In how many ways can the letters of the word “AFLATOON” be arranged if the consonants and vowels must occupy alternate places? -Maths 9th

In how many ways can the letters of the word “AFLATOON” be arranged if the consonants and vowels must occupy alternate places? -Maths 9th
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1 Answer

Answer :

24 ways is the answer
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Related questions

In how many different ways can the letters of the word "POMADE" be arranged in such a way that the vowels occupy only the odd positions? a) 72 b) 144 c) 532 d) 36

Description :  In how many different ways can the letters of the word "POMADE" be arranged in such a way that the vowels occupy only the odd positions? a) 72 b) 144 c) 532 d) 36

Last Answer : Answer: D)  There are 6 different letters in the given word, out of which there are 3 vowels and 3 consonants.  Let us mark these positions as under:  [1] [2] [3] [4] [5] [6]  Now, 3 vowels can be ... of these arrangements = 3P3  = 3!  = 6 ways.  Therefore, total number of ways = 6 x 6 = 36.

In how many different ways can the letters of the word "XANTHOUS" be arranged in such a way that the vowels occupy only the odd positions? a) 2880 b) 4320 c) 2140 d) 5420

Description : In how many different ways can the letters of the word "XANTHOUS" be arranged in such a way that the vowels occupy only the odd positions? a) 2880 b) 4320 c) 2140 d) 5420

Last Answer : Answer: A) There are 8 different letters in the given word "XANTHOUS", out of which there are 3 vowels and 5 consonants.  Let us mark these positions as under:  [1] [2] [3] [4] [5] [6] [7] ... in 5P5 ways = 5! Ways  = 120 ways.  Therefore, required number of ways = 24 x 120 = 2880 ways.

In how many different ways can the letters of the word 'RAPINE' be arranged in such a way that the vowels occupy only the odd positions? A) 32 B) 48 C) 36 D) 60 E) 120

Description : In how many different ways can the letters of the word 'RAPINE' be arranged in such a way that the vowels occupy only the odd positions? A) 32 B) 48 C) 36 D) 60 E) 120

Last Answer : Answer: C)  There are 6 letters in the given word, out of which there are 3 vowels and 3 consonants.  Let us mark these positions as under:  (1) (2) (3) (4) (5) (6)  Now, 3 vowels can be placed at ... .  Number of ways of these arrangements = 3P3 = 3! = 6.  Total number of ways = (6 x 6) = 36

In how many different ways can the letters of the word 'DILUTE' be arranged such that the vowels may appear in the even places? a) 36 b) 720 c) 144 d) 24

Description : In how many different ways can the letters of the word 'DILUTE' be arranged such that the vowels may appear in the even places? a) 36 b) 720 c) 144 d) 24

Last Answer : Answer: A)  There are 3 consonants and 3 vowels in the word DILUTE.  Out of 6 places, 3 places odd and 3 places are even.  3 vowels can arranged in 3 even places in 3p3 ways = 3! = 6 ways.  And then 3 ... 3 places in 3p3 ways = 3! = 6 ways.  Hence, the required number of ways = 6 x 6 = 36.

In how many ways can the letters of the word EDUCATION be rearranged such that the relative positions of the vowels and the consonants remain the same as in the word EDUCATION? a) 9!/4 b) 4!*5! c) 4!*5 d) 9!-4!

Description : In how many ways can the letters of the word EDUCATION be rearranged such that the relative positions of the vowels and the consonants remain the same as in the word EDUCATION? a) 9!/4 b) 4!*5! c) 4!*5 d) 9!-4!

Last Answer : b) 4!*5!

In how many different ways can the letters of the word MULTIPLE be arranged so that the vowels always come together?

Description : In how many different ways can the letters of the word MULTIPLE be arranged so that the vowels always come together? a) 4320 b) 2160 c) 1080 d) 40320 e) 20160

Last Answer : We consider all the three vowels (U, I, E) as one letter, so total number of letters = 6, and three vowels can be arranged in 3! Ways among themselves. However, the letter ‘L’ comes twice. :. Total number of ways = (6! × 3!)/2! = 720 × 3 = 2160 Answer is: b)

In how many different ways can the letters of the word DESIGN be arranged so that the vowels are at the two ends? a) 48 b) 66 c) 12 d) 72 e) 96

Description : In how many different ways can the letters of the word DESIGN be arranged so that the vowels are at the two ends? a) 48 b) 66 c) 12 d) 72 e) 96

Last Answer : Total ways = 4 ! x 2 ! = 4 × 3 × 2 x 2= 48 Answer: a)

In how many different ways can the letters of the word "ZYMOGEN" be arranged in such a way that the vowels always come together? a) 1440 b) 1720 c) 2360 d) 2240

Description : In how many different ways can the letters of the word "ZYMOGEN" be arranged in such a way that the vowels always come together? a) 1440 b) 1720 c) 2360 d) 2240

Last Answer : Answer: A)  The arrangement is made in such a way that the vowels always come together. i.e., "ZYMGN(OE)". Considering vowels as one letter, 6 different letters can be arranged in 6! ways; i.e., 6! = 720 ... in 2! ways; i.e.,2! = 2 ways Therefore, required number of ways = 720 x 2 = 1440 ways.

In how many different ways can the letters of the word 'VINTAGE' be arranged such that the vowels always come together? A) 720 B) 1440 C) 632 D) 364 E) 546

Description : In how many different ways can the letters of the word 'VINTAGE' be arranged such that the vowels always come together? A) 720 B) 1440 C) 632 D) 364 E) 546

Last Answer : Answer: A)  It has 3 vowels (IAE) and these 3 vowels should always come together. Hence these 3 vowels can be grouped and considered as a single letter. That is, VNTG(IAE). Hence we can assume total ... ways to arrange these vowels among themselves 3! = 3 2 1=6 Total number of ways 120 6=720

In how many different ways can the letters of the word 'SPORADIC' be arranged so that the vowels always come together? A) 120 2 B) 1720 C) 4320 D) 2160 E) 2400

Description : In how many different ways can the letters of the word 'SPORADIC' be arranged so that the vowels always come together? A) 120 2 B) 1720 C) 4320 D) 2160 E) 2400

Last Answer : Answer: C) The word 'SPORADIC' contains 8 different letters. When the vowels OAI are always together, they can be supposed to form one letter. Then, we have to arrange the letters SPRDC (OAI). Now, 6 ... be arranged among themselves in 3! = 6 ways. Required number of ways = (720 x 6) = 4320.

In how many different ways can the letters of the word 'POTENCY' be arranged in such a way that the vowels always come together? A) 1360 B) 2480 C) 3720 D) 5040 E) 1440

Description :  In how many different ways can the letters of the word 'POTENCY' be arranged in such a way that the vowels always come together? A) 1360 B) 2480 C) 3720 D) 5040 E) 1440

Last Answer : Answer: E)  The word 'POTENCY' has 7 different letters.  When the vowels EO are always together, they can be supposed to form one letter.  Then, we have to arrange the letters PTNCY (EO).  Now, 6 (5 ... be arranged among themselves in 2! = 2 ways.  Required number of ways = (720 x2)  = 1440.

In how many different ways can the letters of the word 'ABOMINABLES' be arranged so that the vowels always come together? A) 181045 B) 201440 C) 12880 D) 504020 E) 151200

Description :  In how many different ways can the letters of the word 'ABOMINABLES' be arranged so that the vowels always come together? A) 181045 B) 201440 C) 12880 D) 504020 E) 151200

Last Answer : Answer: E)  In the word 'ABOMINABLES', we treat the vowels AOIAE as one letter.  Thus, we have BMNBLS (AOIAE).  This has 7 (6 + 1) letters of which B occurs 2 times and the rest are different. Number of ways arranging these letters = 7! / 2!  = (7×6×5×4×3×2×1) / (2×1) = 2520

The letters of the word ‘SOCIETY’ are placed at random in a row. What is the probability that three vowels come together ? -Maths 9th

Description : The letters of the word ‘SOCIETY’ are placed at random in a row. What is the probability that three vowels come together ? -Maths 9th

Last Answer : There are 7 letters in the word SOCIETY. ∴ Total number of ways of arranging all the 7 letters = n(S) = 7!. When the case of three vowels being together is taken, then the three vowels are considered as one unit, so the ... = 5! 3! ∴ Required probability = \(rac{5! imes3!}{7!}\) = \(rac{1}{7}\)

How many words can be formed from the letters of the word “DAUGHTER” so that the vowels always come together? -Maths 9th

Description : How many words can be formed from the letters of the word “DAUGHTER” so that the vowels always come together? -Maths 9th

Last Answer : The number of words formed from 'DAUGHTER' such that all vowels are together is 4320.

Find how many arrangements can be made with the letters of the word “MATHEMATICS” in which the vowels occur together? -Maths 9th

Description : Find how many arrangements can be made with the letters of the word “MATHEMATICS” in which the vowels occur together? -Maths 9th

Last Answer : (i) There are 11 letters in the word 'MATHEMATICS' . Out of these letters M occurs twice, A occurs twice, T occurs twice and the rest are all different. Hence, the total number of arrangements of ... 4!2!=12. Hence, the number of arrangement in which 4 vowels are together =(10080×12)=120960.

How many words can be formed from the letters of the word “SUNDAY” so that the vowels never come together? -Maths 9th

Description : How many words can be formed from the letters of the word “SUNDAY” so that the vowels never come together? -Maths 9th

Last Answer : Given: The word ‘SUNDAY’ Total number of letters in the word ‘SUNDAY’ is 6. So, number of arrangements of 6 things, taken all at a time is 6P6 = 6! = 6 ... of words using letters of ‘SUNDAY’ starting with ‘N’ and ending with ‘Y’ is 24

Name a five letter word which has three consonants all the same and two different vowels. Every now and then you see this while running a Windows 95/98 on your PC. -Riddles

Description : Name a five letter word which has three consonants all the same and two different vowels. Every now and then you see this while running a Windows 95/98 on your PC. -Riddles

Last Answer : Error! Error! Error!

vowels are worth $50 each,consonants are $40. can you make a word with exactly $200? $600?

Description : vowels are worth $50 each,consonants are $40. can you make a word with exactly $200? $600?

Last Answer : $200- You would have to think of a word that has either 4 vowels or 5 consonants. This is because there is no other way to get to 200 with the numbers given. It is common knowledge that there are ... possible Good trick question, will you answer one for me? what came first the chicken or the egg?

The letters ofthe word ‘NATIONAL’are arranged atrandom. What is the probability that the last letter will be T ? -Maths 9th

Description : The letters ofthe word ‘NATIONAL’are arranged atrandom. What is the probability that the last letter will be T ? -Maths 9th

Last Answer : (c) \(rac{1}{8}\)Let S be the sample space.Then n(S) = Total number of ways in which the letters of word NATIONAL can be arranged= \(rac{8!}{2!\,2!}\) (∵There are 2A s and 2N's in 8 letters)Let E : Event of ... !}{2!\,2!}}\) = \(rac{7!}{8!}\) = \(rac{7!}{8 imes7!}\) = \(rac{1}{8}\).

The letters of the word ‘COCHIN’ are permuted and all the permutations are arranged in alphabetical order as in English dictionary. -Maths 9th

Description : The letters of the word ‘COCHIN’ are permuted and all the permutations are arranged in alphabetical order as in English dictionary. -Maths 9th

Last Answer : answer:

Why do we have vowels and consonants in different categories?

Description : Why do we have vowels and consonants in different categories?

Last Answer : And sometimes Y.

What is the conjunction of vowels with consonants ?

Description : What is the conjunction of vowels with consonants ?

Last Answer : Consonant.

How many vowels and "consonants" are there in Latin?

Description : Please list.

Last Answer : There are 6 vowels, which are: A, E, I, O, U, Y 5 double vowels, which are: AE, AU, EI, EU, EO

In how many ways letter of the world BANKING can be arranged so that vowels always come together? 1)240 2)120 3)720 4)540

Description : In how many ways letter of the world BANKING can be arranged so that vowels always come together? 1)240 2)120 3)720 4)540

Last Answer : 3)720 Exp: [(6!)/(2!)]×(2!)=360×2=720 [(6!)/(2!)]-letters formed, 2!- Vowels.

In how many ways can a mixed doubles game be arranged from amongst 8 married couples if no husband and wife play in the same game? -Maths 9th

Description : In how many ways can a mixed doubles game be arranged from amongst 8 married couples if no husband and wife play in the same game? -Maths 9th

Last Answer : For mixed doubles, we have 2 men and 2 women. 2 men out of 9 can be selected in 9C8 ways Out of 9 women, 2 will be there wife 30 only 7 are remaining for selecting 2 Now out of them 2 possible combinations can be made. So, ... =3024 Hence the answer is 3024.

In how many different ways can the letters of the word DISPLAY be arranged? a) 2601 b) 676 c) 1724 d) 2401 e) 5040

Description : In how many different ways can the letters of the word DISPLAY be arranged? a) 2601 b) 676 c) 1724 d) 2401 e) 5040

Last Answer : 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040 Answer: e)

In how many different ways can the letters of the word “MOBILE” be arranged? 1) 120 2) 240 3) 360 4) 720 5) 36

Description : In how many different ways can the letters of the word “MOBILE” be arranged? 1) 120 2) 240 3) 360 4) 720 5) 36

Last Answer : 4) 720

In how many different ways can any 4 letters of the word 'ABOLISH' be arranged? a) 5040 b) 840 c) 24 d) 120

Description : In how many different ways can any 4 letters of the word 'ABOLISH' be arranged? a) 5040 b) 840 c) 24 d) 120

Last Answer : Answer: B) There are 7 different letters in the word 'ABOLISH'.  Therefore,  The number of arrangements of any 4 out of seven letters of the word = Number of all permutations of 7 letters, taken 4 at a time = ... we have  7p4 = 7 x 6 x 5 x 4 = 840.  Hence, the required number of ways is 840.

In how many different ways can the letters of the word 'GRINDER' be arranged? A) 2520 B) 1280 C) 3605 D) 1807 E) 1900

Description : In how many different ways can the letters of the word 'GRINDER' be arranged? A) 2520 B) 1280 C) 3605 D) 1807 E) 1900

Last Answer : Answer: A)  In these 7 letters, 'R' occurs 2 times, and rest of the letters are different.  Hence, number of ways to arrange these letters  = {7!} / {(2!) }  = {7×6×5×4×3×2×1} / {2×1}  = 2520.

In how many ways can the letters of the word 'NOMINATION' be arranged? A) 237672 B) 123144 C) 151200 D) 150720 E) None of these

Description : In how many ways can the letters of the word 'NOMINATION' be arranged? A) 237672 B) 123144 C) 151200 D) 150720 E) None of these

Last Answer : Answer: C)  The word 'NOMINATION' contains 10 letters, namely  3N, 2O, 1M, 2I,1A, and 1T. Required number of ways = 10 ! / (3!)(2!)(1!)(2!)(1!)(1!)  = 151200

AP and BQ are the bisectors of the two alternate interior angles formed by the intersection of a transversal t with parallel lines l and m (in the given figure). Show that AP || BQ. -Maths 9th

Description : AP and BQ are the bisectors of the two alternate interior angles formed by the intersection of a transversal t with parallel lines l and m (in the given figure). Show that AP || BQ. -Maths 9th

Last Answer : Given In the figure l || m, AP and BQ are the bisectors of ∠EAB and ∠ABH, respectively. To prove AP|| BQ Proof Since, l || m and t is transversal. Therefore, ∠EAB = ∠ABH [alternate interior ... ∠PAB and ∠ABQ are alternate interior angles with two lines AP and BQ and transversal AB. Hence, AP || BQ.

In the given figure, bisectors AP and BQ of the alternate interior angles are parallel, then show that l || m. -Maths 9th

Description : In the given figure, bisectors AP and BQ of the alternate interior angles are parallel, then show that l || m. -Maths 9th

Last Answer : Given, In the figure AP|| BQ, AP and BQ are the bisectors of alternate interior angles ∠CAB and ∠ABF. To show l || m Proof Since, AP|| BQ and t is transversal, therefore ∠PAB = ∠ABQ [alternate interior angles] ⇒ 2 ∠PAB = 2 ∠ABQ [multiplying both sides by 2]

AP and BQ are the bisectors of the two alternate interior angles formed by the intersection of a transversal t with parallel lines l and m (in the given figure). Show that AP || BQ. -Maths 9th

Description : AP and BQ are the bisectors of the two alternate interior angles formed by the intersection of a transversal t with parallel lines l and m (in the given figure). Show that AP || BQ. -Maths 9th

Last Answer : Given In the figure l || m, AP and BQ are the bisectors of ∠EAB and ∠ABH, respectively. To prove AP|| BQ Proof Since, l || m and t is transversal. Therefore, ∠EAB = ∠ABH [alternate interior ... ∠PAB and ∠ABQ are alternate interior angles with two lines AP and BQ and transversal AB. Hence, AP || BQ.

In the given figure, bisectors AP and BQ of the alternate interior angles are parallel, then show that l || m. -Maths 9th

Description : In the given figure, bisectors AP and BQ of the alternate interior angles are parallel, then show that l || m. -Maths 9th

Last Answer : Given, In the figure AP|| BQ, AP and BQ are the bisectors of alternate interior angles ∠CAB and ∠ABF. To show l || m Proof Since, AP|| BQ and t is transversal, therefore ∠PAB = ∠ABQ [alternate interior angles] ⇒ 2 ∠PAB = 2 ∠ABQ [multiplying both sides by 2]

Four boys and three girls stand in a queue for an interview. What is the probability that they will be in alternate positions ? -Maths 9th

Description : Four boys and three girls stand in a queue for an interview. What is the probability that they will be in alternate positions ? -Maths 9th

Last Answer : Total number of ways of arranging 4 boys and 3 girls, i.e., 7 people in a queue (row) = n(S) = 7! Let A : Event in which the 4 boys and 3 girls occupy alternate position. This is possible when the ... {4 imes3 imes2 imes1 imes3 imes2 imes1}{7 imes6 imes5 imes4 imes3 imes2 imes1}\) = \(rac{1}{35}.\)

Sohan wants to show gratitude towards his teacher by giving her a card made by him. He has three pieces of trapezium pasted one above the other as shown in fig. These pieces are arranged in a way that AB||HC || GD || FE. Also BC=CD=DE, and GF=6 cm... -Maths 9th

Description : Sohan wants to show gratitude towards his teacher by giving her a card made by him. He has three pieces of trapezium pasted one above the other as shown in fig. These pieces are arranged in a way that AB||HC || GD || FE. Also BC=CD=DE, and GF=6 cm... -Maths 9th

Last Answer : Given : Sohan wants to show gratitude towards his teacher by giving her a card made by him. He has three pieces of trapezium pasted one above the other as shown in the fig. These pieces are arranged ... length of coloured tape required = 30 cm (b) The values are : Happiness, beauty, Knowledge.

If the median of data (arranged in ascending order) 31, 33, 35, x, x+10, 48, 48, 50 is 40, then find value of x. -Maths 9th

Description : If the median of data (arranged in ascending order) 31, 33, 35, x, x+10, 48, 48, 50 is 40, then find value of x. -Maths 9th

Last Answer : Given data is 31, 33, 35, x, x+10, 48, 48, 50 Number of observation = 8 (even) Median = Value of (8/2)th observation + Value of (8/2+1)th observation / 2 Value of 4th observation + Value of 5th observation / 2 = x + x + 10 / 2 = x + 5 ∴ x + 5 = 40 ⇒ x = 35

If the median of data (arranged in ascending order) 31, 33, 35, x, x+10, 48, 48, 50 is 40, then find value of x. -Maths 9th

Description : If the median of data (arranged in ascending order) 31, 33, 35, x, x+10, 48, 48, 50 is 40, then find value of x. -Maths 9th

Last Answer : Given data is 31, 33, 35, x, x+10, 48, 48, 50 Number of observation = 8 (even) Median = Value of (8/2)th observation + Value of (8/2+1)th observation / 2 Value of 4th observation + Value of 5th observation / 2 = x + x + 10 / 2 = x + 5 ∴ x + 5 = 40 ⇒ x = 35

Sohan wants to show gratitude towards his teacher by giving her a card made by him. He has three pieces of trapezium pasted one above the other as shown in fig. These pieces are arranged in a way that AB||HC || GD || FE. Also BC=CD=DE, and GF=6 cm. He wants to decorate the card by putting up a colored tape on the non-parallel sides of the trapezium.. -Maths 9th

Description : Sohan wants to show gratitude towards his teacher by giving her a card made by him. He has three pieces of trapezium pasted one above the other as shown in fig. These pieces are arranged in a way that ... the card by putting up a colored tape on the non-parallel sides of the trapezium.. -Maths 9th

Last Answer : Let us consider the following lay out of the greeting card. Trapeziums are arranged in such a way that AB || HC || GD || FE. Also BC=CD=DE and GF=6 cm and DE = 4cm. If three parallel lines make equal ... HG+GF+BC+CD+DE = 6+6+6+4+4+4=30 cm. (b) The values are: Happiness, beauty, Knowledge.

A bag contains 30 tickets numbered from 1 to 30. Five tickets are drawn at random and arranged in ascending order -Maths 9th

Description : A bag contains 30 tickets numbered from 1 to 30. Five tickets are drawn at random and arranged in ascending order -Maths 9th

Last Answer : Total number of ways in which 5 tickets can be drawn = n(S) = 30C5. The tickets are arranged in the form T1, T2, T3 (= 20), T4, T5 Where T1, T2 ∈{1, 2, 3, , 19} and T4, T5 ∈{21, 22, , 30 ... {10 imes9}{2}\) x \(rac{5 imes4 imes3 imes2 imes1}{30 imes29 imes28 imes27 imes26}\) = \(rac{285}{5278}.\)

There are 5 red, 4 white and 3 blue marbles in a bag. They are drawn one by one and arranged in a row. Assuming that all the 12 marbles -Maths 9th

Description : There are 5 red, 4 white and 3 blue marbles in a bag. They are drawn one by one and arranged in a row. Assuming that all the 12 marbles -Maths 9th

Last Answer : answer:

The probability that in the random arrangement of the letters of the word ‘UNIVERSITY’the two I‘s do not come together is -Maths 9th

Description : The probability that in the random arrangement of the letters of the word ‘UNIVERSITY’the two I‘s do not come together is -Maths 9th

Last Answer : (b) \(rac{4}{5}\)Let S be the sample space. Then, n(S) = Total number of waysin which the letters of the word UNIVERSITY' can be arranged = \(rac{10!}{2!}\) (∵ There are 2I s) ... ! imes36}{rac{10!}{2!}}\) = \(rac{ ot8! imes36 imes2!}{10 imes9 imes ot8!}\) = \(rac{4}{5}\).

If all the L‘s occur together and also all I‘s occur together, when the letters of the word ‘HALLUCINATION’ are permuted, -Maths 9th

Description : If all the L‘s occur together and also all I‘s occur together, when the letters of the word ‘HALLUCINATION’ are permuted, -Maths 9th

Last Answer : answer:

There are n letters and n addressed envelopes. If the letters are placed in the envelopes at random, -Maths 9th

Description : There are n letters and n addressed envelopes. If the letters are placed in the envelopes at random, -Maths 9th

Last Answer : Total number of ways of placing n letters in n envelopes = n! All the letters can be placed correctly in only 1 way ∴ Probability of placing all the letters in the right envelopes = \(rac{1}{n!}\) ∴ Probability that all the letters are not placed in the right envelope = 1 – \(rac{1}{n!}\) .

All the words that can be formed using the letters A, H, L, U, R are written as in a dictionary -Maths 9th

Description : All the words that can be formed using the letters A, H, L, U, R are written as in a dictionary -Maths 9th

Last Answer : No. of words starting with A are 4!=24 No. of words starting with H are 4!=24 No. of words starting with L are 4!=24 These account for 72 words Next word is RAHLU and the 74th word RAHUL.

Three letters are randomly selected from the 26 capital letters of the English alphabet. -Maths 9th

Description : Three letters are randomly selected from the 26 capital letters of the English alphabet. -Maths 9th

Last Answer : (d) \(rac{36}{1001}\)Required probability = \(rac{P( ext{2 blue balls}) imes{P}( ext{2 red balls})}{P( ext{4 balls out of 14 balls})}\) + \(rac{P( ext{2 green balls}) imes{P}( ext{2 black balls})}{P( ext{4 out of 14 balls})}\)

A letter is taken out at random from ‘ASSISTANT’ and another is taken out from ‘STATISTICS’. The probability that they are same letters is -Maths 9th

Description : A letter is taken out at random from ‘ASSISTANT’ and another is taken out from ‘STATISTICS’. The probability that they are same letters is -Maths 9th

Last Answer : (c) 0.8645Required probability = P(X not defective and Y not defective) = P(\(\bar{X}\)) x P(\(\bar{Y}\))= (1 – P(X)) (1 – P(Y))= \(\bigg(1-rac{9}{100}\bigg)\)\(\bigg(1-rac{5}{100}\bigg)\)= \(rac{91}{100}\) x \(rac{95}{100}\) = \(rac{8645}{10000}\) = 0.8645.

With reference to the area devoted to its cultivation pulses hold the third place in India. Which crops occupy the first and second places respectively? (a) Rice and Wheat (b) Rice and Cotton (c) Jowar and Cotton (d) Jowar and Bajara

Description : With reference to the area devoted to its cultivation pulses hold the third place in India. Which crops occupy the first and second places respectively? (a) Rice and Wheat (b) Rice and Cotton (c) Jowar and Cotton (d) Jowar and Bajara

Last Answer : Ans: (a)

Rationalise the denominator in each of the following and hence evaluate by taking √2 = 1.414, √3 = 1.732 and √5 = 2.236 up to three places of decimal. -Maths 9th

Description : Rationalise the denominator in each of the following and hence evaluate by taking √2 = 1.414, √3 = 1.732 and √5 = 2.236 up to three places of decimal. -Maths 9th

Last Answer : Rationalisation of denominator