__New Pattern Aptitude Questions (Inequality – Data Sufficiency) Asked in SBI PO Mains 2017____:__

Dear Readers, New Pattern Aptitude Questions (Inequality – Data Sufficiency) asked in SBI Po Mains 2017 was given here with Solutions. Aspirants those who are preparing for Bank exams& all other Competitive examination can use this material.

**Directions (1-5) : Each question below contains a statement followed by Quantity I and Quantity II. Find the value of both to find the relationship among them. Mark your answer accordingly.
**

**1.The difference between a two-digit number and the number obtained by interchanging the positions of its digits is 36.**

**Quantity I:** What is the difference between the two digits of that number?

**Quantity II:** What is the difference between the sum and the difference of the digits of the number if the ratio between the digits of the number is 1 : 2?

**I:**Let the ten’s digit be x and unit’s digit be y.

Then, (10x + y) – (10y + x) = 36

9(x – y) = 36

x – y =

**4.**

**II:**Since the number is greater than the number obtained on reversing the digits, so the ten’s digit is greater than the unit’s digit.

Let ten’s and unit’s digits be 2x and x respectively.

Then, (10 x 2x + x) – (10x + 2x) = 36

9x = 36 x = 4.

Required difference = (2x + x) – (2x – x) = 2x =

**8**

Answer :C)

**2.Through T, the mid-point of the side QR of a triangle PQR, a straight line is drawn to meet PQ produced to S and PR at U, so that PU= PS**

**Quantity I:** If length of UR=2 units then the length of QS is ?

**Quantity II:**2√2 units

**2).I.** We have QT = TR and PU=PS.UR = 2 units

We draw RV || PS that meets SU extended at V.

In △ QST and △TVR ∠QTS=∠VTR [Opposite angles]
∠QST=∠TVR [Alternate angles as PS ∥ VR]
QT+ TR

∵ △QST and △TVR are congruent.

∴ QS=VR ——– (i)

Now, ∠QST =∠PUS =∠VUR =∠UVR

∴ In △ UVR

∠VUR =∠RVU

or, RV = UR =2 ——– (ii)

From (i) and (ii)

QS = VR = UR = 2units

**II.** 2√2 = 2.828 units

So, II> I.

Answer: C)

**3. The dimensions of a certain machine are 48″ X 30″ X 52″. If the size of the machine is increased proportionately until the sum of its dimensions equals 156″.**

**Quantity I:** What will be the increase in the shortest side?

**Quantity II:** If sum of its dimensions is decreased by 8” from 156”, then what will be the increase in the longest side?

**3). I.**Sum of present dimension 48+30+52=130.

New dimension =156.

Increase in dimension = 26.

Ratio of dimensions = 48:30:52 = 24:15:26

Therefore increase in the shortest side15*(26)/(24+15+26)=6

**II.**Sum of present dimension 48+30+52=130.

New dimension =156 – 8 = 148.

Increase in dimension = 18.

Ratio of dimensions = 48:30:52 =>24:15:26

Therefore increase in the shortest side = 26*(18) / (24+15+26)= 7.2

Hence I< II.

Answer: A)

**4.A wooden door wedge is in the shape of a sector of a circle of radius 10 cm with angle 24 degree and constant thickness 3 cm.**

**Quantity I:** Find the volume of wood used in making the wedge.

**Quantity II:** 20 π cm^3

**4). I.** The area of the top face of the wedge is the area of a sector of radius 10 cm and angle 24 degree.

Area=24 degree / 360 degree × π ×102 = 20π / 3 = 20.94 cm^2

The volume of the wedge =Area × 3 = 20 π = 62.83 cm^3

**II.** 20 π cm^3

Hence I = II.

Answer: E)

**5. The cost of Type 1 material is Rs. 15 per kg and Type 2 material is Rs.20 per kg. If both Type 1 and Type 2 are mixed in the ratio of 2 : 3.**

**Quantity I:** what is the price per kg of the mixed variety of material?

**Quantity II:**what is the price per kg of the mixed variety of material if both Type 1 and Type 2 are mixed in the ratio of 3 : 2?

**5). I.**Cost Price (CP) of Type 1 material is Rs. 15 per kg

Cost Price (CP) of Type 2 material is Rs. 20 per kg

Type 1 and Type 2 are mixed in the ratio of 2 : 3.

Hence Cost Price(CP) of the resultant mixture

={(15×2) + (20×3)} / (2+3)

= (30+60) / 5 = 90 / 5 = 18

Price per kg of the mixed variety of material = Rs.18.

**II.**Cost Price (CP) of Type 1 material is Rs. 15 per kg

Cost Price (CP) of Type 2 material is Rs. 20 per kg

Type 1 and Type 2 are mixed in the ratio of 3 : 2.

Hence Cost Price(CP) of the resultant mixture

={(15×3) + (20×2)} / (2+3)

= (45+40) / 5 = 85 / 5 = 17

Price per kg of the mixed variety of material = Rs.17.

Hence I> II.

Answer: C)

**Direction (6-10): Each of these questions is followed by three statements. You have to study the question and all the three statements given to decide whether any information provided in the statement(s) is redundant and can be dispensed with while answering the given question.**

**6.At what time will the train reach city X from city Y?**

**Statement I:**The train crosses another train of equal length of 200 metres and running in opposite directions in 15 seconds.

**Statement II:** The train leaves city Y and 7.15 a.m. for city X situated at a distance of 558 km.

**Statement III:**The 200 metres long train crosses a signal pole in 10 seconds.

From the statement I, we get length of the train is 200 metres (Redundant info while comparing with Statement III). The rest of the info given in this statement cannot be used for calculating the speed of the train, because the two trains might run at different speed.

III gives, speed = 200/10 = 20 m/sec = 20 × 18/5 km/hr = 72 km/hr.

II gives, Time taken = 558/72 hrs = 31/4 hrs

= 7 ¾ hrs = 7 hrs 45 min.

So, the train will reach city X at 3 p.m.

Hence II and III only give the answer.

Answer: D)

**7. How much did Rohit get as profit at the year-end in the business done by Nitin, Rohit and Kunal?**

**Statement I:**Kunal invested Rs. 8000 for nine months, his profit was 3/2 times that of Rohit’s and his investment was four times that of Nitin.

**Statement II:** Nitin and Rohit invested for one year in the proportion 1 : 2 respectively.

**Statement III:** The three together got Rs. 1000 as profit at the year end

**7). I and II give:**K = Rs. (8000 x 9) for 1 month = Rs. 72000 for 1 month.

N = Rs. (1/4 × 8000 × 12) for 1 month = Rs. 24000 for 1 month

R = Rs. 48000 for 1 month.

K : N : R = 72000 : 24000 : 48000 = 3 : 1 : 2.

**III gives,**total profit = Rs. 1000.

Rohit’s share = Rs. (1000 × 2/6) = Rs. 333 1/3

Hence I, II and III together gives the answer

Answer: D)

**8. What was the percentage of discount given?**

**Statement I:**23.5% profit was earned by selling a TV for Rs. 12,350.

**Statement II:** If there were no discount, the earned profit would have been 30%.

**Statement III:** The cost price of the TV was Rs. 10,000.

C.P. = Rs. (100/123.5 × 12350) = Rs. 10,000.

II. M.P. = 130% of C.P. = 130% of Rs. 10,000 = Rs. 13,000.

From I and II, discount = Rs. (13000 – 12350) = Rs. 650.

Discount % = (650 / 13000 × 100)% = 5%

Thus, I and II give the answer.

II and III cannot give the answer. Because we require profit percentage with discount and profit percentage without discount. So II and III are not sufficient.

Since III gives C.P. = Rs. 10,000, I and III give the answer.

Therefore, I and II [or] I and III give the answer.

Correct answer is (E)

Answer: E)

**9. In how many days can 10 women finish a work?**

**Statement I:**10 men can complete the work in 6 days.

**Statement II:** 10 men and 10 women together can complete the work in 3 3/7 days

**Statement III:**If 10 men work for 3 days and thereafter 10 women replace them, the remaining work in completed in 4 days

Therefore, 1 man’s 1 day work = 1/60

II. (10 × 24/7) men + (10 × 24/7) women can complete the work in 1 day.

(240/7) men’s 1 day work + (240/7) women’s 1 day work = 1.

(240/7 × 1/60) + (240/7) women’s 1 day work = 1

(240/7) women’s 1 day work = (1 – 4/7) = 3/7

Therefore, 10 women’s 1 day work = (3/7 × 7/240 × 10) = 1/8

So, 10 women can finish the work in 8 days.

III. (10 men’s work for 3 days) + (10 women’s work for 4 days) = 1

(10 x 3) men’s 1 day’s work + (10 x 4) women’s 1 day’s work = 1

30 men’s 1 day’s work + 40 women’s 1 day’s work = 1

Thus, I and III will give us the answer.

And, II and III will give us the answer.

Correct answer is (A).

Answer: A)

**10.What is the capacity of the cylindrical tank?**

**Statement I:**The area of the base is 61,600 sq. cm.

**Statement II:**The height of the tank is 1.5 times the radius.

**Statement III:** The circumference of base is 880 cm<

I gives, πr^2 = 61600. This gives r.

II gives, h = 1.5 r.

Thus, I and II give the answer.

Again, III gives 2πr = 880. This gives r.

So, II and III also give the answer.

Correct answer is (E).

Answer: E)