332354 + Interview Questions in Secondary School in Math

1.	You roll a fair 6-sided die. what is p (not 5)
Answer» ANSWER: p(E) = 5/6 Step-by-step explanation: the dice has ONE FIVE, so the answer is 5/6. this is the answer hope that it HELPS you

1.

You roll a fair 6-sided die. what is p (not 5)

Answer»

ANSWER:

p(E) = 5/6

Step-by-step explanation:

the dice has ONE FIVE, so the answer is 5/6.

this is the answer

hope that it HELPS you

Discussion

2.	The mean of the number of sales of cars over a 3-month period is $87, and the standard deviation is 5. The mean of the commissions is $5225, and the standard deviation is $773. A. calculate the coefficient of variation. B. compare the variation of the two and give your comments regarding the variation.
Answer» Answer: Heya................❤✌ A) Coefficient of Variation (SALES) = (5/87) 100 = 5.74% Coefficient* of Variation ( commotions) = (773/5225)100 = 14.8% B) Commissions* are more VARIABLE & less stable & less uniform than sales. *Explanation* : Coefficient of variation is large INDICATES that the GROUP is more variable & it is less stable or less uniform. If a coefficient of variation is small then it indicates that the group is less variable & it is more stable & more uniform. HOPE it’s helpful....... ☺

2.

The mean of the number of sales of cars over a 3-month period is $87, and the standard deviation is 5. The mean of the commissions is $5225, and the standard deviation is $773. A. calculate the coefficient of variation. B. compare the variation of the two and give your comments regarding the variation.

Answer»

Answer:

Heya................❤✌

A) Coefficient of Variation (SALES) = (5/87) *100

= 5.74%

Coefficient of Variation ( commotions) = (773/5225)*100 = 14.8%

B) Commissions are more VARIABLE & less stable & less uniform than sales.

Explanation : Coefficient of variation is large INDICATES that the GROUP is more variable & it is less stable or less uniform. If a coefficient of variation is small then it indicates that the group is less variable & it is more stable & more uniform.

HOPE it’s helpful....... ☺

Discussion

3.	If a = (3+√7)/2 , then find the value of a² + 1/a²
Answer» $\blue{\bold{\underline{\underline{Answer:}}}}$ $\green{\tt{\therefore{a^{2}+\frac{1}{a^{2}}=\frac{40-9\sqrt{7}}{2}}}}\\$ $\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$ $\green{\underline \bold{<klux>GIVEN</klux> :}} \\ \tt: \implies a = \frac{3 + \sqrt{7} }{ 2} \\ \\ \red{\underline \bold{To \: Find:}} \\ \tt: \implies {a}^{2} + \frac{1}{ {a}^{2} } =?$ • ACCORDING to given QUESTION : $\bold{As \: we \: know \: that} \\ \tt: \implies {a}^{2} = ( \frac{3 + \sqrt{7} }{2})^{2} \\ \\ \tt: \implies {a}^{2} = \frac{ {3}^{2} + {( \sqrt{7}) }^{2} + 2 \times 3 \times \sqrt{7} }{4} \\ \\ \tt: \implies {a}^{2} = \frac{9 + 7 + 6 \sqrt{7} }{4} \\ \\ \tt: \implies {a}^{2} = \frac{16 + 6 \sqrt{7} }{4} \\ \\ \tt: \implies {a}^{2} = \frac{8 + 3 \sqrt{7} }{2} - - - - - (1) \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies \frac{1}{ {a}^{2} } = \frac{1}{( { \frac{3 + \sqrt{7} }{2}) }^{2} } \\ \\ \tt: \implies \frac{1}{ {a}^{2} } = \frac{4}{9 + 7 + 6 \sqrt{7} } \\ \\ \tt: \implies \frac{1}{ {a}^{2} } = \frac{2}{8 + 3 \sqrt{7} } \\ \\ \tt: \implies \frac{1}{ {a}^{2} } = \frac{2}{8 + 3 \sqrt{7} } \times \frac{8 - 3\sqrt{7} }{8 - 3 \sqrt{7} } \\ \\ \tt: \implies \frac{1}{ {a}^{2} } = \frac{16 - 6 \sqrt{7} }{64 - 63} \\ \\ \tt: \implies \frac{1}{ {a}^{2} } = 16 - 6 \sqrt{7} - - - - - (2) \\ \\ \bold{For \: finding \: value } \\ \tt: \implies {a}^{2} + \frac{1}{ {a}^{2} } = \frac{8 + 3 \sqrt{7} }{2} + 16 - 6 \sqrt{7} \\ \\ \tt: \implies {a}^{2} + \frac{1}{ {a}^{2} } = \frac{8 + 3 \sqrt{7} + 32 - 12 \sqrt{7} }{2} \\ \\ \green{\tt: \implies {a}^{2} + \frac{1}{ {a}^{2} } = \frac{40 - 9 \sqrt{7} }{2} }$

3.

If a = (3+√7)/2 , then find the value of a² + 1/a²

Answer»

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{a^{2}+\frac{1}{a^{2}}=\frac{40-9\sqrt{7}}{2}}}}\\$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{<klux>GIVEN</klux> :}} \\ \tt: \implies a = \frac{3 + \sqrt{7} }{ 2} \\ \\ \red{\underline \bold{To \: Find:}} \\ \tt: \implies {a}^{2} + \frac{1}{ {a}^{2} } =?$

• ACCORDING to given QUESTION :

$\bold{As \: we \: know \: that} \\ \tt: \implies {a}^{2} = ( \frac{3 + \sqrt{7} }{2})^{2} \\ \\ \tt: \implies {a}^{2} = \frac{ {3}^{2} + {( \sqrt{7}) }^{2} + 2 \times 3 \times \sqrt{7} }{4} \\ \\ \tt: \implies {a}^{2} = \frac{9 + 7 + 6 \sqrt{7} }{4} \\ \\ \tt: \implies {a}^{2} = \frac{16 + 6 \sqrt{7} }{4} \\ \\ \tt: \implies {a}^{2} = \frac{8 + 3 \sqrt{7} }{2} - - - - - (1) \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies \frac{1}{ {a}^{2} } = \frac{1}{( { \frac{3 + \sqrt{7} }{2}) }^{2} } \\ \\ \tt: \implies \frac{1}{ {a}^{2} } = \frac{4}{9 + 7 + 6 \sqrt{7} } \\ \\ \tt: \implies \frac{1}{ {a}^{2} } = \frac{2}{8 + 3 \sqrt{7} } \\ \\ \tt: \implies \frac{1}{ {a}^{2} } = \frac{2}{8 + 3 \sqrt{7} } \times \frac{8 - 3\sqrt{7} }{8 - 3 \sqrt{7} } \\ \\ \tt: \implies \frac{1}{ {a}^{2} } = \frac{16 - 6 \sqrt{7} }{64 - 63} \\ \\ \tt: \implies \frac{1}{ {a}^{2} } = 16 - 6 \sqrt{7} - - - - - (2) \\ \\ \bold{For \: finding \: value } \\ \tt: \implies {a}^{2} + \frac{1}{ {a}^{2} } = \frac{8 + 3 \sqrt{7} }{2} + 16 - 6 \sqrt{7} \\ \\ \tt: \implies {a}^{2} + \frac{1}{ {a}^{2} } = \frac{8 + 3 \sqrt{7} + 32 - 12 \sqrt{7} }{2} \\ \\ \green{\tt: \implies {a}^{2} + \frac{1}{ {a}^{2} } = \frac{40 - 9 \sqrt{7} }{2} }$

Discussion

4.	1-sin square theta divided by 1+cos theta=cos theta
Answer» Step-by-step EXPLANATION: I HOPE that above PROCEDURE HELPS U

4.

1-sin square theta divided by 1+cos theta=cos theta

Answer»

Step-by-step EXPLANATION:

I HOPE that above PROCEDURE HELPS U

Discussion

5.	❤️ solve this problem if you can
Answer» Answer: I am providing u two solutions,. one is typed and other is solved on paper..... Step-by-step EXPLANATION: = (sec a + TAN a)/(cosec a + COT a) = (1+sin a)sin a/ COS a (1 + cos a) = ( (1-cos a)/sin a) / ( (1-sin a)/cos a) = (cosec a - cot a)/(sec a - tan a) hence proved

5.

❤️ solve this problem if you can

Answer»

Answer:

I am providing u two solutions,. one is typed and other is solved on paper.....

Step-by-step EXPLANATION:

= (sec a + TAN a)/(cosec a + COT a)

= (1+sin a)sin a/ COS a (1 + cos a)

= ( (1-cos a)/sin a) / ( (1-sin a)/cos a)

= (cosec a - cot a)/(sec a - tan a)

hence proved

Discussion

6.	The perimeter of a square field is 168m. find the length of the side of the square
Answer» ANSWER: The area of the SQUARE field = 1764 SQ m. So each side of the field = 1764^0.5 = 42 m. The PERIMETER of the field = 168 m Step-by-step explanation: *MARK* ME AS BRAINLIEST !!

6.

The perimeter of a square field is 168m. find the length of the side of the square

Answer»

ANSWER:

The area of the SQUARE field = 1764 SQ m. So each side of the field = 1764^0.5 = 42 m. The PERIMETER of the field = 168 m

Step-by-step explanation:

MARK ME AS BRAINLIEST !!

Discussion

7.	ONLY DO 11th QUESTION PLEASE
Answer»

Discussion

8.	If the L.C.M and H.C.F of two number are 169 and 8 respectively. One number is 32, find the other number
Answer» Step-by-step EXPLANATION: l.c.mh.c.f=one NUMBERSECOND number 1698= 32second number 169*8 = second number 32 1352 = second number 32 42.25= second number

8.

If the L.C.M and H.C.F of two number are 169 and 8 respectively. One number is 32, find the other number

Answer»

Step-by-step EXPLANATION:

l.c.m*h.c.f=one NUMBER*SECOND number

169*8= 32*second number

169*8 = second number

32

1352 = second number

32

42.25= second number

Discussion

9.	5^x-2=625 what's the answer for this equation
Answer» ANSWER: 5x - 2 = 625 5x = 625 + 2 5x = 627 x = 627/ 5. this is the answer of this QUESTION.

9.

5^x-2=625 what's the answer for this equation

Answer»

ANSWER:

5x - 2 = 625

5x = 625 + 2

5x = 627

x = 627/ 5.

this is the answer of this QUESTION.

Discussion

10.	Sameer bought a jacket for rupees 10 and a pair of shoes for rupees 15 estimate the cost he has pay to the shopkeeper and also find the actual cost
Answer»

Discussion

11.	The mass of a brick is 2 kg 750 gram. what is the total mass of 14 such bricks ?
Answer» Step-by-step EXPLANATION: MASS of one BRICK= 2.75 kg Mass of 14 bricks= 2.7514= 38.5kg Hop*e it will HELP you

11.

The mass of a brick is 2 kg 750 gram. what is the total mass of 14 such bricks ?

Answer»

Step-by-step EXPLANATION:

MASS of one BRICK= 2.75 kg

Mass of 14 bricks= 2.75*14= 38.5kg

Hope it will HELP you

Discussion

12.	12 Notes of rs 5 how many value
Answer» Step-by-step EXPLANATION: ANSWER is Rs60 hope it will HELP you

12.

12 Notes of rs 5 how many value

Answer»

Step-by-step EXPLANATION:

ANSWER is Rs60

hope it will HELP you

Discussion

13.	2х3×5=10 and 7×4×8=36 then 6×5×4=?
Answer» ANSWER: ✨✨✨✨ans:- 29✨✨✨✨ hope U understand✨follow me❤ gm FRIEND

13.

2х3×5=10 and 7×4×8=36 then 6×5×4=?

Answer»

ANSWER:

✨✨✨✨ans:- 29✨✨✨✨

hope U understand✨follow me❤

gm FRIEND

Discussion

14.	If the L.C.M and H.C.F of two number are 160 and 8 respectively. One number is 32, find the other number?
Answer» Answer: 40 is your answer Step-by-step explanation: Hope it helps you... PLEASE MARK me as BRAINLIEST.

14.

If the L.C.M and H.C.F of two number are 160 and 8 respectively. One number is 32, find the other number?

Answer»

Answer:

40 is your answer

Step-by-step explanation:

Hope it helps you...

PLEASE MARK me as BRAINLIEST.

Discussion

15.	5th January is Tuesday which day of that month will fall not occur 5 times
Answer» Step-by-step EXPLANATION: ANSWER is Thursday Hope it will HELP you

15.

5th January is Tuesday which day of that month will fall not occur 5 times

Answer»

Step-by-step EXPLANATION:

ANSWER is Thursday

Hope it will HELP you

Discussion

16.	for making 8 trousers of the same size 14 m 80 CM of cloth is needed. how much cloth will be required for each such trouser ?
Answer» Correct Question - For making 8 pairs of trousers of the same size 14 m and 80 centimeters of CLOTH is needed. How much cloth will be required for each such trouser ? Solution - This question can be easily be solved by various methods . The most easy one is the UNITARY method . Method 1 - In this method , from the given Information , we need to find the VALUE of a single unit, and then find the required value by multiplying the given single unit value. So, In this question - For making 8 trousers of the same size 14 m and 80 centimeters of cloth is needed. So, 8 pairs of trousers require 14 m , 80 cm of cloth, i.e, 1480 cm of Cloth . So, 1 pair of trousers require ( 1480 / 8 ) cm of cloth => ( 185 cm ) of cloth . Method 2 - Proportionality Method - Here , the length of cloth required to make is directly proportional to the number of trousers made . So, 1480 = n × 8 => n = 185 cm . Hence, For making each such trouser, 185 cm or 1.85 m of cloth is required . Answer - For making each such trouser, 185 cm or 1.85 m of cloth is required .

16.

for making 8 trousers of the same size 14 m 80 CM of cloth is needed. how much cloth will be required for each such trouser ?

Answer»

Correct Question -

For making 8 pairs of trousers of the same size 14 m and 80 centimeters of CLOTH is needed.

How much cloth will be required for each such trouser ?

Solution -

This question can be easily be solved by various methods .

The most easy one is the UNITARY method .

Method 1 -

In this method , from the given Information , we need to find the VALUE of a single unit, and then find the required value by multiplying the given single unit value.

So,

In this question -

For making 8 trousers of the same size 14 m and 80 centimeters of cloth is needed.

So,

8 pairs of trousers require 14 m , 80 cm of cloth, i.e, 1480 cm of Cloth .

So,

1 pair of trousers require ( 1480 / 8 ) cm of cloth

=> ( 185 cm ) of cloth .

Method 2 -

Proportionality Method -

Here , the length of cloth required to make is directly proportional to the number of trousers made .

So,

1480 = n × 8

=> n = 185 cm .

Hence, For making each such trouser, 185 cm or 1.85 m of cloth is required .

Answer -

For making each such trouser, 185 cm or 1.85 m of cloth is required .

Discussion

17.	Prove that:- sinA(1+tanA) + cosA(1+cotA) = secA+cosecA.please answer fast need it urgently
Answer» HELLO MATE... YOUR ANSWER IS IN THE IMAGE.... HOPE IT HELPS YOU........ MARK IT BRAINLIEST.

17.

Prove that:- sinA(1+tanA) + cosA(1+cotA) = secA+cosecA.please answer fast need it urgently

Answer»

HELLO MATE...

YOUR ANSWER IS IN THE IMAGE....

HOPE IT HELPS YOU........

MARK IT BRAINLIEST.

Discussion

18.	Evaluate 125×125-25×25 by suitable Identity
Answer» 125×125-25×25 = 15000 I HOPE I WILL HELP YOU

18.

Evaluate 125×125-25×25 by suitable Identity

Answer»

125×125-25×25 = 15000

I HOPE I WILL HELP YOU

Discussion

19.	Please answer meiiiii
Answer» ANSWER: it's in SANSKRIT I can't UNDERSTAND this language Mark me as BRAINLIEST

19.

Please answer meiiiii

Answer»

ANSWER:

it's in SANSKRIT I can't UNDERSTAND this language

Mark me as BRAINLIEST

Discussion

20.	What are a,b and c equal to for this quadratic equationy = 4x^2 + 3x - 5a = -4, b = 3, c = -5a = 4, b = -3, c = 5a = 4, b = 3, c = 5a= 4, b = 3, c= -5
Answer» Hey friend! Here is your answer; A GENERAL polynomial is EXPRESSED in ax^2 + bx + c Here y = 4x^2 + 3x - 5 So, a = 4 b = 3 c = -5 I hope it HELPS you

Discussion

21.	Zeroes of the polynomial p(x)=2x^2-5x+7,find a polynomial whose zeros are 2alpha+3 and 2 Beta+3
Answer» Step-by-step EXPLANATION: readiiiiiiiiit i=sqrt(-1) i²=(-1)

21.

Zeroes of the polynomial p(x)=2x^2-5x+7,find a polynomial whose zeros are 2alpha+3 and 2 Beta+3

Answer»

Step-by-step EXPLANATION:

readiiiiiiiiit

i=sqrt(-1)

i²=(-1)

Discussion

22.	Determine the value of a and such that x2-2x-3 is tga factor of the polynomial x3-3x2+ax-b
Answer» hope it HELPS you... THANKS

22.

Determine the value of a and such that x2-2x-3 is tga factor of the polynomial x3-3x2+ax-b

Answer»

hope it HELPS you... THANKS

Discussion

23.	Polygon in which diagonals intersect externally will have one angle. Less than 90 degrees Two reflex angles Can't be determined Greater than 180 degrees
Answer» It is obvious that if you draw two DIAGONALS in a PLOYGON you will GET a ANGLE of exactly 90° or >90°.

23.

Polygon in which diagonals intersect externally will have one angle. Less than 90 degrees Two reflex angles Can't be determined Greater than 180 degrees

Answer»

It is obvious that if you draw two DIAGONALS in a PLOYGON you will GET a ANGLE of exactly 90° or >90°.

Discussion

24.	Irrational no. between 2/5 and 3/5
Answer» Answer: 21/50, Step-by-step EXPLANATION:

24.

Irrational no. between 2/5 and 3/5

Answer»

Answer:

21/50,

Step-by-step EXPLANATION:

Discussion

25.	Lcm of 10,5,15,20 and convert into like fraction
Answer» ANSWER: LCM is 5 and in FRACTION it will be 10/2

25.

Lcm of 10,5,15,20 and convert into like fraction

Answer»

ANSWER:

LCM is 5 and in FRACTION it will be 10/2

Discussion

26.	A and B are two pipes connected to a tank. A is the filler pipe and B is the ejector. The capacity of the tank is 6000L.The ejector pipe can eject water at the rate 4L/min higher than the filler pipe. The filler pipe can fill the tank 5 min later than the ejector pipe can eject. Find the flow rate of the ejector pipe in L/min(approx.).
Answer» Step-by-step explanation: Given : ejector pipe can eject water at the rate 4L/MIN higher than the filler pipe . The filler pipe can fill the tank 5 min later than the ejector pipe can eject To find : flow rate of the ejector pipe in L/min(approx.). solution:- ejector pipe flow rate = E L/Min Filler Pipe flow rate = E - 4 L/Min Capacity of Tank = 6000 L Time Taken by Ejector pipe = 6000/E Min Time Taken by filler pipe = 6000/(E - 4) Min The filler pipe can fill the tank 5 min later than the ejector pipe can eject. => 6000/(E - 4) = 6000/E + 5 => 6000E = 6000(E - 4) + 5E(E - 4) => 6000E = 6000E - 24000 + 5(E² - 4E) => 5(E² - 4E) - 24000 = 0 => E² - 4E - 4800 = 0 => E = (4 ± √(4² - 4(-4800)) ) /2 => E = 2 + √4801 ( ignoring - ve value) => E = 2 + 69.3 => E = 71.3 flow rate of the ejector pipe = 71.3 L/min I hope* it will HELP you to understand ❤

26.

A and B are two pipes connected to a tank. A is the filler pipe and B is the ejector. The capacity of the tank is 6000L.The ejector pipe can eject water at the rate 4L/min higher than the filler pipe. The filler pipe can fill the tank 5 min later than the ejector pipe can eject. Find the flow rate of the ejector pipe in L/min(approx.).

Answer»

Step-by-step explanation:

Given : ejector pipe can eject water at the rate 4L/MIN higher than the filler pipe . The filler pipe can fill the tank 5 min later than the ejector pipe can eject

To find : flow rate of the ejector pipe in L/min(approx.).

solution:-

ejector pipe flow rate = E L/Min

Filler Pipe flow rate = E - 4 L/Min

Capacity of Tank = 6000 L

Time Taken by Ejector pipe = 6000/E Min

Time Taken by filler pipe = 6000/(E - 4) Min

The filler pipe can fill the tank 5 min later than the ejector pipe can eject.

=> 6000/(E - 4) = 6000/E + 5

=> 6000E = 6000(E - 4) + 5E(E - 4)

=> 6000E = 6000E - 24000 + 5(E² - 4E)

=> 5(E² - 4E) - 24000 = 0

=> E² - 4E - 4800 = 0

=> E = (4 ± √(4² - 4*(-4800)) ) /2

=> E = 2 + √4801 ( ignoring - ve value)

=> E = 2 + 69.3

=> E = 71.3

flow rate of the ejector pipe = 71.3 L/min

I hope it will HELP you to understand ❤

Discussion

27.	25 me kya ghataye ki anasar 50ho jaiya
Answer» Answer: 25-(-25) =50 its EASY.....

27.

25 me kya ghataye ki anasar 50ho jaiya

Answer»

Answer:

25-(-25) =50

its EASY.....

Discussion

28.	Take a=5,b=-3 and c=4, verify that a×b=b×a
Answer» Answer: 5-3=-35 =-15=-15 ................

28.

Take a=5,b=-3 and c=4, verify that a×b=b×a

Answer»

Answer:

5*-3=-3*5

=-15=-15

................

Discussion

29.	Simply (3√3+2√2)(2√3+3√2)
Answer» ANSWER: (3√3+2√2)(2√3+3√2) = 3√3(2√3+3√2)+2√2(2√3+3√2) = 6×6+9√6 + 4√6+6×2 = 36+13√6+12 = 48+13√6

29.

Simply (3√3+2√2)(2√3+3√2)

Answer»

ANSWER: (3√3+2√2)(2√3+3√2)

= 3√3(2√3+3√2)+2√2(2√3+3√2)

= 6×6+9√6 + 4√6+6×2

= 36+13√6+12

= 48+13√6

Discussion

30.	What is the First qaurtile for numbers 15,18,10,20,23,28,12,16 is?
Answer» Answer: 15 is the FIRST LETTER ACCORDING to the REASONING

30.

What is the First qaurtile for numbers 15,18,10,20,23,28,12,16 is?

Answer»

Answer:

15 is the FIRST LETTER ACCORDING to the REASONING

Discussion

31.	A dealer X at hapur(UP) supplies goods woth rs. 50000 to some other dealer Y in the same city. Now dealer Y supplies the same goods to dealer Z in calcutta at profit of rs. 20000 find:a) the output and inout taxes for the dealer Yb) the net gst payable by dealer Y[ the rate of GST at each stage is 28 oercent]
Answer» Given : A dealer X at hapur(UP) supplies goods woth rs. 50000 to some other dealer Y in the same CITY. Now dealer Y supplies the same goods to dealer Z in calcutta at profit of rs. 2000 . [ the rate of GST at each stage is 28 oercent] To FIND : a) the output and inout taxes for the dealer Y b) the net gst payable by dealer Y Solution: A dealer X at hapur(UP) supplies goods woth rs. 50000 to some other dealer Y in the same city. GST charged = 28 % = (28/100)50000 = Rs 14000 CGST = SGST = Rs 14000/2 = Rs 7000 dealer Y pays GST = Rs 14000 to dealer X Y supplies the same goods to dealer Z in calcutta at profit of rs. 20000 Selling price = 50000 + 20000 = Rs 70000 GST charged = 28 % = (28/100)70000 = Rs 19600 IGST = Rs 19600 Output taxes for dealer Y = Rs 19600 Input taxes for dealer Y = Rs 14000 the net GST payable by dealer Y = 19600 - 14000 = Rs 5600 Learn more: A manufacturer buys raw MATERIAL worth 12000 rupees paying GST 5 ... brainly.in/question/8759742 Anil buys a bicycle whose list price is rupees 1800 with additional ... brainly.in/question/10699098

31.

A dealer X at hapur(UP) supplies goods woth rs. 50000 to some other dealer Y in the same city. Now dealer Y supplies the same goods to dealer Z in calcutta at profit of rs. 20000 find:a) the output and inout taxes for the dealer Yb) the net gst payable by dealer Y[ the rate of GST at each stage is 28 oercent]

Answer»

Given : A dealer X at hapur(UP) supplies goods woth rs. 50000 to some other dealer Y in the same CITY. Now dealer Y supplies the same goods to dealer Z in calcutta at profit of rs. 2000 . [ the rate of GST at each stage is 28 oercent]

To FIND : a) the output and inout taxes for the dealer Y

b) the net gst payable by dealer Y

Solution:

A dealer X at hapur(UP) supplies goods woth rs. 50000 to some other dealer Y in the same city.

GST charged = 28 % = (28/100)50000 = Rs 14000

CGST = SGST = Rs 14000/2 = Rs 7000

dealer Y pays GST = Rs 14000 to dealer X

Y supplies the same goods to dealer Z in calcutta at profit of rs. 20000

Selling price = 50000 + 20000 = Rs 70000

GST charged = 28 % = (28/100)70000 = Rs 19600

IGST = Rs 19600

Output taxes for dealer Y = Rs 19600

Input taxes for dealer Y = Rs 14000

the net GST payable by dealer Y = 19600 - 14000 = Rs 5600

Learn more:

A manufacturer buys raw MATERIAL worth 12000 rupees paying GST 5 ...

brainly.in/question/8759742

Anil buys a bicycle whose list price is rupees 1800 with additional ...

brainly.in/question/10699098

Discussion

32.	2x2 +7x/2+3/4 splitting middle term
Answer» ANSWER: PLS MARK it as BRAINLEST answer

32.

2x2 +7x/2+3/4 splitting middle term

Answer»

ANSWER:

PLS MARK it as BRAINLEST answer

Discussion

33.	If a car starts to move from point A at 7:30am and reaches to point B at 1:15pm covering the distance of 1200km; calculate the speed of the car.
Answer» ANSWER: I dont KNOW the answer ASK someone ELSE

33.

If a car starts to move from point A at 7:30am and reaches to point B at 1:15pm covering the distance of 1200km; calculate the speed of the car.

Answer»

ANSWER:

I dont KNOW the answer ASK someone ELSE

Discussion

34.	Find the hcf of 45and 120
Answer» Answer: hii Step-by-step explanation: done with LONG division METHOD the hcf (highest COMMON factor ) is 15 hope it helps MARK as BRAINLIEST

34.

Find the hcf of 45and 120

Answer»

Answer:

hii

Step-by-step explanation:

done with LONG division METHOD

the hcf (highest COMMON factor )

is 15

hope it helps

MARK as BRAINLIEST

Discussion

35.	Find the product of 12,0.6and61/4
Answer» ANSWER: the answer is 109.8 ......

35.

Find the product of 12,0.6and61/4

Answer»

ANSWER:

the answer is 109.8 ......

Discussion

36.	Can of solution is 88cm and hight 14cm and radius of base is 2cm
Answer» ANSWER: what we have to FIND in this QUESTION the AREA

36.

Can of solution is 88cm and hight 14cm and radius of base is 2cm

Answer»

ANSWER:

what we have to FIND in this QUESTION the AREA

Discussion

37.	In order to make a shirt , a length of 25 CM of cloth is needed . how much length of the cloth will be required for 16 such shirt?
Answer» Answer: Given, Lenght required for 1 SKIRT= 25 CM Lenght required for 16 skirts= 16×25= 400 ( I hope that it is helpful. Do SUBSCRIBE to me on YouTube at this CHANNEL: Jamini The SECOND)

37.

In order to make a shirt , a length of 25 CM of cloth is needed . how much length of the cloth will be required for 16 such shirt?

Answer»

Answer:

Given,

Lenght required for 1 SKIRT= 25 CM

Lenght required for 16 skirts= 16×25= 400

( I hope that it is helpful. Do SUBSCRIBE to me on YouTube at this CHANNEL: Jamini The SECOND)

Discussion

38.	Given log10 a = b, express 102b–3 in terms of a.
Answer» Step-by-step EXPLANATION: I HOPE that above PROCEDURE HELPS U

38.

Given log10 a = b, express 102b–3 in terms of a.

Answer»

Step-by-step EXPLANATION:

I HOPE that above PROCEDURE HELPS U

Discussion

39.	Find three rational numbers between-20/50 -10/50
Answer» Answer: THREE rational NUMBERS between $\frac{20}{50}$ and $\frac{-10}{50}$ are $\frac{-1}{50}, \frac{-2}{50} and \frac{-3}{50}$ . (There are a TOTAL of 29 numbers between $\frac{20}{50}$ and $\frac{-10}{50}$ , including 0) #bal #answerwithquality #markmeasbrainliest

39.

Find three rational numbers between-20/50 -10/50

Answer»

Answer:

THREE rational NUMBERS between $\frac{20}{50}$ and $\frac{-10}{50}$ are $\frac{-1}{50}, \frac{-2}{50} and \frac{-3}{50}$ .

(There are a TOTAL of 29 numbers between $\frac{20}{50}$ and $\frac{-10}{50}$ , including 0)

#bal #answerwithquality #markmeasbrainliest

Discussion

40.	Pls answer this fast no wrong answers are taken as joke and will be reported
Answer» Answer: Hello: Step-by-step explanation: Answer: a. 1/2 --- this RATIONAL NUMBER lies between 0 and 1, in the middle position. See figure 1. b. 5/2 --- this rational number lies between 2 and 3, in the middle position. See figure 2. c. 1/3 --- this rational number lies between 0 and 1. See figure 3. d. 4/5 --- this rational number lies between 0 and 1. use the number line to work out -4×3

40.

Pls answer this fast no wrong answers are taken as joke and will be reported

Answer»

Answer:

Hello:

Step-by-step explanation:

Answer:

a. 1/2 --- this RATIONAL NUMBER lies between 0 and 1, in the middle position. See figure 1.

b. 5/2 --- this rational number lies between 2 and 3, in the middle position. See figure 2.

c. 1/3 --- this rational number lies between 0 and 1. See figure 3.

d. 4/5 --- this rational number lies between 0 and 1.

use the number line to work out -4×3

Discussion

41.	A man travels 370 km partly by train and partly by car. If he covers 250 km bytrain and rest by car, if takes him 4 hours. But, if he travels 130 km by train and the rest bycar, he takes 18 minutes longer. Find the speed of the train and that of the car.
Answer» $\huge\underline\mathtt \green{SOLUTION:-}$ $\underline{\star\:\large{\textit{1st Case:}}}$ $:\implies\tt \dfrac{Distance}{Train\:Speed}+\dfrac{Distance}{Car\:Speed}=Time\\\\\\:\implies\tt \dfrac{250}{Train} + \dfrac{(370 - 250)}{Car} = 4\\\\\\:\implies\tt\dfrac{250}{Train} + \dfrac{120}{Car} = 4 \qquad {\sf\dfrac{\quad}{}\:eq.(1)}$ $\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\thicklines\put(1,1){\line(1,0){6.5}}\put(1,1.1){\line(1,0){6.5}}\end{picture}$ $\underline{\star\:\large{\textit{2nd Case:}}}$ $:\implies\tt \dfrac{Distance}{Train\:Speed}+\dfrac{Distance}{Car\:Speed}=Time\\\\\\:\implies\tt \dfrac{130}{Train} + \dfrac{(370 - 130)}{Car} = 4hr \:18min.\\\\\\:\implies\tt\dfrac{130}{Train} + \dfrac{240}{Car} = 4{}^{18}\! /{}_{60}\\\\\\:\implies\tt \dfrac{130}{Train} + \dfrac{240}{Car} = \dfrac{43}{10} \qquad {\sf\dfrac{\quad}{}\:eq.(2)}$ $\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\thicklines\put(1,1){\line(1,0){6.5}}\put(1,1.1){\line(1,0){6.5}}\end{picture}$ $\sf Let\:\:\frac{1}{Train} = x\:\:and\:\:\frac{1}{Car}=y$ ☯ $\underline{\textsf{Multiplying eq.(3) with 2 :}}$ $\longrightarrow\sf 250x +120y = 4\qquad\dfrac{\quad}{}\:eq.(3) \times 2\\\\\longrightarrow\sf 130x +240y = \dfrac{43}{10}\qquad\dfrac{\quad}{} \:eq.(4)$ $\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\thicklines\put(1,1){\line(1,0){6.5}}\put(1,1.1){\line(1,0){6.5}}\end{picture}$ ☯ $\underline{\textsf{Subtracting eq.(4) from eq.(3) :}}$ $\dashrightarrow\tt\:\:500x + 240y = 8\\\\\dashrightarrow\tt\:\:130x + 240y = \dfrac{43}{10} \\ \dfrac{\qquad \qquad \qquad \qquad \qquad \quad}{}\\\dashrightarrow\tt\:\:(500x - 130x) = \bigg(8 - \dfrac{43}{10} \bigg)\\\\\\\dashrightarrow\tt\:\:370x = \dfrac{(80 - 43)}{10}\\\\\\\dashrightarrow\tt\:\:370x = \dfrac{37}{10}\\\\\\\dashrightarrow\tt\:\:x = \dfrac{37}{3700}\\\\\\\dashrightarrow\:\:\underline{\boxed{\red{\tt x =\dfrac{1}{100} = \dfrac{1}{Train}}}}$ $\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\thicklines\put(1,1){\line(1,0){6.5}}\put(1,1.1){\line(1,0){6.5}}\end{picture}$ ☯ $\underline{\textsf{Putting value of x in eq.(3) :}}$ $\dashrightarrow\tt\:\:250x+120y=4\\\\\\\dashrightarrow\tt\:\:250 \times \dfrac{1}{100} + 120y = 4\\\\\\\dashrightarrow\tt\:\:\dfrac{5}{2} + 120y = 4\\\\\\\dashrightarrow\tt\:\:120y = 4 - \dfrac{5}{2}\\\\\\\dashrightarrow\tt\:\:120y =\dfrac{(8 - 5)}{2}\\\\\\\dashrightarrow\tt\:\:120y = \dfrac{3}{2}\\\\\\\dashrightarrow\tt\:\:y = \dfrac{3}{240}\\\\\\\dashrightarrow\:\:\underline{\boxed{\red{\tt y = \dfrac{1}{80} = \dfrac{1}{Car}}}}$ ⠀ $\begin{cases}\textsf{Speed of Train = \textbf{100 Km/hr}}\\\textsf{Speed of Car = \textbf{80 Km/hr}}\end{cases}$

41.

A man travels 370 km partly by train and partly by car. If he covers 250 km bytrain and rest by car, if takes him 4 hours. But, if he travels 130 km by train and the rest bycar, he takes 18 minutes longer. Find the speed of the train and that of the car.

Answer»

$\huge\underline\mathtt \green{SOLUTION:-}$

$\underline{\star\:\large{\textit{1st Case:}}}$

$:\implies\tt \dfrac{Distance}{Train\:Speed}+\dfrac{Distance}{Car\:Speed}=Time\\\\\\:\implies\tt \dfrac{250}{Train} + \dfrac{(370 - 250)}{Car} = 4\\\\\\:\implies\tt\dfrac{250}{Train} + \dfrac{120}{Car} = 4 \qquad {\sf\dfrac{\quad}{}\:eq.(1)}$

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\thicklines\put(1,1){\line(1,0){6.5}}\put(1,1.1){\line(1,0){6.5}}\end{picture}$

$\underline{\star\:\large{\textit{2nd Case:}}}$

$:\implies\tt \dfrac{Distance}{Train\:Speed}+\dfrac{Distance}{Car\:Speed}=Time\\\\\\:\implies\tt \dfrac{130}{Train} + \dfrac{(370 - 130)}{Car} = 4hr \:18min.\\\\\\:\implies\tt\dfrac{130}{Train} + \dfrac{240}{Car} = 4{}^{18}\! /{}_{60}\\\\\\:\implies\tt \dfrac{130}{Train} + \dfrac{240}{Car} = \dfrac{43}{10} \qquad {\sf\dfrac{\quad}{}\:eq.(2)}$

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\thicklines\put(1,1){\line(1,0){6.5}}\put(1,1.1){\line(1,0){6.5}}\end{picture}$

$\sf Let\:\:\frac{1}{Train} = x\:\:and\:\:\frac{1}{Car}=y$

☯ $\underline{\textsf{Multiplying eq.(3) with 2 :}}$

$\longrightarrow\sf 250x +120y = 4\qquad\dfrac{\quad}{}\:eq.(3) \times 2\\\\\longrightarrow\sf 130x +240y = \dfrac{43}{10}\qquad\dfrac{\quad}{} \:eq.(4)$

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\thicklines\put(1,1){\line(1,0){6.5}}\put(1,1.1){\line(1,0){6.5}}\end{picture}$

☯ $\underline{\textsf{Subtracting eq.(4) from eq.(3) :}}$

$\dashrightarrow\tt\:\:500x + 240y = 8\\\\\dashrightarrow\tt\:\:130x + 240y = \dfrac{43}{10} \\ \dfrac{\qquad \qquad \qquad \qquad \qquad \quad}{}\\\dashrightarrow\tt\:\:(500x - 130x) = \bigg(8 - \dfrac{43}{10} \bigg)\\\\\\\dashrightarrow\tt\:\:370x = \dfrac{(80 - 43)}{10}\\\\\\\dashrightarrow\tt\:\:370x = \dfrac{37}{10}\\\\\\\dashrightarrow\tt\:\:x = \dfrac{37}{3700}\\\\\\\dashrightarrow\:\:\underline{\boxed{\red{\tt x =\dfrac{1}{100} = \dfrac{1}{Train}}}}$

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\thicklines\put(1,1){\line(1,0){6.5}}\put(1,1.1){\line(1,0){6.5}}\end{picture}$

☯ $\underline{\textsf{Putting value of x in eq.(3) :}}$

$\dashrightarrow\tt\:\:250x+120y=4\\\\\\\dashrightarrow\tt\:\:250 \times \dfrac{1}{100} + 120y = 4\\\\\\\dashrightarrow\tt\:\:\dfrac{5}{2} + 120y = 4\\\\\\\dashrightarrow\tt\:\:120y = 4 - \dfrac{5}{2}\\\\\\\dashrightarrow\tt\:\:120y =\dfrac{(8 - 5)}{2}\\\\\\\dashrightarrow\tt\:\:120y = \dfrac{3}{2}\\\\\\\dashrightarrow\tt\:\:y = \dfrac{3}{240}\\\\\\\dashrightarrow\:\:\underline{\boxed{\red{\tt y = \dfrac{1}{80} = \dfrac{1}{Car}}}}$

⠀

$\begin{cases}\textsf{Speed of Train = \textbf{100 Km/hr}}\\\textsf{Speed of Car = \textbf{80 Km/hr}}\end{cases}$

Discussion

42.	Sum of two rational numbers is 3 upon 5 if 1 of them is -2.7 find the other one method
Answer» ANSWER: The other NUMBER is 3.3 Explanation: LET the other number be x x+(-2.7)=3/5 x-2.7=3/5 x=3/5+2.7 MULTIPLYING 2.7 WITH 5, x=3+13.5/5 x=16.5/5 THEREFORE, x=3.3

42.

Sum of two rational numbers is 3 upon 5 if 1 of them is -2.7 find the other one method

Answer»

ANSWER: The other NUMBER is 3.3

Explanation:

LET the other number be x

x+(-2.7)=3/5

x-2.7=3/5

x=3/5+2.7

MULTIPLYING 2.7 WITH 5,

x=3+13.5/5

x=16.5/5

THEREFORE, x=3.3

Discussion

43.	former cloth 5 metre long. a piece of length 2 metre 85 cm is cut off .what is the length of the remaining piece
Answer» ANSWER: Hello: Step-by-step explanation: Here is your answer ❤️❤️☺️☺️ Total length of CLOTH = 5 m or 500 cm Cloth that cut off = 2m 85 cm or 285 cm Remaining cloth = 500 -285 cm 215 cm. Hope this HELPS you ☺️☺️❤️❤️ Mark me as brainlist⚠️⚠️✨✨

43.

former cloth 5 metre long. a piece of length 2 metre 85 cm is cut off .what is the length of the remaining piece

Answer»

ANSWER:

Hello:

Step-by-step explanation:

Here is your answer ❤️❤️☺️☺️

Total length of CLOTH = 5 m

or 500 cm

Cloth that cut off = 2m 85 cm

or 285 cm

Remaining cloth = 500 -285 cm

215 cm.

Hope this HELPS you ☺️☺️❤️❤️

Mark me as brainlist⚠️⚠️✨✨

Discussion

44.	Consider the following expression and determine which statements are true.x2+5yz-8Choose 2 answers:A. There are 3 terms.B. The variables are x,y, and z.C. The ceofficiant of x is 2.D. The term 5yz is made up of 2 factors.
Answer» Step-by-step EXPLANATION: A,B options are TRUE hope you will like it

44.

Consider the following expression and determine which statements are true.x2+5yz-8Choose 2 answers:A. There are 3 terms.B. The variables are x,y, and z.C. The ceofficiant of x is 2.D. The term 5yz is made up of 2 factors.

Answer»

Step-by-step EXPLANATION:

A,B options are TRUE

hope you will like it

Discussion

45.	the mass of a cylinder filled with gas is 30 kg 250 gram and the mass of empty cylinder is 14 kg 480 gram . how much is the mass of the gas contained in it
Answer» ANSWER: hhhuufgyuiiuufifjgjfjif

45.

the mass of a cylinder filled with gas is 30 kg 250 gram and the mass of empty cylinder is 14 kg 480 gram . how much is the mass of the gas contained in it

Answer»

ANSWER:

hhhuufgyuiiuufifjgjfjif

Discussion

46.	the length of the minute hand of a clock is 14cm then the area swept by the minute hand is 5 minutes is
Answer» ANSWER: HOPE it MAY HELP U please mark it as a brainlist answer

46.

the length of the minute hand of a clock is 14cm then the area swept by the minute hand is 5 minutes is

Answer»

ANSWER:

HOPE it MAY HELP U

please mark it as a brainlist answer

Discussion

47.	The perimeter of a rectangular field is 95m if the length of the field is 32.5 m find the breadth of the field.
Answer» Answer: HOPE it's clear Step-by-step EXPLANATION: please MARK breanlieat

47.

The perimeter of a rectangular field is 95m if the length of the field is 32.5 m find the breadth of the field.

Answer»

Answer:

HOPE it's clear

Step-by-step EXPLANATION:

please MARK breanlieat

Discussion

48.	An aeroplane covers 1485 kilometre in 1 hour how much distance will it covers in 72 hours ?
Answer» Answer: 106920 kms Step-by-step EXPLANATION: It will COVER the DISTANCE which is=1485×72=106920 KILOMETRES

48.

An aeroplane covers 1485 kilometre in 1 hour how much distance will it covers in 72 hours ?

Answer»

Answer:

106920 kms

Step-by-step EXPLANATION:

It will COVER the DISTANCE which is=1485×72=106920 KILOMETRES

Discussion

49.	Divide p(x) = 2x^3 - 7x^2 + 5x - 8 by (x+2) and verify your remainder by the remainder theorem
Answer» Answer: x+2=0 x = -2 p(x) = 2x^3 - 7x^2 + 5x - 8 = 2(-2)^3-7(-2)^2+5(-2)-8 = 2(-8)-7(4)+(-10)-8 = -16-28-10-8 = -62 is the Remainder.

49.

Divide p(x) = 2x^3 - 7x^2 + 5x - 8 by (x+2) and verify your remainder by the remainder theorem

Answer»

Answer: x+2=0

x = -2

p(x) = 2x^3 - 7x^2 + 5x - 8

= 2(-2)^3-7(-2)^2+5(-2)-8

= 2(-8)-7(4)+(-10)-8

= -16-28-10-8

= -62 is the Remainder.

Discussion

50.	6) find the 10 rational number between-3/4&5/6
Answer» ANSWER: The rational number between -3\4&5\ 6 r. Firstly w find the LCM of 4 and 6 and the LCM is 12 . And then -9\12&10\12. Then -9\12×10\10= -90\120

50.

6) find the 10 rational number between-3/4&5/6

Answer»

ANSWER:

The rational number between -3\4&5\ 6 r. Firstly w find the LCM of 4 and 6 and the LCM is 12 . And then -9\12&10\12. Then -9\12×10\10= -90\120

Discussion

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