Hello and welcome to exampundit. Here is a set of Quantitative Aptitude Quiz for SBI PO 2018 Prelims exam with Timer. The following set consists Miscellaneous problems.
8 Questions on Miscellaneous Problems, 8 Minutes total time.
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The following questions are created as per the recent trends and level of SBI PO Prelims 2018.
Test Name: Misecllaneous for SBI PO Prelims 2018
Duration: 8 min
Level: Moderate
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Question 1 of 8
1. Question
Three circles of equal radii touch each other as shown in the figure given below. The radius of each circle is 2 cm. What is the area of the shaded region?
Correct
Area of the triangle= √3/4 * 4² = 4√3
Area of 3 circles enclosed by the triangle= 3*π* 2²* 60/360 = 2π
Area of the shaded region= 4√3 – 2π = 2 (2√3 – π) cm²
Incorrect
Area of the triangle= √3/4 * 4² = 4√3
Area of 3 circles enclosed by the triangle= 3*π* 2²* 60/360 = 2π
Area of the shaded region= 4√3 – 2π = 2 (2√3 – π) cm²

Question 2 of 8
2. Question
The ratio of the cost price to the marked price of an article is 2:3 and the ratio of the percentage profit to the percentage discount is 3:2. What is the discount percentage?
Correct
Ratio of CP: MP= 2x: 3x
Profit= x
The ratio of % profit: % discount= 3:2
Cp= 200 SP= 300
3x/100*200 + 2x/100* 300= 100
6x + 6x= 100
X= 100/12= 25/3%
Incorrect
Ratio of CP: MP= 2x: 3x
Profit= x
The ratio of % profit: % discount= 3:2
Cp= 200 SP= 300
3x/100*200 + 2x/100* 300= 100
6x + 6x= 100
X= 100/12= 25/3%

Question 3 of 8
3. Question
Babul borrowed Rs. 1200 at 12% rate of interest. He repaid Rs. 500 at the end of the first year. What is the amount required to pay at the end of the second year to discharge his loan which was calculated at compound interest?
Correct
Amount to be paid at the end of 2 years= 1200*12*2/100 + 1200= 1488
Amount left as principal for the second year= 1488500= 988
Amount to be paid after 2nd year= 988+ 988*12/100= 1106.56
Incorrect
Amount to be paid at the end of 2 years= 1200*12*2/100 + 1200= 1488
Amount left as principal for the second year= 1488500= 988
Amount to be paid after 2nd year= 988+ 988*12/100= 1106.56

Question 4 of 8
4. Question
A book consists of 40 pages, 30 lines on each page and 50 characters on each line. If this content is written in another note book consisting of 30 lines per page and 20 characters per line then the required no. of pages will be how much % more than the previous no. of pages?
Correct
40* 30* 50 = x * 30 * 20
X= 100
% increase in the no. of pages= 10040/40 * 100 = 150%
Incorrect
40* 30* 50 = x * 30 * 20
X= 100
% increase in the no. of pages= 10040/40 * 100 = 150%

Question 5 of 8
5. Question
Sourav wants to purchase Lenovo mobile phone. The shopkeepers told him to pay 25% tax if he asked for the bill. Sourav managed to get a discount of 5% on the actual sale price (without tax) of the mobile and paid to the shopkeeper Rs. 5000. In doing so, he managed to avoid paying the 25% tax. What is the amount of discount that he has received on the selling price (inclusive tax)?
Correct
Sp= 100 Sp= 125 (with tax)
New SP = 100 5 = 95
Effective discount= 12595 = 30
When SP is 95 then discount is 30
When SP is 5000, discount will be= 30/95 * 5000=1578.95
Incorrect
Sp= 100 Sp= 125 (with tax)
New SP = 100 5 = 95
Effective discount= 12595 = 30
When SP is 95 then discount is 30
When SP is 5000, discount will be= 30/95 * 5000=1578.95

Question 6 of 8
6. Question
Suhani is thrice as efficient as Raima and Raima can do a piece of work in 27 days. Raima started working and after a few days Suhani joined him. They completed the work in 15 days from the beginning. For how many days did they work together?
Correct
Efficiency of Suhani= 3x Raima= x
Raima can do a work in 27 days
Suhani can do a work= 27/3= 9 days
In one day they both work in= 1/27 + 1/9= 4/27
Suhani joined Raima after (15x) days
15x/27 + 4x/27 = 1
15 – x + 4x= 27
3x= 12
X= 4
Incorrect
Efficiency of Suhani= 3x Raima= x
Raima can do a work in 27 days
Suhani can do a work= 27/3= 9 days
In one day they both work in= 1/27 + 1/9= 4/27
Suhani joined Raima after (15x) days
15x/27 + 4x/27 = 1
15 – x + 4x= 27
3x= 12
X= 4

Question 7 of 8
7. Question
When Rahul Increases his speed from 25 kmph to 30 kmph he takes 2 hr less than his usual time to cover a certain distance. What is the distance covered by Rahul in the usual time?
Correct
25t = 30 (t – 2)
25t = 30t – 60
T= 12
Distance= 25*12= 300
Incorrect
25t = 30 (t – 2)
25t = 30t – 60
T= 12
Distance= 25*12= 300

Question 8 of 8
8. Question
From a group of 5 men and 2 women, 2 persons are selected randomly. Find the probability that at least one woman is selected.
Correct
n(S) = 7!/ 2! * 5!
= 7*6*5! / 2*1*5!
= 21
n(E)= 2!/1!*1! * 5! /1!*4! * 2! /2! * 0!
= 2*5 + 1= 11
Req probability= 11/21
Incorrect
n(S) = 7!/ 2! * 5!
= 7*6*5! / 2*1*5!
= 21
n(E)= 2!/1!*1! * 5! /1!*4! * 2! /2! * 0!
= 2*5 + 1= 11
Req probability= 11/21
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