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The River Goddess rewarded the woodcutter by giving him all the three axes.

a. dishonest honest
b. come go
c. cold hot
d. laugh cry
e. punish reward
f. sad happy

I like the profession of a doctor.

Correct option is 3. River Goddess

The woodcutter was worried to think how he would earn for his family. So, he sarted crying.

While he was drinking water from the river, his axe accidently slipped from his hand and fell into the river.

Correct option is (C) x + y = 23

a. My mother helps sick people. She is a doctor.
b. My uncle fights fires. He is a fire-fighter.
c. My father serves food in a hotel. He is a waiter.
d. My aunt teaches maths. She is a teacher.

Let the number of students in halls A and B are x and y, respectively.

Now, by given condition, x – 10 = y + 10

x - y = 20 …(i)

and (x + 20) = 2(y - 20)

x–2y = -60 …(ii)

On subtracting Eq. (ii) from Eq. (i), we get

(x - y) - (x -2y) = 80

y = 80

On putting y = 80 in Eq. (i), we get

x – 80 = 20

so, x = 100

Hence, 100 students are in hall A and 80 students are in hall B.

The woodcutter was cutting wood with his axe, on a very hot day.

Given:

Father’s age is three times the sum of ages of his two children.

After 5 years his age will be twice the sum of ages of two children.

To find:

The age of father.

Solution:

Let the present age of father be ‘a’ and present sum of age of both sons be b.

Given,

Father’s age is three times the sum of ages of his two children.

⇒ a = 3b

After 5 years his age will be twice the sum of ages of two children.

So age of father will be a + 5 and as there are two sons, age of both sons combined after 5 years will be b + 10.

⇒ a + 5 = 2(b + 10)

Put the value of a in above equation,

⇒ 3b + 5 = 2b + 20

⇒ 3b - 2b = 20 - 15

⇒ b = 15

Thus,

a = 3b = 3 (15) = 45

Hence father's present age is 45 years.

Let the two numbers be x and y.

Then, by first condition, ratio of these two numbers = 5 :6

x/y = 5/6

So y = 6x/5

and by second condition, then, 8 is subtracted from each of the numbers, then ratio becomes 4:5

(x-8)/(y-8) = 4/5

5(x-8) = 4(y-8)

5x – 4y = 8

Now, put the value of y in Eq. (ii), we get

5x – 4 ( 6x/5 ) = 8

So x/5 = 8

And x = 40

Put the value of x in Eq. (i), we get

So, y = (6/5 x 40) = 48

Hence, the required numbers are 40 and 48.

a. Boil the potoes.
b. Mango leaves are green.
c. My mother is fond of watches.
d. The table is four feet long.

Ritesh was making his own National Flag to carry it in the parade of the Independence Day.

a. Each member of the society should work for the prosperity of the country.
b. I love my country.
c. We got freedom from the British Rule on 15 August 1947.
d. We should be ready to sacrifice our lives for our country.

The 24 spokes of the Ashoka Chakra represent the precious 24 hours of the day.

We can do good for the society by helping others, maintaining cleanliness, planting trees and respecting elders.

Saffron stands for sacrifice, white for peace and green for prosperity.

Let the present age (in year) of father and his two children be x, y and z year, respectively.

Now by given condition, x = 2(y + z) …(i)

and after 20 year, (x + 20) = (y + 20) + (z + 20)

so, x = y + z + 20 or y + z = x - 20 …(ii)

On putting the value of (y + z) in Eq, (i) and get the present age of father

So, x = 2(x -20)

x = 40

Hence, the father’s age is 40 yr.

Shubhi wants to give her old toys and clothes to the poor children because it is one of the ways to serve our country.

Let Salim and his daughter’s age be x and y years respectively.

Now, by first condition

Two years ago, Salim was thrice as old as his daughter.

i.e., x - = 3(y - 2)

so, x - 3y = - 4 …(i)

and by second condition, six years later. Salim will be four years older than twice her age.

(x + 6) = 2(y + 6) + 4

x + 6 = 2y + 16

x – 2y = 10 …(ii)

On subtracting Eq. (i) from Eq. (ii), we get

x – 2y = 10

- (x–3y =- 4)

We get, y = 14

Put the value of y in Eq. (ii), we get

x - 2(14) = 10

so, x = 38

Hence, Salim and his daughter’s age are 38 year and 14 year, respectively.

Let the present age of Nuri = x yr

And the present age of Sonu = y yr

Five years ago,

Nuri’s age = (x – 5)yr

Sonu’s age = (y – 5)yr

According to the question,

(x – 5) = 3(y – 5)

⇒ x – 5 = 3y – 15

⇒ x – 3y = – 10 …(i)

After ten years,

Aftab’s age = (x + 10)yr

Daughter’s age = (y + 10)yr

According to the question,

(x + 10) = 2(y + 10)

⇒ x + 10 = 2y + 20

⇒ x – 2y = 10 …(ii)

Now, we can solve this by an elimination method

On subtracting Eq. (ii) from (i) we get

x – 2y – x + 3y = 10 – ( – 10)

⇒ – 2y + 3y = 10 + 10

⇒ y = 20

On putting y = 20 in Eq. (i) we get

x – 3(20) = – 10

⇒ x – 60 = – 10

⇒ x = 50

Hence, the age of Nuri is 50 years and age of Sonu is 20 years.

Let the age of A = x years

And the age of B = y years

Five years ago,

A’s age = (x – 5) years

B’s age = (y – 5) years

According to the question,

(x – 5) = 3(y – 5)

⇒ x – 5 = 3y – 15

⇒ x = 3y – 10 …(i)

Ten years later,

A’s age = (x + 10)

B’s age = (y + 10)

According to the question,

(x + 10) = 2(y + 10)

⇒ x + 10 = 2y + 20

⇒ x = 2y + 10 …(ii)

From Eq. (i) and (ii), we get

3y – 10 = 2y + 10

⇒ 3y – 2y = 10 + 10

⇒ y = 20

On putting the value of y = 20 in Eq. (i), we get

x = 3(20) – 10

⇒ x = 50

Hence, the age of person A is 50 years and Age of B is 20 years.

Given:

Five years ago, Nuri was thrice as old as Sonu.

Ten years later, Nuri will be twice as old as Sonu.

To find:

The age of Nuri and Sonu.

Solution:

Let the present age of Nuri and his son be ‘a’ and ‘b’ respectively.

Given,

Five years ago, Nuri was thrice as old as Sonu.

This implies age of Nuri and Sonu five years ago was

a - 5 and b - 5.

⇒ a – 5 = 3(b – 5)

⇒ a - 5 = 3b – 15

⇒ a = 3b – 15 + 5

⇒ a = 3b – 10 ------- (1)

Ten years later, Nuri will be twice as old as Sonu.

This implies after ten years the age of Nuri and Sonu will be

a + 10 and b + 10.

⇒ a + 10 = 2(b + 10)

⇒ a + 10 = 2b + 20

⇒ a = 2b + 20 - 10

⇒ a = 2b + 10 -------- (2)

Equating (1) and (2), we get

3b – 10 = 2b + 10

⇒ 3b - 2b = 10 + 10

 b = 20

Thus,

a = 60 – 10 = 50

Hence, Nuri's present age is 50 years and Sonu's age is 20 years.

Let the present ages of A and B be ‘a’ and ‘b’ respectively.

Given, ten years later, A will be twice as old as B and five years ago,

A was three times as old as B

⇒ a + 10 = 2(b + 10)

⇒ a = 2b + 10 ------- (1)

Also,

a – 5 = 3(b – 5)

⇒ a = 3b – 10 -------- (2)

Equating eq2 and eq1

⇒ 2b + 10 = 3b – 10

⇒ b = 20

Thus,

a = 2b + 10 = 50

Let’s assume the numerator of the fraction to be x and the denominator of the fraction to be y. 

So, the required fraction is x/y. 

From the question it’s given as,

The sum of the numerator and denominator of the fraction is 12. 

Thus, the equation so formed is, 

x + y = 12 

⇒ x + y – 12 = 0 

And also it’s given in the question as, 

If the denominator is increased by 3, the fraction becomes 1/2. 

Putting this as an equation, we get 

x/ (y+3) = 1/2

⇒ 2x = (y+3)

⇒ 2x – y – 3 = 0 

The two equations are, 

x + y – 12 = 0…… (i) 

2x – y – 3 = 0…….. (ii) 

Adding (i) and (ii), we get 

x + y – 12 + (2x – y – 3) = 0 

⇒ 3x -15 = 0 

⇒ x = 5 

Using x = 5 in (i), we find y 

5 + y – 12 = 0 

⇒ y = 7 

Therefore, the required fraction is \(\frac{5}{7}\).

Correct option is (B) 1, -2

Given that \(a_1,b_1,c_1,a_2,b_2\;and\;c_2\) are consecutive integers in the same order.

Given system of equations is

\( a_1 x + b_1 y + c _1= 0\)

and \(a_2 x + b_2 y + c_2 = 0\)

Then the solution of given system of equations by cross multiplication method :

\(\frac x{b_1c_2-b_2c_1}=\frac y{c_1a_2-c_2a_1}=\frac1{a_1b_2-a_2b_1}\)

\(\Rightarrow\) \(x=\frac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1}\;and\;y=\frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}\)

\(\Rightarrow x=\frac{(a_1+1)(a_1+5)-(a_1+4)(a_1+2)}{a_1(a_1+4)-(a_1+3)(a_1+1)}\)

\(=\frac{(a_1^2+6a_1+5)-(a_1^2+6a_1+8)}{(a_1^2+4a_1)-(a_1^2+4a_1+3)}\)

\(=\frac{5-8}{-3}\)

\(=\frac{-3}{-3}=1\)

and \(y=\frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}\)

\(=\frac{(a_1+2)(a_1+3)-(a_1+5)a_1}{a_1(a_1+4)-(a_1+3)(a_1+1)}\)

\(=\frac{(a_1^2+5a_1+6)-(a_1^2+5a_1)}{(a_1^2+4a_1)-(a_1^2+4a_1+3)}\)

\(=\frac6{-3}=-2\)

Hence, the value of x and y are x = 1 & y = - 2.

Correct option is B) 1, -2

Correct option is A) Rs. 7and Rs. 8

Let the numerator be ‘a’ and denominator be ‘b’.

Given,

Sum of the numerator and denominator of a fraction is 3 less than twice the denominator.

⇒ a + b = 2b – 3

⇒ a = b – 3 ------ (1)

Also,

If the numerator and denominator are decreased by 1, the numerator becomes half the denominator.

⇒ a – 1 = (b – 1)/2

⇒ b = 2a - 1 ------ (2)

Substituting value of b in eq1.

⇒ a = 2a – 1 – 3

⇒ a = 4.

Thus,

b = 2 × 4 – 1 = 7

Correct option is (B) 7

Let the required fraction be \(\frac xy.\)

According to first condition, we have

\(\frac{x-1}{y+1}=\frac12\)

\(\Rightarrow\) 2x - 2 = y + 1    (By cross multiplication)

\(\Rightarrow\) 2x - y - 3 = 0   ___________(1)

According to second condition, we have

\(\frac{x+1}{y-1}=\frac45\)

\(\Rightarrow\) 5 (x+1) = 4 (y - 1)  (By cross multiplication)

\(\Rightarrow\) 5x + 5 = 4y - 4

\(\Rightarrow\) 5x - 4y + 9 = 0   ___________(2)

Multiply equation (1) by 4, we get

8x - 4y - 12 = 0   ___________(3)

Subtract equation (2) from (3), we get

(8x - 4y - 12) - (5x - 4y + 9) = 0 - 0

\(\Rightarrow\) 3x - 21 = 0

\(\Rightarrow\) x = \(\frac{21}3\) = 7

Hence, the numerator of fraction \(\frac xy\) is x = 7.

Correct option is B) 7

Let the numerator be ‘a’ and denominator be ‘b’.

Given,

Sum of the numerator and denominator of a fraction is 12 .

If the denominator is increased by 3, the fraction becomes 1/2.

⇒ a + b = 12 ------ (1)

Also,

a/(b + 3) = 1/2

⇒ 2a – b = 3 ------- (2)

Adding eq1 and eq2

⇒ a + b + 2a – b = 15

⇒ 3a = 15

⇒ a = 5

Thus,

b = 7

Fraction is 5/7.

Correct option is D) 2/3

Let the numerator be a and denominator be b.

Given, numerator of a fraction is 4 less than the denominator.

⇒ a = b – 4 --------- (1)

Also,

If the numerator is decreased by 2 and denominator is increased by 1, then the denominator is eight times the numerator.

⇒ b + 1 = 8(a – 2)

⇒ b = 8a – 17 -------- (2)

Substituting value of b from (2) in (1).

⇒ a = 8a – 17 – 4

⇒ a = 3

Thus,

b = 3 + 4 = 7

Fraction is 3/7.

Let the numerator be ‘a’ and denominator be ‘b’.

Given,

Sum of the numerator and denominator of a fraction is 4 more than twice the numerator.

If the numerator and denominator are increased by 3, they are in the ratio 2 : 3.

⇒ a + b = 2a + 4

⇒ b – a = 4 -------- (1)

Also,

⇒ 3a + 9 = 2b + 6

⇒ 3a – 2b = - 3 ------(2)

Multiplying eq1 by 2 and adding to eq1

⇒ 2b – 2a + 3a – 2b = 8 - 3

⇒ a = 5

Thus b = 9

Fraction is 5/9

Let the numerator be ‘a’ and denominator be ‘b’

Given,

Sum of a numerator and denominator of a fraction is 18.

⇒ a + b = 18 ----- (1)

Also,

If the denominator is increased by 2, the fraction reduces to 1/3.

⇒ a/(b + 2) = 1/3

⇒ 3a - b = 2 ------ (2)

Adding (1) and (2)

⇒ 4a = 20

⇒ a = 5

Thus, b = 13

Fraction is 5/13

Correct option is (C) 48

Let both numbers are a & b.

\(\therefore\) Their sum = a+b,

their difference = a - b and

their product = ab

According to given condition,

a - b : a+b : ab = 1 : 7 : 24

\(\Rightarrow\frac{a-b}{ab}=\frac1{24}\) and \(\frac{a+b}{ab}=\frac7{24}\)

\(\Rightarrow\) \(\frac1b-\frac1a=\frac1{24}\) and \(\frac1b+\frac1a=\frac7{24}\)

\(\Rightarrow\) \((\frac1b-\frac1a)+(\frac1b+\frac1a)=\frac1{24}+\frac7{24}\)

\(\Rightarrow\) \(\frac2b=\frac8{24}=\frac1{3}\)

\(\Rightarrow\) \(b=2\times3=6\)

\(\therefore\) \(\frac16-\frac1a=\frac1{24}\)

\(\Rightarrow\) \(\frac1a=\frac16-\frac1{24}\)

\(=\frac{4-1}{24}=\frac{3}{24}=\frac{1}{8}\)

\(\Rightarrow\) a = 8

\(\therefore\) Product of both numbers = ab

\(=8\times6=48\)

Correct option is B) 24

Let the required fraction be x/y . 

Then, we have: x+2/y = 1/2 

⇒ 2(x + 2) = y

⇒ 2x + 4 = y 

⇒2x - y = -4 ……(i) 

Again, x/y−1 = 1/3 

⇒3x = 1(y – 1) 

⇒3x – y = -1 ……(ii) 

On subtracting (i) from (ii), we get: 

x = (-1 + 4) = 3 

On substituting x = 3 in (i), we get: 

2 × 3 – y = -4 

⇒ 6 – y = -4 

⇒ y = (6 + 4) = 10 

∴ x = 3 and y = 10 

Hence, the required fraction is 3/10

The correct option is : (B) Has no solution.

Explanation:

We have, ax + by = c and lx + my = n

Now, a/l = b/m ≠ c/n (given)

. ^.. The given system of equations has no solution.

Correct option is (C) x = 2, y = 6

Given system of equations is

\(\frac{x+y-8}{2}=\frac{x+2y-14}{3}=\frac{3x+y-12}{11}\)  _________(1)

\(\Rightarrow\) \(\frac{x+y-8}{2}=\frac{x+2y-14}{3}\)

\(\Rightarrow\) \(3(x+y-8)=2(x+2y-14)\)

\(\Rightarrow\) 3x + 3y - 24 = 2x + 4y - 28

\(\Rightarrow\) x - y = -4           _________(2)

From (1), we also have

\(\frac{x+y-8}{2}=\frac{3x+y-12}{11}\)

\(\Rightarrow\) 11x + 11y - 88 = 6x + 2y - 24

\(\Rightarrow\) 5x + 9y = 64     _________(3)

\(\Rightarrow\) 5 (y - 4) + 9y = 64    (From (2))

\(\Rightarrow\) 5y - 20 + 9y = 64

\(\Rightarrow\) 14y = 64+20 = 84

\(\Rightarrow\) y = \(\frac{84}{14}\) = 6

Then from (2), we have

x = y - 4

= 6 - 4 = 2

Hence, the solution of given system of equations is x = 2 & y = 6.

Correct option is C) x = 2, y = 6

Let the numerator be ‘a’ and denominator be ‘b’.

Given,

If 2 is added to the numerator of a fraction, it reduces to 1/2 and if 1 is subtracted from the denominator, it reduces to 1/3

⇒ (a + 2)/b = 1/2

⇒ 2a + 4 = b ------ (1)

Also,

a/(b – 1) = 1/3

⇒ 3a + 1 = b ------- (2)

Subtracting eq2 from eq1

⇒ 2a + 4 – 3a – 1 = b – b

⇒ a = 3

Thus,

b = 2a + 4 = 10

Fraction is 3/10

Correct option is (B) p = -8/3

Given system of equations is

2x – py = 0       __________(1)

3x + 4y = 0       __________(2)

Multiply equation (1) by 3 & equation (2) by 2, we obtain

6x - 3py = 0       __________(3)

and 6x+8y = 0   __________(4)

Subtract equation (3) from (4), we get

(6x+8y) - (6x - 3py) = 0 - 0 = 0

\(\Rightarrow\) (8+3p)y = 0

for non-zero solution (i.e., \(y\neq0),\)

8+3p = 0

\(\Rightarrow\) \(p=\frac{-8}3\)

Hence, for \(p=\frac{-8}3,\) the given system of equations has non zero solution.

Correct option is B) p = -8/3

Correct option is (C) 5/8

Let the original rational number is \(\frac xy.\)

According to first condition, we have

y = x + 3          ___________(1)

According to second condition, we have

\(\frac{x-3}{y+2}=\frac15\)

\(\Rightarrow\) 5x - 15 = y + 2      (By cross multiplication)

\(\Rightarrow\) 5x - 15 = (x + 3) + 2   (From (1))

\(\Rightarrow\) 5x - x = 15 + 5

\(\Rightarrow\) 4x = 20

\(\Rightarrow\) x = \(\frac{20}4\) = 5

\(\therefore\) y = 5+3 = 8            (From (1))

\(\therefore\) Original number \(=\frac xy=\frac58.\)

Correct option is C) 5/8

LCM of two consecutive even numbers = 180 

But, HCF of two consecutive even numbers = 2 

Now, product of the given number = HCF x LCM 

= 2 x 180 

= 360 

To find the two consecutive even numbers, we have to factorize 360.

360 = 2 x 2 x 2 x 3 x 3 x 5 

360 = (2 x 3 x 3) x (2 x 2 x 5) 

= 18 x 20

∴ The two consecutive even numbers whose LCM is 180 are 18 and 20.

(i) No, he was wrong. Pure gold is very soft and is, therefore, not suitable for making jewelry. It is alloyed with either silver or copper to make it hard. But, sometimes jewelers mix a large quantity of copper and silver in gold, to earn more profit.
(ii) Untrustworthiness, cheating, cunning.
(iii) We should always purchase the gold jewelry from a branded shop with proper receipt and Hallmark certificate.
(iv) Government insists on purchasing hallmarked jewellery as it contains the gold and its alloyed metal (i.e., copper or silver) in a fixed ratio.

To polish a gold ornament, it is dipped in a liquid called aqua regia (a mixture of hydrochloric acid and nitric acid). On getting the environment of aqua regia, the outer layer of gold dissolves and the inner shiny layer appears. The dissolving of the layer causes a reduction in the weight of the jewellery. 

(d) provides required life to the aeroplane

According to the hermit the most important business was the King’s digging the beds, for the hermit and caring the wound of the bearded man. These are considered as the most important business to do good that person.

The old father looked weary of his mischieveous son. The son was known for his bad deeds everywhere. He was so tired of his son’s misdeeds that often there used to be bitter quarrels between them. The old man decided not to care for him anymore. One day the son was knocked down by a speeding vehicle. His condition was serious that the father decided to make peace with his son. He poured in all had on the treatment of his son. He looked after him in such way that the sun came round very soon.

The World Environment Day is observed on June 5th every year. Millions of saplings are planted by responsible citizens, the world over. In schools and colleges, students are given the task of picking all non-bio degradable materials like plastic. They are told how to keep the environment free of pollution.