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Question 1 of 10
1. Question
1 pointsCategory: Quantitative Aptitude(1/30)+(1/42)+(1/56)+(1/72)+(1/90)+(1/110)
Correct
Answer – 4) 6/55
Explanation:
According to the questions,
1/30+1/42+1/56+1/72+1/90+1/110
(1/51/6)+ (1/61/7)+ (1/71/8)+ (1/81/9)+ (1/91/10)+ (1/101/11)
(1/51/11)
(115)/55
6/55Incorrect
Answer – 4) 6/55
Explanation:
According to the questions,
1/30+1/42+1/56+1/72+1/90+1/110
(1/51/6)+ (1/61/7)+ (1/71/8)+ (1/81/9)+ (1/91/10)+ (1/101/11)
(1/51/11)
(115)/55
6/55 
Question 2 of 10
2. Question
1 pointsCategory: Quantitative AptitudeCorrect
Answer – 3) 2
Explanation:
Incorrect
Answer – 3) 2
Explanation:

Question 3 of 10
3. Question
1 pointsCategory: Quantitative AptitudeWhat should come in place of question mark (?) in the following number series?
102 186 252 300 (?) 410Correct
Answer – 5) 330
Explanation:
111(1+1+1)^{2}= 1119 = 102
222(2+2+2)^{2}=22236 = 186
333(3+3+3)^{2}=33381 = 252
444(4+4+4)^{2}=444144= 300
555(5+5+5)^{2}=555225= 330
666(6+6+6)^{2}=666256= 410Incorrect
Answer – 5) 330
Explanation:
111(1+1+1)^{2}= 1119 = 102
222(2+2+2)^{2}=22236 = 186
333(3+3+3)^{2}=33381 = 252
444(4+4+4)^{2}=444144= 300
555(5+5+5)^{2}=555225= 330
666(6+6+6)^{2}=666256= 410 
Question 4 of 10
4. Question
1 pointsCategory: Quantitative AptitudeThe average marks of 32 boys of section A of class X is 60 whereas the average marks of 40 boys of section B of class X is 33. What isthe average marks for both the sections combined together ?
Correct
Answer – 2) 45
Explanation:
The average marks of 32 boys= 32 X 60
The average marks of 40 boys= 40 X 33
Therefore, Mixed average= (32 X 60 + 40 X 33)/72
= (1920+1320)/72
= (3240)/72 = 45Incorrect
Answer – 2) 45
Explanation:
The average marks of 32 boys= 32 X 60
The average marks of 40 boys= 40 X 33
Therefore, Mixed average= (32 X 60 + 40 X 33)/72
= (1920+1320)/72
= (3240)/72 = 45 
Question 5 of 10
5. Question
1 pointsCategory: Quantitative AptitudeA principal of Rs. 10,000 after 2 years compounded annually, the rate of interest being 10% per annum during the first year and 12% per annum during the second year (in rupees), what will be the total price after 2 years?
Correct
Answer – 2) Rs 12320
Explanation:
Principal= Rs 10,000
Time= 2 years
Rate of Interest of 1st year= 10%
Rate of interest of 2nd year= 12%
Incorrect
Answer – 2) Rs 12320
Explanation:
Principal= Rs 10,000
Time= 2 years
Rate of Interest of 1st year= 10%
Rate of interest of 2nd year= 12%

Question 6 of 10
6. Question
1 pointsCategory: Quantitative AptitudeBy melting a solid lead sphere of diameter 12cm, three small spheres are made whose diameters are in the ratio 3 : 4 : 5. What is the radius (in cm) of the smallest sphere?
Correct
Answer – 4) 3
Explanation:
Let radius of smallest sphere= 3x
Radius of sphere 2= 4x
Radius of sphere 3= 5x
Volume of Sphere=(4/3) ∏R^{3}
(4/3)∏ {(3x)^{ 3} + (4x)^{ 3} + (5x)^{ 3}} = (4/3)∏(6)^{3}
X^{3}(27 + 64 + 125) = 216
X^{3}(216) = 216
X^{3} = 1
X = 1
Radius of smallest sphere= 3x =3 * 1= 3cmIncorrect
Answer – 4) 3
Explanation:
Let radius of smallest sphere= 3x
Radius of sphere 2= 4x
Radius of sphere 3= 5x
Volume of Sphere=(4/3) ∏R^{3}
(4/3)∏ {(3x)^{ 3} + (4x)^{ 3} + (5x)^{ 3}} = (4/3)∏(6)^{3}
X^{3}(27 + 64 + 125) = 216
X^{3}(216) = 216
X^{3} = 1
X = 1
Radius of smallest sphere= 3x =3 * 1= 3cm 
Question 7 of 10
7. Question
1 pointsCategory: Quantitative AptitudeThe annual income of A and B are in the ratio 4:3 and the ratio of their expenditures is 3:2. If each of them saves Rs 600 in the year, what is the annual income of A?
Correct
Answer – 2) Rs 2400
Explanation:
Therefore, 1 unit= 600
Income of A= 600 x 4= Rs 2400
Income of B= 600 x 3= Rs 1800Incorrect
Answer – 2) Rs 2400
Explanation:
Therefore, 1 unit= 600
Income of A= 600 x 4= Rs 2400
Income of B= 600 x 3= Rs 1800 
Question 8 of 10
8. Question
1 pointsCategory: Quantitative AptitudeIn an examination A got 25% marks more than B, B got 10% less than C and C got 25% more than D. If D got 320 marks out of 500, What were the marks obtained by A?
Correct
Answer – 4) 450
Explanation:
Marks obtained by D = 320
Marks obtained by C = 320 X (125/100) = 400
Marks obtained by A = 320 X (125/100) = 450
Hence, required marks obtained by A = 450Incorrect
Answer – 4) 450
Explanation:
Marks obtained by D = 320
Marks obtained by C = 320 X (125/100) = 400
Marks obtained by A = 320 X (125/100) = 450
Hence, required marks obtained by A = 450 
Question 9 of 10
9. Question
1 pointsCategory: Quantitative AptitudeA container contains 60 kg of milk. From this container 6kg of milk was taken out and replaced by water. This process was repeated further two times. What is the amount of milk left in the container?
Correct
Answer – 3) 43.74 Kg
Explanation:
According to the question,
Initial quantity= 60 kg
Final quantity= Initial quantity ((1(Volume taken out/Initial Quantity)) n
Where, n= number of times volume taken out
Final Quantity = 60 (1(6/60)) 3
= 60 X 9/10 X 9/10 X 9/10
Final Quantity of milk= 43.74 kgIncorrect
Answer – 3) 43.74 Kg
Explanation:
According to the question,
Initial quantity= 60 kg
Final quantity= Initial quantity ((1(Volume taken out/Initial Quantity)) n
Where, n= number of times volume taken out
Final Quantity = 60 (1(6/60)) 3
= 60 X 9/10 X 9/10 X 9/10
Final Quantity of milk= 43.74 kg 
Question 10 of 10
10. Question
1 pointsCategory: Quantitative AptitudeTwo Quantities (I) and (II) are given in each question. On the basis of these equations you have to decide the relation between “X” and “Y” and given answer.
Quantity I : 6x^{2} – 19x + 15=0
Quantity II: 10y^{2} – 29y + 21= 0Correct
Answer – 3) Quantity I ≥ Quantity II
Explanation:
6x^{2} – 19x + 15=0
6x^{2} 9x 10x +15=0
3x (2x3) – 5(2x3) =0
(3x5)(2x3) =0
X= 5/3, 3/210y^{2} – 29y +21=0
10y^{2 }15y 14y +21=0
5y (2y3) 7(2y3) =0
(5y7)(2y3) =0
Y= 7/5, 3/2Incorrect
Answer – 3) Quantity I ≥ Quantity II
Explanation:
6x^{2} – 19x + 15=0
6x^{2} 9x 10x +15=0
3x (2x3) – 5(2x3) =0
(3x5)(2x3) =0
X= 5/3, 3/210y^{2} – 29y +21=0
10y^{2 }15y 14y +21=0
5y (2y3) 7(2y3) =0
(5y7)(2y3) =0
Y= 7/5, 3/2
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