332354 + Interview Questions in Secondary School in Math Page 77 InterviewSolution

3801.	Reprsent 2/3,-4/3,6/3 on number line
Answer» Answer: Given numbers are 5 2 , 5 3 , 5 8 , 5 4 LET's draw number line fro 0 to 2. REF.Image So, the denominator is 5. We divide 0 & 2 into 5 EQUAL parts.

3801.

Reprsent 2/3,-4/3,6/3 on number line

Answer»

Answer:

Given numbers are

5

2

,

5

3

,

5

8

,

5

4

LET's draw number line fro 0 to 2.

REF.Image

So, the denominator is 5.

We divide 0 & 2 into 5 EQUAL parts.

Discussion

3802.	If length of rectangle is (x+ 2)cm and breath is (x+1)cm , find the area of rectangle
Answer» HOPE it HELPS you KINDLY MARK me the BRAINLIEST!

3802.

If length of rectangle is (x+ 2)cm and breath is (x+1)cm , find the area of rectangle

Answer»

HOPE it HELPS you

KINDLY MARK me the BRAINLIEST!

Discussion

3803.	7log base2 (16/15)+5logbase 2 (25/24)
Answer» Answer: GIVEN, → $7 log_{2}( \frac{16}{15} ) + 5 log_{2}( \frac{25}{24} )$ = $log_{2}(( { \frac{16}{15}) }^{7} ) + log_{2}(( { \frac{25}{24}) }^{5} )$ = $log_{2}( ( { \frac{16}{15} )}^{7} \times ( ({ \frac{25}{24}) }^{5} )$ = $log_{2}( \frac{{2}^{28} \times {5}^{10} }{ {2}^{15} \times {<klux>3</klux>}^{12} \times {5}^{7} } )$ = $log_{2}( \frac{ {2}^{13} \times {5}^{3} }{ {3}^{12} } )$ = $13 log_{2}(2) + 3 log_{2}(5) - 12 log_{2}(3)$ = $13 + 3 log_{2}(5) - 12 log_{2}(3)$ FORMULAE used:- 1) LOG(ab) = log(a) + log(B) 2) log(a/b) = log(a) - log(b) 3) log(a^(m)) = mlog(a)

3803.

7log base2 (16/15)+5logbase 2 (25/24)

Answer»

Answer:

GIVEN,

→ $7 log_{2}( \frac{16}{15} ) + 5 log_{2}( \frac{25}{24} )$

= $log_{2}(( { \frac{16}{15}) }^{7} ) + log_{2}(( { \frac{25}{24}) }^{5} )$

= $log_{2}( ( { \frac{16}{15} )}^{7} \times ( ({ \frac{25}{24}) }^{5} )$

= $log_{2}( \frac{{2}^{28} \times {5}^{10} }{ {2}^{15} \times {<klux>3</klux>}^{12} \times {5}^{7} } )$

= $log_{2}( \frac{ {2}^{13} \times {5}^{3} }{ {3}^{12} } )$

= $13 log_{2}(2) + 3 log_{2}(5) - 12 log_{2}(3)$

= $13 + 3 log_{2}(5) - 12 log_{2}(3)$

FORMULAE used:-

1) LOG(ab) = log(a) + log(B)

2) log(a/b) = log(a) - log(b)

3) log(a^(m)) = mlog(a)

Discussion

3804.	What will be the volume of a pyramids, at the right whose left is 62m,40m width 50m height?
Answer» Answer: Pyramid Formulas derived in terms of SIDE length a and height h: Volume of a square pyramid: V = (1/3)a2h. Step-by-step explanation: hope it was HELP full mark as BRAINLIEST answer

3804.

What will be the volume of a pyramids, at the right whose left is 62m,40m width 50m height?

Answer»

Answer:

Pyramid Formulas derived in terms of SIDE length a and height h: Volume of a square pyramid: V = (1/3)a2h.

Step-by-step explanation:

hope it was HELP full

mark as BRAINLIEST answer

Discussion

3805.	If:b=2:3and b:c= 2:3 , then : b:c =_____
Answer» Step-by-step EXPLANATION: a : b : c 2 3 ³ ² 2 3 ----------------------------------- 4 6 9

3805.

If:b=2:3and b:c= 2:3 , then : b:c =_____

Answer»

Step-by-step EXPLANATION:

a : b : c

2 3 ³

² 2 3

-----------------------------------

4 6 9

Discussion

3806.	Answer this with step by step explanation
Answer» Step-by-step explanation: HEAR is your answer in ATTACHMENT PLEASE give me some THANKS

3806.

Answer this with step by step explanation

Answer»

Step-by-step explanation:

HEAR is your answer in ATTACHMENT PLEASE give me some THANKS

Discussion

3807.	Find the acute angle o satisfying 5 tan² tita+ 3 = 9 sec titaPlease answer as soon as possible
Answer» ANSWER: there is UR answer HOPE it will HELP U

3807.

Find the acute angle o satisfying 5 tan² tita+ 3 = 9 sec titaPlease answer as soon as possible

Answer»

ANSWER:

there is UR answer HOPE it will HELP U

Discussion

3808.	10 The scores of 30 students in a class test are given below. Make a frequency table and answer the following questions? 18, 15. 19, 17, 14, 12, 20, 13, 10, 11 11, 19, 12, 20, 13 16, 15, 20, 14, 11, 10 20, 14, 16, 13, 15, 18 15, 15, a) How many children scored 20?
Answer» Answer: 3 Step-by-step EXPLANATION: only 3 STUDENTS have 20 MARKS in the test

3808.

10 The scores of 30 students in a class test are given below. Make a frequency table and answer the following questions? 18, 15. 19, 17, 14, 12, 20, 13, 10, 11 11, 19, 12, 20, 13 16, 15, 20, 14, 11, 10 20, 14, 16, 13, 15, 18 15, 15, a) How many children scored 20?

Answer»

Answer:

3

Step-by-step EXPLANATION:

only 3 STUDENTS have 20 MARKS in the test

Discussion

3809.	For a given circle a cente O and radius OA, if two arcs B and C are drawn on the circumference of the circle from 1 by taking the mecicunement of the radius of the are, what will be the shape of the of the triangle obtained by joining points 1, ОА as B and C.
Answer» ANSWER: OA is a IMPORTANT THING in ENGLISH

3809.

For a given circle a cente O and radius OA, if two arcs B and C are drawn on the circumference of the circle from 1 by taking the mecicunement of the radius of the are, what will be the shape of the of the triangle obtained by joining points 1, ОА as B and C.

Answer»

ANSWER:

OA is a IMPORTANT THING in ENGLISH

Discussion

3810.	Find the smallest five digit number which when divisible by 15 20 30 and 35 leaves remainder 5 in each case
Answer» Answer: So,LCM(15,20,35)=2^2 × 3×5×7 =420. HENCE,REQUIRED number =420+5 =425 Ans .

3810.

Find the smallest five digit number which when divisible by 15 20 30 and 35 leaves remainder 5 in each case

Answer»

Answer:

So,LCM(15,20,35)=2^2 × 3×5×7 =420. HENCE,REQUIRED number =420+5 =425 Ans .

Discussion

3811.	What is the value of x^91826789Where X=9032
Answer» ANSWER: didn't UNDERSTAND your QUESTION ❓❓❓❓❓

3811.

What is the value of x^91826789Where X=9032

Answer»

ANSWER:

didn't UNDERSTAND your QUESTION ❓❓❓❓❓

Discussion

3812.	Round off to the nearest 5
Answer» ANSWER: 5 can be ROUNDED off 10 Hope the answer HELPS you XD

3812.

Round off to the nearest 5

Answer»

ANSWER:

5 can be ROUNDED off 10

Hope the answer HELPS you XD

Discussion

3813.	एक प्रतल व एक रेषा यांचा छेद संच काय असतो
Answer» ANSWER: I don't KNOW how to SOLVE the PROBLEM is this QUESTION

3813.

एक प्रतल व एक रेषा यांचा छेद संच काय असतो

Answer»

ANSWER:

I don't KNOW how to SOLVE the PROBLEM is this QUESTION

Discussion

3814.	When rounded off to the nearest thousand the number 63,634 becomes
Answer» $\huge\tt\fcolorbox{lavenderblush}{plum}{⌗Aŋʂwɛr♡༉}$ ✿ When ROUNDED off to the NEAREST THOUSAND the NUMBER 63,634 BECOMES 64,000. ⠀ ━━━━━━━━━━━━━━━━━━━

3814.

When rounded off to the nearest thousand the number 63,634 becomes

Answer»

$\huge\tt\fcolorbox{lavenderblush}{plum}{⌗Aŋʂwɛr♡༉}$

✿ When ROUNDED off to the NEAREST THOUSAND the NUMBER 63,634 BECOMES 64,000.

⠀ ━━━━━━━━━━━━━━━━━━━

Discussion

3815.	What is the meaning of national numbers in simple language
Answer» ANSWER: You mean Rational NUMBER? Step-by-step explanation: In mathematics, a rational number is a number that can be expressed as the quotient or fraction pq of two integers, a numerator p and a non-zero denominator q. For example, −37 is a rational number, as is every integer. Or A rational number is a number that is of the form p/q where p and q are integers and q is not EQUAL to 0. The SET of rational numbers is denoted by Q. In other words, If a number can be expressed as a fraction where both the numerator and the denominator are integers, the number is a rational number.

3815.

What is the meaning of national numbers in simple language

Answer»

ANSWER:

You mean Rational NUMBER?

Step-by-step explanation:

In mathematics, a rational number is a number that can be expressed as the quotient or fraction pq of two integers, a numerator p and a non-zero denominator q. For example, −37 is a rational number, as is every integer.

Or

A rational number is a number that is of the form p/q where p and q are integers and q is not EQUAL to 0. The SET of rational numbers is denoted by Q. In other words, If a number can be expressed as a fraction where both the numerator and the denominator are integers, the number is a rational number.

Discussion

3816.	Expandsin4  cos2 in a series of cosines ofmultiple of  .
Answer» ANSWER: the use of COMMON people what B is the CHEMICAL laboratory

3816.

Expandsin4  cos2 in a series of cosines ofmultiple of  .

Answer»

ANSWER:

the use of COMMON people what B is the CHEMICAL laboratory

Discussion

3817.	The square root of 0.0484 please tell by step by step.
Answer» Step-by-step explanation: 0.0484^1/2 you NEED to RAISE the NUMBER by HALF

3817.

The square root of 0.0484 please tell by step by step.

Answer»

Step-by-step explanation:

0.0484^1/2 you NEED to RAISE the NUMBER by HALF

Discussion

3818.	1 kg apple 37.75 rs. ka hai to 7.5 kg apples kitne ka hoga?
Answer» Here is your answer mate, HOPE it HELPS you.. 1kg apple price is 37.75 then price of 7.5 KG of apple is 37.75*7.5=283.125

3818.

1 kg apple 37.75 rs. ka hai to 7.5 kg apples kitne ka hoga?

Answer»

Here is your answer mate,

HOPE it HELPS you..

1kg apple price is 37.75

then price of 7.5 KG of apple is 37.75*7.5=283.125

Discussion

3819.	Find the value of Y in 23 x Y – 3 = 89:
Answer» ANSWER: y = 4 Step-by-step explanation: 23Y - 3 = 89 ⇒ 23y = 89 +3 ⇒ y = 92 / 23 = 4

3819.

Find the value of Y in 23 x Y – 3 = 89:

Answer»

ANSWER:

y = 4

Step-by-step explanation:

23Y - 3 = 89

⇒ 23y = 89 +3

⇒ y = 92 / 23 = 4

Discussion

3820.	Where is Warren Occult Museum located?
Answer» The Warren OCCULT Museum is located behind the WARRENS’ former home in Monroe, CONNECTICUT.

Discussion

3821.	Test whether the following numbers are divisible by 3 7203
Answer» ANSWER: YES it is DIVISIBLE by THREE because 7+2+0+3=12 =1+2 =3 Step-by-step EXPLANATION:

3821.

Test whether the following numbers are divisible by 3 7203

Answer»

ANSWER: YES

it is DIVISIBLE by THREE because 7+2+0+3=12

=1+2

=3

Step-by-step EXPLANATION:

Discussion

3822.	What is 11/6 divided by 61/5 ?
Answer» Step-by-step EXPLANATION: $\frac{11}{6} \div \frac{61}{5} \\ = \frac{11}{6} \times \frac{5}{61} \\ = \frac{55}{366}$

3822.

What is 11/6 divided by 61/5 ?

Answer»

Step-by-step EXPLANATION:

$\frac{11}{6} \div \frac{61}{5} \\ = \frac{11}{6} \times \frac{5}{61} \\ = \frac{55}{366}$

Discussion

3823.	Using the number 7 only seven times make total of 100
Answer» ANSWER: 7+7+7+7+7+7+7 +51 this your answer HOPE it is HELPFUL for you

3823.

Using the number 7 only seven times make total of 100

Answer»

ANSWER:

7+7+7+7+7+7+7 +51 this your answer HOPE it is HELPFUL for you

Discussion

3824.	There is only one center of circle true or false
Answer» ANSWER: True Step-by-step EXPLANATION: A CIRCLE has only ONE CENTRE

3824.

There is only one center of circle true or false

Answer»

ANSWER:

True

Step-by-step EXPLANATION:

A CIRCLE has only ONE CENTRE

Discussion

3825.	Which formula cAn be used to determine the efficiency of pile group?
Answer» Answer: Converse Labarre FORMULAE, Seiler-kenney formulae, FELD's RULE are some of the common formulae AVAILABLE for DETERMINING the efficiency of pile group. Step-by-step explanation: Hope it helped☺

3825.

Which formula cAn be used to determine the efficiency of pile group?

Answer»

Answer:

Converse Labarre FORMULAE, Seiler-kenney formulae, FELD's RULE are some of the common formulae AVAILABLE for DETERMINING the efficiency of pile group.

Step-by-step explanation:

Hope it helped☺

Discussion

3826.	Find the unit digit of 7 203×7 899
Answer» Answer: 7 Step-by-step EXPLANATION:

3826.

Find the unit digit of 7 203×7 899

Answer»

Answer:

7

Step-by-step EXPLANATION:

Discussion

3827.	২.একটি স্থূলকোণী ত্রিভুজের কোণের সমষ্টি কত?
Answer» যে কোণের মান ৯০° এর চেয়ে বড় এবং ১৮০° এর চেয়ে ছোট তাকে স্থুলকোণ বলে।

Discussion

3828.	Find the value with the help of logarithem tables. 3614 x 46.282 102.32 x 16.82
Answer» ANSWER: LET x=5872×0.058 Taking LOG both SIDES, we have logx=log(5872×0.058) logx=log5872+log0.058(∵logab=loga+logb) logx=log(58.72×10 2 )+log(58×0 −3 ) logx=log58.72+2+log58−3(∵log10=1) ⇒1+logx=log58.72+log58 Using log table, we get log58.72=1.77 log58=1.76 ∴1+logx=1.77+1.76 ⇒logx=3.53−1 ⇒logx=2.53 ⇒x=Antilog(2.53)=338.84 Therefore, 5872×0.058≈338.84

3828.

Find the value with the help of logarithem tables. 3614 x 46.282 102.32 x 16.82

Answer»

ANSWER:

LET x=5872×0.058

Taking LOG both SIDES, we have

logx=log(5872×0.058)

logx=log5872+log0.058(∵logab=loga+logb)

logx=log(58.72×10

2

)+log(58×0

−3

)

logx=log58.72+2+log58−3(∵log10=1)

⇒1+logx=log58.72+log58

Using log table, we get

log58.72=1.77

log58=1.76

∴1+logx=1.77+1.76

⇒logx=3.53−1

⇒logx=2.53

⇒x=Antilog(2.53)=338.84

Therefore,

5872×0.058≈338.84

Discussion

3829.	Show that sin10sin50sin70=1/8
Answer» Step-by-step explanation: Caste-based trade among the Nakarattars of Tamil Nadu This is not to say that the Nakarattar banking system resembled an economist’s model of Western-style banking systems … the Nakarattars loaned and deposited money with one another in caste-defined social relationships based on business territory, residential location, descent, marriage, and common cult membership. Unlike in MODERN Western banking systems, it was the reputation, decisions, and reserve deposits shared among exchange spheres defined according to these PRINCIPLES, and not a government-controlled central bank, that … assured public confidence in individual Nakarattars as representatives of the caste as a whole. In other words, the Nakarattar banking system was a caste-based banking system. Individual Nakarattars organised their lives around participation in and MANAGEMENT of various communal institutions adapted to the task of accumulating and distributing RESERVES of capital. Source: Rudner Read the extracts from Caste and Capitalism in Colonial India(Rudner 1994) in the box and answer the following questions. 1.What are the significant differences between the Nakarattar bankingsystem and the modern Western banking system, according to the author? 2.What are the DIFFERENT ways in which Nakarattar trading andbanking activities are linked to other social structures? 3.Can you think of examples within the modern capitalist economy whereeconomic activities are similarly ‘embedded’ in social structures?

3829.

Show that sin10sin50sin70=1/8

Answer»

Step-by-step explanation:

Caste-based trade among the Nakarattars of Tamil Nadu This is not to say that the Nakarattar banking system resembled an economist’s model of Western-style banking systems … the Nakarattars loaned and deposited money with one another in caste-defined social relationships based on business territory, residential location, descent, marriage, and common cult membership. Unlike in MODERN Western banking systems, it was the reputation, decisions, and reserve deposits shared among exchange spheres defined according to these PRINCIPLES, and not a government-controlled central bank, that … assured public confidence in individual Nakarattars as representatives of the caste as a whole. In other words, the Nakarattar banking system was a caste-based banking system. Individual Nakarattars organised their lives around participation in and MANAGEMENT of various communal institutions adapted to the task of accumulating and distributing RESERVES of capital. Source: Rudner

Read the extracts from Caste and Capitalism in Colonial India(Rudner 1994) in the box and answer the following questions.

1.What are the significant differences between the Nakarattar bankingsystem and the modern Western banking system, according to the author?

2.What are the DIFFERENT ways in which Nakarattar trading andbanking activities are linked to other social structures?

3.Can you think of examples within the modern capitalist economy whereeconomic activities are similarly ‘embedded’ in social structures?

Discussion

3830.	Saurav gets pocket money from his father every day.Out of the pocket money, he saves 2.75 on first dayand on each succeeding day he increases his savingby 25 paise. Find(i) the amount saved by Saurav on 14th day,(ii) the amount saved by Saurav on 25th day, and(iii) the total amount saved by Saurav in 30 days.(explain step by step in details)
Answer» Answer: i) 5.75 ii) 8.50 iii) 9.75 Step-by-step explanation: I have no idea what paise is but ill try i) FIRST day = 2.25 to find the answer times 25 by each day which is 14: 25 * 14 = 350 = 3.50 2.25 (first day) + 3.50 (the rest) 5.75 therefore, on the 14TH day he has 5.75 ii) first day = 2.25 we do the same as the last question but times by 25: 25 * 25 = 6.25 add 2.25 from the first day with 6.25 that he earned from all of the other days: 2.25 + 6.25 = 8.50 iii) first day = 2.25 this question goes the same way as the other TWO questions where we times 25 by how MANY days which it 30 then add that with the MONEY from the first day, 2.25 25 * 30 = 7.50 2.25 + 7.50 = 9.75 sorry if I got it wrong I have no idea what currency you are using hope this helps though :D

3830.

Saurav gets pocket money from his father every day.Out of the pocket money, he saves 2.75 on first dayand on each succeeding day he increases his savingby 25 paise. Find(i) the amount saved by Saurav on 14th day,(ii) the amount saved by Saurav on 25th day, and(iii) the total amount saved by Saurav in 30 days.(explain step by step in details)

Answer»

Answer:

i) 5.75

ii) 8.50

iii) 9.75

Step-by-step explanation:

I have no idea what paise is but ill try

i)

FIRST day = 2.25

to find the answer times 25 by each day which is 14:

25 * 14 = 350 = 3.50

2.25 (first day) + 3.50 (the rest)

5.75

therefore, on the 14TH day he has 5.75

ii)

first day = 2.25

we do the same as the last question but times by 25:

25 * 25 = 6.25

add 2.25 from the first day with 6.25 that he earned from all of the other days:

2.25 + 6.25 = 8.50

iii)

first day = 2.25

this question goes the same way as the other TWO questions where we times 25 by how MANY days which it 30

then add that with the MONEY from the first day, 2.25

25 * 30 = 7.50

2.25 + 7.50 = 9.75

sorry if I got it wrong

I have no idea what currency you are using

hope this helps though

:D

Discussion

3831.	(a³+b³)/(a²-ab+b²) = ?
Answer» (a³+b³)/(a²-ab+b²) =0.09 0.09 is the ANSWER.

3831.

(a³+b³)/(a²-ab+b²) = ?

Answer»

(a³+b³)/(a²-ab+b²) =0.09

0.09 is the ANSWER.

Discussion

3832.	Find the value of lcube+mcube+ncube w -3lmn when l+m+n=4 and lm+mn+nl=7
Answer» FIRST secdind because of me likf fbjdcvfddasvg ihrhn

3832.

Find the value of lcube+mcube+ncube w -3lmn when l+m+n=4 and lm+mn+nl=7

Answer»

FIRST

secdind because of me likf fbjdcvfddasvg ihrhn

Discussion

3833.	4 example of unlike fraction
Answer» Step-by-step EXPLANATION: MARK me BRAINLIST and FOLLOW

3833.

4 example of unlike fraction

Answer»

Step-by-step EXPLANATION:

MARK me BRAINLIST and FOLLOW

Discussion

3834.	By how much is 120271 is greater than 31456
Answer» Answer: 88815 Step-by-step explanation: 120271 is GREATER than 31456. So, 120271 is greater than 31456 by 88815.

3834.

By how much is 120271 is greater than 31456

Answer»

Answer:

88815

Step-by-step explanation:

120271 is GREATER than 31456. So, 120271 is greater than 31456 by 88815.

Discussion

3835.	Which measure of central tendency is easily affected by extreme values
Answer» Answer: the mean value Explanation: Mean values are easily affected by EXTREME values. MEASURES that are not that affected by extreme values are called RESISTANT. Measures that are affected by extreme values are called SENSITIVE.

3835.

Which measure of central tendency is easily affected by extreme values

Answer»

Answer:

the mean value

Explanation:

Mean values are easily affected by EXTREME values. MEASURES that are not that affected by extreme values are called RESISTANT. Measures that are affected by extreme values are called SENSITIVE.

Discussion

3836.	The squares of odd numbers are _______ numbers
Answer» $\huge\tt\fcolorbox{lavenderblush}{plum}{⌗Aŋʂwɛr♡༉}$ ✿ The SQUARES of ODD NUMBERS are odd numbers. ⠀ ━━━━━━━━━━━━━━━━━━━

3836.

The squares of odd numbers are _______ numbers

Answer»

$\huge\tt\fcolorbox{lavenderblush}{plum}{⌗Aŋʂwɛr♡༉}$

✿ The SQUARES of ODD NUMBERS are odd numbers.

⠀ ━━━━━━━━━━━━━━━━━━━

Discussion

3837.	120 180 75 195 what was the avrage number of boxes of crops harvisted for that month
Answer» Answer: average = 114 Step-by-step EXPLANATION: to FIND the average, ADD all the numbers and DIVIDE by the the AMOUNT of numbers there is which is 5 120 + 180 + 75 + 195 = 570 570/ 5 = 114

3837.

120 180 75 195 what was the avrage number of boxes of crops harvisted for that month

Answer»

Answer:

average = 114

Step-by-step EXPLANATION:

to FIND the average, ADD all the numbers and DIVIDE by the the AMOUNT of numbers there is which is 5

120 + 180 + 75 + 195 = 570

570/ 5 = 114

Discussion

3838.	If the difference between the compound interest and simple interest on a sum of money is Rs. 32 at a rate of 5% pa for 2 yr. What is SI?
Answer» $\large\underline{\sf{Solution-}}$ GIVEN that, $\sf{\rm :\longmapsto\: Time\:_{(annually)}, \: <klux>N</klux> \: = \: 2\: years}$ $\sf{\rm :\longmapsto\: Rate\:_{(interest)} \: r \: = \: 5\% \: per \: annum}$ Let assume that sum of money invested be Rs P. We know that, Simple Interest (SI) on a CERTAIN sum of money of Rs P invested at the rate of r % per annum for n years is $\red{\rm :\longmapsto\:SI = \dfrac{P \times r \times n}{100}}$ and Compound interest (CI) on a certain sum of money of Rs P invested at the rate of r % per annum compounded annually for n years is given by $\red{\rm :\longmapsto\: CI=P\bigg(1+\dfrac{r}{100}\bigg)^{n}-P}$ Now, According to statement, it is given that, $\rm :\longmapsto\: CI \: - \: SI \: = \: 32$ On substituting the values, we get $\rm :\longmapsto\:P\bigg(1+\dfrac{r}{100}\bigg)^{n} - P - \dfrac{Prn}{100} = 32$ $\rm :\longmapsto\:P\bigg(1+\dfrac{5}{100}\bigg)^{2} - P - \dfrac{P \times 5 \times 2}{100} = 32$ $\rm :\longmapsto\:P\bigg(1+\dfrac{1}{20}\bigg)^{2} - P - \dfrac{P }{10} = 32$ $\rm :\longmapsto\:P\bigg(\dfrac{20 + 1}{20}\bigg)^{2} - P - \dfrac{P }{10} = 32$ $\rm :\longmapsto\:P\bigg(\dfrac{21}{20}\bigg)^{2} - P - \dfrac{P }{10} = 32$ $\rm :\longmapsto\:P\bigg(\dfrac{441}{400}\bigg) - P - \dfrac{P }{10} = 32$ $\rm :\longmapsto\:\dfrac{441P - 400P - 40P}{400} = 32$ $\rm :\longmapsto\:\dfrac{441P - 440P}{400} = 32$ $\rm :\longmapsto\:\dfrac{P}{400} = 32$ $\bf\implies \:P = 12800$ So, Sum of money invested = Rs 12800 Rate of interest, r = 5 % per annum Time, n = 2 years So, $\red{\rm :\longmapsto\:SI = \dfrac{P \times r \times n}{100}}$ On substituting the values, we get $\rm :\longmapsto\:SI = \dfrac{12800 \times 5 \times 2}{100}$ $\bf\implies \:SI \: = \: Rs \: 1280$ Additional Information :- 1. Amount on a certain sum of money of Rs P invested at the rate of r % per annum for n years compounded annually is $\boxed{ \rm{ Amount = P\bigg(1+\dfrac{r}{100}\bigg)^{n}}}$ 2. Amount on a certain sum of money of Rs P invested at the rate of r % per annum for n years compounded semi - annually is $\boxed{ \rm{ Amount = P\bigg(1+\dfrac{r}{200}\bigg)^{2n}}}$ 3. Amount on a certain sum of money of Rs P invested at the rate of r % per annum for n years compounded QUARTERLY is $\boxed{ \rm{ Amount = P\bigg(1+\dfrac{r}{400}\bigg)^{4n}}}$

3838.

If the difference between the compound interest and simple interest on a sum of money is Rs. 32 at a rate of 5% pa for 2 yr. What is SI?

Answer»

$\large\underline{\sf{Solution-}}$

GIVEN that,

$\sf{\rm :\longmapsto\: Time\:_{(annually)}, \: <klux>N</klux> \: = \: 2\: years}$

$\sf{\rm :\longmapsto\: Rate\:_{(interest)} \: r \: = \: 5\% \: per \: annum}$

Let assume that sum of money invested be Rs P.

We know that,

Simple Interest (SI) on a CERTAIN sum of money of Rs P invested at the rate of r % per annum for n years is

$\red{\rm :\longmapsto\:SI = \dfrac{P \times r \times n}{100}}$

and

Compound interest (CI) on a certain sum of money of Rs P invested at the rate of r % per annum compounded annually for n years is given by

$\red{\rm :\longmapsto\: CI=P\bigg(1+\dfrac{r}{100}\bigg)^{n}-P}$

Now, According to statement, it is given that,

$\rm :\longmapsto\: CI \: - \: SI \: = \: 32$

On substituting the values, we get

$\rm :\longmapsto\:P\bigg(1+\dfrac{r}{100}\bigg)^{n} - P - \dfrac{Prn}{100} = 32$

$\rm :\longmapsto\:P\bigg(1+\dfrac{5}{100}\bigg)^{2} - P - \dfrac{P \times 5 \times 2}{100} = 32$

$\rm :\longmapsto\:P\bigg(1+\dfrac{1}{20}\bigg)^{2} - P - \dfrac{P }{10} = 32$

$\rm :\longmapsto\:P\bigg(\dfrac{20 + 1}{20}\bigg)^{2} - P - \dfrac{P }{10} = 32$

$\rm :\longmapsto\:P\bigg(\dfrac{21}{20}\bigg)^{2} - P - \dfrac{P }{10} = 32$

$\rm :\longmapsto\:P\bigg(\dfrac{441}{400}\bigg) - P - \dfrac{P }{10} = 32$

$\rm :\longmapsto\:\dfrac{441P - 400P - 40P}{400} = 32$

$\rm :\longmapsto\:\dfrac{441P - 440P}{400} = 32$

$\rm :\longmapsto\:\dfrac{P}{400} = 32$

$\bf\implies \:P = 12800$

So, Sum of money invested = Rs 12800

Rate of interest, r = 5 % per annum

Time, n = 2 years

So,

$\red{\rm :\longmapsto\:SI = \dfrac{P \times r \times n}{100}}$

On substituting the values, we get

$\rm :\longmapsto\:SI = \dfrac{12800 \times 5 \times 2}{100}$

$\bf\implies \:SI \: = \: Rs \: 1280$

Additional Information :-

1. Amount on a certain sum of money of Rs P invested at the rate of r % per annum for n years compounded annually is

$\boxed{ \rm{ Amount = P\bigg(1+\dfrac{r}{100}\bigg)^{n}}}$

2. Amount on a certain sum of money of Rs P invested at the rate of r % per annum for n years compounded semi - annually is

$\boxed{ \rm{ Amount = P\bigg(1+\dfrac{r}{200}\bigg)^{2n}}}$

3. Amount on a certain sum of money of Rs P invested at the rate of r % per annum for n years compounded QUARTERLY is

$\boxed{ \rm{ Amount = P\bigg(1+\dfrac{r}{400}\bigg)^{4n}}}$

Discussion

3839.	I have Bolt with property class 4.6Calculate proof load and yeild strength ?
Answer» ANSWER: PLEASE MARK me as BRAINLIST

3839.

I have Bolt with property class 4.6Calculate proof load and yeild strength ?

Answer»

ANSWER:

PLEASE MARK me as BRAINLIST

Discussion

3840.	2-2(2-(2-(2-2-2))) BODMAS SOLVE
Answer» ANSWER: Answer is zero solve the BRACKETS that are in the INNERMOST SIDE at FIRST

3840.

2-2(2-(2-(2-2-2))) BODMAS SOLVE

Answer»

ANSWER:

Answer is zero

solve the BRACKETS that are in the INNERMOST SIDE at FIRST

Discussion

3841.	27a cube - bcube+8c cube + 18 ABC
Answer» Step-by-step explanation: 27a³ -b³ +8c³ +18ABC = (3A)³ -b³+(2C)³ + 18ABC

3841.

27a cube - bcube+8c cube + 18 ABC

Answer»

Step-by-step explanation:

27a³ -b³ +8c³ +18ABC

= (3A)³ -b³+(2C)³ + 18ABC

Discussion

3842.	if S1,S2,S3 be the sum of n terms of three A.P ,then the first term of each being 1 and respectI've common difference 1,2,3 prove that S1+S2=2S2
Answer» $\large\underline{\sf{Solution-}}$ Given three AP series of n terms having FIRST term 1 and common difference 1, 2 and 3 respectively. Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ, ↝ Sum of n terms of an arithmetic sequence is, $\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{S_n\:=\dfrac{n}{2} \bigg(2 \:a\:+\:(n\:-\:1)\:d \bigg)}}}}}} \\ \end{gathered}$ Wʜᴇʀᴇ, Sₙ is the sum of n terms of AP. a is the first term of the sequence. n is the no. of terms. d is the common difference. CASE :- 1 First term, a = 1 Common difference, d = 1 Number of terms = n Sum of n terms is $\rm :\longmapsto\:S_1\:=\dfrac{n}{2} \bigg(2 \:(1)\:+\:(n\:-\:1)\:(1) \bigg)$ $\rm :\longmapsto\:S_1\:=\dfrac{n}{2} \bigg(2 \:+\:(n\:-\:1) \bigg)$ $\rm :\longmapsto\:S_1\:=\dfrac{n}{2} \bigg(2 \:+\:n\:-\:1 \bigg)$ $\bf :\longmapsto\:S_1\:=\dfrac{n}{2} \bigg(1 \:+\:n\: \bigg) - - - (1)$ Case :- 2 First term, a = 1 Common difference, d = 2 Number of terms = n Sum of n terms is $\rm :\longmapsto\:S_2\:=\dfrac{n}{2} \bigg(2 \:(1)\:+\:(n\:-\:1)\:(2) \bigg)$ $\rm :\longmapsto\:S_2\:=\dfrac{n}{2} \bigg(2 \:+\:2n\:-\:2\: \bigg)$ $\rm :\longmapsto\:S_2\:=\dfrac{n}{2} \bigg(\:2n\: \bigg)$ $\bf :\longmapsto\:S_2\:= {n}^{2} - - - - (2)$ Case :- 3 First term, a = 1 Common difference, d = 3 Number of terms = n Sum of n terms is $\rm :\longmapsto\:S_3\:=\dfrac{n}{2} \bigg(2 \:(1)\:+\:(n\:-\:1)\:(3) \bigg)$ $\rm :\longmapsto\:S_3\:=\dfrac{n}{2} \bigg(2 \:+\:3n\:-\:3\: \bigg)$ $\bf :\longmapsto\:S_3\:=\dfrac{n}{2} \bigg(\:3n\:-\:1\: \bigg) - - - - (3)$ Now, CONSIDER $\rm :\longmapsto\:S_1 + S_3$ $\rm \: = \: \: \dfrac{n}{2} \bigg(\:n\: + \:1\: \bigg) + \dfrac{n}{2} \bigg(\:3n\:-\:1\: \bigg)$ $\rm \: = \: \: \dfrac{n}{2} \bigg(n \: + \: 1 \: + \:3n\:-\:1\: \bigg)$ $\rm \: = \: \: \dfrac{n}{2} \bigg(4n \: \bigg)$ $\rm \: = \: \: {2n}^{2}$ $\rm \: = \: \: 2S_2$ Hence, Proved Additional INFORMATION :- ↝ nᵗʰ term of an arithmetic sequence is, $\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \\ \end{gathered}$ Wʜᴇʀᴇ, aₙ is the nᵗʰ term. a is the first term of the sequence. n is the no. of terms. d is the common difference.

3842.

if S1,S2,S3 be the sum of n terms of three A.P ,then the first term of each being 1 and respectI've common difference 1,2,3 prove that S1+S2=2S2

Answer»

$\large\underline{\sf{Solution-}}$

Given three AP series of n terms having FIRST term 1 and common difference 1, 2 and 3 respectively.

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ Sum of n terms of an arithmetic sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{S_n\:=\dfrac{n}{2} \bigg(2 \:a\:+\:(n\:-\:1)\:d \bigg)}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

Sₙ is the sum of n terms of AP.

a is the first term of the sequence.

n is the no. of terms.

d is the common difference.

CASE :- 1

First term, a = 1

Common difference, d = 1

Number of terms = n

Sum of n terms is

$\rm :\longmapsto\:S_1\:=\dfrac{n}{2} \bigg(2 \:(1)\:+\:(n\:-\:1)\:(1) \bigg)$

$\rm :\longmapsto\:S_1\:=\dfrac{n}{2} \bigg(2 \:+\:(n\:-\:1) \bigg)$

$\rm :\longmapsto\:S_1\:=\dfrac{n}{2} \bigg(2 \:+\:n\:-\:1 \bigg)$

$\bf :\longmapsto\:S_1\:=\dfrac{n}{2} \bigg(1 \:+\:n\: \bigg) - - - (1)$

Case :- 2

First term, a = 1

Common difference, d = 2

Number of terms = n

Sum of n terms is

$\rm :\longmapsto\:S_2\:=\dfrac{n}{2} \bigg(2 \:(1)\:+\:(n\:-\:1)\:(2) \bigg)$

$\rm :\longmapsto\:S_2\:=\dfrac{n}{2} \bigg(2 \:+\:2n\:-\:2\: \bigg)$

$\rm :\longmapsto\:S_2\:=\dfrac{n}{2} \bigg(\:2n\: \bigg)$

$\bf :\longmapsto\:S_2\:= {n}^{2} - - - - (2)$

Case :- 3

First term, a = 1

Common difference, d = 3

Number of terms = n

Sum of n terms is

$\rm :\longmapsto\:S_3\:=\dfrac{n}{2} \bigg(2 \:(1)\:+\:(n\:-\:1)\:(3) \bigg)$

$\rm :\longmapsto\:S_3\:=\dfrac{n}{2} \bigg(2 \:+\:3n\:-\:3\: \bigg)$

$\bf :\longmapsto\:S_3\:=\dfrac{n}{2} \bigg(\:3n\:-\:1\: \bigg) - - - - (3)$

Now, CONSIDER

$\rm :\longmapsto\:S_1 + S_3$

$\rm \: = \: \: \dfrac{n}{2} \bigg(\:n\: + \:1\: \bigg) + \dfrac{n}{2} \bigg(\:3n\:-\:1\: \bigg)$

$\rm \: = \: \: \dfrac{n}{2} \bigg(n \: + \: 1 \: + \:3n\:-\:1\: \bigg)$

$\rm \: = \: \: \dfrac{n}{2} \bigg(4n \: \bigg)$

$\rm \: = \: \: {2n}^{2}$

$\rm \: = \: \: 2S_2$

Hence, Proved

Additional INFORMATION :-

↝ nᵗʰ term of an arithmetic sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

aₙ is the nᵗʰ term.

a is the first term of the sequence.

n is the no. of terms.

d is the common difference.

Discussion

3843.	If a =3 b = 4 c = 5 then prove that a+( b+c ) =(a+b) +c
Answer» Answer: the answer is 12 Step-by-step explanation: by using distribute METHOD we can prove that a+(b+C)=(a+b)+c 3+(4+5)=(3+4)+5 12. = 12 so here is it proved both SIDES are EQUAL. HOPE it will help you

3843.

If a =3 b = 4 c = 5 then prove that a+( b+c ) =(a+b) +c

Answer»

Answer:

the answer is 12

Step-by-step explanation:

by using distribute METHOD

we can prove that a+(b+C)=(a+b)+c

3+(4+5)=(3+4)+5

12. = 12

so here is it proved both SIDES are EQUAL.

HOPE it will help you

Discussion

3844.	A society collection ₹ 12,35,820 for charity fund in one year and ₹ 28,25,320 in another year. If they used ₹ 26,72,500 out of the total amount collected in the two years , then how much money are they left with now ?
Answer» Step-by-step explanation: one year COLLECT money = 1235820 second year collect money = 2825320 they used money in work = 2672500 they had LEFT Money now= 4061140 - 2672500 = 1388640 RUPEES

3844.

A society collection ₹ 12,35,820 for charity fund in one year and ₹ 28,25,320 in another year. If they used ₹ 26,72,500 out of the total amount collected in the two years , then how much money are they left with now ?

Answer»

Step-by-step explanation:

one year COLLECT money = 1235820

second year collect money = 2825320

they used money in work = 2672500

they had LEFT Money now= 4061140 - 2672500

= 1388640 RUPEES

Discussion

3845.	Differencebetween mean median in fibre prime number
Answer» Answer: The mean is the average, found by ADDING the NUMBERS and dividing by the number of numbers, in this case, 5. So the mean is , or . The median is the middle number in a set. Step-by-step explanation: HOPE this is useful for you.

3845.

Differencebetween mean median in fibre prime number

Answer»

Answer:

The mean is the average, found by ADDING the NUMBERS and dividing by the number of numbers, in this case, 5. So the mean is , or . The median is the middle number in a set.

Step-by-step explanation:

HOPE this is useful for you.

Discussion

3846.	dienesh is x years old .his uncle is three times as old as dinesh .in ten years time, remesh will be twice as old as dinesh
Answer» ANSWER: so popular ONLINE SEND me the link below Step-by-step explanation: he has been CHECKED the website and

3846.

dienesh is x years old .his uncle is three times as old as dinesh .in ten years time, remesh will be twice as old as dinesh

Answer»

ANSWER:

so popular ONLINE SEND me the link below

Step-by-step explanation:

he has been CHECKED the website and

Discussion

3847.	I want limits paper practice set 1
Answer» ANSWER: ALKALINE Step-by-step explanation: please MARK me as BRAINLIST

3847.

I want limits paper practice set 1

Answer»

ANSWER:

ALKALINE

Step-by-step explanation:

please MARK me as BRAINLIST

Discussion

3848.	If p(x) =2x+1, find the value of p(1) ka answer
Answer» ANSWER: PUT x=1 in the equation p(1)=2*1+1 =2+1 =3

3848.

If p(x) =2x+1, find the value of p(1) ka answer

Answer»

ANSWER:

PUT x=1 in the equation

p(1)=2*1+1

=2+1

=3

Discussion

3849.	If p(x)=2x+ 1 ,find the value of p ( 1)Option is 2,1,5,3
Answer» <P>Answer: p(1) = 3 is the CORRECT OPTION Step-by-step explanation: p(x)=2x+ 1 Put x = 1 p(1)=2(1)+ 1 p(1)=2+1 p(1)=3

3849.

If p(x)=2x+ 1 ,find the value of p ( 1)Option is 2,1,5,3

Answer»

<P>Answer:

p(1) = 3 is the CORRECT OPTION

Step-by-step explanation:

p(x)=2x+ 1

Put x = 1

p(1)=2(1)+ 1

p(1)=2+1

p(1)=3

Discussion

3850.	Cosec 2 theta - 1 is equal to
Answer» ANSWER: cosec^2 THETA - 1 = cot^2 theta

3850.

Cosec 2 theta - 1 is equal to

Answer»

ANSWER:

cosec^2 THETA - 1 = cot^2 theta

Discussion

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if S1,S2,S3 be the sum of n terms of three A.P ,then the first term of each being 1 and respectI've common difference 1,2,3 prove that S1+S2=2S2

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If a =3 b = 4 c = 5 then prove that a+( b+c ) =(a+b) +c

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