MCQ Answers

1. The average age of 3 girls is 20 years and their ages are in the proportion 3:5:7. The age of the youngest girl is Answer: 12 years Solution: Sum of ages = 3*20 = 60. Let the ages be 3x,5x,7x, Adding, 15x=60, x=4. Hence youngest girl's age = 3x = 12 years 2. The numerator and the denominator of a fraction are in the ratio 3 : 4. When both are increased by the same value, the ratio becomes 4 : 5. Which of the following could be the value added? Answer: All of these Solution: Let the numerator be 3x. Then the denominator is 4x. If y be the quantity added, then (3x + y)/(4x + y) = 4/5 (15x + 5y) = (16x + 4y), x = y. So for every value of x there is a value of y. 3. Monthly earnings of two persons are in the ratio 4:5 and their monthly expenses are in the ratio 7:9. If each saves Rs.5000 per month, their respective monthly incomes are Answer: Rs. 40000 and Rs.50000 Solution: Let the monthly incomes be Rs.4x and Rs.5x respectively and monthly expenses be Rs.7y and Rs.9y respectively....

1. Read the following passage and answer the questions below Virtually everything astronomers known about objects outside the solar system is based on the detection of photons—quanta of electromagnetic radiation. Yet there is another form of radiation that permeates the universe: neutrinos. With (as its name implies) no electric charge, and negligible mass, the neutrino interacts with other particles so rarely that a neutrino can cross the entire universe, even traversing substantial aggregations of matter, without being absorbed or even deflected. Neutrinos can thus escape from regions of space where light and other kinds of electromagnetic radiation are blocked by matter. Furthermore, neutrinos carry with them information about the site and circumstances of their production: therefore, the detection of cosmic neutrinos could provide new information about a wide variety of cosmic phenomena and about the history of the universe. But how can scientists detect a particle that interacts s...

 1. Between 250 and 750,how many integers are divisible by 11? Answer: 46 Solution: The integers are like 253, 264, ...., 748 Here d=11 and a1 = 253, An=748. So, 748 = 253 + 11(n-1), Solving n=46 2. Three terms in a geometric progression have their sum as 49 and the product as 2744. Find the smallest among these terms. Answer: 7 Solution: Let the terms be x/d, x, xd. Product = x^3 = 2744. So x=14. Sum = 14/d + 14 + 14d = 49, Solving d=1/2. The terms are 28, 14, 7. 3. Find the sum of 20 + 40 + 60 + 80 + 100 + ........ + 2000 Answer: 101000 Solution: 20 + 40 + 60 + 80 + 100 + ........ + 2000 can be written as 20 (1+2+3+ .....+100) = 20 * 100/2 (1+100) = 101000   4. If the sum of 3rd and 15th terms of an arithmetic progression is equal to the sum of 6th, 11th and 13th terms of the same progression, which term of the series should be equal to zero? Answer: 12th term Solution: Let the initial term be a and the difference between the terms be d. 3rd term = a+2d and 15th term = ...

1. A dishonest seller uses 860 grams instead of 1kg weight scale. Find his actual profit or loss if the seller says he sells his items on 5% gain on cost price. Answer: Gain of 22.09% Solution: Let x be the proper weight and y be the actual weight used. Let p be the gain percentage mentioned. The shortcut to calculate the actual gain percentage is = (100+p)(x/y) -100. So gain % = (100+5) * (1000/860)  - 100 = 22.09% 2. A shopkeeper proposes to sell at cost price but has a faulty balance which shows 1000 grams for 800 grams. What is his profit percentage? Answer: 25% Solution: Difference = 1000 - 800 = 200 grams. Profit percentage = (200/800) * 100 = 25% 3. A manufaturer supplies 50 tables at the marked price of 42 tables to the retailer. The retailer sells them at a 10% discount on the marked price. What is the approximate profit or loss percentage of the retailer? Answer: Profit of 7.14% Solution: Let marked price of a table be x. C.P of 50 tables for the retailer = 42x. S.P of 50...

1. Choose the option to fill in the blanks. A discount series of 10%, 20% and 40% is equal to a single discount of _ . Answer: 56.8% Solution: Let price be x. Final price = x * 0.9 * 0.8 * 0.6 = 0.432x. Hence effective discount = 1 - 0.432x = 0.568x. As % it is 56.8% 2. By selling an article for Rs.21, a man lost such that the percentage loss was equal to the cost price. The cost price of the article was Answer: Rs.70 Solution: Let c.p be x. Loss = x-21 100 * (x-21)/x = x, Solving x=30 or 70. 3. The C.P of 15 articles is equal to the S.P of 10 articles. Find the percentage of the profit. Answer: 50% Solution: 15*c.p = 10*s.p, s.p = 1.5c.p Profit = 0.5c.p, Gain % = 100 * 0.5c.p/c.p = 50% 4. Two articles are bought at the same price. One is sold at 20% profit and the other is sold at 10% loss. Find the overall profit/loss percentage. Answer: 5% profit Solution: Let C.P of one article be x. Total C.P = 2x Total S.P = 1.2x + 0.9x = 2.1x. Hence Profit = 0.1x Profit % = 100 * 0.1x/2x = 5% 5....

 1. In a class 52% of the students are girls of which 25% are interested in tennis. What is the probability that a randomly selected student from the class is a girl who is interested in tennis? Answer: 13/100 Solution: Assume 100 students are in the class. Num of girls = 52 Girls interested in tennis = 52 * 25/100 = 13 Reqd probability = 13/100 2. On a highway, the probability of seeing an ambulance during a twenty-minutes period is 11/36. What is the probability of not seeing an ambulance in a ten-minutes period? Answer: 5/6 Solution: Probability of not seeing an ambulance in 20 mins period = 1 - 11/36 = 25/36 Probability of not seeing a lorry in 10 mins period = Square root of 25/36 = 5/6 3. Find the probability that 2 men selected at random were born in the same month. Answer: 1/12 Solution: The first person can be born in any of the 12 months. The second person also being born in the same month = 1/12. 4. Four boys and three girls stand in queue for an interview. The probabili...

1. A bomber drops 4 bombs to destroy a target. Two bombs are enough to destroy the target. If the chance of a bomb hitting the target is 0.4, find the probability that the target is not destroyed. Answer: 297/625 Solution: Probability of a bomb hitting the target = 0.4 = 2/5, not hitting = 3/5 The target is not destroyed only when just one or no bomb hits it. Probability that all bombs miss = (3/5)^4 = 81/625 Probability that just one bomb hits = 4C1 * (2/5) * (3/5)^3 = 216/625 Probability that the target is NOT destroyed = 81/625 + 216/625 = 297/625 2. A classroom has 3 electric bulb holders. From a collection of 10 bulbs of which 4 are defective, 3 are selected at random and put in the holders. Find the probability that all bulbs are glowing. Answer: 1/6 Solution: Number of ways to pick all three bulbs as working = 6C3 = 20 Total number of ways to pick 3 bulbs = 10C3 = 120 Hence probability that all bulbs are glowing = 20/120 = 1/6 3. In a company there are 5 engineers out of 20 work...