# Quantitative Aptitude Questions (Data Interpretation) for SBI PO Mains 2018 Day-106

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Practice Aptitude Questions (Data Interpretation) Day-106

maximum of 10 points

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Directions (Q. 1 – 5) Study the following information carefully and answer the given questions:

The table shows the number of balls in three different colours, and then one ball is taken from each box and given the probability of getting one green colour ball.

 Box name White balls Green balls Black balls Probability of getting one green ball Box 1 15 10 _ 1/5 Box 2 12 _ 12 2/5 Box 3 10 15 _ 1/3 Box 4 _ 10 16 1/5 Box 5 15 _ 15 2/5

Note: Find the missing values in the given table

1) If 2x number of green balls taken from box 1 and placed it into box 3, again 5x number of black balls taken from box 3 and placed it into box 1. If we pick one ball from box 3 then the probability of that ball is white ball is 5/21, then find the value of โxโ

a) 2

b) 3

c) 1

d) 4

e) 5

2) If all the balls of box 2, box 3 and box 5 is placed into box 6 and some white balls taken and placed it into box 4. If we pick two balls in box 4 then the probability of getting 2 white balls is 87/308, and then find the number of white balls taken from box 6

a) 4

b) 6

c) 2

d) 5

e) 8

3) Total number of green colour balls in all the box together is what percentage more/less than the total number of white colour balls in all the box together?

a) 5%

b) 9%

c) 10%

d) 7%

e) 13%

4) If three balls are taken from box 2, what is the probability of getting one green ball in box 2 or one black ball in box 2?

a) 1119/1235

b) 1199/1235

c) 129/1235

d) 119/1235

e) 111/1235

5) If two balls are taken from box 1 or box 5, what is the probability of getting both are white balls?

a) 23/245

b) 24/245

c) 21/245

d) 6/245

e) 1/235

Directions (Q. 6 – 10): Study the following graph carefully and answer the given questions

The table given below show different discount rates are given for different products for different shop. Selling price is same for a particular product for all shops. (MP= market price, CP=Cost price, SP= selling price)

 Shops Item 1 Item 2 Item 3 A 8 16 6 B 10 12 20 C 12 8 12 D 5 12 10 E 15 6 20

6) If the sum of MP of item 1 for shops B, D and E is 97100 then find MP of item 1 for shop B?ย ย ย ย ย ย ย ย ย ย ย ย

a) Rs.30600

b) Rs.32300

c) Rs.34200

d) Rs.32800

e) Rs.35400

7) Difference between MP of item 2 in Shop D and shop A is Rs.100 then find the approximate MP of item 2 for Shop C?

a) Rs.2105

b) Rs.2200

c) Rs.2100

d) Rs.2010

e) Rs.2050

8) If the ratio of MP of item 3 in shop B, D and E is 9: 8: 9 and the difference between the marked price of item 3 in shop D and E is Rs.1100, then find the SP of item 3

a) Rs.8800

b) Rs.7920

c) Rs.7540

d) Rs.7250

e) Rs.8450

9) If difference between MP and SP for item 1 in shop B is Rs.250, then find the approximate average MP of item 1 in shop D and shop E?

a) Rs.2507.5

b) Rs.2560.5

c) Rs.2440.25

d) Rs.2640.75

e) Rs.2250.5

10) If the MP of all the items in shop B is same, then find the ratio of SP of item 1, item 2 and item 3?

a) 42: 45: 46

b) 41: 45: 46

c) 45: 44: 40

d) 39: 32: 31

e) None of these

Directions (Q. 1-5):

Box 1:

Let take number of black balls be x

Given,

(10C1)/((25+x) C1 )=1/5

=>10/(25+x)=1/5

=>50=25+x

=>x=50-25=25

Similarly find missing balls in the table

Total number of balls in box 3 =45+2x-5x=45-3x

Given,

(10C1)/((45-3x) C1 )=5/21

=>10/(45-3x)=5/21

=>42=45-3x

=>3x=3=>x=1

Total number of balls in box 6 is white-37, green-51, and black-47

Total number of balls in box 4 = 50+x

Given,

((24+x) C2)/((50+x) C2 )=87/308

=>(24+x)(23+x)/(50+x)(49+x) =87/308

Simplify the above equation, we get x =6

Required percentage

= [(10+16+15+10+20) โ (15+12+10+24+15)]/(15+12+10+24+15) *100

= (71-76)/76 *100

= 5/76*100 = 6.57% = 7%

Required probability = ((16C1ร24C2) + (12C1ร28C2)) / (40C3)

=>1119/1235

Required probability = 1/2ร((15C2)/(50C2)+(15C2)/(50C2))

=>1/2ร(21/245+21/245)

=>21/245

Direction (6-10)

6. Answer: bย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย

Sum of MP of shop B, D and E = (SP/90*100+SP/95*100+SP/85*100) = 97100

SP (100/90+100/95+100/85) =97100

SP = 97100/ (100/90+100/95+100/85)

SP = 97100/[100 *(1/90+1/95+1/85)]

SP = 971 /[(323+306+342)/29070)

SP = Rs.29070

MP of item 1 for shop B = 29070/90*100 = Rs.32300

Difference = (SP/84*100 โ SP/88*100) = 100

SP* 100 (1/84 -1/88) =100

SP = (84*88)/(88-84)

SP = 88 * 84 /4 = Rs. 1848

MP of item 2 in shop C = 1848/92 *100 =2009 = Rs. 2010

MP of item 3 in shop B = (1100/(9x-8x))*9 = Rs.9900

SP of item 3 = 9900 *80/100 = Rs.7920

Given,

MP of item 1 in Shop B = 250/10*100 = Rs.2500

SP of item 1 in shop B = 2500*90/100 = Rs.2250

MP of item 1 in shop D = 2250/95 *100 =Rs.2368

MP of item 1 in shop E = 2250/85*100 = Rs.2647

Required average = (2368+2647)/2 =Rs.2507.5

Required ratio = MP* 90/100: MP* 88/100: MP *80/100

= 90: 88: 80 = 45: 44: 40

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 โ20-20โ 11.00 AM Vocabulary (Based on The Hindu) 12.00 PM General Awareness โ20-20โ 1.00 PM English Language โ20-20โ 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 โ20-20โ 7.00 PM English Language (New Pattern Questions) 8.00 PM