Quantitative Aptitude Questions (Data Interpretation) for SBI PO Mains 2018 Day-88

Dear Readers, SBI is conducting Online Examination for the recruitment of probationary officer. To enrich your preparation here we have providing new series of Data Interpretation – Quantitative Aptitude Questions. Candidates those who are appearing in SBI PO Exams can practice these Quantitative Aptitude average questions daily and make your preparation effective.

Click “Start Quiz” to attend these Questions and view Solutions

Click here to view Quantitative Aptitude Questions in Hindi

Directions (1-5): The following table shows different items sold by the shopkeeper, cost price/kg of items, percentage mark-up on cost price, marked price/kg and percentage discount offered on marked price.

(Note: Some values are missing, you need to calculate those values if required.)

Items	Cost Price/Kg	% Mark-up price	Marked Price/Kg	% Discount
A	–	–	–	9%
B	500	20%	–	–
C	–	10%	275	–
D	–	–	800	18%
E	750	–	–	–

 

1. Shopkeeper gives two successive discounts of 12% and 13% instead of single discount of 18% on item D. If percentage Mark-up price for the item is 33 1/3%, then his profit is decreased by what percent as compared to previous profit (approx)?

1. 37 %
2. 48 %
3. 78 %
4. 73 %
5. 69 %

2. How much quantity of item A is sold by shopkeeper if cost price of item A is Rs. 150 less than the twice of the marked price of item C and marked price of item A is 5/8 times of marked price of item D and total profit earned by selling item A is Rs.1320.

1. 26 kg
2. 25 kg
3. 20 kg
4. 24 kg
5. 28 kg

3. To earn more profit shopkeeper mixes 5kg of cheaper quality of item C (Costs Rs.120/kg) with the 15kg of pure item C and sold the mixture at a discount which is 10% more than the original discount offered on item C. If original discount offered on item C is 14%, then find the new profit % on selling the whole quantity of item C (approx).

1. 14 %
2. 21 %
3. 13 %
4. 18 %
5. 7 %

4. If 1kg of item B is spoiled out of total 10 kg, then what is the total profit or loss (in Rs.) made by shopkeeper on selling a remaining quantity of the item, if a discount of 8% is given on marked price of the item.

1. Rs.32
2. Rs.85
3. Rs.25
4. Rs.35
5. Rs.44

5. Find the profit percentage of item E, if it marks-up 20% above the cost price and allows 17% discount

1. 0.4% profit
2. 4% loss
3. 0.4% loss
4. 4% profit
5. 0.04% profit

Directions (Q. 6-10) Read the following information carefully and answer the following questions carefully. The pie-chart below shows percentage distance travelled by different cars and the table shows speed of different cars.

Note: All the questions are independent from other questions.

6. If the time taken by the car C to reach its destination is 4 hours then find the time taken by the car F to reach its destination if the speed of car F is 20% more than the speed of car C?

1. 7(5/8)
2. 11(2/5)
3. 2(8/21)
4. 2(5/7)
5. 6(4/7)

7. If the difference between the distance travelled by car D and B is 20km while the difference between the time taken by car D and B to reach their destination is 5 hours and the speed of car B is twice the speed of car D then find the speed of car B?

1. 72 kmph
2. 11 kmph
3. 33 kmph
4. 20 kmph
5. 64 kmph

8. If the total time taken by car A and E to travel their respective distance is 9(1/4) hours then find the distance travelled by car C?

1. 155 km
2. 105 km
3. 197 km
4. 205 km
5. 175 km

9. The total distance travelled by all the cars is 1250km then find the approximate difference between the time taken by car C and A?

1. 10 hrs
2. 12 hrs
3. 7 hrs
4. 20 hrs
5. 5 hrs

10. If car E started with its usual speed but after covering half of the distance its speed reduced by 30% and the total time taken by car E to cover the total distance is 8.5 hours then find the time taken by car C to cover its respective distance?

1. 12 hrs
2. 17 hrs
3. 9 hrs
4. 5 hrs
5. 7 hrs

Answers:

1). Answer: c

Cost Price of item D = MP×100/((100 + %Mark-up) )

= 800×100/((100 + 33 1/3) )

= 800 * (100/(400/3))

= 800 * ¾

= Rs.600

Original Selling Price = MP × ((100 – %Discount))/100

= 800 × ((100 – 18))/100

= Rs. 800 *82/100

= Rs. 656

Original Profit = SP – CP = Rs.656-600 = Rs.56

New Selling Price = MP × ((100 – %Discount))/100 × ((100 – %Discount))/100

= 800 × ((100 – 12))/100 × ((100 – 13))/100

= Rs. 800 * 88/100 * 87/100 = Rs.612.48

New Profit = SP – CP = 612.48 – 600 = Rs.12.48

% Decrease in Profit = ((56-12.48))/56×100 = 78% approximately

2). Answer: d

Cost price of item A = 2 * Marked price of item C – 150 = Rs.400

Marked price of item A =   5/8× Marked price of item D = Rs.800 * 5/8 = Rs.500

Selling Price of item A = MP×((100 – %Discount))/100

= 500×((100 – 9))/100

= Rs.455

Profit/kg = SP – CP = 455 – 400 = Rs.55

Quantity Sold = (Total Profit)/(Profit/kg)  =  1320/55  = 24 kg

3). Answer: e

Cost Price of pure item C = MP×100/((100 + %Mark-up))

= 275×100/((100 + 10) )

= 275 *100/110 = Rs.250

Cost Price/kg of the mixture = (Total Cost Price)/(Total Quantity)

= > (5×120 + 15×250)/(5 + 15)  = Rs.217.5

New Discount = 14% + 10% * 14 = 14+1.4 = 15.4%

New Selling price = = MP×((100 – %Discount))/100

= 275×((100 – 15.4))/100

= Rs.232.65

New Profit % = (New Selling Price – New Cost Price)/(New Cost Price)×100

= (232.65- 217.5)/217.5×100

= 6.9 = 7% (Approximately)

4). Answer: a

Selling Price of item B = CP × ((100 + %Mark-up))/100 × ((100 – %Discount))/100

= 500 × ((100 + 20))/100 × ((100 – 8))/100

= Rs.552

1kg of item B is spoiled out of total 10 kg, so only 9 kg is available for sale.

Total Profit = Total Selling Price – Total Cost Price

= 9 × 552– 10 × 500

= 4968 – 5000 = – 32

= Rs. 32 (loss)

5). Answer: c

Selling Price of item E = CP * (100+% Marks-up)/100×((100 – %Discount))/100

Selling price of item E = 750* (100+20)/100 * (100 – 17)/100 = Rs. 747

Loss = CP – SP = 750 – 747 = Rs.3

Loss percentage = (3/750) * 100 = 0.4%

 

Direction (6-10)

6). Answer: c

Distance travelled by Car C= 4*35= 140km

Distance travelled by car F= 140*10/14= 100km

Speed of car F = 35*120/100= 42km/hr

Time taken by car F to cover the distance

= 100/42= 2(8/21) hours.

7). Answer: d

Distance travelled by car B

= 20*15/5

= 60km

Distance travelled by car D

= 20*20/5

= 80km

Let the speed of car D be x

Speed of car B= 2x

So,

80/x- 60/2x= 5

160-60/2x= 5

100= 10x

x= 10kmph

Speed of car B= 20kmph

8). Answer: b

Let the total distance travelled by all the cars be x

So,

Distance travelled by car A= 25*x/100= x/4

Distance travelled by car E= 16*x/100= 4x/25

Time taken by car A= x/4 ÷ 30= x/120

Time taken by car E= 4x/25 ÷40= x/250

Now,

x/ 120 + x/250= 37/4

(25x + 12x)/3000= 37/4

37x/3000 = 37/4

X= 3000/4 = 750

x= 750 km

Distance travelled by car C= 14*750/100= 105km

9). Answer: e

Distance travelled by car C

= 1250*14/100

= 175 km

Distance travelled by car A

= 1250*25/100

= 312.5 km

Time taken by car C= 175/35= 5 hours

Time taken by car A= 312.5/30= 10 hours

Required difference= 10-5= 5 hours

10). Answer: e

Let the total distance travelled by all the car be x

So,

Distance travelled by car E= 16*x/100= 4x/25

Distance travelled with 40kmph speed= 4x/25÷2= 2x/25

Reduced speed= 40*70/100= 28kmph

Distance travelled with 28kmph speed= 2x/25

Now,

2x/25÷40 + 2x/25 ÷ 28= 8.5

2x/1000 + 2x/700 = 8.5

14x+ 20x/7000= 8.5

34x= 8.5*7000

x= 8.5*7000/34= 1750km

Distance travelled by car C= 14*1750/100= 245km

Time taken by car C= 245/35= 7hours

Daily Practice Test Schedule | Good Luck

Topic	Daily Publishing Time
Daily News Papers & Editorials	8.00 AM
Current Affairs Quiz	9.00 AM
Quantitative Aptitude “20-20”	11.00 AM
Vocabulary (Based on The Hindu)	12.00 PM
General Awareness “20-20”	1.00 PM
English Language “20-20”	2.00 PM
Reasoning Puzzles & Seating	4.00 PM
Daily Current Affairs Updates	5.00 PM
Data Interpretation / Application Sums (Topic Wise)	6.00 PM
Reasoning Ability “20-20”	7.00 PM
English Language (New Pattern Questions)	8.00 PM

Click Here for More Quantitative Aptitude Questions