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 Total number of questions : 20.
 Time alloted : 30 minutes.
 Each question carry 1 mark, no negative marks.
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1.  Father is aged three times more than his son Ronit. After 8 years, he would be two and a half times of Ronit's age. After further 8 years, how many times would he be of Ronit's age? 

Your Answer: Option (Not Answered) Correct Answer: Option A Explanation: Let Ronit's present age be x years. Then, father's present age =(x + 3x) years = 4x years.
8x + 16 = 5x + 40 3x = 24 x = 8.
Learn more problems on : Problems on Ages Discuss about this problem : Discuss in Forum 
2.  Q is as much younger than R as he is older than T. If the sum of the ages of R and T is 50 years, what is definitely the difference between R and Q's age? 

Your Answer: Option (Not Answered) Correct Answer: Option D Explanation: Given that: 1. The difference of age b/w R and Q = The difference of age b/w Q and T. 2. Sum of age of R and T is 50 i.e. (R + T) = 50. Question: R  Q = ?. Explanation: R  Q = Q  T (R + T) = 2Q Now given that, (R + T) = 50 So, 50 = 2Q and therefore Q = 25. Question is (R  Q) = ? Here we know the value(age) of Q (25), but we don't know the age of R. Therefore, (RQ) cannot be determined. Learn more problems on : Problems on Ages Discuss about this problem : Discuss in Forum 
3.  The last day of a century cannot be 

Your Answer: Option (Not Answered) Correct Answer: Option C Explanation: 100 years contain 5 odd days. Last day of 1^{st} century is Friday. 200 years contain (5 x 2) 3 odd days. Last day of 2^{nd} century is Wednesday. 300 years contain (5 x 3) = 15 1 odd day. Last day of 3^{rd} century is Monday. 400 years contain 0 odd day. Last day of 4^{th} century is Sunday. This cycle is repeated. Last day of a century cannot be Tuesday or Thursday or Saturday. Learn more problems on : Calendar Discuss about this problem : Discuss in Forum 
4.  The captain of a cricket team of 11 members is 26 years old and the wicket keeper is 3 years older. If the ages of these two are excluded, the average age of the remaining players is one year less than the average age of the whole team. What is the average age of the team? 

Your Answer: Option (Not Answered) Correct Answer: Option A Explanation: Let the average age of the whole team by x years. 11x  (26 + 29) = 9(x 1) 11x  9x = 46 2x = 46 x = 23. So, average age of the team is 23 years. Learn more problems on : Average Discuss about this problem : Discuss in Forum 
5.  Six bells commence tolling together and toll at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together ? 

Your Answer: Option (Not Answered) Correct Answer: Option D Explanation: L.C.M. of 2, 4, 6, 8, 10, 12 is 120. So, the bells will toll together after every 120 seconds(2 minutes).
Learn more problems on : Problems on H.C.F and L.C.M Discuss about this problem : Discuss in Forum 
6.  If 2994 ÷ 14.5 = 172, then 29.94 ÷ 1.45 = ? 

Your Answer: Option (Not Answered) Correct Answer: Option C Explanation:
= 17.2 Learn more problems on : Decimal Fraction Discuss about this problem : Discuss in Forum 
Insert the missing number.  
7.  2, 4, 12, 48, 240, (....) 

Your Answer: Option (Not Answered) Correct Answer: Option B Explanation: Go on multiplying the given numbers by 2, 3, 4, 5, 6. So, the correct next number is 1440. Learn more problems on : Odd Man Out and Series Discuss about this problem : Discuss in Forum 
8.  The true discount on Rs. 2562 due 4 months hence is Rs. 122. The rate percent is: 

Your Answer: Option (Not Answered) Correct Answer: Option C Explanation: P.W. = Rs. (2562  122) = Rs. 2440. S.I. on Rs. 2440 for 4 months is Rs. 122.
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Each of the questions given below consists of a question followed by three statements. You have to study the question and the statements and decide which of the statement(s) is/are necessary to answer the question.  
9. 


Your Answer: Option (Not Answered) Correct Answer: Option C Explanation: I. Material cost = Rs. 2.50 per m^{2} II. Labour cost = Rs. 3500. III. Total cost = Rs. 14,500. Let the area be A sq. metres. Material cost = Rs. (14500  3500) = Rs. 11,000.
Thus, all I, II and III are needed to get the answer. Correct answer is (C). Learn more problems on : Area Discuss about this problem : Discuss in Forum 
Find out the wrong number in the series.  
10.  64, 71, 80, 91, 104, 119, 135, 155 

Your Answer: Option (Not Answered) Correct Answer: Option E Explanation: Go on adding 7, 9, 11, 13, 15, 17, 19 respectively to obtain the next number. So, 135 is wrong. Learn more problems on : Odd Man Out and Series Discuss about this problem : Discuss in Forum 
11.  A, B and C can complete a piece of work in 24, 6 and 12 days respectively. Working together, they will complete the same work in: 

Your Answer: Option (Not Answered) Correct Answer: Option C Explanation:
Learn more problems on : Time and Work Discuss about this problem : Discuss in Forum 
12.  4 men and 6 women can complete a work in 8 days, while 3 men and 7 women can complete it in 10 days. In how many days will 10 women complete it? 

Your Answer: Option (Not Answered) Correct Answer: Option B Explanation: Let 1 man's 1 day's work = x and 1 woman's 1 day's work = y.
Hence, 10 women will complete the work in 40 days. Learn more problems on : Time and Work Discuss about this problem : Discuss in Forum 
13.  A works twice as fast as B. If B can complete a work in 12 days independently, the number of days in which A and B can together finish the work in : 

Your Answer: Option (Not Answered) Correct Answer: Option A Explanation: Ratio of rates of working of A and B = 2 : 1. So, ratio of times taken = 1 : 2.
So, A and B together can finish the work in 4 days. Learn more problems on : Time and Work Discuss about this problem : Discuss in Forum 
Find out the wrong number in the given sequence of numbers.  
14.  6, 13, 18, 25, 30, 37, 40 

Your Answer: Option (Not Answered) Correct Answer: Option D Explanation: The differences between two successive terms from the beginning are 7, 5, 7, 5, 7, 5. So, 40 is wrong. Learn more problems on : Odd Man Out and Series Discuss about this problem : Discuss in Forum 
15.  A, B and C jointly thought of engaging themselves in a business venture. It was agreed that A would invest Rs. 6500 for 6 months, B, Rs. 8400 for 5 months and C, Rs. 10,000 for 3 months. A wants to be the working member for which, he was to receive 5% of the profits. The profit earned was Rs. 7400. Calculate the share of B in the profit. 

Your Answer: Option (Not Answered) Correct Answer: Option B Explanation: For managing, A received = 5% of Rs. 7400 = Rs. 370. Balance = Rs. (7400  370) = Rs. 7030. Ratio of their investments = (6500 x 6) : (8400 x 5) : (10000 x 3) = 39000 : 42000 : 30000 = 13 : 14 : 10
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16.  Find a positive number which when increased by 17 is equal to 60 times the reciprocal of the number. 

Your Answer: Option (Not Answered) Correct Answer: Option A Explanation: Let the number be x.
x^{2} + 17x  60 = 0 (x + 20)(x  3) = 0 x = 3. Learn more problems on : Problems on Numbers Discuss about this problem : Discuss in Forum 
17.  A can contains a mixture of two liquids A and B is the ratio 7 : 5. When 9 litres of mixture are drawn off and the can is filled with B, the ratio of A and B becomes 7 : 9. How many litres of liquid A was contained by the can initially? 

Your Answer: Option (Not Answered) Correct Answer: Option C Explanation: Suppose the can initially contains 7x and 5x of mixtures A and B respectively.
252x  189 = 140x + 147 112x = 336 x = 3. So, the can contained 21 litres of A. Learn more problems on : Alligation or Mixture Discuss about this problem : Discuss in Forum 
18.  Tea worth Rs. 126 per kg and Rs. 135 per kg are mixed with a third variety in the ratio 1 : 1 : 2. If the mixture is worth Rs. 153 per kg, the price of the third variety per kg will be: 

Your Answer: Option (Not Answered) Correct Answer: Option C Explanation: Since first and second varieties are mixed in equal proportions.
So, the mixture is formed by mixing two varieties, one at Rs. 130.50 per kg and the other at say, Rs. x per kg in the ratio 2 : 2, i.e., 1 : 1. We have to find x. By the rule of alligation, we have:
x  153 = 22.50 x = 175.50 Learn more problems on : Alligation or Mixture Discuss about this problem : Discuss in Forum 
19.  In how many different ways can the letters of the word 'CORPORATION' be arranged so that the vowels always come together? 

Your Answer: Option (Not Answered) Correct Answer: Option D Explanation: In the word 'CORPORATION', we treat the vowels OOAIO as one letter. Thus, we have CRPRTN (OOAIO). This has 7 (6 + 1) letters of which R occurs 2 times and the rest are different.
Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged
Required number of ways = (2520 x 20) = 50400. Learn more problems on : Permutation and Combination Discuss about this problem : Discuss in Forum 
20.  In how many different ways can the letters of the word 'LEADING' be arranged in such a way that the vowels always come together? 

Your Answer: Option (Not Answered) Correct Answer: Option C Explanation: The word 'LEADING' has 7 different letters. When the vowels EAI are always together, they can be supposed to form one letter. Then, we have to arrange the letters LNDG (EAI). Now, 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways. The vowels (EAI) can be arranged among themselves in 3! = 6 ways. Required number of ways = (120 x 6) = 720. Learn more problems on : Permutation and Combination Discuss about this problem : Discuss in Forum 