# Accenture Previous Years Question Bank 10

### Set A

1.If the cost price is 25% of selling price. Then what is the profit percent.
a.  150%
b.  200%
c.  300%
d.  350%
2.A book was sold for Rs 27.50 with a profit of 10%. If it were sold for Rs. 25.75, then would have been percentage of profit and loss ?
a.  2% Profit
b.  3% Profit
c.  2% Loss
d.  3% Loss
3.In terms of percentage profit, which among following the best transaction.
a.  C.P. 36, Profit 17
b.  C.P. 50, Profit 24
c.  C.P. 40, Profit 19
d.  C.P. 60, Profit 29
4.A pair of articles was bought for Rs. 37.40 at a discount of 15%. What must be the marked price of each of the articles ?
a.  Rs15
b.  Rs 20
c.  Rs 22
d.  Rs 25
5.A shopkeeper fixes the marked price of an item 35% above its cost price. The percentage of discount allowed to gain 8% is
a.  18%
b.  20%
c.  22%
d.  24%
6.A fruit seller sells mangoes at the rate of Rs.9 per kg and thereby loses 20%. At what price per kg, he should have sold them to make a profit of 5%
a.  Rs 8.81
b.  Rs 9.81
c.  Rs 10.81
d.  Rs 11.81
7.A producer of tea blends two varieties of tea from two tea gardens one costing Rs 18 per kg and another Rs 20 per kg in the ratio 5 : 3. If he sells the blended variety at Rs 21 per kg, then his gain percent is
a.  12%
b.  13%
c.  14%
d.  15%
8.A shopkeeper sold an article for Rs 2564.36. Approximately what was his profit percent if the cost price of the article was Rs 2400
a.  4%
b.  5%
c.  6%
d.  7%
9.A man gains 20% by selling an article for a certain price. If he sells it at double the price, the percentage of profit will be.
a.  130%
b.  140%
c.  150%
d.  160%
10.If the cost price of 12 pens is equal to the selling price of 8 pens, the gain percent is ?
a.  12%
b.  30%
c.  50%
d.  60%
1- (C)(Explanation:Let the S.P = 100 then C.P. = 25 , Profit = 75,  Profit% = 75/25 * 100 = 3005.
2-(B)
3-(D)
4-(C)(Explanation:As question states that rate was of pair of articles,So rate of One article = 37.40/2 = Rs. 18.70
Let Marked price = Rs X then 85% of X = 18.70 => X = 1870/85 = 22
5-(B)(Explanation:Let the cost price = Rs 100 then, Marked price = Rs 135 Required gain = 8%,
So Selling price = Rs 108 Discount = 135 – 108 = 27, Discount% = (27/135)*100 = 20%
6-(D)(Explanation: 85 : 9 = 105 : x, x= (9?105/85) = Rs 11.81
7-(A)(Explanation: Suppose he bought 5 kg and 3 kg of tea,  Cost Price = Rs. (5 x 18 + 3 x 20) = Rs. 150.
Selling price = Rs. (8 x 21) = Rs. 168. Profit = 168 – 150 = 18, So, Profit % = (18/150) * 100 = 12%
8-(D)(Explanation: Gain % = (164.36*100/2400) = 6.84 % = 7% approx
9-(B)(Explanation:Let the C.P. = x, Then S.P. = (120/100)x = 6x/5  New S.P. = 2(6x/5) = 12x/5
Profit = 12x/5 – x = 7x/5  Profit% = (Profit/C.P.) * 100 => (7x/5) * (1/x) * 100 = 140 %
10-(C)(Explanation:Friends, we know we will need gain amount to get gain percent, right. So lets get gain first.
Let the cost price of 1 pen is Re 1  Cost of 8 pens = Rs 8 Selling price of 8 pens = 12
Gain = 12 – 8 = 4 Gain%=(GainCost*100)%=(48*100)%=50%

### Set B

1.    A man saves Rs 200 at the end of each year and lends the money at 5% compound interest. How much will it become at the end of 3 years.
(A) 661.01
(B) 662.01
(C) 663.01
(D) 664.01

2.    A rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered. If the area of the field is 680 sq.ft, how many feet of fencing will be required?
(A) 90 feet
(B) 85 feet
(C) 88 feet
(D) 84 feet

3.    The difference between simple interest and compound interest on Rs. 1200 for one year at 10% per annum reckoned half – yearly is :
(A) Rs. 2
(B) Rs. 3
(C) Rs. 4
(D) Rs. 5

4.   The Diagonals of two squares are in the ratio of 2:5. find the ratio of their areas.
(A) 5 : 2
(B) 5 : 2
(C) 4 : 25
(D) 5 : 4

5.    A starts business with Rs. 3500 and after 5 months, B joins with A as his partner. After a year, the profit is divided in the ratio 2:3. What is B’s contribution in the capital
(A) 1000
(B) 10000
(C) 12000
(D) 9000

6.   In a set of three numbers, the average of first two numbers is 2, the average of the last two numbers is 3, and the average of the first and the last numbers is 4. What is the average of three numbers?
(A) 2
(B) 2.5
(C) 3
(D) 3.5
7.   A vendor sells 60 percent of apples he had and throws away 15 percent of the remainder. Next day he sells 50 percent of the remainder and throws away the rest. What percent of his apples does the vendor throw?
(A) 17
(B) 23
(C) 77
(D) None of these
8.    When processing flower-nectar into honeybees’ extract, a considerable amount of water gets reduced. How much flower-nectar must be processed to yield 1kg of honey, if nectar contains 50% water, and the honey obtained from this nectar contains 15% water?
(A) 1.5 kgs
(B) 1.7 kgs
(C) 3.33 kgs
(D) None of these
9.   Gauri went to the stationers and bought things worth Rs. 25, out of which 30 paise went on sales tax on taxable purchases. If the tax rate was 6%, then what was the cost of the tax free items?
(A) Rs. 15
(B) Rs. 15.70
(C) Rs. 19.70
(D) Rs. 20
10.  If the price of petrol increases by 25%, by how much must a user cut down his consumption so that his expenditure on petrol remains constant?
(A) 25%
(B) 16.67%
(C) 20%
(D) 33.33%
1.    (B)
2.    (C) :
We are given with length and area, so we can find the breadth.
as Length x Breadth = Area
=> 20 x Breadth = 680
Area to be fenced = 2B + L = 2 x 34 + 20
= 88 feet
3.    (B) :
S. I = Rs. ((1200 ×10 ×1)/100) = Rs. 120
(C) I. = Rs.   = Rs. 123.
Difference = Rs. 3
4.    (C) :
Let the diagonals of the squares be 2x and 5x.
Then ratio of their areas will be
Area of square=(1/2) x Diagonal2
(1/2 ) x 2x^2 : (1/2) x 5x^2
4x^2 :25x^2 = 4:25
5.    (D) :
Let B contribution is x.
(3500 x 12) /7x=2/3 =>14x=126000=>x=Rs9000
6.    (C)
7.    (B)
8.    (B)
Flower-nectar contains 50% of non-water part.
In honey this non-water part constitutes 85% (100-15).
Therefore 0.5 X Amount of flower-nectar = 0.85 X Amount of honey = 0.85 X 1 kg
Therefore amount of flower-nectar needed = .85/.5*1= 1.7 kgs
9.    .(C)
Let the amount taxable purchases be Rs. x.
Then, 6% of x=30/100
x=30/100*100/6 = 5
Cost of tax free items = Rs. [25 – (5 + 0.30)] = Rs. 19.70
10. (C)

### Set C

1. In what ratio must tea worth Rs. 60 per kg be mixed with tea worth Rs. 65 a Kg such that by selling the mixture at Rs. 68.20 a Kg ,there can be a gain 10%?
A. 3 : 2
B. 2 : 3
C. 4 : 3
D. 3 : 4
2. A container contains a mixture of two liquids P and Q in the ratio 7 : 5. When 9 litres of mixture are drawn off and the container is filled with Q, the ratio of P and Q becomes 7 : 9. How many litres of liquid P was contained in the container initially?
A. 23
B. 21
C. 19
D. 17
3. A vessel is filled with liquid, 3 parts of which are water and 5 parts of syrup. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?
A. 1/3
B. 1/4
C. 1/5
D. 1/6
4. In what ratio must water be mixed with milk costing Rs.12 per litre in order to get a mixture worth of Rs.8 per litre?
A. 1 : 3
B. 2 : 2
C. 1 : 2
D. 3 : 1
5. In what ratio must tea at Rs.62 per Kg be mixed with tea at Rs. 72 per Kg so that the mixture must be worth Rs. 64.50 per Kg?
A. 1 : 2
B. 2 : 1
C. 3 : 1
D. 1 : 3
6. In what ratio must a grocer mix two varieties of pulses costing Rs.15 and Rs. 20 per kg respectively to obtain a mixture worth Rs.16.50 per Kg?
A. 1 : 2
B. 2 : 1
C. 3 : 7
D. 7 : 3
7. A merchant has 1000 kg of sugar part of which he sells at 8% profit and the rest at 18% profit. He gains 14% on the whole. The Quantity sold at 18% profit is
A. 300
B. 400
C. 600
D. 500
8. A dishonest milkman sells his milk at cost price but he mixes it with water and thereby gains 25%. What is the percentage of water in the mixture?
A. 25%
B. 20%
C. 22%
D. 24%
9. In what ratio must water be mixed with milk to gain 50/3% on selling the mixture at cost price?
A. 6 : 1
B. 1 : 6
C. 1 : 4
D. 4 : 1
10. In what ratio must rice at Rs.7.10 be mixed with rice at Rs.9.20 so that the mixture may be worth Rs.8 per Kg?
A. 5 : 4
B. 2 : 1
C. 3 : 2
D. 4 : 3
1.( A)Explanation :Cost Price(CP) of 1 Kg mixture = Rs. 68.20
Profit = 10%
Cost Price(CP) of 1 Kg mixture= (100/(100+Profit%))×SP
=100(100+10)×68.20
=100/110×68.20=682/11
= Rs.62
By the rule of alligation:
CP of 1 kg tea of 1st kind CP of 1 kg tea of 2nd kind
60                                           65
Mean Price
62
65 – 62 = 3                                           62 – 60 = 2
Hence required ratio = 3 : 2
2.( B)Explanation :Let’s initial quantity of P in the container be 7x
and initial quantity of Q in the container be 5x
Now 9 litres of mixture are drawn off from the container Quantity of P in 9 litres of the mixtures drawn off = 9×7/12=63/12 =21/4
Quantity of Q in 9 litres of the mixtures drawn off  =9×5/12=45/12=15/4
Hence Quantity of P remains in the mixtures after 9 litres is drawn off =7-21/4
Quantity of Q remains in the mixtures after 9 litres is drawn off =5x-15/4
Since the container is filled with Q after 9 litres of mixture is drawn off,Quantity of Q in the mixtures =5x-15/4+9=5x+21/4
Given that the ratio of P and Q becomes  7 : 9
(7x-21/4) : (5x+21/4)=7:9
(7x-21/4)/(5x+21/4)=7/9
63x-(9×21/4)=35x+(7×21/4)
28x=(16×21/4)
x=(16×21)/(4×28)
litres of P contained in the container initially = 7x=(7×16×21)/(4×28)
=(16×21)/(4×4) =21
3.( C)Explanation :Let the quantity of the initial liquid in the vessel = 8 litre
and quantity of water in the initial liquid = 3 litre
andquantity of syrup in the initial liquid = 5 litre
Let x litre of the mixture is drawn off and replaced with waterQuantity of water in the new mixture = 3-3x/8+x
Quantity of syrup in the new mixture = 5-5x/8
Given that in the new mixture, quantity of water = quantity of syrup 3-3x/8+x = 5-5x/8
10x/8=2
5x/4=2
x=8/5 :8/5litre
Initially we assumed that the quantity of the initial liquid in the vessel = 8 litre for
the ease of calculations. For that 8/5 litre of the mixture to be drawn off and replaced with water so that the mixture may be half water and half syrup
Now, if the initial liquid in the vessel = 1 litre, quantity of the mixture to be drawn off and replaced with water so that the mixture may be half water and half syrup
=8/5×1/8=1/5
It means 1/5 of the mixture has to be drawn off and replaced with water so that the mixture may be half water and half syrup
4.(C)Explanation :By the rule of alligation, we have
Cost Price of 1 litre of water Cost Price of 1 litre of milk
0                                         12
Mean Price
8
12-8=4                                                        8-0=8
Required Ratio = 4 : 8 = 1 : 2
5.(C)Explanation :By the rule of alligation, we have
Cost of 1 kg of 1st kind tea Cost of 1 kg tea of 2nd kind tea
62                                          72
Mean Price
64.5
72-64.5=7.5                                       64.5-62=2.5
Required Ratio = 7.5 : 2.5 = 3 : 1
6.(D)Explanation :By the rule of alligation,we have
CP of 1 kg of 1st variety pulse CP of 1 kg of 2nd variety pulse
15                                             20
Mean Price
16.5
20-16.5 = 3.5                                      16.5-15=1.5
Required Ratio = 3.5 : 1.5 = 35 : 15 = 7 : 3
7.(C)Explanation :By the rule of alligation,we have
Profit% by selling 1st part Profit% by selling 2nd part
8                                         18
Net % profit
14
18-14=4                                                  14-8=6
=> Quantity of part1 : Quantity of part2 = 4 : 6 = 2 : 3
Total quantity is given as 1000Kg
So Quantity of part2 (Quantity sold at 18% profit) = 1000×3/5=600Kg
8.( B)Explanation :If a trader professes to sell his goods at cost price, but uses false weights, then Gain%=[Error/(True Value-Error)×100]%
Here Gain= 25%
Here error = quantity of water he mixes in the milk = x
Here the true value = true quantity of milk = T
So the formula becomes 25 = x/(T-x)×100
1 =x/(T-x)×4
T-x=4x
T= 5x
percentage of water in the mixture =x/T×100=x/5x×100=1/5×100=20%
9.(B).Explanation :Let CP of 1 litre milk = Rs.1
SP of 1 litre mixture = CP of 1 litre milk = Rs.1
Gain=50/3%
CP of 1 litre mixture = 100/(100+Gain%)×SP
=100/(100+50/3)×1=100/(350/3)=300/350=6/7
By the rule of alligation, we have
CP of 1 litre water                            CP of 1 litre milk
0                                                           1
CP of 1 litre mixture
6/7
1 – 6/7 = 1/7                                  6/7 – 0 = 6/7
Quantity of water : Quantity of milk = 1/7 : 6/7 = 1 : 6
10.( D)Explanation :By the rule of alligation, we have
CP of 1 kg Rice of 1st kind          CP of 1 kg Rice of 2nd kind
7.1                                                   9.2
Mean Price
8
9.2 – 8 = 1.2                                               8 – 7.1 = .9
Required ratio = 1.2 : .9 = 12 : 9 = 4 : 3