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Question 1 of 10

1. Question

1 points

Category: Quantitative Aptitude

Directions (1-5) In the given line chart the amount (in kg) of three different varieties of mangoes (Hapus, Kesar, Jardalu) sold by different shops (A, B, C, D and E) is given. Read the chart carefully and answer the following questions.

What is the total amount of Jardalu mango sold by shop E and shop B together?

Directions (1-5) In the given line chart the amount (in kg) of three different varieties of mangoes (Hapus, Kesar, Jardalu) sold by different shops (A, B, C, D and E) is given. Read the chart carefully and answer the following questions.

Total Hapus mango sold by all shops together is approximately what percent of total mangoes sold by all shops together?

Correct

Answer – 3) 27%
Explanation –
Total Hapus mangoes sold = 85+65+85+75+55=365kg
Total mangoes sold = 220+325+275+260+255 = 1335kg
Percentage = 365/1335 × 100 = 73/267 × 100 = 27.3%

Incorrect

Answer – 3) 27%
Explanation –
Total Hapus mangoes sold = 85+65+85+75+55=365kg
Total mangoes sold = 220+325+275+260+255 = 1335kg
Percentage = 365/1335 × 100 = 73/267 × 100 = 27.3%

Question 3 of 10

3. Question

1 points

Category: Quantitative Aptitude

Directions (1-5) In the given line chart the amount (in kg) of three different varieties of mangoes (Hapus, Kesar, Jardalu) sold by different shops (A, B, C, D and E) is given. Read the chart carefully and answer the following questions.

Total quantity of mangoes sold by shop A in what percent of total quantity of mangoes sold by shop C?

Correct

Answer – 2) 80%
Explanation –
Total mangoes sold by shop A = 220kg
Total mangoes sold by shop C = 275kg
220/275 × 100 = 4/5 × 100 = 80%

Incorrect

Answer – 2) 80%
Explanation –
Total mangoes sold by shop A = 220kg
Total mangoes sold by shop C = 275kg
220/275 × 100 = 4/5 × 100 = 80%

Question 4 of 10

4. Question

1 points

Category: Quantitative Aptitude

Directions (1-5) In the given line chart the amount (in kg) of three different varieties of mangoes (Hapus, Kesar, Jardalu) sold by different shops (A, B, C, D and E) is given. Read the chart carefully and answer the following questions.

What is the difference between total Hapus mangoes sold and total Kesar mangoes sold by all shop together?

Correct

Answer – 4) 30kg
Explanation –
Total Hapus mangoes sold = 85+65+85+75+55=365kg
Total Kesar mangoes sold = 55+95+60+105+80=395kg
Difference = 30kg

Incorrect

Answer – 4) 30kg
Explanation –
Total Hapus mangoes sold = 85+65+85+75+55=365kg
Total Kesar mangoes sold = 55+95+60+105+80=395kg
Difference = 30kg

Question 5 of 10

5. Question

1 points

Category: Quantitative Aptitude

Directions (1-5) In the given line chart the amount (in kg) of three different varieties of mangoes (Hapus, Kesar, Jardalu) sold by different shops (A, B, C, D and E) is given. Read the chart carefully and answer the following questions.

What is the ratio of total Jardalu mangoes sold by shop B to total Jardalu mangoes sold by shop D?

Correct

Answer – 1. 33/16
Explanation –
Jardalu mangoes sold by B = 325-(95+65) = 165kg
Jardalu mangoes sold by D = 260 –(105+75) = 80kg
Ratio = 165/80 = 33/16

Incorrect

Answer – 1. 33/16
Explanation –
Jardalu mangoes sold by B = 325-(95+65) = 165kg
Jardalu mangoes sold by D = 260 –(105+75) = 80kg
Ratio = 165/80 = 33/16

Question 6 of 10

6. Question

1 points

Category: Quantitative Aptitude

A metallic sphere of diameter 12cm is melted and drawn into a wire of uniform cross-section. If the radius of the wire is 2mm, find its length.

Correct

Answer -1) 72 meter
Explanation –
Radius of metallic wire = 12/2 = 6cm
Radius of wire = 2mm = .2cm
Volume of metallic sphere = 4/3 × π × (6)^{3 }
Volume of metallic sphere = Volume of drawn wire
4/3 × π × (6)^{3= }π × (.2) ^{2 }× L
L = 7200cm = 72m

Incorrect

Answer -1) 72 meter
Explanation –
Radius of metallic wire = 12/2 = 6cm
Radius of wire = 2mm = .2cm
Volume of metallic sphere = 4/3 × π × (6)^{3 }
Volume of metallic sphere = Volume of drawn wire
4/3 × π × (6)^{3= }π × (.2) ^{2 }× L
L = 7200cm = 72m

Question 7 of 10

7. Question

1 points

Category: Quantitative Aptitude

Anil scores 30% marks in an examination and fails by 25 marks, while his friend Bunny who scores 55% marks, gets 50 marks more than the minimum marks required to pass the examination. Find the minimum marks required to pass the examination.

Correct

Answer -2) 115
Explanation-
Let the minimum marks required to pass the examination is P
Marks scored by Anil = P – 25
Marks scored by Bunny = P + 50
Difference in marks scored by Anil and Bunny = (P+50) – (P-25) = 75
Difference in percentage of marks scored by Anil and Bunny =55% – 30% =25%
25% = 75
1% = 3
Minimum marks required to pass the examination =
P+50 = 55 × 3 (since 1% = 3)
P = 115

Incorrect

Answer -2) 115
Explanation-
Let the minimum marks required to pass the examination is P
Marks scored by Anil = P – 25
Marks scored by Bunny = P + 50
Difference in marks scored by Anil and Bunny = (P+50) – (P-25) = 75
Difference in percentage of marks scored by Anil and Bunny =55% – 30% =25%
25% = 75
1% = 3
Minimum marks required to pass the examination =
P+50 = 55 × 3 (since 1% = 3)
P = 115

Question 8 of 10

8. Question

1 points

Category: Quantitative Aptitude

P, Q and R are partners in a business. P, invested for 4 months and received 12.5% of the profit, Q invested for 6 months and received 1/3rd of the profit. R had invested Rs. 780 for 8 months and received rest of the profit. How much money did P invested?

Correct

Answer -5) 360
Explanation
Take 12.5% as 1/8
Profit of R = 1- (1/8 + 1/3) = 13/24
Ration of profit is 1/8 : 1/3 : 13/24
= 3:8:13
P’s investment
= (P × 4)/ 780 × 8 = 3/13
P = Rs. 360

Incorrect

Answer -5) 360
Explanation
Take 12.5% as 1/8
Profit of R = 1- (1/8 + 1/3) = 13/24
Ration of profit is 1/8 : 1/3 : 13/24
= 3:8:13
P’s investment
= (P × 4)/ 780 × 8 = 3/13
P = Rs. 360

Question 9 of 10

9. Question

1 points

Category: Quantitative Aptitude

A toy is sold at 20% profit. If its cost price is increased by Rs. 100 and at the same time if its selling price is also increase by Rs. 60, the percentage of profit decreases to 50/3%. Find the initial selling price.

Correct

Answer: 2) Rs. 2040
Explanation –
Let the initial cost price is 100x
Initial selling price becomes 120x
Now, cost price becomes 100x + 100
And selling price becomes 120x +60
According to question,
50/3 = (SP – CP)/CP ×100
50/3 = [(120x+60) – (100x+100)] / (100x+100) × 100
50/3 = (20x-40)/ (100x+100) × 100
X = 17
Initial selling price = 120x = 120 × 17 = Rs. 2040

Incorrect

Answer: 2) Rs. 2040
Explanation –
Let the initial cost price is 100x
Initial selling price becomes 120x
Now, cost price becomes 100x + 100
And selling price becomes 120x +60
According to question,
50/3 = (SP – CP)/CP ×100
50/3 = [(120x+60) – (100x+100)] / (100x+100) × 100
50/3 = (20x-40)/ (100x+100) × 100
X = 17
Initial selling price = 120x = 120 × 17 = Rs. 2040

Question 10 of 10

10. Question

1 points

Category: Quantitative Aptitude

Sam can row his boat at a speed of 5 km/hr in still water. If he takes 30minutes more to row the boat 2625 meter upstream than to return downstream, find the speed of the stream.

Correct

Answer -4) 2 km/hr
Explanation –
30minutes = ½ hour
2625 meter = 2.625 km
Let the speed of the stream = x km/hr
Speed of the boat upstream = (5 – x) km/hr
Speed of the boat downstream = (5 + x) km/hr
Time taken for 2.625 km upstream = 2.625 / (5-x)
Time taken for 2.625 km downstream = 2.625 / (5+x)
According to question,
2.625 / (5-x) – 2.625 / (5+x) = ½
2×2+21x-50 = 0
2×2+25x-4x-50 = 0
X (2x+25)-2(2x+25) =0
X = 2, x = -25/2 (speed cannot be negative)
X = 2km/hr

Incorrect

Answer -4) 2 km/hr
Explanation –
30minutes = ½ hour
2625 meter = 2.625 km
Let the speed of the stream = x km/hr
Speed of the boat upstream = (5 – x) km/hr
Speed of the boat downstream = (5 + x) km/hr
Time taken for 2.625 km upstream = 2.625 / (5-x)
Time taken for 2.625 km downstream = 2.625 / (5+x)
According to question,
2.625 / (5-x) – 2.625 / (5+x) = ½
2×2+21x-50 = 0
2×2+25x-4x-50 = 0
X (2x+25)-2(2x+25) =0
X = 2, x = -25/2 (speed cannot be negative)
X = 2km/hr

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