Explore topic-wise InterviewSolutions in .

The correct option is (a) 9.

The correct option is (b) 423.

The correct option is (d) 163.

If on multiplying a number with 100,0 takes unit or tens place and original number shifted from unit or tens place ahead on multiplying with 1000,0 takes unit, tens or hundredth place and original number shifted to ahead from unit, tens or hundredth place.

yes, 25 = 100/4 can be expressed.

(i) 50 x 5 

= 50 x 10/2 

= 50/ x 10

= 25 x 10 

= 250

(ii) 500 x 5 

= 500 x 10/2 

= 500/2 x 10

= 250 x 10 

= 2500

\(\frac{1}{8}\) x \(\frac{3}{5}\)

= \(\frac{1 \times 3}{8 \times 5}\)

= \(\frac{3}{40}\)

3 1/4 x 4

= 13/4 x 4

= 13 x 4/4

= 13 x 1

= 13

2 1/5 x 5

= 11/5 x 5

= 11 x 5/5

= 11 x 1

= 11

3 1/2 x 4

= 7/2 x 4

= 7 x 4/2

= 7 x 2

= 14

4 1/3 x 6

= 13/3 x 6

= 13 x 6/3

= 13 x 2

= 26

Base for number 7 = 10 

∴ Deviation = 10 – 7 = 3 

Base for number 93 = 100 

∴ Deviation = 100 – 93 = 7

(iv) In 46, eknunen poorven of 6 will be 36

1024

(i) Making pairs from right to left, first pair = 24

(ii) Extreme digit of first pair = 4

∴ Extreme digit in possible square root may be either 2 or 8.

(iii) Greatest square root within the maximum limit 10 = 3.

∴ Possible square root may be either 32 or 38.

Product of 3 and 8 = 3 x 8 = 24

(iv) ∵ 10 < 24

∴ Smallest number will be square root.

Hence square root = 32

Those two digits whose sum is 10, are called parammitra of each other. 

∴ Parammitra digit of 2 = 10 – 2 = 8

Eknunen poorven of digit 6 in 2675 2675 = 1675

169

∵ Unit place of 169 is 9

∴ 169 is a perfect square number

Sum of digits = 1 + 6 + 9 = 16

Extreme digit (charam ank) of square root will be either 1 or 3 and tens place digit will be 1.

Thus hence square root of 169 = 13

Ekadhiken poorven of digit 1 in 125 = 0125 = 1125

(i) Eknunen of 8 will be 7

324

(i) Unit place of 324 is 4. So it is a perfect square number.

(ii) Sum of digits = 3 + 2 + 4 = 9 which is perfect square.

(iii) This number will have two digits in square root.

(iv) Making pair of 2-2 in 324 from right to left, second pair has only one digit 3.

∴ Tens place in square root of 324 will be 1.

(v) ∵ Extreme digit of number 324 is 3.

∴ Extreme digit of square root will be either 1 or 8.

Hence, the square root of 324 is 18.

2025

(i) Making pair from right to left first pair is 25 and another is 20.

(ii) Extreme digit of first pair is 5 therefore extreme digit of its square root will be 5.

(iii) The greatest square root within the maximum limit of 20 is 4.

Therefore possible square root will be 45 Hence square root of 2025 = 45

576

(i) Unit place digit in 576 is 6. So the given number is a perfect square.

(ii) Sum of digits = 5 + 7 + 6 = 18 which is not perfect square.

(iii) Square root of this number will have 2 digits.

(iv) On making the pairs of 2 – 2 from right to left, second pair has only one digit 5, therefore tens place digit of square root will be 2.

(v) Extreme digits of number will be 2.

So the extreme digit of its square root will be either 2 or 8 and extreme digit for tens place is 5 which is in between 4 – 8, therefore tens place digits in square root will be 2.

∴ Square root of 576 may be either 24 or 28.
Perfect square root will be 24.

Hence square root = 24

(i) 3 bar 5

Hint : 

(a) Write the parammitra digit 5 of

= 3 bar 5 the positive value of bar 5 Le., 5. 

= \(\underline{3}\)5 

(b) Put one less sign on poorven of 

= 25 bar 5 i.e„ 3.

(c) Write \(\underline{3}\) = 1 

(d) Thus, we got general number of 3 bar 5 

(ii) 5 bar 4 

Hint : 

= 5 bar 4 (a) Write the parammitra digit 6 of the 

= \(\underline{5}\)6 positive value of i.e. bar 4. 

= 46 

(b) Put one less sign on poorven of bar 4 i.e., 5. 

(c) Write \(\underline{5}\) = 4

(d) Thus, we got general number of 5 bar 4 

(iii) 13 bar 2 

Hint :

13 bar 2 (a) Write the parammitra digit 8 of 

= 1\(\underline{3}\)8 the positive value of i.e., bar 2. 

= 128 

(b) Put one less sign on poorven of bar 2 i.e., 3.

(c) Write \(\underline {3}\) = 2

(d) Thus, we got general number of 13 bar 2. 

(iv) 5 bar 42

Hint : 

= 5 bar 42 (a) Write the parammitra digit of 

\(\underline {5}\)6 bar 2 positive value 4 of (at the tens place) i.e., 6. 

= 46 bar 2 

(b) Put one less sign on poorven of 

= 4\(\underline {6}\)8 bar 4 i.e., 5. 

= 458 

(c) Write \(​\underline {5}​\) = 4 

(d) Write the parammitra digit of positive value 2 of bar 2 (at the ones place) i.e., 8. 

(e) Put one less sign on poorven of bar 2 i.e., 6. 

(f) Write \(​\underline {6}\) = 5

(g) Thus, we got general number of 5 bar 42. 

(v) 6 bar 23 

Hints : 

= 6 bar 23 

(a) Write the parammitra digit of 

= \(​\underline {6}​\)8 bar 3 positive value 2 of bar 2 (at the tens place) i.e., 8. 

= 58 bar 3

= 5\(​\underline {8}\)7 (b) Pat one less sign on poorven of 

= 577 bar 2 i.e., 6. 

(c) Write \(​\underline {6}​\) = 5

(d) Write the parammitra digit of positive value of bar 3 of 3 (at the ones place), i.e., 7. 

(e) Put one less sign on poorven of bar 3 i.e., 8. 

(f) Write \(​\underline {8}​\) = 7

(g) Thus, we got general number of 6 bar 23.

(i) 8 

Hints : 

(a) Put vinkulum line on the parammitra digit of 8 i.e.,2 = bar 2 

(b) Put one more sign on poorven digit of 8 i.e., 0 = bar \(\dot{0}\)2

(c) Write \(\dot{0}\) = 1. 

(d) Thus, we got the vinkulum of 8 = bar 2

(ii) 27 

Hints : 

= bar 87 

(a) Digit 7 would be as it is and 

\(\dot{0}\) 8 bar 7 = 7 vinkulum line would be on param 

= 187 mitra digit of 2 i.e., 8. 

(b) One more sign on the poorven of 8 i.e., 0. 

(c) Write \(\dot{0}\) = 1, 

(d) Thus, we got vinkulum of 27 

(iii) 82 

Hint : 

= 2 bar 2 

(a) Digit 2 would be as it is and vinkulum line would be on 

= \(\dot{0}\) 2 bar 2 parammitra digit of 8 i.e., 2. 

= 1 2 bar 2 

(b) One more sign on the poorven digit of 8 ie., 0. 

(c) Write \(\dot{0}\) = 1 

(d) Thus, we get vinkulum of 82. 

(iv) 78 

Hint : 

= 3 bar 8 (a) Digit 8 would be as it is and vinkulum line would be on 

= \(\dot{0}\) 3 bar 8 parammitra digit of 7 i.e., 3. 

= 1 3 bar 8 

(b) One more sign on the poorven digit of 7 ie., 0

(c) Write \(\dot{0}\) = 1. 

(d) Thus, we got vinkulum of 78. 

(v) 96 

Hint : 

= 1 bar 6 (a) Digit 6 would be as it is and 

= \(\dot{0}\) 1 bar 6 vinkulum line would be on parammitra digit of 9 i.e., 1. 

= 1 1 bar 6 (b) One more sign on the poorven digit of 9 Le., 0. 

(c) Write \(\dot{0}\) = 1. 

(d) Thus, we got vinkulum of 96

(45 + 55) + (67 + 33) + (38 + 62)

= 100 + 100 + 100 

= 300

9025

(i) Making pairs from right to left, first pair is 25 and another pair is 90.

(ii) Extreme digit of first pair = 5 therefore extreme digit of possible square root will be 5.

(iii) Greatest square root within the maximum limit 90 is 9.

Hence square root of 9025 = 95

3025

(i) Making pair from right to left, first pair is 25 and another pair is 30.

(ii) Extreme digit of first pair is 5 therefore extreme digit of its square roots will be 5.

(iii) The greatest square root within the maximum limit of 30 is 5. Therefore possible square root will be 55.

Hence square root of 3025 = 55.

(ii) Sign of vinkulum is - (minus)

(iii) Ekadhiken number of 7 will be 8

(iv) In 46, ekadhiken poorven of 4 will be 146

(ii) In 289, ekadhiken of 2 will be 389

(i) one less 

(ii) 1 

(iii) negative 

(iv) 10, 10, 10, 

(v) vinkulum.

441

(i) Unit place digit of 441 is 1. Therefore 441 may be a perfect square number.

(ii) Sum of digit of 441 = 4 + 4 + 1 = 9 which is perfect square.

(iii) There may be two digits in square root of 441.

(iv) Making pair of 2 – 2 digits from right to left, second pair has only one digit 4. Therefore tens place digits in square root will be 2.

(v) Extreme digit of number is 1.

Therefore extreme digit in square root will be either 1 or 9 and for tens place is 2 which belongs a group 1 – 3 so the tens place will be 1.

(vi) Thus square root of 441 may be either 21 or 29.

(vii) On multiplying 2 with (2+1=3)

2 x 3 = 6

Here 4 digit of second pair < Product 6

∴ Smaller number out of 21 or 29 will be taken as square root

Hence square root of 441 = 21

(iii) Parammitra digit of 4 will be 7

On filling the blanks :

(ii) Meaning of ekadhiken is one more

17² = 17 + 7/7² 

Base = 10, deviation = +7 

= 24/49 = 289

(i) 93

93² = 9(93 + 3)/(3)²

= 9(96)/9 Base = 10

= 864/9 Sub-base = 10 × 9

8649 deviations = + 3

(ii) 206

206² = 2(206 + 06)/(06)²

= 2(212)/36   Base = 100

= 424/36   Sub-base = 100 × 2

= 42436   deviation = +06

(iii) 211

211² =2(211 + 11)/(11)²

= 2(222)₁21 Base = 100

= 444/₁21 Sub-base = 100 x 2

= 44521 deviation = +11

(iv) 405

405² = 4(405 + 05)/(05)²

= 4(410) / 25 Base = 100

= 1640 / 25 Sub-base = 100 × 4

= 164025 deviation = +05

(i) 45

45² = 4 × 5 / 5 × 5 

= 20 / 25 = 2025

(ii) 85

85² = 8 × 9 / 5 × 5 

= 72 / 25 = 7225

(iii) 115

115² = 11 × 12 / 5 × 5 

= 132 / 25 = 13225

(iv) 125

125² = 12 × 13 / 5 × 5 

= 156 / 25 = 15625

(i) 23

23² = (23 + 3)(23 – 3) + 3²

= 26 × 20 + 3²

= 520 + 9

= 529

(ii) 38

38² = (38 + 8)(38 – 8) + (8)²

= (46 × 30) + 64

= 1380 + 64

= 1444

(iii) 69

69² = (69 + 9)(69 – 9) + 9²

= (78)(60) + 81

= 4680 + 81

= 4761

(iv) 89

89² = (89 + 9)(89 – 9) + 9²

= (98)(80) + 81

= 7840 + 81

= 7921

103 × 197 = 1 × 2/03 × 97 

= 2/0291 = 20291

31(1/6) x 31(5/6)

=  31 x 32(1/6 x 5/6)

= 992(5/36)

\(= 992\frac{5}{36}\)

(i) 54 × 56

54 × 56 = 54 + 4 + 56 + 6 

= 5(54 + 6)/4 × 6 = 5 × 60/24 

= 300/24 = 3024

(ii) 108 × 112

108 × 112 = 108 + 08 × 112 + 12 

= 1(108 + 12)/08 × 12 

= 1 × 120/90 = 12096

(iii) 137 × 9999

L.H.S = 0137 – 1 = 0136 

R.H.S. = 9999 – 0136 

= 9863 137 × 9999 

= 0137 – 1/9999 – 0136 

= 1369863

Both the numerators of fractions due same = m. 

By formula 2x + 1 + 3x + 4 = 0 

or 5x + 5 = 0 

or x = – 1

Constant terms in both sides are same, 

so, x = 0

Sum of numerators of both sides 

= 3x + 4 + x + 1 = 4x + 5 …(i) 

Sum of denominators of both sides 

= 6x + 7 + 2x + 3 = 8x + 10 …(ii) 

Ratio of (i) by (ii) = 1 : 2 

According to question any sum equation to zero.

From 4x + 5 = 0 

or 8x + 10 = 0

⇒ x = -5/4

Sum of the denominator of both sides are equal 

= 2x – 17 

According to formula 2x – 17 = 0 

⇒ x = 17/2 = 8(1/2)

3 groups makes when two digit number is multiplied by two digit number.

1. (5x + 7)/(2x + 1) = (x + 1)/(3x + 5)

Difference between numerator and denominator 

= 5x + 7 – 2x – 1 

= 3x + 6 = 3(x + 2) …(i) 

Difference between numerator and denominator is

R.H.S. = 3x + 5 – x – 1 

= 2x + 4 = 2(x + 2) …(ii) 

Ratio of (i) and (ii) is 3 : 2 

Thus, by formula 3(x + 2) = 0 

⇒ x + 2 = 0 

⇒ x = -2

2. (3x + 6)/(6x + 3) = (5x + 4)/(2x + 7)

Difference between numerator and denominator in L.H.S. 

= 6x + 3 – 3x – 6 = 3x – 3 …(i) 

Difference numerator and denominator is L.H.S. 

= 5x + 4 – 2x – 7 = 3x – 3 …(ii) 

Difference numerator and denominator in R.H.S. 3x – 3 = 0 

⇒ 3x = 3 

⇒ x = 1 

Thus, x = 1

(i) and (ii) are equal, so by formula 

3x + 6 + 5x + 4 = 8x + 10 …(i) 

Adding denominators of both sides 

6x + 3 + 2x + 7 = 8x + 10 …(ii) 

(i) and (ii) are equal so by formula. 

8x + 10 = 0 

⇒ 8x = – 10 

⇒ x = -10/8 = -5/4

thus, x = -5/4 and x = 1

3. 1/(x + 2) + 1/(x + 6) = 1/(x + 1) + 1/(x + 7)

Sum of denominators of L.H.S. 

= x + 2 + x + 6 

= 2x + 8 …(i) 

Sum of denominators of R.H.S. 

= x + 1 + x + 7 

= 2x + 8 …(ii) 

(i) and (ii) are equal so by formula 

2x + 8 = 0 

⇒ 2x = -8 

⇒ x = -4

4. 1/(x - 4) + 1/(x - 6) = 1/(x - 2) + 1/(x - 8)

Sum of denominators of L.H.S. 

= x – 4 + x – 6 

= 2x – 10 …(i) 

Sum of denominators of R.H.S. 

= x – 2 + x – 8 

= 2x – 10 …(ii) 

(i) and (ii) are same so by formula 

2x – 10 = 0 

⇒ 2x = 10 

⇒ x = 5

1. (2x + 1) + (x + 3) = 5x + 4

(2x + 1) + (x + 3) = 5x + 4 

independent term of both sides are same (= 4) 

So, x = 0.

2. a(x – 1) + b(x – 1) = c(x – 1) + d(x – 1)

Here 

a(x – 1) + b(x – 1) = c(x – 1) + d(x – 1) 

Thus, (x – 1) is common factor is both sides, then x – 1 = 0 

⇒ x = 1

3. (x + 1)(x + 9) = (x + 3)(x + 3)

(x + 1)(x + 9) = (x + 3)(x + 3) 

Here independent term in both sides are same (= 9) then x = 0

4. x/2 + x/3 = x/4 + x/1.

Numerator (x) is same in both sides 

Sum of denominators in R.H.S. = 2 + 3 = 5 

Sum of denominators in R.H.S. = 4 + 1 = 5 

Thus, According to formula, x = 0