**Dear Aspirants,** **Our SSC Crackers team** is providing a new series of Quantitative Aptitude Questions for **Upcoming Exam **so the aspirants can practice it on a daily basis. These questions are framed by our skilled experts after understanding your needs thoroughly. Aspirants can practice these new series questions daily to familiarize with the exact exam pattern and make your preparation effective.

**1) M and N agreed to complete a work for Rs 1800. M alone can do it in 8 days and N alone can do it in 12 days. With the help of O, They can finish it in 4 days, Then N’s share will be**

a) 360

b) 600

c) 1200

d) 720

**2) The marked price of a mobile phone is Rs. 5000 which is sold at Rs. 3740 after two successive discounts. If the second rate of discount is 15%, then what was the first rate of discount?**

a) 12%

b) 10%

c) 14%

d) 16 %

**3) The average age of 41 students of a class and their teacher is 24 years. When the teacher’s age is excluded the average age is reduced by 1 year. What is the age of the teacher?**

a) 50 years

b) 55 years

c) 60 years

d) 65 years

**4) The cost of eight apples and five mangoes is Rs. 194. The cost of five guavas and six mangoes is Rs. 183. The cost of seven apples and nine guavas is Rs. 226. What is the cost of three mangoes, two guavas and one apple?**

a) Rs. 84

b) Rs. 95

c) Rs. 97

d) Rs. 103

**5) A sum of money at compound interest amounts to Rs 578.40 in 2 years and to Rs 614.55 in 3 years. The rate of interest per annum is**

a) 4%

b) 5%

c) 6¼ %

d) 8⅓ %

**6) Paras can complete 1/18 ^{th} of a task in three days. Manasvi can do 1/3^{rd} of the same task in 9 days. If they start doing the task together and after completion of 50% of the task, only Paras completes the remaining task alone, then what would be the total number of days taken to complete the task?**

a) 27 days

b) 11 days

c) 36 days

d) 45 days

**7) Ranjit sells a house worth Rs 3,25,000 to Manikaran Kumar at a profit of 6%. Ranjit again purchases the same house from Manikaran at a loss of 4%. How much does Ranjit gain or lose from the entire transaction?**

a) 22,560

b) 19,500

c) 23,984

d) None of these

**8) A class comprises of 40 students. The average of the percentage marks obtained by 28 students of the class is 55%. When the marks of rest of the students are added to the previous total marks, the average of the percentage marks of the class becomes 62.5%. What is the average score of the students whose marks were added later?**

a) 7.5%

b) 15%

c) 80%

d) 12.5%

**9) A shopkeeper marks up his goods by 20% and then gives a discount of 20%. Besides he cheats both his suppliers and customer by 100gm per kg. i.e. he takes 1100g from his supplier and sells only 900g to his customer. What is his net profit percentage?**

a) 17.5%

b) 16.66%

c) 18.22%

d) 17.33%

**10) In a marathon, Jyoti ran for 29 seconds while for the rest of the distance, she preferred to brisk walk at a speed of 2 m/s for 35 seconds. If the total distance to be covered was 215 m, find the speed at which she ran.**

a) 3 m/s

b) 4 m/s

c) 5 m/s

d) 7 m/s

**Answers :**

**1) Answer: B**

M’s 1 day work = 1/8

N’s 1 day work = 1/12

So, O’s 1 day work = 1/4 – (1/8 + 1/12) = 1/24

Ratio of their shares = 1/8: 1/12: 1/24 or 3: 2: 1

Hence, N’s share = 2/6 x 1800 = 600

**2) Answer: A **

The total discount on giving two successive discounts equals Rs. 1260 = (5000 – 3740) which is 25.2% of the marked price.

Let the first discount be Y% and if the second discount is 15%, we can write:

Y + 15 – [(Y × 15) /100] = 25.2

On solving this equation, we get Y = 12%

**3) Answer: D **

Sum of age of 41 students and the teacher = 24 x 42 = 1008

Let the age of the teacher be N years.

Total age of students = 1008 – N

Average age of students = (1008 – N)/41 = 23

=> 1008 – N = 943

=> N = 65

**4) Answer: C **

Let the cost of one apple, one mango and one guava be Rs. A, Rs. M and Rs. G respectively

According to the given question, the following equations can be formed-

8A + 5M = 194 — (1)

5G + 6M = 183 — (2)

7A + 9G = 226 — (3)

On solving the three equations, we get: A = 13, M = 18 and G = 15

Thus, the cost of three mangoes, two guavas and one apple (3M + 2G + 1A) = Rs. 97

**5) Answer: C**

∵ A_{1} = Rs578.40, n_{1} = 2 years, A_{2} = 614.55, n_{2} = 3 years

A = P(1 + r/100)^{n}

578.40 = P(1+r/100)^{2}

614.55 = P(1+r/100)^{3}

Dividing, (1+r/100) = 614.55/578.40

r/100 = 614.55/578.40 – 1 = 36.55/578.40

r = 100 x 36.55/578.40 = 6¼ %

**6) Answer: C **

Paras can do 1/18^{th} of a task in three days, which implies it takes 54 days for him to complete the task (or 1/54^{th} of the task done by him in 1 day).

Similarly, it takes 27 days for Mansavi to complete the task (or 1/27^{th} of the task done by her in 1 day).

If they both start doing the task together, 1/54 + 1/27 = 1/18^{th} of the task shall be done in 1 day or the task shall get completed in 18 days, implying that half of the work shall be done in 9 days.

Since Paras can complete the task alone in 54 days, half of the task shall be completed by him alone in 27 days.

Thus, 36 days (9 + 27) will be taken to complete the task.

**7) Answer: D **

Price of the house = Rs3,25,000

Money received by Ranjit from Manikaran= 106% of Rs 3,25,000=3,44,500.

Therefore, price at which Manikaran sells the house back to Ranjit = 96% of Rs 3,44,500

= Rs 3,30,720

Thus, Ranjit gains Rs (3,44,500 -3,30,720) = Rs 13,780

**8) Answer: C **

If the average score of 28 students of the class is 55%, then the total marks of 28 students will be 28 × 55 = 1540 (assuming the percentage score to be marks out of 100).

Similarly, the total marks of 40 students will be 40 × 62.5 = 2500. Then, the total marks of the 12 students whose marks were added later = 2500 – 1540 = 960.

Consequently, average score of the students whose marks were added later = 960/ 12 = 80%

**9) Answer: D**

Let actual CP per kg be 100

Since he buys 1100g and sells only 900g, projected CP = 100 x 11/9 = 1100/9

MP = 1.2CP and discount = 20%

Hence SP = 1100/9 x 1.2 x 0.8 = Rs 117.33

Therefore, gain % = 17.33%

**10) Answer: C**

Distance covered by walking = (speed × time) = 2 × 35 = 70 m

Distance covered by running = 215 – 70 = 145 m

Now, 145 = Running speed × 29

Thus, running speed = 145/ 29 = 5 m/s