1.A jar contains a mixture of two liquids A and B in the ratio 3:1 . When 15 ltrs of this mixture is drawn off and replaced with 9 ltrs of B,then ratio between A and B becomes 3:4. How many ltrs of liq. A was initially present in the jar ?
Sol. Let assume A = 3x ltrs and B = x ltrs.
Now 15 ltrs is drawn and 9 ltrs of B is added,
A = 3x - (3/4)*15 = (12x-45)/4
B = x - (1/4)*15 + 9 = (4x+21)/4
Now this ratio is equal to 3:4 , thus,
3/4 = (12x-45)/(4x+21)
x = 6.75
Thus A = 3x = 3*6.75 = 20.25 ltrs
2. Two equal glasses filled with mixture of alcohol and water in the ratio 2:1 and 1:1 respectively emptied in third glass. Find the ratio of alcohol and water in the third glass ?
Sol. For 1st glass,
Lets assume, alcohol = 2x and water = x
For 2nd glass,
Alcohol = y and water = y
Now as both glasses are equal, thus,
3x =2y
x = 2y/3
Now for 3rd glass,
Alcohol = 2x + y and water = x + y
Ratio of this will be,
(2x+y)/(x+y)
Put x=2y/3 and solve,
=7/5
Thus alcohol:water is 7:5.
Directions( 3-7):Find the missing numbers in the given series
3.12, 19, 35, 59, 90, _
Sol. 19-12 = 7
35-19 = 16
59-35 = 24
90-59 = 31
Now,
16-7 = 9
24-16= 8
31-24= 7
x-31 = 6
x = 37
NOW, 90 + 37 = 127
4. 4, 10, 27, 112, 555, _
Sol. 4*2 + 2 = 10
10*3 - 3 = 27
27*4 + 4 = 112
112*5 - 5 = 555
= 555*6 + 6 = 3336
5. 19, 23, 14, 30, 5, _
Sol. 19 + 2^2 = 23
23 - 3^2 = 14
14 + 4^2 = 30
30 - 5^2 = 5
5 + 6^2 = 41
6. 5, 12, 33, 136, 675, _
Sol. 5*2 + 2 = 12
12*3 - 3 = 33
33*4 + 4 = 136
136*5 - 5 = 675
675*6 + 6 = 4056
7. 14, 6, 4, 4, 8, _
A. 17
B. 12
C. 16
D. 15
Sol. 14*0.5 - 1 = 6
6*0.5 + 1 = 4
4*1.5 - 2 = 4
4*1.5 + 2 = 8
8*2.5 - 3 = 17
8. A seller mark the price 50% above the cost price and give 10% discount on an item.While selling he cheats customer by giving 20% less in weight .Find his overall profit percent(approximate)?
A. 70%
B. 65%
C. 68.75%
D. 64.8%
E. None of these
Sol. Let cost price be Rs 100
Then marked price will be Rs 150
He gives 10% discount, so,
= 0.9*150 = Rs 135
Now this will be our selling price = Rs 135
As he cheats and gives 20% less in weight,
Cost Price becomes = 0.8*100 = Rs 80
Profit % = [(135-80)/80]*100 = 68.75 %.
9. There are 81 liters pure milk in a container. One-third of milk is replaced by water in the container. Again one-third of mixture is extracted and equal amount of water is added. What is the ratio of milk to water in the new mixture?
A. 4:5
B. 3:5
C. 3:4
D. 1:2
E. None of these
Sol. Amount of milk left after two withdrawls,
= 81*[(81-27)/81]^2
= 36 ltrs
Amount of water left = (81-36) = 45 ltrs
Ratio milk:water ,
= 36/45
= 4:5.
10. A rectangle is circumscribed by a circle. Ratio of length and breadth of the rectangle is 4:3, then the ratio of circumference of the circle to perimeter of the rectangle is ?
A. 127/140
B. 137/140
C. 147/140
D. 157/140
E. None of these
Sol. Given,
L/B = 4/3
B= 3L/4
Also,
L^2 + B^2 = (2R)^2 (where R is the radius of the circle)
L^2 + (3L/4)^2 = (2R)^2
2R = 5L/4
Now ratio of circumference:perimeter wil be,
= (2R*3.14)/2(L + 3L/4)
= (5L*3.14/4)/2(7L/4)
= 157/140
= 157:140
11. Find the probability of getting a multiple of 2 or 6 from numbers [1,2,3,.......,200] ?
A. 1/3
B. 1/2
C. 2/3
D. 1/4
E. 3/4
Sol. As 6 is a multiple of 2 thus all the numbers multiple of 6 are already included in the set of multiples of 2.
No. of multples of 2 = 100
Total numbers = 200
Probability = 100/200 = 1/2.
12. The average marks in Science subject of a class of 20 students is 68. If the marks of two students were misread as 48 and 65 of the actual marks 72 and 61 respectively, then what would be the correct average?
(a) 68.5
(b) 69
(c) 69.5
(d) 70
(e) 66
Sol. Total marks in science/20 = 68
Total marks in science = 68*20
Now net correction in tota marks is = (72+61)-(48+65) = 20
Correct total marks = 68*20 + 20
Thus correct avg will be,
= (68*20 + 20)/20
= 69 marks.
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