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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
In How Many Different Ways Can The Letters Of The Word 'happyholi' Be Arranged? |
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Answer» The GIVEN word HAPPYHOLI has 9 letters These 9 letters can e arranged in 9! WAYS. But here in the given word letters H & P are repeated TWICE each Therefore, NUMBER of ways these 9 letters can be arranged is 9!/2! x 2! = 9 x 8 x 7 x 6 x 5 x 4 x 3/2 = 90,720 ways. The given word HAPPYHOLI has 9 letters These 9 letters can e arranged in 9! ways. But here in the given word letters H & P are repeated twice each Therefore, Number of ways these 9 letters can be arranged is 9!/2! x 2! = 9 x 8 x 7 x 6 x 5 x 4 x 3/2 = 90,720 ways. |
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| 2. |
In How Many Different Ways Could Couples Be Picked From 6 Men And 9 Women ? |
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Answer»
Number of WOMENS = 5 Different ways COULD couples be picked = 6C1×9C1 = 9 x 6 = 54 ways. Number of mens = 8 Number of womens = 5 Different ways could couples be picked = 6C1×9C1 = 9 x 6 = 54 ways. |
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| 3. |
In How Many Ways Can Letter Of The Word Railings Arrange So That R And S Always Come Together? |
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Answer» The number of WAYS in which the letters of the word RAILINGS can be arranged such that R & S always come together is Count R & S as only 1 SPACE or letter so that RS or SR can be arranged => 7! X 2! But in the word RAILINGS, I repeated for 2 times => 7! x 2!/2! = 7! ways = 5040 ways. The number of ways in which the letters of the word RAILINGS can be arranged such that R & S always come together is Count R & S as only 1 space or letter so that RS or SR can be arranged => 7! x 2! But in the word RAILINGS, I repeated for 2 times => 7! x 2!/2! = 7! ways = 5040 ways. |
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| 4. |
In A Bag, There Are 8 Red, 7 Blue And 6 Green Flowers. One Of The Flower Is Picked Up Randomly. What Is The Probability That It Is Neither Red Nor Green? |
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Answer» TOTAL number of flowers = (8+7+6) = 21. Let E = event that the flower drawn is neither red nor GREEN. = event TAHT the flower drawn is blue. --> n(E)= 7 --> P(E)= 7/21=1/3. Total number of flowers = (8+7+6) = 21. Let E = event that the flower drawn is neither red nor green. = event taht the flower drawn is blue. --> n(E)= 7 --> P(E)= 7/21=1/3. |
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| 5. |
How Many Words Can Be Formed With Or Without Meaning By Using Three Letters Out Of K, L, M, N, O Without Repetition Of Alphabets? |
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Answer» Given letters are k, L, m, N, o = 5 number of letters to be in the WORDS = 3 Total number of words that can be FORMED from these 5 letters taken 3 at a time without repetation of letters = 5P3 ways. ⇒ 5P3 = 5 X 4 x 3 = 60 words. Given letters are k, l, m, n, o = 5 number of letters to be in the words = 3 Total number of words that can be formed from these 5 letters taken 3 at a time without repetation of letters = 5P3 ways. ⇒ 5P3 = 5 x 4 x 3 = 60 words. |
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| 6. |
To Fill 8 Vacancies There Are 15 Candidates Of Which 5 Are From St. If 3 Of The Vacancies Are Reserved For St Candidates While The Rest Are Open To All, Find The Number Of Ways In Which The Selection Can Be Done ? |
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Answer» ST CANDIDATES vacancies can be filled by 5C3 ways = 10 Remaining vacancies are 5 that are to be filled by 12 =>12C5= (12x11x10x9x8)/(5x4x3x2x1) = 792 Total number of filling the vacancies = 10 x 792 = 7920. ST candidates vacancies can be filled by 5C3 ways = 10 Remaining vacancies are 5 that are to be filled by 12 =>12C5= (12x11x10x9x8)/(5x4x3x2x1) = 792 Total number of filling the vacancies = 10 x 792 = 7920. |
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| 7. |
How Many More Words Can Be Formed By Using The Letters Of The Given Word 'creativity'? |
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Answer» The number of letters in the GIVEN WORD CREATIVITY = 10 Here T & I letters are repeated => Number of WORDS that can be formed from CREATIVITY = 10!/2!x2! = 3628800/4 = 907200. The number of letters in the given word CREATIVITY = 10 Here T & I letters are repeated => Number of Words that can be formed from CREATIVITY = 10!/2!x2! = 3628800/4 = 907200. |
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| 8. |
In How Many Different Ways Can The Letters Of The Word 'abysmal' Be Arranged ? |
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Answer» Total number of letters in the WORD ABYSMAL are 7 Number of ways these 7 letters can be ARRANGED are 7! ways But the LETTER is REPEATED and this can be arranged in 2! ways Total number of ways ARRANGING ABYSMAL = 7!/2! = 5040/2 = 2520 ways. Total number of letters in the word ABYSMAL are 7 Number of ways these 7 letters can be arranged are 7! ways But the letter is repeated and this can be arranged in 2! ways Total number of ways arranging ABYSMAL = 7!/2! = 5040/2 = 2520 ways. |
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| 9. |
If (1 × 2 × 3 × 4 ........ × N) = N!, Then 15! - 14! - 13! Is Equal To ___? |
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Answer»
= (15 × 14 × 13!) - (14 × 13!) - (13!) = 13! (15 × 14 - 14 - 1) = 13! (15 × 14 - 15) = 13! X 15 (14 - 1) = 15 × 13 × 13!. 15! - 14! - 13! = (15 × 14 × 13!) - (14 × 13!) - (13!) = 13! (15 × 14 - 14 - 1) = 13! (15 × 14 - 15) = 13! x 15 (14 - 1) = 15 × 13 × 13!. |
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| 10. |
In How Many Different Ways Can The Letters Of The Word 'therapy' Be Arranged So That The Vowels Never Come Together? |
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Answer» Given word is THERAPY. Number of letters in the given word = 7 These 7 letters can be ARRANGED in 7! ways. Number of vowels in the given word = 2 (E, A) The number of ways of ARRANGEMENT in which vowels come together is 6! x 2! ways HENCE, the required number of ways can the letters of the word 'THERAPY' be arranged so that the vowels never come together = 7! - (6! x 2!) ways = 5040 - 1440 = 3600 ways. Given word is THERAPY. Number of letters in the given word = 7 These 7 letters can be arranged in 7! ways. Number of vowels in the given word = 2 (E, A) The number of ways of arrangement in which vowels come together is 6! x 2! ways Hence, the required number of ways can the letters of the word 'THERAPY' be arranged so that the vowels never come together = 7! - (6! x 2!) ways = 5040 - 1440 = 3600 ways. |
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| 11. |
There Are Four Hotels In A Town. If 3 Men Check Into The Hotels In A Day Then What Is The Probability That Each Checks Into A Different Hotel? |
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Answer» TOTAL cases of checking in the HOTELS = 4 X 4 x 4 = 64 ways. Cases when 3 men are checking in different hotels = 4×3×2 = 24 ways. Required probability =24/64 = 3/8. Total cases of checking in the hotels = 4 x 4 x 4 = 64 ways. Cases when 3 men are checking in different hotels = 4×3×2 = 24 ways. Required probability =24/64 = 3/8. |
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| 12. |
In A Simultaneous Throw Of Two Dice, What Is The Probability Of Getting A Doublet ? |
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Answer» In a simultaneous throw of TWO dice, n(S) = 6 x 6 = 36 Let E = event of getting a doublet = { (1,1), (2,2), (3,3), (4,4), (5,5), (6,6)} P(E)=n(E)n(S)=6/36=1/6. In a simultaneous throw of two dice, n(S) = 6 x 6 = 36 Let E = event of getting a doublet = { (1,1), (2,2), (3,3), (4,4), (5,5), (6,6)} P(E)=n(E)n(S)=6/36=1/6. |
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| 13. |
A Number X Is Chosen At Random From The Numbers -3, -2, -1, 0, 1, 2, 3. What Is The Probability That |x|<2? |
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Answer»
To GET |X|+2) take X={-1,0,1} => P(|X|<2) = Favourable CasesTotal Cases = 3/7. X can take 7 values. To get |X|+2) take X={-1,0,1} => P(|X|<2) = Favourable CasesTotal Cases = 3/7. |
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| 14. |
What Is The Probability Of Getting At Least One Six In A Single Throw Of Three Unbiased Dice? |
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Answer» Find the number of cases in which NONE of the digits show a '6'. i.e. all three dice show a number other than '6', 5×5×5=125 cases. Total possible outcomes when three dice are thrown = 216. The number of outcomes in which at LEAST one die shows a '6' = Total possible outcomes when three dice are thrown - Number of outcomes in which none of them show '6'. =216-125=91 The required probability = 91/256. Find the number of cases in which none of the digits show a '6'. i.e. all three dice show a number other than '6', 5×5×5=125 cases. Total possible outcomes when three dice are thrown = 216. The number of outcomes in which at least one die shows a '6' = Total possible outcomes when three dice are thrown - Number of outcomes in which none of them show '6'. =216-125=91 The required probability = 91/256. |
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| 15. |
An Unbiased Die Is Tossed.find The Probability Of Getting A Multiple Of 3? |
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Answer» <P>Here S = {1,2,3,4,5,6} Let E be the event of getting the MULTIPLE of 3 Then, E = {3,6} P(E) = n(E)/n(S) = 2/6 = 1/3. Here S = {1,2,3,4,5,6} Let E be the event of getting the multiple of 3 Then, E = {3,6} P(E) = n(E)/n(S) = 2/6 = 1/3. |
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| 16. |
In A Single Throw Of Two Dice , Find The Probability That Neither A Doublet Nor A Total Of 8 Will Appear? |
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Answer» n(S) = 36 A = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)} B = { (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) } n(A)=6, n(B)=5, n(ANB)=1 Required probability = P(A?B) = P(A)+P(B)-P(AnB) = 6/36+5/36-1/36 = 5/18. n(S) = 36 A = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)} B = { (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) } n(A)=6, n(B)=5, n(AnB)=1 Required probability = P(A?B) = P(A)+P(B)-P(AnB) = 6/36+5/36-1/36 = 5/18. |
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| 17. |
Two Brother X And Y Appeared For An Exam. The Probability Of Selection Of X Is 1/7 And That Of B Is 2/9. Find The Probability That Both Of Them Are Selected? |
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Answer» LET A be the EVENT that X is selected and B is the event that Y is selected. P(A) = 1/7, P(B) = 2/9. Let C be the event that both are selected. P(C) = P(A) × P(B) as A and B are independent events: = (1/7) × (2/9) = 2/63. Let A be the event that X is selected and B is the event that Y is selected. P(A) = 1/7, P(B) = 2/9. Let C be the event that both are selected. P(C) = P(A) × P(B) as A and B are independent events: = (1/7) × (2/9) = 2/63. |
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| 18. |
Three Houses Are Available In A Locality. Three Persons Apply For The Houses. Each Applies For One House Without Consulting Others. The Probability That All The Three Apply For The Same House Is? |
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Answer» One person can select one HOUSE out of 3= 3C1 ways =3. HENCE, three persons can select one house out of 3 in 3 x 3 x 3 =9. Therefore, probability that all THRE apply for the same house is 1/9. One person can select one house out of 3= 3C1 ways =3. Hence, three persons can select one house out of 3 in 3 x 3 x 3 =9. Therefore, probability that all thre apply for the same house is 1/9. |
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| 19. |
A Bag Contains 2 Red, 3 Green And 2 Blue Balls. Two Balls Are Drawn At Random. What Is The Probability That None Of The Balls Drawn Is Blue? |
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Answer» Total number of balls = (2 + 3 + 2) = 7. Let S be the sample space. Then, N(S) = Number of WAYS of drawing 2 balls out of 7 =7C2 = 21 Let E = EVENT of drawing 2 balls, none of which is blue. n(E) = Number of ways of drawing 2 balls out of (2 + 3) balls =5C2= 10 Therefore, P(E) = n(E)/n(S) = 10/ 21. Total number of balls = (2 + 3 + 2) = 7. Let S be the sample space. Then, n(S) = Number of ways of drawing 2 balls out of 7 =7C2 = 21 Let E = Event of drawing 2 balls, none of which is blue. n(E) = Number of ways of drawing 2 balls out of (2 + 3) balls =5C2= 10 Therefore, P(E) = n(E)/n(S) = 10/ 21. |
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| 20. |
In A Box, There Are 8 Red, 7 Blue And 6 Green Balls. One Ball Is Picked Up Randomly. What Is The Probability That It Is Neither Red Nor Green? |
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Answer» Total number of balls = (8 + 7 + 6) = 21. LET E = EVENT that the BALL drawn is neither red nor green = event that the ball drawn is blue. n(E) = 7. P(E) = n(E)/n(S) = 7/21 = 1/3. Total number of balls = (8 + 7 + 6) = 21. Let E = event that the ball drawn is neither red nor green = event that the ball drawn is blue. n(E) = 7. P(E) = n(E)/n(S) = 7/21 = 1/3. |
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| 21. |
Arjun Bought A T.v With 20% Discount On The Labelled Price. Had He Bought It With 25% Discount, He Would Have Saved Rs. 500. At What Price Did He Buy The T.v? |
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Answer» LET labeled price was 100. Arjun bought T.V with 20% DISCOUNT, then Cost price of the T.V =labeled price - 20% oflabeled price = 100 - (20% of 100) = 80 Rs. If Arjunbought it at 25% discount, then Cost price of the T.V =100 - 25% of 100 = 75 Rs. Given, If Arjun bought T.V with 25% discount he would have saved Rs.500. So, Difference between Cost price for 20% and 25% discount = 500 => 80 - 75 = 500 => 5 parts = 500 => 1 PART = 500/5 =>1 part = 100 Then, the actual Cost price of the T.V =80 parts = 100 *80 = 8000. Correct Answer is Rs. 8000. Let labeled price was 100. Arjun bought T.V with 20% discount, then Cost price of the T.V =labeled price - 20% oflabeled price = 100 - (20% of 100) = 80 Rs. If Arjunbought it at 25% discount, then Cost price of the T.V =100 - 25% of 100 = 75 Rs. Given, If Arjun bought T.V with 25% discount he would have saved Rs.500. So, Difference between Cost price for 20% and 25% discount = 500 => 80 - 75 = 500 => 5 parts = 500 => 1 part = 500/5 =>1 part = 100 Then, the actual Cost price of the T.V =80 parts = 100 *80 = 8000. Correct Answer is Rs. 8000. |
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| 22. |
Find The Cost Price Of An Article Which Is Sold At A Loss Of 25% For Rs. 480? |
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Answer» We know that the article was SOLD at loss of 25% it MEANS that only 75% of the cost price is PAID. Which means, 75% of C.P = 480 => (75/100) x C.P = 480 => C.P = 480 x 4/3 = 160 x 4 = 640 Rs Cost price of an article = Rs 640. We know that the article was sold at loss of 25% it means that only 75% of the cost price is paid. Which means, 75% of C.P = 480 => (75/100) x C.P = 480 => C.P = 480 x 4/3 = 160 x 4 = 640 Rs Cost price of an article = Rs 640. |
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| 23. |
A Shopkeeper Buys Bananas At 15 For Rs 12 And Sells At 12 For Rs.15. Find His Gain Or Loss Percent? |
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Answer» Given, the C.P of 15 bananas = RS. 12 => C.P of 1 banana = Rs. 12/15 => C.P of 12 bananas = Rs. (12 /15)* 12 = Rs. 144/15 = Rs. 48/ 5 Thus,C.P of 12 bananas = Rs. 48/ 5 Given,S,P of 12 bananas = Rs. 15 Gain = (S.P - C.P)of 12 bananas = 15- (48/5) = (75- 48) / 5 = 27/ 5 =>Gain = Rs. 27/ 5 PERCENTAGE Gain = (Gain/C.P of 12 bananas) * 100% = (27/5) / (48/5) * 100% = (27/48) * 100% = 56.25% Thus,Gain % = 56.25%. Given, the C.P of 15 bananas = Rs. 12 => C.P of 1 banana = Rs. 12/15 => C.P of 12 bananas = Rs. (12 /15)* 12 = Rs. 144/15 = Rs. 48/ 5 Thus,C.P of 12 bananas = Rs. 48/ 5 Given,S,P of 12 bananas = Rs. 15 Gain = (S.P - C.P)of 12 bananas = 15- (48/5) = (75- 48) / 5 = 27/ 5 =>Gain = Rs. 27/ 5 Percentage Gain = (Gain/C.P of 12 bananas) * 100% = (27/5) / (48/5) * 100% = (27/48) * 100% = 56.25% Thus,Gain % = 56.25%. |
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| 24. |
A Girl Bought A Book For Rs.450 And Sold It At 20% Profit. By Using That Amount She Bought Another Book And Sold It At 5% Loss. Then Overall Profit Amount Is? |
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Answer»
Cost Price of 1st book = Rs.450. Profit % on SELLING 1st book = 20% Selling price of 1st book = {(100 + Profit%) / 100} * C.P = {(100 + 20) / 100} * 450 = (120 / 100) * 450 = 12 * 45 = 540 Rs. Cost price of 2nd book = Rs.540 Loss % on selling 2nd book = 5% Selling price of 2nd book = {(100 - Loss%) / 100} * C.P = {(100 - 5) / 100} * 540 = (95 / 100) * 540 = 513 Rs. OVERALL profit = Selling price of 2nd book - Cost Price of 1st book = Rs.513 - Rs.450 = Rs. 63. Given Cost Price of 1st book = Rs.450. Profit % on selling 1st book = 20% Selling price of 1st book = {(100 + Profit%) / 100} * C.P = {(100 + 20) / 100} * 450 = (120 / 100) * 450 = 12 * 45 = 540 Rs. Cost price of 2nd book = Rs.540 Loss % on selling 2nd book = 5% Selling price of 2nd book = {(100 - Loss%) / 100} * C.P = {(100 - 5) / 100} * 540 = (95 / 100) * 540 = 513 Rs. Overall profit = Selling price of 2nd book - Cost Price of 1st book = Rs.513 - Rs.450 = Rs. 63. |
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| 25. |
A Fruits Vender Cost Price Of 20 Apples Is Equal To Selling Price Of 16 Apples, What Is The Profit/loss Percentage? |
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Answer» Here using the following formula If cost price of “x “objects is same as selling price of “y’ objects, profit / LOSS PERCENTAGE [ ( x – y ) 100 ] / y Now x = 20 and y = 15 Profit / loss percent = [ ( 20 – 16) 100 ] / 16 = 400/16 = + 25% ( “+ve sig means profit and -ve sign means loss) Profit percent = 25%. Here using the following formula If cost price of “x “objects is same as selling price of “y’ objects, profit / loss percentage [ ( x – y ) 100 ] / y Now x = 20 and y = 15 Profit / loss percent = [ ( 20 – 16) 100 ] / 16 = 400/16 = + 25% ( “+ve sig means profit and -ve sign means loss) Profit percent = 25%. |
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| 26. |
The Cost Price Of The Two Articles Were Same And They Sold One At 25% Profit And Another Sold At 20% Loss. Now Find The Overall Profit/ Loss In This Transaction? |
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Answer» One SIMPLE formula while at same cost price Net profit / loss PERCENT = ( x + y ) / 2 Here to be take ” +ve” sign for profit and “-ve ” sign for Loss & same rule ALSO applicable for final answer. So x = 25 & y = -20 = ( 25 – 20) / 2 = + 2.5% Profit = 2.5%. One simple formula while at same cost price Net profit / loss percent = ( x + y ) / 2 Here to be take ” +ve” sign for profit and “-ve ” sign for Loss & same rule also applicable for final answer. So x = 25 & y = -20 = ( 25 – 20) / 2 = + 2.5% Profit = 2.5%. |
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| 27. |
Naveen Purchased A Cycle At Cost Of ?2000 And It Sold Ram At 20% Profit. Now Ram Spent ?100 For Its Repair And Sold To Jhon At 10% Loss. Find The Cost Price Of Jhon? |
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Answer» The Cost Price of NAVEEN = 2000 SELL price of Naveen = 20% profit on 2000 = 2000 x 120 / 100 = 2400 Ram cost price = 2400 + 100 = 2500 (because Ram spent ? 100 for its REPAIR ) Sell price of Ram = 10% loss on 2500 = 2500 x 90 / 100 = 2250 The Cost price of Jhon =2250. The Cost Price of Naveen = 2000 Sell price of Naveen = 20% profit on 2000 = 2000 x 120 / 100 = 2400 Ram cost price = 2400 + 100 = 2500 (because Ram spent ? 100 for its repair ) Sell price of Ram = 10% loss on 2500 = 2500 x 90 / 100 = 2250 The Cost price of Jhon =2250. |
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| 28. |
A Pair Of Articles Was Bought For Rs. 37.40 At A Discount Of 15%. What Must Be The Marked Price Of Each Of The Articles ? |
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Answer» As question states that rate was of PAIR of articles, So rate of One article = 37.40/2 = Rs. 18.70 then 85% of X = 18.70 => X = 1870/85 = 22. As question states that rate was of pair of articles, So rate of One article = 37.40/2 = Rs. 18.70 Let Marked price = Rs X then 85% of X = 18.70 => X = 1870/85 = 22. |
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| 29. |
A Producer Of Tea Blends Two Varieties Of Tea From Two Tea Gardens One Costing Rs 18 Per Kg And Another Rs 20 Per Kg In The Ratio 5 : 3. If He Sells The Blended Variety At Rs 21 Per Kg, Then His Gain Percent Is? |
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Answer» Suppose he BOUGHT 5 kg and 3 kg of tea. Cost Price = Rs. (5 x 18 + 3 x 20) = Rs. 150. Selling price = Rs. (8 x 21) = Rs. 168. Profit = 168 - 150 = 18 So, Profit % = (18/150) * 100 = 12%. Suppose he bought 5 kg and 3 kg of tea. Cost Price = Rs. (5 x 18 + 3 x 20) = Rs. 150. Selling price = Rs. (8 x 21) = Rs. 168. Profit = 168 - 150 = 18 So, Profit % = (18/150) * 100 = 12%. |
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| 30. |
If The Cost Price Of 12 Items Is Equal To The Selling Price Of 16 Items, The Loss Percent Is? |
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Answer» Let the COST Price of 1 item = Re. 1 Cost Price of 16 items = 16 Selling Price of 16 items = 12 Loss = 16 - 12 = RS 4 Loss % = (4/16)* 100 = 25%. Let the Cost Price of 1 item = Re. 1 Cost Price of 16 items = 16 Selling Price of 16 items = 12 Loss = 16 - 12 = Rs 4 Loss % = (4/16)* 100 = 25%. |
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| 31. |
A Man Gains 20% By Selling An Article For A Certain Price. If He Sells It At Double The Price, The Percentage Of Profit Will Be? |
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Answer»
Then S.P. = (120/100)x = 6x/5 New S.P. = 2(6x/5) = 12x/5 Profit = 12x/5 - x = 7x/5 Profit% = (Profit/C.P.) * 100 => (7x/5) * (1/x) * 100 = 140 %. Let the C.P. = x, Then S.P. = (120/100)x = 6x/5 New S.P. = 2(6x/5) = 12x/5 Profit = 12x/5 - x = 7x/5 Profit% = (Profit/C.P.) * 100 => (7x/5) * (1/x) * 100 = 140 %. |
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| 32. |
Mr. Rajan Invested Rs 1,00,000 In Us Stock Markets When The Gbpinr Rate Was 75. After One Year His Investment Appreciated By 20% In Gbp Terms. He Sold Of His Investments And Repatriated The Money To India At The Then Existing Rate Of 80. What Was Real Returns In Inr? |
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Answer» Money invested by Rajan before 1 year was = Rs. 100000 Money in UK pounds @ 75 is = 100000/75 = 1333.33 Pounds Now, after 1 year invested amount was appreciated by 20% => 20% of 1333.33 = 266.66 Total investment becomes = 1333.33 + 266.66 = 1600 Pounds This 1600 Pounds @ Indian currency at 80 = 1600 X 80 = Rs. 1,28,000 Hence, Rajan's investment of Rs. 1,00,000 becomes Rs. 1,28,000 in 1 year THEREFORE, his profit % = [(128000 - 100000)/100000] x 100 = 28%. Money invested by Rajan before 1 year was = Rs. 100000 Money in UK pounds @ 75 is = 100000/75 = 1333.33 Pounds Now, after 1 year invested amount was appreciated by 20% => 20% of 1333.33 = 266.66 Total investment becomes = 1333.33 + 266.66 = 1600 Pounds This 1600 Pounds @ Indian currency at 80 = 1600 x 80 = Rs. 1,28,000 Hence, Rajan's investment of Rs. 1,00,000 becomes Rs. 1,28,000 in 1 year Therefore, his profit % = [(128000 - 100000)/100000] x 100 = 28%. |
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| 33. |
Cost Price Of 22 Articles Is Same As The Selling Price Of 18 Articles, Find The Profit Percentage? |
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Answer» LET the cost price of 1 article = Rs. 1 From the given DATA, Then, the SELLING price of 1 article = 22/18 = 11/9 Then, PROFIT = SP - CP = 11/9 - 1 = 2/9 Required, profit % = Profit/CP x 100 = [(2/9)/1] x 100 = 200/9 = 22.222%. Let the cost price of 1 article = Rs. 1 From the given data, Then, the selling price of 1 article = 22/18 = 11/9 Then, Profit = SP - CP = 11/9 - 1 = 2/9 Required, profit % = Profit/CP x 100 = [(2/9)/1] x 100 = 200/9 = 22.222%. |
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| 34. |
The Least Number Which Should Be Added To 2497 So That The Sum Is Exactly Divisible By 5, 6, 4 And 3 Is? |
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Answer» L.C.M. of 5, 6, 4 and 3 = 60. On DIVIDING 2497 by 60, the remainder is 37. Number to be ADDED = (60 – 37) = 23. L.C.M. of 5, 6, 4 and 3 = 60. On dividing 2497 by 60, the remainder is 37. Number to be added = (60 – 37) = 23. |
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| 35. |
The H.c.f. And L.c.m. Of Two Numbers Are 11 And 385 Respectively. If One Number Lies Between 75 And 125, Then That Number Is? |
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Answer» Product of numbers = 11 x 385 = 4235. Let the numbers be 11a and 11b. Then 11a x 11b = 4235 ab = 35 Now, co-primes with product 35 are (1, 35) and (5, 7). So, the numbers are (11 x 1, 11 x 35) and (11 x 5, 11 x 7). Since one NUMBER lies between 75 and 125, the suitable pair is (55, 77) Hence, required number = 77. Product of numbers = 11 x 385 = 4235. Let the numbers be 11a and 11b. Then 11a x 11b = 4235 ab = 35 Now, co-primes with product 35 are (1, 35) and (5, 7). So, the numbers are (11 x 1, 11 x 35) and (11 x 5, 11 x 7). Since one number lies between 75 and 125, the suitable pair is (55, 77) Hence, required number = 77. |
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| 36. |
The Least Number Of Five Digits Which Is Exactly Divisible By 12, 15 And 18, Is? |
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Answer» LEAST number of 5 digits is 10000. L.C.M. of 12, 15 and 18 is 180. On DIVIDING 10000 by 180, the remainder is 100. Required number = 10000 + (180 - 100) = 10080. Least number of 5 digits is 10000. L.C.M. of 12, 15 and 18 is 180. On dividing 10000 by 180, the remainder is 100. Required number = 10000 + (180 - 100) = 10080. |
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| 37. |
The Least Number Which Is A Perfect Square And Is Divisible By Each Of The Numbers 16, 20 And 24, Is? |
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Answer» The LEAST number divisible by 16, 20, 24 = L.C.M. of 16, 20, 24 = 240 = 2 x 2 x 2 x 2 x 3 x 5 To make it a PERFECT SQUARE, it must be multiplied by 3 x 5 Required number = 240 x 3 x 5 = 3600. The least number divisible by 16, 20, 24 = L.C.M. of 16, 20, 24 = 240 = 2 x 2 x 2 x 2 x 3 x 5 To make it a perfect square, it must be multiplied by 3 x 5 Required number = 240 x 3 x 5 = 3600. |
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| 38. |
Let The Least Number Of Six Digits, Which When Divided By 4, 6, 10 And 15, Leaves In Each Case The Same Remainder Of 2, Be N. The Sum Of The Digits In N Is? |
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Answer» LEAST NUMBER of 6 DIGITS is 100000. L.C.M. of 4, 6, 10 and 15 = 60. On dividing 100000 by 60, the remainder obtained is 40. Least number of 6 digits divisible by 4, 6, 10 and 15 = 100000 + (60 - 40) = 100020. N = (100020 + 2) = 100022. Sum of digits in N = (1 + 2 + 2) = 5. Least number of 6 digits is 100000. L.C.M. of 4, 6, 10 and 15 = 60. On dividing 100000 by 60, the remainder obtained is 40. Least number of 6 digits divisible by 4, 6, 10 and 15 = 100000 + (60 - 40) = 100020. N = (100020 + 2) = 100022. Sum of digits in N = (1 + 2 + 2) = 5. |
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| 39. |
The Product Of The L.c.m. And H.c.f. Of Two Numbers Is 24. The Difference Of Two Numbers Is 2. Find The Numbers? |
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Answer» Let the numbers be x and (x + 2). Then, x(x + 2) = 24 x = 4 So, the numbers are 4 and 6. Let the numbers be x and (x + 2). Then, x(x + 2) = 24 x2+ 2x - 24 = 0 (x - 4) (x + 6) = 0 x = 4 So, the numbers are 4 and 6. |
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| 40. |
The Least Number, Which When Divided By 48, 60, 72, 108 And 140 Leaves 38, 50, 62, 98 And 130 As Remainders Respectively, Is? |
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Answer» Here (48 - 38) = 10, (60 - 50) = 10, (72 - 62) = 10, (108 - 98) = 10 & (140 - 130) = 10. Required number = (L.C.M. of 48, 60, 72, 108, 140) ? 10 = 15120 - 10 = 15110. Here (48 - 38) = 10, (60 - 50) = 10, (72 - 62) = 10, (108 - 98) = 10 & (140 - 130) = 10. Required number = (L.C.M. of 48, 60, 72, 108, 140) ? 10 = 15120 - 10 = 15110. |
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| 41. |
The Largest Four ? Digit Number Which When Divided By 4, 7 Or 13 Leaves A Remainder Of 3 In Each Case, Is? |
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Answer» Greatest NUMBER of 4 digits is 9999, L.C.M. of 4, 7 and 13 = 364 On DIVIDING 9999 by 364, REMAINDER OBTAINED is 171. Greatest number of 4 digits DIVISIBLE by 4, 7 and 13 = (9999 ? 171) = 9828 Hence, required number = (9828 + 3) = 9831. Greatest number of 4 digits is 9999, L.C.M. of 4, 7 and 13 = 364 On dividing 9999 by 364, remainder obtained is 171. Greatest number of 4 digits divisible by 4, 7 and 13 = (9999 ? 171) = 9828 Hence, required number = (9828 + 3) = 9831. |
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| 42. |
H.c.f. Of 3240, 3600 And A Third Number Is 36 And Their L.c.m. Is 24x 35 X 52 X 72 The Third Number Is? |
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Answer» 3240 = 23x 34x 5; 3600 = 24x32 X 52; H.C.F. = 36 = 22x32 SINCE H.C.F. is the PRODUCT of lowest powers of common factors, so the third number must have (22x32) as its factors Since L.C.M. is the product of highest powers of common prime factors, so the third number must have 35AND 72 as its factors. Third number = 22x35x72. 3240 = 23x 34x 5; 3600 = 24x32 x 52; H.C.F. = 36 = 22x32 Since H.C.F. is the product of lowest powers of common factors, so the third number must have (22x32) as its factors Since L.C.M. is the product of highest powers of common prime factors, so the third number must have 35and 72 as its factors. Third number = 22x35x72. |
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