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

1 pointsCategory: Quantitative Aptitude**Find the wrong term in the given series**

44, 51, 65, 85, 114, 149Correct**Answer -3) 85**

Explanation –

44 + 7 = 51

51+14 = 65

65 + 21 =86

86 + 28 = 114

114 + 35 = 149Incorrect**Answer -3) 85**

Explanation –

44 + 7 = 51

51+14 = 65

65 + 21 =86

86 + 28 = 114

114 + 35 = 149 - Question 2 of 10
##### 2. Question

1 pointsCategory: Quantitative Aptitude**Find the wrong term in the given series**

59, 58, 50, 23, -41, -164, -382Correct**Answer -5. -164**

**Explanation**

59 – 1^{3}= 58

58 – 2^{3}= 50

50 – 3^{3 }= 23

23 – 4^{3}= -41

-41 – 5^{3}= -166-166 – 6

^{3}= -382Incorrect**Answer -5. -164**

**Explanation**

59 – 1^{3}= 58

58 – 2^{3}= 50

50 – 3^{3 }= 23

23 – 4^{3}= -41

-41 – 5^{3}= -166-166 – 6

^{3}= -382 - Question 3 of 10
##### 3. Question

1 pointsCategory: Quantitative Aptitude**Find the approximate value of x?**

223.96 / 7.97 + 11.93 × 15.97 = x + 29.48 ×4 – 5.32 ×6Correct**Answer -3) 130**

Explanation

223.96 / 7.97 + 11.93 × 15.97 = x + 29.48 ×4– 5.32 ×6

224/8 +12×16=x+29.5 ×4 – 5.33×6

28+192 = x +118-32

220+32 -118 = x

134 = xIncorrect**Answer -3) 130**

Explanation

223.96 / 7.97 + 11.93 × 15.97 = x + 29.48 ×4– 5.32 ×6

224/8 +12×16=x+29.5 ×4 – 5.33×6

28+192 = x +118-32

220+32 -118 = x

134 = x - Question 4 of 10
##### 4. Question

1 pointsCategory: Quantitative Aptitude**Find the approximate value of x?**

**X= (8.12 +18.88)**^{2 }– (14.95 – 12.01)^{2}Correct**Answer -3. 721**

**Explanation –**

X= (8.12 +18.88)^{2 }– (14.95 – 12.01)^{2}

X = 27^{2}– 3^{2}

X= 729-9

X =720Incorrect**Answer -3. 721**

**Explanation –**

X= (8.12 +18.88)^{2 }– (14.95 – 12.01)^{2}

X = 27^{2}– 3^{2}

X= 729-9

X =720 - Question 5 of 10
##### 5. Question

1 pointsCategory: Quantitative Aptitude**6x**^{2 }-43x+ 72 = 0

**6y**^{2 }– 35y + 36 = 0Correct**Answer- 5. x = y or no relation can be established.**

**Explanation –**

6x^{2 }-43x+ 72 = 0

6x^{2 }-27x-16x+ 72 = 0

3x(2x-9)-8(2x-9) = 0

(3x-8)(2x-9) = 0

**X = 8/3, 9/2**6y

^{2 }– 35y + 36 = 0

6y^{2 }– 8y -27y + 36 = 0

2y(3y-4)-9(3y-4) = 0

(2y-9)(3y-4) = 0

**Y = 9/2, 4/3**Incorrect**Answer- 5. x = y or no relation can be established.**

**Explanation –**

6x^{2 }-43x+ 72 = 0

6x^{2 }-27x-16x+ 72 = 0

3x(2x-9)-8(2x-9) = 0

(3x-8)(2x-9) = 0

**X = 8/3, 9/2**6y

^{2 }– 35y + 36 = 0

6y^{2 }– 8y -27y + 36 = 0

2y(3y-4)-9(3y-4) = 0

(2y-9)(3y-4) = 0

**Y = 9/2, 4/3** - Question 6 of 10
##### 6. Question

1 pointsCategory: Quantitative Aptitude**x**^{2}-18x+77 = 0

**y**^{2}+4y-77 = 0Correct**Answer- 2. x****≥****y**

**Explanation**

x^{2}-18x+77 = 0

x^{2}-7x-11x+77 = 0

x(x-7)-11(x-7)= 0

x =7, 11y

^{2}+4y-77=0

y^{2}+11y-7y-77 = 0

y (y+11)-7(y+11) = 0

y = -11, 7Incorrect**Answer- 2. x****≥****y**

**Explanation**

x^{2}-18x+77 = 0

x^{2}-7x-11x+77 = 0

x(x-7)-11(x-7)= 0

x =7, 11y

^{2}+4y-77=0

y^{2}+11y-7y-77 = 0

y (y+11)-7(y+11) = 0

y = -11, 7 - Question 7 of 10
##### 7. Question

1 pointsCategory: Quantitative Aptitude**Directions (7-8): In each of the following questions, a question is followed by two or more statements. Read all the statements and find that which statements are required to answer the question.****Find the number of students in School P?**

1. The number of students in school P is 40% of that of school Q, the difference of students of school P and Q is 660.

2. The ratio of the number of boys and girls in school P is 7:8 and the difference between them is 44.Correct**Answer – 4) Either 1 or 2.**

Explanation –

From 1

Let students in school Q = 100x

Students in school P = 40% of 100x = 40x

Difference = 100x – 40x = 60x

60x = 660

X = 11

From 2

Let number of boys = 7x and no. of girls = 8x , total student = 7x+8x = 15x

Difference = 8x – 7x = x

X = 44Incorrect**Answer – 4) Either 1 or 2.**

Explanation –

From 1

Let students in school Q = 100x

Students in school P = 40% of 100x = 40x

Difference = 100x – 40x = 60x

60x = 660

X = 11

From 2

Let number of boys = 7x and no. of girls = 8x , total student = 7x+8x = 15x

Difference = 8x – 7x = x

X = 44 - Question 8 of 10
##### 8. Question

1 pointsCategory: Quantitative Aptitude**Directions (7-8): In each of the following questions, a question is followed by two or more statements. Read all the statements and find that which statements are required to answer the question.****What is the percent profit earned by Raghu on selling a shirt?**

1. MRP of the shirt is 20% more than the cost price.

2. Cost price of the shirt is Rs. 320

3. A discount of 15% was offered on MRP.Correct**Answer- 1) Only 1 and 3.**

Explanation-

Let the CP of shirt = 100x, MRP = 120x, discount = 15% of 120x = 18x, SP = 120x -18x = 102x

Profit = 102x -100x = 2x

Profit %age of 2x/100x × 100 = 2%Incorrect**Answer- 1) Only 1 and 3.**

Explanation-

Let the CP of shirt = 100x, MRP = 120x, discount = 15% of 120x = 18x, SP = 120x -18x = 102x

Profit = 102x -100x = 2x

Profit %age of 2x/100x × 100 = 2% - Question 9 of 10
##### 9. Question

1 pointsCategory: Quantitative Aptitude**Directions- Q (9-10): Each question contains two sub- questions i.e., Quantity 1 and Quantity 2. Quantity 1 will be the answer of the first question and Quantity 2 will be the answer of second question. Compare the two quantities –****Monthly salary of Disha is Rs. 72000. She spends 25 % on clothes, X% on perfume, y% on makeup and her saving was Rs. 14400.**

**Quantity 1 Amount spent on perfume and makeup.**

**Quantity 2 Rs. [(55)**^{2 }×10 + 9350]Correct**Answer – 5) Quantity 1 = Quantity 2 or no relation can be established.**

Explanation

**Quantity 1**

Saving = 14400 = 20% of 72000

25% of clothes

Therefore, 55% is spent on perfume and makeup.

55% of 72000 = 39600

**Quantity 2**

55×55×10 + 9350 = 39600Incorrect**Answer – 5) Quantity 1 = Quantity 2 or no relation can be established.**

Explanation

**Quantity 1**

Saving = 14400 = 20% of 72000

25% of clothes

Therefore, 55% is spent on perfume and makeup.

55% of 72000 = 39600

**Quantity 2**

55×55×10 + 9350 = 39600 - Question 10 of 10
##### 10. Question

1 pointsCategory: Quantitative Aptitude**Directions- Q (9-10): Each question contains two sub- questions i.e., Quantity 1 and Quantity 2. Quantity 1 will be the answer of the first question and Quantity 2 will be the answer of second question. Compare the two quantities –****The ratio of the present age of Nikita to that of Arush is 3:7. Arush is 8 years younger than Rohit. Rohit’s age after 7 years will be 36 years.**

Quantity 1: Present age of Rupali, who is 5 years younger than Arush.

Quantity 2: Present age of Nitin who is 7 years older than Nikita.Correct**Answer- 5) Quantity 1 = Quantity 2 or no relation can be established.**

Explanation –

Let the present age of nikita = 3x, present age of arush = 7x

Rohit’s present age = 36-7 = 29

Arush’s present age = 29-8 = 21

7x = 21

X = 3

3x = 9

Q1

Present age of Rupali = 21-5 = 16

Q2

Present age of Nitin = 9+ 7 = 16Incorrect**Answer- 5) Quantity 1 = Quantity 2 or no relation can be established.**

Explanation –

Let the present age of nikita = 3x, present age of arush = 7x

Rohit’s present age = 36-7 = 29

Arush’s present age = 29-8 = 21

7x = 21

X = 3

3x = 9

Q1

Present age of Rupali = 21-5 = 16

Q2

Present age of Nitin = 9+ 7 = 16

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