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Question 1 of 10
1. Question
1 points
Category: Quantitative Aptitude
Directions.Q(1-5) Refer to the table and answer the given questions
Data related to number of tourists in two National Parks (A & B) during 4 days.
In National Park A, during March 30% of male tourists and 70% of female tourists were from College X. What was the number of tourists (male + female) in National Park A from College X during in March ?
Correct
Answer – 3) 161
Explanation:
In National Park A, number of male tourists from College X =350 *60/100 *30/100 = 63
Similarly, female = 98
So Total tourists = 63 + 98 = 161
Incorrect
Answer – 3) 161
Explanation:
In National Park A, number of male tourists from College X =350 *60/100 *30/100 = 63
Similarly, female = 98
So Total tourists = 63 + 98 = 161
Question 2 of 10
2. Question
1 points
Category: Quantitative Aptitude
Directions.Q(1-5) Refer to the table and answer the given questions
Data related to number of tourists in two National Parks (A & B) during 4 days.
Number of tourists (male + female) in National Park A and B increased by 5% and 15% respectively from April to June. If the total number of female tourists in National Park A & B together of female tourists in National Parks A & B together was equal during April and June, what was the total number of male tourists in National Parks A & B together in June?
Correct
Answer – 1) 268
Explanation:
Total number of tourists in National Park A in June = 400 x 15/100=420
Similarly, in National Park B = 138
Total number of male tourists in June
= [420 – (400*65/100)+ 138-(120*25/100)]
= 268
Incorrect
Answer – 1) 268
Explanation:
Total number of tourists in National Park A in June = 400 x 15/100=420
Similarly, in National Park B = 138
Total number of male tourists in June
= [420 – (400*65/100)+ 138-(120*25/100)]
= 268
Question 3 of 10
3. Question
1 points
Category: Quantitative Aptitude
Directions.Q(1-5) Refer to the table and answer the given questions
Data related to number of tourists in two National Parks (A & B) during 4 days.
In National Park B, the respective ratio of female tourists during February and May was 2 : 3. If female tourists constituted 50% of the number of tourists (male + female) during May, what was the number of tourists (male + female) in National Park B in May?
Correct
Answer – 4) 282
Explanation:
Number of female tourists in National Park B in February =188 * 50/100=94
Similarly, in May =94 *3/2=141
Total number of tourists (male + female) in National Park B in May = 141 * (100/50)=282
Incorrect
Answer – 4) 282
Explanation:
Number of female tourists in National Park B in February =188 * 50/100=94
Similarly, in May =94 *3/2=141
Total number of tourists (male + female) in National Park B in May = 141 * (100/50)=282
Question 4 of 10
4. Question
1 points
Category: Quantitative Aptitude
Directions.Q(1-5) Refer to the table and answer the given questions
Data related to number of tourists in two National Parks (A & B) during 4 days.
In National Park A, what is the difference between the total number of male tourists during January & February together and the during March and April together?
Correct
Answer – 5) 181
Explanation:
Number of male tourists in National Park A in January and February together
=(190 x 30/100)+ (280 x 40/100) = 169
=(350 * 60/100)+(400 * 35/100) = 350
Required difference = 350-169 = 181
Incorrect
Answer – 5) 181
Explanation:
Number of male tourists in National Park A in January and February together
=(190 x 30/100)+ (280 x 40/100) = 169
=(350 * 60/100)+(400 * 35/100) = 350
Required difference = 350-169 = 181
Question 5 of 10
5. Question
1 points
Category: Quantitative Aptitude
Directions.Q(1-5) Refer to the table and answer the given questions
Data related to number of tourists in two National Parks (A & B) during 4 days.
What is the average number of male tourists in National Park B during January, March and April?
Correct
Answer – 1) 134
Explanation:
Number of male tourists in National Park A in March and April together
Required average = 134
Incorrect
Answer – 1) 134
Explanation:
Number of male tourists in National Park A in March and April together
Required average = 134
Question 6 of 10
6. Question
1 points
Category: Quantitative Aptitude
Find out the wrong number in the series given below
Aravind sells a mobile to Bala at 4/5th the rate of profit at which Bala sells it to Chand. Further Chand sells it to Dinesh at half the rate of profit at which Aravind sold it to Bala. If Chand earns a profit of 10% by selling it to Dinesh for Rs. 2805. What is the cost price of mobile from Bala?
Correct
Answer – 5) Rs. 2040
Explanation:
Chand’s profit = 10%.
Aravind’s profit = 20%
Bala profit = 25%
Let cost price for Aravind = Rs. 100
So cost price for Bala = 120
Cost price for Chand = 150
Cost price for Dinesh = 165
The cost price for Dinesh = Rs. 2805(17 times of 165)
Cost price for Bala = 120 x 17 = Rs. 2040.
Incorrect
Answer – 5) Rs. 2040
Explanation:
Chand’s profit = 10%.
Aravind’s profit = 20%
Bala profit = 25%
Let cost price for Aravind = Rs. 100
So cost price for Bala = 120
Cost price for Chand = 150
Cost price for Dinesh = 165
The cost price for Dinesh = Rs. 2805(17 times of 165)
Cost price for Bala = 120 x 17 = Rs. 2040.
Question 9 of 10
9. Question
1 points
Category: Quantitative Aptitude
Anu and Beulah entered a venture investing Rs.16000 and Rs.12000, respectively. After 3 months Anu takes out Rs. 5000, while Beulah puts in Rs. 5000 more. After 3 months more, Chitra joins the business with a capital of Rs. 21000. After a year, they earned a profit of Rs. 13200. By what value does the share of Beulah exceeds the share of Chitra?
Correct
Answer – 2) Rs. 1800
Explanation:
Anu’s share : Beulah’s share : Chitra’s share
= [16000 x 3 + (16000 – 5000) x 9] : [12000 x 3 + (12000 + 5000) x 9] : (21000 x 6)
= (16 x 3 + 11 x 9) : (12 x 3 + 17 x 9) : (21 x 6)
= 147 : 189 : 126
= 7 : 9 : 6
Hence, Beulah’s share exceeds Chitra’s share by = {13200/(7 + 9 + 6)} x (9 – 6)
= (13200 x 3)/22 = Rs. 1800
Incorrect
Answer – 2) Rs. 1800
Explanation:
Anu’s share : Beulah’s share : Chitra’s share
= [16000 x 3 + (16000 – 5000) x 9] : [12000 x 3 + (12000 + 5000) x 9] : (21000 x 6)
= (16 x 3 + 11 x 9) : (12 x 3 + 17 x 9) : (21 x 6)
= 147 : 189 : 126
= 7 : 9 : 6
Hence, Beulah’s share exceeds Chitra’s share by = {13200/(7 + 9 + 6)} x (9 – 6)
= (13200 x 3)/22 = Rs. 1800
Question 10 of 10
10. Question
1 points
Category: Quantitative Aptitude
A father divided his property between his two daughters Amit and Bharat. Amit invests the amount at CI of 8% per annum and Bharat invests the amount at 10% per annum SI. At the end of 2 year, the interest received by Bharat is Rs. 1336 more than the interest received by Amit. Find the share of Amit in the father’s property of Rs. 25000.?
Correct
Answer – 4) Rs. 10000
Explanation:
Suppose Amit gets P and Bharat gets (25000 – P).
Interest received by Amit at the rate of 8% per annum CI
= P(1 + 8/100)^2 – P
= P(27/25)^2 – P
= 104P/625
Interest received by Bharat at the rate of 10% per annum SI
= [(25000 – P) x 10 x 2]/100 = (25000 – P)/5
(25000 – P)/5 = 104P/(625 + 1336)
P = 10000
Incorrect
Answer – 4) Rs. 10000
Explanation:
Suppose Amit gets P and Bharat gets (25000 – P).
Interest received by Amit at the rate of 8% per annum CI
= P(1 + 8/100)^2 – P
= P(27/25)^2 – P
= 104P/625
Interest received by Bharat at the rate of 10% per annum SI
= [(25000 – P) x 10 x 2]/100 = (25000 – P)/5
(25000 – P)/5 = 104P/(625 + 1336)
P = 10000
_____________________
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