Practice Quantitative Aptitude Questions For SBI PO Prelims& NIACL 2017 (Application Problems) Answers Updated

Practice Quantitative Aptitude Questions For SBI PO Prelims& NIACL 2017 (Application Problems):

Dear Readers, Important Practice Aptitude Questions for Upcoming Exams was given here with Solutions. Aspirants those who are preparing for the Bank Examination and other Competitive Examination can use this.

 

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1).A bag contains 6 red balls, 11 yellow balls and 5 pink balls. If two balls are drawn at random from the bag, one after another, what is the probability that the first ball is red and the second ball is yellow?
a)  1/14
b)  2/7
c)  5/7
d)  3/14
e)  None of these
 
2).The sum of the radius and the height of a cylinder is 19m. The total surface area bf the cylinder is 1672 m2, what is the volume  of the cylinder? (in m3)
a)  3080
b)  2940
c)  3220
d)  2660
e)  2800
 
3).The ratio of the speed of the boat upstream to the speed of the boat downstream is 2 : 3. What is the speed of the boat in still water if it covers 42 km downstream in 2 hours 20 minutes? (in km/h)
a)  13.5
b)  15
c)  18
d)  20
e)  16
 
4).35 men complete a piece of work in 16 days and 20 women complete the same piece of work in 30 days. What is the ratio of the amount of work done by 40 men in 1 day to the amount of work done by 15 women in 1 day?
a)  10:7
b)  20:7
c)  7:20
d)  21:20
e)  7 : 10
 
5).A man sold an article for Rs. 6800 and incurred a loss. Had he sold the article for Rs.7850, his gain would have been equal to half of the amount of loss that he incurred. At what price should he sell the article to have 20% profit?
a)  Rs.7500
b)  Rs.9000
c)  Rs.10680
d)  Rs. 9600
e)  Rs.11500
 
 6). A bought a certain quantity of bananas at a total cost of Rs. 1500. He sold 1/3 of these bananas at 25% loss. If he earns an overall profit of 10%, at what percentage profit did A sell the rest of the bananas?
a)  25%
b)  27.5%
c)  30%
d)  15%
e)  18%
 
7).A tank has two inlets: P and Q. P alone takes 6 hours and Q alone takes 8 hours to fill the empty tank completely when there is no leakage. A leakage was caused which would empty the full tank completely in ‘X’ hours when no inlet is open. Now, when only inlet P was opened, it took 15 hours to fill the empty tank completely. How much time will Q alone take to fill the empty tank completely? (in hours)
a)  40 hours
b)  20 hours
c)  30 hours
d)  10 hours
e)  25 hours
 
8).At present, the ratio of the ages of A to B is 3 : 8; and that of A to C is 1 : 4. Three years ago, the sum of the ages of A, B and C was 83 years. What is the present age (in years) of C?
a)  32
b)  12
c)  48
d)  54
e)  15
 
9).The sum invested in Scheme B is thrice the sum invested in Scheme A. The investment in Scheme A is made for 4 years at 8% p.a. simple interest and in Scheme B for 2 years at 13% p.a. simple interest. The total interest earned from both the schemes is Rs.1320. How much amount was invested?
a)  Rs. 1200
b)  Rs. 1140
c)  Rs. 960
d)  Rs. 1500
e)  Rs. 840
 
10).Kim and Om are travelling from point A to B, Which are 400 km apart. Travelling at a certain speed Kim takes one hour more than Om to reach point B. If Kim doubles her speed she will take 1 hour 30 mins less than Om to reach point B. At what speed was Kim Driving from point A to B? (in kmph)
a)  90 kmph
b)  70 kmph
c)  80 kmph
d)  160 kmph
e)  100 kmph

Solution:
 
1). B) Total of balls = 6 + 11 + 5 = 22
n(S) =22C2 = (21×22) / 2 = 231
Now, n(E) =6C1 x11C1 = 6 x 11 = 66
P(E) = n(E)/n(S) = 66/231 = 6/21 = 2/7
 
2). A)Let the radius of the cylinder be r and height be h.
Then, r + h = 19        …..(i)
Again, total surface area of cylinder = (2ᅲrh + 2ᅲr2)
Now, 2ᅲr(h + r) = 1672
or, 2ᅲr x 19 = 1672
or, 38ᅲr = 1672 , ᅲr = (1672/38) = 44m, r = (44 × 7) / 22 = 14
Height = 19 – 14 = 5m
Volume of cylinder = ᅲr2h = (22/7) x 14 x 14 x 5 =14m
= 22 × 2 × 14 × 5 = 3080m3
 
3). B)Let the speed of the boat in still water be x and that of the current be y.
Then, downstream speed = x + y and upstream speed = x – y
Now, downstream speed = 42 / [2 20/60] = (42 × 3) / 7 =18 km
x+y=18
Again, 3 : 18, 2 : 12
(As ratio of downstream to upstream is 2 : 3)
 x – y = 12 Solving (i) and (ii), we get
(x+y=18) + (x – y =12) = 2x =30
x = 15 kmph
Hence speed of the boat 15 kmph
 
4). B)35 men complete the work in 16 days.
1 man completes the work in 16 x 35 days,
32 men complete the work in (16×35)/40 = 14 days.
Again, 20 women complete the same piece of work in 30 days.
1 woman completes the same piece of work in 20 × 30 days.
15 women can complete the work in (20×30)/15 = 40 days.
Ratio = 1/14 : 1/40 = 40 : 14 = 20 : 7
 
5). B)Let the cost price be x.
Then, loss = (x – 6800)
Again, profit = (7850 – x)
Now, (7850 – x) = (x – 6800)/2 or, 15700 – 2x = x – 6800
or, 3x = 15700 + 6800 = 22500
x = 22500/3 = 7500
Selling price = (7500×120)/100 = Rs. 9000
 
6). B)Total CP = 1500
Total SP = 1500 + 10% of 1500 = 1500 + 150 = 1650
CP of 1/3 of bananas = 1500/3 = Rs.500
SP of 1/3 of bananas at 25% loss
= 500 – [ (500 x25 / 100)] = 500 – 125 = 375
SP of the rest of bananas = 1650 – 375 = 1275
Now, CP of the test of bananas = 1500 – 500 = 1000
Profit on the rest of bananas = 1275 -1000 = 275
% of profit on the rest of bananas = (275/1000)×100 = 27.5%
 
7). A)(1/P) – (1/X) = (1/15)
Or, (1/6) – (1/X) = (1/15) (P = 6 hours)
Or, (1/X) = (1/6) – (1/15) = (10-4)/60 = 1/10
x = 10 hours
Now,
(1/Q) – (1/10) = (1/8) – (1/10) = (5-4)/40 = 1/40
Hence, Q fills the tank in 40 hours.
 
8). C)According to the question, A : B = 3 : 8
A : C = 1 : 4
B : A = 8 : 3
A : C = 1 : 4
8 : 3 : 12
Sum = 8x + 3x 12x = 23x
Now, 23x = 92
x = 4
Hence the present age of C = 12x = 12 x 4 = 48 years
 
9). A)Let the amount invested in scheme A be Rs.x and that in B be Rs. 3x.
Then, [(x × 4 × 8)/100] [(3x × 2 × 13) /100] = 1320
Or, (32x/100) + (78x/100) = 1320
110x/100 = 1320
x = (1320 x 100) / 110 = Rs. 1200
 
10). D)
Let the speed of Kim be x and that of Om be y.
Then, (400/x) – (400/y) = 1
Let 1/x = u and 1/y = v
 400u — 400v = 1     …(i)
Again, (400/y) – (400/2y) = 3/2
400v – 200u = (3/2)
Or, 800v – 400u = 3      …(ii)
Solving (i) and (ii), we get
(400u – 400v =1) + (-400u + 800v) = 400v = 4
v = (4/400) = (1/100) km
y = 100 km
now, (400/x) – (400/100) = 1
or, (400/x) = 5

x = 80 kmph

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