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- Question 1 of 10
##### 1. Question

1 pointsCategory: Quantitative Aptitude**Directions (Q1-5). In the questions given below, there are three statements i.e. l, ll and lll. Read them carefully and decide which of the statement (s) are sufficient to answer the question.****What is the selling price of the article?**

(l) Despite giving a discount of 50%, shopkeeper earns a profit of 18%.

(ll) The difference between the discount given and profit earned is Rs 400.

(lll) The profit calculated on selling price is 25%.Correct**Answer-2) Only l and ll are sufficient to answer the question.**

Explanation-

By using l and ll,

Let the cost price of the article be 100x,

Then, selling price = 118x

Marked price = 118x/50 × 100 = 236x

Discount given = 236x – 118x = 118x

Profit = 18x

Difference between discount given and profit earned = 118x – 18x = 100x

100x = Rs 400

Therefore, cost price = Rs 400Incorrect**Answer-2) Only l and ll are sufficient to answer the question.**

Explanation-

By using l and ll,

Let the cost price of the article be 100x,

Then, selling price = 118x

Marked price = 118x/50 × 100 = 236x

Discount given = 236x – 118x = 118x

Profit = 18x

Difference between discount given and profit earned = 118x – 18x = 100x

100x = Rs 400

Therefore, cost price = Rs 400 - Question 2 of 10
##### 2. Question

1 pointsCategory: Quantitative Aptitude**Directions (Q1-5). In the questions given below, there are three statements i.e. l, ll and lll. Read them carefully and decide which of the statement (s) are sufficient to answer the question.****Is n<m?**

**(l) n = 1/m**^{-5}

**(ll) m<0**

**(lll) |n| > |m|**Correct**Answer-2) Only l and ll are sufficient to answer the question.**

**Explanation-**

By using statement l, we have,

n = m^{5}

by statement ll

since m<0, this means m is in negative. Therefore, m^{5}will be a negative number as well and its magnitude will be more than that of m.

Therefore, by using statement l and ll, we can answer the question.

If we use statement lll, we will not know if m and n are negative or not.Incorrect**Answer-2) Only l and ll are sufficient to answer the question.**

**Explanation-**

By using statement l, we have,

n = m^{5}

by statement ll

since m<0, this means m is in negative. Therefore, m^{5}will be a negative number as well and its magnitude will be more than that of m.

Therefore, by using statement l and ll, we can answer the question.

If we use statement lll, we will not know if m and n are negative or not. - Question 3 of 10
##### 3. Question

1 pointsCategory: Quantitative Aptitude**Directions (Q1-5). In the questions given below, there are three statements i.e. l, ll and lll. Read them carefully and decide which of the statement (s) are sufficient to answer the question.****What are the minimum passing marks required to pass the exam?**

**(l) A student scored 40% marks in the exam and failed by 7 marks. Another student scored 48% marks and passed the exam by 21 marks.**

**(ll) Difference between the maximum marks and marks required to pass the exam is 203.**

**(lll) A student scored 33% marks and failed by 31.5 marks. Had he scored 59.5 more marks, his percentage of marks would have been 50%.**Correct**Answer- 4) Either l alone or lll alone.**

Explanation-

Let the maximum marks are 100x.

By using statement l alone, we have,

48x – 40x = 21 + 7 = 28

8x = 28, 100x = 2800/8 = 350

Marks required to pass the exam = 48% of 350 – 21 = 168 – 21 = 147

By using statement lll alone, we have,

50x – 33x = 59.5

17x = 59.5, x = 3.5, 100x = 350

Marks required to pass the exam = 33% of 350 + 31.5 = 115.5 + 31.5 = 147

By statement ll only, the answer will not be possible because from that information, maximum marks cannot be known.Incorrect**Answer- 4) Either l alone or lll alone.**

Explanation-

Let the maximum marks are 100x.

By using statement l alone, we have,

48x – 40x = 21 + 7 = 28

8x = 28, 100x = 2800/8 = 350

Marks required to pass the exam = 48% of 350 – 21 = 168 – 21 = 147

By using statement lll alone, we have,

50x – 33x = 59.5

17x = 59.5, x = 3.5, 100x = 350

Marks required to pass the exam = 33% of 350 + 31.5 = 115.5 + 31.5 = 147

By statement ll only, the answer will not be possible because from that information, maximum marks cannot be known. - Question 4 of 10
##### 4. Question

1 pointsCategory: Quantitative Aptitude**What is the share of C in the profit at the end of the year?**

(l) A invested the amount for three months less than that of B, while C joined the business 2 months after A. One of them participated for the whole year.

(ll) Profit of B in the profit is Rs 630.

(lll) Ratio of investments of A and B is 2:1 respectively.Correct**Answer-3) None of these.**

Explanation-

No information has been given about the amount invested by C and without that we will not be able to calculate the profit share of C.Incorrect**Answer-3) None of these.**

Explanation-

No information has been given about the amount invested by C and without that we will not be able to calculate the profit share of C. - Question 5 of 10
##### 5. Question

1 pointsCategory: Quantitative Aptitude**x, y and z are three positive integers. What is the value of y?**

**(l) x + y + z = 90**

**(ll) z = 59 – √y**

**(lll) z = 29 + y**Correct**Answer- 5) ll and lll together.**

**Explanation-**

By using ll and lll together,

Let √y = m, then, y = m^{2}

Therefore, z = 59 – m and z = 29 + m^{2}

29 + m^{2}= 59 – m

We can easily solve this equation and the value of y will be known.It cannot be calculated by using statement (l) and (ll) because √y will make a different variable there and therefore, there will be four variables to solve.

Incorrect**Answer- 5) ll and lll together.**

**Explanation-**

By using ll and lll together,

Let √y = m, then, y = m^{2}

Therefore, z = 59 – m and z = 29 + m^{2}

29 + m^{2}= 59 – m

We can easily solve this equation and the value of y will be known.It cannot be calculated by using statement (l) and (ll) because √y will make a different variable there and therefore, there will be four variables to solve.

- Question 6 of 10
##### 6. Question

1 pointsCategory: Quantitative AptitudeDirections (Q6-10).There are three bags A, B and C. Each bag has three type of balls having different colors i.e. Red, Blue and Black. In bag C, the number of blue balls are 7 more than that of red balls. In bag B, the number of blue colored balls in bag B are 20% more than that of red colored balls in the bag C. Total number of blue balls in all the three bags together are 72. Total number of red and black colored balls in bag A are 25. Total number of balls in bag C are 80. When one ball is picked at random from bag A, the probability of getting a blue ball is 21/46. Number of red balls in bag C are double the number of black balls in bag A.

**Number of blue balls in bag B are:**Correct**Answer- 2) 24**

Explanation-

Let the number of red balls in bag C are x.

Then, number of blue balls in bag C = x + 7

Number of blue balls in bag B = 120% of x = 1.2x

Let the number of blue balls in bag A are y.

Number of red and black balls in bag A = 25

Total number of balls in bag A = 25 + y

According to question, y/(25 + y) = 21/46

46y = 21 × 25 – 21y

25 y = 21 × 25, y = 21

Therefore, number of blue balls in bag A = 21

Since total number of blue balls are 72, we have,

1.2x + x + 7 + 21 = 72

2.2x = 44, x = 20

Number of red balls in bag C = 20

Number of blue balls in bag C = 20 + 7 = 27

Number of blue balls in bag B = 1.2 × 20 = 24

No of black balls in bag C = 80 – (20 + 27) = 80 – 47 = 33

Number of black balls in bag A = 20/2 = 10

Number of red balls in bag A = 25 – 10 = 15Incorrect**Answer- 2) 24**

Explanation-

Let the number of red balls in bag C are x.

Then, number of blue balls in bag C = x + 7

Number of blue balls in bag B = 120% of x = 1.2x

Let the number of blue balls in bag A are y.

Number of red and black balls in bag A = 25

Total number of balls in bag A = 25 + y

According to question, y/(25 + y) = 21/46

46y = 21 × 25 – 21y

25 y = 21 × 25, y = 21

Therefore, number of blue balls in bag A = 21

Since total number of blue balls are 72, we have,

1.2x + x + 7 + 21 = 72

2.2x = 44, x = 20

Number of red balls in bag C = 20

Number of blue balls in bag C = 20 + 7 = 27

Number of blue balls in bag B = 1.2 × 20 = 24

No of black balls in bag C = 80 – (20 + 27) = 80 – 47 = 33

Number of black balls in bag A = 20/2 = 10

Number of red balls in bag A = 25 – 10 = 15 - Question 7 of 10
##### 7. Question

1 pointsCategory: Quantitative AptitudeDirections (Q6-10).There are three bags A, B and C. Each bag has three type of balls having different colors i.e. Red, Blue and Black. In bag C, the number of blue balls are 7 more than that of red balls. In bag B, the number of blue colored balls in bag B are 20% more than that of red colored balls in the bag C. Total number of blue balls in all the three bags together are 72. Total number of red and black colored balls in bag A are 25. Total number of balls in bag C are 80. When one ball is picked at random from bag A, the probability of getting a blue ball is 21/46. Number of red balls in bag C are double the number of black balls in bag A.

**Number of black balls in bag C are how much more than the number of red balls in bag A?**Correct**Answer- 5) 18**

Explanation-

Required difference = 33 – 15 = 18 (watch the solution of question 1)Incorrect**Answer- 5) 18**

Explanation-

Required difference = 33 – 15 = 18 (watch the solution of question 1) - Question 8 of 10
##### 8. Question

1 pointsCategory: Quantitative AptitudeDirections (Q6-10).There are three bags A, B and C. Each bag has three type of balls having different colors i.e. Red, Blue and Black. In bag C, the number of blue balls are 7 more than that of red balls. In bag B, the number of blue colored balls in bag B are 20% more than that of red colored balls in the bag C. Total number of blue balls in all the three bags together are 72. Total number of red and black colored balls in bag A are 25. Total number of balls in bag C are 80. When one ball is picked at random from bag A, the probability of getting a blue ball is 21/46. Number of red balls in bag C are double the number of black balls in bag A.

**Two balls are picked at random from bag A. What is the probability that one is red and other is black?**Correct**Answer- 4) 10/69**

Explanation-

Number of red balls in bag A = 15

Number of black balls in bag A = 10

Total number of balls in bag A = 15 + 21 + 10 = 46

Required probability = 15/46 × 10/45 + 10/46 × 15/45 = 5/69 + 5/69 = 10/69Incorrect**Answer- 4) 10/69**

Explanation-

Number of red balls in bag A = 15

Number of black balls in bag A = 10

Total number of balls in bag A = 15 + 21 + 10 = 46

Required probability = 15/46 × 10/45 + 10/46 × 15/45 = 5/69 + 5/69 = 10/69 - Question 9 of 10
##### 9. Question

1 pointsCategory: Quantitative Aptitude**If a ball is picked randomly from bag B, the probability of getting a red ball is 12/55 and probability of getting a black ball is 19/55. Number of blue balls in bag B are how much more than that of black balls?**Correct**Answer- 1) 5**

Let the number of red balls and black balls in the bag B are x and y respectively.

Number of blue balls in bag B = 24

According to the question,

x/(x + y + 24) = 12/55

y/(x + y + 24) = 19/55

by using hit and trial method, we have,

x = 12, y = 19

Required difference = 24 – 19 = 5Incorrect**Answer- 1) 5**

Let the number of red balls and black balls in the bag B are x and y respectively.

Number of blue balls in bag B = 24

According to the question,

x/(x + y + 24) = 12/55

y/(x + y + 24) = 19/55

by using hit and trial method, we have,

x = 12, y = 19

Required difference = 24 – 19 = 5 - Question 10 of 10
##### 10. Question

1 pointsCategory: Reasoning**Number of blue balls in bag C are how much percent more than the number of Red balls in bag A?**Correct**Answer- 4) 80%**

Explanation-

Required percentage = (27 – 15)/15 × 100 = 1200/15 = 80%Incorrect**Answer- 4) 80%**

Explanation-

Required percentage = (27 – 15)/15 × 100 = 1200/15 = 80%

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