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The fee structure for the Medicaps University BCom Hons course is a composition of various components such as tuition fees, examination fees, hostel fees, etc. As per the structure, the total tuition fee to pursue this course ranges from INR 2.1 Lacs to INR 3.75 lakh.

Note: The above-mentioned fee is as per the official sources. However, it is indicative and subject to change.

The solid sphere is melted and recast into cones. The volume of material is the same before and after casting.

Volume of the sphere = $\left(\frac{4}{3}\right)\pi r^3$ . (1)

Volume of the right circular cone = $\frac{1}{3}\pi R^{2}H$ . (2)

Diameter of the base of the cone

= $\sqrt[]{2}$ slant height

= $\sqrt[]{2}$ S (say). (3)

4R = 2S = 2 (H² + R²)

2R² = 2H

R = H. (4)

Volume of the cone = $\frac{1}{3}\pi R^{2}H$

$=\frac{1}{3}\pi r^2.(r)$  (since, R = r)

$=\frac{1}{3}\pi r^3$

From equations (1) and (4), it can be seen that the melted material creates exactly 4 cones of the specified dimensions. No material is left over.

(6x² + 43x – 49) – (5x² + 2x – 7) = 0

x² + 41x – 42 = 0

x² + 42x – x – 42 = 0

x (x + 42) – 1 (x + 42) = 0

(x – 1) (x + 42) = 0

x = 1 satisfy both the equations. So, there is only one common root.

Q can be 0 or 5, but 65 is not divisible by 4.

So, Q must be 0.

Now, sum of the digit should be divisible by 9.

So, P is 8.

Let y gold coins were given to 1st son.

Number of gold coins given to 2nd son = $\frac{5}{7}y$

Number of gold coins given to 3rd son = $\frac{3}{7}y$

So,  $y-\frac{3}{7}y=60\Rightarrow \frac{4y}{7}=60$

$\Rightarrow y=\frac{60\times 7}{4}=15\times 7=105$

Hence, the required number of gold coins:

$=105+\frac{5}{7}\times 105+\frac{3}{7}\times 105$

= 105 + 75 + 45 = 225

The sum of three numbers is given; the product will be maximum if the numbers are equal.

So, if a + b + c is defined, abc will be maximum when all three terms are equal. In this instance, however, with a, b, c being distinct integers, they cannot all be equal. So, we need to look at a, b, c to be as close to each other as possible.

a = 10, b = 10. c = 11 is one possibility, but a, b, c have to be distinct. So, this can be ruled out.

The close options are,

a, b, c = 9, 10, 12; product = 1080

a, b, c = 8, 11, 12; product = 1056

Maximum product = 1080

Coefficient of x49 = − Sum of the roots of f (x)

= − (1 + 3 + 7 +9+ . + 99)

− (50)²= −2500

(15, 3)! = Product of 3 consecutive natural numbers starting from 15, which is at least divisible by 3!

Hence, H = 3!

Consider the vowels A, E as one unit. Besides this, we have 2R, 1N, 1G that is a total of 5 things out of which the 2 R are identical.

Required number of ways $=\frac{5!}{2!}$

Now A, A and E can be arranged within themselves in ways. $\frac{3!}{2!}$

Required answer = $\frac{5!\times 3!}{2!\times 2!}$

 $\frac{lo{g}_{y}8x}{lo{g}_{y}x}=\frac{4}{3}$

Log_x 8x = $\frac{4}{3}$

Log_x 8 + log_xx = $\frac{4}{3}$

Log_x 8 = $\frac{1}{3}$

x = 8³

x = 512

Option (b) and (c) both satisfies the given condition, but 997918 is the larger of the two.

Since x > 0 then at x = 1, we get y = 2 and for the rest values we get y > 2

Also for 0 < x < 1 ; x+ $\frac{1}{x}$ > 2

and for x > 1 ; x+ $\frac{1}{x}$ > 2

Hence, y ≥ 2

We know that,

$a^2+a^2={\left(4\sqrt[]{2}\right)}^2,a^2=16,a=4$

Now, 2 (area of square) = (square of diagonal)

So, for new area 32 (16 × 2)

2 × 32 = d²

8 = d

Final ratio of wages = 5 × 4 : 7 × 3 : 9 × 2 = 20 : 21 : 18

Total wages received by P = $\frac{20}{59}\times 5900=2000$

Daily wage of P = $\frac{2000}{5}=400$

Let h = height of the container,

Given, h = 2r

$\therefore$ Volume v = πr²h = πr² (2r) = 2πr³

$\therefore$ Absolute error = (1.02)³ × 2πr³ − 2πr³ × 13

= 2πr³ [ (1.02)² − 1] = 2πr³ [1.0612 − 1]

= 0.0612 × 2πr³

$\therefore$ % error $=\frac{0.0612\times 2\pi r^3}{2\pi r^3}\times 100\mathrm{\%}=6.12\mathrm{\%}$