1. A can buy goods at the rate of Rs. 20 per piece. A sells the first good for Rs. 2, second one for Rs. 4, third for Rs. 6…and so on. If he wants to make an overall profit of at least 40%, what is the minimum number of goods he should sell ?
A. 32
B. 24
C. 27
D. 18
E. None of these
Sol. Let us assume he buys n goods.
Total CP = 20n
Total SP = 2 + 4 + 6 + 8 ….n terms
Total SP should be at least 40% more than total CP
2 + 4 + 6 + 8 ….n terms = 1.4 ×20 n
2 (1 + 2 + 3 + ….n terms) = 28n
n(n + 1) = 28n
n^2 + n = 28n
n^2 - 27n = 0
n = 27
He should sell a minimum of 27 goods.
2. In a class, if 50% of the girls were boys, then there would be 50% more boys than girls. What percentage of the overall class is boys ?
A. 30%
B. 40%
C. 20%
D. 50%
E. None of these
Sol. Number of boys should be 1.5 times the number of girls. When 50% of the girls are taken as boys, let the number of girls = x
Number of boys = 1.5x
Total number of students = 2.5x
Original number of girls = 2x (50% of boys = x)
Original number of boys = 0.5x
Boys form 20% of the overall class.
3. 4 men and 6 women complete a task in 24 days. If the women are at least half as efficient as the men, but not more efficient than the men, what is the range of the number of days for 6 women and 2 men to complete a task ?
A. 27 to 33.6 days
B. 30 to 36.6 days
C. 30 to 33.6 days
D. 24 to 27.6 days
E. None of these
Sol. 4m and 6w finish in 24 days.
In one day, 4m + 6w = 1/24 of task.
In these questions, just substitute extreme values to get the whole range
If a woman is half as efficient as man
4m + 3m = 1/24, 7m = 1/24, m = 1/168
6w + 2m = 3m + 2m = 5m, 5m will take 168/5 days = 33.6 days
If a woman is as efficient as a man
4m + 6w finish in 24 days
10m finish 1/24 of task in a day
6w + 2m = 8m, 8m will take 240/8 = 30 days to finish the task.
So, the range = 30 to 33.6 days. The new team will take 30 to 33.6 days to finish the task.
4. How many 5-digit numbers can be formed by using the digits 2, 3 and 4 so that each digit appears at least once ?
A. 200
B. 240
C. 180
D. 150
E. None of these
Sol. Suppose the 3 digits are x, y and z. Since each of these digits will appear at least once in the 5-digit number, there are 2 possibilities – (1) one of the digits appears 3 times and the other 2 digits appear once each; (2) two of the digits appear twice each and the third digit appears once.
Consider the first possibility. The digit that appears 3 times can be chosen in 3C1 = 3 ways. The digits can now be arranged in (5!/3!) = 20 ways. So, 3 × 20 = 60 numbers can be formed.
Consider the second possibility. The 2 digits that appear 2 times can be chosen in 3C2 = 3 ways. The digits can now be arranged in (5!/2!2!) = 30 ways. So, 3 × 30 = 90 numbers can be formed.
Thus, a total of 60 + 90 = 150 numbers can be formed.
5. Madhu, Shilpa and Mehr along with seven other girls had reached the finals of Miss Timbuktu 2003. Intially all the girls had an average weight of 47.9 kgs. After the elimination of Madhu, the average weight got reduced to 46 kgs. If the average weight of the rest of the girls is 3kgs less than the average weight of Madhu, Shilpa and Mehr together then:
What is the average weight of the rest of the girls, excluding Madhu, Shilpa and Mehr?
A. 45
B. 47
C. 48
D. 46
E. None of these
Sol. Let the weights of Madhu, Shilpa and Mehr be M1, S and M2 respectively. Also let the total weight of the rest be k kgs.
Then M1+S+M2+K = 47.9 × 10 = 479
Let average weight of M1, S and M2 be x then average weight of other 7 girls = x - 3.
3x + 7(x-3) = 479
x = 50.
So average weight of rest of the girls = 50 – 3 = 47.
5. If the radius of a sphere is doubled, then its volume is increased by:
A. 100%
B. 200%
C. 700%
D. 800%
E. None of these
Sol.
6. A ball of lead, 4 cm in diameter, is covered with gold. If the volume of the gold and lead are equal, then the thickness of gold is approximately [given∛2=1.259]
A. 5.038 cm
B. 5.190 cm
C. 1.038 cm
D. 0.518 cm
E. None of these
Sol.
7. A conical cup is filled with ice cream. The ice cream forms a hemispherical shape on its open top. The height of the hemispherical part is 7 cm. The radius of the hemispherical part equals the height of the cone. Then the volume of the ice cream is [Ï€=22/7]
A. 1078 m^3
B. 1708 m^3
C. 7108 m^3
D. 7180 m^3
E. None of these
Sol.
8. Rs.5200 was partly invested in Scheme A at 10% pa CI for 2 years and Partly invested in Scheme B at 10% pa SI for 4 years. Both the schemes earn equal interests. How much was invested in Scheme A ?
A.Rs.1790
B.Rs.2200
C.Rs.3410
D.Rs.2670
E.None of these
Sol. Amount invested in Scheme B = X
Amount invested in Scheme A = 5200 – x
X*10*4/100 = (5200-x)*21/100……………………[(1-10/100)2-1] = 21/100
40x/100 = (5200-x)*21/100
2x/5 = (5200-x)*21/100
200x = 5200*21*5 – x*5*21
200x = 546000 – 105x
305x = 546000
X = 1790
Scheme A = 5200 – 1790 = 3410.
9. A sum of Rs.3,50,500 is to be paid back in 2 equal annual instalments. How much is each instalment, if the rate of interest charged 4% per annum compound annually ?
A.Rs.1,90,450
B.Rs.1,65,876
C.Rs.1,76,545
D.Rs.1,85,834
E.None of these
Sol. Total value of all the 3 instalments
[X*100/104] + [X*100*100/104] = 3,50,500
X*25/26 + x*625/676 = 3,50,500
X*25/26[1+25/26] = 3,50,500
X*25/26[51/26] = 3,50,500
X = 3,50,500*676/25*51 = 1,85,833.72 = 1,85,834.
10. The difference between the total simple interest and the total compound interest compounded annually at the same rate of interest on a sum of money at the end of two years is Rs. 350. What is definitely the rate of interest per cent per annum?
A.9,300
B.7600
C.12000
D.Data inadequate
E.None of these
Sol. Difference = Pr2/(100)2
= (350×100×100)/(P×r2)
P is not given. Thus data inadequate.
