125876 + Interview Questions in General QA

1.	The value of \(sin\left(sin^{-1}\frac{1}{2}+cos^{-1}\frac{1}{2}\right)\)= ?A. 0 B. 1 C. -1 D. none of these
Answer» Correct Answer is 1 Now, let x = sin\(\left(sin^{-1}\frac{1}{2}+cos^{-1}\frac{1}{2}\right)\) \(\Rightarrow x=sin\left(\frac{\pi}{2}\right)\) \(\left(\because sin^{-1}\theta+cos^{-1}\theta=\frac{\pi}{2}\right)\) ⇒ x = 1 \(\left(\because sin\left(\frac{\pi}{2}\right)=1\right)\)

1.

The value of \(sin\left(sin^{-1}\frac{1}{2}+cos^{-1}\frac{1}{2}\right)\)= ?A. 0 B. 1 C. -1 D. none of these

Answer»

Correct Answer is 1

Now, let x = sin\(\left(sin^{-1}\frac{1}{2}+cos^{-1}\frac{1}{2}\right)\)

\(\Rightarrow x=sin\left(\frac{\pi}{2}\right)\) \(\left(\because sin^{-1}\theta+cos^{-1}\theta=\frac{\pi}{2}\right)\)

⇒ x = 1 \(\left(\because sin\left(\frac{\pi}{2}\right)=1\right)\)

2.	Define total internal reflection of light.
Answer» When light travels from a denser medium to a rarer medium, if the angle of incidence is greater than the critical angle, there is no refraction of light and all the light is reflected in the denser medium. This is called total internal reflection of light.

Discussion

3.	Define critical angle.
Answer» When light travels from a denser medium to a rarer medium, the angle of incidence for which the angle of refraction becomes 90°, is called the critical angle.

Discussion

4.	If f = {(5, 2), (6, 3)}, g = {(2, 5), (3, 6)}, write f o g.
Answer» f o g = {(2, 2), (3, 3)}

Discussion

5.	The relation f is defined by f(x) = {(x2, 0 ≤ x ≤ 3),(3x, 3 ≤ x ≤ 10) The relation g is defined by g(x) = {(x2, 0 ≤ x ≤ 2), (3x, 2 ≤ x ≤ 10) Show that/is a function and g is not a function.
Answer» Since f(x) is unique for 0 ≤ x ≤ 10. f(x) is a function. g(2) = 2² =4 and g(2) = 3(2) = 6 ∴ z has two images under g. ∴ g is not a function.

5.

The relation f is defined by f(x) = {(x2, 0 ≤ x ≤ 3),(3x, 3 ≤ x ≤ 10) The relation g is defined by g(x) = {(x2, 0 ≤ x ≤ 2), (3x, 2 ≤ x ≤ 10) Show that/is a function and g is not a function.

Answer»

Since f(x) is unique for 0 ≤ x ≤ 10.

f(x) is a function. g(2) = 2² =4 and g(2) = 3(2) = 6

∴ z has two images under g.

∴ g is not a function.

Discussion

6.	Let f = {(1, 1), (2, 3), (0, –1), (–1, –3)} be a function from Z to Z definedby f(x) = ax + b, for some integers a, b. Determine a, b.
Answer» f = {(1, 1), (2, 3), (0, –1), (–1, –3)} and f(x) = ax + b (1, 1) ∈ f ⇒ f(1) = 1 ⇒ a × 1 + b = 1 ⇒ a + b = 1 (0, –1) ∈ f ⇒ f(0) = –1 ⇒ a × 0 + b = –1 ⇒ b = –1 On substituting b = –1 in a + b = 1, We obtain a + (–1) = 1 ⇒ a = 1 + 1 = 2. Thus, the respective values of a and b are 2 and –1.

Discussion

7.	If f (x) = ax + b, where a and b are integers, f (–1) = – 5 and f (3) = 3, then a and b are equal to A. a = – 3, b = –1 B. a = 2, b = – 3 C. a = 0, b = 2 D. a = 2, b = 3
Answer» Given: f (x) = ax + b, where a and b are integers, f (–1) = – 5 and f (3) = 3 To find: the values of a and b Explanation: we have f (x) = ax + b Put x = -1 in above equation we get f (-1) = a(-1) + b But it is also given f(-1) = -5, so above equation becomes -5 = -a + b ⇒ b = a-5………(i) we have f (x) = ax + b Put x = 3 in above equation we get f (3) = a(3) + b But it is also given f(3) = 3, so above equation becomes 3 = 3a + b Now substituting the value of b from equation (i), we get 3 = 3a + a-5 ⇒ 4a = 5 + 3 ⇒ 4a = 8 ⇒ a = 2 Substituting the value of a in equation (i), we get b = a-5 = 2-5 = -3 Hence the values of a and b are 2 and -3 respectively. Hence the correct answer is option (B)

7.

If f (x) = ax + b, where a and b are integers, f (–1) = – 5 and f (3) = 3, then a and b are equal to A. a = – 3, b = –1 B. a = 2, b = – 3 C. a = 0, b = 2 D. a = 2, b = 3

Answer»

Given: f (x) = ax + b, where a and b are integers, f (–1) = – 5 and f (3) = 3

To find: the values of a and b

Explanation: we have f (x) = ax + b

Put x = -1 in above equation we get

f (-1) = a(-1) + b

But it is also given f(-1) = -5, so above equation becomes

-5 = -a + b

⇒ b = a-5………(i)

we have f (x) = ax + b

Put x = 3 in above equation we get

f (3) = a(3) + b

But it is also given f(3) = 3, so above equation becomes

3 = 3a + b

Now substituting the value of b from equation (i), we get

3 = 3a + a-5

⇒ 4a = 5 + 3

⇒ 4a = 8

⇒ a = 2

Substituting the value of a in equation (i), we get

b = a-5 = 2-5 = -3

Hence the values of a and b are 2 and -3 respectively.

Hence the correct answer is option (B)

Discussion

8.	Find whether the following operations are binary operation or not (i) on N defined by a b = ab (ii) * on Z defined by a * b = ab
Answer» (i) For a, b ∈ N, a * b = a^b is a unique natural number * is a binary operation on N. (ii) Let a,b ∈ Z⁺, a * b = a^b ∈ Z⁺ * is a binary operation on Z⁺.

8.

Find whether the following operations are binary operation or not (i) on N defined by a b = ab (ii) * on Z defined by a * b = ab

Answer»

(i) For a, b ∈ N, a * b = a^b is a unique natural number

* is a binary operation on N.

(ii) Let a,b ∈ Z⁺, a * b = a^b ∈ Z⁺

* is a binary operation on Z⁺.

Discussion

9.	Let f be the subset of Z × Z defined by f = {(ab, a + b): a, b ∈ Z}. Is f a function from Z to Z: justify your answer.
Answer» The relation f is defined as f = {(ab, a + b): a, b ∈ Z} We know that a relation f from a set A to a set B is said to be a function if every element of set A has unique images in set B. Since 2, 6, –2, –6 ∈ Z, (2 × 6, 2 + 6), (–2 × –6, –2 + (–6)) ∈ f i.e., (12, 8), (12, –8) ∈ f It can be seen that the same first element i.e., 12 corresponds to two different images i.e., 8 and –8. Thus, relation f is not a function.

Discussion

10.	Let f be the subset of Z x Z defined by f ={(ab,a +b):a,b ∈ z) Is f a function from Z to Z?
Answer» Given f = {(ab, a + b):a,b ∈ Z) If a = 1 and b = 4 = ab = 4 and a + b = 5 ∴ (4,5) ∈ f If a = 2 and b = 2 ⇒ ab = 4 and a + b = 4 ∴ (4,4) ∈ f ∴ The element 4 has two images, so f is not a function.

10.

Let f be the subset of Z x Z defined by f ={(ab,a +b):a,b ∈ z) Is f a function from Z to Z?

Answer»

Given f = {(ab, a + b):a,b ∈ Z)

If a = 1 and b = 4 = ab = 4 and a + b = 5

∴ (4,5) ∈ f If a = 2 and b = 2 ⇒ ab = 4 and a + b = 4

∴ (4,4) ∈ f

∴ The element 4 has two images, so f is not a function.

Discussion

11.	In the set N of natural numbers, define the binary operation * by m * n = g.c.d (m, n), m, n ∈ N. Is the operation * commutative and associative?
Answer» The operation is clearly commutative since m * n = g.c.d (m, n) = g.c.d (n, m) = n * m ∀m, n ∈ N. It is also associative because for l, m, n ∈ N, we have l * (m * n) = g. c. d (l, g.c.d (m, n)) = g.c.d. (g. c. d (l, m), n) = (l * m) * n.

11.

In the set N of natural numbers, define the binary operation * by m * n = g.c.d (m, n), m, n ∈ N. Is the operation * commutative and associative?

Answer»

The operation is clearly commutative since

m * n = g.c.d (m, n) = g.c.d (n, m) = n * m ∀m, n ∈ N.

It is also associative because for l, m, n ∈ N, we have

l * (m * n) = g. c. d (l, g.c.d (m, n))

= g.c.d. (g. c. d (l, m), n)

= (l * m) * n.

Discussion

12.	\(\left\| -\left( \frac { 3 }{ -7 } \right) \right\|\) का मान है :
Answer» \(\left\| -\left( \frac { 3 }{ -7 } \right) \right\|\) = \(\|\frac{3}{7}\|\) = \(\frac{3}{7}\)

12.

\(\left| -\left( \frac { 3 }{ -7 } \right) \right|\) का मान है :

Answer»

\(\left| -\left( \frac { 3 }{ -7 } \right) \right|\)

= \(|\frac{3}{7}|\)

= \(\frac{3}{7}\)

Discussion

13.	State whether the statements are true (T) or false (F). Each prime factor appears 3 times in its cube.
Answer» True. We know that, a perfect cube can always be expressed as the product of triplets of prime factors.

13.

State whether the statements are true (T) or false (F). Each prime factor appears 3 times in its cube.

Answer»

True.

We know that, a perfect cube can always be expressed as the product of triplets of prime factors.

Discussion

14.

Multiply the given integers and complete the table given below.(-5) × (-3) = 15(-6) × (-3) =.....(-7) × (-3) =.....(-5) × (-2) = 15(-6) × (-3) =.....(-7) × (-3) =.....(-5) × (-1) = 15(-6) × (-3) =.....(-7) × (-3) =.....(-5) × 0 = 0(-6) × 0 =.....(-7) × 0 =.....(-5) × 1 = -5(-6) × 1 =.....(-7) × 1 =.....(-5) × 2 = -10(-6) × 2 =.....(-7) × 2 =.....(-5) × 3 = -15(-6) × 3 =.....(-7) × 3 =.....(-5) × 4 = -20(-6) × 4 =.....(-7) × 4 =.....

Answer»

(-6) × (-3) =	18	(-7) × (-3) =	21
(-6) × (-2) =	12	(-7) × (-2) =	14
(-6) × (-1) =	6	(-7) × (-1) =	7
(-6) × 0 =	0	(-7) × 0 =	0
(-6) × 1 =	-6	(-7) × 1 =	-7
(-6) × 2 =	-12	(-7) × 2 =	-14
(-6) × 3 =	-18	(-7) × 3 =	-21
(-6) × 4 =	-24	(-7) × 4 =	-28

Discussion

15.	State whether the statements are true (T) or false (F).The square root of 1521 is 31.
Answer» False As, the square of 31 = (31)² = 31 x 31 = 961

15.

State whether the statements are true (T) or false (F).The square root of 1521 is 31.

Answer»

False

As, the square of 31 = (31)²

= 31 x 31

= 961

Discussion

16.	Find the squares of the following numbers using the identity (a-b)2= a2-2ab+b2(i) 395(ii) 995(iii)495(iv) 498
Answer» (i) 395 We know, (a-b)²= a²-2ab+b² 395 = (400-5)² = (400)² + 5² – 2 (400) (5) = 160000 + 25 – 4000 = 156025 (ii) 995 We know, (a-b)²= a²-2ab+b² 995 = (1000-5)² = (1000)² + 5² – 2 (1000) (5) = 1000000 + 25 – 10000 = 990025 (iii) 495 We know, (a-b)²= a²-2ab+b² 495 = (500-5)² = (500)²+ 5² – 2 (500) (5) = 250000 + 25 – 5000 = 245025 (iv) 498 We know, (a-b)²= a²-2ab+b² 498 = (500-2)² = (500)² + 2² – 2 (500) (2) = 250000 + 4 – 2000 = 248004

16.

Find the squares of the following numbers using the identity (a-b)2= a2-2ab+b2(i) 395(ii) 995(iii)495(iv) 498

Answer»

(i) 395

We know, (a-b)²= a²-2ab+b²

395 = (400-5)²

= (400)² + 5² – 2 (400) (5)

= 160000 + 25 – 4000

= 156025

(ii) 995

We know, (a-b)²= a²-2ab+b²

995 = (1000-5)²

= (1000)² + 5² – 2 (1000) (5)

= 1000000 + 25 – 10000

= 990025

(iii) 495

We know, (a-b)²= a²-2ab+b²

495 = (500-5)²

= (500)²+ 5² – 2 (500) (5)

= 250000 + 25 – 5000

= 245025

(iv) 498

We know, (a-b)²= a²-2ab+b²

498 = (500-2)²

= (500)² + 2² – 2 (500) (2)

= 250000 + 4 – 2000

= 248004

Discussion

17.	Observe the following pattern1 = \(\frac{1}{2}\){1 x (1 + )}1 + 2 = \(\frac{1}{2}\){2 x (2 + )}1 + 2 + 3 + 4 = \(\frac{1}{2}\){3 x (3 x 1)}1 + 2 + 3 + 4 = \(\frac{1}{2}\){4 x (4 x 1)}And find the values of each of the following:(i) 1+2+3+4+5+……….+50(ii) 31+32+……..+50
Answer» R.H.S = \(\frac{1}{2}\)[No.of terms in L.H.S x (No. of terms + 1)] (Therefore, only when L.H.S starts with 1) Therefore. (i) 1 + 2 + 3 +…..50 = \(\frac{1}{2}\)[50 x (50 + 1)} = 25 × 51 = 1275 (ii) 31 + 32 +…..+50 = (1 + 2 + 3 + …. + 50) – (1 + 2 + ….. 30) = 1275 - [\(\frac{1}{2}\)(30 x 30 + 1)] = 1275 – 465 = 810

17.

Observe the following pattern1 = \(\frac{1}{2}\){1 x (1 + )}1 + 2 = \(\frac{1}{2}\){2 x (2 + )}1 + 2 + 3 + 4 = \(\frac{1}{2}\){3 x (3 x 1)}1 + 2 + 3 + 4 = \(\frac{1}{2}\){4 x (4 x 1)}And find the values of each of the following:(i) 1+2+3+4+5+……….+50(ii) 31+32+……..+50

Answer»

R.H.S = \(\frac{1}{2}\)[No.of terms in L.H.S x (No. of terms + 1)] (Therefore, only when L.H.S starts with 1)

Therefore.

(i) 1 + 2 + 3 +…..50 = \(\frac{1}{2}\)[50 x (50 + 1)}

= 25 × 51

= 1275

(ii) 31 + 32 +…..+50 = (1 + 2 + 3 + …. + 50) – (1 + 2 + ….. 30)

= 1275 - [\(\frac{1}{2}\)(30 x 30 + 1)]

= 1275 – 465

= 810

Discussion

18.	Find the squares of the following numbers using the identity (a - b)2 = a2 - 2ab + b2 :(i) 395 (ii) 995 (iii)495 (iv) 498 (v) 99 (vi) 999 (vii)599
Answer» (i) 395 395 = (400 – 5)² = (400)²+ 5² – 2 (400) (5) = 160000 + 25 – 4000 = 156025 (ii) 995 995 = (1000 – 5)² = (1000)² + 5² – 2 (1000) (5) = 1000000 + 25 – 10000 = 990025 (iii)495 495 = (500 – 5)² = (500)² + 5² – 2 (500) (5) = 250000 + 25 – 5000 = 245025 (iv) 498 498 = (500 – 2)² = (500)² + 2² – 2 (500) (2) = 250000 + 4 – 2000 = 248004 (v) 99 99 = (100 – 1)² = (100)² + 1² – 2 (100) (1) = 10000 + 1 – 200 = 9799 (vi) 999 999 = (1000 – 1)² = (1000)² + 1² – 2 (1000) (1) = 1000000 + 1 – 2000 = 998001 (vii)599 (600 – 1)² = (600)² + 1² – 2 (600) (1) = 360000 + 1 – 1200 = 358801

18.

Find the squares of the following numbers using the identity (a - b)2 = a2 - 2ab + b2 :(i) 395 (ii) 995 (iii)495 (iv) 498 (v) 99 (vi) 999 (vii)599

Answer»

(i) 395

395 = (400 – 5)²

= (400)²+ 5² – 2 (400) (5)

= 160000 + 25 – 4000

= 156025

(ii) 995

995 = (1000 – 5)² = (1000)² + 5² – 2 (1000) (5)

= 1000000 + 25 – 10000

= 990025

(iii)495

495 = (500 – 5)² = (500)² + 5² – 2 (500) (5)

= 250000 + 25 – 5000

= 245025

(iv) 498

498 = (500 – 2)² = (500)² + 2² – 2 (500) (2)

= 250000 + 4 – 2000

= 248004

(v) 99

99 = (100 – 1)²

= (100)² + 1² – 2 (100) (1)

= 10000 + 1 – 200 = 9799

(vi) 999

999 = (1000 – 1)²

= (1000)² + 1² – 2 (1000) (1)

= 1000000 + 1 – 2000

= 998001

(vii)599

(600 – 1)² = (600)² + 1² – 2 (600) (1)

= 360000 + 1 – 1200

= 358801

Discussion

19.	Observe the following pattern12 = \(\frac{1}{6}\)[1 x (1 + 1) x (2 x 1 + 1)12 + 22 = \(\frac{1}{2}\)[2 x (2 + 1) x (2 x 2 + 1)]12 + 22 + 32 = \(\frac{1}{6}\)[3 x (3 + 1) x (2 x 3 + 1)]12 + 22 + 32 + 42 = \(\frac{1}{6}\)[4 x (4 + 1) x (2 x 4 + 1)]And find the values of each of the following.11 + 22 + 32 + 42 + .........+ 10251 + 62 + 72 + 82 + 92 + 102 + 112 + 122
Answer» R.H.S = \(\frac{1}{6}\)[ (No.of terms in L.H.S) x (No. + 1) x (2 x No. + 1)] (i) 1² + 2² + 3² + 4² + …… + 10² = \(\frac{1}{6} \) [10 (10 + 1) × (2 × 10 + 1)] = \(\frac{1}{6}\)[2310] = 385 (ii) 5² + 6² +….. + 12² = 1² + 2² + ….. 12² – (1² + 2² + 3³ + 4²) \(\frac{1}{6} \)[12 × (12 + 1) × (2 × 12 + 1)] -\(\frac{1}{6}\) [4 ×(4 + 1) × (2 × 4 + 1)] = 650 – 30 = 620

19.

Observe the following pattern12 = \(\frac{1}{6}\)[1 x (1 + 1) x (2 x 1 + 1)12 + 22 = \(\frac{1}{2}\)[2 x (2 + 1) x (2 x 2 + 1)]12 + 22 + 32 = \(\frac{1}{6}\)[3 x (3 + 1) x (2 x 3 + 1)]12 + 22 + 32 + 42 = \(\frac{1}{6}\)[4 x (4 + 1) x (2 x 4 + 1)]And find the values of each of the following.11 + 22 + 32 + 42 + .........+ 10251 + 62 + 72 + 82 + 92 + 102 + 112 + 122

Answer»

R.H.S = \(\frac{1}{6}\)[ (No.of terms in L.H.S) x (No. + 1) x (2 x No. + 1)]

(i) 1² + 2² + 3² + 4² + …… + 10² = \(\frac{1}{6} \) [10 (10 + 1) × (2 × 10 + 1)]

= \(\frac{1}{6}\)[2310]

= 385

(ii) 5² + 6² +….. + 12² = 1² + 2² + ….. 12² – (1² + 2² + 3³ + 4²)

\(\frac{1}{6} \)[12 × (12 + 1) × (2 × 12 + 1)] -\(\frac{1}{6}\) [4 ×(4 + 1) × (2 × 4 + 1)]

= 650 – 30

= 620

Discussion

20.	Solve i. (- 96) ÷ 16 ii. 98 ÷ (- 28) iii. (- 51) ÷ 68 iv . 38 ÷ (- 57) v . (- 85) ÷ 20
Answer» i. (- 96) ÷ 16 = -96/16 = (-1 x 96)/16 = (-1) x (6) = -6 ii. 98 ÷ (-28) = 98/(-28) = 98/(-1)( 28) = 7/(-1 x 2) = -7/2 iii. (-51) ÷ 68 = -51/68 = (-1 x 51)/68 = (-1 x 3)/4 = -3/4 iv. 38 ÷ (-57) = 38/(-57) = 38/(-1 x 57) = 2/(-1 x 3) = -2/3 v. -85 ÷ 20 = -85/20 = (-1 x 85)/20 = (-1 x 85)/20 = (-1 x 17)/4 = -17/4

20.

Solve i. (- 96) ÷ 16 ii. 98 ÷ (- 28) iii. (- 51) ÷ 68 iv . 38 ÷ (- 57) v . (- 85) ÷ 20

Answer»

i. (- 96) ÷ 16

= -96/16 = (-1 x 96)/16

= (-1) x (6) = -6

ii. 98 ÷ (-28)

= 98/(-28) = 98/(-1)( 28)

= 7/(-1 x 2) = -7/2

iii. (-51) ÷ 68

= -51/68 = (-1 x 51)/68

= (-1 x 3)/4

= -3/4

iv. 38 ÷ (-57)

= 38/(-57) = 38/(-1 x 57)

= 2/(-1 x 3) = -2/3

v. -85 ÷ 20

= -85/20 = (-1 x 85)/20

= (-1 x 85)/20

= (-1 x 17)/4

= -17/4

Discussion

21.	Multiply: 1. (-5) × (-7) 2. (-9) × (6) 3. (9) × (-4) 4. (8) × (-7) 5. (-124) × (-1) 6. (-12) × (-7) 7. (-63) × (-7) 8. (-7) × (15)
Answer» 1. 35 2. -54 3. -36 4. -56 5. 124 6. 84 7. 441 8. -105

21.

Multiply: 1. (-5) × (-7) 2. (-9) × (6) 3. (9) × (-4) 4. (8) × (-7) 5. (-124) × (-1) 6. (-12) × (-7) 7. (-63) × (-7) 8. (-7) × (15)

Answer»

1. 35

2. -54

3. -36

4. -56

5. 124

6. 84

7. 441

8. -105

Discussion

22.	Solve i. (-150) ÷ (-25) ii. 100 ÷ 60 iii. 9 ÷ (- 54) iv. 78 ÷ 65 v. (-5) ÷ (-315)
Answer» i. (-150) ÷ (-25) = -150/(-25) = (-1 x 150)/(-1 x 25) = 6 ii. 100 ÷ 60 = 100/60 = 5/3 iii. 9 ÷ (-54) = 9/-54 = 9/(-1 x 54) = -1/6 iv 78 ÷ 65 = 78/65 = 6/5 v. (-5) ÷ (-315) = -5/-315 = (-1 x 5)/(-1 x 315) = 1/63

22.

Solve i. (-150) ÷ (-25) ii. 100 ÷ 60 iii. 9 ÷ (- 54) iv. 78 ÷ 65 v. (-5) ÷ (-315)

Answer»

i. (-150) ÷ (-25)

= -150/(-25) = (-1 x 150)/(-1 x 25) = 6

ii. 100 ÷ 60

= 100/60 = 5/3

iii. 9 ÷ (-54)

= 9/-54 = 9/(-1 x 54) = -1/6

iv 78 ÷ 65

= 78/65 = 6/5

v. (-5) ÷ (-315)

= -5/-315 = (-1 x 5)/(-1 x 315) = 1/63

Discussion

23.	In the previous class, we have learnt to add and subtract integers. Using those methods, fill in the blanks below.1. 5 + 7 = __ 2. 10 + (-5) = __ 3. -4 + 3 = __ 4. (-7) + (-2) = __ 5. (+8) – (+ 3) = __ 6. (+8) – (-3) = __
Answer» 1. 12 2. 5 3. -1 4. -9 5. 5 6. 11

23.

In the previous class, we have learnt to add and subtract integers. Using those methods, fill in the blanks below.1. 5 + 7 = 2. 10 + (-5) = 3. -4 + 3 = 4. (-7) + (-2) = 5. (+8) – (+ 3) = 6. (+8) – (-3) =

Answer»

1. 12

2. 5

3. -1

4. -9

5. 5

6. 11

Discussion

24.	Multiply: 1. (-5) × (-7) 2. (-9) × (6) 3. (9) × (-4) 4. (8) × (-7) 5. (-124) × (-1) 6. (-12) × (-7) 7. (-63) × (-7) 8. (-7) × (15)
Answer» 1. 35 2. -54 3. -36 4. -56 5. 124 6. 84 7. 441 8. -105

24.

Multiply: 1. (-5) × (-7) 2. (-9) × (6) 3. (9) × (-4) 4. (8) × (-7) 5. (-124) × (-1) 6. (-12) × (-7) 7. (-63) × (-7) 8. (-7) × (15)

Answer»

1. 35

2. -54

3. -36

4. -56

5. 124

6. 84

7. 441

8. -105

Discussion

25.	Write three divisions of integers such that the fractional form of each will be -5/7.
Answer» 1. -5/7 = (-5 x2)/(7 x 2) = -10/14 = (-10 ÷ 14) 2. -5/7 = (-5 x (-5))/(7 x (-5)) = 25/(-35) = 25 ÷ (-35) 3. -5/7 = (-5 x 7)/7 x 7 = (-35)/49 = (-35 ÷ 49)

25.

Write three divisions of integers such that the fractional form of each will be -5/7.

Answer»

1. -5/7 = (-5 x2)/(7 x 2) = -10/14 = (-10 ÷ 14)

2. -5/7 = (-5 x (-5))/(7 x (-5)) = 25/(-35) = 25 ÷ (-35)

3. -5/7 = (-5 x 7)/7 x 7 = (-35)/49 = (-35 ÷ 49)

Discussion

26.	Write three divisions of integers such that the fractional form of each will be 24/5.
Answer» 1. 24/5 = (24 x 1)/(5 x 1) = 24/5 = 24 ÷ 5 2. 24/5 = (24 x 2)/(5 x 2) = 48/10 = 48 ÷ 10 3. 24/5 = (24 x (-10))/(5 x (-10)) = -240/(-50) = (-240) ÷ (-50)

26.

Write three divisions of integers such that the fractional form of each will be 24/5.

Answer»

1. 24/5 = (24 x 1)/(5 x 1) = 24/5 = 24 ÷ 5

2. 24/5 = (24 x 2)/(5 x 2) = 48/10 = 48 ÷ 10

3. 24/5 = (24 x (-10))/(5 x (-10)) = -240/(-50) = (-240) ÷ (-50)

Discussion

27.	-1 और 1 के बीच की परिमेय संख्या ज्ञात कीजिए।
Answer» -1 और 1 के बीच की परिमेय संख्या \(=\frac{1}{2}\times(-1+1) \) \(= \frac{1}{2}\times0\) \(= 0\)

27.

-1 और 1 के बीच की परिमेय संख्या ज्ञात कीजिए।

Answer»

-1 और 1 के बीच की परिमेय संख्या

\(=\frac{1}{2}\times(-1+1) \)

\(= \frac{1}{2}\times0\)

\(= 0\)

Discussion

28.	The fish in the pond below, carry some numbers. (Choose any 4 pairs and carry out four multiplications with those numbers. Now, choose four other pairs and carry out divisions with these numbers.Examples: i. (-13) × (-15) = 195ii. (-24) ÷ 9 = (-24)/9 = -8/3
Answer» 1. (-13) × 9 = -117 2. 12 × 13 = 156 3. 9 × (-37) = -333 4. (-15) × (-8) = 120 5. -28 ÷ 12 = (-28)/12 = (-1 x 28)/12 = -7/3 6. 12 ÷ 9 = 12/9 = 4/3 7. 9 ÷ (-24) = 9/(-24) = 8/(-1 x 24) = (-3/8) 8. -18 ÷ (-27) = (-18)/(-27) = (-1 x 18)/(-1 x 27) = 2/3 Note: Problems 2, 3 and 4 have many answers. Students may write answers other than the ones given

28.

The fish in the pond below, carry some numbers. (Choose any 4 pairs and carry out four multiplications with those numbers. Now, choose four other pairs and carry out divisions with these numbers.Examples: i. (-13) × (-15) = 195ii. (-24) ÷ 9 = (-24)/9 = -8/3

Answer»

1. (-13) × 9 = -117

2. 12 × 13 = 156

3. 9 × (-37) = -333

4. (-15) × (-8) = 120

5. -28 ÷ 12 = (-28)/12 = (-1 x 28)/12 = -7/3

6. 12 ÷ 9 = 12/9 = 4/3

7. 9 ÷ (-24) = 9/(-24) = 8/(-1 x 24) = (-3/8)

8. -18 ÷ (-27) = (-18)/(-27) = (-1 x 18)/(-1 x 27) = 2/3

Note: Problems 2, 3 and 4 have many answers. Students may write answers other than the ones given

Discussion

29.	Write three divisions of integers such that the fractional form of each will be 24/5.
Answer» 1. \(\frac{24}{5} = \frac{24 \times 1}{5 \times 1} = \frac{24}{5} = 24 \div 5\) 2. \(\frac{24}{5} = \frac{24 \times 2}{5 \times 2} = \frac{48}{10} = 48 \div 10\) 3. \(\frac{24}{5} = \frac{24 \times (-10)}{5 \times (-10)} = \frac{-240}{-50} = (-240) \div (-50)\)

29.

Write three divisions of integers such that the fractional form of each will be 24/5.

Answer»

1. \(\frac{24}{5} = \frac{24 \times 1}{5 \times 1} = \frac{24}{5} = 24 \div 5\)

2. \(\frac{24}{5} = \frac{24 \times 2}{5 \times 2} = \frac{48}{10} = 48 \div 10\)

3. \(\frac{24}{5} = \frac{24 \times (-10)}{5 \times (-10)} = \frac{-240}{-50} = (-240) \div (-50)\)

Discussion

30.	Write three divisions of integers such that the fractional form of each will be -5/7.
Answer» 1. \(\frac{-5}{7} = \frac{-5 \times 2}{7 \times 2} = \frac{-10}{14} = (-10) \div 14\) 2. \(\frac{-5}{7} = \frac{-5 \times (-5)}{7 \times (-5)} = \frac{25}{-35} = 25 \div (-35)\) 3. \(\frac{-5}{7} = \frac{-5 \times 7}{7 \times 7} = \frac{-35}{49} = (-35) \div 49\)

30.

Write three divisions of integers such that the fractional form of each will be -5/7.

Answer»

1. \(\frac{-5}{7} = \frac{-5 \times 2}{7 \times 2} = \frac{-10}{14} = (-10) \div 14\)

2. \(\frac{-5}{7} = \frac{-5 \times (-5)}{7 \times (-5)} = \frac{25}{-35} = 25 \div (-35)\)

3. \(\frac{-5}{7} = \frac{-5 \times 7}{7 \times 7} = \frac{-35}{49} = (-35) \div 49\)

Discussion

31.	The fish in the pond below, carry some numbers. (Choose any 4 pairs and carry out four multiplications with those numbers. Now, choose four other pairs and carry out divisions with these numbers.)Examples: i. (-13) × (-15) = 195ii. (-24) ÷ 9 = -24/9 = -8/3
Answer» 1. (-13) × 9 = -117 2. 12 × 13 = 156 3. 9 × (-37) = -333 4. (-15) × (-8) = 120 5. (−28) ÷ 12 = -28/12 \(=\frac{(-1)\times (28)}{12}\) = -7/3 6. 12 ÷ 9 = 12/9 = 4/3 7. 9 ÷ (−24) = 9/-24 \(=\frac{9}{(-1)\times (24)}\) = -3/8 8. (−18) ÷ (−27) = -18/-27 \(=\frac{(-1)\times (18)}{(-1)\times (27)}\) = 2/3

31.

The fish in the pond below, carry some numbers. (Choose any 4 pairs and carry out four multiplications with those numbers. Now, choose four other pairs and carry out divisions with these numbers.)Examples: i. (-13) × (-15) = 195ii. (-24) ÷ 9 = -24/9 = -8/3

Answer»

1. (-13) × 9 = -117

2. 12 × 13 = 156

3. 9 × (-37) = -333

4. (-15) × (-8) = 120

5. (−28) ÷ 12 = -28/12 \(=\frac{(-1)\times (28)}{12}\) = -7/3

6. 12 ÷ 9 = 12/9 = 4/3

7. 9 ÷ (−24) = 9/-24 \(=\frac{9}{(-1)\times (24)}\) = -3/8

8. (−18) ÷ (−27) = -18/-27 \(=\frac{(-1)\times (18)}{(-1)\times (27)}\) = 2/3

Discussion

32.	Write two prime numbers whose difference is 10.
Answer» Two prime numbers whose difference is 10 are (7, 17); (13, 23); (31, 41); (43, 53); (61, 71); (79, 89);…….

Discussion

33.	Write seven consecutive composite numbers less than 100.
Answer» Consecutive composite numbers are 8, 9; 9, 10; 14, 15; 15, 16; 20, 21; 21, 22; 24, 25; 25, 26; 26, 27; 27, 28; 32, 33; 33, 34; 34, 35; 35, 36;…….

Discussion

34.	Write 10 odd and 10 even composite numbers.
Answer» Odd composite numbers are 9, 15, 21, 25, 27, 33, 35, 39, 45, 49. Even composite numbers are 4, 6, 8, 10, 12, 14, 16, 18, 20, 22. Except 2, every even number is a composite number.

34.

Write 10 odd and 10 even composite numbers.

Answer»

Odd composite numbers are 9, 15, 21, 25, 27, 33, 35, 39, 45, 49.

Even composite numbers are 4, 6, 8, 10, 12, 14, 16, 18, 20, 22.

Except 2, every even number is a composite number.

Discussion

35.	If the difference of any two prime numbers is 2, then they are said to be A) Twin primesB) Co-primes C) Perfect numbers D) Exact numbers
Answer» Correct option is A) Twin primes

Discussion

36.	Every prime number other than 2 is ……………… A) even B) composite C) odd D) A and B
Answer» Correct option is C) odd

Discussion

37.	The number which is divisible by 1 and itself is called ……………… A) Composite number B) Prime number C) Even number D) Odd number
Answer» B.Prime Number Ex: 5,7,17... 5 = 5×1 1 and itself . B) Prime number

37.

The number which is divisible by 1 and itself is called ……………… A) Composite number B) Prime number C) Even number D) Odd number

Answer»

B.Prime Number

Ex: 5,7,17...

5 = 5×1

1 and itself .

B) Prime number

Discussion

38.	Smallest even prime number is ……………….. A) 4 B) 3 C) 2D) 5
Answer» Correct option is C) 2

Discussion

39.	Number 1 is …………….A) Prime B) CompositeC) Both A and B D) Neither A nor B
Answer» Correct option is B) Composite

Discussion

40.

Complete the table given below.-3\(\frac{3}{5}\)-17\(\frac{-5}{11}\)5Natural Numbersx✓Integers✓Rational Numbers✓

Answer»

	-3	\(\frac{3}{5}\)	-17	\(\frac{-5}{11}\)	5
Natural Numbers	x	x	x	x	✓
Integers	✓	x	✓	x	✓
Rational Numbers	✓	✓	✓	✓	✓

Discussion

41.	L.C.M means ………………
Answer» Least Common Multiple

Discussion

42.	G.C.D means ………………
Answer» Greatest Common Divisor

Discussion

43.

Discuss the characteristics of various groups of numbers in class and complete the table below. In front of each group, write the inference you make after carrying out the operations of addition, subtraction, multiplication and division, using a (✓) or a (x).Remember that you cannot divide by zero.Group of NumbersAdditionSubtractionMultiplicationDivisionNatural Numbers✓x(7- 10 =-3)✓x(3 ÷ 5 = 3/5)IntegersRational Numbers

Answer»

Group of Numbers	Addition	Subtraction	Multiplication	Division
Natural Numbers	✓	x (7- 10 =-3)	✓	x (3 ÷ 5 = 3/5)
Integers	✓	✓	✓	x (4 ÷ 9 = 4/9)
Rational Numbers	✓	✓	✓	✓

Discussion

44.	Use the signs and numbers in the boxes and form an expression such that its value will be 112. [0, 1, 2, 3, 4, 5, 6, 7, 8, 9][+ x ÷ -]
Answer» {3 + (6 x 7) + (9 ÷ 3)} + {- 8 + 8 x 9} Note: The above problem has many solutions. Students may write solution other than the one given.

44.

Use the signs and numbers in the boxes and form an expression such that its value will be 112. [0, 1, 2, 3, 4, 5, 6, 7, 8, 9][+ x ÷ -]

Answer»

{3 + (6 x 7) + (9 ÷ 3)} + {- 8 + 8 x 9}

Note: The above problem has many solutions. Students may write solution other than the one given.

Discussion

45.	Answer the following questions: 1. Write all the natural numbers between 2 and 9. 2. Write all the integers between -4, and 5. 3. Which rational numbers are there between 1/2 and 3/4 ?
Answer» 1. 3, 4, 5, 6, 7, 8 2. -3, -2, -1, 0, 1, 2, 3, 4 3. 1/2 = (1 x 2)/(2 x 2) = 2/4 = (2 x 10)/(4 x 10) = 20/40 3/4 = (3 x 10)/(4 x 10) = 30/40 ∴The rational numbers between 1/2 and 3/4 are 21/40, 22/40, 25/40, 27/40 etc.

45.

Answer the following questions: 1. Write all the natural numbers between 2 and 9. 2. Write all the integers between -4, and 5. 3. Which rational numbers are there between 1/2 and 3/4 ?

Answer»

1. 3, 4, 5, 6, 7, 8

2. -3, -2, -1, 0, 1, 2, 3, 4

3. 1/2 = (1 x 2)/(2 x 2) = 2/4 = (2 x 10)/(4 x 10) = 20/40

3/4 = (3 x 10)/(4 x 10) = 30/40

∴The rational numbers between 1/2 and 3/4 are 21/40, 22/40, 25/40, 27/40 etc.

Discussion

46.	Show that the lines \(\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1}\) and \(\frac{x}{1}=\frac{y}{2}=\frac{z}{3}\) are perpendicular to each other.
Answer» The Cartesian equation of the lines are \(\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z-0}{1}\) and \(\frac{x-0}{1}=\frac{y-0}{2}=\frac{z-0}{3}\), as we know their direction ratios we can find weather they are perpendicular or not, for proving the lines to be perpendicular we can only consider the numerator of the dot product when we use cos θ = numerator cos θ = 7 × 1 + 2 × (– 5) + 1 × 3 cos θ = 7 – 10 + 3 cos θ = 0 θ = \(\frac{\pi}{2}\) Therefore, the lines are perpendicular.

46.

Show that the lines \(\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1}\) and \(\frac{x}{1}=\frac{y}{2}=\frac{z}{3}\) are perpendicular to each other.

Answer»

The Cartesian equation of the lines are \(\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z-0}{1}\) and \(\frac{x-0}{1}=\frac{y-0}{2}=\frac{z-0}{3}\), as we know their direction ratios we can find weather they are perpendicular or not, for proving the lines to be perpendicular we can only consider the numerator of the dot product when we use cos θ = numerator

cos θ = 7 × 1 + 2 × (– 5) + 1 × 3

cos θ = 7 – 10 + 3

cos θ = 0

θ = \(\frac{\pi}{2}\)

Therefore, the lines are perpendicular.

Discussion

47.

Discuss the characteristic of various groups of numbers in class and complete the table below. in front of each group, write the inference you make after carrying out the operations if addition,subtraction, multiplication, and division, using a (✓) or an (x). Remember that you cannot divide by zero.Group of numbers AdditionSubtractionMultiplication Division NaturalNumbers✓x(7- 10 =-3)✓x(3 ÷ 5 = 3/5)IntegersRational Numbers

Answer»

Group of numbers	Addition	Subtraction	Multiplication	Division
Natural Numbers	✓	x (7- 10 =-3)	✓	x (3 ÷ 5 = 3/5)
Integers	✓	✓	✓	x (4 ÷ 9 = 4/9)
Rational Numbers	✓	✓	✓

Discussion

48.

Show that the lines \(\frac{x-5}{7}=\frac{Ψ+2}{-5}=\frac{z}{1}\) and \(\frac{x}{1}=\frac{y}{2}=\frac{z}{3}\) are perpendicular to each other.

Answer»

The Cartesian equation of the lines are \(\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z-0}{1}\) and \(\frac{x}{1}=\frac{y}{2}=\frac{z}{3}\) and we need to find that weather the lines are perpendicular or not, so we will use the dot product equation, as we know the direction ratios of both the lines.

a₁a₂+ b₁b₂+ c₁c₂ = (7)(1) + (–5)(2) + (1)(3) = 7 – 10 + 3 = 0

Hence the given lines are perpendicular because,

cos θ = 0

\(\theta=\frac{\pi}{2}\)

Discussion

49.

Write three rational numbers that lie between the two given numbers.i. 2/7, 6/7ii. 4/5, 2/3iii. -(2/3), 4/5iv. 7/9, -(5/9)

Answer»

i. 2/7, 6/7

The three numbers lying between 2/7 and 6/7 are 3/7, 4/7, 5/7

ii. 4/5, 2/3

4/5 = 24/30, 2/3 = 20/30

The three numbers between 4/5 and 2/3 are 21/30, 22/30, 23/30

iii. -(2/3), 4/5

-2/3 = (-10)/15, 4/5 = 12/15

The three numbers between -2/3 and 4/5 are -9/15, -7/15, 4/15

iv. 7/9, -(5/9)

The three numbers between 7/9 and -5/9 are 6/9, 0, -4/9

Discussion

50.	HCF means ……………….. A) Highest Common Factor B) Highest Continued FactorC) Hight of the Common Factor D) Highest composite factor
Answer» A) Highest Common Factor

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

The value of \(sin\left(sin^{-1}\frac{1}{2}+cos^{-1}\frac{1}{2}\right)\)= ?A. 0 B. 1 C. -1 D. none of these

Define total internal reflection of light.

Define critical angle.

If f = {(5, 2), (6, 3)}, g = {(2, 5), (3, 6)}, write f o g.

The relation f is defined by f(x) = {(x2, 0 ≤ x ≤ 3),(3x, 3 ≤ x ≤ 10) The relation g is defined by g(x) = {(x2, 0 ≤ x ≤ 2), (3x, 2 ≤ x ≤ 10) Show that/is a function and g is not a function.

Let f = {(1, 1), (2, 3), (0, –1), (–1, –3)} be a function from Z to Z definedby f(x) = ax + b, for some integers a, b. Determine a, b.

If f (x) = ax + b, where a and b are integers, f (–1) = – 5 and f (3) = 3, then a and b are equal to A. a = – 3, b = –1 B. a = 2, b = – 3 C. a = 0, b = 2 D. a = 2, b = 3

Find whether the following operations are binary operation or not (i) *on N defined by a * b = ab (ii) * on Z defined by a * b = ab

Let f be the subset of Z × Z defined by f = {(ab, a + b): a, b ∈ Z}. Is f a function from Z to Z: justify your answer.

Let f be the subset of Z x Z defined by f ={(ab,a +b):a,b ∈ z) Is f a function from Z to Z?

In the set N of natural numbers, define the binary operation * by m * n = g.c.d (m, n), m, n ∈ N. Is the operation * commutative and associative?

\(\left| -\left( \frac { 3 }{ -7 } \right) \right|\) का मान है :

State whether the statements are true (T) or false (F). Each prime factor appears 3 times in its cube.

State whether the statements are true (T) or false (F).The square root of 1521 is 31.

Find the squares of the following numbers using the identity (a-b)2= a2-2ab+b2(i) 395(ii) 995(iii)495(iv) 498

Observe the following pattern1 = \(\frac{1}{2}\){1 x (1 + )}1 + 2 = \(\frac{1}{2}\){2 x (2 + )}1 + 2 + 3 + 4 = \(\frac{1}{2}\){3 x (3 x 1)}1 + 2 + 3 + 4 = \(\frac{1}{2}\){4 x (4 x 1)}And find the values of each of the following:(i) 1+2+3+4+5+……….+50(ii) 31+32+……..+50

Find the squares of the following numbers using the identity (a - b)2 = a2 - 2ab + b2 :(i) 395 (ii) 995 (iii)495 (iv) 498 (v) 99 (vi) 999 (vii)599

Solve i. (- 96) ÷ 16 ii. 98 ÷ (- 28) iii. (- 51) ÷ 68 iv . 38 ÷ (- 57) v . (- 85) ÷ 20

Multiply: 1. (-5) × (-7) 2. (-9) × (6) 3. (9) × (-4) 4. (8) × (-7) 5. (-124) × (-1) 6. (-12) × (-7) 7. (-63) × (-7) 8. (-7) × (15)

Solve i. (-150) ÷ (-25) ii. 100 ÷ 60 iii. 9 ÷ (- 54) iv. 78 ÷ 65 v. (-5) ÷ (-315)

In the previous class, we have learnt to add and subtract integers. Using those methods, fill in the blanks below.1. 5 + 7 = __ 2. 10 + (-5) = __ 3. -4 + 3 = __ 4. (-7) + (-2) = __ 5. (+8) – (+ 3) = __ 6. (+8) – (-3) = __

Multiply: 1. (-5) × (-7) 2. (-9) × (6) 3. (9) × (-4) 4. (8) × (-7) 5. (-124) × (-1) 6. (-12) × (-7) 7. (-63) × (-7) 8. (-7) × (15)

Write three divisions of integers such that the fractional form of each will be -5/7.

Write three divisions of integers such that the fractional form of each will be 24/5.

-1 और 1 के बीच की परिमेय संख्या ज्ञात कीजिए।

The fish in the pond below, carry some numbers. (Choose any 4 pairs and carry out four multiplications with those numbers. Now, choose four other pairs and carry out divisions with these numbers.Examples: i. (-13) × (-15) = 195ii. (-24) ÷ 9 = (-24)/9 = -8/3

Write three divisions of integers such that the fractional form of each will be 24/5.

Write three divisions of integers such that the fractional form of each will be -5/7.

The fish in the pond below, carry some numbers. (Choose any 4 pairs and carry out four multiplications with those numbers. Now, choose four other pairs and carry out divisions with these numbers.)Examples: i. (-13) × (-15) = 195ii. (-24) ÷ 9 = -24/9 = -8/3

Write two prime numbers whose difference is 10.

Write seven consecutive composite numbers less than 100.

Write 10 odd and 10 even composite numbers.

If the difference of any two prime numbers is 2, then they are said to be A) Twin primesB) Co-primes C) Perfect numbers D) Exact numbers

Every prime number other than 2 is ……………… A) even B) composite C) odd D) A and B

The number which is divisible by 1 and itself is called ……………… A) Composite number B) Prime number C) Even number D) Odd number

1 and itself .

Smallest even prime number is ……………….. A) 4 B) 3 C) 2D) 5

Number 1 is …………….A) Prime B) CompositeC) Both A and B D) Neither A nor B

Complete the table given below.-3\(\frac{3}{5}\)-17\(\frac{-5}{11}\)5Natural Numbersx✓Integers✓Rational Numbers✓

L.C.M means ………………

G.C.D means ………………

Use the signs and numbers in the boxes and form an expression such that its value will be 112. [0, 1, 2, 3, 4, 5, 6, 7, 8, 9][+ x ÷ -]

Answer the following questions: 1. Write all the natural numbers between 2 and 9. 2. Write all the integers between -4, and 5. 3. Which rational numbers are there between 1/2 and 3/4 ?

Show that the lines \(\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1}\) and \(\frac{x}{1}=\frac{y}{2}=\frac{z}{3}\) are perpendicular to each other.

Show that the lines \(\frac{x-5}{7}=\frac{Ψ+2}{-5}=\frac{z}{1}\) and \(\frac{x}{1}=\frac{y}{2}=\frac{z}{3}\) are perpendicular to each other.

Write three rational numbers that lie between the two given numbers.i. 2/7, 6/7ii. 4/5, 2/3iii. -(2/3), 4/5iv. 7/9, -(5/9)

HCF means ……………….. A) Highest Common Factor B) Highest Continued FactorC) Hight of the Common Factor D) Highest composite factor

Find whether the following operations are binary operation or not (i) on N defined by a b = ab (ii) * on Z defined by a * b = ab

In the previous class, we have learnt to add and subtract integers. Using those methods, fill in the blanks below.1. 5 + 7 = 2. 10 + (-5) = 3. -4 + 3 = 4. (-7) + (-2) = 5. (+8) – (+ 3) = 6. (+8) – (-3) =