Numerical Estimation MCQ Quiz in मल्याळम - Objective Question with Answer for Numerical Estimation - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

• Condition 1: If the house number is a multiple of 3, then it is a number from 50 to 59.
  • Possibilities: 51, 54, 57
  • But if one of these is the right answer, then it should be a multiple of 4 and 6 which is not the case here.
• Condition 2: If the house number is NOT a multiple of 4, then it is a number from 60 to 69.
  • Possibilities: 61, 63, 65, 66, 67, 69
  • But if one of these is the right answer, it should be multiple of six (i.e. 66) and not a multiple of 3 (but 66 is multiple of 3). So it is not a required sample space.
• Condition 3: If the house number is NOT a multiple of 6, then it is a number from 70 to 79.
  • Possibilities: 70, 71, 73, 74, 75, 76, 77, 79
  • But if one of these is the right answer, it should be multiple of 4 (i.e. 76) and not a multiple of 3 (i.e. 70, 71, 73, 74, 76, 77, 79).

So 76 is the correct answer.

(x % of y) + (y % of x) 

\(= \frac{x}{{100}} \times y + \frac{y}{{100}} \times x\)

= 2xy/100

From the geometric series formula,

\(\mathop \sum \limits_{n = 0}^\infty a{r^n} = \frac{a}{{1 - r}}\)

Substitute a = 1,

\(\mathop \sum \limits_{n = 0}^\infty {r^n} = \frac{1}{{1 - r}}\)

On differentiating both sides with respect to r, we get

\(\mathop \sum \limits_{n = 0}^\infty n{r^{n - 1}} = \frac{1}{{{{\left( {1 - r} \right)}^2}}}\)

Substitute r = 1/3

\(\mathop \sum \limits_{n = 0}^\infty n{\left( {\frac{1}{3}} \right)^{n - 1}} = \frac{1}{{{{\left( {1 - \frac{1}{3}} \right)}^2}}} = \frac{9}{4}\)

\( \Rightarrow 1 + \frac{2}{3} + \frac{3}{9} + \frac{4}{{27}} + \frac{5}{{81}} + \ldots = \frac{9}{4} = 2.25\)

Concept:

Area of circle = \(\frac{π}{4}d^2\)

Area of square = a²

Perimeter of circle = πd

Perimeter of square = 4a

Calculation:

Given:

Area of square and circle is equal.

\(\frac{π}{4}d^2=a^2\)

\(\frac{d}{a}=\sqrt{\frac{4}{\pi}}\)

\(\frac{Perimeter\;of\;circle}{Perimeter\;of\;square}=\frac{\pi{d}}{4a}\Rightarrow\frac{\pi}{4}\sqrt{\frac{4}{\pi}}=\sqrt{\frac{\pi}{4}}\;\;(<1)\)

∴ perimeter of square is greater than the perimeter of circle.

Scenario:

We observe that each letter is written the times as the letter’s serial number in alphabets.

The series consists of alphabets in the following manner:

A (1^st in series), 2 times B (2^nd, 3^rd), 3 times C (4^th,5^th, 6^th), 4 times D (7^th, 8^th, 9^th, 10^th) and so on….…

Each letter is written the times as the letter’s serial number in alphabets. Also, If the serial number of letters in alphabets is n, after writing n times the number of letters in the series follows the pattern \(\frac{n(n+1)}{2}\).

Example: After writing last D which is 10^th number in the series and 4^th number in alphabets written 4 times, the pattern satisfies, i.e.

\(10=\frac{4(4+1)}{2}\)

Analysis:

Now, the 240^th number may be the end of a letter, or may be repetition of a letter at any position, but still, the following equality will hold true:

\(\frac{n(n+1)}{2}≤ 240\)

n (n + 1) ≤ 480

The maximum value of n for which this equality holds true is n = 21, i.e.

\(\frac{21(21+1)}{2} = 231\)

∴ The 21^st number in alphabets after writing 21 times will give 231^st number in the series which is U. V is 22^nd number in the alphabets which is written 22 times.

But since 240 - 231 = 9, after “U”, the 9^th letter in series will be “V” as “V” is written 22 times.

∴ The 240^th letter is “V”.

Let Rs. 100,000 is divided into two parts x and y respectively.

Part I: Scheme – 1 (10% profit)

After profit in Scheme – 1, he will get 1.10x rupees

Scheme – 2 (12% profit)

After profit in Scheme – 2, he will get 1.12y rupees

Total amount that he will get from Scheme 1 and Scheme 2 after applying profits of 10% and 12% respectively = 1.10x + 1.12y     …1)

Part II: When profit percentages are interchanged

Scheme – 1: (12% profit), Scheme – 2: (10% profit)

Total amount after applying these profit percentages = 1.12x + 1.10y     …2)

Given that,

1.10x + 1.12y – 1.12x – 1.10y = 120

0.02y – 0.02x = 120

2y – 2x = 12000

y – x = 6000     …3)

y + x = 100000     …4)

Solving 3) and 4)

2y = 106000

Y = 53000, x = 47000

90% of area of circular sheet = surface area of the cone (only the conical part)

⇒ 0.9 × π R² = π.r. l

where R radius of circular sheet, r is radius of cone, l is slant length of cone.

Given, R = 30 cm, l = R = 30 cm

⇒ 0.9 × π × 30 × 30 = π × r × 30

⇒ r = 27 cm 

now height of the cone h  \(= \sqrt {{{l}^2} - {{r}^2}}= \sqrt {{{30}^2} - {{27}^2}}\)

= 13.08 cm

Given:

PR = PS = QS = QT = RT 

Then ABCDEF will be regular pentagon

∠EAB = Sum of angles of pentagon/5

⇒ ∠EAB = 540°/5 = 108° 

⇒ ∠PAB = 180° – 108° = 72°

⇒ θ  + ∠PAB + ∠PBA = 180

⇒ θ  + 72 + 72 = 180°

⇒ θ  = 36°

Option (a) is correct.

Explanation:

Minutes gained in every 24 hours = 15 = 0.25 hour

⇒ every 1 hour, 1/96 hours are gained.

When the clock shows 2 PM on 15th July of the same year, let's assume x hours are spent in an error-free clock.

So for x hours, the gain will be x/96 = 0.0104x

So, the total time gone in the faulty clock will be x + 0.0104x = 1.0104x

The total time spent on the faulty clock is given by  

No. of hours between 9 AM on 11th July and 2 PM on 15th July = 24 × 4 + 5 = 101

Now, equating both

1.0104x = 101 ⇒ x = 99.96 hours

So the correct time spent is 99.96 hours = 4 days + 3 + 0.96 hours

⇒ 4 days + 3 hrs + 57.62 min

Calculating from 9 AM, 11th July, the correct time to the nearest minute = approximately 12.58 PM.

M has two times efficiency as that of E.

But E works 12 hrs per day and M works for 6 hrs per day i.e. half of working hrs by E.

So the work done per day by both M and E is same.

M worked half of the days that E worked for.