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(c): A + B + C = 12540 .… (1)

$A=\frac{3}{7}(B+C)$

$\frac{A}{B+C}=\frac{3}{7}$

So, A’s Share = $\frac{3}{10}\times 12540=3762$ $\frac{}{}$

$B=\frac{2}{9}(A+C)$

$\frac{B}{A+C}=\frac{2}{9}$

So, B’s Share = $\frac{2}{11}\times 12540=2280$

To secure admission to FIEM for the BTech course, students must submit multiple documents for the applied course. Apart from this, candidates can find the list of basic documents to be kept ready while applying for the course:

• ID proof
• Address proof
• Self-attested copies of Class 10 and Class 12. 
• Two passport-size photos
• Supporting documents for the Reserved category

The BITSAT 2025 entrance exam is meant to take place in the month of May. The exact date of the BITSAT 2025 entrance exam lies somewhere between 26th May and 30th May 2025. The admit cards for the exam will be issued somewhere around 23rd May 2025. Candidates can visit the official website of BITS Pilani Goa to learnmore about the BITSAT 2025 entrance exam timeline.

(c):Ratio of efficiency of P & Q is 3 : 2.

Ratio of efficiency of Q & R is 5 : 4.

Ratio of efficiency of P, Q & R is 15 : 10 : 8

Total work done = 15 * 11 = 165

Time require by P, Q, R = 165/33 = 5 days

(c):Let x be the percentage of water in the resultant mixture.

Using the alligation rule, we have

$\frac{30}{25}=\frac{(40-x)}{(x-20)}$

6 (x – 20) = 5 (40 – x)

6x – 120 = 200 – 5x

11x = 320

x = $\frac{320}{11}=29.09\mathrm{\%}$

The marking distribution of the GATE Data Science question paper is as follows -

• General aptitude - 15 marks
• Subject-based questions - 85 marks
• Total marks allotted - 100 marks

The GATE DA question paper marking scheme is as follows -

• Questions are either of 1 or 2 marks
• For 1-mark MCQ, 1/3 mark will be deducted for an incorrect response.
• For 2- mark MCQ, 2/3rd mark will be deducted for an incorrect response. 
• There is no negative marking for MSQs and NATs. 

(c):Let the individual price of jeans be X and Y.

X + Y = 560.

Using the profit and loss percentages of the individual jeans:

1.15X + 0.9Y = 560.

Solving the two equations we get,

X = 224 and Y = 336

Suppose the position-time graph is a straight line, in this case, the velocity is constant. This means that there is no acceleration.
If the graph is curved, velocity is changing, which means that there is acceleration. If the graph is concave, the slopes will get more positive with time. This means that there is positive acceleration. If the graph is cap-shaped, the slope will become more negative with time. This is known as negative acceleration.

(b):Vande Bharat takes 6 hours from New Delhi to Bhopal. Duronto takes 4 hours from Bhopal to New Delhi. Let the distance between Delhi and Bhopal be ‘d’ kms.

Speed of Vande Bharat = $\frac{d}{6}km/hr$ $$

Speed of Duronto = $\frac{d}{4}km/hr$ $$

They will meet after = $\frac{d}{\frac{d}{6}+\frac{d}{4}}=2\frac{2}{5}hours$ $$

= 2 hours & 24 minutes

Hence, required time = (7 + 2.24)

= 9 : 24 a.m.

Factorised from = 2⁴ × 3⁴ × 5² × 7²

Total no. of factor = 5× 5 × 3 × 5

= 225 factors

113^rd factor = $\sqrt[]{N}$

 i.e. 7 × 5 × 3² × 2² = 1260

150/- wax; 8.5/- thread; 12.5/- coloured paper; 9/- silver powder

Profit = 40%.

Net input = 150 + 8.5 + 12.5 + 9 = 180/-

Net profit = 180 * .4 = 72/-

$$ $\frac{1}{2}$  the candles = $\frac{}{}$ $\frac{1}{2}$   (180) = 90/-

At 50% profit = 45.

90 + 45 = 135/-

Let the remaining $\frac{1}{2}$ $\frac{}{}$ = x

Net price = (72 + 180) = 135 + x

x = 117

Profit on second $\frac{1}{2}$ $\frac{}{}$ = 117 – 90 = 27

$\frac{}{}$ $\frac{27}{90}*100$  = 30%

No, To gain admission into BITS Pilani Goa, candidates must appear for the BITSAT exam. The BITSAT exam is the entrance exam for BE admissions into BITS Pilani. The admission of a candidate depends on the score they obtain in the BITSAT exam. Candidates can register for the BITSAT exam from the end of January till the end of April. The BITSAT exam is scheduled in the month of May.

P (1.05)2 − P = 369 or P = 3600

The exam duration of the GATE exam is three hours. The language of the GATE question paper is English, comprising General Aptitude (GA) and the candidate's selected subject. 

The GATE question format are as follows -

• Multiple Choice Questions (MCQs)
• Multiple Select Questions (MSQs)
• Numerical Answer Type (NATs)

Divine College of Pharmacy placements 2025 will be updated here once out. The college is visited by top companies for placements such as Alkem, Novartis, and others. The below-mentioned are the general placement statistics:

The mode of Divine College of Pharmacy course admissions is online. Admission to various courses is based on merit. Candidates willing to get admission must take part in the Divine College of Pharmacy admission process before the last date of application. Candidates must know their Divine College of Pharmacy eligibility criteria and apply accordingly. Apply as soon as the admission window opens.

ECE International is an association which is specifically dedicated towards international students. It aims to create a welcoming atmosphere for international students to help them easily settle in and promote cultural diversity.  The association organises various events including social, cultural and educational. Students are also provided assistance in a hope to provide them a comfortable stay during their study period.