Q: Watson bought a book for Rs. 240 and sold to Johny at a profit of 50%. Johny wants to sell this book to Shekar so that he earns a profit of 25%. But Shekar is adamant at buying this book at a discount of 10%. What marked price should Johny quote to Shekar so as to achieve desired profit ?
Solution: Watson bought the book for Rs. 240 and sold to Johny at a profit of 50%. S.P = C.P(1 + P%/100) => S.P for Watson = C.P for Johny = 240(1 + 50/100) = 240 x 1.5 = Rs. 360 Let Johny quoted the marked price of the book as Rs. M We know, SP = M.P(1 - Discount(%)/100) Here discount = 10% to Shekar, S.P for Johny = M(1 - 10/100) = 0.9M But Johny want to earn 25% profit, => S.P = C.P(1 + P%/100) => 0.9M = 360(1 + 25/100) => M = (360x1.25)/0.9 => M = Rs. 500 Therefore, Johny should quote Rs. 500 as the marked price of the book to get 25% profit and allowing 10% discount to Shekar.
Q: A shopkeeper gives 20% discount on the marked price of a book. He provides 1 pair of books free with the sale of 9 pair of books. In the whole transaction, he gets profit of 26%. Find the percentage increase in marked price from the cost price.
Solution: Given,Let the cost price of single book be Rs. 100. The cost price of (9 + 1) = 10 pair i.e. 20 books = Rs. (100 × 20) = Rs. 2000.He gets profit of 26%. So, the selling price of 9 pair i.e. 18 books = Rs. 2000 × (126/100) = Rs. 2520 Then,the selling price of single book = Rs. 2520/18 = Rs. 140 He gives 20% discount on the marked price of a book. That means, when the selling price is Rs. 80 then the marked price is Rs. 100. ∴When the selling price of single book is Rs. 140, the marked price = Rs. 140 × (100/80) = Rs. 175 ∴The percentage increase in marked price from the cost price = (175 –100)% = 75%
Q: If b2 + 1⁄b2 = 1, then value of b3 + 1⁄b3 is ,
Solution: E = (b + 1⁄b)(b2 - 1 + 1⁄b2)
Q: A student scores 55% marks in 8 papers of 100 marks each. He scores 15% of his total marks in English. How much does he score in English?
Solution: We have to find the marks scored by the student in English. Given that his score in English is 15% of his total marks. So let us first find the total marks scored by the student in all the 8 papers. He scored 55% in 8 papers of 100 marks each. Since each paper is of 100 marks, total marks of the exam = (100 × 8) = 800 Thus, the total marks scored by him in all the 8 papers = 55% of 800 = (55/100)× 800 = (55 × 8) = 440 Thus, the marks scored by the student in English = 15% of 440 = 15/100× 440 = 66
Q: A software engineer creates a LAN game where an 8 digit code made up of 1, 2, 3, 4, 5, 6, 7, 8 has to be decided on as a universal code. There is a condition that each number has to be used and no number can be repeated. What is the probability that the first 4 digits of the code are even number?
Solution: Given that an 8 digit code is made up of the digits 1, 2, 3, 4, 5, 6, 7, 8 using all without repeating any number.
Total number of possible codes = 8!
(Because 8 numbers can be arranged in 8 positions in 8! ways)
We have to find the probability that the first four digits of the code are even numbers.
Out of the given 8 numbers, exactly 4 are even numbers.
So, these 4 even numbers will occupy the first 4 positions in 4! ways
The remaining 4 odd numbers will occupy the remaining 4 positions (i.e. the last four positions) in 4! ways
To get an 8 digit code both the first 4 positions and the last 4 positions are to be occupied.
Thus, the total number of codes in which the first four digits are even numbers = 4! × 4!
We know that the probability of an event = Number of favourable outcomes / Total number of possible outcomes
Thus, the required probability = Number of codes with the first four digits as even numbers / Total number of possible codes
Q: A teacher has to choose the maximum different groups of three students from a total of six students. Of these groups, in how many groups there will be included in a particular student?
Solution: If students are A, B, C, D, E and F; we can have 6C3 groups in all. However, if we have to count groups in which a particular student (say A) is always selected we would get 5C2 = 10 ways of doing it.
Q: The sum of two numbers is 528 and their H.C.F is 33. The number of pairs of numbers satisfying the above condition is
Solution: Let the required numbers be 33a and 33b. Then 33a +33b= 528 => a+b = 16. Now, co-primes with sum 16 are (1,15) , (3,13) , (5,11) and (7,9). Therefore, Required numbers are ( 33 x 1, 33 x 15), (33 x 3, 33 x 13), (33 x 5, 33 x 11), (33 x 7, 33 x 9) The number of such pairs is 4
Q: Find the largest number of three digits exactly divisible by 15, 18, 27 and 30.
Solution:
Q: A woman fills a bucket in 6 minutes. 1845 buckets have to be filled from 8 am. to 9:30 am. How many woman employees should be employed for this task?
Solution: We know that, M1 × D1W1 = M2 × D2W2 Here M1 = 1, D1 = 6 min, W1 = 1 and M2 = M, D2 = 90 min, W2 = 1845 1 × 61 = M × 901845 => M = 123
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