-
1
1. In a two digit number, the digit in the units place is more
than twice the digit in ten's place by
1. If the digits in the units place and the ten's place are
interchanged, the difference between the
newly formed number and the original number is less than the
original number by 1, what is the
original number? (JBL AEO 15)
Solution>>>
Let, tens digit = x, so units digit = 2x+1, the number is =
10x+2x+1 & the alternate number = 10(2x+1)+x
According to question,
10(2x+1)+x{10x +(2x+1)} = 10x+(2x+1)-1 Or, 20x+10+x-10x-2x-1 =
12x
Or, 9x+9 = 12x
Or, x = 3
So, the number is = 10x3+2x3+1=37 Ans.
2. A, B & C Started a business each investing tk 20,000/=,
after Five months A withdraw tk
5000, B withdraw tk 4000 and C invest tk 6000 more. At the end
of the year a total profit of tk
69,900/= was recorded. Find the share of each. (JBL AEO 15)
Solution>>>
Let us consider, k=1000tk
then, ratio is, A:B:C=20k*5+15k*7: 20k*5+16k*7:20k*5+26k*7
=205k:212k+282k
=205:212:282
sum of the ratio=205+212+282=699
profit,
A=205*69900/699
=20500
B=212*69900/699
=21200
C=282*69900/699
=28200 Ans.
3. A machine P can print 1 lakh books in 8 hour, Q can same in
10 hour & R can print them in 12
hour. All the machine r started at 9 AM, while machine P in
closed at 11 am and the remaining
two machine complete the work. Approximately at what time the
work will be finished? (JBL
AEO 15)
Solution>>>
(P + Q + R)'s 1 hour's work = (1/8 + 1/10 + 1/12) = 37/120
Work done by P, Q and R in 2 hours = (37 x 2)/120 = 37/60
Remaining work = (1 - 37/60) = 23/60
(Q + R)'s 1 hour's work = (1/10 + 1/12) = 11 /60.
Now, 11/60 work is done by Q and R in 1 hour.
So, 23/60 work will be done by Q and R in (60 /11x 23/60) =
23/11 hours = 2 hours.
So, the work will be finished approximately 2 hours after 11
A.M., i.e., around 1 P.M. Ans.
-
2
4. a, b, c, d, e are 5 consecutive numbers in increasing order,
deleting one of them from the set
decreased the sum of the remaining numbers by 20% of the sum of
5. Which one of the number
is deleted from the set? (BB AD 14)
Solution>>>
Let, the consecutive numbers are,
1, 1+1=2, 1+2=3, 1+3=4 & 1+4=5
So, total = 1+2+3+4+5 = 15
Deleting 1 of the 5 numbers from the set, then decreased 20% of
the sum.
20% of the sum = (15 x 20)/100 = 3
So, the deleted number is the 3rd from the set Ans.
5. Rahim bought 2 varieties of rice costing tk 5 & 6 per kg
each. Then he sold the mixture at
tk7/kg, making profit of 20%. What was the ratio of the mixture?
(BB AD 14)
Solution>>>
Let, Rahim bought xkg rice at tk 5, so his cost = 5x tk & in
same way other variety cost = 6y tk
According to ques.
(5x + 6y) + (5x + 6y)20/100 = 7(x + y)
Or, (5x + 6y)(1 + 1/5) = 7x + 7y
Or, (5x + 6y)6/5 = 7x + 7y
Or, 30x + 36y = 35x + 35y
Or, y = 5x
Or, x/y = 1/5, so the ratio is x:y = 1:5 Ans.
Another Way:
Profit = (7x+7y) (5x+6y)tk = (2x + y)tk According to the
question,
2x + y = (5x+6y)20/100
or, 10x + 5y = 5x + 6y
or, 5x = y
Or, x/y = 1/5.
So, X:Y = 1:5 Ans.
6. A team of 2 men and 5 women completed 1/4th of a job in 3ds.
After that another man joined
them and they all complete the next 1/4th of the job in 2ds. How
many men can complete the
whole job in 4 ds? (BB AD 14, Pubali SO 13)
Solution>>>
Here, in 3ds, 2 men & 5 women do part
Again, in 2ds, 3 men & 5 women do part
or in 1d 3 men & 5 women do 1/8 part
or, in 3d 3 men & 5 women do 3/8 part
so, 1 mans 3 day work = 1/4 3/8 = 1/8 part
-
3
now, 1/8 part work done in 3ds by 1 man
1/8 part work done in 1d by 1x3 men
1(whole) part work done in 1d by 3x8 men
1(whole) part work done in 4ds by 3x8/4 men = 6 men Ans.
7. A trader bought some mangoes for tk150/dozon and equal number
of apples for tk 100/doz. If
he sells all the fruits tk 140/dozon, what will be his
profit/loss in percentage? (RKUB Senior,
2014)
Solution>>>
Here, cost price of 1dozon mango = 150tk & cost price of
1dozon apple = 100tk
So, cost price (1dozon mango + 1dozon apple) = 150 + 100 =
250tk
Cell price (1dozon mango + 1dozon apple) = 140 x 2 = 280tk
Profit = 280 250 = 30tk Percentage of profit = (30 x 100)/250 =
12% Ans.
8. Two partners A & B have 70% and 30% share in a business.
After sometimes, a third partner
C joined by investing tk 10 lakh and thus having 20% share in
the business. What is the
percentage of share of A now in the business? (RKUB Senior,
2014)
Solution>>>
Ratio of A & B, A:B = 70%:30% = 7:3
After joining C, he got 20% share, so then total share of
A&B = 100-20 = 80%
Now As share is = (7 x 80%)/10 = 56% Ans. Another way:
20% share = 10lakh, 100% share = (10 x 100)/20 = 50 lakh
Here, share of A&B = 50 10 = 40lakh As investment = (7 x
40)/10 = 28 lakh So, As percentage after joining C = (28 x 100)/50
= 56% Ans.
9. The average weight of three men A, B and C is 84kg. Another
man D joins the group and the
average becomes 80kg. Another man E, his wt is 3 kg more than
that of D, replaces A. Now
average wt of B, C, D & E becomes 79kg. What is the weight
of A? (RKUB Senior, 2014)
Solution>>>
A+B+C = 84 x 3 = 252kg
A+B+C+D = 80 x 4 = 320kg
So, D = 320 252 = 68kg, E = 68 + 3 = 71kg B+C+D+E = 4 x 79 =
316
Or, B+C+68+71 = 316 or, B+C = 316 (68+71) = 177 So, A = 252 177
= 75 Ans.
10. A rectangle PQRS inscribed in a circle and PQ = 6. If the
area of the rectangular region is 48,
what is the area of the circular region? (RKUB Senior, 2014)
-
4
Solution>>>
Here, area of PQRS = 48, PQ = 6, so, PQ x QR = 48 or, QR = 8
Now, if we connect P to R through a line, a right triangle PQR
is formed.
From Pythagoras theory we found,
PR^2 = PQ^2 + QR^2
Or, PR = (6^2 + 8^2)^1/2 = 10
PR = Diameter of the circle = 10
So, area of the circle = x (10/2)^2 = 25 Ans.
11. If 5 is added to the sum of two digits of a number
consisting of two digits, the sum will be
three times the digits of the tens place. Moreover, if the place
of the digits is interchanged, the
number thus found will be 9 less than the original number. Find
the number. (RB SO, 2013)
Solution>>>
Let, ten's digit = x & unit's digit = y, the number is =
10x+y.
According to the question,
x+y+5 = 3*x
or, y = 2x-5 (1) again, 10y+x = 10x+y-9
or, 9y = 9x-9
or, y = x-1
or, 2x-5 = x-1 [from (1)],
so, x = 4, & y= 4-1 = 3.
The number is = 10*4+3 = 43 (ans)
12. If a^2 - root3a + 1 = 0, what is the value of a^3 + 1/a^3?
(RB SO, 2013)
Solution>>>
a^2 - root3a + 1 = 0
or, a(a+1/a - root3) = 0
or, a+1/a - root3 = 0
or, a + 1/a = root3..............(1)
now, a^3 + 1/a^3
= (a+1/a)^3 - 3a. 1/a (a+1/a)
= (root3)^3 - 3* root3* root3 [from (1)]
= 3root3 - 3root3
= 0. (ans)
13.
3 1200sq.m 50cm ? (SBL SO 14)
-
5
Solution>>>
, = x, = 3x,
, 3x*x = 1200, or, x = 20m & = 60m
= 2(60+20) = 160m =
, , 4A = 160 or, A = 40m & = 40^2 = 1600sq.m
= 50cm = 0.5m, = 0.5x0.5 = 0.25sq.m
= 1600/0.25 = 6400 Ans.
14. &
? (SBL SO 14)
Solution>>>
= =
= / =
= * =
, = =
= ( * )/ = .
. % Ans.
15. 20%
3 2
? (SBL SO 14)
Solution>>>
, = x, = x*20/100 = x/5
3 = x + 3*x/5 = 8x/5
, = y, 2 = y + 2*y/5 = 7y/5
,
=
8x/5 = 7y/5
, x/y = 7/8 Ans.
-
6
16. 13.5% 5 8375 ?
10400 ? (SBL JO 14)
Solution>>>
, P = 100*A/100+rt, A= , r= , t=
, = (100*8375)/100+13.5*5 = 5000
, 10500 = 10400-5000 = 5400
5000 1 = 5000*13.5/100 = 675
675 1
5400 = 5400/675 = 8 Ans.
17.
, ? (SBL JO 14)
Solution>>>
= * = .
= ( + * )*( + * )= * = .
= - = .
= ( * ) = Ans.
18. Mr. Zaman defeated Mr. Younus in a vote where the ratio of
their vote was 4:3, total number
of voters was 581 of which 91 didnt vote. Find out the margin of
defeat.
Solution>>>
Casting of vote = 581-91=490, vote for Zaman = 490*4/7 = 280,
Yonus = 210
So, margin of defeat = 280-210=70 Ans.
19. A mixture of 20kg spirit and water contains 10% water. How
much water is to be added to
increase the water up to 25%?
Solution>>>
Here, in 20kg solution amount of water = 20*10/100 = 2kg &
amount of spirit = 20-2 = 18kg
Now, in new solution, amount of water will be changed but amount
of spirit will be the same.
Percentage of water will be = 25% & percentage of spirit
will be = 100-25 = 75%
So, 75% spirit = 18kg, or, 100% spirit = 18*100/75 = 24kg.
Water should be added = 24-20 = 4kg. Ans.
Same problem: % % ? Ans:
-
7
20. 15 years hence A will be twice of his son, 5 years ago he
was 4th times of his son, what is
their present age?
Solution>>>
Let us consider, present age of father = x & son = y
5 yrs ago fathers age, x-5 = 4(y-5) or, x = 4y-15
Again, after 15yrs fathers age,
x+15 = 2(y+15)
4y-15+15 = 2y+30
Y = 15, so father x = 4*15-15 = 45 Ans.
21. A series has 3 numbers a, ar, ar^2. In the series, the first
term is twice of the second term.
What is the ratio of the sum of the first 2 terms to the sum of
the last 2 terms? (BB AD 12)
Solution:
Let the 3rd term, ar^2 = x, so, 2nd term = 2x, 1st term = 4x
So, a + ar = 4x + 2x = 6x
ar + ar^2 = 2x + x = 3x
ratio of first two terms and last two terms = 6x : 3x = 2:1
Ans.
22.Two alloys A and B are composed of two basic elements. The
ratios of the compositions of
the two basic elements in the two alloy are 5:3 and 1:2. A new
alloy X is formed by mixing the
two alloys A & B in the ratio 4:3. What is the ratio of the
composition of the two basic elements
in alloy X? (BB AD 12)
Solution:
Let the amount of A = 4x and amount of B = 3x in alloy X
amount of A in the new alloy X = (5/8)*4x + (1/3)*3x =
(7x)/2
amount of B in the new alloy X = (3/8)*4x + (2/3)*3x= (7x)/2
so ratio A to B = [(7x)/2 ] / [ 7x/2] = 1/1 = 1:1 Ans.
23. A trade where selling an item was asking for such a price
that would enable him to offer a
10% discount and still make a profit of 20%. If the cost of the
product was tk 50, what was his
asking price? (BB Officer 01)
Solution:
If the cost price was 50tk then @20% profit selling price is =
50 + 50*20/100 = 60tk
@ 10% discount
If the selling price is 90 then asking price is 100
If the selling price is 1 then asking price is 100/90
If the selling price is 60 then asking price is (100*60)/90 =
66.67 Ans.
-
8
24. Mr. X pays 10% tax on all income over 60000tk bt he does not
pay any tax on interest on
postal saving certificate. In 2000 he paid 7500 as tax & he
earned 12000 as interest on postal
savings account. What is his net income in 2000? (BB Officer
01)
solution:
Mr. X gives 10% taxes upon his income, so,
If tax is 10tk then income is 100tk
If tax is 7500tk the income is (100*7500)/10 = 75000tk
His total income in 2000 = 75000 + 60000 + 12000 = 147000tk
So, his net income in 2000 is = 147000 7500 = 139500tk Ans.
25. A candidate answered all 22q in a test & received 63.5
marks. If the total marks were derived
by adding 3.5 marks each correct answer and deducting 1 mark for
each incorrect answer, how
many q did the student answer correctly? (BB Officer 01)
Solution:
If all the answers were right he might got = 22*3.5 = 77
So, his deducted number is = 77 63.5 = 13.5 Number deducted for
each wrong answer = 1+3.5 = 4.5
Quantity of given wrong answer = 13.5/4.5 = 3
So, the candidate answered correctly = 22 3 = 19 q. Ans.
26. in an organization 30% of all employees live over 10miles
away from the place of work &
60% of worker who live who live over 10miles use company
transport. If 40% of employees of
the company use company transport, what percent of the employees
live 10miles or less from
work and use company transport? (BB Officer 01)
Solution:
Let total employees = 100, employees live over 10 miles = 30
So, employees live in 10 mile or less = 100 30 = 70 Total 40% or
40 employees use transport
Now, employees live over 10miles using transport = 60% of 30 =
(60*30)/100 = 18
So, employees live in 10miles or less using transport = 40 18 =
22 Here, total 70 employee live in 10miles or less and among them
22 use company transport
So, percentage of using transport of employee who live in
10miles or less = (22*100)/70 =
31*3/7% Ans.
27. Mr. Reach sold two properties P1 & P2 for tk 50000 each.
He sold property P1 for 20% more
then what he paid for it & sold P2 less than 20% what he
paid for it. What was his total gain or
loss, if any, on the scale of two properties? (BB Officer
01)
Solution:
In case of P1 @ 20% profit
If selling price is 120tk then buying price 100tk
If selling price is 50000tk then cost price = (50000*100)/120 =
500000/12 tk = 125000/3
-
9
In case of P2 @ 20% loss
If selling price is 80tk then buying price 100tk
If selling price is 50000tk then cost price = (50000*100)/80 =
62500tk
Now, total sell price = 50000*2 = 100000tk
Total cost price = 125000/3 + 62500 = 312500/3 = 104166.67tk
So, loss = 104166.67 100000 = 4166.67tk Ans.
28. Mr. X has a investable amount of tk 100000, he will invest
the amount for two years. He has
two options. He can invest at simple interest rate of 12% per
annum, alternatively he can invest
at compound rate of 10% (semi annually). Calculate the earnings
at two option and advice him.
(BB Officer 01)
Solution:
At simple interest, Mr. X can get = (12*100000*2)/100 [formula,
I = rpt/100]
= 24000tk
At 10 compound amount = 100000(1+10/100)^4, [formula, CA =
P(1+r/100)^t](t=4 because of
semi annually)
= 121550.6tk
So, compound interest = 121550.6 100000 = 21550.6tk So, simple
interest is better option for Mr. X Ans.
29. A total of 50 employees work in a branch. Of them 22 have
taken the accounting course, 15
have taken finance, 14 marketing. 9 of them taken exactly 2 of
the courses, 1 of them has taken
all. How many of the 50 employees have taken none of this? (BB
AD 01)
Solution:
Here, 1 of the employee has taken all of the courses
9 of the employee have taken exactly 2 of the courses
Number of employee have taken only Accounting = 22 (9+1) = 12
Numbers of employee have taken only Finance = 15 (9+1) = 5 Numbers
of employee have taken only Marketing = 14 1 = 13
Numbers of total employee have taken 1, 2 or 3 courses =
12+5+13+9+1 = 40
So, employees who have not taken any course = 50 40 = 10
Ans.
30. 3 partners A, B & C start a business. Twice the
investment of A is equal to thrice the capital
of B is 4 times the capital of C. Find the share of each out of
a profit of tk. 297000. (Rupali SO
13)
Solution
let, 2A = 3B = 4B = x.
A = x/2,
B = x/3 &
C = x/4.
-
10
Ratio = A:B:C = x/2 : x/3 : x/4
= 1/2 : 1/3 : 1/4 = 6:4:3.
A's share of profit = 297000*6/13
= tk. 137077.
B's share of profit = 297000*4/13
= tk. 91385.
C's share of profit = 29000*3/13
= tk. 68538. Ans.
31. in a flight of 600 km, an aircraft was slowed down due to
bad weather. Its average speed for
the trip was reduced by 200 km/hr & the time of flight
increased by 30 minutes. Find out the
duration of the flight. (Rupali SO 13)
solution
let, usual speed of the flight is X km/hr.
According to the question,
[600/(x-200)] - 600/x = 30mnts = 1/2hr,
12000/(x^2 - 200x) = 1/2,
x^2 - 200x = 24000,
x^2 - 200x + 1000 = 25000,
(x-100)^2 = (500)^2,
x-100 = 500,
x = 600.
Duration of the flight
= (600/600 hr + 1/2 hr)
= (1+1/2)hr = 3/2 hour Ans.
32. 2 pipes A & B can fill a tank in 36 minutes & 45
minutes respectively. Waste pipe C can
empty the tank in 30 minutes. 1st A & B are opened. After 7
minutes, C is also opened. In how
much time, the tank is full? (Rupali O 13)
solution:
In 1 min,A & B together can fill =(1/36+1/45)=1/20 part.
In 7 min's they can fill=7/20 part.
Remaining =(1-7/20)=13/20 part.
IN 1 min C can empties=1/30 part.
After 7 min's,every 1 min tank fill=(1/20-1/30)=1/60 part.
SO, Tank fill 1/60 part in 1 min.
Tank fill remaining 13/20 part in=60*13/20=39 min's.
so total time require to fill tank=(7+39)=46 minutes. Ans.
33. If 5 is added to the sum of two digits of a number
consisting of two digits, the sum will be
three times the digits of the tens place. Moreover, if the
places of the digits are interchanged, the
number thus found will be 9 less than the original number. Find
the number. (Rupali SO 13)
-
11
Solution
Let, ten's digit = x
& unit's digit = y
The number is = 10x+y.
According to the question,
x+y+5 = 3*x,
y = 2x-5..........(1)
& 10y+x = 10x+y-9,
9y = 9x-9,
y = x-1,
2x-5 = x-1 [from (1)],
x = 4, & y= 4-1 = 3.
The number is = 10*4+3 = 43 Ans.
34. If a^2 - root3a + 1 = 0, what is the value of a^3 + 1/a^3 ?
(Rupali SO 13)
Solution
a^2 - root3a + 1 = 0,
a(a+1/a - root3) = 0,
a+1/a -root3 = 0,
a + 1/a = root3..............(1)
now, a^3 + 1/a^3
= (a+1/a)^3 - 3a. 1/a (a+1/a)
= (root3)^3 - 3* root3* root3 [from (1)]
= 3root3 - 3root3
= 0. Ans.
35. Akti ayotokar ghor er doyrgho prostho opekkha 14 m beshi.
Ghortir porishima 72 m hole
ghortir porishima r soman porishimar kono borgokhetrer khetrofol
koto? (Sonali SO 13)
Solution
Ayotokar ghor er porishima = borgokhetrer porishima = 72 m.
Borgokhetrer bahur doyrgho = 72/4 m = 18 m.
Borgokhetrer khetrofol = (18)^2
= 324 sq.m Ans.
36. Sud-Asol e 5 years e tk. 525 & 8 years e tk. 660 hoi.
Shotkora sud er har & asol koto? (Sonali
SO 13)
Solution
(8-5) = 3 years e sud = (660-525) = tk. 135
1 ,, ,, ,, = tk. 135/3
5 ,, ,, ,, = tk.(135*5)/3 = tk. 225.
Asol = tk. (525-225) = tk. 300.
Tk. 300 er 5 years e sud tk. 225
tk. 1 ,, 1 ,, ,, =tk. 225/300*5
-
12
tk. 100 ,, 1 ,, ,, = tk. (225*100)/1500 = 15%.
Suder har 15% & asol tk. 300 Ans.
37. A simple interest rate of a bank was reduced to 5% from 7%.
As a consequences Mr. Bs income was reduced by tk 2100 in 5 yrs.
How much is Mr. Bs initial deposit in the bank? (BB AD 01)
Solution:
Here, reduced interest rate = 7% - 5% = 2%
In 5 yrs interest reduced = 2100 tk
In 1 yr interest reduced = 2100/5 = 420 tk
Now, if interest reduced 2 tk then deposit is 100 tk
If interest reduced 1 tk then deposit is 100/5 tk
If interest reduced 2100 tk then deposit is (100*2100)/5 tk
= 21000 tk Ans.
38. Mr. A purchased a house for tk 1000000 tk in 1995, he spent
100000 tk for routine
maintenance & upkeep of the house. 1n 1999 he sold the house
for 25% of more then what he
paid for it. He paid 5% of the proceeds as gain tax & he has
to pay 50% of his net profit to the
broker, what is his net income? (BB AD 01)
Solution:
Here, purchasing cost + routine maintenance & upkeep
cost
= (1000000 + 100000) = 1100000 tk
Profit gained @ 25% of total cost = (1100000*25)/100 = 275000
tk
5% gain tax = (5*275000)/100 = 13750 tk
So, net profit after deducting gain tax = (275000 13750) tk =
261250 tk
Net income after deducting brokers commission = 261250/2
= 130625 tk Ans.
39. A trader sells on an average 18 pencils and 12 pens per day.
The profit comes from pencil is
1/3rd of the profit made from selling a pen. If he makes profits
of tk 900 in a month by selling
pencils, how much profit does he make per month by selling pens?
The trader sells 30ds in a
month. (BB AD 01)
Solution:
Let, Profit from 1 pen = x tk so, 12*30 pen = 360x tk.
Profit from 1 pencil = x/3 tk, 18*30 pencil = 540x/3 = 180x
tk.
According to ques. 180x = 900 or, x = 5 tk
So, profit from pen = 360*5 = 1800 tk. Ans.
-
13
40. A can do a piece of work in 10 days, while B alone can do it
in 15 days. They work together
for 5 days and the rest of the work is done by C in 2 days. If
they get Tk.450 for the whole work,
how should they divide the money? (BASIC-2014)
Solution:
A's 1 day work=1/10, his 5 days work = 5/10 = 1/2 of the
work
B's 1 day work =1/15, his 5 days work = 5/15 = 1/3 of the
work
1 day work of (A+B) =1/10+1/15 = 1/6 so, 5 days work = 5/6
Remaining work=1-5/6=1/6
So, C has done 1/6 of the work
So, As money = 450/2 = 225 tk, B = 450/3 = 150 tk, C = 450/6 =
75 tk Ans.
41. A sum of Tk. 1260 is borrowed from a money lender at 10% p.a
compounded annually. If the
amount is to be paid in two equal annual installments, find the
annual installments. (BASIC-
2014)
Solution:
We know, compound amount = P(1+r/100)^t
= 1260(1+10/100)^2
= 1260(1.1) ^2
= 1260*1.21 =1524.6
So, annual installment = 1524.62=762.3tk. Ans.
42. Rahim, Karim & Gazi 3jon 1ti kaj korte pare jothakrome
15, 6 & 10 days e. Tahara akotre
3jon oi kajti koto din e sesh korbe? (SBL O 13)
Solution:
/
/ /
= / + / + / = ( + + )/ = /
, = /
, , = Ans.
43. How much interest will tk 1000 earn in 1 year @ an annual
interest rate of 20% if interest
rate is compounded every 6 months? (BB AD 10, Rupali SO 13)
Solution:
We know, compound amount = P(1+r/100)^t
= 1000 (1+10/100)^2 [t=2 bcz interest rate is compounded
in every 6 months]
= 1000 * (1 + 1/10) ^2
= 1000*(1.1)^2 = 1000*1.21 = 1210 tk.
So, interest = 1210 1000 = 210 tk. Ans.
-
14
44. A train went 300 miles from X to city Y @ an average speed
of 80mph. At what speed did it
travel on the way back if it its average speed for the whole
trip was 100mph?
Solution:
The train, started its journey from X to Y, than came back to X
again
So, total distance covered by it = 300*2 = 600 mile
Here, average speed of total journey is 100 mph, so, total time
spent = 600/100 = 6 hr.
Time req. to go from X to Y at 80mph = 300/80 = 3.75 hrs.
Time req. to comeback from Y to X = 6 3.75 = 2.25 hrs. So, speed
of travelling on way back = 300/2.25 = 133.33 mph Ans.
45. If 6 men and 8 boys can do a piece of work in 10 days while
26 men and 48 boys can do the
same in 2 days, the time taken by 15 men and 20 boys in doing
the same type of work will be?
Solution :
Let 1 man's 1 day's work = x and 1 boy's 1 day's work = y.
Then, 6x + 8y = 1/10
and 26x + 48y = 1/2.
Solving these two equations, we get: x =1/ 100 and y = 1/200
.
(15 men + 20 boy)'s 1 day's work =15/100 + 20/200 = 1/ 4
15 men and 20 boys can do the work in 4 days. Ans.
46. An article is sold at 20% profit. Its cost price is
increased by tk. 50 and at the same time if its
selling price is also increased by tk. 30, the percentage of
profit decreases by 3.33333. find the
cost price? (Pubali SO 13)
Solution: Cost = p
Profit = 0.2p
Selling price = 1.2p
New:
Cost = p+50 [cost increases by 50 tk]
Profit = (20 - 3.33)% of (p+50) = 16.67% of (p+50)= 0.1667p +
8.33
Selling price = 1.2p + 30 [s.p. increases by 30 tk]
According to the question:
Cost + profit = selling price
p+50 +( 0.1667p + 8.33) = 1.2p + 30
0.0333p = 28.33
p = 850 tk Ans.
47. A picnic was attended by 240 persons. There were 20 more men
than women and 20 more
adults than children. How many men were there in the picnic?
Solution:
First condition : Men+Women+Children =240
Second Condition: Men=Women+ 20
-
15
Women= Men-20-----(1)
Third Condition: (Men+Women)=Children+ 20
Men+Women-20=Children
Men+(Men-20)-20=Children [putting the value of eq (1)]
2Men-40=Children....(2)
Putting the value of eq. (1) and (2) in condition 1, we get
,
Men+(Men-20)+(2Men-40)=240
Men= 75 Ans.
48. The ratio of the number of boys and girls in a school is
3:2. If 20% of the boys and 25% of
girls are scholarship holders, what % of the students does not
get scholarship?
Solution:
Let total Student 100
Boys = 100*3/5= 60
Scholarship holder = 60*20%= 12
Girls = 100*2/5=40
Scholarship Holder = 40*25%= 10
Total Scholarship holder = 12+10= 22
Does not get Scholarship: 100-22=78% Ans.
49. A salesman's commission is 5% on all sales up to tk 10000
and 4% on all sales exceeding
this. He remits tk 31100 to his parent company after deducting
his commission. Find the total
sales.
Solution :
let total sales =s
First condition : 10000x5%= 500
Second Condition : (s-10000)x4%=0.04s- 400
According to the question ,
S-(500+0.04S-400)= 31100
S-0.04S-100=31100
0.96S= 31100+100=31200
S=31200/0.96
S=32500 tk Ans.
50. A,B and C do a job alone in 20,30,and 60 days respectively.
In how many days can A do the
job if he is assisted by B & C?(BB AD 13)
Solution:
(A+B+C) work 1 days=1/20+1/30+1/60 = 1/10
With help of B & C A do 1/10 part work in 1 day
,, ,, ,, ,, ,, ,, 1 (whole) part in 10 days Ans
-
16
51. A bus is traveling with 52 passengers. When it arrives at a
stop, Y passengers get off and 4
get I at the next stop one-third of the passengers get off and 3
get on. There are 25 passengers.
Find out how many passengers got off at the first stop? (BB AD
13)
Solution:
According to q,
(52-Y+4) *2/3 + 3 = 25
Or, Y= 23 Ans.
52. An Eskimo leaves its igloo and travels 3 km north, then 8 km
east and finally 3 km north to
reach the north pole. How many km does he have two travels to
return to his igloo in a straight
line? (BB AD-13)
Solution:
Here, 1st go A to B -3 Km
Then go B To C -8Km
Finally, go C to D -3km
A to D = ?
Now using the formula of Pythagoras, we get
AD^2 = AE^2 + ED^2
Or, AD^2 = 8^2 + 6^2
Or, AD=10 Ans.
53. If sugar price reduced 25/4%, then one can buy 1kg more
sugar at 120tk. Find the rate of
original and reduced price. (JBL Cash 15)
Solution:
Let, the original rate = xtk/kg, so in 120 tk sugar can be found
= 120/x kg
Now, 25/4% discount in x tk = 25x/4*100 = x/16 tk
So, discount rate = x - x/16 = 15x/16 tk per kg and in 120 tk we
can found = 120*16/15x kg
sugar
According to question,
120*16/15x - 120/x = 1 Or, 120/15x = 1, or x = 8 tk/kg
So, discount price is = 15*8/16 = 7.5 tk/kg Ans.
54. If 2 men and 3 boys can do a piece of work in 10 days; and
if 3 men and 2 boys can do the
same piece of work in 8 days, then 2 men and 1 boy can do that
work in how many days? (JBL
Cash 15)
Solution:
In I day
2 men and 3 boy do = 1/10 part (1) 3 men and 2 boy do = 1/8 part
.. (2) _____________________________________
-
17
5 men and 5 boy do = 9/40 part
So, 1 man and 1 boy do = 9/200 part in 1 day . (3)
We find from (2) (3) 2 men and 1 boy do = 1/8 9/200 = (25 9)/200
= 2/25 part So, 2 men and 1 boy do 2/25 part work in 1 day
2 men and 1 boy do 1 part or whole work in 25/2 days = 12.5 days
Ans.
Alternate way:
(2M+3B)10=(3M+3B)8
So, 4M=14B
We can say 2 men = 7 boy
2 men and 3 boys or 10 boys can do the work in 10 days
So 1 boys can do the work in 10*10 days
So 2 men and 1 boy or 8 boys can do the work in 10*10/8 days
So 12.5 days Ans.
55. A total amount of 1550 tk was invested in two parts. One
part is 8% rate and the other part is
6% rate. If the annual income is tk 106, then how much money was
invested in each part? (JBL
Cash 15)
Solution:
Let, 8% invest is = x
6% invest is =1550 x The interest on tk x = 8x/100 tk
And interest on tk (1550 x) = 6(1550 x)/100 tk
According to the question,
(1550 - x)*6/100 + 8x/100 = 106
=> (1550-x)*6 + 8x = 106*100
=>9300 - 6x + 8x = 10600
=>2x = 10600 - 9300
=>x = 650
8% invest is = 650
6% invest is = 1550 650 = 900 Ans.
56. A and B together can do a piece of work in 12 days, which B
and C together can do in 16
days. A has been working at it for 5 Days and B for 7 Days, C
finishes in 13 days. In how many
days C alone will do the work?
Solution:
5 days work of A and B=5/12 so B's (7-5) or 2 days remain which
is done with C so 2 days work
of B&C=2/16=1/8 Now C's remain (13-2) or 11 days Now remain
part=1-(5/12+1/8)=11/24
which is done by C in 11 days so whole part needs for
C=11*24/11=24 days
-
18
57. A,B and C enter into partnership by investing in the ratio
of 3:2:4.After one year, B invests
another tk 270000.At the end of three years profits are shared
in the ratio of 3:4:5. Find the initial
investment of each.
Solution:
Say, A,B,C initial investment=3x,2x,4x
ATQ,
3x*3:[2x*1+(2x+270000)2]:4x*3=3:4:5
9x:2x+4x+540000:12x=3:4:5
9x:6x+540000:12x=3:4:5
Now,
9x/6x+540000=3/4
36x=18x+1620000
18x=1620000
x=90000
so A's initial investment=3*90000=270000 tk
B's " " =2*90000=180000 "
C's " " =4*90000=360000 " (ans)
58. A & B can do a work in 12 days where B & C can do
the job in 16 days. A worked for 5
days, B worked for 7 days and C did the remaining job in 13
days. How many days would
require for C to do the job alone? (BCBL 2015)
Solution:
A&B do in 12 days =1 work
" " " 1 day =1/12 "
" " " 5 days =5/12 "
Again B&C do in 16 days= 1 work
" " " 1 day =1/16 "
" " " 2 days=2/16=1/8 work
Work remaining=1-5/12-1/8=11/24 work
11/24 work completed by C in = 11 days
1 " " " C " =11*24/11~24 days (ans)
59. Mr. Nader drove for Mymensingh to Dhaka @ 60 miles/hr.
Returning on the same route
there was a lot of traffic and he was only able to drive @ 40
miles/hr. If the return trip took 1hr
longer, what is the distance between Dhaka & Mymensingh?
(BASIC AM 12)
Solution:
Let the distance between Dhaka & Mymensingh = X km
We know, Speed = Distance/Time or, Time = Distance/Speed
According to question,
x/40 x/60 = 1 or, (60x 40x)/2400 = 1 or, 20x = 2400
or, x = 120 miles Ans.
-
19
60. A businessman before closing his shop, counted the money
kept in the cash box and found
there were X number 50 paisa coin, X number of 1 tk notes, X
number of 2 tk notes and X
number of 5 tk notes. Apart from this there is nothing in the
box. The next day when he opened
the shop he founds that the cash box had been stolen. As he was
insured, he got tk 1615 which is
95% of the stolen money from the insurance company. How many 2
tk notes were in the box?
(BASIC AM 12)
Solution:
Here, 95% of stolen money = 1615 tk
So, 100% of the stolen money = (1615*100)/95 = 1700tk
According to question,
0.5x + 1x + 2x + 5x = 1700
Or, x(0.5 + 1 +2 + 5) = 1700
Or, x = 1700/8.5 = 200 Ans.
61. %
? (BKB CO 12)
Solution:
%, = = %
= + =
, % =
% = ( * )/ = Ans.
62. % %
? (BKB CO 12)
Solution:
= + =
, =
%
=
= ( * )/ =
%
=
-
20
= ( * )/ =
, = + =
= = Ans.
63. A person spends 1/3rd of the money with him on food, 1/5th
of the remaining on education,
1/4th of the remaining on treatment. Now he is left with tk 200.
How much did he have with him
in the beginning? (One Bank Officer 12)
Solution:
Let us consider initial money = x tk
1/3rd of the money is spent for food, so remaining money = x x/3
= 2x/3 1/5th of the remaining is spent for education = 1/5*2x/3 =
2x/15,
Remaining money = 2x/3 2x/15 = 8x/15 1/4th of the remaining
money spent on treatment = * 8x/15 = 2x/15
Remaining money = 8x/15 2x/15 = 2x/5 Acq. to ques. 2x/5 = 200
or, x = (200*5)/2 = 500 tk Ans.
64. Suppose 81p + 62q = 138 and 62p + 81q = 5, find out the
value of p & q. (One Bank Officer
12)
Solution:
Here,
81p + 62q = 138 (1) 62p + 81q = 5. (2) We find from (1)*62 and
2*81
5022p + 3844q = 8556 . (4) 5022p + 6561q = 405 (5) Now, we can
find from (4) (5) - 2717q = 8151 or, q = -3
So, p = 4 Ans.
65. in a business Piku invested tk 6500 for 6 months, Qazi
invested 8400 for 5 months, Raj
invested 10000 for 3 months. Piku wants to be working member for
which he will receive 5% of
profit. If the total profit earned is tk 7400, what is the share
of Qazi in profit? (Jamuna O 12)
Solution:
Ratio of Piku, Qazi & Rajs investment = (6500*6) : (8400*5)
: (10000*3) = 39000 : 42000 : 30000 = 13:14:10
As a working member Piku will receive = 5% of 7400 =
(5*7400)/100 = 370 tk
So, remaining = 7400 370 = 7030 tk So, Qazis share in profit =
(7030*24)/37 = 2660 tk Ans.
66. A bag contains tomatoes that are green or red. The ratio of
green tomatoes in the bag is 4:3.
When 5 green and 5 red tomatoes are removed, ratio becomes 3:2.
How many red tomatoes were
originally in the bag? (Jamuna MTO 14)
-
21
Solution:
Let, Green tomatoes = 4x, Red tomatoes = 3x
According to Q, 4x-5/3x-5=3/2 or, 8x 10 = 9x 15 or, x=5 Red
tomatoes originally in the bag are 15 Ans.
67. Kalim is asked to write a study guide for a text book. For
his work the publishing company is
giving him a choice of a onetime payment of tk 13375 or tk 2000
plus 10% royalty per copy
sold. If the proposed royalty rate of tk 3.25 per copy sold, how
many study guide to be sold for
the total income received by Kalim to be the same from either
choice? (Jamuna MTO 14)
Solution:
Here, 10 % of royalties =3.25tk/copy
Amount of royalties kalim has to receive = 13375-2000
=11375tk
So, copy to be sold = 13375/3.25 = 3500 nos. Ans.
68. In a certain store, the profit was 320% of the cost. If the
cost increases by 25% but the selling
price remains constant, approximately what percentage of the
selling price is the profit now?
(Jamuna O 12)
Solution:
Let, cost price is = 100tk, profit = 320 tk, so selling price =
100+320 = 420tk
Then, new cost price = 125 tk, selling price = 420 tk, so profit
= 420 125 = 295 tk Profit is the percentage of new selling price =
(295*100)/420 = 1475/21 = 70% approximately
Ans.
69. A can do a work in 10 days, B can do it in 15 days. They
work together for 5 days, rest of the
work is done by C in 2 days. If they get tk 4500 for whole work,
how should they divide money?
(Basic Officer 14)
Solution:
Here, A do in 1 day = 1/10, so in 5 days =
Similarly, B don in 5 days = 1/3
(A+B) 5 days work = + 1/3 = 5/6
So, 2 days work of C = 1 5/6 = 1/6 So, share of A = 4500*1/2 =
2250 tk
Share of B = 4500*1/3 = 1500 tk
Share of C = 4500*1/6 = 750 tk Ans.
70. There are 2 examination room A and B. if 10 students are
sent from A to B, then the number
of students in each room is the same. If 20 candidates are sent
B to A, then the number of
students in A is double of the number of the students in B.
Number of the students in A? (NRB O
14)
Solution:
Let us consider, there are x students in the room A and y
students in the room B.
According to the question,
-
22
x 10 = y + 10 or, y = x 20 . (1) Again, x + 20 = 2(y 20) . (2)
Solving for x on (2)
X + 20 = 2(x 40) Or, x + 20 = 2x 80 Or, x = 100
So, there are 100 students in room A Ans.
71. Amin and Sajal are friends. Each has some money. If Amin
gives Tk 30 to Sajal, then Sajal
will have twice the money left with Amin. But, if Sajal gives Tk
10 to Amin, then Amin will
have thrice as much as is left with Sajal. How much money does
each have? (NRB SEO 14)
Solution:
Let, Amin has Tk x Sajal has Tk y Sajals money will be twice of
Amin, if Amin gives Tk 30 to Sajal. So according to the question,
2(x-30)=y+30
2x-60= y+30
2x-y=90.. (i) If Sajal gives taka 10 to Amin, according to the
question,
3(y-10)=x+10
3y-30=x+10
3y-x=40... (ii) -x+3y=40
-2x+6y=80. (iii)
Adding equation (i) & (iii) we can get,
2x-y =90
-2x+6y=80
5y=170
y=34
Substituting the value of y in equation (i), we can get
2x-34=90
2x=124
x=62
Amin has Taka 62 & Sajal has Taka 34. (Ans)
72. A contract is to be completed in 46 days and 117 men were
set to work, each working 8
hours a day. After 33 days, 4/7 of the work is completed. How
many additional men may be
employed so that the work may be completed in time, each man now
working 9 hours a day?
(NRB SEO 14)
Solution:
After 33 days time remains = (46-33) = 13 days
4/7th of the work is completed, so works
remains=1-4/7=(7-4)/7=3/7
-
23
In 33 days 8 hours 4/7th of work is done by 117 men
1 1 1 (117x7x33x8)/4 13 9 3/7 (117x7x33x8x3)/(4x19x9x7)=198 Men
Additional fund required = 198-117 =81 men. (Ans)
73. Mr. Jones gave 40% of the money he had, to his wife. He also
gave 20% of the remaining
amount to each of his three sons. Half of the amount now left
was spent on miscellaneous items
and the remaining amount of Tk. 12000 was deposited in the Bank.
How much money did Mr.
Jones have initially? (NRB SEO 14)
Solution:
Jones gave 40% of money to his wife,
Remaining amount (100-40) = 60%
Each son got 20% of the remaining (60x20%)=(60x20/100)=12%
Three Son get= (12+12+12) =36%
Money spent on miscellaneous items was half of the amount
left.
Amount left=(60-36)=24%
Spent of miscellaneous items=24%/2 =12%
Remaining amount is deposited to bank =24%-12%=12%
Accordingly to the question,
12%=12000
1%=12000/12
100%=(12000x100)/12
=10000
Ans: 10000 Taka.
74. Robi was 4 times as old as his son 8 years ago. After 8
years, Robi will be twice as old as his
son. What are their present ages? (NRB O 14)
Solution:
Robi was 4 times old than his son 8 years ago
Let, 8 years ago sons age was x years So, Robis age was 4x years
At present sons age = (x+8) years
Robis age = (4x+8) years After 8 years sons age will be= x+8+8 =
(x+16) years
8 Robis =4x+8+8 = (4x+16) years
After 8 years Robi will be twice of his son
So, 4x+16=2(x+16)
4x+16= 2x+32
2x=16
X=8
At present sons age = (8+8) = 16 years
-
24
Robis age = (4x8)+8 =40 years Ans: Robis age 40 years, sons age
16 years
75. In an examination, 80%of the students passed in English, 85%
in Mathematics, and 75% in
both English and Mathematics. If 40 students failed in both the
subjects, find the total number of
students? (NRB O 14)
Solution:
Let us consider, total students = 100
Passed in English, E = 80
Passed in Mathematics, M = 85
Passed in both subject, EM = 75
According to Venn theory we find,
Total students = E + M - EM + EuM(Both Failed)
Or,100 = 80 + 85 -75 + Both Failed
Or, Both Failed = 100 90 = 10%
If 10 students failed in both subjects, so total students
=100
1 100/10 40 (100/10)x40 = 400 Ans.
76. A, B and C enter into a partnership by investing in ratio
3:2:4. After one year, B invests
another Tk 270000. At the end of three year profits are shared
in the ratio of 3:4:5. Find the
initial investment of each. (NRB O 14)
Solution:
Let us consider, invest of A, B & C for 1 year is 3x, 2x
& 4x
3 year invest of A = 3x*3 = 9x, B= 3*2x + 270000*2 = 6x +
540000, C = 3x*4 = 12x
Now, according to the question,
6x + 540000/9x = 4/3
Or, 36x = 18x + 1620000
Or, 18x = 1620000
Or, x = 90000
So, investment of A = 3*90000 = 270000 tk
Investment of B = 2*90000 = 180000 tk
Investment of C = 4*90000 = 360000 tk Ans.
77. In a country, 60% of the male citizen and 70% of the female
citizen are eligible to vote. 70%
of the male & 60% of female citizen is eligible to cast
their vote. What fraction of citizens voted
during their election? (JBL EO 12).
Solution:
Let, total number of male = x
Total number of female = y
So, male voted = x*(60/100)*(70/100)
And female voted = y*(70/100)*(60/100)
-
25
So, fraction of the citizen voted = x*(60/100)*(70/100) +
y*(70/100)*(60/100)
x+y
= (x+y)(60*70)/100*100
(x+y)
= 60*70/10000
= 21/50 Ans.
78. A boy purchased some chocolates from a shop for tk 120. In
the next shop he found that the
price of per piece chocolate is tk3 less than that charged @the
previous shop, as such he could
have purchased 2 more chocolates. How many chocolates did he buy
from the first shop? (JBL
EO 12).
Solution:
Let, the price of one piece chocolate in 1st shop = x
So, total chocolate = 120/x
Price of one piece chocolate in 2nd shop = x 3 Total chocolate
120/(x 3) According to the question,
120/(x 3) - 120/x = 2 Or, (120x 120x + 360)/x(x 3) = 2 Or,
360/x^2 3x = 2 Or, 2x^2 6x 360 = 0 Or, x^2 3x 180 = 0 Or, x^2 15x +
12x 180 = 0 Or, (x + 12)(x 15) = 0 So, x = 15
Number of chocolates bought in the first shop = 120/15 = 8 nos.
Ans.
79. A borrower pays 8% interest/yr on the first 600tk he borrows
and 7%/yr on the part of the
loan in excess of tk 600. How much interest will the borrower
pay on a loan of tk 6,000 for one
year? (PJO Cash 12)
Solution:
Out of tk 6000, the first 600 is charged with 8% interest &
the rest amount is charged with 7%
interest.
So, 8% interest for first 600 tk = 600*8/100 = 48 tk
And, 7% interest for next (6000 600) = 5400tk is = 5400*7/100 =
378tk So, total interest for 6000tk for 1 year = 48 + 378 = 426 tk
Ans.
80. Ripon, Liton and Pintu started a business jointly with a
total amount of taka 280. Ripon paid
taka 45 more than Liton & Liton paid tk 70 less than Pintu.
If the company made a profit of tk
56, how much profit should Liton receive? (PJO Cash 12)
Solution:
Let investment of Liton = xtk
-
26
So, investment of Ripon = (x + 45) tk & Pintu = (x +70)
tk
According to the question,
x + (x + 45) + (x +70) = 280
or, 3x + 115 = 280
or, 3x = 165
or, x = 55
so, share of Liton = 45 tk, Ripon = 55 + 45 = 100 tk, Pintu = 55
+ 70 = 125 tk
L:R:P = 55:100:125 = 11:20:25
Profit will be received by Liton = (11/56)56 = 11 tk Ans.
81. A worker is paid taka x per hour for the first 5 hrs he
works each day. He is paid tk y per hour for each hr he works in
excess of 5 hrs. During one week, he works 8 hrs on Saturday,
11
hrs on Sunday, 12 hrs on Monday, 10 hrs on Tuesday, 9 hrs on
Wednesday. What is the average
daily wage for five days of week? (PJO Cash 12)
Solution:
Day wise earning of the worker:
Saturday = 5x + 3y
Sunday = 5x + 6y
Monday = 5x + 7y
Tuesday = 5x + 5y
Wednesday = 5x + 4y
Sum of five days = 5x*5 + y(3 + 6 + 7 + 5 + 4) = 25x + 25y
Average earning = 25(x + y)/5 = 5(x + y) Ans.
82. Reena took a loan of tk 1200 at simple interest for as many
years as the rate of interest. If she
paid tk 432 as interest at the end of the loan period, what was
the rate of interest? (IFIC O 12)
Solution:
We know in case of simple interest, I = rpt/100
Here, interest: I = 432tk, principal: P = 1200tk, t = time, r =
rate of interest
According to question, rate = time or, r = t
So, I = rpt/100
Or, rt = 100*I/p
Or, r^2 = (100*432)/1200 = 36
Or, r = 6
So, rate of interest is 6% Ans.
83. Two students Rahim & Karim, appeared at an examination.
Rahim scored 9 marks more than
Karim did and Rahims mark was 56% of the sum of their marks.
What was Karims score? (IFIC O 12)
Solution:
Let, Karims marks = x So, Rahims mark = x + 9 Their total marks
= x + x + 9 = 2x + 9
-
27
According to question
(2x + 9)56/100 = x + 9
Or, 112x + 504 = 100x + 900
Or, 112x 100x = 900 504 Or, 12x = 396
Or, x = 33 Ans.
84. Abu started a business investing tk 70,000. Robu joined him
after 6 months with an amount
of tk 1,05,000 and Sabu joined them with tk 1,40,000 after
another 6 months. In what ratio the
amount of profit earned should be distributed among them after 3
years? (DBBL PO 12)
Solution:
Here, ratio of investment of Abu, Robu & Sabu is
(70,000*36):(1,05,000*30):(1,40,000*24)
= 25,20,000:31,50,000:33,60,000
= 252:315:336
= 12:15:16
So, the profit earned by Abu, Robu & Sabu should be
distributed to the following proportion
12:15:16 Ans.
85. A trader mixes 26kg of rice priced at tk20/kg with 30kg rice
of another variety priced at
tk36/kg & sells the mixture at tk30/kg. What is the
percentage of profit or loss? (DBBL PO 12)
Solution:
Here, profit = {30*(26 + 30)} (26*20 + 30*36) = 1680 1600 = 80
tk
So, percentage of profit = (80*100)/1600 = 5% Ans.
86. If zakir loses 8 pounds, he will weight twice as much his
sister. Together they are now
weighs 278 pounds. What is zakirs present weight? (SEBL PO
12)
Solution:
Let, zakirs wt. = x & his sisters wt. = y According to
question, x + y = 278 .. (1) Again, x 8 = 2y or, x = 2y + 8 . (2)
So, we can get from equation (1) by solving (2)
2y + 8 + y = 278
Or, 3y = 278 8 Or, y = 270/3 = 90
So, zakirs wt. = 278 90 = 188lb Ans.
87. A seller incurs a loss of 15% when a table is sold @
tk10,200. At what price the table should
be sold to make a profit of 35% (STBL AO 12)
-
28
Solution:
At 15% loss,
If tk85 is the sold price then the buying price is tk100
tk1 ,, ,, ,, ,, ,, ,, ,, ,, ,, 100/85
10200 ,, ,, ,, ,, ,, ,, ,, ,, ,, 100*10200/85 = 12,000tk
Now, at 35% profit,
If tk100 is the buying price then the sold price is tk135
tk1 ,, ,, ,, ,, ,, ,, ,, ,, ,, 135/100
12,000 ,, ,, ,, ,, ,, ,, ,, ,, ,, 135*12,000/100 = 16,200tk
Ans.
88. You want to make a garden in front of your house. The length
of the rectangular region is
greater than its breadth by 20 meters. If the perimeter of the
land is 200m and gardening cost is
tk20 for each square meter of land, how much will be the total
gardening cost? (STBL AO 12)
Solution:
Let, the breadth = x meter & length = x + 20
According to question, the perimeter
2(x + 20 + x) = 200
Or, 4x = 200 40 Or, x = 160/40 = 40
So, breadth = 40m & length = 40 + 20 = 60m
Area of the garden = 40*60 = 2400sqm
So, money required to make the garden = 2400*20 = 48000 tk
Ans.
89. A basketball team has won 15 games and lost 9. If these
games represent 50/3% of the total
games to be played, then how many more games must the team win
to average 75% for the
season? (Premier MTO 12)
Solution:
Here, games already played = 15 + 9 = 24
These are the 50/3% of the total games have to be played
So, total games = (24*3*100)/50 = 144
And 75% of 144 = (144*75)/100 = 108
So, the team has to win more = 108 15 = 93 games Ans.
90. A person earns yearly interest of tk 920 by investing tk x
at 4% & tk y at 5% simple interest.
If he invest tk x at 5% & tk y at 4% his yearly interest
earning will be reduced by tk 40. Find out
his investment. (Premier MTO 12)
Solution:
The reduced interest due to rate change = 920 40 = 880 According
to 1st condition,
(x*4%*1)+(y*5%*1) = 920
Or, 0.04x+0.05y=920 (1) Again, 2nd condition,
-
29
(X*5%*1)+(y*4%*1) = 880
Or, 0.05x+0.04y = 880 (2) By solving this two equation with
(1)*50 - (2)*40 we find
2x + 2.5y = 46,000 .. (3) 2x + 1.6y = 35,200 .. (4) 2.5y 1.6y =
10,800 Or, 0.9y = 10,800
Or, y = 12,000
We can put this figure of y in equation (3) 2x + 2.5*12000 =
46000
Or, 2x = 46000 30000 = 16000 Or, x = 16000/2 = 8000
So, his investment at 4% rate is 8000tk and 12000tk at 5% rate
of interest Ans.