HCL Previous Years Question Bank 10

Find solutions for all the questions at the end of each set:

Set A



1.The proportion of acid and water in three samples is 2 : 1 ,3 : 2 and 5 : 3. A mixture containing equal quantities of all three samples is made. The ratio of water and acid in the mixture is : 
(a) 120 : 133  
(b) 227 : 133   
(c) 227 : 120   
(d) 133 : 227


2.40 litres of a mixture of milk and water contains 10% of water, the water to be added, to make the water content 20% in the new mixture is : 
(a) 6 litres  
(b) 6.5 litres  
(c) 5.5 litres  
(d) 5 litres

3.Zinc and copper are in the ratio 5 : 3 in 400 gm of an alloy. How much of copper (in grams) should be added to make the ratio 5 : 4 ? 
(a) 50   
(b) 66    
(c) 72    
(d) 200

4.Two vessels A and B contain milk and water mixed in the ratio 8 : 5 and 5 : 2 respectively. The ratio in which these two mixtures be mixed to get a new mixture containing 69 3/13% milk is: 
(a) 3 : 5  
(b) 5 : 2 
(c) 5 : 7  
(d) 2 : 7

5.A barrel contains a mixture of wine and water in the ratio 3 : 1. How much fraction of the mixture must be drawn off and substituted by water so that the ratio of wine and water in the resultant mixture in the resultant mixture in the barrel becomes 1 : 1 ? 
(a) 1/4  
(b) 1/3  
(c) 3/4   
(d) 2/3 

6.A jar contained a mixture of two liquids A and B in the ratio 4 : 1 when 10 litres of the mixture was taken out and 10 litres of liquid B was poured into the jar, this ratio become 2 : 3. The quantity of liquid A contained in the jar initially was 
(a) 4litres  
(b) 8litres 
(c) 16litres  
(d) 40litres

7.A and B are tow alloys of gold and copper prepared by mixing metals in the ratio  5 : 3 and 5 : 11 respectively. Equal quantities of these alloys are melted to form a third alloy C.  The ratio of gold and copper in the alloy C is 
(a) 25 : 33   
(b) 33 : 25 
(c) 15 : 17  
(d) 17 : 15

8.A mixture contains wine and water in the ratio 3 : 2 and another mixture contains them in the ratio 4 : 5. How many litres of the latter must be mixed with 3 litres of the former so that the resulting mixture may contain equal quantities of wine and water? 
(a) 5 2/5  
(b) 5 2/3  
(c) 4 1/2   
(d) 3 3/4

9.An alloy contains zinc, copper and tin in the ratio 2 : 3 : 1 and another contains copper, tin and lead in the ratio 5 : 4 : 3. If equal weights of both alloys are melted together to form a third alloy, then the weight of lead per kg in the new alloy will be
(a)  1/2  kg  
(b)  1/8  kg  
(c)  3/14 kg   
(d)  7/9  kg

10.An alloy contains copper, zinc and nickel in the ratio of 5 : 3 : 2. The quantity of nickel in kg that must be added to 100 kg of this alloy to have the new ratio 5 : 3 : 3 is 
(a) 8  
(b) 10   
(c) 12    
(d) 15

ANSWERS AND SOLUTION
1.(b) Required ratio
= (2/3+3/5+5/8) :(1/3+2/5+3/8)
= ((80+72+75 )/120) : ((40+48+45 )/120)
= 227 : 133

2.(a) In 400 gm of alloy,
Zinc = 5/8 * 400 = 250 gm.
Copper = 3/8 * 40 = 150 gm.
If x gm of copper be mixed, then 250/(150 + x) = 5/4
=> 750 + 5x = 1000
=> 5x = 1000 – 750 = 250
x = 50 gm.

3.(d) Water content in 40 litres of mixture = 40 * 10/100 = 4 litres
so Milk content = 40 – 4  = 36 litres
Let x litres of water is mixed.
so (4+x )/(40+x )=20/100 ? (4+x )/(40+x )=1/5
=> 20 + 5x = 40 + x
4x = 20
X = 5 litres

4.(d)
5.(b) Let the barrel contain 4 liters  of mixture.
so Wine = 3 litres
Water = 1 litre
Let x litre mixture is taken out.
so Wine in (4 – x) litres mixtures = 3/4(4 –x)
On adding x litres water, water in mixture
= (4 – x) * 1/4 + x
= 1 – x/4 + x
= (4 – x+ 4x)/4 = (4 + 3x)/4  
? 3/4 (4-x)= (4+3x )/4
=> 3 – 3x/4 = 1 + 3x/4
=> 2 = 6x/4
x  = 2 * 4/6 = 4/3
Required answer
= 4/3//4  = 1/3

6.(d) Let the initial quantity of liquids A and B in the jar be 4x and x litres respectively.
After taking out 10 litres of the mixture.
Liquied A = 4x – 4/5 * 10 = (4x – 8) litres
Liquied B = 4x – 1/5 * 10 = (4x -2) litres
After pouring 10 litres of liquied B,  (4x-8 )/(4x-2+10 )= 2/3
12x – 24 = 8x + 16
4x = 40
X = 40/4 = 10
so quantity of liquid A = 4x
= 4 x 10 = 40 litres

7.(c) Let 1 kg of each of the alloys A and B  be mixed together.
In alloy A,
Quantity of gold = 5/8 kg.
Quantity of copper = 3/8 kg.
In alloy B,
Quantity of gold = 5/16 kg.
Quantity of copper = 11/16 kg.
so Required ratio = (5/8+5/16) :(3/8+11/16)
= 15/16 : 17/16 = 15 : 17

8.(a)
9.(b) Quantity of glass in 1 kg of first alloy = zero
Quantity of glass in 1 kg.of second alloy = 3/12 = 1/4 kg
so Quantity of glass in 1 kg of new alloy = 1/8 kg
 
10.(b) Let x kg of nickel be mixed.
so (20+x)/(100+x)=3/11
=> 220 + 11x = 300 +  3x
=> 11x – 3x = 300 – 220
=> 8x = 80
=> X = 10 kg.

Find solutions for all the questions at the end of each set:

Set B


1.The marked price of a watch is Rs. 1000. A retailer buys it at Rs. 810 after getting two successive discounts of 10% and another rate which is illegible. What is the second discount rate? 
(a) 15%   
(b) 10%   
(c) 8%  
(d) 6.5% 
 
2.A fan is listed at Rs. 1500 and a discount of 20% is offered on the list price. What additional discount must be offered to the customer to bring the net price to Rs. 1104?
(a) 8%   
(b) 10%   
(c) 12%  
(d) 15%
 
3.The marked price of watch was Rs. 820. A man bought the watch for Rs. 570. 72 after getting two successive discounts, of which the first was 20%. The second discount was 
(a) 18%   
(b) 15%   
(c) 13%  
(d) 11%
 
4.A purchased a dining table, marked at Rs. 3,000 at a successive discounts of 10% and 15% respectively. He gave Rs. 105 as transportation charge and sold it at Rs. 3,200. What is his gain percentage? 
(a) 22 1/3%   
(b) 25%   
(c) 33 1/3%  
(d) 37 17/24%
 
5.If a dining table with marked price Rs. 6,000 was sold to a customer for Rs. 5, 500, then the rate of discount allowed on the table is 
(a) 10%   
(b) 8%    
(c) 8 1/3%  
(d) 9%
 
6.A discount of 15% on the article is the same as discount of 20% on a second article. The costs of the two articles can be: 
(a) Rs. 85, Rs. 60  
(a) Rs. 60, Rs. 40 
(a) Rs. 40, Rs. 20 
(a) Rs. 80, Rs. 60

7.A shopkeeper gives 12 percent additional discount after giving an initial discount of 20 percent on the marked price of a radio. If the sale price of the radio is Rs. 704, the marked price is 
(a) Rs. 844.80  
(b) Rs. 929.28 
(c) Rs. 1,044.80 
(d) Rs. 1,000

8.If an electricity bill is paid before due date, one gets a reduction of 4% on the amount of the bill. By paying the bill before due date a person got a reduction of Rs. 13. The amount of his electricity bill was 
(a) Rs. 125  
(b) Rs. 225  
(c) Rs. 325  
(d) Rs. 425
 
9.During a month- long annual sale, a shopkeeper sells his goods at a discount of 50%. But in the last week, he offers an additional discount of 40%. If the original price of a shirt is Rs.x then the price, in rupees, during the last week of the sale will be 
(a) 90% of x 
(b) 70% of x  
(c) 30 % of x   
(d) 10% of x 
 
10.The cost of manufacture of a tape recorder is Rs. 1,500. The manufacturer fixes the marked price 20% above the cost of manufacture and allows a discount in such a way as to get a profit of 8%. The rate of discount is 
(a) 12%   
(b) 8%    
(c) 20%  
(d) 10%

ANSWERS AND SOLUTION

1.(b) Price after 10% first discount
 =  1000 * (100-10)/100
= 1000 * 90/100 = Rs. 900
Given :
Price after second discount
= Rs. 810
so Second discount
= 900 – 810 = Rs. 90
so Percentage second discount
= 90 * 100/900 = 10%

2.(a) After a discount of 20%
Listed price = 80% of Rs. 1500
= Rs. (1500 * 80/100)= 1200
Difference
= Rs. (1200 – 1104) = Rs. 96
Let x% of 1200 = 96
x = 96 * 100/1200 = 8
so Second discount = 8%

3.(c) Total discount
= Rs. (820 – 570.72)
= Rs. 249.28
First discount = 820 * 20/100
= Rs. 164
so Second discount
Rs. (249.28 – 164)
= Rs. 85.28
Price of the article after first discount
= Rs. (820 – 164) = Rs. 656
If the second discount be x% then
x% of 656 = 85.28 
=> x = 85.28 * 100 /656 = 13%

4.(c)C.P. for A
= 3000 * 90/100 * 85/100
= Rs. 2295
Actual C.P. = 2295 + 105
= Rs. 2400
so Gain percent = 800/2400 * 100
= 100/3 = 33 1/3%

5.(c)Discount = 6000 = 5500
= Rs. 500
If discount = x%, then
6000 * x /100 = 500
=> x = 500/60 = 25/3 = 8 1/3 %

6.(d) 15% of Rs. 80 = 80 * 15/  100
= Rs. 12
and 20% of 60 = (60 ×20)/100
= Rs. 12
Therefore, 15% of 80 and 20% of 60 are same.
Hence the cost prices should be Rs. 80 and Rs. 60.

7.(d)  Let the marked price of the radio be Rs.  x.
According to the question.
x* 80/100 * 88/100 = 704
so x = (704 ×100 ×100)/(80 ×88)
= Rs. 1000

8.(c) Let the amount of the bill be Rs. x.
4x/100 = 13
x = 1300/4 = Rs. 325

9.(c)Single equivalent discount
= (50+40-(50 ×40)/100)%
= 70%
so Required price of shirt
= 30% of x

10.(4) Marked price of tape recorder
= 1500 * 120/100 = Rs. 1800
Gain = 1500 * 8/100 = Rs. 120
Discount = 1800 -  (1500 + 120)
= Rs. 180
so Discount per cent = x%, then (1800 × x)/100 = 180.
x = 10%

Find solutions for all the questions at the end of each set:

Set C


1.The age of father 3 years ago was 8 times the age of his son. Presently the father’s age is 5 times that of the son. Find their present ages. 
(a) 35 
(b) 30 
(c) 38 
(d) 40


2.The present age of A is 9 times the present age of B. 3 years hence A’s age will be 5 times that of B. Find their present ages. 
(a) 25 
(b) 23 
(c) 19 
(d) 27

3.The ratio of Ram’s present age to Mohan’s present age is 5 : 6.  8 years hence this ratio changes to 7 : 8 Find their present ages. 
(a) 18 
(b) 19 
(c) 20 
(d) 21

4.The sum of present ages of Rohit and Mohit is 30 years. 5 years hence the ratio of their ages will be Rohit : Mohit : : 5 : 3. Find their present ages. 
(a) 18 
(b) 19 
(c) 21 
(d) 20

5.The average age of a man and his son is 35 years. The ratio of their ages is 5 : 2 respectively. What is the son’s age? 
(a) 50 years 
(b) 35 years 
(c) 15 years 
(d) 20 years 

6.The ages of Aarzoo and Arnav are in the ratio of 11 : 13 respectively. After 7 years the ratio of their ages will be 20 : 23. What is the difference in years between their ages? 
(a) 4 years 
(b) 7 years 
(c) 6 years 
(d) 5 years

 7.The ratio between the ages of Ram and Mohan is 4 : 5 and that between Mohan and Anil is 5 : 6. If sum of the ages of three be 90 years. How old is Mohan? 
(a) 24 years 
(b) 20 years 
(c) 30 years 
(d) 25 years

8.The age of father is three times the age of his son at present. Four years hence the ratio between the ages of father and son becomes 13 : 5 respectively. What is the present age of father? 
(a) 48 years 
(b) 42 years 
(c) 52 years 
(d) Cannot be determined 

9.Harsha is 40 years old and Rita is 60 years old. How many years ago  was the ratio of their ages 3 : 5 ? 
(a) 10 years 
(b) 20 years 
(c) 37 years 
(d) 5 years

10. Present ages of Radha and sudha are in the ratio of 7 : 9 respectively. Five years ago ratio of their ages was 3 : 4. What will be Sudha’s age after 3 years from now? 
(a) 48 years 
(b) 42 years 
(c) 43 years 
(d) 38 years 

Answers and Solution:
1.(a) Let the present age of son be x years.
Then, the present age of father = 5x years.
3 years ago,
Son’s age = (x -3) years
Father’s age = (5x – 3) years
But 3 years ago the father’s age was 8 times that of his son.
Therefore, we can write.
8 (x -3) = 5x -3
or, 8x  - 24 = 5x -3
or, 3x = 21
or, x = 7 years
and 5x = 35
We get the present age of son = 7 years
and the present age of father = 35 years

2.(d) Let the present age of B be x years.
Then, the present age of A = 9x years
3 years hence,
B’s age = (9x + 3) years.
But it is given that 3 years hence A’s age will be 5 times the age of B.
so 9x + 3 = 5(x + 3 )
Or, 9x + 3 = 5x + 15
Or, 4x = 12
Or, x =3
B’s present age = 3 years
A’s present age = 9 * 3 = 27 years
3.(c)
4.(d)
5.(d) Let the present ages of father and son be 5x and 2x years respectively.
According to the question.
=> 5x + 2x = 2 * 35
=> 7x = 70
=> X = 70/7 = 10
so son's age = 2x =  2 x 10 = 20

6.(c)Let the present ages of Aarzoo and Arnav be 11x and 13x years respectively,
According to the question,
=> (11x+7)/(13x+7)=20/23
=> 260x + 140 = 253x + 161
=> 260x – 253x = 161 – 140
=> 7x = 21
=> X = 21/7 = 3
Difference between their ages
= 13x – 11x = 2x = 2 * 3 = 6 years

7.(c) Ram : Mohan = 4 : 5
Mohan : Anil = 5 : 6
Ram  : Mohan : Anil = 4 : 5 : 6
so Mohan’s age = 5/15 * 90 = 30 years
 
8.(a) Present age of son = x years then father’s present ages =3x years
According to the question
(3x+4)/(x+4)=13/5
=> 15x + 20 = 13x + 52 = 2x = 32
so X = 16
so Present age of father = 3 * 16 = 48 years

9.(a)Let x years ago the radio of their ages was 3 : 5
so According to the question
(40-x )/(60-x )=3/5
Or, 200 – 5x = 180 – 3x
or, 2x = 20
so X = 10 years
10.(a)



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