Aptitude & reasoning

Age Problems

 

Important Formulas on “Problems on Ages”:

 

1. If the current age is x, then ntimes the age is nx.
2. If the current age is x, then age nyears later/hence = x+ n.
3. If the current age is x, then age nyears ago = x– n.
4. The ages in a ratio a: bwill be ax and bx.

5. If the current age is x, then	1	of the age is	x	.
	n		n

Example:

A problem with one variable: How old is Al?

Many single-variable algebra word problems have to do with the relations between different people’s ages. For example:

Al’s father is 45. He is 15 years older than twice Al’s age. How old is Al?

We can begin by assigning a variable to what we’re asked to find. Here this is Al’s age, so let Al’s age = x.

We also know from the information given in the problem that 45 is 15 more than twice Al’s age. How can we translate this from words into mathematical symbols? What is twice Al’s age?

Well, Al’s age is x, so twice Al’s age is 2x, and 15 more than twice Al’s age is 15 + 2x.That equals 45, right? Now we have an equation in terms of one variable that we can solve for x: 45 = 15 + 2x.

original statement of the problem:	45 = 15 + 2x
subtract 15 from each side:	30 = 2x
divide both sides by 2:	15 = x

Since x is Al’s age and x = 15, this means that Al is 15 years old.

It’s always a good idea to check our answer:

twice Al’s age is 2 x 15:	30
15 more than 30 is 15 + 30:	45

This should be the age of Al’s father, and it is.

 

 

Questions:

Level-I:

 

1.	Father is aged three times more than his son Ronit. After 8 years, he would be two and a half times of Ronit’s age. After further 8 years, how many times would he be of Ronit’s age?
	A. 2 times B. 2 1 times 2 C. 2 3 times 4 D. 3 times

 

2.	The sum of ages of 5 children born at the intervals of 3 years each is 50 years. What is the age of the youngest child?
	A. 4 years B. 8 years C. 10 years D. None of these

 

3.	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, the son’s age five years back was:
	A. 14 years B. 19 years C. 33 years D. 38 years

 

4.	A is two years older than B who is twice as old as C. If the total of the ages of A, B and C be 27, the how old is B?
	A. 7 B. 8 C. 9 D. 10 E. 11

 

5.	Present ages of Sameer and Anand are in the ratio of 5 : 4 respectively. Three years hence, the ratio of their ages will become 11 : 9 respectively. What is Anand’s present age in years?
	A. 24 B. 27 C. 40 D. Cannot be determined E. None of these

 

6.	A man is 24 years older than his son. In two years, his age will be twice the age of his son. The present age of his son is:
	A. 14 years B. 18 years C. 20 years D. 22 years

 

7.	Six years ago, the ratio of the ages of Kunal and Sagar was 6 : 5. Four years hence, the ratio of their ages will be 11 : 10. What is Sagar’s age at present?
	A. 16 years B. 18 years C. 20 years D. Cannot be determined E. None of these

 

8.	The sum of the present ages of a father and his son is 60 years. Six years ago, father’s age was five times the age of the son. After 6 years, son’s age will be:
	A. 12 years B. 14 years C. 18 years D. 20 years

 

9.	At present, the ratio between the ages of Arun and Deepak is 4 : 3. After 6 years, Arun’s age will be 26 years. What is the age of Deepak at present ?
	A. 12 years B. 15 years C. 19 and half D. 21 years

 

10.	Sachin is younger than Rahul by 7 years. If their ages are in the respective ratio of 7 : 9, how old is Sachin?
	A. 16 years B. 18 years C. 28 years D. 24.5 years E. None of these

 

 

11.	Level-II:         The present ages of three persons in proportions 4 : 7 : 9. Eight years ago, the sum of their ages was 56. Find their present ages (in years).
	A. 8, 20, 28 B. 16, 28, 36 C. 20, 35, 45 D. None of these

 

12.	Ayesha’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. 2 years B. 4 years C. 6 years D. 8 years

 

13.	A person’s present age is two-fifth of the age of his mother. After 8 years, he will be one-half of the age of his mother. How old is the mother at present?
	A. 32 years B. 36 years C. 40 years D. 48 years

 

14.	Q is as much younger than R as he is older than T. If the sum of the ages of R and T is 50 years, what is definitely the difference between R and Q’s age?
	A. 1 year B. 2 years C. 25 years D. Data inadequate E. None of these

 

15.	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. The ratio of their present ages is:
	A. 5 : 2 B. 7 : 3 C. 9 : 2 D. 13 : 4

 

16.	What is Sonia’s present age? I. Sonia’s present age is five times Deepak’s present age.  II. Five years ago her age was twenty-five times Deepak’s age at that time.
	A. I alone sufficient while II alone not sufficient to answer B. II alone sufficient while I alone not sufficient to answer C. Either I or II alone sufficient to answer D. Both I and II are not sufficient to answer E. Both I and II are necessary to answer

 

17.	Average age of employees working in a department is 30 years. In the next year, ten workers will retire. What will be the average age in the next year? I. Retirement age is 60 years.  II. There are 50 employees in the department.
	A. I alone sufficient while II alone not sufficient to answer B. II alone sufficient while I alone not sufficient to answer C. Either I or II alone sufficient to answer D. Both I and II are not sufficient to answer E. Both I and II are necessary to answer

 

 

18.	Divya is twice as old as Shruti. What is the difference in their ages? I. Five years hence, the ratio of their ages would be 9 : 5.  II. Ten years back, the ratio of their ages was 3 : 1.
	A. I alone sufficient while II alone not sufficient to answer B. II alone sufficient while I alone not sufficient to answer C. Either I or II alone sufficient to answer D. Both I and II are not sufficient to answer E. Both I and II are necessary to answer

 

 

 

Answers:

Level-I:

 

Answer:1 Option A

 

Explanation:

Let Ronit’s present age be x years. Then, father’s present age =(x + 3x) years = 4x years.

	(4x + 8) =	5	(x + 8)
		2

8x + 16 = 5x + 40

3x = 24

x = 8.

Hence, required ratio =	(4x + 16)	=	48	= 2.
	(x + 16)		24

 

 

Answer:2 Option A

 

Explanation:

Let the ages of children be x, (x + 3), (x + 6), (x + 9) and (x + 12) years.

Then, x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 50

5x = 20

x = 4.

Age of the youngest child = x = 4 years.

 

 

 

Answer:3 Option A

 

Explanation:

Let the son’s present age be x years. Then, (38 – x) = x

2x = 38.

x = 19.

Son’s age 5 years back (19 – 5) = 14 years.

 

Answer:4 Option D

 

Explanation:

Let C’s age be x years. Then, B’s age = 2x years. A’s age = (2x + 2) years.

(2x + 2) + 2x + x = 27

5x = 25

x = 5.

Hence, B’s age = 2x = 10 years.

 

Answer:5 Option A

 

Explanation:

Let the present ages of Sameer and Anand be 5x years and 4x years respectively.

Then,	5x + 3	=	11
	4x + 3		9

9(5x + 3) = 11(4x + 3)

45x + 27 = 44x + 33

45x – 44x = 33 – 27

x = 6.

Anand’s present age = 4x = 24 years.

 

Answer:6 Option D

 

Explanation:

Let the son’s present age be x years. Then, man’s present age = (x + 24) years.

(x + 24) + 2 = 2(x + 2)

x + 26 = 2x + 4

x = 22.

 

Answer:7 Option A

 

Explanation:

Let the ages of Kunal and Sagar 6 years ago be 6x and 5x years respectively.

Then,	(6x + 6) + 4	=	11
	(5x + 6) + 4		10

10(6x + 10) = 11(5x + 10)

5x = 10

x = 2.

Sagar’s present age = (5x + 6) = 16 years.

 

Answer:8 Option D

 

Explanation:

Let the present ages of son and father be x and (60 –x) years respectively.

Then, (60 – x) – 6 = 5(x – 6)

54 – x = 5x – 30

6x = 84

x = 14.

Son’s age after 6 years = (x+ 6) = 20 years..

 

Answer:9 Option B

 

Explanation:

Let the present ages of Arun and Deepak be 4x years and 3x years respectively. Then,

4x + 6 = 26        4x = 20

x = 5.

Deepak’s age = 3x = 15 years.

 

Answer:10 Option D

 

Explanation:

Let Rahul’s age be x years.

Then, Sachin’s age = (x – 7) years.

	x – 7	=	7
	x		9

9x – 63 = 7x

2x = 63

x = 31.5

Hence, Sachin’s age =(x – 7) = 24.5 years.

 

Answer:11 Option B

 

Explanation:

Let their present ages be 4x, 7x and 9x years respectively.

Then, (4x – 8) + (7x – 8) + (9x – 8) = 56

20x = 80

x = 4.

Their present ages are 4x = 16 years, 7x = 28 years and 9x = 36 years respectively.

 

Answer:12 Option C

 

Explanation:

Mother’s age when Ayesha’s brother was born = 36 years.

Father’s age when Ayesha’s brother was born = (38 + 4) years = 42 years.

Required difference = (42 – 36) years = 6 years.

 

Answer:13 Option C

 

Explanation:

Let the mother’s present age be x years.

Then, the person’s present age =		2	x		years.
		5

 

		2	x + 8		=	1	(x + 8)
		5				2

2(2x + 40) = 5(x + 8)

x = 40.

 

Answer:14 Option D

 

Explanation:

Given that:

1. The difference of age b/w R and Q = The difference of age b/w Q and T.

2. Sum of age of R and T is 50 i.e. (R + T) = 50.

Question: R – Q = ?.

Explanation:

R – Q = Q – T

(R + T) = 2Q

Now given that, (R + T) = 50

So, 50 = 2Q and therefore Q = 25.

Question is (R – Q) = ?

Here we know the value(age) of Q (25), but we don’t know the age of R.

Therefore, (R-Q) cannot be determined.

 

Answer:15 Option B

 

Explanation:

Let the ages of father and son 10 years ago be 3x and x years respectively.

Then, (3x + 10) + 10 = 2[(x + 10) + 10]

3x + 20 = 2x + 40

x = 20.

Required ratio = (3x + 10) : (x + 10) = 70 : 30 = 7 : 3.

 

 

Answer:16  Option E

 

Explanation:

I. S = 5D     D =	S	….(i)
	5

1. S – 5 = 25 (D – 5)    S = 25D – 120 ….(ii)

Using (i) in (ii), we get S =		25 x	S		– 120
			5

4S = 120.

S = 30.

Thus, I and II both together give the answer. So, correct answer is (E).

 

Answer:17 Option E

 

Explanation:

1. Retirement age is 60 years.
2. There are 50 employees in the department.

Average age of 50 employees = 30 years.

Total age of 50 employees = (50 x 30) years = 1500 years.

Number of employees next year = 40.

Total age of 40 employees next year (1500 + 40 – 60 x 10) = 940.

Average age next year =	940	years = 23	1	years.
	40		2

Thus, I and II together give the answer. So, correct answer is (E).

 

Answer:18   Option C

 

Explanation:

Let Divya’s present age be D years and Shruti’s present age b S years

Then, D = 2 x S        D – 2S = 0 ….(i)

I.	D + 5	=	9	….(ii)
	S + 5		5

 

II.	D – 10	=	3	….(iii)
	S – 10		1

From (ii), we get : 5D + 25 = 9S + 45        5D – 9S = 20 ….(iv)

From (iii), we get : D – 10 = 3S – 30        D – 3S = -20 ….(v)

Thus, from (i) and (ii), we get the answer.

Also, from (i) and (iii), we get the answer.

I alone as well as II alone give the answer. Hence, the correct answer is (C).

 

• Percentage

  Percent means for every 100

  So, when percent is calculated for any value, it means that we calculate the value for every 100 of the reference value.

  percent is denoted by the symbol %. For example, x percent is denoted by x%

• x%=x/100

  Example : 25%=25/100=1/4

• To express x/y as a percent,we have x/y=(x/y×100)%

  Example : 1/4=(1/4×100)%=25%

• If the price of a commodity increases by R%, the reduction in consumptionso as not to increase the expenditure = [R/(100+R)×100]%
• If the price of a commodity decreases by R%, the increase in consumptionso as not to decrease the expenditure = [R/(100−R)×100]%
• If the population of a town = P and it increases at the rate of R% per annum, thenPopulation after n years = P((1+R)/100))n
• If the population of a town = P and it increases at the rate of R% per annum, thenPopulation before n years = P((1+R)/100))n
• If the present value of a machine = P and it depreciates at the rate of R% per annum,

ThenValue of the machine after n years = P((1-R)/100))n

• If the present value of a machine = P and it depreciates at the rate of R% per annum,

ThenValue of the machine before n years = P((1-R)/100))n

 

Solved Examples

Level 1

1.    If A = x% of y and B = y% of x, then which of the following is true?
A. None of these	B. A is smaller than B.
C. Relationship between A and B cannot be determined.	D. If x is smaller than y, then A is greater than B.
E. A is greater than B.

 

Answer : Option A

Explanation :

A = xy/100 ………….(Equation 1)

B = yx/100……………..(Equation 2)

From these equations, it is clear that A = B

 

 

2.If 20% of a = b, then b% of 20 is the same as:
A. None of these	B. 10% of a
C. 4% of a	D. 20% of a

 

Answer :Option C

Explanation :

20% of a = b

=> b = 20a/100

b% of 20 = 20b/100=(20a/100) × 20/100

=(20×20×a)/(100×100)=4a/100 = 4% of a

 

3.Two numbers A and B are such that the sum of 5% of A and 4% of B is two-third of the sum of 6% of A and 8% of B. Find the ratio of A : B.
A. 2 : 1	B. 1 : 2
C. 1 : 1	D. 4 : 3

 

 

Answer :Option D

Explanation :

5% of A + 4% of B = 2/3(6% of A + 8% of B)

5A/100+4B/100=2/3(6A/100+8B/100)

⇒5A+4B=2/3(6A+8B)

⇒15A+12B=12A+16B

⇒3A=4B

⇒AB=43⇒A:B=4:3

4.The population of a town increased from 1,75,000 to 2,62,500 in a decade. What is the average percent increase of population per year?
A. 4%	B. 6%
C. 5%	D. 50%

 

Answer :Option C

Explanation :

Increase in the population in 10 years = 2,62,500 – 1,75,000 = 87500

% increase in the population in 10 years = (87500/175000)×100=8750/175=50%

Average % increase of population per year = 50%/10=5%

 

5.Three candidates contested an election and received 1136, 7636 and 11628 votes respectively. What percentage of the total votes did the winning candidate get?
A. 57%	B. 50%
C. 52%	D. 60%

 

Answer :Option A

Explanation :

Votes received by the winning candidate = 11628

Total votes = 1136 + 7636 + 11628 = 20400

Required percentage = (11628/20400)×100=11628/204=2907/51=969/17=57%

 

6.A fruit seller had some oranges. He sells 40% oranges and still has 420 oranges. How many oranges he had originally?
A. 420	B. 700
C. 220	D. 400

 

Answer :Option B

Explanation :

He sells 40% of oranges and still there are 420 oranges remaining

=> 60% of oranges = 420

⇒(60×Total Oranges)/100=420

⇒Total Oranges/100=7

⇒ Total Oranges = 7×100=700

7.A batsman scored 110 runs which included 3 boundaries and 8 sixes. What percent of his total score did he make by running between the wickets?
A. 499/11 %	B. 45 %
C. 500/11 %	D. 489/11 %

 

Answer :Option C

Explanation :

Total runs scored = 110

Total runs scored from boundaries and sixes = 3 x 4 + 8 x 6 = 60

Total runs scored by running between the wickets = 110 – 60 = 50

Required % = (50/110)×100=500/11%

 

8.What percentage of numbers from 1 to 70 have 1 or 9 in the unit’s digit?
A. 2023%	B. 20%
C. 21%	D. 2223%

 

 

Answer :Option B

Explanation :

Total numbers = 70

Total numbers in 1 to 70 which has 1 in the unit digit = 7

Total numbers in 1 to 70 which has 9 in the unit digit = 7

Total numbers in 1 to 70 which has 1 or 9 in the unit digit = 7 + 7 = 14

Required percentage = (14/70)×100=140/7=20%

 

Level 2

 

1.In an election between two candidates, one got 55% of the total valid votes, 20% of the votes were invalid. If the total number of votes was 7500, what was the number of valid votes that the other candidate got?
A. 2800	B. 2700
C. 2100	D. 2500

 

Answer :Option B

Explanation :

Total number of votes = 7500

Given that 20% of Percentage votes were invalid

=> Valid votes = 80%

Total valid votes = (7500×80)/100

1^st candidate got 55% of the total valid votes.

Hence the 2^nd candidate should have got 45% of the total valid votes
=> Valid votes that 2nd candidate got = (total valid votes ×45)/100

=7500×(80/100)×(45/100)=75×(4/5)×45=75×4×9=300×9=2700

 

2.In a competitive examination in State A, 6% candidates got selected from the total appeared candidates. State B had an equal number of candidates appeared and 7% candidates got selected with 80 more candidates got selected than A. What was the number of candidates appeared from each State?
A. 8200	B. 7500
C. 7000	D. 8000

 

Answer :Option D

Explanation :

State A and State B had an equal number of candidates appeared.

In state A, 6% candidates got selected from the total appeared candidates

In state B, 7% candidates got selected from the total appeared candidates

But in State B, 80 more candidates got selected than State A

From these, it is clear that 1% of the total appeared candidates in State B = 80

=> total appeared candidates in State B = 80 x 100 = 8000

=> total appeared candidates in State A = total appeared candidates in State B = 8000

 

3.In a certain school, 20% of students are below 8 years of age. The number of students above 8 years of age is 2/3 of the number of students of 8 years of age which is 48. What is the total number of students in the school?
A. 100	B. 102
C. 110	D. 90

 

 

Answer :Option A

Explanation :

Let the total number of students = x

Given that 20% of students are below 8 years of age

then The number of students above or equal to 8 years of age = 80% of x —–(Equation 1)

Given that number of students of 8 years of age = 48 —–(Equation 2)

Given that number of students above 8 years of age = 2/3 of number of students of 8 years of age

=>number of students above 8 years of age = (2/3)×48=32—–(Equation 3)

From Equation 1,Equation 2 and Equation 3,
80% of x = 48 + 32 = 80

⇒80x/100=80

⇒x100=1⇒x=100

4.In an examination, 5% of the applicants were found ineligible and 85% of the eligible candidates belonged to the general category. If 4275 eligible candidates belonged to other categories, then how many candidates applied for the examination?
A. 28000	B. 30000
C. 32000	D. 33000

 

Answer :Option B

Explanation :

Let the number of candidates applied for the examination = x

Given that 5% of the applicants were found ineligible.

It means that 95% of the applicants were eligible (∴ 100% – 5% = 95%)

Hence total eligible candidates = 95x/100

Given that 85% of the eligible candidates belonged to the general category

It means 15% of the eligible candidates belonged to other categories(∴ 100% – 85% = 15%)
Hence Total eligible candidates belonged to other categories

=(total eligible candidates×15)/100=(95x/100)×(15/100)

=(95x×15)/(100×100)

Given that Total eligible candidates belonged to other categories = 4275

⇒(95x×15)/(100×100)=4275

⇒(19x×15)/(100×100)=855

⇒(19x×3)/(100×100)=171

⇒(x×3)/(100×100)=9

⇒x/(100×100)=3

⇒x=3×100×100=30000

 

5.A student multiplied a number by 3/5 instead of 5/3.What is the percentage error in the calculation?
A. 64%	B. 32%
C. 34%	D. 42%

 

Answer :Option A

Explanation :

Let the number = 1

Then, ideally he should have multiplied 1 by 5/3.

Hence the correct result was 1 x (5/3) = (5/3)

By mistake, he multiplied 1 by 3/5.

Hence the result with the error = 1 x (3/5) = (3/5)

Error = 5/3−3/5=(25−9)/15=16/15

percentage error = (Error/True Value)×100={(16/15)/(5/3)}×100

=(16×3×100)/(15×5)=(16×100)/(5×5)=16×4=64%

 

6.The price of a car is Rs. 3,25,000. It was insured to 85% of its price. The car was damaged completely in an accident and the insurance company paid 90% of the insurance. What was the difference between the price of the car and the amount received ?
A. Rs. 76,375	B. Rs. 34,000
C. Rs. 82,150	D. Rs. 70,000

 

Answer :Option A

Explanation :

Price of the car = Rs.3,25,000

Car insured to 85% of its price

=>Insured price=(325000×85)/100

Insurance company paid 90% of the insurance
⇒Amount paid by Insurance company =(Insured price×90)/100

=325000×(85/100)×(90/100)=325×85×9=Rs.248625

Difference between the price of the car and the amount received

= Rs.325000 – Rs.248625 = Rs.76375

7.If the price of petrol increases by 25% and Benson intends to spend only an additional 15% on petrol, by how much % will he reduce the quantity of petrol purchased?
A. 8%	B. 7%
C. 10%	D. 6%

 

 

Answer :Option A

Explanation :

Assume that the initial price of 1 Litre petrol = Rs.100 ,Benson spends Rs.100 for petrol,

such that Benson buys 1 litre of petrol

After the increase by 25%, price of 1 Litre petrol = (100×(100+25))/100=Rs.125

Since Benson spends additional 15% on petrol,

amount spent by Benson = (100×(100+15))/100=Rs.115

Hence Quantity of petrol that he can purchase = 115/125 Litre

Quantity of petrol reduced = (1−115/125) Litre

Percentage Quantity of reduction = ((1−115/125))/1×100=(10/125)/×100=(10/5)×4=2×4=8%

8.30% of the men are more than 25 years old and 80% of the men are less than or equal to 50 years old. 20% of all men play football. If 20% of the men above the age of 50 play football, what percentage of the football players are less than or equal to 50 years?
A. 60%	B. 70%
C. 80%	D. 90%

 

Answer :Option C

Explanation :

Let total number of men = 100

Then

80 men are less than or equal to 50 years old

(Since 80% of the men are less than or equal to 50 years old)

=> 20 men are above 50 years old (Since we assumed total number of men as 100)

20% of the men above the age of 50 play football

⇒Number of men above the age of 50 who play football = (20×20)/100=4

Number of men who play football = 20 (Since 20% of all men play football)
Percentage of men who play football above the age of 50 = (4/20)×100=20%

=>Percentage of men who play football less than or equal to the age 50 = 100%−20%=80%

 

 

 

Introduction:

There are four main directions – East, West, North and South as shown below:

 

 

 

 

There are four cardinal directions – North-East (N-E), North-West (N-W), South-East (S-E), and South-West (S-W) as shown below:

 

 

 

Key points

 

1. At the time of sunrise if a man stands facing the east, his shadow will be towards west.
2. At the time of sunset the shadow of an object is always in the east.
3. If a man stands facing the North, at the time of sunrise his shadow will be towards his left and at the time of sunset it will be towards his right.
4. At 12:00 noon, the rays of the sun are vertically downward hence there will be no shadow

 

 

 

 

 

 

 

 

 

Practice Questions

Type 1:

Siva starting from his house, goes 5 km in the East, then he turns to his left and goes 4 km. Finally he turns to his left and goes 5 km. Now how far is he from his house and in what direction?

Solution:

From third position it is clear he is 4 km from his house and is in North direction.

 

 

 

 

 

 

Type 2:

Suresh starting from his house, goes 4 km in the East, then he turns to his right and goes 3 km. What minimum distance will be covered by him to come back to his house?

Solution:

 

Type 3:

One morning after sunrise Juhi while going to school met Lalli at Boring road crossing. Lalli’s shadow was exactly to the right of Juhi. If they were face to face, which direction was Juhi facing?

 

Solution: In the morning sunrises in the east.

So in morning the shadow falls towards the west.

Now Lalli’s shadow falls to the right of the Juhi. Hence Juhi is facing South.

 

 

 

 

Type 4: Hema starting from her house walked 5 km to reach the crossing of Palace. In which direction she was going, a road opposite to this direction goes to Hospital. The road to the right goes to station. If the road which goes to station is just opposite to the road which IT-Park, then in which direction to Hema is the road which goes to IT-Park?

Solution:

From II it is clear that the road which goes to IT-Park is left to Hema.

 

 

 

 

 

 

 

 

Questions

 

Level-1

 

1.	One morning Udai and Vishal were talking to each other face to face at a crossing. If Vishal’s shadow was exactly to the left of Udai, which direction was Udai facing?
	A. East B. West C. North D. South

2.	Y is in the East of X which is in the North of Z. If P is in the South of Z, then in which direction of Y, is P?
	A. North B. South C. South-East D. None of these

3.	If South-East becomes North, North-East becomes West and so on. What will West become?
	A. North-East B. North-West C. South-East D. South-West

4.	A man walks 5 km toward south and then turns to the right. After walking 3 km he turns to the left and walks 5 km. Now in which direction is he from the starting place?
	A. West B. South C. North-East D. South-West

5.	Rahul put his timepiece on the table in such a way that at 6 P.M. hour hand points to North. In which direction the minute hand will point at 9.15 P.M. ?
	A. South-East   B. South   C. North D. West
6.	Rasik walked 20 m towards north. Then he turned right and walks 30 m. Then he turns right and walks 35 m. Then he turns left and walks 15 m. Finally he turns left and walks 15 m. In which direction and how many metres is he from the starting position?
	A. 15 m West B. 30 m East C. 30 m West D. 45 m East

7.

Two cars start from the opposite places of a main road, 150 km apart. First car runs for 25 km and takes a right turn and then runs 15 km. It then turns left and then runs for another 25 km and then takes the direction back to reach the main road. In the mean time, due to minor break down the other car has run only 35 km along the main road. What would be the distance between two cars at this point?

A. 65 km B. 75 km C. 80 km D. 85 km

8.	Starting from the point X, Jayant walked 15 m towards west. He turned left and walked 20 m. He then turned left and walked 15 m. After this he turned to his right and walked 12 m. How far and in which directions is now Jayant from X?
	A. 32 m, South B. 47 m, East C. 42 m, North D. 27 m, South

 

9.	One evening before sunset Rekha and Hema were talking to each other face to face. If Hema’s shadow was exactly to the right of Hema, which direction was Rekha facing?
	A. North B. South C. East D. Data is inadequate

10.	A boy rode his bicycle Northward, then turned left and rode 1 km and again turned left and rode 2 km. He found himself 1 km west of his starting point. How far did he ride northward initially?
	A. 1 km B. 2 km C. 3 km D. 5 km

 

 

Answers:

1Answer: Option C

Explanation:

 

2Answer: Option D

Explanation:

P is in South-West of Y.

 

3Answer: Option C

Explanation:

It is clear from the diagrams that new name of West will become South-East.

 

4Answer: Option D

Explanation:

Hence required direction is South-West.

 

5Answer: Option D

Explanation:

At 9.15 P.M., the minute hand will point towards west.

 

6Answer: Option D

Explanation:

 

7Answer: Option A

Explanation:

 

 

 

 

8Answer: Option A

Explanation:

 

9Answer: Option B

Explanation:

In the evening sun sets in West. Hence then any shadow falls in the East. Since Hema’s shadow was to the right of Hema. Hence Rekha was facing towards South.

 

10Answer: Option B

Explanation:

The boy rode 2 km. Northward

 

 

Level – 2

 

Dev, Kumar, Nilesh, Ankur and Pintu are standing facing to the North in a playground such as given below:

1. Kumar is at 40 m to the right of Ankur.
2. Dev is are 60 m in the south of Kumar.
3. Nilesh is at a distance of 25 m in the west of Ankur.
4. Pintu is at a distance of 90 m in the North of Dev

 

 

1.	Which one is in the North-East of the person who is to the left of Kumar?
	A. Dev B. Nilesh C. Ankur D. Pintu

2.	If a boy starting from Nilesh, met to Ankur and then to Kumar and after this he to Dev and then to Pintu and whole the time he walked in a straight line, then how much total distance did he cover?
	A. 215 m B. 155 m C. 245 m D.   185 m

 

 

Directions to Solve

Each of the following questions is based on the following information:

1. Six flats on a floor in two rows facing North and South are allotted to P, Q, R, S, T and U.
2. Q gets a North facing flat and is not next to S.
3. S and U get diagonally opposite flats.
4. R next to U, gets a south facing flat and T gets North facing flat.

 

 

3.	If the flats of P and T are interchanged then whose flat will be next to that of U?
	A. P B. Q C. R D. T

4.	Which of the following combination get south facing flats?
	A. QTS B. UPT C. URP D. Data is inadequate

5.	The flats of which of the other pair than SU, is diagonally opposite to each other?
	A. QP B. QR C. PT D. TS
6.	Whose flat is between Q and S?
	A. T B. U C. R D. P

 

 

Directions to Solve

Each of the following questions is based on the following information:

1. 8-trees → mango, guava, papaya, pomegranate, lemon, banana, raspberry and apple are in two rows 4 in each facing North and South.
2. Lemon is between mango and apple but just opposite to guava.
3. Banana is at one end of a line and is just next in the right of guava or either banana tree is just after guava tree.
4. Raspberry tree which at one end of a line, is just diagonally opposite to mango tree.

 

 

	7 .Which of the following statements is definitely true?
	A. Papaya tree is just near to apple tree. B. Apple tree is just next to lemon tree. C. Raspberry tree is either left to Pomegranate or after. D. Pomegranate tree is diagonally opposite to banana tree.

8	Which tree is just opposite to raspberry tree?
	A. Papaya B. Pomegranate C. Papaya or Pomegranate D. Data is inadequate
9	Which tree is just opposite to banana tree?
	A. Mango B. Pomegranate C. Papaya D. Data is inadequate

 

 

Answer: 1 Option D

Explanation:

Ankur is in the left of Kumar. Hence Pintu is in North-East of Ankur

 

 

Answer: 2 Option A

Explanation:

Required distance = 25 m + 40 m + 60 m + 90 m

Required distance = 215 m

 

 

Answer:3 Option C

Explanation:

Hence flat R will be next to U.

 

 

Answer:4 Option C

 

Explanation:

Hence URP flat combination get south facing flats.

 

Answer:5 Option A

 

Explanation:

Hence QP is diagonally opposite to each other.

 

 

 

 

 

 

Answer:6 Option A

 

Explanation:

Hence flat T is between Q and S.

 

Answer: 7 Option B

 

Explanation:

 

 

Answer:8 Option C

 

Explanation:

 

 

 

 

 

 

 

Answer:9 Option A

 

Explanation:

Profit and loss

 

IMPORTANT FACTS

Cost Price:

The price, at which an article is purchased, is called its cost price, abbreviated as C.P.

 

Selling Price:

The price, at which an article is sold, is called its selling prices, abbreviated as S.P.

 

Profit or Gain:

If S.P. is greater than C.P., the seller is said to have a profit or gain.

 

Loss:

If S.P. is less than C.P., the seller is said to have incurred a loss.

 

IMPORTANT FORMULAE

1. Gain = (S.P.) – (C.P.)
2. Loss = (C.P.) – (S.P.)
3. Loss or gain is always reckoned on C.P.
4. Gain Percentage: (Gain %)

Gain % =		Gain x 100
		C.P.

5. Loss Percentage: (Loss %)

Loss % =		Loss x 100
		C.P.

6. Selling Price: (S.P.)

SP =		(100 + Gain %)	x C.P
		100

7. Selling Price: (S.P.)

SP =		(100 – Loss %)	x C.P.
		100

8. Cost Price: (C.P.)

C.P. =		100	x S.P.
		(100 + Gain %)

9. Cost Price: (C.P.)

C.P. =		100	x S.P.
		(100 – Loss %)

10. If an article is sold at a gain of say 35%, then S.P. = 135% of C.P.
11. If an article is sold at a loss of say, 35% then S.P. = 65% of C.P.
12. When a person sells two similar items, one at a gain of say x%, and the other at a loss of x%, then the seller always incurs a loss given by:

Loss % =		Common Loss and Gain %		2	=		x		2	.
		10					10

13. If a trader professes to sell his goods at cost price, but uses false weights, then

Gain % =		Error	x 100	%.
		(True Value) – (Error)

 

Questions:

Level-I:

 

 

1.	Alfred buys an old scooter for Rs. 4700 and spends Rs. 800 on its repairs. If he sells the scooter for Rs. 5800, his gain percent is:
	A. 4 4 % 7 B. 5 5 % 11 C. 10% D. 12%

 

2.	The cost price of 20 articles is the same as the selling price of x articles. If the profit is 25%, then the value of xis:
	A. 15 B. 16 C. 18 D. 25

 

3.	If selling price is doubled, the profit triples. Find the profit percent.
	A. 66 2 3 B. 100 C. 105 1 3 D. 120

 

4.	In a certain store, the profit is 320% of the cost. If the cost increases by 25% but the selling price remains constant, approximately what percentage of the selling price is the profit?
	A. 30% B. 70% C. 100% D. 250%

 

 

5.	A vendor bought toffees at 6 for a rupee. How many for a rupee must he sell to gain 20%?
	A. 3 B. 4 C. 5 D. 6

 

6.	The percentage profit earned by selling an article for Rs. 1920 is equal to the percentage loss incurred by selling the same article for Rs. 1280. At what price should the article be sold to make 25% profit?
	A. Rs. 2000 B. Rs. 2200 C. Rs. 2400 D. Data inadequate

 

7.	A shopkeeper 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. Rs. 18.20 B. Rs. 70 C. Rs. 72 D. Rs. 88.25

 

8.	A man buys a cycle for Rs. 1400 and sells it at a loss of 15%. What is the selling price of the cycle?
	A. Rs. 1090 B. Rs. 1160 C. Rs. 1190 D. Rs. 1202

 

9.	Sam 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. What was his percentage profit?
	A. 3.5 B. 4.5 C. 5.6 D. 6.5

 

10.	Some articles were bought at 6 articles for Rs. 5 and sold at 5 articles for Rs. 6. Gain percent is:
	A. 30% B. 33 1 % 3 C. 35% D. 44%
11.	Level-II:     On selling 17 balls at Rs. 720, there is a loss equal to the cost price of 5 balls. The cost price of a ball is:
	A. Rs. 45 B. Rs. 50 C. Rs. 55 D. Rs. 60

 

 

12.	When a plot is sold for Rs. 18,700, the owner loses 15%. At what price must that plot be sold in order to gain 15%?
	A. Rs. 21,000 B. Rs. 22,500 C. Rs. 25,300 D. Rs. 25,800

 

13.	100 oranges are bought at the rate of Rs. 350 and sold at the rate of Rs. 48 per dozen. The percentage of profit or loss is:
	A. 14 2 % gain 7 B. 15% gain C. 14 2 % loss 7 D. 15 % loss

 

14.	A shopkeeper sells one transistor for Rs. 840 at a gain of 20% and another for Rs. 960 at a loss of 4%. His total gain or loss percent is:
	A. 5 15 % loss 17 B. 5 15 % gain 17 C. 6 2 % gain 3 D. None of these

 

 

15.	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. His profit percent is:
	A. No profit, no loss B. 5% C. 8% D. 10% E. None of these

 

16. A man buys an article for Rs. 27.50 and sells it for Rs 28.60. Find his gain percent
17. 1%
18. 2%
19. 3%
20. 4%

 

 

17. A TV is purchased at Rs. 5000 and sold at Rs. 4000, find the lost percent.
18. 10%
19. 20%
20. 25%
21. 28%

 

 

18. In terms of percentage profit, which among following the best transaction.
    1. P. 36, Profit 17
    2. P. 50, Profit 24
    3. P. 40, Profit 19
    4. P. 60, Profit 29

 

 

 

 

Answer:1 Option B

 

Explanation:

Cost Price (C.P.) = Rs. (4700 + 800) = Rs. 5500.

Selling Price (S.P.) = Rs. 5800.

Gain = (S.P.) – (C.P.) = Rs.(5800 – 5500) = Rs. 300.

Gain % =		300	x 100	%	= 5	5	%
		5500				11

 

Answer:2 Option B

 

Explanation:

Let C.P. of each article be Re. 1 C.P. of x articles = Rs. x.

S.P. of x articles = Rs. 20.

Profit = Rs. (20 – x).

		20 – x	x 100 = 25
		x

2000 – 100x = 25x

125x = 2000

x = 16.

 

 

Answer:3 Option B

 

Explanation:

Let C.P. be Rs. x and S.P. be Rs. y.

Then, 3(y – x) = (2y – x)    y = 2x.

Profit = Rs. (y – x) = Rs. (2x – x) = Rs. x.

Profit % =		x	x 100	% = 100%

 

 

Answer:4 Option B

 

Explanation:

Let C.P.= Rs. 100. Then, Profit = Rs. 320, S.P. = Rs. 420.

New C.P. = 125% of Rs. 100 = Rs. 125

New S.P. = Rs. 420.

Profit = Rs. (420 – 125) = Rs. 295.

Required percentage =		295	x 100	%	=	1475	% = 70% (approximately).
		420				21

 

 

Answer:5 Option C

 

Explanation:

C.P. of 6 toffees = Re. 1

S.P. of 6 toffees = 120% of Re. 1 = Rs.	6
	5

 

For Rs.	6	, toffees sold = 6.
	5

 

For Re. 1, toffees sold =		6 x	5		= 5.
			6

 

 

Answer:6 Option A   Explanation: Let C.P. be Rs. x. Then, 1920 – x x 100 = x – 1280 x 100 x x 1920 – x = x – 1280 2x = 3200 x = 1600  Required S.P. = 125% of Rs. 1600 = Rs. 125 x 1600 = Rs 2000. 100

 

 

Answer:7 Option C

 

Explanation:

C.P. = Rs.		100	x 392		= Rs.		1000	x 392		= Rs. 320
		122.5					1225

Profit = Rs. (392 – 320) = Rs. 72.

 

Answer:8 Option C

 

Explanation:

S.P. = 85% of Rs. 1400 = Rs.		85	x 1400		= Rs. 1190
		100

 

 

 

Answer:9 Option C

 

Explanation:

Cost Price of 1 toy = Rs.		375		= Rs. 31.25
		12

Selling Price of 1 toy = Rs. 33

So, Gain = Rs. (33 – 31.25) = Rs. 1.75

Profit % =		1.75	x 100	%	=	28	% = 5.6%
		31.25				5

 

 

 

Answer:10 Option D

 

Explanation:

Suppose, number of articles bought = L.C.M. of 6 and 5 = 30.

C.P. of 30 articles = Rs.		5	x 30		= Rs. 25.
		6

 

S.P. of 30 articles = Rs.		6	x 30		= Rs. 36.
		5

 

Gain % =		11	x 100	% = 44%.
		25

 

 

Answer:11 Option D

 

Explanation:

(C.P. of 17 balls) – (S.P. of 17 balls) = (C.P. of 5 balls)

C.P. of 12 balls = S.P. of 17 balls = Rs.720.

C.P. of 1 ball = Rs.		720		= Rs. 60.
		12

 

 

Answer:12 Option C

 

Explanation:

85 : 18700 = 115 : x

x =		18700 x 115		= 25300.
		85

Hence, S.P. = Rs. 25,300.

 

Answer:13 Option A

 

Explanation:

C.P. of 1 orange = Rs.		350		= Rs. 3.50
		100

 

S.P. of 1 orange = Rs.		48		= Rs. 4
		12

 

Gain% =		0.50	x 100	%	=	100	% = 14	2	%
		3.50				7		7

 

 

 

Answer:14 Option B

 

Explanation:

C.P. of 1st transistor = Rs.		100	x 840		= Rs. 700.
		120

 

C.P. of 2nd transistor = Rs.		100	x 960		= Rs. 1000
		96

So, total C.P. = Rs. (700 + 1000) = Rs. 1700.

Total S.P. = Rs. (840 + 960) = Rs. 1800.

Gain % =		100	x 100	%	= 5	15	%
		1700				17

 

 

 

Answer:15 Option B

 

Explanation:

C.P. of 56 kg rice = Rs. (26 x 20 + 30 x 36) = Rs. (520 + 1080) = Rs. 1600.

S.P. of 56 kg rice = Rs. (56 x 30) = Rs. 1680.

Gain =		80	x 100	% = 5%.
		1600

 

Answer:16 Option D

 

Explanation:

So we have C.P. = 27.50
S.P. = 28.60

Gain = 28.60 – 27.50 = Rs. 1.10

Gain%=(Gain/Cost∗100)%=(1.10/27.50∗100)%=4%

 

 

 

 

Answer:17 Option B

 

Explanation:

We know, C.P. = 5000
S.P. = 4000
Loss = 5000 – 4000 = 1000
Loss%=(Loss/Cost∗100)%=(1000/5000∗100)%=20%

 

 

Answer:18 Option D

 

Explanation:

Hint: Calculate profit percent as

Profit% = (profit/cost) * 100

Square Root & Cube Root

 

Step 1: First of all group the number in pairs of 2 starting from the right.

 

Step 2: To get the ten’s place digit, Find the nearest square (equivalent or greater than or less than) to the first grouped pair from left and put the square root of the square.

 

Step 3: To get the unit’s place digit of the square root

 

Remember the following

If number ends in	Unit’s place digit of the square root
1	1 or 9(10-1)
4	2 or 8(10-2)
9	3 or 7(10-3)
6	4or 6(10-4)
5	5
0	0

 

Lets see the logic behind this for a better understanding

We know,

1²=1

2²=4

3²=9

4²=16

5²=25

6²=36

7²=49

8²=64

9²=81

10²=100

 

Now, observe the unit’s place digit of all the squares.

Do you find anything common?

 

We notice that,

Unit’s place digit of both 1² and 9² is 1.

Unit’s place digit of both 2² and 8² is 4

Unit’s place digit of both 3² and 7² is 9

Unit’s place digit of both 4² and 6² is 6.

Step 4: Multiply the ten’s place digit (found in step 1) with its consecutive number and compare the result obtained with the first pair of the original number from left.

 

Remember,

If first pair of the original number > Result obtained on multiplication then  select the greater number  out of the two numbers as the unit’s place digit of the square root.

 

If firstpair of the original number < the result obtained on multiplication,then select the lesser number out of the two numbers as the unit’s place digit of the square root.

 

 

Let us consider an example to get a better understanding of the method

 

 

Example 1: √784=?

Step 1: We start by grouping the numbers in pairs of two from right as follows

7 84

 

Step 2: To get the ten’s place digit,

We find that nearest square to first group (7) is 4 and √4=2

Therefore ten’s place digit=2

 

Step 3: To get the unit’s place digit,

We notice that the number ends with 4, So the unit’s place digit of the square root should be either 2 or 8(Refer table).

 

Step 4: Multiplying the ten’s place digit of the square root that we arrived at in step 1(2) and its consecutive number(3) we get,

2×3=6
ten’s place digit of original number > Multiplication result
7>6
So we need to select the greater number (8) as the unit’s place digit of the square root.
Unit’s place digit =8

Ans:√784=28

 

 

 

Cube roots of perfect cubes

It may take two-three minutes to find out cube root of a perfect cube by using conventional method. However we can find out cube roots of perfect cubes very fast, say in one-two seconds using Vedic Mathematics.

We need to remember some interesting properties of numbers to do these quick mental calculations which are given below.

 

Points to remember  for speedy  calculation of cube roots

1. To calculate cube root of any perfect cube quickly, we need to remember the cubes of 1 to 10 which is given below.

13	=	1
23	=	8
33	=	27
43	=	64
53	=	125
63	=	216
73	=	343
83	=	512
93	=	729
103	=	1000

2. From the above cubes of 1 to 10, we need to remember an interesting property.

13 = 1	=>	If last digit of the perfect cube = 1, last digit of the cube root = 1
23 = 8	=>	If last digit of the perfect cube = 8, last digit of the cube root = 2
33 = 27	=>	If last digit of the perfect cube = 7, last digit of the cube root = 3
43 = 64	=>	If last digit of the perfect cube = 4, last digit of the cube root = 4
53 = 125	=>	If last digit of the perfect cube =5, last digit of the cube root = 5
63 = 216	=>	If last digit of the perfect cube = 6, last digit of the cube root = 6
73 = 343	=>	If last digit of the perfect cube = 3, last digit of the cube root = 7
83 = 512	=>	If last digit of the perfect cube = 2, last digit of the cube root = 8
93 = 729	=>	If last digit of the perfect cube = 9, last digit of the cube root = 9
103 = 1000	=>	If last digit of the perfect cube = 0, last digit of the cube root = 0

 

It’s very easy to remember the relations given above because

1	->	1	(Same numbers)
8	->	2	(10’s complement of 8 is 2 and 8+2 = 10)
7	->	3	(10’s complement of 7 is 3 and 7+3 = 10)
4	->	4	(Same numbers)
5	->	5	(Same numbers)
6	->	6	(Same numbers)
3	->	7	(10’s complement of 3 is 7 and 3+7 = 10)
2	->	8	(10’s complement of 2 is 8 and 2+8 = 10)
9	->	9	(Same numbers)
0	->	0	(Same numbers)

 

Also see
8 ->  2 and 2 ->  8
7 -> 3 and 3-> 7

 

 

 

 

 

Questions

Level-I

1.	The cube root of .000216 is:
	A. .6 B. .06 C. 77 D. 87

 

2.	What should come in place of both x in the equation x = 162 . 128 x
	A. 12 B. 14 C. 144 D. 196

 

3.	The least perfect square, which is divisible by each of 21, 36 and 66 is:
	A. 213444 B. 214344 C. 214434 D. 231444

 

4.	1.5625 = ?
	A. 1.05 B. 1.25 C. 1.45 D. 1.55

 

5.	If 35 + 125 = 17.88, then what will be the value of 80 + 65 ?
	A. 13.41 B. 20.46 C. 21.66 D. 22.35
6.	If a = 0.1039, then the value of 4a2 – 4a + 1 + 3a is:
	A. 0.1039 B. 0.2078 C. 1.1039 D. 2.1039

 

7.	If x = 3 + 1 and y = 3 – 1 , then the value of (x2 + y2) is: 3 – 1 3 + 1
	A. 10 B. 13 C. 14 D. 15

 

8.	A group of students decided to collect as many paise from each member of group as is the number of members. If the total collection amounts to Rs. 59.29, the number of the member is the group is:
	A. 57 B. 67 C. 77 D. 87

 

9.	The square root of (7 + 35) (7 – 35) is
	A. 5 B. 2 C. 4 D. 35

 

 

 

10.	If 5 = 2.236, then the value of 5 – 10 + 125 is equal to: 2 5
	A. 5.59 B. 7.826 C. 8.944 D. 10.062       Level-II
11.	625 x 14 x 11 is equal to: 11 25 196
	A. 5 B. 6 C. 8 D. 11

 

12.	0.0169 x ? = 1.3
	A. 10 B. 100 C. 1000 D. None of these

 

13.	3 – 1 2 simplifies to: 3
	A. 3 4 B. 4 3 C. 4 3 D. None of these

 

14.	How many two-digit numbers satisfy this property.: The last digit (unit’s digit) of the square of the two-digit number is 8 ?
	A. 1 B. 2 C. 3 D. None of these

 

15.	The square root of 64009 is:
	A. 253 B. 347 C. 363 D. 803

 

 

16. √29929 = ?

A. 173 B. 163 C. 196 D. 186

 

 

 

 

 

 

17. √106.09 = ?
A.	10.6
B.	10.5
C.	10.3
D.	10.2

 

18.  ?/√196 = 5
A.	76
B.	72
C.	70
D.	75

 

Answers

Level-I

 

Answer:1 Option B

 

Explanation:

(.000216)1/3	=		216		1/3
			106

 

=		6 x 6 x 6		1/3
		102 x 102 x 102

 

=	6
	102

 

=	6
	100

= 0.06

 

Answer:2 Option A

 

Explanation:

Let	x	=	162
	128		x

Then x² = 128 x 162

= 64 x 2 x 18 x 9

= 8² x 6² x 3²

= 8 x 6 x 3

= 144.

x = 144 = 12.

 

Answer:3 Option A

 

Explanation:

L.C.M. of 21, 36, 66 = 2772.

Now, 2772 = 2 x 2 x 3 x 3 x 7 x 11

To make it a perfect square, it must be multiplied by 7 x 11.

So, required number = 2² x 3² x 7² x 11² = 213444

 

Answer:4 Option B

 

Explanation:

1|1.5625( 1.25

|1

|——-

22| 56

| 44

|——-

245| 1225

| 1225

|——-

|    X

|——-

1.5625 = 1.25.

 

 

Answer:5 Option D

 

Explanation:

35 + 125 = 17.88

35 + 25 x 5 = 17.88

35 + 55 = 17.88

85 = 17.88

5 = 2.235

80 + 65 = 16 x 5 + 65

= 45 + 65

= 105 = (10 x 2.235) = 22.35

 

 

 

Answer:6 Option C

 

Explanation:

4a² – 4a + 1 + 3a = (1)² + (2a)² – 2 x 1 x 2a + 3a

= (1 – 2a)² + 3a

= (1 – 2a) + 3a

= (1 + a)

= (1 + 0.1039)

= 1.1039

 

Answer:7 Option C

 

Explanation:

x =	(3 + 1)	x	(3 + 1)	=	(3 + 1)2	=	3 + 1 + 23	= 2 + 3.
	(3 – 1)		(3 + 1)		(3 – 1)		2

 

y =	(3 – 1)	x	(3 – 1)	=	(3 – 1)2	=	3 + 1 – 23	= 2 – 3.
	(3 + 1)		(3 – 1)		(3 – 1)		2

x² + y² = (2 + 3)² + (2 – 3)²

= 2(4 + 3)

= 14

 

Answer:8 Option C

 

Explanation:

Money collected = (59.29 x 100) paise = 5929 paise.

Number of members = 5929 = 77

 

 

Answer:9 Option B

 

Explanation:

(7 + 35)(7 – 35)

=

(7)2 – (35)2

= 49 – 45  = 4  = 2

 

 

Answer:10 Option B

 

Explanation:

5	–	10	+ 125	=	(5)2 – 20 + 25 x 55
2		5			25

 

=	5 – 20 + 50
	25

 

=	35	x	5
	25		5

 

=	355
	10

 

=	7 x 2.236
	2

 

= 7 x 1.118

 

= 7.826

 

 

Level-II

Answer:11 Option A

 

Explanation:

Given Expression =	25	x	14	x	11	= 5.
	11		5		14

 

 

 

Answer:12 Option B

 

Explanation:

Let 0.0169 x x = 1.3.

Then, 0.0169x = (1.3)² = 1.69

x =	1.69	= 100
	0.0169

 

 

 

Answer:13 Option C

 

Explanation:

	3 –	1		2	= (3)2 +		1		2	– 2 x 3 x	1
		3					3				3

 

= 3 +	1	– 2
	3

 

= 1 +	1
	3

 

=	4
	3

 

 

 

Answer:14 Option D

 

Explanation:

A number ending in 8 can never be a perfect square.

 

 

Answer:15 Option A

 

Explanation:

2 |64009( 253      |4      |———-45  |240      |225      |———-503| 1509      |  1509      |———-      |     X      |———-

64009 = 253.

 

 

Answer:16 Option A

 

Explanation:
√29929 = So, √29929 = 173

 

 

Answer:17 Option C

 

Answer:18 Option C

 

Speed has no sense of direction unlike the velocity. Relative speed is the speed of one object as observed from another moving object. Questions on train are the classic examples of relative speed and in all these questions it is assumed that trains move parallel to each other – whether in the same direction or the opposite direction. Thus, we shall see how the relative speed is calculated and using it we come to know the time taken by the trains to cross each other and some other like aspects.

Important Formulas – Problems on Trains

1. x km/hr = (x×5)/18 m/s

 

2. y m/s = (y×18)/5 km/hr

 

3. Speed = distance/time, that is, s = d/t

 

4. velocity = displacement/time, that is, v = d/t

 

5. Time taken by a train x meters long to pass a pole or standing man or a post
   = Time taken by the train to travel x meters.

 

6. Time taken by a train x meters long to pass an object of length y meters

= Time taken by the train to travel (x + y) metres.

 

7. Suppose two trains or two objects are moving in the same direction at v1 m/s and v2 m/s where v1 > v2,

then their relative speed = (v1 – v2) m/s

 

8. Suppose two trains or two objects are moving in opposite directions at v1 m/s and v2 m/s ,

then their relative speed = (v1+ v2) m/s

 

9. Assume two trains of length x metres and y metres are moving in opposite directions at v1 m/s and v2 m/s, Then

The time taken by the trains to cross each other = (x+y) / (v1+v2) seconds

 

10. Assume two trains of length x metres and y metres are moving in the same direction at at v1 m/s and v2 m/s where v1 > v2, Then

The time taken by the faster train to cross the slower train = (x+y) / (v1-v2) seconds

 

11. Assume that two trains (objects) start from two points P and Q towards each other at the same time and after crossing they take p and q seconds to reach Q and P respectively. Then,

A’s speed: B’s speed = √q: √p

 

 

Solved Examples

Level 1

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 Answer : Option C	D. 120 metres

 

Explanation :

Speed of the train, v = 40 km/hr = 40000/3600 m/s = 400/36 m/s

Time taken to cross, t = 18 s

Distance Covered, d = vt = (400/36)× 18 = 200 m

Distance covered is equal to the length of the train = 200 m

2.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

 

Answer : Option C

Explanation :

v = 240/24 (where v is the speed of the train) = 10 m/s

t = (240+650)/10 = 89 seconds

3.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

 

Answer : Option A

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

4.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

 

Answer : Option A

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

5.Two trains, one from P to Q and the other from Q to P, start simultaneously. After they meet, the trains reach their destinations after 9 hours and 16 hours respectively. The ratio of their speeds is
A. 2 : 3	B. 2 :1
C. 4 : 3	D. 3 : 2

 

Answer : Option C

Explanation :

Ratio of their speeds = Speed of first train : Speed of second train

= √16: √ 9

= 4:3

6.Train having a length of 270 meter 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

 

Answer : Option D

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

7.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.

 

Answer : Option B

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

8.Two trains are running in opposite directions in 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

 

Answer : Option B

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

 

Level 2

1.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

 

Answer : Option B

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

2.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 the train, 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

 

Answer : Option A

Explanation :

Length of the train, l = 150m

Speed of the man, Vm= 2 km/hr

Relative speed, Vr = total distance/time = (150/3) m/s = (150/3) × (18/5) = 180 km/hr

Relative Speed = Speed of train, Vt – Speed of man (As both are moving in the same direction)

=> 180 = Vt – 2 => Vt = 180 + 2 = 182 km/hr

3.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

 

Answer : Option D

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

4.A jogger is running at 9 kmph alongside a railway track in 240 meters ahead of the engine of a 120 meters long train . The train is running at 45 kmph in the same direction. How much time does it take for the train to pass the jogger?
A. 46	B. 36
C. 18	D. 22

 

Answer : Option B

Explanation :

Distance to be covered = 240+ 120 = 360 m

Relative speed = 36 km/hr = 36×10/36 = 10 m/s

Time = distance/speed = 360/10 = 36 seconds

5.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

 

Answer : Option C

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

6.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

 

Answer : Option C

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

7.A train is traveling at 48 kmph. It crosses another train having half of its length, traveling 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

 

Answer : Option D

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

8.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

 

Answer : Option B

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

9.Two trains are running at 40 km/hr and 20 km/hr respectively in the same direction. If the fast train completely passes a man sitting in the slower train in 5 seconds, the length of the fast train is :
A. 19 m	B. 2779 m
C. 1329 m	D. 33 m

 

Answer : Option B

Explanation :

Relative speed = 40-20 = 20 km/hr = 200/36 m/s = 100/18 m/s

Time = 5 s

Distance = speed × time = (100/18) × 5 = 500/18 m = 250/9 = 2779 m = length of the fast train

 

This topic has many practical application in day to day life. In engineering stage it is used in surveying. The basic purpose is to find the unknown variables by observing the angle of the line of sight. This is done by using some the fact that in a right angled triangle the ratio of any two sides is a function of the angle between them. From exam point of view this is one of the more tough sections and tedious to some extent. So an aspirant must thoroughly solve all the questions given here.

Important Formulas

1. Trigonometric Basics

sinθ=oppositeside/hypotenuse=y/r

cosθ=adjacentside/hypotenuse=x/r

tanθ=oppositeside/adjacentside=y/x

cosecθ=hypotenuse/oppositeside=r/y

secθ=hypotenuse/adjacentside=r/x

cotθ=adjacentside/oppositeside=x/y

From Pythagorean theorem, x2+y2=r2 for the right angled triangle mentioned above

 

2. Basic Trigonometric Values

 

θ in degrees	θ in radians	sinθ	cosθ	tanθ
0°	0	0	1	0
30°	π/6	1/2	3/√2	1/√3
45°	π/4	1/√2	1/√2	1
60°	π/3	3/√2	1/2	√3
90°	π/2	1	0	Not defined

 

3. Trigonometric Formulas

Degrees to Radians and vice versa

360°=2π radian

 

Trigonometry – Quotient Formulas

tanθ=sinθ/cosθ

cotθ=cosθ/sinθ

 

Trigonometry – Reciprocal Formulas

cosecθ=1/sinθ

secθ=1/cosθ

cotθ=1/tanθ

 

Trigonometry – Pythagorean Formulas

sin2θ+cos2θ=1

sec2θ−tan2θ=1

cosec2θ−cot2θ=1

 

4. Angle of Elevation

Suppose a man from a point O looks up at an object P, placed above the level of his eye. Then, angle of elevation is the angle between the horizontal and the line from the object to the observer’s eye (the line of sight).

i.e., angle of elevation =  AOP

5. Angle of Depression

Suppose a man from a point O looks down at an object P, placed below the level of his eye. Then, angle of depression is the angle between the horizontal and the observer’s line of sight

i.e., angle of depression =  AOP

6. Angle Bisector Theorem

Consider a triangle ABC as shown above. Let the angle bisector of angle A intersect side BC at a point D. Then BD/DC=AB/AC

(Note that an angle bisector divides the angle into two angles with equal measures.
i.e., BAD = CAD in the above diagram)

7. Few Important Values to memorize

√2=1.414, √3=1.732, √5=2.236

 

Solved Examples

Level 1

1.The angle of elevation of a ladder leaning against a wall is 60º and the foot of the ladder is 12.4 m away from the wall. The length of the ladder is:
A. 14.8 m	B. 6.2 m
C. 12.4 m	D. 24.8 m

 

 

Answer : Option D

Explanation :

Consider the diagram shown above where PR represents the ladder and RQ represents the wall.

cos 60° = PQ/PR

1/2=12.4/PR

PR=2×12.4=24.8 m

2.From a point P on a level ground, the angle of elevation of the top tower is 30º. If the tower is 200 m high, the distance of point P from the foot of the tower is:
A. 346 m	B. 400 m
C. 312 m	D. 298 m

 

 

Answer : Option A

Explanation :

tan 30°=RQ/PQ

1/√3=200/PQ

PQ=200√3=200×1.73=346 m

3.The angle of elevation of the sun, when the length of the shadow of a tree is equal to the height of the tree, is:
A. None of these	B. 60°
C. 45°	D. 30°

 

 

Answer : Option C

Explanation :

Consider the diagram shown above where QR represents the tree and PQ represents its shadow

We have, QR = PQ
Let QPR = θ

tan θ = QR/PQ=1 (since QR = PQ)

=> θ = 45°

i.e., required angle of elevation = 45°

4.An observer 2 m tall is 103√ m away from a tower. The angle of elevation from his eye to the top of the tower is 30º. The height of the tower is:
A. None of these	B. 12 m
C. 14 m	D. 10 m

 

 

Answer : Option B

Explanation :

SR = PQ = 2 m

PS = QR = 10√3m

tan 30°=TS/PS

1/3=TS/10√3

TS=10√3/√3=10 m

TR = TS + SR = 10 + 2 = 12 m

5.From a tower of 80 m high, the angle of depression of a bus is 30°. How far is the bus from the tower?
A. 40 m	B. 138.4 m
C. 46.24 m	D. 160 m

 

 

Answer : Option B

Explanation :

Let AC be the tower and B be the position of the bus.

Then BC = the distance of the bus from the foot of the tower.

Given that height of the tower, AC = 80 m and the angle of depression, DAB = 30°

ABC = DAB = 30° (Because DA || BC)

tan 30°=AC/BC=>tan 30°=80/BC=>BC = 80/tan 30°=80/(1/√3)=80×1.73=138.4 m

i.e., Distance of the bus from the foot of the tower = 138.4 m

6.Find the angle of elevation of the sun when the shadow of a pole of 18 m height is 6√3 m long?
A. 30°	B. 60°
C. 45°	D. None of these

 

 

Answer : Option B

Explanation :

Let RQ be the pole and PQ be the shadow

Given that RQ = 18 m and PQ = 6√3 m

Let the angle of elevation, RPQ = θ

From the right  PQR,

tanθ=RQ/PQ=18/6√3=3/√3=(3×√3)/( √3×√3)=3√3/3=√3

θ=tan−1(3√)=60°

 

Level 2

1.A man on the top of a vertical observation tower observers a car moving at a uniform speed coming directly towards it. If it takes 8 minutes for the angle of depression to change from 30° to 45°, how soon after this will the car reach the observation tower?
A. 8 min 17 second	B. 10 min 57 second
C. 14 min 34 second	D. 12 min 23 second

 

 

Answer : Option B

Explanation :

Consider the diagram shown above. Let AB be the tower. Let D and C be the positions of the car

Then, ADC = 30° , ACB = 45°

Let AB = h, BC = x, CD = y

tan 45°=AB/BC=h/x

=>1=h/x=>h=x——(1)

tan 30°=AB/BD=AB/(BC + CD)=h/(x+y)

=>1/√3=h/(x+y)

=>x + y = √3h

=>y = √3h – x

=>y = √3h−h(∵ Substituted the value of x from equation 1 )

=>y = h(√3−1)

Given that distance y is covered in 8 minutes
i.e, distance h(√3−1) is covered in 8 minutes

Time to travel distance x
= Time to travel distance h (∵ Since x = h as per equation 1).

Let distance h is covered in t minutes

since distance is proportional to the time when the speed is constant, we have

h(√3−1)∝8—(A)

h∝t—(B)

(A)/(B)=>h(√3−1)/h=8/t

=>(√3−1)=8/t

=>t=8/(√3−1)=8/(1.73−1)=8/.73=800/73minutes ≈10 minutes 57 seconds

2.The top of a 15 metre high tower makes an angle of elevation of 60° with the bottom of an electronic pole and angle of elevation of 30° with the top of the pole. What is the height of the electric pole?
A. 5 metres	B. 8 metres
C. 10 metres	D. 12 metres

 

 

Answer : Option C

Explanation :

Consider the diagram shown above. AC represents the tower and DE represents the pole

Given that AC = 15 m , ADB = 30°, AEC = 60°

Let DE = h

Then, BC = DE = h, AB = (15-h) (∵ AC=15 and BC = h), BD = CE

tan 60°=AC/CE=>√3=15/CE=>CE = 15√3— (1)

tan 30°=AB/BD=>1/√3=(15−h)/BD

=>1/√3=(15−h)/(15/√3)(∵ BD = CE and Substituted the value of CE from equation 1)

=>(15−h)=(1/√3)×(15/√3)=15/3=5

=>h=15−5=10 m

i.e., height of the electric pole = 10 m

 

3.Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30º and 45º respectively. If the lighthouse is 100 m high, the distance between the two ships is:
A. 300 m	B. 173 m
C. 273 m	D. 200 m

 

 

Answer : Option C

Explanation :

Let BD be the lighthouse and A and C be the positions of the ships.
Then, BD = 100 m,  BAD = 30° ,  BCD = 45°

tan 30° = BD/BA⇒1/√3=100/BA

⇒BA=100√3

tan 45° = BD/BC

⇒1=100/BC

⇒BC=100

Distance between the two ships = AC = BA + BC
=100√3+100=100(√3+1)=100(1.73+1)=100×2.73=273 m

4.From the top of a hill 100 m high, the angles of depression of the top and bottom of a pole are 30° and 60° respectively. What is the height of the pole?
A. 52 m	B. 50 m
C. 66.67 m	D. 33.33 m

 

Answer : Option C

Explanation :

Consider the diagram shown above. AC represents the hill and DE represents the pole

Given that AC = 100 m

XAD = ADB = 30° (∵ AX || BD )
XAE = AEC = 60° (∵ AX || CE)

Let DE = h

Then, BC = DE = h, AB = (100-h) (∵ AC=100 and BC = h), BD = CE

tan 60°=AC/CE

=>√3=100/CE=>CE = 100/√3— (1)

tan 30°=AB/BD=>1/√3=(100−h)/BD

=>1/√3=(100−h)/(100/√3)(∵ BD = CE and Substituted the value of CE from equation 1 )

=>(100−h)=1/√3×100/√3=100/3=33.33=>h=100−33.33=66.67 m

i.e., the height of the pole = 66.67 m

5.A vertical tower stands on ground and is surmounted by a vertical flagpole of height 18 m. At a point on the ground, the angle of elevation of the bottom and the top of the flagpole are 30° and 60° respectively. What is the height of the tower?
A. 9 m	B. 10.40 m
C. 15.57 m	D. 12 m

 

 

Answer : Option A

Explanation :

Let DC be the vertical tower and AD be the vertical flagpole. Let B be the point of observation.

Given that AD = 18 m, ABC = 60°, DBC = 30°

Let DC be h.

tan 30°=DC/BC

1/√3=h/BC

h=BC√3—— (1)

tan 60°=AC/BC

√3=(18+h)/BC

18+h=BC×√3—— (2)

(1)/(2)=>h/(18+h)=(BC/√3)/(BC×√3)=1/3

=>3h=18+h=>2h=18=>h=9 m

i.e., the height of the tower = 9 m

6.A balloon leaves the earth at a point A and rises vertically at uniform speed. At the end of 2 minutes, John finds the angular elevation of the balloon as 60°. If the point at which John is standing is 150 m away from point A, what is the speed of the balloon?
A. 0.63 meter/sec	B. 2.16 meter/sec
C. 3.87 meter/sec	D. 0.72 meter/sec

 

 

Answer : Option B

Explanation :

Let C be the position of John. Let A be the position at which balloon leaves the earth and B be the position of the balloon after 2 minutes.

Given that CA = 150 m, BCA = 60°

tan 60°=BA/CA

√3=BA/150

BA=150√3

i.e, the distance travelled by the balloon = 150√3meters

time taken = 2 min = 2 × 60 = 120 seconds

Speed = Distance/Time=150√3/120=1.25√3=1.25×1.73=2.16 meter/second

7. The angles of depression and elevation of the top of a wall 11 m high from top and bottom of a tree are 60° and 30° respectively. What is the height of the tree?
A. 22 m	B. 44 m
C. 33 m	D. None of these

 

 

Answer : Option B

Explanation :

Let DC be the wall, AB be the tree.

Given that DBC = 30°, DAE = 60°, DC = 11 m

tan 30°=DC/BC

1/√3=11/BC

BC = 11√3 m

AE = BC =11√3 m—— (1)

tan 60°=ED/AE

√3=ED/11√3[∵ Substituted the value of AE from (1)]

ED =11√3×√3=11×3=33

Height of the tree = AB = EC = (ED + DC) = (33 + 11) = 44 m

 

8. Two vertical poles are 200 m apart and the height of one is double that of the other. From the middle point of the line joining their feet, an observer finds the angular elevations of their tops to be complementary. Find the heights of the poles.
A. 141 m and 282 m	B. 70.5 m and 141 m
C. 65 m and 130 m	D. 130 m and 260 m

 

 

Answer : Option B

Explanation :

Let AB and CD be the poles with heights h and 2h respectively

Given that distance between the poles, BD = 200 m

Let E be the middle point of BD.

Let AEB = θ and CED = (90-θ) (∵ given that angular elevations are complementary)

Since E is the middle point of BD, we have BE = ED = 100 m

From the right  ABE,
tanθ=AB/BE and tanθ=h/100

h = 100tanθ—— (1)

From the right  EDC,

tan(90−θ)=CD/ED

cotθ=2h/100[∵tan(90−θ)=cotθ]

2h =100cotθ—— (2)

(1) × (2) => 2h2=1002[∵tanθ×cotθ=tanθ×1/tanθ=1]

=>√2h=100

=>h=100/√2=(100×√2)/( √2×√2)=50√2=50×1.41=70.5

2h=2×70.5=141

i.e., the height of the poles are 70.5 m and 141 m.

9. To a man standing outside his house, the angles of elevation of the top and bottom of a window are 60° and 45° respectively. If the height of the man is 180 cm and he is 5 m away from the wall, what is the length of the window?
A. 8.65 m	B. 2 m
C. 2.5 m	D. 3.65 m

 

 

Answer : Option D

Explanation :

Let AB be the man and CD be the window

Given that the height of the man, AB = 180 cm, the distance between the man and the wall, BE = 5 m,
DAF = 45° , CAF = 60°

From the diagram, AF = BE = 5 m

From the right  AFD, tan45°=DF/AF

1=DF/5

DF = 5—— (1)From the right  AFC, tan60°=CF/AF

√3=CF/5

CF=5√3—— (2)

Length of the window = CD = (CF – DF)

=5√3−5[∵ Substitued the value of CF and DF from (1) and (2)]=5(√3−1)=5(1.73−1)=5×0.73=3.65 m

10.The elevation of the summit of a mountain from its foot is 45°. After ascending 2 km towards the mountain upon an incline of 30°, the elevation changes to 60°. What is the approximate height of the mountain?
A. 1.2 km	B. 0.6 km
C. 1.4 km	D. 2.7 km

 

 

Answer : Option D

Explanation :

Let A be the foot and C be the summit of a mountain.

Given that CAB = 45°

From the diagram, CB is the height of the mountain. Let CB = x

Let D be the point after ascending 2 km towards the mountain such that
AD = 2 km and given that DAY = 30°

It is also given that from the point D, the elevation is 60°

i.e., CDE = 60°

From the right  ABC,

tan45°=CB/AB

=>1=x/AB[∵ CB = x (the height of the mountain)]

=>AB = x—— (eq:1)

From the right  AYD,

sin30°=DY/AD

=>1/2=DY/2(∵ Given that AD = 2)

=> DY=1—— (eq:2)

cos30°=AY/AD=>√3/2=AY/2(∵ Given that AD = 2)=> AY=√3—— (eq:3)

From the right  CED, tan60°=CE/DE=>tan60°=(CB – EB)/YB∵ [CE = (CB – EB) and DE = YB)]

=>tan60°=(CB – DY)/(AB – AY)[ ∵ EB = DY and YB = (AB – AY)]

=>tan60°=(x – 1)/(x -√3)∵ [CB = x, DY = 1(eq:2), AB=x (eq:1) and AY = 3√(eq:3)]

=>√3=(x – 1)/(x -√3)=>x√3−3=x−1=>x(√3−1)=2=>0.73x=2=>x=2/0.73=2.7

i.e., the height of the mountain = 2.7 km

 

Pipes and Cistern

 

1. Inlet:

A pipe connected with a tank or a cistern or a reservoir, that fills it, is known as an inlet.

 

Outlet:

A pipe connected with a tank or cistern or reservoir, emptying it, is known as an outlet.

 

2. If a pipe can fill a tank in xhours, then:

part filled in 1 hour =	1	.
	x

3. If a pipe can empty a tank in yhours, then:

part emptied in 1 hour =	1	.
	y

4. If a pipe can fill a tank in xhours and another pipe can empty the full tank in y hours (where y > x), then on opening both the pipes, then

the net part filled in 1 hour =		1	–	1		.
		x		y

5. If a pipe can fill a tank in xhours and another pipe can empty the full tank in y hours (where x > y), then on opening both the pipes, then

the net part emptied in 1 hour =		1	–	1		.
		y		x

 

 

Questions:

 

Level-I:

 

1.	Three pipes A, B and C can fill a tank from empty to full in 30 minutes, 20 minutes, and 10 minutes respectively. When the tank is empty, all the three pipes are opened. A, B and C discharge chemical solutions P,Q and R respectively. What is the proportion of the solution R in the liquid in the tank after 3 minutes?
	A. 5 11 B. 6 11 C. 7 11 D. 8 11

 

2.	Pipes A and B can fill a tank in 5 and 6 hours respectively. Pipe C can empty it in 12 hours. If all the three pipes are opened together, then the tank will be filled in:
	A. 1 13 hours 17 B. 2 8 hours 11 C. 3 9 hours 17 D. 4 1 hours 2

 

3.	A pump can fill a tank with water in 2 hours. Because of a leak, it took 2 hours to fill the tank. The leak can drain all the water of the tank in:
	A. 4 1 hours 3 B. 7 hours C. 8 hours D. 14 hours
4.	Two pipes A and B can fill a cistern in 37 minutes and 45 minutes respectively. Both pipes are opened. The cistern will be filled in just half an hour, if the B is turned off after:
	A. 5 min. B. 9 min. C. 10 min. D. 15 min.

 

5.	A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. The time required by the first pipe is:
	A. 6 hours B. 10 hours C. 15 hours D. 30 hours
6.	Two pipes can fill a tank in 20 and 24 minutes respectively and a waste pipe can empty 3 gallons per minute. All the three pipes working together can fill the tank in 15 minutes. The capacity of the tank is:
	A. 60 gallons B. 100 gallons C. 120 gallons D. 180 gallons

 

7.	A tank is filled in 5 hours by three pipes A, B and C. The pipe C is twice as fast as B and B is twice as fast as A. How much time will pipe A alone take to fill the tank?
	A. 20 hours B. 25 hours C. 35 hours D. Cannot be determined E. None of these

 

8.	Two pipes A and B together can fill a cistern in 4 hours. Had they been opened separately, then B would have taken 6 hours more than A to fill the cistern. How much time will be taken by A to fill the cistern separately?
	A. 1 hour B. 2 hours C. 6 hours D. 8 hours

 

9.	Two pipes A and B can fill a tank in 20 and 30 minutes respectively. If both the pipes are used together, then how long will it take to fill the tank?
	A. 12 min B. 15 min C. 25 min D. 50 min

 

10.	Two pipes A and B can fill a tank in 15 minutes and 20 minutes respectively. Both the pipes are opened together but after 4 minutes, pipe A is turned off. What is the total time required to fill the tank?
	A. 10 min. 20 sec. B. 11 min. 45 sec. C. 12 min. 30 sec. D. 14 min. 40 sec.
11.	Level-II:   One pipe can fill a tank three times as fast as another pipe. If together the two pipes can fill the tank in 36 minutes, then the slower pipe alone will be able to fill the tank in:
	A. 81 min. B. 108 min. C. 144 min. D. 192 min.

 

12.	A large tanker can be filled by two pipes A and B in 60 minutes and 40 minutes respectively. How many minutes will it take to fill the tanker from empty state if B is used for half the time and A and B fill it together for the other half?
	A. 15 min B. 20 min C. 27.5 min D. 30 min

 

13.	A tap can fill a tank in 6 hours. After half the tank is filled, three more similar taps are opened. What is the total time taken to fill the tank completely?
	A. 3 hrs 15 min B. 3 hrs 45 min C. 4 hrs D. 4 hrs 15 min

 

14.	Three taps A, B and C can fill a tank in 12, 15 and 20 hours respectively. If A is open all the time and B and C are open for one hour each alternately, the tank will be full in:
	A. 6 hours B. 6 2 hours 3 C. 7 hours D. 7 1 hours 2

 

15.	Three pipes A, B and C can fill a tank in 6 hours. After working at it together for 2 hours, C is closed and A and B can fill the remaining part in 7 hours. The number of hours taken by C alone to fill the tank is:
	A. 10 B. 12 C. 14 D. 16
16.	How much time will the leak take to empty the full cistern? I. The cistern is normally filled in 9 hours.  II. It takes one hour more than the usual time to fill the cistern because of la leak in the bottom.
	A. I alone sufficient while II alone not sufficient to answer B. II alone sufficient while I alone not sufficient to answer C. Either I or II alone sufficient to answer D. Both I and II are not sufficient to answer E. Both I and II are necessary to answer

 

17.	How long will it take to empty the tank if both the inlet pipe A and the outlet pipe B are opened simultaneously? I. A can fill the tank in 16 minutes.  II. B can empty the full tank in 8 minutes.
	A. I alone sufficient while II alone not sufficient to answer B. II alone sufficient while I alone not sufficient to answer C. Either I or II alone sufficient to answer D. Both I and II are not sufficient to answer E. Both I and II are necessary to answer

 

18.	If both the pipes are opened, how many hours will be taken to fill the tank? I. The capacity of the tank is 400 litres. II. The pipe A fills the tank in 4 hours.  III. The pipe B fills the tank in 6 hours.
	A. Only I and II B. Only II and III C. All I, II and III D. Any two of the three E. Even with all the three statements, answer cannot be given.

 

 

Answers:

 

Level-I:

 

Answer:1 Option B

 

Explanation:

Part filled by (A + B + C) in 3 minutes = 3		1	+	1	+	1		=		3 x	11		=	11	.
		30		20		10					60			20

 

Part filled by C in 3 minutes =	3	.
	10

 

Required ratio =		3	x	20		=	6	.
		10		11			11

 

Answer:2 Option C

 

Explanation:

Net part filled in 1 hour		1	+	1	–	1		=	17	.
		5		6		12			60

 

The tank will be full in	60	hours i.e., 3	9	hours.
	17		17

 

 

Answer:3 Option D

 

Explanation:

Work done by the leak in 1 hour =		1	–	3		=	1	.
		2		7			14

Leak will empty the tank in 14 hrs.

 

 

Answer:4 Option B

 

Explanation:

Let B be turned off after x minutes. Then,

Part filled by (A + B) in x min. + Part filled by A in (30 –x) min. = 1.

x		2	+	1		+ (30 – x).	2	= 1
		75		45			75

 

	11x	+	(60 -2x)	= 1
	225		75

11x + 180 – 6x = 225.

x = 9.

 

 

Answer:5 Option C

 

Explanation:

Suppose, first pipe alone takes x hours to fill the tank .

Then, second and third pipes will take (x -5) and (x – 9) hours respectively to fill the tank.

	1	+	1	=	1
	x		(x – 5)		(x – 9)

 

	x – 5 + x	=	1
	x(x – 5)		(x – 9)

(2x – 5)(x – 9) = x(x – 5)

x² – 18x + 45 = 0

(x – 15)(x – 3) = 0

x = 15.    [neglecting x = 3]

 

 

Answer:6 Option C

 

Explanation:

Work done by the waste pipe in 1 minute =	1	–		1	+	1
	15			20		24

 

=		1	–	11
		15		120

 

= –	1	.    [-ve sign means emptying]
	40

 

Volume of	1	part = 3 gallons.
	40

Volume of whole = (3 x 40) gallons = 120 gallon

 

 

Answer:7 Option C

 

Explanation:

Suppose pipe A alone takes x hours to fill the tank.

Then, pipes B and C will take	x	and	x	hours respectively to fill the tank.
	2		4

 

	1	+	2	+	4	=	1
	x		x		x		5

 

	7	=	1
	x		5

x = 35 hrs.

 

Answer:8 Option C

 

Explanation:

Let the cistern be filled by pipe A alone in x hours.

Then, pipe B will fill it in (x + 6) hours.

	1	+	1	=	1
	x		(x + 6)		4

 

	x + 6 + x	=	1
	x(x + 6)		4

x² – 2x – 24 = 0

(x -6)(x + 4) = 0

x = 6.     [neglecting the negative value of x]

 

 

Answer:9 Option A

 

Explanation:

Part filled by A in 1 min =	1	.
	20

 

Part filled by B in 1 min =	1	.
	30

 

Part filled by (A + B) in 1 min =		1	+	1		=	1	.
		20		30			12

Both pipes can fill the tank in 12 minutes.

 

 

Answer:10 Option D

 

Explanation:

Part filled in 4 minutes = 4		1	+	1		=	7	.
		15		20			15

 

Remaining part =		1 –	7		=	8	.
			15			15

 

Part filled by B in 1 minute =	1
	20

 

	1	:	8	:: 1 : x
	20		15

 

x =		8	x 1 x 20		= 10	2	min = 10 min. 40 sec.
		15				3

The tank will be full in (4 min. + 10 min. + 40 sec.) = 14 min. 40 sec.

 

Level-II:

 

Answer:11 Option C

 

Explanation:

Let the slower pipe alone fill the tank in x minutes.

Then, faster pipe will fill it in	x	minutes.
	3

 

	1	+	3	=	1
	x		x		36

 

	4	=	1
	x		36

x = 144 min.

 

 

 

Answer:12 Option D

 

Explanation:

Part filled by (A + B) in 1 minute =		1	+	1		=	1	.
		60		40			24

Suppose the tank is filled in x minutes.

Then,	x		1	+	1		= 1
	2		24		40

 

	x	x	1	= 1
	2		15

x = 30 min.

 

Answer:13 Option B

 

Explanation:

Time taken by one tap to fill half of the tank = 3 hrs.

Part filled by the four taps in 1 hour =		4 x	1		=	2	.
			6			3

 

Remaining part =		1 –	1		=	1	.
			2			2

 

	2	:	1	:: 1 : x
	3		2

 

x =		1	x 1 x	3		=	3	hours i.e., 45 mins.
		2		2			4

So, total time taken = 3 hrs. 45 mins.

 

Answer:14 Option C

 

Explanation:

(A + B)’s 1 hour’s work =		1	+	1		=	9	=	3	.
		12		15			60		20

 

(A + C)’s hour’s work =		1	+	1		=	8	=	2	.
		12		20			60		15

 

Part filled in 2 hrs =		3	+	2		=	17	.
		20		15			60

 

Part filled in 6 hrs =		3 x	17		=	17	.
			60			20

 

Remaining part =		1 –	17		=	3	.
			20			20

 

Now, it is the turn of A and B and	3	part is filled by A and B in 1 hour.
	20

Total time taken to fill the tank = (6 + 1) hrs = 7 hrs.

 

Answer:15 Option C

 

Explanation:

Part filled in 2 hours =	2	=	1
	6		3

 

Remaining part =		1 –	1		=	2	.
			3			3

 

(A + B)’s 7 hour’s work =	2
	3

 

(A + B)’s 1 hour’s work =	2
	21

C’s 1 hour’s work = { (A + B + C)’s 1 hour’s work } – { (A + B)’s 1 hour’s work }

=		1	–	2		=	1
		6		21			14

C alone can fill the tank in 14 hours.

 

Answer:16 Option E

 

Explanation:

1. Time taken to fill the cistern without leak = 9 hours.

Part of cistern filled without leak in 1 hour =	1
	9

1. Time taken to fill the cistern in presence of leak = 10 hours.

Net filling in 1 hour =	1
	10

 

Work done by leak in 1 hour =		1	–	1		=	1
		9		10			90

Leak will empty the full cistern in 90 hours.

Clearly, both I and II are necessary to answer the question.

Correct answer is (E).

 

 

 

 

Answer:17 Option E

 

Explanation:

I. A’s 1 minute’s filling work =	1
	16

 

II. B’s 1 minute’s filling work =	1
	8

 

(A + B)’s 1 minute’s emptying work =		1	–	1		=	1
		8		16			16

Tank will be emptied in 16 minutes.

Thus, both I and II are necessary to answer the question.

Correct answer is (E).

 

Answer:18 Option B

 

Explanation:

II. Part of the tank filled by A in 1 hour =	1
	4

 

III. Part of the tank filled by B in 1 hour =	1
	6

 

(A + B)’s 1 hour’s work =		1	+	1		=	5
		4		6			12

 

A and B will fill the tank in	12	hrs = 2 hrs 24 min.
	5

So, II and III are needed.

Correct answer is (B).

Fractions
Any unit can be divided into any numbers of equal parts, one or more of this parts is called fraction of that unit. e.g. one-forth (1/4), one-third (1/3), three-seventh (3/7) etc.

The lower part indicates the number of equal parts into which the unit is divided, is called denominator. The upper part, which indicates the number of parts taken from the fraction is called the numerator. The numerator and the denominator of a fraction are called its terms.

• A fraction is unity, when its numerator and denominator are equal.
• A fraction is equal to zero if its numerator is zero.
• The denominator of a fraction can never be zero.
• The value of a fraction is not altered by multiplying or dividing the numerator and the denominator by the same number.e.g. 2/3 = 2/6 = 8/12 = (2/4)/(3/4)
• When there is no common factor between numerator and denominator it is called in its lowest terms.e.g. 15/25 = 3/5
• When a fraction is reduced to its lowest term, its numerator and denominator are prime to each other.
• When the numerator and denominator are divided by its HCF, fraction reduces to its lowest term.

Proper fraction: A fraction in which numerator is less than the denominator. e.g. 1/4, 3/4, 11/12 etc.

 

Improper Fraction:  A fraction in which numerator is equal to or more than the denominator. e.g. 5/4, 7/4, 13/12 etc.

 

Like fraction: Fractions in which denominators are same is called like fractions.

e.g. 1/12, 5/12, 7/12, 13/12 etc.

 

Unlike fraction: Fractions in which denominators are not same is called, unlike fractions.

e.g. 1/12, 5/7, 7/9 13/11 etc.

 

Compound Fraction: Fraction of a fraction is called a compound fraction.

e.g. 1/2 of 3/4 is a compound fraction.

 

Complex Fractions: Fractions in which numerator or denominator or both are fractions, are called complex fractions.

 

Continued fraction: Fraction that contain additional fraction is called continued fraction.

e.g.

 

 

 

Rule: To simplify a continued fraction, begin from the bottom and move upwards.

 

Decimal Fractions: Fractions in which denominators are 10 or multiples of 10 is called, decimal fractions. e.g. 1/10, 3/100, 2221/10000 etc.

 

Recurring Decimal: If in a decimal fraction a digit or a set of digits is repeated continuously, then such a number is called a recurring decimal. It is expressed by putting a dot or bar over the digits. e.g.

 

 

Pure recurring decimal: A decimal fraction in which all the figures after the decimal point is repeated is called a pure recurring decimal.

 

Mixed recurring decimal: A decimal fraction in which only some of the figures after the decimal point is repeated is called a mixed recurring decimal.

 

Conversion of recurring decimal into proper fraction: 

CASE I: Pure recurring decimal

 

Write the repeated digit only once in the numerator and put as many nines as in the denominator as the number of repeating figures. e.g.

 

CASE II: Mixed recurring decimal

In the numerator, take the difference between the number formed by all the digits after the decimal point and that formed by the digits which are not repeated. In the denominator, take the number formed as many nines as there are repeating digits followed by as many zeros as is the number of non-repeating digits. e.g.

 
Questions

Level-I

1.	Evaluate : (2.39)2 – (1.61)2 2.39 – 1.61
	A. 2 B. 4 C. 6 D. 8

 

2.	What decimal of an hour is a second ?
	A. .0025 B. .0256 C. .00027 D. .000126

 

3.	The value of (0.96)3 – (0.1)3 is: (0.96)2 + 0.096 + (0.1)2
	A. 0.86 B. 0.95 C. 0.97 D. 1.06

 

4.	The value of 0.1 x 0.1 x 0.1 + 0.02 x 0.02 x 0.02 is: 0.2 x 0.2 x 0.2 + 0.04 x 0.04 x 0.04
	A. 0.0125 B. 0.125 C. 0.25 D. 0.5

 

5.	If 2994 ÷ 14.5 = 172, then 29.94 ÷ 1.45 = ?
	A. 0.172 B. 1.72 C. 17.2 D. 172
6.	When 0.232323….. is converted into a fraction, then the result is:
	A. 1 5 B. 2 9 C. 23 99 D. 23 100

 

7.	.009 = .01 ?
	A. .0009 B. .09 C. .9 D. 9

 

8.	The expression (11.98 x 11.98 + 11.98 x x + 0.02 x 0.02) will be a perfect square for x equal to:
	A. 0.02 B. 0.2 C. 0.04 D. 0.4

 

9.	(0.1667)(0.8333)(0.3333) is approximately equal to: (0.2222)(0.6667)(0.1250)
	A. 2 B. 2.40 C. 2.43 D. 2.50

 

10.	3889 + 12.952 – ? = 3854.002
	A. 47.095 B. 47.752 C. 47.932 D. 47.95
11.	Level-II     0.04 x 0.0162 is equal to:
	A. 6.48 x 10-3 B. 6.48 x 10-4 C. 6.48 x 10-5 D. 6.48 x 10-6

 

12.	4.2 x 4.2 – 1.9 x 1.9 is equal to: 2.3 x 6.1
	A. 0.5 B. 1.0 C. 20 D. 22

 

13.	If 144 = 14.4 , then the value of x is: 0.144 x
	A. 0.0144 B. 1.44 C. 14.4 D. 144

 

 

 

14.	The price of commodity X increases by 40 paise every year, while the price of commodity Y increases by 15 paise every year. If in 2001, the price of commodity X was Rs. 4.20 and that of Y was Rs. 6.30, in which year commodity X will cost 40 paise more than the commodity Y ?
	A. 2010 B. 2011 C. 2012 D. 2013

 

15.	Which of the following are in descending order of their value ?
	A. 1 , 2 , 3 , 4 , 5 , 6 3 5 7 5 6 7 B. 1 , 2 , 3 , 4 , 5 , 6 3 5 5 7 6 7 C. 1 , 2 , 3 , 4 , 5 , 6 3 5 5 6 7 7 D. 6 , 5 , 4 , 3 , 2 , 1 7 6 5 7 5 3
16.	Which of the following fractions is greater than 3 and less than 5 ? 4 6
	A. 1 2 B. 2 3 C. 4 5 D. 9 10

 

17.	The rational number for recurring decimal 0.125125…. is:
	A. 63 487 B. 119 993 C. 125 999 D. None of these

 

18.	617 + 6.017 + 0.617 + 6.0017 = ?
	A. 6.2963 B. 62.965 C. 629.6357 D. None of these

 

19.	The value of 489.1375 x 0.0483 x 1.956 is closest to: 0.0873 x 92.581 x 99.749
	A. 0.006 B. 0.06 C. 0.6 D. 6

 

20.	0.002 x 0.5 = ?
	A. 0.0001 B. 0.001 C. 0.01 D. 0.1

 

 

 

 

 

Answers

Level-I

Answer:1 Option B

 

Explanation:

Given Expression =	a2 – b2	=	(a + b)(a – b)	= (a + b) = (2.39 + 1.61) = 4.
	a – b		(a – b)

 

Answer:2 Option C

 

Explanation:

Required decimal =	1	=	1	= .00027
	60 x 60		3600

 

 

Answer:3 Option A

 

Explanation:

Given expression	= (0.96)3 – (0.1)3 (0.96)2 + (0.96 x 0.1) + (0.1)2
	= a3 – b3 a2 + ab + b2
	= (a – b)   = (0.96 – 0.1)   = 0.86

Answer:4 Option B

 

Explanation:

Given expression =	(0.1)3 + (0.02)3	=	1	= 0.125
	23 [(0.1)3 + (0.02)3]		8

 

 

 

 

Answer:5 Option C

 

Explanation:

29.94	=	299.4
1.45		14.5

 

=		2994	x	1		[ Here, Substitute 172 in the place of 2994/14.5 ]
		14.5		10

 

=	172
	10

= 17.2

 

 

Answer:6 Option C

 

Explanation:

0.232323… = 0.23 =	23
	99

 

Answer:7 Option C

 

Explanation:

Let	.009	= .01;     Then x =	.009	=	.9	= .9
	x		.01		1

 

 

Answer:8 Option C

 

Explanation:

Given expression = (11.98)² + (0.02)² + 11.98 x x.

For the given expression to be a perfect square, we must have

11.98 x x = 2 x 11.98 x 0.02 or x   = 0.04

 

Answer:9 Option D

 

Explanation:

Given expression	= (0.3333) x (0.1667)(0.8333) (0.2222) (0.6667)(0.1250)
	= 3333 x 1 x 5 6 6 2222 2 x 125 3 1000
	= 3 x 1 x 5 x 3 x 8 2 6 6 2
	= 5 2
	= 2.50

 

Answer:10 Option D

 

Explanation:

Let 3889 + 12.952 – x = 3854.002.

Then x = (3889 + 12.952) – 3854.002

= 3901.952 – 3854.002

= 47.95.

 

Level-II

Answer:11 Option B

 

Explanation:

4 x 162 = 648. Sum of decimal places = 6.
So, 0.04 x 0.0162 = 0.000648 = 6.48 x 10^-4

 

Answer:12 Option B

 

Explanation:

Given Expression =	(a2 – b2)	=	(a2 – b2)	= 1.
	(a + b)(a – b)		(a2 – b2)

 

 

Answer:13 Option A

 

Explanation:

144	=	14.4
0.144		x

 

	144 x 1000	=	14.4
	144		x

 

x =	14.4	= 0.0144
	1000

 

 

Answer:14 Option B

 

Explanation:

Suppose commodity X will cost 40 paise more than Y after z years.

Then, (4.20 + 0.40z) – (6.30 + 0.15z) = 0.40

0.25z = 0.40 + 2.10

z =	2.50	=	250	= 10.
	0.25		25

X will cost 40 paise more than Y 10 years after 2001 i.e., 2011.

 

 

 

Answer:15 Option D

Answer:16 Option C

 

Explanation:

3	= 0.75,	5	= 0.833,	1	= 0.5,	2	= 0.66,	4	= 0.8,	9	= 0.9.
4		6		2		3		5		10

Clearly, 0.8 lies between 0.75 and 0.833.

	4	lies between	3	and	5	.
	5		4		6

 

 

 

Answer:17 Option C

 

Explanation:

0.125125… = 0.125 =	125
	999

 

 

Answer:18 Option C

 

Explanation:

617.00

6.017

0.617

+  6.0017

——–

629.6357

———

 

Answer:19 Option B

 

Explanation:

489.1375 x 0.0483 x 1.956		489 x 0.05 x 2
0.0873 x 92.581 x 99.749		0.09 x 93 x 100

 

=	489
	9 x 93 x 10

 

=	163	x	1
	279		10

 

=	0.58
	10

= 0.058  0.06.

 

Answer:20 Option B

 

Explanation:

2 x 5 = 10.

Sum of decimal places = 4

0.002 x 0.5 = 0.001