A question paper has two parts A and B, each containing 12 questions. If a student needs to choose 10 from part A and 8 from part B, in how many ways can he do that? A) 32670 B) 36020 C) 41200 D) 29450

1 Answer

Answer :

Answer: A)

 Number of ways to choose 10 questions from part A = 12C10

 Number of ways to choose 8 questions from part B = 12C8

 Total number of ways= 12C10 × 12C8

 = 12C2 × 12C4 [∵ nCr = nC(n-r)]

 ={12×11}/{2×1}X{12×11×10×9} / {4×3×2×1}

 =66×495

=32670

Related questions

Description :  Pramoth has 12 friends and he wants to invite 7 of them to a party. How many times will 4 particular friends never attend the party? A) 8 B) 7 C) 12 D) 15 

Last Answer : Answer: A)  Remove the 4 particular friends and invite 7 friends from the remaining 8 (12-4) friends . this can be done in 8C7 ways.  Therefore , required number of ways = = 8C7  = 8C1  = 8

Description : An shopkeeper has 15 models of cup and 9 models of saucer. In how many ways can he make a pair of cup and saucer? A) 100 B) 80 C) 110 D) 135

Last Answer : Answer: D) He has 15 patterns of cup and 9 model of saucer A cup can be selected in 15 ways. A saucer can be selected in 9 ways. Hence one cup and one saucer can be selected in 15×9 ways =135 ways

Description :  In a birthday party, every person shakes hand with every other person. If there was a total of 66 handshakes in the party, how many persons were present in the party? A) 9 B) 8 C) 7 D) 12

Last Answer : Answer: D)  Assume that in a party every person shakes hand with every other person  Number of hand shake = 66  Total number of hand shake is given by nC2  Let n = the total number of persons ... cannot take negative value of n  So, n = 12  Therefore number of persons in the party = 12

Description : In how many ways can a group of 10 men and 5 women be made out of a total of 12 men and 10 women? A) 16632 B) 15290 C) 25126 D) 34845 E) 38135

Last Answer : Answer: A)  Required number of ways = 12C10 x 10C5  = 66 × 252 = 16632

Description :  In how many ways can a team of 6 persons be formed out of a total of 12 persons such that 3 particular persons should not be included in any team? A) 56 B) 112 C) 84 D) 128

Last Answer : Answer: C) Three particular persons should not be included in each team. i.e., we have to select remaining 6- 3= 3 persons from 12-3 = 9 persons.  Hence, required number of ways = 9C3  = {9×8 × 7} / {3 × 2 × 1} = 504 / 6 = 84

Description : Find out the number of ways in which 12 Bangles of different types can be worn in 2 hands? A) 1260 B) 2720 C) 1225 D) 4096

Last Answer : Answer: D)  The first bangle can be worn in any of the 2 hands (2 ways). Similarly each of the remaining 11 bangles also can be worn in 2 ways. Hence total number of ways=2×2×2×2×2×2×2×2×2×2×2×2  =2^12  =4096

Description : There are 7 periods in each working day of a college. In how many ways can one organize 6 subjects such that each subject is allowed at least one period? A) 33200 B) 15120 C) 10800 D) 43600

Last Answer : Answer: B)  6 subjects can be arranged in periods in 7P6 ways.  Remaining 1 period can be arranged in 6P1 ways.  Two subjects are alike in each of the arrangement. So we need to divide by 2! to avoid over counting. ... of arrangements = (7P6 x 6P1)/2!  = 5040 6 / 2 = 30240 / 2 = 15120

Description : In how many ways can 8 different balls be distributed in 6 different boxes can contain any number of balls except that ball 4 can only be put into box 4 or 5 ? A) 2×5^6 B) 2×6^7 C) 2×5^4 D) 2×4^7

Last Answer : Answer: B)  1st ball can be put in any of the 6 boxes.  2nd ball can be put in any of the 6 boxes.  3rd ball can be put in any of the 6 boxes.  Ball 4 can only be put into box 4 or box 5. Hence, 4th ball ... put in any of the 6 boxes.  Hence, required number of ways = 6 6 6 2 6 6 6 6  = 2 6^7

Description : A bowl contains 8 violet, 6 purple and 4 magenta balls. Three balls are drawn at random. Find out the number of ways of selecting the balls of different colours? A) 362 B) 2 48 C) 122 D) 192

Last Answer : Answer: D)  1 violet ball can be selected is 8C1 ways.  1 purple ball can be selected in 6C1 ways.  1 magenta ball can be selected in 4C1 ways.  Total number of ways = 8C1 × 6C1 × 4C1  = 8×6×4  = 192

Description : In how many ways can 8 different ballons be distributed among 7 different boxes when any box can have any number of ballons? A) 5^4-1 B) 5^4 C) 4^5-1 D) 7^8

Last Answer : Answer: D) Here n = 7, k = 8. Hence, required number of ways = n^k  =7^8 

Description :  In how many ways can 10 stones can be arranged to form a bangles? A) 267720 B) 284360 C) 125380 D) 181440

Last Answer : Answer: D) Number of arrangements possible = {1} / {2} X (10-1)!  = {1} / {2}X 9! = {1} / {2} X 9×8×7× 6×5×4×3×2×1 = {1 / 2 } ×362880 = 181440

Description : Gowthaman needs 25% to pass. If he scored 424 marks and falls short by 26 marks, what was the maximum marks he could have got? a) 2725 b) 2650 c) 1800 d) 1750 e) 989 

Last Answer : Answer: C  If Gowthaman had scored 26 marks more, he could have scored 25%  Therefore, Mike required 424 + 26 = 450 marks  Let the maximum marks be x.  Then 25 % of x = 450  (25/100) × x = 450  x = (450 × 100)/25 x = 45000/25 x = 1800

Description : A school has 9 maths teachers and 6 science teachers. In how many ways can a team of 4 maths teachers be formed from them such that the team must contain exactly 1 science teacher? A) 800 B) 720 C) 680 D) 504 

Last Answer : Answer: D) The team should have 4 maths teachers. But the team must contain exactly 1 science teacher. Hence, select 3 maths teachers from 9 maths teachers and select 1 science teachers from 6 science teachers.  Number of ways this can be ...  ={9 8 7}/{3 2 1}X6  = 504 / 6 6  = 84 6 =504 

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

Description : A box contains 2 pink balls, 3 brown balls and 4 blue balls. In how many ways can 3 balls be drawn from the box, if at least one brown ball is to be included in the draw? A) 32 B) 48 C) 64 D) 96 E) None

Last Answer : Answer: C) 

Description :  In how many different ways can 6 apple and 6 orange form a circle such that the apple and the orange alternate? A) 82880 B) 86400 C) 71200 D) 63212

Last Answer : Answer: B)  6 apples can be arranged in (6-1)! Ways  Now there are 6 positions in which 6 orange can be placed.  This can be done in 6! ways. Required number of ways = (6-1)! × 6!  = 5! × 6!  = 120 × 720  = 86400

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.

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.

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.

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.

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

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.

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

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.

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

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.

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.

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

Description : In an examination it is required to get 672 aggregate marks to pass. A student gets 70% marks and is declared failed by 126 marks. What are the maximum aggregate marks a student can get? a) 780 b) 840 c) 741 d) 805 e) 983

Last Answer : Answer: A Difference = 672-126 = 546 According to the question, 70% of total aggregate = 546 Total aggregate marks = 546 × 100 /70  = 54600/70  = 780

Description : In an examination, 900 students appeared. Out of these students; 56 % got first division, 27 % got second division and the remaining just passed. Assuming that no student failed; find the number of students who just passed. a) 225 b) 153 c) 245 d) 148 e) 298

Last Answer : Answer: B The number of students with first division = 56 % of 900  = 56/100 × 900 = 50400/100  = 504 And, the number of students with second division = 27 % of 900  = 27/100 × 900  =24300/100  = 243 Therefore, the number of students who just passed = 900 – (504 + 243)  = 900- 747  = 153

Description : Two students appeared at an examination. One of them secured 18marks more than the other and his marks was 72% of the sum of their marks. What are the marks obtained by them? a) 12.5,23.3 b) 26.7,16.0 c) 13.3,14.2 d) 11.45, 29.45 e) 29.8,15.4 

Last Answer : Answer: D  Let the marks secured by them be x and (x + 18)  Then sum of their marks = x + (x + 18) = 2x + 18  Given that (x + 18) was 72% of the sum of their marks  =>(x+18) = 72/100(2x+18)  => ... 11x = 126 x = 11.45  Then (x + 18) = 11.45 + 18 = 29.45  Hence their marks are 11.45 and 29.45

Description : Chenna dhal is now being sold at Rs. 70 a kg. Last month, is rate was Rs. 80 per kg. By how much percent should a family reduce its consumption so as to keep the expenditure fixed? a)12.5 b)21.8 c)23 d)18

Last Answer : Answer: A  Let a family's monthly consumption of chenna dhal be x kg.  To keep the expenditure fixed,  their consumption for this month should be 70x/80 = 7x/8.  Reduction in consumption = x/8 = 12.5% of x

Description : The Shopkeeper increased the price of a product by 75% so that customer finds it difficult to purchase the required amount. But somehow the customer managed to purchase only 140% of the required amount. What is the net difference in the ... on that product? a) 12.5 b) 26.0 c) 13.5 d) 17.5 e) 19.8

Last Answer : Answer: D  Quantity×Rate=Price  1x1=1  1.4x1.75=2.45  Decrease in price = (0.175/1) × 100  = 17.5%

Description : There are 10 orange, 2 violet and 4 purple balls in a bag. All the 16 balls are drawn one by one and arranged in a row. Find out the number of different arrangements possible. A) 25230 B) 23420 C) 120120 D) 27720

Last Answer : Answer: C)  Number of different arrangements possible  = {16!} / {10! 2! 4!}  = {16×15×14×13×12×11×10×9×8×7×6×5×4×3×2 } /  {(10×9×8×7×6×5×4×3×2 ) (2) (4×3×2)}}  = {16×15×14×13×12×11} / {(2)(4×3×2)}  = {8×5×7×13×3×11}  = 120120

Description : How many 5-letter code words are possible using last 10 letter of the English alphabet , if no letter can be repeated ? a) 30240 b) 25440 c) 45640 d) 32940

Last Answer : Answer: A)  The number of 5 letter code words out of the last 10 letters of the English alphabets are = 10× 9× 8 × 7× 6  = 80 × 63× 6  = 30240 ways.

Description : On a test consisting of 500 questions, Dhivya answered 80% of the first 250 questions correctly. What percent of the other 250 questions does she need to answer correctly for her grade on the entire exam to be 60% ? a) 20 b) 60 c) 45 d) 80 e) 40

Last Answer : Answer: E  60% of 500 = 300  80% of 250 = 200 No. of correct answers in remaining 250 questions = 300 – 200  = 100  Percentage = 100 x 100/ 250  = 40%

Description : Vasavi spends 50% of her monthly income on grocery, clothes and education in the ratio of 8 : 4 : 10 respectively. If the amount spent on clothes is 2770/–, what is Vasavi's monthly income? a) 15235 b) 65000 c) 55400 d) 30470 e) 98700

Last Answer : Answer: D Ratio of Expenses = 8: 4: 10 therefore amount spend on clothes, i.e. 4x = 2770  x = 692.5  Total exp = (8 + 4 + 10)x  = 22x. = 22 × 692.5 = 15235  Monthly income be x.  50% of x = 15235  x = 15235 × 100/ 50  X = 30470 

Description : How many 3-letter words can be formed with or without meaning from the letters A , G , M , D , N , and J , which are ending with G and none of the letters should be repeated? a) 20 b) 18 c) 25 d) 27

Last Answer : Answer: A) Since each desired word is ending with G, the least place is occupied with G. So, there is only 1 way. The second place can now be filled by any of the remaining 5 letters (A , M , D , N , J ... letters. So, there are 4 ways to fill. Required number of words = (1 x 5 x 4) = 20.

Description : How many 3 letters words (with or without meaning) can be formed out of the letters of the word, "PLATINUM", if repetition of letters is not allowed? a) 742 b) 850 c) 990 d) 336 

Last Answer : Answer: D)  The word PLATINUM contains 8 different letters. Required number of words = number of arrangements of 8 letters taking 3 at a time.  = 8p3  = 8 x 7 x 6  = 56×6  = 336

Description : How many different possible permutations can be made from the word ‘WAGGISH’ such that the vowels are never together? A) 3605 B) 3120 C) 1800 D) 1240 E) 2140

Last Answer : Answer: C)  The word ‘WAGGISH’ contains 7 letters of which 1 letter occurs twice  = 7! / 2! = 2520 No. of permutations possible with vowels always together = 6! * 2! / 2!  = 1440 / 2 = 720 No. of permutations possible with vowels never together = 2520-720  = 1800.

Description : How many arrangements can be made out of the letters of the word 'BIGBOSS' ? A) 9240 B) 2772 C) 1260 D) 1820 E) 2800

Last Answer : Answer: C)  The word 'BIGBOSS' has 7 letters  In these 7 letters, B(2) , I(1), G(1) , O(1),S(2)  Hence, number of ways to arrange these letters  = {7!} / (2!)(1!)(1!)(2!)}  = 5040/4  = 1260

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 

Description : Forty percent of Mouli's annual salary is equal to 160% of Surya's annual salary. Surya's monthly salary is 80%of Gowthaman's monthly salary. If Gowthaman's annual salary is ` 12 lacs, what is Mouli's ... income and in some place monthly income is given.) a) 180000 b) 1200000 c) 320000 d) 250000

Last Answer : Answer: C Gowthaman’s monthly salary = 12,00,000/12  = 1,00,000 Surya’s monthly salary = 1,00,000 x 80/100 = 80,000 Mouli’s monthly salary = 80,000x 1600/40  =3,20,000

Description : Mithun went to a shop and bought things worth Rs. 75, out of which 90 Paise went on sales tax on taxable purchases. If the tax rate was 18%, then what was the cost of the tax free items? a) 12.5 b) 69.80 c) 19.7 d) 34.5 e) 69.1

Last Answer : Answer: E  Total cost of the items he purchased = Rs.75  Given that out of this Rs.75, 90 Paise is given as tax  => Total tax incurred = 90 Paise  = Rs.90/100  Let the cost of the tax free items = x  Given that tax rate = ... −0.9 −x) = 90  ⇒ (75 − 0.9 − x) = 5  x = 75 − 0.9 - 5  X=69.1 

Description : What percent of a day in 12 hours? a) 25 b) 60 c) 45 d) 20 e) 50

Last Answer : Answer: E  Total hours in a day = 24 Required percent = 12/24 ×100  = 50%

Description : A shopkeeper bought 1800 blackberry and 1200 blueberry. He found 45% of blackberry and 24% of blueberry were rotten. Find the percentage of fruits in good condition. a) 63.4 b) 32.9 c) 48.5 d) 56.3

Last Answer :  Answer: A Total number of fruits shopkeeper bought = 1800 + 1200 = 1000 Number of rotten blackberry = 45% of 1800  = 45/100 1800  = 81000/100  = 810 Number of rotten blueberry = 24% of 1200 ... Therefore Percentage of fruits in good condition = (1902/3000 100)  = (190200/3000)  = 63.4%

Description : mouli had $ 3600 left after spending 40 % of the money he took for shopping. How much money did he take along with him? a) 2500 b) 6000 c) 4500 d) 3000 e) 9800 

Last Answer : Answer: B  Let the money he took for shopping be x.  Money he spent = 40 % of x  = 40/100 x  = 4/10 x  Money left with him = x - 4/10 x = (10x - 4x)/10 = 6x/10  But money left with ... 6x/10 = 3600  x = 3600 10/6  x = 36000/6  x = 6000  Therefore, the money he took for shopping is 6000.

Description : In a competitive examination in Pondicherry, 18% candidates got selected from the total appeared candidates. Tamilnadu and Pondicherry had an equal number of candidates appeared and in Tamilnadu 21% candidates got selected ... appeared from each State? a) 25000 b) 84000 c) 24000 d) 700000 e) 9800

Last Answer : Answer: C  Pondichery and Tamilnadu had an equal number of candidates appeared  In Pondicherry, 18% candidates got selected from the total appeared candidates  In tamilnadu, 21% candidates ... > total appeared candidates in pondicherry = total appeared candidates in tamilnadu = 24000

Description : The value of a machine depreciates at the rate of 20% every year. It was purchased 6 years ago. If its present value is Rs. 17,496, its purchase price was : a) 25023.4 b) 67040.0 c) 34171.8 d) 27337.5

Last Answer : Answer: C  Purchase price = 17,496 / (1 -20/ 100)³  = 17,496 × 10/8 × 10/8 ×10/8  = 17496 × 1.25 × 1.25 ×1.25  = 17496 ×1.953  = 34171.8

Description : Find the number which when decreased by 24 % becomes 594. a) 625.8 b) 360.9 c) 465.7 d) 781.5 e) 398.5

Last Answer : Answer: D Let the number be m. Decrease = 24 % of m  = 24/100 × m = 6m /25 Therefore, decrease number = m – 6m/25 = (25m – 6m)/25 = 19m/25 According to the question 19m/25 = 594  19m = 594 × 25  19m = 14850  m = 14850/19  m =781.5