Quantitative Aptitude

Sample Paper 9

Explanation:

Question 1: The sum of ratio should be a number which can fully divide 40. In case of option (d), the sum of ratio = 5 + 4 = 9. It is obvious that 40 is not divisible by 9.

In case of option (a): 3 + 2 = 5 and 5 can divide 40

In case of option (b): 3 + 1 = 4 and 4 can divide 40

In case of option (c): 13 + 7 = 20 and 20 can divide 40

Question 2: First discount @15% = 10,000 × 15% 1500

Price after first discount = 10,000 – 1500 = 8500

Second discount @20% = 8500 × 20% = 1700

Price after second discount = 8500 – 1700 = 6800

Question 3: The principal becomes 7875 after 3 years 9 months.

This means 6300 becomes 7875 after 1 year 9 months or after 21 months.

Interest after 21 months = 7875 – 6300 = 1575

Since, interest of 21 months = 1575

Hence, interest of 12 months = (1575 ÷ 21) × 12 = 900

6300 is the sum of principal amount and two years’ interest

Interest for 2 years = 900 × 1800

So, principal = 6300 – 1800 = 4500

As interest for 1 year = 900

So, rate of interest = (900 ÷ 4500) × 100 = 20%

Question 4: Work done by Ramkhelawan in 1 day = 1/24

Work done by Gurran in 1 day = 1/48

So, Work done both Ramkhelawan and Gurran in 1 day = (1/24) + (1/48)

= (24 + 48)/(24 × 48)

So, time to complete the work = (24 × 48)/(24 + 48) = (24 × 48)/72 = 16 days

Question 5: Sample space for toss of coin is as follows:

(HHH), (TTT), (HTT), (TTH), (THH), (HHT), (THT), (HTH)

Total number of events = 8

Number of favourable event = 3

So, probability of getting only one head = 3/8

Question 6: When the 1st pen is drawn, probability of it being a defective pen = 4/12 = 1/3

After that 11 pens are left out of which 3 pens are defective.

So, when the 2nd pen is drawn, probability of it being a defective pen = 3/11

So, out of 12 pens, probability of defective pens when 2 pens are drawn

= 1/3 × 3/11 = 1/11

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