seconds. Its length is

A. 150 m

B. 300 m

C. 400 m

D. 180 m

Explanation :

Assume x is the length of the train and v is the speed x/v = 15
=> v = x/15
(x+100)/v = 25
=> v = (x+100)/25
Ie, x/15 = (x+100)/25
=> 5x = 3x+ 300
=> x = 300/2 = 150

24. A train , 800 meter long is running with a speed of 78 km/hr. It crosses a tunnel in 1 minute. What is the length of the tunnel (in meters)?

A. 440 m

B. 500 m

C. 260 m

D. 430 m

Explanation :

Distance = 800+x meter where x is the length of the tunnel
Time = 1 minute = 60 seconds
Speed = 78 km/hr = 78×10/36 m/s = 130/6 = 65/3 m/s
Distance/time = speed
(800+x)/60 = 65/3
=> 800+x = 20×65 = 1300
=> x = 1300 – 800 = 500 meter

25. Two trains each 500 m long, are running in opposite directions on parallel tracks. If their speeds are 45 km/ hr and 30 km/hr respectively, the time taken by the slower train to pass the driver of the faster one is

A. 50 sec

B. 58 sec

C. 24 sec

D. 22 sec

Explanation :

Relative speed = 45+30 = 75 km/hr = 750/36 m/s = 125/6 m/s
We are calculating the time taken by the slower train to pass the driver of the faster one
Hence the distance = length of the smaller train = 500 m
Time = distance/speed = 500/(125/6) = 24 s

26. Two trains are running in opposite directions at the same speed. The length of each train is 120 meter. If they cross each other in 12 seconds, the speed of each train (in km/hr) is

A. 42

B. 36

C. 28

D. 20

Explanation :

Distance covered = 120+120 = 240 m
Time = 12 s
Let the speed of each train = v. Then relative speed = v+v = 2v
2v = distance/time = 240/12 = 20 m/s
Speed of each train = v = 20/2 = 10 m/s
= 10×36/10 km/hr = 36 km/hr

27. A train 108 m long is moving at a speed of 50 km/hr . It crosses a train 112 m long coming from opposite direction in 6 seconds. What is the speed of the second train?

A. 82 kmph

B. 76 kmph

C. 44 kmph

D. 58 kmph

Explanation :

Total distance = 108+112 = 220 m
Time = 6s
Relative speed = distance/time = 220/6 m/s = 110/3 m/s
= (110/3) × (18/5) km/hr = 132 km/hr
=> 50 + speed of second train = 132 km/hr
=> Speed of second train = 132-50 = 82 km/hr

28. How many seconds will a 500-meter long train moving with a speed of 63 km/hr, take to cross a man walking with a speed of 3 km/hr in the direction of the train?

A. 42

B. 50

C. 30

D. 28

Explanation :

Distance = 500m
Speed = 63 -3 km/hr = 60 km/hr = 600/36 m/s =
50/3 m/s
Time taken = distance/speed= 500/(50/3) = 30 s

Categories

Maths : Problems on Age (Solved in Steps)

1. A man’s age is 125% of what it was 10 years ago, but 83 1/3 % of what it will be after ten 10 years. What is his present age?

A.70

B.60

C.50

D.40

Explanation :

Let the age before 10 years = x
Then 125x/100 = x + 10
=>125x = 100x + 1000
=> x = 1000/25 = 40
Present age = x + 10 = 40 +10 = 50

2. Ten years ago, R was half of S in age. If the ratio of their present ages is 3:4, what will be the total of their present ages?

A. 45

B. 40

C. 35

D. 30

Explanation :

Let the present age of R and S be 3x and 4x respectively. Ten years ago, R was half of S in age
=> (3x – 10) = ½ (4x – 10)
=> 6x – 20 = 4x – 10
=> 2x = 10
=> x = 5
total of their present ages = 3x + 4x = 7x = 7 × 5 = 35

3. Father is aged three times more than his son Anil. After 8 years, he would be two and a half times of Anil’s age. After further 8 years, how many times would he be of Anil’s age?

A. 4 times

B. 5 times

C. 2 times

D. 3 times

Explanation :

Assume that Anil’s present age = x.
Then father’s present age = 3x + x = 4x
After 8 years, fathers age = 2 ½ times of Anils’ age
=> (4x+8) = 2 ½ × (x+8)
=> 4x + 8 = (5/2) ×(x + 8)
=> 8x + 16 = 5x+ 40
=> 3x = 40-16 = 24
=> x = 24/3 = 8
After further 8 years,
Anil’s age = x + 8+ 8 = 8+8+8 = 24
Father’s age = 4x + 8 + 8 = 4 × 8 + 8 + 8 = 48
Father’s age/Anil’s age = 48/24 = 2

4. A man is 24 years older than his son. In two years, his age will be twice the age of his son. What is the present age of his son?

A.23

B.22

C.21

D.20

Explanation :

Let the present age of the son = x years
Then present age the man = (x+24) years
Given that in 2 years, man’s age will be twice the age of his son
=> (x+24) +2 = 2(x+2)
=> x = 22

5. Present ages of Kiran and Syam are in the ratio of 5 : 4 respectively. Three years hence, the ratio of their ages will become 11 : 9 respectively. What is Syam’s present age in years?

A.28

B.27

C.26

D.24

Explanation :

Ratio of the present age of Kiran and Syam = 5 : 4
=> Let the present age of Kiran = 5x
Present age of Syam = 4x
After 3 years, ratio of their ages = 11:9
=> (5x + 3) : (4x + 3) = 11 : 9
=> (5x+3) / (4x+3) = 11/9
=> 9(5x + 3) = 11(4x + 3)
=> 45x + 27 = 44x + 33
=> x = 33-27 =6
Syam’s present age = 4x = 4×6 = 24

6. The sum of ages of 5 children born at the intervals of 3 years each is 50 years. Find out the age of the youngest child?

A. 6

B.5

C.4

D.3

Explanation :

Let the age of the youngest child = x
Then the ages of 5 children can be written as x,
(x+3), (x+6),(x+9) and (x+12)
X + (x+3) + (x+6) + (x+9) + (x+12) = 50
=> 5x + 30 =50
=> 5x = 20
=> x = 20/5 = 4

7. A is two years older than B who is twice as old as C. The total of the ages of A, B and C is 27. How old is B?

A.10

B.9

C.8

D.7

Explanation :

Let the age of C = x. Then
Age of B = 2x
Age of A = 2 + 2x
The total age of A,B and C =27
=> (2+2x) + 2x + x = 27
=> 5x = 25
=> 25/5 = 5
B’s age = 2x = 2×5 = 10

8. The Average age of a class of 22 students is 21 years. The average increased by 1 when the teacher’s age also included. What is the age of the teacher?

A. 48

B. 45

C. 43

D. 44

Explanation :

Total age of all students = 22×21
Total age of all students + age of the teacher = 23 × 22
Age of the teacher = 23×22 – 22×21 = 22(23-21) = 22×2 = 44

9. A father said to his son, “I was as old as you are at the present at the time of your birth”. If the father’s age is 38 years now, what was the son’s age five
years back?

A. 20 years

B. 18 years

C. 14 years

D. 22 years

Explanation :

Let the son’s present age be x years.
Then, (38 – x) = x
=> 2x = 38
=> x = 38/2 = 19
Son’s age 5 years back = 19-5 = 14

10. Ayisha’s age is 1/6th of her father’s age. Ayisha ‘s father’s age will be twice the age of Shankar’s age after 10 years. If Shankar’s eight birthdays was celebrated two years before, then what is Ayisha ‘s present age?

A. 10 years

B. 18 years

C. 8 years

D. 5 years

Explanation :

Consider Ayisha’s present age = x
Then her father’s age = 6x
Given that Ayisha ‘s father’s age will be twice the age of Shankar’s age after 10 years
=> Shankar’s age after 10 years = ½ (6x + 10) =
3x + 5
Also given that Shankar’s eight birthdays was celebrated two years before =>
Shankar’s age after 10 years = 8 + 12 = 20
=> 3x + 5 = 20
=> x = 15/3 = 5
=> Ayisha ‘s present age = 5 years

11.The sum of the present ages of a son and his father is 60 years. Six years ago, father’s age was five times the age of the son. After 6 years, what will be son’s age?

A 23 years

B. 22 years

C. 21 years

D. 20 years

Explanation :

Let the present age of the son = x, then
Present age of the father = 60-x
Six years ago father’s age was 5 times the age of the son
=> (60-x)-6 = 5(x-6)
=> 84 = 6x
=> x = 84/6 = 14
Son’s age after 6 years = x + 6 = 14 + 6 = 20

12.Kiran is younger than Bineesh by 7 years and their ages are in the respective ratio of 7 : 9, how old is Kiran?

A. 25

B. 24.5

C. 24

D. 23.5

Explanation :

Let the ages of Kiran and Bineesh are 7x and 9x respectively
7x = 9x-7
=> x = 7/2 = 3.5
Kiran’s age = 7x = 7 × 3.5 = 24.5

13.Six years ago, the ratio of the ages of Vimal and Saroj was 6 : 5. Four years hence, the ratio of their ages will be 11 : 10. What is Saroj’s age at present?

A. 18

B. 17

C. 16

D. 15

Explanation :

Given that , six years ago, the ratio of the ages of Vimal and Saroj = 6 : 5
Hence we can assume that
The age of Vimal six years ago = 6x
The age of Saroj six years ago = 5x
After 4 years, the ratio of their ages = 11 : 10
=> (6x+10) / (5x + 10) = 11/10
=> 10(6x + 10) = 11(5x + 10)
=> 5x = 10
=> x = 10/5 = 2
Saroj’s present age = 5x + 6 = 5×2 +6 = 16

14.One year ago sunu’s & Manu’s age was 6:7 respectievely. Four years hence, the ratio would become 7:8. How old is sunu ?

A. 31

B. 36

C. 37

D. 30

Explanation :

Let sunu’s & manu’s ages one year ago be 6x & 7x years respectively. Then sunu’s age four years hence
= (6x+1)+4=6x+5 years
Manu’s age four years hence= (7x+1)+4=7x+5 years
ie 6x+5/7x+5=7/8
=> 8(6x+5)=7(7x+5)
=> 48x+40=49x+35
=> 49x-48x=40-35
=> x=5
hence sunu’s present age is=6x+1 = 6×5+1=30+1=31 years

16.My brother is 3 years elder to me. My father was 28 years of age when my sister was born while my mother was 26 years of age when I was born. If my sister was 4 years of age when my brother was born, then what was the age my father when my brother was born?

A. 35

B. 34

C. 33

D.32

Explanation :

Let my age = x
Then My brother’s age = x + 3
My mother’s age = x + 26
My sister’s age = (x + 3) + 4 = x + 7
My Father’s age = (x + 7) + 28 = x + 35
=> age my father when my brother was born = x + 35 – (x + 3) = 32

17.The present ages of A,B and C are in proportions 4 : 7 : 9. Eight years ago, the sum of their ages was 56. What are their present ages (in years)?

A. Insufficient data

B. 16, 30, 40

C. 16, 28 40

D. 16, 28, 36

Explanation :

Let’s take the present age of A, B and C as 4x, 7x and 9x respectively
Then (4x – 8) + (7x – 8) + (9x – 8) = 56
=> 20x = 80
=> x = 4
Hence the present age of A, B and C are 4×4,
7×4 and 9×4 respectively
ie, 16,28 and 36 respectively.

18.A person’s present age is two-fifths of the age of his mother. After 8 years, he will be one-half of the age of his mother. What is the present age of the mother?

A.60

B. 50

C. 40

D.30

Explanation :

Let the present age of the person = x .
Then present age of the mother = 5x/2
Given that , after 8 years, the person will be onehalf of the age of his mother.
=> (x + 8) = (1/2)(5x/2 + 8)
=> 2x + 16 = 5x/2 + 8
=> x/2 = 8
=> x = 16
Present age of the mother = 5x/2 = 5×16/2 = 40

19.A is as much younger than B and he is older than C. If the sum of the ages of B and C is 50 years, what is definitely the difference between B and A’s age?

B. 3

C. 2

D. 5

Explanation :

Age of C < Age of A < Age of B
Given that sum of the ages of B and C is 50 years.
=> Let’s take B’s age = x and C’s age = 50-x
Now we need to find out B’s age – A’s age. But we cannot find out this with the given data.

20.Sobha’s father was 38 years of age when she was born while her mother was 36 years old when her brother four years younger to her was born. What is the difference between the ages of her parents?

A. 6

B. 5

C. 4

D. 12

Explanation :

Let Sobha’s age = x and her brother’s age = x-4
Then Sobha’s father’s age = x + 38
Sobha’s mother’s age = (x-4) + 36
=> Sobha’s father’s age – Sobha’s mother’s age = (x + 38) – [(x-4) + 36]
= x + 38 –x +4 – 36 = 6

21.The age of father 10 years ago was thrice the age of his son. Ten years hence, father’s age will be twice that of his son. What is the ratio of their present ages?

A. 7 : 3

B. 3 : 7

C. 9 : 4

D.4 : 9

Explanation :

Let the age of the son before 10 years = x and age of the father before 10 years = 3x
Now we can write as (3x + 20) = 2(x + 20)
=> x = 20
=> Age of Father the son at present = x + 10 = 20 + 10 = 30
Age of the father at present = 3x + 10 = 3×20 + 10 = 70
Required ratio = 70 : 30 = 7 : 3

22.The ages of two persons differ by 16 years. 6 years ago, the elder one was 3 times as old as the younger one. What are their present ages of the elder person?

A.10

B. 20

C. 30

D.40

Explanation :

Let’s take the present age of the elder person = x and the present age of the younger person = x – 16 (x – 6) = 3 (x-16-6)
=> x – 6 = 3x – 66
=> 2x = 60
=> x = 60/2 = 30

23.The present age of a father is 3 years more than three times the age of his son. Three years hence, father’s age will be 10 years more than twice the age of the son. What is father’s present age?

A.30

B.31

C. 32

D.33

Explanation :

Let the present age the son = x.
Then present age of the father = 3x + 3
Given that , three years hence, father’s age will be 10 years more than twice the age of the son
=> (3x+3+3) = 2(x + 3) +10
=> x = 10
Father’s present age = 3x + 3 = 3×10+ 3 = 33

24.Kamal was 4 times as old as his son 8 years ago. After 8 years, Kamal will be twice as old as his son. Find out the present age of Kamal.

A.40

B.38

C. 42

D.36

Explanation :

Let the age of the son before 8 years = x.
Then age of Kamal before 8 years ago = 4x
After 8 years, Kamal will be twice as old as his son
=> 4x + 16 = 2(x + 16)
=> x = 8
Present age of Kamal = 4x + 8 = 4×8 +8 = 40

25.If 6 years are subtracted from the present age of Ajay and the remainder is divided by 18, then the present age of Rahul is obtained. If Rahul is 2 years younger to Denis whose age is 5 years, then what is Ajay ‘s present age?

A.50

B.60

C. 55

D.62

Explanation :

Present age of Denis = 5 years
Present age of Rahul = 5-2 = 3
Let the present age of Ajay = x
Then (x-6)/18 = present age of Rahul = 3
=> x- 6 = 3×18 = 54
=> x = 54 + 6= 60

26.The ratio of the age of a man and his wife is 4:3. At the time of marriage the ratio was 5:3 and After 4 years this ratio will become 9:7. How many years ago were they married?

A.8

B.10

C. 11

D.12

Explanation :

Let the present age of the man and his wife be 4x and 3x respectively.
After 4 years this ratio will become 9:7
=> (4x + 4)/ (3x + 4) = 9/7
=> 28x + 28 = 27x + 36
=> x = 8
=> Present age of the man = 4x = 4×8 = 32
Present age of his wife = 3x = 3×8 = 24
Assume that they got married before t years.
Then (32 – t) / (24 – t) = 5/3
=> 96 – 3t = 120 – 5t
=> 2t = 24
=> t = 24/2 = 12

27.The product of the ages of Syam and Sunil is 240. If twice the age of Sunil is more than Syam’s age by 4 years, what is Sunil’s age?

A.16

B.14

C. 12

D.10

Explanation :

Let the age of Sunil = x and age of Syam = y.
Then xy = 240 —(1) 2x = y + 4
=> y = 2x – 4
=> y = 2(x – 2) —(2)
Substituting equation (2) in equation (1). We get x × 2(x-2) = 240
=> x (x-2) = 240/2
=> x (x -2) = 120 —(3)
We got a quadratic equation to solve. Always time is precious and objective tests measures not only how accurate you are but also how fast you are. We can either solve this quadratic equation in the traditional way. But the more easy way is just substitute the values given in the choices in the quadratic equation (equation 3 ) and see which choice satisfy the equation.
Here the option A is 10. If we substitute that value in the quadratic equation, x(x-2) = 10 × 8 which is not equal to 120
Now try option B which is 12. If we substitute that value in the quadratic equation, x(x-2) = 12 × 10 = 120.
See, we got that x = 12.
Hence Sunil’s age = 12

28.One year ago, the ratio of Sooraj’s and Vimal’s age was 6: 7 respectively. Four years hence, this ratio would become 7: 8. How old is Vimal?

A.32

B.34

C. 36

D.38

Explanation :

Let take the age of Sooraj and Vimal , 1 year ago as 6x and 7x respectively. Given that, four years hence, this ratio would become 7: 8.
=> (6x + 5)/(7x + 5) = 7/8
=> 48x + 40 = 49x + 35
=> x = 5
Vimal’s present age = 7x + 1 = 7×5 + 1 = 36

29.The total age of A and B is 12 years more than the total age of B and C. C is how many years younger than A?

A.10

B.11

C. 12

D.13

Explanation :

Given that A+B = 12 + B + C
=> A – C = 12 + B – B = 12
=> C is younger than A by 12 years

30.Sachin’s age after 15 years will be 5 times his age 5 years back. Find out the present age of Sachin?

A.10

B.11

C. 12

D.13

Explanation :

Let the present age of Sachin = x
Then (x+15) = 5(x-5)
=> 4x = 40
=> x = 10

31.Sandeep’s age after six years will be three-seventh of his father’s age. Ten years ago the ratio of their ages was 1 : 5. What is Sandeep’s father’s age at present?

A.30

B.40

C. 50

D.60