1.Clock A loses 1 minute a day and clock B gains 1 1/2 minutes a day. If clock B is 30 minutes ahead of clock A, how many days will clock B take to be 45 minutes ahead of clock A?
(1) 10
(2) 6
(3) 18
(4) 8
2.Which of the following equations are equivalent?
(a) 2p^2 - (3p^2 + 5) + 9 -2p^2 (p^2 - 1)
(b) (2p^2 + 3)^2
(c) (2p^2 - 3)^2
(d) 4p^2 + 12p^2 + 9
(1) Only a and b
(2) Only b and c
(3) Only b and d
(4) All except c
3.The population‘ of a town decreases every year due to death and migration by 5%. The population was 1 .20,000 last year. What will be the expected population at the end of the expected population at the end" of the next year?
(1)1,09,744
(2) 1,02,885
(3) l,O8,3O0
(4) Data inadequate.
4.Which of the following numbers is / are completely divisible by thirteen?
(a) 138138
(b) 121211
(c) 124214
(d) 120120
(1) All are divisible
(2) Only a and b
(3) Only b and d
(4) Only a and d
5.A certain tank can be filled by pipes A and B separately in 4 and 5 minutes respectively. Whereas, pipe C can empty it in 3 minutes. How long will it take to fill or empty the 4/5 full tank. If all the three pipes start together?
(1) 1 5/7 minutes to fill
(2) 6 6/7 minutes to fill
(3) 1 5/7 minutes to fill
(4) 6 6/7 minutes to fill
6.What will be the difference the interest obtained by investing Rs. 500/- for two years at the rate of 10% p.a. compounded yearly and Rs. 800/- at the rate of 10% p.a. compounded half-yearly for one year‘?
(1) Rs. 20
(2) Rs. 27
(3) Rs. 60
(4) Rs. 23
7.When a dealer heard that the price of certain product was about to increase Rs. 5 per pack, he purchased a number of packs of this product for Rs. 4,500/-. Had he pur- chased it at the new price, he would have obtained 10 fewer packs. How many packs did he buy’?
(1) 90
(2) 100
(3) 50
(4) 125
8.How many kilograms of coffee price at Rs. 33 per kg should be mixed with the coffee at Rs. 27 per kg so that the 60 kg of mixture costs Rs. 32 per kg?
(1) 30
(2) 20
(3) 50
(4) 10
9.If A and B work together, they complete a work in 3 days, B and C together complete the same work in 6 days, while A and C complete that work in 8 days. If there of them many will they complete the woek?
(1) More than 3 days
(2) More than 2 days
(3) Less than 3 days
(4) More than 4 days
10.If a rectangular space 8 cm long and 6 cm wide is increased in its lenght and width by 25%, then what will be the percentage increase in the area of original rectangular space?
(1) 25
(2) 50
(3) 36
(4) 56 1/4
SOLUTION
1.(2) According to question.
Clock B gains per hours to
A = (1/24 +1/16)
= (2 + 3) /48 = 5/48 minutes
Now.
Clock B gains 5/48 minutes in a hour
So Clock B gains 15 minutes in
48/5 * 15 hour
= (48 * 15)/(5 * 24) days = 6 days.
2.(5) (a) = 2P^2 (3p^2 + 5) + 9 - 2p^2 (P^2 - 1)
= 6p^4 + 10p^2 + 9 - 2p^4 + 2p^2
= 4p^2 + 12p^2 + 9
= (2p^2 + 3)^2
(b) = (2p^2 + 3)^2
(c) = (2p^2 - 3)^2
(d) = 4p^4 + 12p^2 + 9
= (2p^2 + 3)^2
so (a) = (b) = (d)
Therefore, the required answer is option (4).
3(2)According to question.
Population of town (last year) = 120,000
Present population of town
= 100000 - 5% of 120000
= 120000 - (5 * 120000)/100
= 120000 - 60000
= 114000
so Next year population of town
= 114000 - 5% of 114000
= 114000 - (5 * 114000)/100
= 114000 - 5700 = 108300
4.(4) (a) = 138138 ÷ 13 = 10626
(b) = 121211 ÷ 13 = 9323 12/13
(c) = 124214 ÷ 13 = 9554 12/13
(d) = 120120 ÷ 13 = 9240
Therefore, the required, answer is option (4).
5(1) A's work in one minute = 1/4
B's work in one minute = 1/5
C's work in one minute = 1/3
so (A+B+C)'s work in one minute
= (1/4 + 1/5 - 1/3)
= (15 + 12 - 20)/60 = 7/60
so Time taken to fill 1-4/5
i.e. 1/5 part of tank
= 60/7 * 1/5 = 12/7 = 1 5/7 minutes
6.(4) From the Formula
A = P (1 + 10/100 )^t
in first case
A1 = 500 (1 + 10/100 )^2
= 500 * 110/100 * 110/100 = Rs. 605
so Compound Interest
= 605 - 500 = Rs. 105
so In second case
r = 10/2 = 5% half yearly
t = 1 * 2 = 2 half yearly.
so A2 = 800 (1 + 5/100 )^2
= 800 * 105/100 * 105/100
= 800 * 21/20 * 21/20 = Rs. 882
so compound Interest
= Rs. 882 = Rs. 800
= Rs. 82
so Required difference
= 105 - 82 = Rs. 23
7.(2) Let the number of purchased pack = x
so According to question.
4500/x = 4500/x -10 - 5
or, 4500 [1/(x )-1/(x - 10)] = -5
or, 900 [1/(x - 10) - 1/(x )] = 1
or, (900 [x - (x - 10) ])/((x - 10)x)= 1
or, 900 * 10 = x^2 - 10x
or , x^2 - 10x - 9000 = 0
or x^2 - 100x + 90x - 9000
= 0
or, x(0 - 100) + 90 (x -100)
= 0
or, (x + 90) (x - 100) = 0
or, x = - 90 and 100
Since the value of x should not be negative, therefore, the required answer is 100.
8.(3)
9.(1) work done by 2(A + B + C) in one day
= (1/3 +1/6+1/8 )
= (8 + 4 + 3)/24 = 15/24 = 5/8 part
so (A + B + C)'s one day work
= 5/8*2 = 5/16 work
so Whole work can be done by
(A + B + C) in 16/5 days i.e. 3 1/5 days
Therefore, the required answer is option (1).
10.(4)
1.(2) According to question.
Clock B gains per hours to
A = (1/24 +1/16)
= (2 + 3) /48 = 5/48 minutes
Now.
Clock B gains 5/48 minutes in a hour
So Clock B gains 15 minutes in
48/5 * 15 hour
= (48 * 15)/(5 * 24) days = 6 days.
2.(5) (a) = 2P^2 (3p^2 + 5) + 9 - 2p^2 (P^2 - 1)
= 6p^4 + 10p^2 + 9 - 2p^4 + 2p^2
= 4p^2 + 12p^2 + 9
= (2p^2 + 3)^2
(b) = (2p^2 + 3)^2
(c) = (2p^2 - 3)^2
(d) = 4p^4 + 12p^2 + 9
= (2p^2 + 3)^2
so (a) = (b) = (d)
Therefore, the required answer is option (4).
3(2)According to question.
Population of town (last year) = 120,000
Present population of town
= 100000 - 5% of 120000
= 120000 - (5 * 120000)/100
= 120000 - 60000
= 114000
so Next year population of town
= 114000 - 5% of 114000
= 114000 - (5 * 114000)/100
= 114000 - 5700 = 108300
4.(4) (a) = 138138 ÷ 13 = 10626
(b) = 121211 ÷ 13 = 9323 12/13
(c) = 124214 ÷ 13 = 9554 12/13
(d) = 120120 ÷ 13 = 9240
Therefore, the required, answer is option (4).
5(1) A's work in one minute = 1/4
B's work in one minute = 1/5
C's work in one minute = 1/3
so (A+B+C)'s work in one minute
= (1/4 + 1/5 - 1/3)
= (15 + 12 - 20)/60 = 7/60
so Time taken to fill 1-4/5
i.e. 1/5 part of tank
= 60/7 * 1/5 = 12/7 = 1 5/7 minutes
6.(4) From the Formula
A = P (1 + 10/100 )^t
in first case
A1 = 500 (1 + 10/100 )^2
= 500 * 110/100 * 110/100 = Rs. 605
so Compound Interest
= 605 - 500 = Rs. 105
so In second case
r = 10/2 = 5% half yearly
t = 1 * 2 = 2 half yearly.
so A2 = 800 (1 + 5/100 )^2
= 800 * 105/100 * 105/100
= 800 * 21/20 * 21/20 = Rs. 882
so compound Interest
= Rs. 882 = Rs. 800
= Rs. 82
so Required difference
= 105 - 82 = Rs. 23
7.(2) Let the number of purchased pack = x
so According to question.
4500/x = 4500/x -10 - 5
or, 4500 [1/(x )-1/(x - 10)] = -5
or, 900 [1/(x - 10) - 1/(x )] = 1
or, (900 [x - (x - 10) ])/((x - 10)x)= 1
or, 900 * 10 = x^2 - 10x
or , x^2 - 10x - 9000 = 0
or x^2 - 100x + 90x - 9000
= 0
or, x(0 - 100) + 90 (x -100)
= 0
or, (x + 90) (x - 100) = 0
or, x = - 90 and 100
Since the value of x should not be negative, therefore, the required answer is 100.
8.(3)
9.(1) work done by 2(A + B + C) in one day
= (1/3 +1/6+1/8 )
= (8 + 4 + 3)/24 = 15/24 = 5/8 part
so (A + B + C)'s one day work
= 5/8*2 = 5/16 work
so Whole work can be done by
(A + B + C) in 16/5 days i.e. 3 1/5 days
Therefore, the required answer is option (1).
10.(4)
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