Interview Master In India: ARITHMATICAL

APTITUDE | ARITHMATICAL | TRUE DISCOUNT | PROBLEMS BASED ON DISCOUNTS 3

DIRECTIONS : Problems based on discounts .

11.	The simple interest and the true discount on a certain sum for a given time and at a given rate are Rs. 85 and Rs. 80 respectively. The sum is
	A. Rs.1800	B. Rs.1450
	C. Rs.1360	D. Rs.6800

Solution

Sum	=S.I x T.D/(S.I)-(T.D)
	= (85x80/85-80)
	= Rs.1360.

12.	The true discount on a bill due 9 months hence at 12% per annum is Rs.540. Find the amount of the bill and its present worth.
	A. Rs.2500	B. Rs. 5000
	C. Rs. 5500	D. Rs. 6000

Solution

Let amount be Rs.x then,
x×R×T/100+(R×T)	=T.D
	= x×12×3/4/100+(12×3/4)
	= 540
	x=(540×109/9)
	= Rs. 6540
Amount	= Rs. 6540.
P.W	= Rs.(6540 - 540)
P.W	= Rs. 6000.

APTITUDE | ARITHMATICAL | TRUE DISCOUNT | PROBLEMS BASED ON DISCOUNTS 2

DIRECTIONS : Problems based on discounts .

6.	The present worth of Rs. 2310 due 2 1/2 years hence, the rate of interest being 15% per annum, is:
	A. Rs.1680	B. Rs.1750
	C. Rs.1840	D. Rs.1443.75

Solution

P.W.= Rs			=[100 x 2310/100 +(15 x5/2)]
P.W.= Rs			=Rs. 1680.

7.	The true discount on Rs. 1760 due after a certain time at 12% per annum is Rs. 160. The time after which it is due iss
	A. 6 months	B. 8 months
	C. 9 months	D. 10 months

Solution

S.I. on Rs. 1600 at 12% is Rs. 160.
P.W	= Rs. (1760 -160)
P.W	= Rs.1600
Time	= (100 x 160/1600 x 12)
	=5/6 years.
	=(5/6 x12) months
	=10 months.

8.	Find the present worth of Rs. 930 due 3 years hence at 8% per annum. Also find the discount?
	A. Rs.500	B. Rs.625
	C. Rs. 750	D. Rs. 890

Solution

P.W				=100 x Amount/100+(RxT)
				= Rs[100x930/100+(8x3)]
				= Rs.(100x930/124)
				= Rs.750.

9.	A man purchased a cow for Rs. 3000 and sold it the same day for Rs. 3600, allowing the buyer a credit of 2 years. If the rate of interest be 10% per annum, then the man has a gain of
	A. 0%	B. 5%
	C. 7.5P%	D. 10%

Solution

C.P			= Rs. 3000.
S.P			=Rs[3600x100/100+(10x2)]
S.P			=3000
So, Gain is 0 %.

10.	If the true discount on s sum due 2 years hence at 14% per annum be Rs. 168, the sum due is
	A. Rs.768	B. Rs.968
	C. Rs.1960	D. Rs.2400

Solution

P.W	=100 x T.D/R x T
	= 100 x 168/14 x 2
	= 600.
Sum	= (P.W + T.D)
	Rs. (600 + 168)
	= Rs. 768.

APTITUDE | ARITHMATICAL | TRUE DISCOUNT | PROBLEMS BASED ON DISCOUNTS 1

DIRECTIONS : Problems based on discounts .

1.	The true discount on a certain sum of money due 3 years hence is Rs.250 and the simple interest on the same sum for the time and at the same rate is Rs.375. Find the sum ?
	A. Rs. 500	B. Rs.750
	C. Rs. 1000	D. Rs. 1250

Solution

T.D	= Rs.250
S.I	= Rs.375
Sum due	= S.I×T.D/(S.I)-(T.D)
	= Rs.(375×250/375-250 )
	= Rs.750.

2.	A man buys a watch for Rs.1950 in cash and sells it for Rs.2200 at a credit of 1 year. If the rate of interests is 10% per annum, the man
	A. gain Rs.55	B. gain Rs.30
	C. loses Rs.30	D. gains Rs.50

Solution

S.P			=P.W. of Rs.2200 due 1 year
			= Rs.[2200×100/100+(10×1)]
			= Rs.2000.
Gain			=Rs.(2000 - 1950)
Gain			= Rs.50.

3.	Rs. 20 is the true discount on Rs. 260 due after a certain time. What will be the true discount on the same sum due after half of the former time, the rate of interest being the same?
	A. Rs. 10	B. Rs. 10.40
	C. Rs. 15.20	D. Rs. 13

Solution

S.I.on Rs.(260-20) for a given time	= Rs. 20
S.I. on Rs. 240 for half the time	= Rs. 10.
T.D. on Rs. 250	= Rs. 10.
T.D. on Rs. 260	= Rs.(10/250×260)
T.D. on Rs. 260	= Rs. 10.40

4.	A trader owens a merchant Rs.10,028 due 1` year hence. The trader wants to settle the account after 3 months.If the rate of interest is 12% per annum , how much cash should he pay?
	A. Rs.9025.20	B. Rs.9200
	C. Rs.9600	D. Rs.9560

Solution

Required money			P.W of Rs.10028 due 9 months
			Rs.[10028x100/100+(12x9/12)]
			= Rs.9200.

5.	The true discount on Rs.2562 due 4 months hence is Rs.122. The rate percent is
	A. 12 %	B. 13 %
	C. 15 %	D. 14 %

Solution

S.I on Rs.2440 for 4 months is Rs.12
P.W	= Rs.(2562 - 122)
P.W	= Rs. 2440.
Rate	= (100 x 122/2440x1/3)%
Rate	= 15%.

APTITUDE | ARITHMATICAL | TRUE DISCOUNT | IMPORTANT FACTS AND FORMULAE

Important facts and Formulae

	Suppose a man has to pay Rs. 156 after 4 years and the rate of interest is 14% per annum. Clearly, Rs. 100 at 14% will amount to R. 156 in 4 years. So, the payment of Rs. now will clear off the debt of Rs. 156 due 4 years hence. We say that : Sum due = Rs. 156 due 4 years hence; Present Worth (P.W.) = Rs. 100; True Discount (T.D.) = Rs. (156 - 100) = Rs. 56 = (Sum due) - (P.W.) We define :T.D. = Interest on P.W.;Amount = (P.W.) + (T.D.)

	Let rate = R% per annum and Time = T years. Then,

1.	P.W. = 100 x Amount / 100 + (R x T) = 100 x T.D / R x T

2.	T.D = (P.W.) x R x T / 100 = Amount x R x T / 100 + (R x T)

3.	Sum = (S.I.) x (T.D.) / (S.I.) - (T.D.)

4.	(S.I.) - (T.D.) = S.I. on T.D.

5.	When the sum is put at compound interest, then P.W. = Amount / (1 + R /100)^T

APTITUDE | ARITHMATICAL | TIME AND WORK | PROBLEMS BASED ON TIME AND WORK 5

DIRECTIONS : Problems based on time and work .

21.	A, B and C are employed to do a piece of work for Rs.529. A and B together are supposed to do 19/23 of the work and Band C together 8/23 of the work. What amount should A be paid?
	A. Rs.315	B. Rs.345
	C. Rs.355	D. Rs.375

Solution

Work done by A	=(1 - 8/23)
Work done by A	= 15/23
A : (B+C)	=(15/23 : 8/23)
A : (B+C)	= 15 : 8
So, A's share=Rs.(15/23 x 529)	= Rs.345.

APTITUDE | ARITHMATICAL | TIME AND WORK | PROBLEMS BASED ON TIME AND WORK 4

DIRECTIONS : Problems based on time and work .

16.	Worker A takes 8 hours to do a job. Worker B takes 10 hours to do the same job. How long should it take both A and B, working together but indepently, to do the same job?
	A. 2x4/9	B. 4x4/9
	C. 5x4/9	D. 4x2/9

Solution

A's 1 day's work		= 1/8
B's 1 day's work		= 1/10
(A+B)'s 1 day's work		= (1/8 + 1/10)
		= 9/40.
Both A and B will finish the work in		= 40 / 9
		‹=›4x4/9

17.	A and B together can complete a work in 12 days. A alone can complete it in 20 days. If B does the work only for half a day daily, then in how many days A and B together will complete the work?
	A. 10 days	B. 11 days
	C. 15 days	D. 20 days

Solution

B's 1 day's work	= (1/12 - 1/20)
	= 2/60
	= 1/30
Now, (A+B)'s 1 day's work	(1/12 +1/60)
	= 4/60
	= 1/15.
So A and B together will complete the work in 15 days.

18.	A can do a work in 15 days and B in 20 days. If they work on it together for 4 days, then the fraction of the work that is left is
	A. 1/4	B. 1/10
	C. 7/15	D. 8/15

Solution

A's 1 day's work		= 1/15
B's 1 day's work		= 1/20
(A+B)'s 1 day's work		= (1/15 + 1/20)
		= 7/60.
(A+B)'s 4 day's work		= (7/60x4)
		= 7/15.
Remaining Work		= (1-7/15)
		= 8/15.

19.	A and B together can complete a piece of work in 4 days. If A alone can complete the same work in 12 days. in how many days can b alone complete that work?
	A. 1/3 days	B. 1/6 days
	C. 1/9 days	D. 1/12 days

Solution

Hence, B alone can complete the work in 6 days.
(A+B)'s 1 day's work		= 1/4
A's 1 day's work		= 1/12
B's 1 day's work		= (1/4 - 1/12)
B's 1 day's work		= 1/6.

20.	A and B together can do a piece of work in 30 days.A having worked for 16 days, B finishes the remaining work in 44 days. In how many days shall B finish the whole work alone?
	A. 30 days	B. 40 days
	C. 60 days	D. 70 days

Solution

Therefore B's 1 day's work= 1/60.
A's 1 day's work	= x
B's 1 day's work	= y
Then, x+y = 1/30 and 16x+44y = 1.
Solving these two equations, we get		x=1/10 and y=1/60

APTITUDE | ARITHMATICAL | TIME AND WORK | PROBLEMS BASED ON TIME AND WORK 3

DIRECTIONS : Problems based on time and work .

11.	Ronald and Elan are working on an assignment. Ronald takes 6 hours to type 32 pages on a computer, while Elan takes 5 hours to type 40 pages. How much time will they take, working together on two different computers to type an assignment of 110 pages?
	A. 7 hours 30 minutes	B. 8 hours
	C. 8 hours 15 minutes	D. 8 hours 25 minutes

Solution

Number of pages typed by ronald in 1 hour	= 32/6
Number of pages typed by ronald in 1 hour	= 16/3.
Number of pages typed by elan in 1 hour	= 40/5
Number of pages typed by elan in 1 hour	= 8.
Number of pages typed by both in 1 hour	= (16/3+8)
Number of pages typed by both in 1 hour	=40/3.
Time taken by both to type 110 pages	= (110×3/40)hrs
	= 8×1/4hrs
	= 8 hrs 15 min.

12.	A can finish a work in 18 days and B can do the same in half the time taken by A. Then, working together , what part of the same work they can finish in a day?
	A. 1/6	B. 1/9
	C. 2/5	D. 2/7

Solution

A's 1 day's work	= 1/18
and B's 1 day's work	= 1/9
(A+B)'s 1 day's work	= (1/18+1/9)
(A+B)'s 1 day's work	= 1/6

13.	P can complete a work in 12 days working 8 hours a day,Q can complete the same work in 8 days working 10 hours a day. If both P and Q work together, working 8 hours a day, in how many days can they complete the work?
	A. 5x5/11	B. 5x6/11
	C. 6x5/11	D. 6x6/11

Solution

P can complete the work in (12x 8) hrs		= 96 hrs.
Q can complete the work in (8x 10) hrs		= 80 hrs.
P's 1 hour's work	= 1/96
Q's 1 hrs work	= 1/80
(P+Q)'s 1 hour's work	= (1/96+1/80)
(P+Q)'s 1 hour's work	= 11/480.
So, both P and Q will finish the work in		=(480/11) hrs.
Number of days of 8 hours each	=(480/11x1/8) hrs.
	= 60/11 days
	=5x5/11 days.

14.	A,B, and C can complete a piece of work in 24,6 and 12 days respectively. Working together , they will complete the same work in
	A. 1/24 day	B. 7/24 day
	C. 3x3/7 days	D. 4 days

Solution

(A+B+C)'s 1 day's work	= (1/24+1/6+1/12)
(A+B+C)'s 1 day's work	= 7/24.
So, A,B and C together will complete the job in 24/7		= 3x3/7 days.

15.	A can lay railway track between two given stations in 16 days and B can do the same job in 12 days. With the help of C, they did the job in 4 days only. Then, C alone can do the job in
	A. 9x1/5 days	B. 9x2/5 days
	C. 9x3/5 days	D. 10 days

Solution

(A+B+C)'s 1 day's work		= 1/4
A's 1 day's work	= 1/16
B's 1 day's work	= 1/12
C's 1 day's work			= 1/4-(1/16+1/12)
			= (1/4 - 7/48)
			= 5/48.
So,C alone can do the work in 48/5			=9x3/5 days.