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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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