81279 + Interview Questions in Mathematics QA in Mathematics Page 1439 InterviewSolution

71901.	2. Teena borrowed Rs 5000 from Meena at 8% per annum interest. She returned the morafter 146 days. What interest did she pay? Find the amount she paid.
Answer» 8% per anum = 8% for 365 daysthen for 146 days,365days---------8%146days---------? (1468)/365 = 3.2% is the interest she paid then total amaount paid after 146 days is5000+(50003.2)/100 = 5000 + 160 = Rs 5160

71901.

2. Teena borrowed Rs 5000 from Meena at 8% per annum interest. She returned the morafter 146 days. What interest did she pay? Find the amount she paid.

Answer»

8% per anum = 8% for 365 daysthen for 146 days,365days---------8%146days---------?

(146*8)/365 = 3.2% is the interest she paid

then total amaount paid after 146 days is5000+(5000*3.2)/100 = 5000 + 160 = Rs 5160

Discussion

71902.	Ex. 29: A sum of money doubles itself in 4 yrs at a simple interest. Inhow many yrs will it amount to 8 times itself?
Answer» Principal amount = P, Rate of Interest = R, Time period =T, Simple Interest = SI As per given conditionSI = PRT/100P = PR4/100 R = 25% Amount will be 8 times, then SI = 7P7P = P25T/100T = 7*4T = 28 Therefore, in 28 years amount will become 8 times of itself

71902.

Ex. 29: A sum of money doubles itself in 4 yrs at a simple interest. Inhow many yrs will it amount to 8 times itself?

Answer»

Principal amount = P, Rate of Interest = R, Time period =T, Simple Interest = SI

As per given conditionSI = P*R*T/100P = P*R*4/100

R = 25%

Amount will be 8 times, then SI = 7P7P = P*25*T/100T = 7*4T = 28

Therefore, in 28 years amount will become 8 times of itself

Discussion

71903.	A and B are friends and B is 2 years youngerthan A. A's father D is twice as old as A andB is twice as his sister C. The age of D andC differ by 40 years. The ages of 'A' and 'B'are respectively(A) 26 yrs., 24 yrs. (B) 25 yrs., 23 yrs.(C) 30 yrs., 28 yrs. (D) of these
Answer» Solution By the given conditionA-B=2 (i)Now as D is twice of 'A'D's age = 2A yearsand B is twice of his sister CC's age = B/2 yearsSo D is older than C2A-(B/2)=40 = 4A-B=80 (ii)solve (i) and (ii) eq.A=26B= 24

71903.

A and B are friends and B is 2 years youngerthan A. A's father D is twice as old as A andB is twice as his sister C. The age of D andC differ by 40 years. The ages of 'A' and 'B'are respectively(A) 26 yrs., 24 yrs. (B) 25 yrs., 23 yrs.(C) 30 yrs., 28 yrs. (D) of these

Answer»

Solution

By the given conditionA-B=2 (i)Now as D is twice of 'A'D's age = 2A yearsand B is twice of his sister CC's age = B/2 yearsSo D is older than C2A-(B/2)=40 = 4A-B=80 (ii)solve (i) and (ii) eq.A=26B= 24

Discussion

71904.	A person deposited a certain sum in a bank oncompound interest. The sum becomes * 900in 3 yrs and 1000 in 4 yrs. the rate of interestis
Answer»

Discussion

71905.	10.Prove that one of every three consecutive positive integers is divisible by 3.
Answer» Ans :- Let three consecutive positive integers be n, n + 1 and n + 2 Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2 ∴ n = 3p or 3p + 1 or 3p + 2, where p is some integer. If n = 3p, then n is divisible by 3. If n = 3p + 1, then n + 2 = 3p + 1 + 2 = 3p + 3 = 3(p + 1) is divisible by 3. If n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3 So, we can say that one of the numbers among n, n + 1 and n + 2 is always divisible by 3. thanks

71905.

10.Prove that one of every three consecutive positive integers is divisible by 3.

Answer»

Ans :- Let three consecutive positive integers be n, n + 1 and n + 2

Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2

∴ n = 3p or 3p + 1 or 3p + 2, where p is some integer.

If n = 3p, then n is divisible by 3.

If n = 3p + 1, then n + 2 = 3p + 1 + 2 = 3p + 3 = 3(p + 1) is divisible by 3.

If n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3

So, we can say that one of the numbers among n, n + 1 and n + 2 is always divisible by 3.

thanks

Discussion

71906.	Fihd ulc iuu4. Find the ratio of the circumferences of two circles whose radii are in the ratio 4: 5.
Answer»

Discussion

71907.	5(y^4z^3) - 146y^2z^4
Answer»

Discussion

71908.	By selling goods for 153, a man loses 10%. For how much should hesell them to gain 20%?
Answer» Selling at Rs 153 gives→10%loss ⟹cost = 90°of 153 cost price=153×100/90=170 Gaining20%⟹170×120/100=204

71908.

By selling goods for 153, a man loses 10%. For how much should hesell them to gain 20%?

Answer»

Selling at Rs 153 gives→10%loss

⟹cost = 90°of 153

cost price=153×100/90=170

Gaining20%⟹170×120/100=204

Discussion

71909.	2. In her game on Wednesday, Siomara scored 32points. On Saturday, Siomara made 6 two-pointfield goals, 4 three-point field goals, and 8 freethrows that are worth one point each. How manypoints did Siomara score in the two gamesaltogether?
Answer» she scored 50 points 6+8+4=50

Discussion

71910.	B)Solve any two of the following sub-questionsâĄABCD is a parallelogram. The ratio of04A and < B is 5 : 4, Find their measures.
Answer»

Discussion

71911.	2-Divède1000amoamong2:3-5
Answer» 1st : 2nd : 3rd= 2:3:5sum of ratio= 2+3+5=101st = 1000×2/10= 2002nd= 1000×3/10=3003rd= 1000×5/10= 500 1st person 200 টাকা ,2nd 300 টাকা and 3rd 500টাকা পাবে 2x+3x+5x=100010x=1000x=1000/10x=100now 2x=2x100=2003x=3×100=3004x=4×100=400 1000 among 2:3:5 2 + 3 + 5 = 10 So shares = (2/10)×1000 , (3/10)×1000 , (5/10)×1000 = 200, 300, 500 500 is the correct answer as solution is given to you

71911.

2-Divède1000amoamong2:3-5

Answer»

1st : 2nd : 3rd= 2:3:5sum of ratio= 2+3+5=101st = 1000×2/10= 2002nd= 1000×3/10=3003rd= 1000×5/10= 500

1st person 200 টাকা ,2nd 300 টাকা and 3rd 500টাকা পাবে

2x+3x+5x=100010x=1000x=1000/10x=100now 2x=2x100=2003x=3×100=3004x=4×100=400

1000 among 2:3:5

2 + 3 + 5 = 10 So shares = (2/10)×1000 , (3/10)×1000 , (5/10)×1000

= 200, 300, 500

500 is the correct answer as solution is given to you

Discussion

71912.	3. Three consecutive positive integers are such that the sum of the square of tfirst and the product of other two is 46, find the integers.
Answer»

Discussion

71913.	Prove that the product of three consecutive positive integers is divisible by 6
Answer»

Discussion

71914.	PART-AThe product of three consecutive positive integers is always divisible by(A) 6(B) 8(C) 9(D) 10
Answer»

Discussion

71915.	3. "The product of three consecutive positive integers is divisible by 6". Is this statement true orfalse"? Justify your answer.
Answer» I need small answer which can be easy to understand Let three consecutive integers be, n, n + 1 and n + 2. Whenever a number is divided by 3 the remainder obtained is either 0 or 1 or 2. let n = 3p or 3p + 1 or 3p + 2, where p is some integer. If n = 3p, then n is divisible by 3. If n = 3p + 1, then n + 2 = 3p + 1 + 2 = 3p + 3 = 3(p + 1) is divisible by 3. If n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3. So that n, n + 1 and n + 2 is always divisible by 3. ⇒ n (n + 1) (n + 2) is divisible by 3. Similarly, whenever a number is divided 2 we will get the remainder is 0 or 1. ∴ n = 2q or 2q + 1, where q is some integer. If n = 2q, then n and n + 2 = 2q + 2 = 2(q + 1) are divisible by 2. If n = 2q + 1, then n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2. So that n, n + 1 and n + 2 is always divisible by 2. ⇒ n (n + 1) (n + 2) is divisible by 2. But n (n + 1) (n + 2) is divisible by 2 and 3. ∴ n (n + 1) (n + 2) is divisible by 6. yes this is the true answer yes dude

71915.

3. "The product of three consecutive positive integers is divisible by 6". Is this statement true orfalse"? Justify your answer.

Answer»

I need small answer which can be easy to understand

Let three consecutive integers be, n, n + 1 and n + 2. Whenever a number is divided by 3 the remainder obtained is either 0 or 1 or 2.

let n = 3p or 3p + 1 or 3p + 2, where p is some integer.

If n = 3p, then n is divisible by 3.

If n = 3p + 1, then n + 2 = 3p + 1 + 2 = 3p + 3 = 3(p + 1) is divisible by 3.

If n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3.

So that n, n + 1 and n + 2 is always divisible by 3.

⇒ n (n + 1) (n + 2) is divisible by 3. Similarly, whenever a number is divided 2 we will get the remainder is 0 or 1.

∴ n = 2q or 2q + 1, where q is some integer.

If n = 2q, then n and n + 2 = 2q + 2 = 2(q + 1) are divisible by 2.

If n = 2q + 1, then n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2.

So that n, n + 1 and n + 2 is always divisible by 2.

⇒ n (n + 1) (n + 2) is divisible by 2.

But n (n + 1) (n + 2) is divisible by 2 and 3.

∴ n (n + 1) (n + 2) is divisible by 6.

yes this is the true answer

yes dude

Discussion

71916.	24) prove thatedochtan+A-OSATSIMAamo
Answer» LHS: tan(pi/4 + A) = tan pi/4 + tanA/ 1 - tan pi/4.tan A = 1 + tanA/ 1 - 1*tanA = 1 + sinA/cosA / 1 - sin A/cosA = (cos A + sin A) / (cos A - sin A) = RHS Hence proved

71916.

24) prove thatedochtan+A-OSATSIMAamo

Answer»

LHS: tan(pi/4 + A)

= tan pi/4 + tanA/ 1 - tan pi/4.tan A

= 1 + tanA/ 1 - 1*tanA

= 1 + sinA/cosA / 1 - sin A/cosA

= (cos A + sin A) / (cos A - sin A)

= RHS

Hence proved

Discussion

71917.	Rs 900at 7% per annum for a year compounded annually.
Answer» thank you

Discussion

71918.	TimeInterestPrincipalRate p.a.Rs. 250(ii) Rs. 900(ii) Rs. 7007%109012%4 yrs11months146 days
Answer» hit like if you find it useful

Discussion

71919.	By selling a chair for Rs 360, Aman loses 10%. At what price must he sell to gain 5%
Answer»

Discussion

71920.	ticket number 1 to 20 are mixed up then a ticket is drawn what is the probability that the ticket drawn has a number which multiple of 3 or 5
Answer»

Discussion

71921.	))Solve any four of the following sub-questions.If the class mark is 10 and class width is 6, then find the class.
Answer» Let lower term of class be X and upper term be Y class mark = X + Y/ 210 = X + Y/220 = X + Y .... (1) class width = Y - X 6 = Y - X .... (2) On subtracting equation (2) from (1) 20 - 6 = X + Y - Y + X14 = 2X X = 14/2 X = 7 = lower limit on putting X in equation (2) 6 = Y - X6 = Y - 7Y = 7 + 6 = 13 = upper limit Ans. Class = 7 - 13

71921.

))Solve any four of the following sub-questions.If the class mark is 10 and class width is 6, then find the class.

Answer»

Let lower term of class be X and upper term be Y

class mark = X + Y/ 210 = X + Y/220 = X + Y .... (1)

class width = Y - X 6 = Y - X .... (2)

On subtracting equation (2) from (1)

20 - 6 = X + Y - Y + X14 = 2X X = 14/2 X = 7 = lower limit

on putting X in equation (2)

6 = Y - X6 = Y - 7Y = 7 + 6 = 13 = upper limit

Ans. Class = 7 - 13

Discussion

71922.	4 What is a regular polygon?
Answer» A regular polygon is a polygon that is equiangular and equilateral. ok

71922.

4 What is a regular polygon?

Answer»

A regular polygon is a polygon that is equiangular and equilateral.

ok

Discussion

71923.	4. What is the measure of each interior angle of a regular hexagon?
Answer» To find the interior angle of anyregularpolygon, you can use the formula: 180(n-2)/n where n is the amount of sides in the regular polygon. A hexagon has 6 sides, so 180(6-2)/6=120.

71923.

4. What is the measure of each interior angle of a regular hexagon?

Answer»

To find the interior angle of anyregularpolygon, you can use the formula: 180(n-2)/n

where n is the amount of sides in the regular polygon.

A hexagon has 6 sides, so 180(6-2)/6=120.

Discussion

71924.	10. Find three consecutive positive integers whose product is equal to sixteen times theirsum.
Answer»

Discussion

71925.	27. Find three consecutive positive integers whoseproduct is equal to sixteen times their sum.OR
Answer» Let m,(m + 1),(m + 2) be the consecutive integers. As per given condition,m(m + 1)(m+2) = 16(m+m+1+m+2)(m²+2m)(m+1) = 16(3m+3)(m²+2m)(m+1) = 48(m+1)m² + 2m = 48m² + 2m - 48 = 0m² + 8m - 6m - 48 = 0m(m + 8) - 6(m + 8) = 0(m + 8)(m - 6) = 0 m = -8 OR. m = 6 (m) is positive integer, so m≠-8, thus m=6 THEREFORE, THE POSITIVE INTEGERS AREm = 6m + 1 = 6 + 1 = 7m + 2 = 6 + 2 = 8 i.e., 6,7,8 are the consecutive positive integers.

71925.

27. Find three consecutive positive integers whoseproduct is equal to sixteen times their sum.OR

Answer»

Let m,(m + 1),(m + 2) be the consecutive integers.

As per given condition,m(m + 1)(m+2) = 16(m+m+1+m+2)(m²+2m)(m+1) = 16(3m+3)(m²+2m)(m+1) = 48(m+1)m² + 2m = 48m² + 2m - 48 = 0m² + 8m - 6m - 48 = 0m(m + 8) - 6(m + 8) = 0(m + 8)(m - 6) = 0

m = -8 OR. m = 6

(m) is positive integer, so m≠-8, thus m=6

THEREFORE, THE POSITIVE INTEGERS AREm = 6m + 1 = 6 + 1 = 7m + 2 = 6 + 2 = 8

i.e., 6,7,8 are the consecutive positive integers.

Discussion

71926.	Fill in the blanks :(i) The greatest negative integer is .....
Answer» answers well, zero is neither positive nor negative. it is the integer that divides the number line into negative on the left and positive on the right. The largest negative integer is −1. The smallest positive integer is 1. The greatest negative Integer is the of the least positive Integer. the greatest negative integer is -1 the greatest negative integer is 1 the greatest negative integer is _1

71926.

Fill in the blanks :(i) The greatest negative integer is .....

Answer»

answers

well, zero is neither positive nor negative. it is the integer that divides the number line into negative on the left and positive on the right.

The largest negative integer is −1.

The smallest positive integer is 1.

The greatest negative Integer is the of the least positive Integer.

the greatest negative integer is -1

the greatest negative integer is 1

the greatest negative integer is _1

Discussion

71927.	NA3.Find three Conseldive positiveeven integers whose Sumisao.
Answer» Let the positive even number be x,x + 2, x + 4 A.T.Qx+x+2+x+4=903x+6=903x=90-6x=84/3x=28Now,x=28x+2=28+2=30x+4=28+4=32 28, 30, 32 is the correct answer

71927.

NA3.Find three Conseldive positiveeven integers whose Sumisao.

Answer»

Let the positive even number be x,x + 2, x + 4 A.T.Qx+x+2+x+4=903x+6=903x=90-6x=84/3x=28Now,x=28x+2=28+2=30x+4=28+4=32

28, 30, 32 is the correct answer

Discussion

71928.	29. Divide Rs.672 in the ratio 3:3916753 నిపుచిలో రెండు భాగమే
Answer» 5 + 3 = 8672 × (5/8) = 420672 × (3/8) = 252Thus, 420 and 252 are the two parts. thus 420 and 252 are the two parts the answer is 420 and 252

71928.

29. Divide Rs.672 in the ratio 3:3916753 నిపుచిలో రెండు భాగమే

Answer»

5 + 3 = 8672 × (5/8) = 420672 × (3/8) = 252Thus, 420 and 252 are the two parts.

thus 420 and 252 are the two parts

the answer is 420 and 252

Discussion

71929.	16. Divide Rs 380 between A. B, and C. so that B nay have Rs 30more than A, and C may have Rs,20 myre than AB.
Answer»

Discussion

71930.	Divide Rs 900 in the ratio of 2:4.
Answer» Divide Rs 900 in the ratio of 2:4.. Ask for details; Follow; Report. by Rajendragandhi369 1 minute ago... Divide Rs 900 in the ratio of 2:4.. Ask for details; Follow; Report. by Rajendragandhi369 1 minute ago. 2+4=6One share = 2/6 x 900 = 300Second share = 4/6 x 900 = 600 hope this will help you like my answer and MARK IT AS BEST ANSWER ratio 2:4lets the no. is 2x, 4x 2x+ 4x = 900 6x = 900 x =900/6 x = 150then no 2x = 2×150=300 4x = 4×150 = 600 2:42x,4x2x+4x=9006x=900x= 900/6x= 1502x= 2×150= 3004x= 4×150= 600 300 and 600 is the correct answer of the given question 2+4=6=2/6×900=300=2/6×900=600ans:-ratio=300,600

71930.

Divide Rs 900 in the ratio of 2:4.

Answer»

Divide Rs 900 in the ratio of 2:4.. Ask for details; Follow; Report. by Rajendragandhi369 1 minute ago...

Divide Rs 900 in the ratio of 2:4.. Ask for details; Follow; Report. by Rajendragandhi369 1 minute ago.

2+4=6One share = 2/6 x 900 = 300Second share = 4/6 x 900 = 600

hope this will help you like my answer and MARK IT AS BEST ANSWER

ratio 2:4lets the no. is 2x, 4x 2x+ 4x = 900 6x = 900 x =900/6 x = 150then no 2x = 2×150=300 4x = 4×150 = 600

2:42x,4x2x+4x=9006x=900x= 900/6x= 1502x= 2×150= 3004x= 4×150= 600

300 and 600 is the correct answer of the given question

2+4=6=2/6×900=300=2/6×900=600ans:-ratio=300,600

Discussion

71931.	Divide Rs150 in the ratio 2 : 3.
Answer»

Discussion

71932.	Divide Rs 840 in the ratio of 3:2:5.
Answer» numbers are 3x,2x,5xso 3x+2x+5x=10x=840so x=84so numbers are 252,168,420

Discussion

71933.	& Divide Rs 360 between A & B in ratio 7:8
Answer» A:B= 7:8total parts = 7+8= 15A's part = 7/15360= 247168.002Now B` s part = 8/15360= 248= 191.9

Discussion

71934.	Q.11 The class marks of the distribution are 37,42, 47, 52 and$) Determine the class sizeand the class limits of the last class mark.
Answer» Like if you find it useful

Discussion

71935.	\begin{array} { l } { \text { The class marks of distribution are: } } \\ { 6,10,14,18,22,26,30 } \\ { \text { Find the class size and the class interval. } } \end{array}
Answer»

Discussion

71936.	C. Form the antonyms of the following words by adding un, dis, in-, im-or ir1. like2. fit3. regular4. happy5. possible6. complete
Answer» dislikeunfitirregularunhappyimpossibleincomplete 1 dislike2 unfit3 irregular4 unhappy5 impossible6 incomplete

71936.

C. Form the antonyms of the following words by adding un, dis, in-, im-or ir1. like2. fit3. regular4. happy5. possible6. complete

Answer»

dislikeunfitirregularunhappyimpossibleincomplete

1 dislike2 unfit3 irregular4 unhappy5 impossible6 incomplete

Discussion

71937.	1%)The sum of two integers is +20. Dividing thelarger integer by the smaller integer gives aquotient of -3. What are the two integers?
Answer» we get 30 and - 10

Discussion

71938.	The sum of two llitbthe other.78.79. Sum of two integers is -80. If one of the integers is -90, thother.
Answer» thanks

Discussion

71939.	Find three consecutive positive even integers whosesum is 132.6.
Answer» Let the three consecutive even integers be n-2,n and n+2. We have n-2+n+n+2=132 3n=132n=44. The three consecutive even integers are 42,44 and 46.

71939.

Find three consecutive positive even integers whosesum is 132.6.

Answer»

Let the three consecutive even integers be n-2,n and n+2.

We have n-2+n+n+2=132

3n=132n=44.

The three consecutive even integers are 42,44 and 46.

Discussion

71940.	2. The greatest negative integer smaller than ainteger smaller than a whole number is
Answer» Zero is less than every positive integer, and greater than every negative integer. thnku

71940.

2. The greatest negative integer smaller than ainteger smaller than a whole number is

Answer»

Zero is less than every positive integer, and greater than every negative integer.

thnku

Discussion

71941.	Find three consecutive positive even integers whose sum is 90
Answer»

Discussion

71942.	6. Show that the line segments joining the mid-points of the opposite sides of aquadrilateral bisect each other.
Answer»

Discussion

71943.	Flg. 8.31Show that the line segments joining the mid-points of the opposite sides ofquadrilateral bisect each other.
Answer»

Discussion

71944.	divide Rs 60 in the ratio of 1:2 between kirti and kiran
Answer» 20,40 is the correct answer by me 1:2 is the correct answer of this question 20, 40 is the correct answer of the given question 20+40correct answer for this question let ratio=x and 2x3x=60x=20so,kirti:kiran=20:40 20:40 is the correct answer 30, 40 is the best answer let ratio =x and 2x3x=60x=20so kirti:Kiran =20:40

71944.

divide Rs 60 in the ratio of 1:2 between kirti and kiran

Answer»

20,40 is the correct answer by me

1:2 is the correct answer of this question

20, 40 is the correct answer of the given question

20+40correct answer for this question

let ratio=x and 2x3x=60x=20so,kirti:kiran=20:40

20:40 is the correct answer

30, 40 is the best answer

let ratio =x and 2x3x=60x=20so kirti:Kiran =20:40

Discussion

71945.	हक 0+ 5600l-§inf ——un—(6 ) 1+sin2tan (4 Tकक S T भू e
Answer» thank you thank u so much

71945.

हक 0+ 5600l-§inf ——un—(6 ) 1+sin2tan (4 Tकक S T भू e

Answer»

thank you

thank u so much

Discussion

71946.	Find the class size of the class interval 3.5-6.5? -
Answer» Class size will be6.5-3.5= 3thanks the class size will be =6.5-3.5=3

71946.

Find the class size of the class interval 3.5-6.5? -

Answer»

Class size will be6.5-3.5= 3thanks

the class size will be =6.5-3.5=3

Discussion

71947.	The class marks of distribution are:6,10,14,18,22,26,30Find the class size and the class interval
Answer»

Discussion

71948.	6. 11-63 का मान क्या होगा ?un
Answer» second option is correct option 2 is the correct answer of the given question option (2) is the right answer. (2) number hi sahi hai option 2 is the correct answer second option is correct [57/5]÷[19/3][57/5]×[3/19](57×3)/(5×19)171/95is correct answer

71948.

6. 11-63 का मान क्या होगा ?un

Answer»

second option is correct

option 2 is the correct answer of the given question

option (2) is the right answer.

(2) number hi sahi hai

option 2 is the correct answer

second option is correct

[57/5]÷[19/3][57/5]×[3/19](57×3)/(5×19)171/95is correct answer

Discussion

71949.	6. What should be added to(b) 0.2ond+3to get 4?45un
Answer» first add the three values i.e.(1/3 + 1/4 + 1/5)LCM = 60=(20/60 + 15/60 + 12/60)=47/60Now to get denominator of 4 to 60, multiply 4 by 60i.e. (4/1 × 60/60)=240/60Now subtract 47/60 from 240/60=(240/60) - (47/60)=193/60i.e. 3 whole 13 by 60 ( could not type in numerals) should be added to (1/3 + 1/4 + 1/5) to get 4. Hope it helps, please like it🤗😀

71949.

6. What should be added to(b) 0.2ond+3to get 4?45un

Answer»

first add the three values i.e.(1/3 + 1/4 + 1/5)LCM = 60=(20/60 + 15/60 + 12/60)=47/60Now to get denominator of 4 to 60, multiply 4 by 60i.e. (4/1 × 60/60)=240/60Now subtract 47/60 from 240/60=(240/60) - (47/60)=193/60i.e. 3 whole 13 by 60 ( could not type in numerals) should be added to (1/3 + 1/4 + 1/5) to get 4.

Hope it helps, please like it🤗😀

Discussion

71950.	6 An un contains 11 oranges, 8 mangoes and13 apples. A fruit is drawn at random.What is the probability of drawinga) an apple of a mango?6) not an slange
Answer» A)probability of drawing an apple or a mango =no. Of apple or mangoes /total no of fruits =21/32B)21/32 A) probability of drawing. an apple or a mango =no. of apple or mangoes / total no of fruits =21/32B) 21/32 answer

71950.

6 An un contains 11 oranges, 8 mangoes and13 apples. A fruit is drawn at random.What is the probability of drawinga) an apple of a mango?6) not an slange

Answer»

A)probability of drawing an apple or a mango =no. Of apple or mangoes /total no of fruits =21/32B)21/32

A) probability of drawing. an apple or a mango =no. of apple or mangoes / total no of fruits =21/32B) 21/32 answer

Discussion

Explore topic-wise InterviewSolutions in .

2. Teena borrowed Rs 5000 from Meena at 8% per annum interest. She returned the morafter 146 days. What interest did she pay? Find the amount she paid.

Ex. 29: A sum of money doubles itself in 4 yrs at a simple interest. Inhow many yrs will it amount to 8 times itself?

A and B are friends and B is 2 years youngerthan A. A's father D is twice as old as A andB is twice as his sister C. The age of D andC differ by 40 years. The ages of 'A' and 'B'are respectively(A) 26 yrs., 24 yrs. (B) 25 yrs., 23 yrs.(C) 30 yrs., 28 yrs. (D) of these

A person deposited a certain sum in a bank oncompound interest. The sum becomes * 900in 3 yrs and 1000 in 4 yrs. the rate of interestis

10.Prove that one of every three consecutive positive integers is divisible by 3.

Fihd ulc iuu4. Find the ratio of the circumferences of two circles whose radii are in the ratio 4: 5.

5*(y^4*z^3) - 146*y^2*z^4

By selling goods for 153, a man loses 10%. For how much should hesell them to gain 20%?

2. In her game on Wednesday, Siomara scored 32points. On Saturday, Siomara made 6 two-pointfield goals, 4 three-point field goals, and 8 freethrows that are worth one point each. How manypoints did Siomara score in the two gamesaltogether?

B)Solve any two of the following sub-questionsâĄABCD is a parallelogram. The ratio of04A and &lt; B is 5 : 4, Find their measures.

2-Divède1000amoamong2:3-5

3. Three consecutive positive integers are such that the sum of the square of tfirst and the product of other two is 46, find the integers.

Prove that the product of three consecutive positive integers is divisible by 6

PART-AThe product of three consecutive positive integers is always divisible by(A) 6(B) 8(C) 9(D) 10

3. "The product of three consecutive positive integers is divisible by 6". Is this statement true orfalse"? Justify your answer.

24) prove thatedochtan+A-OSATSIMAamo

Rs 900at 7% per annum for a year compounded annually.

TimeInterestPrincipalRate p.a.Rs. 250(ii) Rs. 900(ii) Rs. 7007%109012%4 yrs11months146 days

By selling a chair for Rs 360, Aman loses 10%. At what price must he sell to gain 5%

ticket number 1 to 20 are mixed up then a ticket is drawn what is the probability that the ticket drawn has a number which multiple of 3 or 5

))Solve any four of the following sub-questions.If the class mark is 10 and class width is 6, then find the class.

4 What is a regular polygon?

4. What is the measure of each interior angle of a regular hexagon?

10. Find three consecutive positive integers whose product is equal to sixteen times theirsum.

27. Find three consecutive positive integers whoseproduct is equal to sixteen times their sum.OR

Fill in the blanks :(i) The greatest negative integer is .....

NA3.Find three Conseldive positiveeven integers whose Sumisao.

29. Divide Rs.672 in the ratio 3:3916753 నిపుచిలో రెండు భాగమే

16. Divide Rs 380 between A. B, and C. so that B nay have Rs 30more than A, and C may have Rs,20 myre than AB.

Divide Rs 900 in the ratio of 2:4.

Divide Rs150 in the ratio 2 : 3.

Divide Rs 840 in the ratio of 3:2:5.

&amp; Divide Rs 360 between A &amp; B in ratio 7:8

Q.11 The class marks of the distribution are 37,42, 47, 52 and$) Determine the class sizeand the class limits of the last class mark.

\begin{array} { l } { \text { The class marks of distribution are: } } \\ { 6,10,14,18,22,26,30 } \\ { \text { Find the class size and the class interval. } } \end{array}

C. Form the antonyms of the following words by adding un, dis, in-, im-or ir1. like2. fit3. regular4. happy5. possible6. complete

1%)The sum of two integers is +20. Dividing thelarger integer by the smaller integer gives aquotient of -3. What are the two integers?

The sum of two llitbthe other.78.79. Sum of two integers is -80. If one of the integers is -90, thother.

Find three consecutive positive even integers whosesum is 132.6.

2. The greatest negative integer smaller than ainteger smaller than a whole number is

Find three consecutive positive even integers whose sum is 90

6. Show that the line segments joining the mid-points of the opposite sides of aquadrilateral bisect each other.

Flg. 8.31Show that the line segments joining the mid-points of the opposite sides ofquadrilateral bisect each other.

divide Rs 60 in the ratio of 1:2 between kirti and kiran

हक 0+ 5600l-§inf ——un—(6 ) 1+sin2tan (4 Tकक S T भू e

Find the class size of the class interval 3.5-6.5? -

The class marks of distribution are:6,10,14,18,22,26,30Find the class size and the class interval

6. 11-63 का मान क्या होगा ?un

6. What should be added to(b) 0.2ond+3to get 4?45un

6 An un contains 11 oranges, 8 mangoes and13 apples. A fruit is drawn at random.What is the probability of drawinga) an apple of a mango?6) not an slange

5(y^4z^3) - 146y^2z^4

B)Solve any two of the following sub-questionsâĄABCD is a parallelogram. The ratio of04A and < B is 5 : 4, Find their measures.

& Divide Rs 360 between A & B in ratio 7:8