Ask questions which are clear, concise and easy to understand.
Ask QuestionPosted by Manish Garg 7 years, 6 months ago
- 3 answers
Posted by Harsh Vardhan Singh 7 years, 6 months ago
- 1 answers
Posted by Charanjeet Singh 7 years, 6 months ago
- 1 answers
Ammu Ammu 7 years, 6 months ago
Posted by Sumanpreet Kaur 7 years, 6 months ago
- 0 answers
Posted by Mann Jain 7 years, 6 months ago
- 4 answers
Posted by Jyoti Prada Parida 5 years, 8 months ago
- 0 answers
Posted by Areeb Khan 6 years, 6 months ago
- 1 answers
Sia ? 6 years, 6 months ago
Given 2x + 3y = 7 and 2ax + {tex}( \alpha + \beta ){/tex}y = 28.
We know that the condition for a pair of linear equations to be consistent and having infinite number of solutions is
{tex}\frac { a _ { 1 } } { a _ { 2 } } = \frac { b _ { 1 } } { b _ { 2 } } = \frac { c _ { 1 } } { c _ { 2 } }{/tex}
{tex}\frac { 2 } { 2 a } = \frac { 3 } { a + \beta } = \frac { 7 } { 28 }{/tex}
{tex}\frac { 2 } { 2 \alpha } = \frac { 7 } { 28 }{/tex}
{tex}= 2 \alpha \times 7 = 28 \times 2{/tex}
{tex}\alpha = 4{/tex}
{tex}\frac { 3 } { \alpha + \beta } = \frac { 7 } { 28 }{/tex}
{tex}= 7 ( \alpha + \beta ) = 28 \times 3{/tex}
or, {tex}a + \beta = 12{/tex}
or, {tex}\beta = 12 - \alpha{/tex}
or, {tex}\beta = 12 - 4{/tex}
or, {tex}\beta = 8{/tex}
Posted by Palak Chawla 7 years, 6 months ago
- 1 answers
Posted by Mahendra Sahu 6 years, 6 months ago
- 1 answers
Sia ? 6 years, 6 months ago
- 867 = 255 {tex}\times{/tex} 3 + 102
255 = 102 {tex}\times{/tex} 2 + 51
102 = 51 {tex}\times{/tex} 2 + 0
{tex}\Rightarrow{/tex} HCF = 51 - 616 = 32{tex}\times{/tex}19 + 8
32 = 8 {tex}\times{/tex}4 + 0
{tex}\Rightarrow{/tex}HCF = 8
Posted by Mahendra Sahu 6 years, 6 months ago
- 1 answers
Sia ? 6 years, 6 months ago
The LCM of 28 and 32
28 = 2× 2 × 7=22×7
32 = 2 × 2 × 2 × 2 × 2=25
LCM = 25 × 7 = 224
Smallest no: which leaves remainder 8 and 12 when divided by 28 and 32
= LCM of 8 & 12 = 224 - 20 = 204
Therefore, 204 is smallest number which when divided by 28 and 32 leaves remainder 8 and 12 respectively.
Posted by Mahendra Sahu 6 years, 6 months ago
- 1 answers
Sia ? 6 years, 6 months ago
We have to find the greatest number that divides 445, 572 and 699 and leaves remainders of 4, 5 and 6 respectively. This means when the number divides 445, 572 and 699 leaves remainders 4, 5 and 6 is that
445 - 4 = 441, 572 - 5 = 567 and 699 - 6 = 693 are completely divisible by the required number.For the highest number which divides the above numbers can be calculated by HCF .
Therefore, the required number is the H.C.F. of 441, 567 and 693 Respectively.
First, consider 441 and 567.
By applying Euclid’s division lemma, we get
567 = 441 {tex}\times{/tex} 1 + 126
441 = 126 {tex}\times{/tex} 3 + 63
126 = 63 {tex}\times{/tex} 2 + 0.
Therefore, H.C.F. of 441 and 567 = 63
Now, consider 63 and 693
again we have to apply Euclid’s division lemma, we get
693 = 63 {tex}\times{/tex} 11 + 0.
Therefore, H.C.F. of 441, 567 and 693 is 63
Hence, the required number is 63. 63 is the highest number which divides 445,572 and 699 will leave 4,5 and 6 as remainder respectively.
Posted by Nishant Shishodia 7 years, 6 months ago
- 1 answers
Posted by Shatrughan Sinha 6 years, 6 months ago
- 1 answers
Sia ? 6 years, 6 months ago
Let the radius of the hemispherical dome be r and the total height of the building be h.
Since, the internal diameter of the dome is equal to its total height
2r = h
{tex}\Rightarrow r = \frac { h } { 2 }{/tex}
Let H be the height of the cylindrical position.
{tex}\Rightarrow H = h - r = h - \frac { h } { 2 } = \frac { h } { 2 }{/tex}
Volume of air inside the building = Volume of air inside the dome + Volume of air inside the cylinder
{tex}\Rightarrow 41 \frac { 19 } { 21 } = \frac { 2 } { 3 } \pi r ^ { 3 } + \pi r ^ { 2 } H{/tex}
{tex}\Rightarrow \frac { 880 } { 21 } = \pi r ^ { 2 } \left( \frac { 2 } { 3 } r + H \right){/tex}
{tex}\Rightarrow \frac { 880 } { 21 } = \frac { 22 } { 7 } \times \left( \frac { h } { 2 } \right) ^ { 2 } \left( \frac { 2 } { 3 } \times \frac { h } { 2 } + \frac { h } { 2 } \right){/tex}
{tex}\Rightarrow \frac { 880 \times 7 } { 22 \times 21 } = \frac { h ^ { 2 } } { 4 } \times \left( \frac { h } { 3 } + \frac { h } { 2 } \right){/tex}
{tex}\Rightarrow \frac { 40 \times 4 } { 3 } = h ^ { 2 } \times \left( \frac { 5 h } { 6 } \right){/tex}
{tex}\Rightarrow \frac { 40 \times 4 \times 6 } { 3 \times 5 } = h ^ { 3 }{/tex}
{tex}\Rightarrow h ^ { 3 } = 64{/tex}
{tex}\Rightarrow h = 4{/tex}
Thus, the height of the building is 4 m.
Posted by Shatrughan Sinha 7 years, 6 months ago
- 5 answers
Kannu Kranti Yadav 7 years, 6 months ago
Kannu Kranti Yadav 7 years, 6 months ago
Kannu Kranti Yadav 7 years, 6 months ago
Posted by Kunal Rajput 7 years, 6 months ago
- 1 answers
Sia ? 6 years, 6 months ago
Check formulae in notes : <a href="https://mycbseguide.com/cbse-revision-notes.html">https://mycbseguide.com/cbse-revision-notes.html</a>
Posted by Alankrita Pathak 7 years, 6 months ago
- 2 answers
Yuv Raj 7 years, 6 months ago
Yuv Raj 7 years, 6 months ago
Posted by Amit Shivhare 7 years, 6 months ago
- 4 answers
Posted by Nishi K 7 years, 6 months ago
- 3 answers
Posted by Nishi K 7 years, 6 months ago
- 1 answers
Posted by Ananya Shrivastava 7 years, 6 months ago
- 1 answers
Lucky Arya 7 years, 6 months ago
Posted by Rishav Shrivastava 7 years, 6 months ago
- 1 answers
Sia ? 6 years, 6 months ago
Check CBSE updates here : <a href="https://mycbseguide.com/blog/">https://mycbseguide.com/blog/</a>
Posted by Piyush Mandloi 7 years, 6 months ago
- 0 answers
Posted by Satwika Gautam 7 years, 6 months ago
- 2 answers
Posted by Priya Bharti 7 years, 6 months ago
- 1 answers
Posted by Aarti Makkar 5 years, 8 months ago
- 0 answers
Posted by Aarti Makkar 7 years, 6 months ago
- 0 answers
Posted by Rahul Kumar 6 years, 6 months ago
- 1 answers
Sia ? 6 years, 6 months ago
Since, there are 60 minutes gap between 2 PM & 3 PM.
Time needed by minutes hand after t minutes past 2 PM to show 3 PM = (60 - t) minutes
According to the question ;
60 - t = {tex}\frac{t^2}{4}{/tex} - 3
{tex}\Rightarrow{/tex} 63 = {tex}\frac{t^2}{4}{/tex} + t
{tex}\Rightarrow{/tex} 63 = {tex}\frac{t^2 + 4t}{4}{/tex}
{tex}\Rightarrow{/tex} 252 = t2 + 4t
{tex}\Rightarrow{/tex} t2 + 4t - 252 = 0
{tex}\Rightarrow{/tex} t2 + 18t - 14t - 252 = 0
{tex}\Rightarrow{/tex} t(t + 18) - 14(t + 8) = 0
{tex}\Rightarrow{/tex} (t + 18)(t - 14) = 0
{tex}\Rightarrow{/tex} t + 18 = 0 or t - 14 = 0
{tex}\Rightarrow{/tex} t = -18 or t = 14
Since time cannot be negative, t {tex}\neq{/tex} -18
Hence, t = 14 minutes.
myCBSEguide
Trusted by 1 Crore+ Students
Test Generator
Create papers online. It's FREE.
CUET Mock Tests
75,000+ questions to practice only on myCBSEguide app
Kartik Tiwari 7 years, 6 months ago
1Thank You