Aptitude miscellaneous
- 2 km 5 m is equal to
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As we know that ,
∵ 1000 metre = 1 km∴ 5 metre = 5 = 0.005 km 1000
Correct Option: C
As we know that ,
∵ 1000 metre = 1 km∴ 5 metre = 5 = 0.005 km 1000
∴ 2 km 2 metre = (2 + 0.005) km = 2.005 km
- For Rs. 25,500, a furniture shop sells 3 computer tables and 5 chairs OR 2 computer tables and 9 chairs. If one wants to buy a set of only 1 computer table and 1 chair, how much does he need to pay?
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Let C.P. of 1 computer table = Rs. p
C.P. of 1 chair = Rs. q
According to the question,
3p + 5q = 25500 ..... (i)
2p + 9q = 25500 ..... (ii)
Clearly, cost of 1 computer table = cost of 4 chairs From equation (i),
3 × 4q + 5q = 25500 ⇒ 17q = 25500⇒ q = 25500 = 1500 17
Correct Option: D
Let C.P. of 1 computer table = Rs. p
C.P. of 1 chair = Rs. q
According to the question,
3p + 5q = 25500 ..... (i)
2p + 9q = 25500 ..... (ii)
Clearly, cost of 1 computer table = cost of 4 chairs From equation (i),
3 × 4q + 5q = 25500 ⇒ 17q = 25500⇒ q = 25500 = 1500 17
∴ Cost of 1 computer table = 4 × 1500 = Rs. 6000
∴ Required answer = 6000 + 1500 = Rs. 7500
- Every Sunday, Gin jogs 3 miles. For he rest of the week, each day he jogs 1 mile more than the previous day. How many miles Gin jogs in 2 weeks?
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As per the given in question ,
Total distance covered in two weeks = 2 (3 + 4 + 5 + 6 + 7 + 8 + 9) miles
Total distance covered in two weeks = 2 × 42Correct Option: C
As per the given in question ,
Total distance covered in two weeks = 2 (3 + 4 + 5 + 6 + 7 + 8 + 9) miles
Total distance covered in two weeks = 2 × 42 = 84 miles
Therefore , Gin jogs 84 miles in 2 weeks .
- Three runners A, B and C run a race, with runner A finishing 12 metres ahead of runner B and 18 metres ahead of runner C, while runner B finishes 8 metres ahead of runner C. Each runner travels the entire distance at a constant speed. The length of the race is
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Let the length of the race be y km.
When A finishes the race,
∴ A : B : C = y : (y – 12) : (y – 18)
When B covers 12 km. , C covers 10 km.∴ y - 12 = 12 = 6 y - 18 10 5
Correct Option: B
Let the length of the race be y km.
When A finishes the race,
∴ A : B : C = y : (y – 12) : (y – 18)
When B covers 12 km. , C covers 10 km.∴ y - 12 = 12 = 6 y - 18 10 5
⇒ 6y – 108 = 5y – 60
⇒ 6y – 5y = 108 – 60
⇒ y = 48 km.
Hence , required length of the race be 48 km.
- An hour-long test has 60 problems. If a student completes 30 problems in 25 minutes, then the required seconds he has taken on average for computing each of the remaining problems is
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From the question,
The student is required to answer 30 questions in 35 minutes.∴ Required average time = 35 × 60 seconds 30 Correct Option: A
From the question,
The student is required to answer 30 questions in 35 minutes.∴ Required average time = 35 × 60 seconds = 70 seconds 30