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(i) 2x + 3y = 9.35bar

2x + 3y - 9.35bar = 0

Comparing this equation with ax + by + c = 0, 

a = 2, b = 3,c = 9.35bar

(ii)  x - (y/5) - 10 = 0

Comparing this equation with ax + by + c = 0,

a = 1, b = -1/5, c = -10

(iii)  -2x + 3y = 6

− 2x + 3 y − 6 = 0

Comparing this equation with ax + by + c = 0,

a = −2, b = 3, c = −6

Let us consider,

First number = x and

Second number = y

(Assume y is smaller)

As per the statement,

x – y = 14 …..(1)

and x² – y² = 448

(x + y) (x – y) = 448

(x + y) x 14 = 448

x + y = 32 ……(2)

Adding (1) and (2),

2x = 46 or x = 23

Subtracting (1) from (2)

2y = 18 or y = 9

Answer:

Numbers are 23 and 9

Let the larger number be x and the smaller number be y. 

Then, we have: 

x – y = 14 or x = 14 + y ……….(i) 

x² – y² = 448 ………(ii) 

On substituting x = 14 + y in (ii) we get 

(14 + y)² – y² = 448 

⇒ 196 + y² + 28y – y² = 448 

⇒ 196 + 28y = 448 

⇒ 28y = (448 – 196) = 252 

⇒ y = 252/28 = 9 

On substituting y = 9 in (i), we get: 

x = 14 + 9 = 23 

Hence, the required numbers are 23 and 9.

Let us consider, a fraction x/y, where x is numerator and y is denominator.

As per the statement,

(x + 2)/y = ½

x/y – 1 = 1/3

2x + 4 = y …(1)

And, 3x = y – 1 …(2)

Using Substitution method:

Using (1) in (2)

3x = 2x + 4 – 1

3x = 2x + 3

x = 3

Form (1): y = 2(3) + 4 = 10

Answer: Required fraction is 3/10

Let the cost of each pencil be Rs. x and that of each pen be Rs. y. 

Then, we have: 

5x + 7y = 195 ………(i) 

7x + 5y = 153 ………(ii) 

Adding (i) and (ii), we get: 

12x + 12y = 348 

⇒ 12(x + y) = 348 

⇒ x + y = 29 ………(iii) 

Subtracting (i) from (ii), we get: 

2x – 2y = –42

⇒ 2(x – y) = –42 

⇒ x – y = –21 ………(iv) 

On adding (iii) and (iv), we get: 

4 + y = 29 

⇒ y = (29 – 4) = 25 

Hence, the cost of each pencil is Rs. 4 and the cost of each pen is Rs. 25.

i) 2x + 3y = 9.35

OR 2x + 3y – 9.35 = 0 

If we compare this to ax + by + c = 0, 

Here, a = 2, b = 3, c = 9.35. 

(ii) x – \(\frac{y}{5}\)– 10 = 0 

If we compare this to ax + by + c = 0, 

Here, a = 1, b = \(\frac{-1}{5}\), c=10. 

(iii) -2x + 3y = 6 

OR -2x + 3y – 6 = 0 

If we compare this to ax + by + c = 0, 

Here, a = -2, b = 3, c = -6. 

(iv) x = 3y x – 3y = 0 

x – 3y + 0 = 0 

If we compare this to 

ax + by + c = 0, 

Here, a = 1, b = -3, c = 0. 

(v) 2x = -5y 

2x + 5y = 0 

2x + 5y + 0 = 0 

If we compare this to ax + by + c = 0, 

Here, a = 2, b = 5, c = 0. 

(vi) 3x + 2 = 0 

3x + 0y + 2 = 0 

If we compare this to ax + by + c = 0, 

Here, a = 3, b = 0, c = 2. 

(vii) y – 2 = 0 

0x + 1y – 2 = 0 

If we compare this to ax + by + c = 0, 

Here, a = 0, b = 1, c = -2. 

(viii) 5 = 2x 

-2x + 5 = 0 

-2x + 0y + 5 = 0 

If we compare this to ax + by + c = 0, 

Here, a = -2, b = 0, c = 5.

Let the cost of 1 pen and 1 pencil are ₹x and ₹y respectively. 

Then as per the question 

5x + 8y = 120 …….(i) 

8x + 5y = 153 …….(ii) 

Adding (i) and (ii), we get 

13x + 13y = 273 

⇒ x + y = 21 …….(iii) 

Subtracting (i) from (ii), we get 

3x – 3y = 33 

⇒ x – y = 11 ………(iv) 

Now, adding (iii) and (iv), we get 

2x = 32 ⇒ x = 16 

Substituting x = 16 in (iii), we have 

16 + y = 21 ⇒ y = 5 

Hence, the cost of 1 pen and 1 pencil are respectively ₹16 and ₹5.

(i) 3x + 4y = 7 

At x = 1 

3 + 4y = 7 

⇒ y = 1 

Thus, 

x = 1, y = 1 is a solution 

At x = 0, 

0 + 4y = 7 

⇒ y = 7/4 

Thus, 

x = 0, y = 7/4 is a solution

(ii) x = 6y 

At, 

y = 0 

⇒ x = 0 

Thus, 

x = 0, y = 0 is a solution. 

At y = 1, 

⇒ x = 6 

Thus, 

x = 6, y = 1 is a solution

(iii) x + πy  = 4 

At x = 0, 

πy = 4 

⇒ y = 4/π 

Thus, 

x = 0, y = 4/π is a solution 

At y = 0, 

⇒ x + 0 = 4 

⇒ x = 4 

Thus, 

x = 4, y = 0 is a solution

(iv) \(\frac{2}{3}\)x-y = 4 

At x = 0, 

⇒ 0 – y = 4 

⇒ y = - 4 

Thus, 

x = 0, y = 4 is a solution 

At x = 3, 

⇒ 2 – y = 4 

⇒ y = -2 

Thus, 

x = 3, y = - 2 is a solution

Let x be the first number and y be second number.

x – y = 5

x² – y² = 65 …(2)

Now, by dividing (2) by (1) we get:

x + y = 13 …(3)

On adding (1) and (2) we get

2x = 18

or x = 9

From (3): 9 + y = 13

or y = 4

Two numbers are 4 and 9.

Let x be the first number and y be the second number.

As per statement,

x + y = 80 and

x – 4y = 5

on subtracting both the equations, we get

y = 15

From (1): x + 15 = 80

x = 80 – 15 = 65

Answer:

Required numbers are 15 and 65.

Let the cost of pen b Rs. x 

If the cost of Notebook is Rs. y. 

Cost of 1 notebook is twice the cost of a pen. 

∴ y = 2x -2x + y = 0 

If this is written in the form of 

ax + by + c =, it is 

-2x + y + 0 = 0 

Here, a = -2, b = 1, c = 0.

Let the cost of one fountain pen be Rs. ‘x’ and the cost of one ball pen is Rs. ‘y’ 

Given, 

Cost of ball pen is Rs. 5 less than half of the cost of fountain pen.

⇒ y = (\(\frac{x}{2}\)) – 5 

⇒ 2y = x – 10 

⇒ x – 2y – 10 = 0

Let the cost of each pencil be Rs. x and that of each pen be Rs. y. 

Then, we have: 

5x + 7y = 195 ………(i) 

7x + 5y = 153 ………(ii) 

Adding (i) and (ii), we get: 

12x + 12y = 348 

⇒ 12(x + y) = 348 

⇒ x + y = 29 ………(iii) 

Subtracting (i) from (ii), we get: 

2x – 2y = –42

⇒ 2(x – y) = –42 

⇒ x – y = –21 ………(iv) 

On adding (iii) and (iv), we get: 

4 + y = 29 ⇒ y = (29 – 4) = 25 

Hence, the cost of each pencil is Rs. 4 and the cost of each pen is Rs. 25.

Let the cost of one pen is ₹ x and cost of one pencil is ₹ y, then

As per statement,

5x + 8y = 120 …(1)

8x + 5y = 153 …(2)

Adding both the equations, we get

13x + 13y = 273

x + y = 21 …(3)

On subtracting (1) from (2),

3x – 3y = 33

x – y = 11 …….(iv)

Again,

Adding (3) and (iv),

2x = 32 or x = 16

On subtracting,

2y = 10 or y = 5

Answer:

Cost of 1 pen = ₹ 16

and cost of 1 pencil = ₹ 5

Let the cost of a fountain pen = x 

Let the cost of ball pen = y 

Then y = x – 5 or x – y – 5 = 0

Correct option is (C) x + y = 24

Let both numbers are x and y.

\(\therefore\) Their sum = x+y

\(\therefore\) x + y = 24    \((\because\) Given that sum of both numbers is 24)

C) x + y = 24

Given: that cost of a pencil = ₹2 

Cost of a ball point pen = ₹15 

Let the number of pencils purchased = x 

Let the number of pens purchased = y 

Then the total cost of x – pencils = 2x 

Then the total cost of y – pens = 15y 

By problem 2x + 15y = 100

Let Sindhu’s marks = x 

Bhargavi’s marks = y 

Then by problem y = 2x + 10 or 2x – y + 10 = 0

For rounding off to tenths place, we look at the hundredths place. Here the digit is 5. So, the digit at the tenths place (9) will be increased by 1 (i.e., it will become 9 + 1) Hence, rounding off 87.952 to tenths place, we get 88.0 (Note: Do not write it as 88.)

The correct answer is 12.104, 12.122, 12.142, 12.214, 12.401

The correct answer is 889/80cm

Rank method of correlation is used to find correlation between two qualitative characteristics.

Spearman’s rank correlation useful 

• To find the correlation between two qualitative characteristics, such as Honesty, Intelligence, Taste etc. 
• This method is easy and simple as compared to karl pearsons method when the data has no repetitions.

We know, 

Empirical formula: Mode = 3 Median – 2 Mean 

Since, ratio of mode and median of a certain data is 6:5. 

⇒ \(\frac{Mode}{Median}\) = \(\frac{6}{5}\) or Mode = \(\frac{(6\, Median)}{5} \)

Now, 

\(\frac{(6\, Median)}{5} \) = 3 Median – 2 Mean 

\(\frac{(6\, Median)}{5} \) – 3 Median = – 2 Mean 

or \(\frac{9}{10}\) (Median) = Mean 

or \(\frac{Mean}{Median}\) = \(\frac{9}{10}\) or 9:10.

Empirical formula: Mode = 3 median – 2 mean

Since, ratio of mean and median of a certain data is 2:3, then mean = 2x and median = 3x 

Mode = 3(3x) – 2(2x) 

= 9x – 4x

= 5x 

Therefore, 

Mode: Mean = 5x:2x or 5: 2

f(x) = ax + b

f’(x) = a + 0

f’(x) = a

For, f(x) to be decreasing, it must have

f’(x) < 0

\(\therefore\) a < 0

\(\therefore\) a ∈ (-∞, 0)

Hence the required set of values is a ∈ (-∞, 0).

1. False
2. True
3. False
4. True
5. False.

1. Mode

2. Mode

3. Mean

4. Median

5. Mean

6. Weighted mean

7. Median.

f(x) = log_ax

Domain of the above mentioned function is (0, ∞)

Let x₁, x₂∈ (0, ∞) such that x₁ < x₂.

\(\because\) the function here is a logarithmic function, so either a > 1 or 1 > a > 0.

Case – 1

Let a > 1

x₁ < x₂

\(\therefore\) log_ax₁ < log_ax₂

\(\therefore\) f(x₁) < f(x₂)

\(\therefore\) x₁ < x₂ & f(x₁) < f(x₂), ∀ x₁, x₂∈ (0, ∞)

Hence, f(x) is increasing on (0, ∞).

Case – 2

Let 1 > a > 0

x₁ < x₂

\(\therefore\) log_ax₁ > log_ax₂

\(\therefore\) f(x₁) > f(x₂)

\(\therefore\) x₁ < x₂ & f(x₁) > f(x₂), ∀ x₁, x₂∈ (0, ∞)

Hence, f(x)is decreasing on (0, ∞).

Thus, for 1 > a > 0, f(x) is decreasing in its domain.

f(x) = log_a x

Let x₁, x₂∈ (0, ∞) such that x₁ < x₂.

\(\because\) the function here is a logarithmic function, so either a > 1 or 1 > a > 0.

Case – 1

Let a > 1

x₁ < x₂

\(\therefore\) log_ax₁ < log_ax₂

\(\therefore\) f(x₁) < f(x₂)

\(\therefore\) x₁ < x₂ & f(x₁) < f(x₂), ∀ x₁, x₂∈ (0, ∞)

Hence, f(x) is increasing on (0, ∞).

Case – 2

Let, 1 > a > 0

x₁ < x₂

\(\therefore\) log_ax₁ > log_ax₂

\(\therefore\) f(x₁) > f(x₂)

\(\therefore\) x₁ < x₂ & f(x₁) > f(x₂), ∀ x₁, x₂∈ (0, ∞)

Thus, for a > 1, f(x) is increasing in its domain.

3 Median = Mode + 2 Mean

3 Median = 12 + 2 (24)

3 Median = 12 + 48

3 Median = 60

Median = 20

Correct answer is D.

Mean \(=\frac{(7+5+13+x+9)}{5}\)

\(10=\frac{(34+x)}{5}\)

50 - 34 = x

x = 16

Arithmetic mean of given table:

\(\overline X=\frac{\sum fX}{\sum f}\)

Or, \(\overline X=\frac{3265}{69}\)

\(\overline X=60.37\)

True

Consider the progression a, a + d, a + 2d, … and sequence of prime number 2, 3, 5, 7, 11,…

Clearly, progression is a sequence but sequence is not progression because it does not follow a specific pattern

Let A.P be a, a + d, a + 2d, …, a + (n – 1)d

Taking first and last term

a₁ + a_n = a + a + (n – 1)d

a₁ + a_n = 2a + (n – 1)d …(i)

Taking second and second last term

a₂ + a_n–2 = (a + d) + [a + (n – 2)d]

= a + d + a + nd – 2d

= 2a + nd – d

= 2a + (n – 1)d

= a₁ + a_n [from (i)]

Taking third term from the beginning and the third from the end

a₃ + a_n-3 = (a + 2d) + [a + (n – 3)d]

= a + 2d + a + nd – 3d

= 2a + nd – d

= 2a + (n – 1)d

= a₁ + a_n [from (i)]

From the above pattern, we see that the sum of terms equidistant from the beginning and end in an A.P. is equal to first term + last term.

Answer is (C) 72°

Answer is (C) ∠A and side BC

Answer is (C) 50°

Answer is (B) Isosceles

In ∆PQR

PQ = PR

∠PQR = ∠PRQ …(i) (angles opposite to equal sides of a triangle are equal)

In ∆PSQ

Ext. ∠PQR > ∠PSQ …(ii)

From (i) and (ii), we get

∠PRQ > ∠PSQ

⇒ PS > PR (side opposite to greater angle is longer)

⇒ PS > PQ (∵ PQ = PR)

Answer is AD > AB

In ∆OBC and ∆OAD

∵ ∠B = ∠A = 90°

(as AD and BC are perpendiculars)

and OB = OA (given)

and ∠BOC = ∠AOD

(vertically opposite angles)

∴ ∆OBC ≅ ∠OAD

(by AAS congruence rule)

=> OA = OB (by c.p.c.t)

=> CD bisects AB

Hence proved.

In right angled triangles ABC and CDE

∠B = ∠D = 90° (given)

∠ACB = ∠DCE (vertically opposite angles)

and BC = CD (given)

ΔABC = ΔCDE (by ASA congruency property)

AB = DE (by c.p.c.t)

⇒ Yes, AB and DE are equal.

CD = CA

⇒ ∠CAD = ∠ADC = 20°

⇒ ∠ACD = 180° – (20° + 20°) = 140°

∠ACB = 180° – 140° = 40°

But AB also equal to AC

⇒ ∠ABC = ∠ACB = 40°

Answer is (A) Isosceles triangle

Answer is (A) RHS property

(3) 12

A = {a, b, p}, B = {2, 3}, C = {p, q, r, s} 

n (A ∪ C) x B 

A ∪ C = {a, b, p, q, r, s} 

(A ∪ C) x B = {a, 2), (a, 3), (b, 2), (b, 3), (p, 2), (p, 3), (q, 2), (q, 3), (r, 2), (r, 3), (s, 2), (s, 3)}

n [(A ∪ C) x B] = 12

In ∆ABC and ∆ABD

∠ABC = ∠ABD (given)

BC = BD (given)

AB = AB (common side)

⇒ ∆ABC ≅ ∆ABD (By SAS congruency property)