1. If 3/x - 2x+10/x+5 < 0 what is the range of x ?
A. x < –5 , x > 3/2
B. x < 3/2, x > –3
C. –5 < x < 3
D. –3 < x <5/2
E. Can not be determined
Sol. The given expression can be rewritten as 3x + 15 – 2x^2 – 10x < 0
?15 – 2x^2 – 7x < 0 ? 2x^2 + 7x – 15 > 0.
The roots of this quadratic are 3/2 and –5. So,
The range of values satisfying the inequality is x > 3/2 and x < –5.
2. In a garden, there are 5 rose plants for every 7 marigold plants. There are 3 marigold plants for every 11 pegunias and 2 pegunias for every 9 cactus plants. What is the minimum number of plants in the garden?
A.113
B.793
C.988
D.919
E.None of these
Sol. If the number of marigolds is 21, then there are 15 roses and 77 pegunias.
If there are 154 pegunias, then there are 693 cactus plants.
Consequently, the number of roses and marigolds will be 30 and 42 respectively.
Thus the minimum number of plants is 30 + 42 + 154 + 693 = 919
3. The ages of a husband and his wife x years ago were x + x/2 and x + x/4 respectively. If the husband is 6 years older than his wife, what is the average of their current ages?
A.24 years
B.57 years
C.48 years
D.45 years
E.None of these
Sol. We know that x years ago, the difference between the ages was 6 years.
So, x/2 – x/4 = 6, which on solving yields x = 24.
The ages of the husband and his wife 24 years ago were 24 + 12 = 36 and 24 + 6 = 30 respectively.
The current ages of the husband and his wife are 36 + 24 = 60 and 30 + 24 = 54 respectively.
Thus the average of their current ages is (60 + 54)/2 = 57 years.
4. It takes Gayatri 3 hours to ride her bicycle up a hill and back down. Her average speed while riding the bicycle down the hill is 35 kmph faster than her average speed while riding the bicycle up the hill. If the distance from the bottom of the hill to the top is 40 km, what is Gayatri's average speed while riding her bicycle uphill?
A.17.8 kmph
B.52.8 kmph
C.25.6 kmph
D.14.4 kmph
E.None of these
Sol. Lets assume Gayatri's uphill speed is S kmph then her downhill speed will be (S+35)kmph.
The time taken by her to ride uphill and back downhill is,
40/S + 40/(S+35) = 3
Solving we get the quadratic,
3S^2 +25S - 1400 = 0
Solving this we get,
S=17.8kmph
5. The height of a ball t seconds after it is thrown up is given by the function H(t) = 15t – 4.9t2 + 3. For how much time is the height of the ball more than 12 m?
A.2.22 seconds
B.1.43 seconds
C.2.8 seconds
D.8.6 seconds
E.None of these
Sol. If the height of the ball is more than 12 m, then 15t – 4.9t^2 + 3 > 12.
We can rewrite this as 4.9t^2 –15t + 9 < 0 or 49t^2 – 150t + 90 < 0.
Solving this inequality, we get 0.81 < t < 2.24.
Thus, the height of the ball will be greater than 12 m for 2.24 – 0.81 = 1.43 seconds.
6. If 3 <= n< = 10, what is the probability that the interior angle of a regular n-sided polygon will be greater than 110°?
A.1/4
B.3/8
C.2/3
D.5/8
E.None of these
Sol. We can form a total of 8 polygons with the number of sides ranging from 3 to 10.
Each angle of a regular polygon is (n – 2) × 180°/n.
So, (n – 2) × 180°/n > 110° then 180n – 360 > 110n.
Solving this we get n > 5.1.
So, n can take values 5 different values, 6, 7, 8, 9 and 10.
Thus the required probability is 5/8.
7. The ratio of the marked price to the cost price of a mobile is 5 : 4. The discount percentage offered before it was sold and profit/loss percentage made on it were in the ratio 3 : 4. Find the profit/loss percentage?
A.12.9%
B.14.48%
C.13.8%
D.Cannot be determined
E.None of these
Sol. Let the marked price = 5x and cost price = 4x
Discount percentage = 3y and profit/loss percentage = 4y. Then, SP = 5x(100 – 3y/100) = 4x (100+4y/100)
500 – 15y = 400 + 16y
y = 100/31. Hence, profit percentage = 4y = 400/31, y = 12.9%
8. The sum of the roots of two quadratic equations are equal. The roots of the first equation are in the ratio 3:7 and the roots of the second equation are in the ratio 2:3. What is the minimum possible difference in the product of the roots of the first equation and the products of the roots of the second equation, given that the roots are integers in both cases?
A.15
B.30
C.3
D.12
E.None of these
Sol. Let the roots of the two quadratic equations be 3k, 7k and 2p, 3p respectively.
Given 10k = 5p ? p = 2k.
Product of roots of the first equation is 21k^2 and that of the second equation is 6p^2 i.e. , 24k^2
As the roots are integers least value of k = 1. ? Least difference of the product of roots = 3.
9. A sequence of positive integers includes the number 68 and has arithmetic mean 56. When 68 is removed the arithmetic mean of the remaining numbers is 55. What is the largest possible number in the sequence?
A.109
B.217
C.325
D.649
E.None of these
Sol. 56n – 68 = 55(n – 1), n = 13 To have the largest possible number in the solution we have to consider all remaining 11 numbers as 1.
So largest possible number = 13 × 56 – 68 – 11 = 649
10. 30 soldiers can dig 10 trenches of size 8 × 3 × 3 ft in half a day working 8 hours per day. How many hours will 20 soldiers take to dig 18 trenches of size 6 × 2 × 2 ft working 10 hours per day?
A.6
B.3.6
C.36
D.18
E.None of these
Sol. Work involved in case 1: 8×3×3×10 cu ft.
This is done in 30×4 = 120 man hours.
So in one man hour, 8×3×3×10/120 = 6 cu ft of digging is done.
Work involved in case 2 = 18×6×2×2.
Man hours required = 18×6×2×2/6 = 18×2×2
Hours required = 18×2×2/20 = 3.6 hours
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