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

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

1 Answer

Answer :

4) 720
by

Related questions

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 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 '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 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 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 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 '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

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)

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 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 "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.

How many words can be formed by using all letters of the word 'CABIN'? A) 720 B) 24 C) 120 D) 60 E) None

Description : How many words can be formed by using all letters of the word 'CABIN'? A) 720 B) 24 C) 120 D) 60 E) None

Last Answer : Answer: C)  The word 'CABIN' has 5 letters and all these 5 letters are different.  Total number of words that can be formed by using all these 5 letters  = 5P5  = 5!  = 5×4×3×2×1  = 120 

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)

Two trains are running in opposite directions with the same speed. If the length of each train is 120 metres and they cross each other in 12 seconds, then the speed of each train (in km/hr) is: A.12 kmph B.24 kmph C.36 kmph D.48 kmph E.None of these

Description : Two trains are running in opposite directions with the same speed. If the length of each train is 120 metres and they cross each other in 12 seconds, then the speed of each train (in km/hr) is: A.12 kmph B.24 kmph C.36 kmph D.48 kmph E.None of these

Last Answer : Answer – C (36 kmph) Explanation – Let the speed of each train be x m/sec. Then, relative speed of the two trains = 2x m/sec. So, 2x = (120 + 120)/12 ⇔ 2x = 20 ⇔ x = 10. ∴ Speed of each train = 10 m/sec = [10 x 18/5] km/hr = 36 km/hr

Two train travel in opposite directions at 36 kmph and 45 kmph and a man sitting in slower train passes the faster train in 8 seconds. Then length of the faster train is: A.120 m B.140 m C.160 m D.180 m E.None of these

Description : Two train travel in opposite directions at 36 kmph and 45 kmph and a man sitting in slower train passes the faster train in 8 seconds. Then length of the faster train is: A.120 m B.140 m C.160 m D.180 m E.None of these

Last Answer : Answer – D (180 m) Explanation – When OPP. direction-PLUS Relative speed = (36 + 45) km/hr = [81 x 5/18] m/sec = [45/2] m/sec. Length of train = [45/2 x 8] m = 180 m.

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

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

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

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)

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%

Va and Vb are negative sequence component voltages the difference angle between Va and Vb with respect to Va is a) 240 b) 120 c) 180 d) 360

Description : Va and Vb are negative sequence component voltages the difference angle between Va and Vb with respect to Va is a) 240 b) 120 c) 180 d) 360

Last Answer : Ans: 120

On a certain day, the ratio of the passenger in the 1st class and the second class travelling by train is 1:3. the ratio of the fares collected from each first class and second class passengers is 30:1. If the total amount collected from all the passengers is Rs 1,320. Find the amount in Rs, collected from the second class passengers. A) 240 B) 360 C) 480 D) 120 E) None

Description : On a certain day, the ratio of the passenger in the 1st class and the second class travelling by train is 1:3. the ratio of the fares collected from each first class and second class passengers is 30:1. If ... in Rs, collected from the second class passengers. A) 240 B) 360 C) 480 D) 120 E) None

Last Answer : Answer: D Let the number of passengers travelling by first class and second class be x and 3x respectively. Lets the fares collected from each of the first class and second class passengers be 30y and y ... 3xy=33xy=1320 xy=40. Total amount collected from the second class =3xy=3 40= Rs 120.

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 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 "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 '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 '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 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

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)

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)

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

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

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)

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 ______.

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 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

A letter lock consists of four rings each marked with five different letters. The number of distinct unsuccessful attempts to open the lock is at the most -. a) 625 b) 676 c) 576 d) 624 e) 575

Description : A letter lock consists of four rings each marked with five different letters. The number of distinct unsuccessful attempts to open the lock is at the most -. a) 625 b) 676 c) 576 d) 624 e) 575

Last Answer : d Since each ring consists of five different letters, the total number of attempts possible with the four rings is =5 * 5 * 5 * 5 = 625. Of these attempts, one of them is a successful attempt. Maximum number of unsuccessful attempts = 625 - 1 = 624.

The sum of ages of family members (both children and parents) is 360 years.The total ages of children and parents are in the ratio 2:1 and the ages of wife and husband are in the ratio 5:7.What will be the age of husband? A.65 B.75 C.72 D.70

Description : The sum of ages of family members (both children and parents) is 360 years.The total ages of children and parents are in the ratio 2:1 and the ages of wife and husband are in the ratio 5:7.What will be the age of husband? A.65 B.75 C.72 D.70

Last Answer : Answer : D (70) ExplanationSum of ages is 360 years.The ratio of children and parents ages is 2:1. Total age of parents = 360 x 1 / 3 = 120 years Ratio of wife and husband age is 5:7. Therefore, the age of husband = 120 x 7 / 12 = 70 Hence the age of husband is 70 years.

The sum of ages of family members (both children and parents) is 360 years.The total ages of children and parents are in the ratio 2:1 and the ages of wife and husband are in the ratio 5:7.What will be the age of husband? A.65 B.75 C.72 D.70

Description : The sum of ages of family members (both children and parents) is 360 years.The total ages of children and parents are in the ratio 2:1 and the ages of wife and husband are in the ratio 5:7.What will be the age of husband? A.65 B.75 C.72 D.70

Last Answer : Answer : D (70) ExplanationSum of ages is 360 years.The ratio of children and parents ages is 2:1. Total age of parents = 360 x 1 / 3 = 120 years Ratio of wife and husband age is 5:7. Therefore, the age of husband = 120 x 7 / 12 = 70 Hence the age of husband is 70 years.

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

5, 8, 12, 18, 24, 30, 36, 42, ____ a) 52 b) 46 c) 48 d) 54 e) 56

Description : 5, 8, 12, 18, 24, 30, 36, 42, ____ a) 52 b) 46 c) 48 d) 54 e) 56

Last Answer : Each term in the series is the addition of successive prime numbers. Like Prime numbers are2, 3, 5, 7, 11, 13, 17, 19, 23, 29 and so on. So first term is addition of (2 + 3). Ie 5, then second term is (3 + 5) ie ... +17)ie. 30, then (17+19)ie.36, then (19+23)ie., 42, then (23+29)i.e. 52. Answer: a)

Ram is 3 times as old as his brother. Ram’s father is 3 times older than Ram. In 9 years , Ram will be 1.5 times the age of his brother. What will be the age of Ram’s father when Ram is 1.5 times the age of his brother? a) 27 b) 36 c) 45 d) 54 e) 63

Description : Ram is 3 times as old as his brother. Ram’s father is 3 times older than Ram. In 9 years , Ram will be 1.5 times the age of his brother. What will be the age of Ram’s father when Ram is 1.5 times the age of his brother? a) 27 b) 36 c) 45 d) 54 e) 63

Last Answer : Let the age of Ram be R, brother be B R=3B And (R + 9 ) =1.5(B + 9) 3B + 9 =1.5B + 13.5 1.5B =4.5 or B = 3 years R= 9 years. Father current age = 4R(Since father is 3 times older than Ram, his ... R + 3 R, and not 3R) Father age when Ram is 1.5 times the age of his brother = 4R + 9 = 45. Answer : c

Ram is 3 times as old as his brother. Ram’s father is 3 times older than Ram. In 9 years , Ram will be 1.5 times the age of his brother. What will be the age of Ram’s father when Ram is 1.5 times the age of his brother? a) 27 b) 36 c) 45 d) 54 e) 63

Description : Ram is 3 times as old as his brother. Ram’s father is 3 times older than Ram. In 9 years , Ram will be 1.5 times the age of his brother. What will be the age of Ram’s father when Ram is 1.5 times the age of his brother? a) 27 b) 36 c) 45 d) 54 e) 63

Last Answer : Let the age of Ram be R, brother be B R=3B And (R + 9 ) =1.5(B + 9) 3B + 9 =1.5B + 13.5 1.5B =4.5 or B = 3 years R= 9 years. Father current age = 4R(Since father is 3 times older than Ram, his ... R + 3 R, and not 3R) Father age when Ram is 1.5 times the age of his brother = 4R + 9 = 45. Answer : c

At the time of marriage, a man was 6 years older to his wife. 12 years after their marriage, his age is 6/5 times the age of his wife. What was wife’s age at the time of marriage? a) 18 years b) 24 years c) 30 years d) 36 years e) None of these

Description : At the time of marriage, a man was 6 years older to his wife. 12 years after their marriage, his age is 6/5 times the age of his wife. What was wife’s age at the time of marriage? a) 18 years b) 24 years c) 30 years d) 36 years e) None of these

Last Answer : ratio of their ages, 12 years hence = 6 : 5  Difference in ages = 1 ratio = 6 years wife’s age (12 years hence) = 5 ratio = 5 × 6 years = 30 years wife’s age at the time of marriage = 30 – 12 = 18 years Answer: a)

1595 ÷ 25 × 36.5 a) 2459 b) 2329 c) 2359 d) 2429 e) 2349

Description : 1595 ÷ 25 × 36.5 a) 2459 b) 2329 c) 2359 d) 2429 e) 2349

Last Answer : Answer: b)