Welcome to Online QuantTest in AffairsCloud.com. We are starting Cumulus Course for SBI PO 2019 Prelims Exam and we are creating sample questions in Quant section, this type of Question will be asked in SBI PO 2019.
Q (1-5) Rama has monthly salary of Rs.32000 and his expenditure in terms of percentage on different things in different months is given in bar chart.
Find the ratio of total amount left after spending on food, studies and rent in (May and June) to the ratio of total amount left after spending on food, studies and rent in (April and July).
This 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 – Quantity: 1 Age of father, if the age of son is 1/7th of his father’s and after 6 years son will be 11 years old. Quantity: 2 Age of father, if the ratio of the ages of daughter and her father is 3:8 and after 3 years the ratio of their ages will be 3:7
Correct
Answer – 1) Quantity1 >Quantity 2
Explanation –
Quantity: 1
Since, after 6 years the age of son will be 11 years
Therefore, present age of son = (11-6) = 5 years
Father’s present age = Son’s present age × 7
Father’s present age = 5 × 7 =35years
Quantity: 2
Let the present age of father is f
And present age of daughter is d
According to question,
d / f = 3/8
3f = 8d …… (equation1)
Also, after 3 years
(d + 3)/ (f+3) = 3/7
7d + 21 = 3f+9….. (equation2)
Putting 3f = 8d in equation 2
7d + 21 = 8d + 9
d = 12
Put d = 12 in equation1
3f = 8×12
f = 32
Incorrect
Answer – 1) Quantity1 >Quantity 2
Explanation –
Quantity: 1
Since, after 6 years the age of son will be 11 years
Therefore, present age of son = (11-6) = 5 years
Father’s present age = Son’s present age × 7
Father’s present age = 5 × 7 =35years
Quantity: 2
Let the present age of father is f
And present age of daughter is d
According to question,
d / f = 3/8
3f = 8d …… (equation1)
Also, after 3 years
(d + 3)/ (f+3) = 3/7
7d + 21 = 3f+9….. (equation2)
Putting 3f = 8d in equation 2
7d + 21 = 8d + 9
d = 12
Put d = 12 in equation1
3f = 8×12
f = 32
Question 7 of 10
7. Question
1 points
Category: Quantitative Aptitude
A batsman played 44 matches in his career, his average in first 7 matches is 39, from 8th match to 19th match 41, from 20th match to 29th match 28 and his average from 30th match to 44th match is 49. Find his approximate average for all 44 matches of his career.
Correct
Answer – 4) 40.5
Explanation-
Average = total runs score / total matches played
Runs scored in first 7 matches = 7 × 39
Runs scored from 8th match to 19th match = 12 × 41
Runs scored from 20th match to 29th matches = 10 × 28
Runs scored from 30th match to 44th match = 15 × 49
Average = (7 × 39 + 12 × 41 + 10 × 28 + 15 × 49) / 44
Average = 1780 / 44 = 40.45
Incorrect
Answer – 4) 40.5
Explanation-
Average = total runs score / total matches played
Runs scored in first 7 matches = 7 × 39
Runs scored from 8th match to 19th match = 12 × 41
Runs scored from 20th match to 29th matches = 10 × 28
Runs scored from 30th match to 44th match = 15 × 49
Average = (7 × 39 + 12 × 41 + 10 × 28 + 15 × 49) / 44
Average = 1780 / 44 = 40.45
Question 8 of 10
8. Question
1 points
Category: Quantitative Aptitude
A sum of money is lent out at compound interest at 20% per annum for 2 years. It would fetch Rs. 1928 more if interest is compounded half-yearly. Find the sum.
Correct
Answer – 3) Rs. 80000
Explanation-
Let the principal = P
Interest obtained on 20% per annum for 2 years = 20 + 20 + 20×20/100 = 44%
Interest obtained on half-yearly = 10 + 10 + 10×10/100 = 21
Again, 21+21+ 21×21/100 = 46.41%
Difference = 46.41% – 44% = 2.41%
2.41% = 1928
1% = 800
100% = 80000
Incorrect
Answer – 3) Rs. 80000
Explanation-
Let the principal = P
Interest obtained on 20% per annum for 2 years = 20 + 20 + 20×20/100 = 44%
Interest obtained on half-yearly = 10 + 10 + 10×10/100 = 21
Again, 21+21+ 21×21/100 = 46.41%
Difference = 46.41% – 44% = 2.41%
2.41% = 1928
1% = 800
100% = 80000
Question 9 of 10
9. Question
1 points
Category: Quantitative Aptitude
How many metal sphere can be made out of a solid metal cone of 28cm high and 6cm radius, the diameter of each sphere is 3cm?
Correct
Answer – 2) 74 Explanation –
Number of sphere = Volume of cone / Volume of each sphere
Number of sphere = (1/3 × π × 6 × 6 × 28) / 4 / 3 × π × (1.5)2
= 74.66
= 74 spheres can be made
Incorrect
Answer – 2) 74 Explanation –
Number of sphere = Volume of cone / Volume of each sphere
Number of sphere = (1/3 × π × 6 × 6 × 28) / 4 / 3 × π × (1.5)2
= 74.66
= 74 spheres can be made
Question 10 of 10
10. Question
1 points
Category: Quantitative Aptitude
The manufacturer makes a profit of 28% on an article, the wholesale dealer makes a profit of 25%, and the retailer makes a profit of 35%. Find the approximate manufacturing cost of the article if the retailer sold it for Rs. 250.
