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allam vishwaja 4 years ago

help me for placements

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rohit ranjan 4 years ago

please give more details ?

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nandy ram 4 years ago

during a period of decline in a stock market prices a stock sold at ₹50/share on the first day,₹40/share on the second day and ₹25 on the third day.If a invester bought 100,120&180 shares on the respective 3 days the average price paid per share will be?.

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Devendra Marghade 4 years ago

Avg. Price = (100*50 + 120*40 + 180*25) / (100+120+180) = 35.75

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Karakavalasa sai manohar 4 years ago

Help me to get score in tcs nqt 2021

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TAVVA PAVAN KUMAR 4 years ago

On Feb 18 I have TCS NQT can some one help me friends

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Meghavath Madhu Naik 4 years ago

A man sets out by cycle from Delhi to Rohtak, and at the same time another man starts from Rohtak by cycle to Delhi. After passing each other they complete their journeys in 3 1/3and 4 4/5 hours respectively. At what rates does the second
man cycle if the first cycles at 8 km per hour?

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G.S.Kantharao 4 years ago

Let A and B are the persons.
A travelled x km,B travelled y km before they meet.
Time taken for A and B to reach their destinations are 10/3h,24/5h
Speeds are y/(10/3),x/(24/5)
At the time they meet they travelled same time => x/(3y/10)=y/5x/24)
10x/3y = 24y/5x=>50x^2=72y^2
=>x^2/y^2=72/50
x/y=√(1.44)=1.2
=>x=1.2y
Total distance =(x+y)=(x+x/1.2)=2.2x/1.2=11x/6
Now time taken for B=(11x/6)/(5x/24)
=11*24/6*5=44/5=8h 48m

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JACK KISSER 4 years ago

ORDER RESEARCH CHEMICALS FROM https://www.highgradeschemicals.com/

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Amanjot kaur Saroa 4 years ago

If the sum of the first 43 terms of an arithmetic progression is 2967, then find the sum of the 11th and the 33rd terms of the same series.

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Varun 4 years ago

First term = a, Number of terms=n, Difference = d
nth term = a+(n-1)d
11th term = a+10d, 33rd term = a+32d
Sum of 11th & 33rd= 2a+42d = 2(a+21d)=2(22nd term)

Middle term = Total sum/n
43/2 th term = 22nd term = 2967/43 = 69

=>Sum of 11th & 33rd = 2(69) = 138

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G.S.Kantharao 4 years ago

Sum of first 43 terms in A.P=43(2a+42d)/2=2967
11th term +33rd terms =(a+10d)+(a+32d)=(2a+42d)
43(2a+42d)/2=2967=>(2a+42d)=2967*2/43=138

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Amar Kumar Yadav 4 years ago

solution of quastion

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Renu 4 years ago

Keshav runs a factory of 210 employes in each 3 shift 80 emp are required no one is allowed to work more than 16 hrsin a day how many emp are required to work 16 hrs in a day

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Varun 4 years ago

80*3 = 240 employees needed
We have 210
So 240-210 = 30 employees need to work double shift
Cross check: 180*8+30*16=1920 hours
240*8 = 1920 hours
So 30 is the answer

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naurah 4 years ago

We know that the vertices of a quadrilateralare 2, 3, 5, and 6 cm, respectively,from a point P. What is the largest possible area of this quadrilateral?

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Varun 4 years ago

Lets divide the area into 4 triangles
Area of triangle = 1/2 * base * height
So total area for right angle triangles are:
0.5*(2*3+3*5+5*6+6*2) = 31.5cm^2
If we diminish or expand any of the portions we eventually fall into same bucket I guess, so I think 31.5cm^2 should be the answer, all suggestions are welcome. Thanks.

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PRATIBHA PRAKASH GAVANALE 4 years ago

give me apportunity

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Varun 4 years ago

What exactly apportunity means? Can you be more specific?

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DIVYADEEP 4 years ago

Twenty five sets of problems on Data Interpretation– one each for the DI sections of 25
CATALYST tests were prepared by the AMS research team. The DI section of each CATALYST
contained 50 questions of which exactly 35 questions were unique, i.e. they had not been used in
the DI section of any of the other 24 CATALYSTs. What could be the maximum possible number
of questions prepared for the DI sections of all the 25 CATALYSTs put together?
(Can anyone help by explaining the solution to this question)

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Varun 4 years ago

Let me try to understand the question:
25 set of papers
50 in each with 35 distinct questions
Did I understand the question correctly?

If yes, the answer is: 35*25 = 875 questions.

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Fatima Rahim 4 years ago

Why 0!=1 ?
Please help me

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Devendra Marghade 4 years ago

n! = n*(n-1)!
1! = 1*(1-1)!
1! = 1*0!
As 1! = 1, hence 0! = 1

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Billal Hossain 4 years ago

We know that n! = n(n-1)(n-2)(n-3).....3(2)(1) which means that
n! = n(n-1)!
Dividing both sides of the equation by n, we have
(n - 1)! = n!/n
Using this fact, we can check the following pattern.
4! = 5!/5 = {(5)(4)(3)(2)(1)}/5 = 24
3! = 4!/4 = {(4)(3)(2)(1)}/4 = 6
2! = 3!/3 = {(3)(2)(1)}/3 = 2
1! = 2!/2 = {(2)(1)}/2=1

Now, we go to 0!
0! = 1!/1 = 1

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Asif iqbal Chaudhry 4 years ago

Like and subscribe my Youtube channel
Asif Iqbal
Maths class 10th and 9th Lectures
https://youtu.be/s7qPmsPs5bA

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Neha PARMAR 4 years ago

Ten years ago,the average of family of four members was 24 years.Three children having been born,the average age of family is same today.What are the present ages of children if two children are identically twins and differ by two years from the younger one?

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AMIT GARG 4 years ago

24×4 =96 + 4×10 = 136
current age = 24×7 = 168
hence children age = 32
2A + A - 2 = 32
A = 34/3 = 11.33 not possible
data is wrong as child can not be older than 10 years

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ritik rathore 4 years ago

hi ,my self Ritik Rathore want to place in a core organisation of civil engineering and groom my self .

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