1.There are some boys and girls in a
room. The square of the number of the girls is less than the square of the
number of boys by 28. If there were two more girls, the number of boys would
have been the same as that of the girls. The total number of the boys and girls
in the room are
(a) 56
(b) 14
(c) 10
(d) 7
2.The H.C.F of two numbers, each
having three digits, is 17 and their L.C.M. is 714. The sum of the numbers will
be:
(a) 289
(b) 391
(c) 221
(d) 731
3.The first period of a class starts
at 10 : 30 hours and fourth ends at 13 : 45 hours. If periods are of equal
duration and after each period a break of 5 minutes is given to the students,
the exact duration of each period is :
(a) 35 minutes
(b) 42 minutes
(c) 45 minutes
(d) 40 minutes
4.The smallest positive integer n,
for which 864n is a perfect cube, is :
(a) 1
(b) 2
(c) 3
(d) 4
5.A, B and C can complete a work in
10, 12 and 15 days respectively. The started the work to gether. But A left the
work before 5 days of its completion. B also left the work 2 days after A left.
In how many days was the work completed?
(a) 4
(b) 5
(c) 7
(d) 8
6.A can complete a piece of work in
10 days. B in 15 days and C in 20 days. A and C worked together for two days
and then A was replaced by B. In how many days, altogether, was the work
completed?
(a) 12
(b) 10
(c) 6
(d) 8
7.Two pipes A and B can fill a water
tank in 20 and 24 minutes respectively and a third pipe C can empty at the rate
of 3 drums per minute. If A, B and C opened together fill the tank in 15
minutes, the capacity (in drums) of the tank is :
(a) 180
(b) 150
(c) 120
(d) 60
8.If 10 men or 20 boys can make 260
mats in 20 days, then how many mats will be made by 8 men and 4 boys in 20
days?
(a) 260
(b) 240
(c) 280
(d) 520
9.A fan is listed at Rs. 1,400 and
the discount offered is 10%. What additional discount must be given to bring
the net selling price to Rs. 1,200?
(a) 16 2/3 %
(b) 5%
(c) 4 16/21 %
(d) 6%
10.A man goes from A to B at a
uniform speed of 12 kmph and returns with a uniform speed of 4 kmph His average
speed (in kmph) for the whole journey is :
(a) 8
(b) 7.5
(c) 6
(d) 4.5
ANSWERS AND
SOLUTION :
1.(b) Let the number of boys and girls in the room be x and y respectively.
According to the question,
x^2 = y^2 + 28
=> x^2 – y^2 = 28 …………………..(i)
and
x = y + 2
=> x – y = 2 ……………………….(ii)
On dividing equation (i) by equation (ii), we have
(x^2- y^2)/(x-y)=28/2
=> ((x+y)(x-y))/(x-y)=14
=> x + y = 14
so Total number of boys and girls = 14
2.(c) Let the numbers be 17x and 17y where x and y are co-prime.
LCM of 17x and 17y = 17 xy
According to the question.
17xy = 714
=> xy = 714/17 = 42 = 6 * 7
=> x = 6 and y = 7
or x = 7 and y = 6
so first number = 17x
= 17 * 6 = 102
Second number = 17y
= 17 * 7 = 119
so Sum of the numbers
= 102 + 119 = 221
3.(c) Time interval between the first period and last (fourth) period = 1:45 pm. – 10 : 30 am.
= 3 hours 15 minutes
= Break after each period
= 5 minutes
so Total break = 15 minutes
so Time period of classes = 3 hours 15 minutes – 15 minutes.
= 3 hours
so Exact duration of each period = 3/4 hour = 45 minutes.
4.(b)
5.(c) Let the work be completed in x days.
According to the question,
(x-5 )/10+(x - 3)/12 + x/15 = 1
=>(6x - 30 + 5x – 15 + 4x)/60 = 1
=> 15x – 45 = 60
=> 15x = 105 => x = 105/15 = 7
Hence, the work will be completed in 7 days.
6.(d) Work done by (A +C) in 2 days = 2 (1/10+1/20)
= 2 ((2 + 1)/20)=6/20=3/10
Remaining work = 1 – 3/10 = 7/10
(B + C)’s 1 day’s work
= 1/15 + 1/20 = (4 + 3)/60 = 7/60
so Time taken by (B + C) to finish 7/10 parts of the work
= 60/7 * 7/10 = 6 days
so Total time = 2 + 6 = 8 days
7.(c) Let the capacity of the tank be x drum.
Quantity of water filled in the tank in 1 minute when all the pipes A, B and C are opened simulataneously
= x/20 + x/24 -3
According to the question.
x/20 + x/24 – 3 = x/15.
=> x/20 + x/24 -x/15 = 3
=>(6x+5x-8x)/120 = 3
=> 3x = 3 * 120
=> x = (3 * 120)/3 = 120 drums
8.(a) 10 men = 20 days
so 1 man = 2 boys
so 8 men + 4 boys
= (16 + 4) boys = 20 days
Hence 8 men and 5 boys will make 260 mates in 20 days.
9.(c) Marked price of the fan = Rs. 1400
SP after allowing a discount of 10% = 90% of 1400
= (1400 * 90)/100 = Rs. 1260
Second discount
= Rs. (1260 – 1200) = Rs. 60
Let the second discount be x%
so X% of 1260 = 60
=> x = (60 * 100)/1260 = 100/21 = (4) 16/21%
10.(c) If two equal distances are covered at two unequal speeds of x kmph and y kmph, then average speed
= (2xy/(x+y)) kmph
= ((2 ×12 ×4)/(12+4 )) =96/16 = 6 kmph
1.(b) Let the number of boys and girls in the room be x and y respectively.
According to the question,
x^2 = y^2 + 28
=> x^2 – y^2 = 28 …………………..(i)
and
x = y + 2
=> x – y = 2 ……………………….(ii)
On dividing equation (i) by equation (ii), we have
(x^2- y^2)/(x-y)=28/2
=> ((x+y)(x-y))/(x-y)=14
=> x + y = 14
so Total number of boys and girls = 14
2.(c) Let the numbers be 17x and 17y where x and y are co-prime.
LCM of 17x and 17y = 17 xy
According to the question.
17xy = 714
=> xy = 714/17 = 42 = 6 * 7
=> x = 6 and y = 7
or x = 7 and y = 6
so first number = 17x
= 17 * 6 = 102
Second number = 17y
= 17 * 7 = 119
so Sum of the numbers
= 102 + 119 = 221
3.(c) Time interval between the first period and last (fourth) period = 1:45 pm. – 10 : 30 am.
= 3 hours 15 minutes
= Break after each period
= 5 minutes
so Total break = 15 minutes
so Time period of classes = 3 hours 15 minutes – 15 minutes.
= 3 hours
so Exact duration of each period = 3/4 hour = 45 minutes.
4.(b)
5.(c) Let the work be completed in x days.
According to the question,
(x-5 )/10+(x - 3)/12 + x/15 = 1
=>(6x - 30 + 5x – 15 + 4x)/60 = 1
=> 15x – 45 = 60
=> 15x = 105 => x = 105/15 = 7
Hence, the work will be completed in 7 days.
6.(d) Work done by (A +C) in 2 days = 2 (1/10+1/20)
= 2 ((2 + 1)/20)=6/20=3/10
Remaining work = 1 – 3/10 = 7/10
(B + C)’s 1 day’s work
= 1/15 + 1/20 = (4 + 3)/60 = 7/60
so Time taken by (B + C) to finish 7/10 parts of the work
= 60/7 * 7/10 = 6 days
so Total time = 2 + 6 = 8 days
7.(c) Let the capacity of the tank be x drum.
Quantity of water filled in the tank in 1 minute when all the pipes A, B and C are opened simulataneously
= x/20 + x/24 -3
According to the question.
x/20 + x/24 – 3 = x/15.
=> x/20 + x/24 -x/15 = 3
=>(6x+5x-8x)/120 = 3
=> 3x = 3 * 120
=> x = (3 * 120)/3 = 120 drums
8.(a) 10 men = 20 days
so 1 man = 2 boys
so 8 men + 4 boys
= (16 + 4) boys = 20 days
Hence 8 men and 5 boys will make 260 mates in 20 days.
9.(c) Marked price of the fan = Rs. 1400
SP after allowing a discount of 10% = 90% of 1400
= (1400 * 90)/100 = Rs. 1260
Second discount
= Rs. (1260 – 1200) = Rs. 60
Let the second discount be x%
so X% of 1260 = 60
=> x = (60 * 100)/1260 = 100/21 = (4) 16/21%
10.(c) If two equal distances are covered at two unequal speeds of x kmph and y kmph, then average speed
= (2xy/(x+y)) kmph
= ((2 ×12 ×4)/(12+4 )) =96/16 = 6 kmph
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