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

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
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1 Answer

Answer :

3)720

Exp: [(6!)/(2!)]×(2!)=360×2=720 [(6!)/(2!)]-letters formed, 2!-

Vowels.
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Related questions

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 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 '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 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 '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

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 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 '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 '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 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)

729 656 595 ? 494 a) 540 b) 530 c) 500 d) 529 e) 541

Description : 729 656 595 ? 494 a) 540 b) 530 c) 500 d) 529 e) 541

Last Answer : Difference of Difference 9. Answer: a)

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

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

Last Answer : 24 ways is the answer

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.

worker is allowed 60 hours to complete the job on a guaranteed wage of Rs. 10 per hour. Under the Rowan Plan, he gets an hourly wage of Rs. 12 per hour. For the same saving in time, how much he will get under the Halsey Plan? (a) Rs. 720 (b) Rs. 540 (c) Rs. 600 (d) Rs. 900

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21, 23, 29, 53, 173, _____ a) 210 b) 893 c) 720 d) 253 e) 293

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Last Answer : Difference between each successive tam is factorial of number starting from 2, like 23-21 = 2, then 29-23 = 6 means 3!, then 53-29 = 24 means 4!, then next difference is 173-53 = 120 means 5!, so next difference would be 6! i.e. 720 so term will be 173+720 = 893. Answer: b)

A train travelling at 36 kmph completely crosses another train having half its length and travelling in the opposite direction at 54 kmph, in 12 seconds. If it also passes a railway platform in 112 minutes, the length of the platform is : a) 560 metres b) 620 metres c) 700 meres d) 750 metres e) 720 meters

Description : A train travelling at 36 kmph completely crosses another train having half its length and travelling in the opposite direction at 54 kmph, in 12 seconds. If it also passes a railway platform in 112 minutes, the length ... is : a) 560 metres b) 620 metres c) 700 meres d) 750 metres e) 720 meters

Last Answer : Answer: c)

Amar sold a camera at 20% profit to Bhavan. Bhavan sold it to Chetan at 30% loss. Chetan bought the camera for Rs.840. find the cost price of Chetan, had Amar sold the camera to Bhavan at 30% loss and Bhavan sold it to Chetan at 20% profit (in Rs.) a)1160 b)1080 c)840 d)720

Description : Amar sold a camera at 20% profit to Bhavan. Bhavan sold it to Chetan at 30% loss. Chetan bought the camera for Rs.840. find the cost price of Chetan, had Amar sold the camera to Bhavan at 30% loss and Bhavan sold it to Chetan at 20% profit (in Rs.) a)1160 b)1080 c)840 d)720

Last Answer : c)840

The marked price of a watch was Rs. 720.A man brought the same for Rs.550.80 after getting two successive discounts the first being 10% .What was the second discount rate? A. 12% B. 14% C. 15% D. 18%

Description : The marked price of a watch was Rs. 720.A man brought the same for Rs.550.80 after getting two successive discounts the first being 10% .What was the second discount rate? A. 12% B. 14% C. 15% D. 18%

Last Answer : Answer – C. 15% Explanation – Let the second discount rate be x% Then, (100 – x)% of 90% of 720 = 550.80 => (100 – x)/100 x 90/100 x 720 = 550.80 = (100-x) = [55080/(9 x 72) = 85 => x= 15 second discount rate =15%

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.

Out of 5 women and 4 men, a committee of three members is to be formed in such a way that at least one member is woman. In how many different ways can this be done? 1) 60 2) 100 3) 120 4) 90 5) 80

Description : Out of 5 women and 4 men, a committee of three members is to be formed in such a way that at least one member is woman. In how many different ways can this be done? 1) 60 2) 100 3) 120 4) 90 5) 80

Last Answer : 5) 80

In how many ways can 6 girls be seated in a rectangular order? A) 60 B) 120 C) 5040 D) 720

Description : In how many ways can 6 girls be seated in a rectangular order? A) 60 B) 120 C) 5040 D) 720

Last Answer : Answer: B)  Number of arrangements possible = (6-1)!  = 5! = 5×4×3×2×1 = 120

Range of wavelengths of ultraviolet ∙ A. 670 nm to 10^3 nm ∙ B. 440 nm to 540 nm ∙ C. 110 nm to 240 nm ∙ D. 10 nm to 390 nm

Description : Range of wavelengths of ultraviolet ∙ A. 670 nm to 10^3 nm ∙ B. 440 nm to 540 nm ∙ C. 110 nm to 240 nm ∙ D. 10 nm to 390 nm

Last Answer : D. 10 nm to 390 nm

ange of wavelength of visible light ∙ A. 670 nm to 10^3 nm ∙ B. 440 nm to 540 nm ∙ C. 110 nm to 240 nm ∙ D. 390 nm to 770 nm

Description : ange of wavelength of visible light ∙ A. 670 nm to 10^3 nm ∙ B. 440 nm to 540 nm ∙ C. 110 nm to 240 nm ∙ D. 390 nm to 770 nm

Last Answer : D. 390 nm to 770 nm

19 + (12 ÷ 4 × (240 × 1 ÷ 180)) × 6 ÷ (101 + 98 ÷ 7 - 113) – (7 ÷ 2 + 14 – 3.5 – 9 × 8.5) × 4 a) 201 b) 241 c) -241 d) -281 e) 281

Description : 19 + (12 ÷ 4 × (240 × 1 ÷ 180)) × 6 ÷ (101 + 98 ÷ 7 - 113) – (7 ÷ 2 + 14 – 3.5 – 9 × 8.5) × 4 a) 201 b) 241 c) -241 d) -281 e) 281

Last Answer : 240 1 180 = 4/3 12 4 (240 1 180) = 3 (4/3) = 4 101 + 98 7 - 113 = 101 + 14 - 113 = 2 19 + (12 4 (240 1 180)) 6 (101 + 98 7 - 113) = 19 + 4 6 2 = 19 + 4 3 = 31 7 2 ... 180)) 6 (101 + 98 7 - 113) - (7 2 + 14 - 3.5 - 9 8.5) 4 = 31 - (-250) = 281 Answer: e)

In a class of 24 students, each student shakes hand with everyone els. In another class of 32 students, each student shakes hand with all other students. What is the difference between total number of hand shakes in the 2 classes? a) 180 b) 200 c) 22 d) 240 e) 260

Description : In a class of 24 students, each student shakes hand with everyone els. In another class of 32 students, each student shakes hand with all other students. What is the difference between total number of hand shakes in the 2 classes? a) 180 b) 200 c) 22 d) 240 e) 260

Last Answer : 1 st class : Total Hand shakes = 24 × 23 /2 = 12 × 23 =276 2nd class :Total Hand shakes =32 × 31 / 2 = 16 × 31 = 496 Difference =220 Answer : c

In a class of 24 students, each student shakes hand with everyone els. In another class of 32 students, each student shakes hand with all other students. What is the difference between total number of hand shakes in the 2 classes? a) 180 b) 200 c) 22 d) 240 e) 260

Description : In a class of 24 students, each student shakes hand with everyone els. In another class of 32 students, each student shakes hand with all other students. What is the difference between total number of hand shakes in the 2 classes? a) 180 b) 200 c) 22 d) 240 e) 260

Last Answer : 1 st class : Total Hand shakes = 24 × 23 /2 = 12 × 23 =276 2nd class :Total Hand shakes =32 × 31 / 2 = 16 × 31 = 496 Difference =220 Answer : c

5 6 16 57 ? 1245 7506 a) 240 b) 248 c) 244 d) 250 e) 260

Description : 5 6 16 57 ? 1245 7506 a) 240 b) 248 c) 244 d) 250 e) 260

Last Answer : Answer: b)

256 128 ? 192 96 240 80 a) 64 b) 156 c) 96 d) 128 e) 176

Description : 256 128 ? 192 96 240 80 a) 64 b) 156 c) 96 d) 128 e) 176

Last Answer : The series is ×0.5, ÷1, ×1.5, ÷2, ×2.5.... Answer: d)

185% of 400 + 35% of 240 = ?% of 1648 a) 85 b) 75 c) 125 d) 50 e) None of these

Description : 185% of 400 + 35% of 240 = ?% of 1648 a) 85 b) 75 c) 125 d) 50 e) None of these

Last Answer : 740 + 84 = 16.48 × ? 824 = 16.48 × ? ? = 50 Answer: d)

There are two sets of five numbers each – set P and set Q. P is a set of consecutive even numbers and Q is a set of consecutive numbers. The sum of the numbers of P is 230. The second least number in Q is 33 less than twice the least number in P. Find the sum of the numbers in Q. 1 : 260 2 : 250 3 : 240 4 : 270 5 : None of these

Description : There are two sets of five numbers each - set P and set Q. P is a set of consecutive even numbers and Q is a set of consecutive numbers. The sum of the numbers of P is 230. The second least number in Q is 33 less ... the sum of the numbers in Q. 1 : 260 2 : 250 3 : 240 4 : 270 5 : None of these

Last Answer : 1 : 260

(84111)^1/2= ? 1) 240 2) 270 3) 330 4) 290 5) 310

Description : (84111)^1/2= ? 1) 240 2) 270 3) 330 4) 290 5) 310

Last Answer : 4) 290

All the words of OLIVE are arranged in dictionary order. Find the position of the word LIVEO. a) 49 b) 55 c) 57 d) 59 e) 61

Description : All the words of OLIVE are arranged in dictionary order. Find the position of the word LIVEO. a) 49 b) 55 c) 57 d) 59 e) 61

Last Answer : words starting with E = 4! = 24 words starting with I = 4! = 24 words starting with LE = 3! = 6 words starting with LIE = 2! = 2, LIO = 2! = 2 the first word with starting with LIV will be LIVEO so rank = 24 + 24 + 6 + 2 + 2 + 1 = 59 Answer: d)

In a pie chart, the central angle for a component value of 240, when the total value is 720, is ______.

Description : In a pie chart, the central angle for a component value of 240, when the total value is 720, is ______.

Last Answer : In a pie chart, the central angle for a component value of 240, when the total value is 720, is ______.

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

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.

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 = \(rac{5! imes3!}{7!}\) = \(rac{1}{7}\)

Liquid "X" is at equilibrium with its vapor in a cylinder and piston apparatus. When the volume of the space above the liquid is 50 ml and the temperature 25!, the vapor pressure of "X" is 120 torr. What will the vapor pressure of "X" be when the volume above the liquid is 100 ml and the temperature 25!C? Some liquid is always present. w) 30 torr x) 60 torr y) 120 torr z) 240 torr

Description : Liquid "X" is at equilibrium with its vapor in a cylinder and piston apparatus. When the volume of the space above the liquid is 50 ml and the temperature 25!, the vapor pressure of "X" is 120 torr. What will the ... 25!C? Some liquid is always present. w) 30 torr x) 60 torr y) 120 torr z) 240 torr

Last Answer : ANSWER: Y -- 120 TORR

The value for the ΔU of a system is -120 J. If the system is known to have absorbed 420 J of heat, how much work was done?  a. -540 J  b. -640 J  c. -740 J  d. -840 J formula: ΔU = q +w

Description : The value for the ΔU of a system is -120 J. If the system is known to have absorbed 420 J of heat, how much work was done?  a. -540 J  b. -640 J  c. -740 J  d. -840 J formula: ΔU = q +w

Last Answer : -540 J

n a chemical factory where dyes are made, wet cake at 30 OC consisting of 60% moisture is put in a dryer to obtain an output having only 5% moisture, at atmospheric pressure. In each batch about 120 kgs of material is dried. a. The quantity of moisture removed per batch. b. What is the total quantity ( sensible & latent) of heat required to evaporate the moisture, if the latent heat of water is 540 kcal/kg at atmospheric conditions, Ignore heat absorbed by the solids c. Find the quantity of steam required for the drying process (per batch), if steam at 4 kg/cm2 is used for generating hot air in the dryer and the dryer efficiency is 80%. Latent heat of steam at 4 kg/cm2 is 520 kcal/kg.

Description : n a chemical factory where dyes are made, wet cake at 30 OC consisting of 60% moisture  is put in a dryer to obtain an output having only 5% moisture, at atmospheric pressure. In  each batch about 120 ... dryer and the dryer efficiency  is 80%. Latent heat of steam at 4 kg/cm2 is 520 kcal/kg.

Last Answer : Given that   Qty of material dried per batch - 120 Kgs  Moisture at inlet - 60% a. The quantity of moisture removed per batch.  Water quantity in a wet batch - 120 x 0.6 = 72 Kgs.  Quantity of bone dry ... - 48/0.95 = 50.5 Kgs.  Equivalent water in a dry batch - 50.5 - 48 = 2.5 Kgs.

Ashwini has a certain number of Pencils she wishes to distribute in the class. when she tries to distribute 7 pencils to each student, 3 pencils are left out. if the class had 3 less students, each student would have got 2 pencils more, and no pencils would have been left out. How many pencils does Ashwini have? a) 90 b) 120 c) 140 d) 160 e) None of these

Description : Ashwini has a certain number of Pencils she wishes to distribute in the class. when she tries to distribute 7 pencils to each student, 3 pencils are left out. if the class had 3 less students, each student would have ... How many pencils does Ashwini have? a) 90 b) 120 c) 140 d) 160 e) None of these

Last Answer : Let number of students be X. so, total Pencils = 7X + 3 7X + 3 = (7+2) (X-3) or., 7X + 3 = 9X – 27 or X = 15 total Pencils = 7X + 3 = 108 Answer: e)

24+(72 ÷ 48×8+14) × 36 ÷ (3×4) + 36 ÷ 3 ×4 a) 120 b) 140 c) 160 d) 180 e) None of these

Description : 24+(72 ÷ 48×8+14) × 36 ÷ (3×4) + 36 ÷ 3 ×4 a) 120 b) 140 c) 160 d) 180 e) None of these

Last Answer : 72÷48×8+14 = (72/48) × 8 + 14 = (3/2) × 8 + 14 = 12 +14 = 26 24 + (72 ÷ 48 × 8 + 14) ×36 ÷ (3 × 4) + 36 ÷ 3 × 4 = 24 + 26 × 36÷12 + 36 ÷ 3 × 4 = 24 + 26 × 3 + 12 × 4 = 24 + 78 + 48 = 150 Answer: e)

A dealer sold a radio at a loss of 2.5%. Had he sold it for Rs. 120 more, he would have gained 7.5%. In order to gain 12.5% after a 25% discount, the marked price should be: a) 1750 b) 1800 c) 1857.75 d) 1925 e) None of these

Description : A dealer sold a radio at a loss of 2.5%. Had he sold it for Rs. 120 more, he would have gained 7.5%. In order to gain 12.5% after a 25% discount, the marked price should be: a) 1750 b) 1800 c) 1857.75 d) 1925 e) None of these

Last Answer : 120 = 10% of the CP (2.5% loss to 7.5% profit) Or CP = Rs. 1200 To gain 12.5%, SP = 1200 × 1.125 = Rs. 1350 Rs. 1350 = 25% discounted of Marked Price Or marked Price = 1350/0.75 = Rs. 1800 Answer: b)

In racing over a given distance, A can beat B by 20m, B can beat C by 10m, A can beat C by 28m. The racing distance is? a) 58 m b) 116 m c) 100 m d) 120 m e) cannot be determined

Description : In racing over a given distance, A can beat B by 20m, B can beat C by 10m, A can beat C by 28m. The racing distance is? a) 58 m b) 116 m c) 100 m d) 120 m e) cannot be determined

Last Answer : Let the uniform speeds of A,B,C be vA,vB,vC, respectively. Let t1 be the time taken for the race between A and B and t2 be the time taken for race between B and C Since A covers the distance in time t1, race between A and C ... ) = VB/ VC = d/(d-10) Thus (d-20)(d-10)=d(d-28).Hence, d=100. Answer: c)

121 ? 113 138 89 170 49 a) 125 b) 110 c) 122 d) 120 e) 133

Description : 121 ? 113 138 89 170 49 a) 125 b) 110 c) 122 d) 120 e) 133

Last Answer : The series is +1^2 , -3^2 , +5^2 , - 7^2 , .... Answer: c)

It takes eight hours for a 600 km journey, if 120 km is done by train and the rest by car. It takes 20 minutes more, if 200 km is done by train and the rest by car. The ratio of the speed of the train to that of the cars is? a) 2 : 3 b) 3: 2 c) 3 : 4 d) 4 : 3 e) None of these

Description : It takes eight hours for a 600 km journey, if 120 km is done by train and the rest by car. It takes 20 minutes more, if 200 km is done by train and the rest by car. The ratio of the speed of the train to that of the cars is? a) 2 : 3 b) 3: 2 c) 3 : 4 d) 4 : 3 e) None of these

Last Answer : Let the speed of the train be x km/hr and that of the car be y km/hr. Then, 120/x+480/y= 8 1/x+4/y=1/15 ....(i) And, 200/x+400/y=25/3 1/x+2/y=1/24 ....(ii) Solving (i) and (ii), we get: x = 60 and y = 80. Ratio of speeds = 60 : 80 = 3 : 4. Answer: c)

The price of two dozens of oranges is Rs.120/-more than the price of one kg. of mango. The price of one dozen of apple is Rs.80/- more than the price of one kg of mango. Which of the following represents the ratios of the price of one orange and one apple respectively? a) 4 : 5 b) 3 : 2 c) 7 : 9 d) Cannot be determined

Description : The price of two dozens of oranges is Rs.120/-more than the price of one kg. of mango. The price of one dozen of apple is Rs.80/- more than the price of one kg of mango. Which of the following represents the ... apple respectively? a) 4 : 5 b) 3 : 2 c) 7 : 9 d) Cannot be determined e) None of these

Last Answer : Because no. of mangoes is unknown. So ratio cannot be calculated. Answer: d)

4725,1050,300,120,80,? 1 : 40 2 : 160 3 : 120 4 : 110 5 : 60

Description : 4725,1050,300,120,80,? 1 : 40 2 : 160 3 : 120 4 : 110 5 : 60

Last Answer : Answer-2 (160) Explanation- 4725 / 4.5=1050 1050/3.5=300 300/2.5=120 120/1.5=80 80/0.5=160

A 270 m long train running at the speed of 120 kmph crosses another train running in opposite direction at the speed of 80 kmph in 9 seconds. What is the length of the other train? A.180 B.230 C.245 D.235 E.None of these

Description : A 270 m long train running at the speed of 120 kmph crosses another train running in opposite direction at the speed of 80 kmph in 9 seconds. What is the length of the other train? A.180 B.230 C.245 D.235 E.None of these

Last Answer : Answer – B (230) Explanation – Relative speed = total lenghths /time [120 +80] x5/18 ={ 270 + other train length (say L) } / 9 sec (x5/18 to convert km/hr to m/sec ) 200 x (5/18) x9 = 270 + L 500 = 270 +L L=230 m Note: always do cutting ,avoid solving exact