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1. (c)
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
2. (c)
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.
3. (a)
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.

4. (e)
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

5. (c)
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.

6. (d)
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

7. (d)
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

8. (e)
I. S.P. = Rs. 12350, Gain = 23.5%
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.

9. (a)
I. (10 x 6) men can complete the work in 1 day.
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 × 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.
10. (e)
Capacity = πr^2h.
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.

 

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