Test Series - w Quants

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Test Number 7/24

Q: A shopkeeper marks his goods 50% more than the cost price and allows a discount of 25%. His profit or loss percentage is:
A. 25.5%
B. 25%
C. 12.5%
D. 37.5%
Solution: 12.5%
Q: A dealer sold an article at 6% loss. Had he sold it for Rs. 64 more, he would have made a profit of 10%. Then the cost of the article is
A. Rs.200
B. Rs.400
C. Rs.464
D. Rs.164
Solution: Rs.400
Q: Raj sells a machine for Rs 51 lakhs at a loss. Had he sold it for Rs 60 lakh, his gain would have been 8 times the earlier loss. What is the cost price of the machine?
A. Rs 59 lakhs
B. Rs 52 lakhs
C. Rs 66.375 lakhs
D. Rs 45 lakhs
Solution: Rs 52 lakhs
Q: Two pipes A and B can separately fill a cistern in 10 and 15 minutes respectively. A person opens both the pipes together when the cistern should have been was full he finds the waste pipe open. He then closes the waste pipe and in another 4 minutes the cistern was full. In what time can the waste pipe empty the cistern when fill?
A. 8 min
B. 10 min
C. 9 min
D. 7 min
Solution: 1/10 + 1/15 = 1/6 * 4 = 2/3

1 - 2/3 = 1/3

1/10 + 1/15 - 1/x = 1/3

x = 8
Q: The ages of Krish and Vaibhav are in the proportion of 3 : 5. After 9 years, the proportion of their ages will be 3 : 4. Then the current age of Vaibhav is:
A. 10
B. 15
C. 18
D. 13
Solution: Krish’s age = 3A and Vaibhav’s age = 5A
(3A+9)/(5A+9) = 3/4
=> 4 (3A + 9) = 3 (5A + 9)
=> A = 3
Therefore, Vaibhav’s age = 15 years
Q: The present ages of Pranay and Sai are in the ratio of 15 : 17 respectively. After 6 years, the respective ratio between the age of Pranay and Sai will be 9 : 10. What will be the age of Pranay after 6 years?
A. 42
B. 30
C. 24
D. 36
Solution: Let present age of Pranay = 15p
That of Sai = 17p
ATQ,
15p + 6/17p + 6 = 9/10
=> 150p + 60 = 153p + 54
=> p = 2
Therefore, Age of Pranay after 6 yrs = 15 x 2 + 6 = 36 years.
Q: A shopkeeper has a job to print certain number of documents and there are three machines P, Q and R for this job. P can complete the job in 3 days, Q can complete the job in 4 days and R can complete the job in 6 days. How many days the shopkeeper will it take to complete the job if all the machines are used simultaneously ?
A. 3/2 days
B. 2 days
C. 4/3 days
D. 4 days
Solution: Let the total number of documents to be printed be 12.
 The number of documents printed by P in 1 day = 4.
 The number of documents printed by Q in 1 day = 3.
 The number of documents printed by R in 1 day = 2.
 Thus, the total number of documents that can be printed by all the machines working simultaneously in a single day = 9.
 
Therefore, the number of days taken to complete the whole work = 12/9 = 4/3 days.

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