# HCL Previous Years Question Bank 13

### Set A

1. A and B are two alloys of gold and copper prepared by mixing metals in the ratios 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

2. Rs. 561 are divide among A, B and C such that A’s share is Rs. 120 more than B’s share and Rs. 120 less than C’s. B’s share will be
(a) Rs. 73
(b) Rs. 80
(c) Rs. 67
(d) Rs. 76

3. In an alloy, the ratio of copper and zinc is 5 : 2. If 1.250 kg of zinc is mixed in 17 kg 500 gm of alloy, then the ratio of copper and zinc will be
(a) 2 : 1
(b) 2 : 3
(c) 3 : 2
(d) 1 : 2

4. If 378 coins consist of rupees, 50 paise and 25 paise coins whose values are in the ratio of 13 : 11 : 7, the number of 50 paise coins will be
(a) 132
(b) 128
(c) 136
(d) 133

5. Rs. 500 was invested at 12% per annum simple interest and a certain sum of money invested at 10% per annum simple interest. If the sum of the interests on both the sums after 4 years is Rs. 480, the later sum of money is
(a) Rs. 450
(b) Rs. 750
(c) Rs. 600
(d) Rs. 550

6. A money lender finds that due to a fall in the annual rate of interest from 8% to 7(3/4), his yearly income diminishes by Rs. 61.50. His capital is
(a) Rs. 22400
(b) Rs. 23800
(c) Rs. 24600
(d) Rs. 26000

7. The average age of 40 students of a class is 15 years. When 10 new students are admitted, the average age is increased by 0.2 years. The average age of new students is
(a) 15.2 years
(b) 16 years
(c) 16.2 years
(d) 16.4 years

8. A man travels a distance of 24 km at 6 km/hr another distance of 24 km at 8 km/hr and a third distance of 24 km at 12 km/hr. His average speed for whole journey (in km/hr) is
(a) 8(2/3)
(b) 8
(c) 2(10/13)
(d) 9

9. A cricketer has a certain average of runs for his 8 innings. In the ninth inning he scores 100 runs, thereby increasing his average by 9 runs. His new average of runs is
(a) 20
(b) 24
(c) 28
(d) 32

10. The diameter of the iron ball used for the shot-put game is 14 cm. It is melted and then a solid cylinder of height 2(1/3)cm is made. What will be the diameter of the base of the cylinder?
(a) 14 cm
(b) 28 cm
(c) 14/3 cm
(d) 28/3 cm

Solutions  :

1.  (c);       Alloys ‘A’                                       Alloy ‘B’
Gold       :     Copper                     Gold       :     Copper
5          :          3                              5          :         11
Total ( 5 + 3 = 8)                             Total (5 + 11 = 16)
or, (5 × 2) + (3 × 2) = 16
to equalize
[L.C.M. of 8 & 16 = 16 unit of each alloy]
The ratio of gold and copper in the alloy C
= ((5 × 2) + 5)/((3 × 2) + 11) = 15/17
⇒ 15 : 17

2. (c); A = B + 120 = C – 120
A = x (say)
B = x – 120
C = x + 120
A + B + C = 561
x + (x – 120) + (x + 120) = 561
3x  = 561
x = 561/3 = 187
B’s share = 187 – 120 = 67

3. (a); In 17 kg 500 gm of Alloy
Copper             :               Zinc
5                  :                  2
5/7 ×  17500 cm     :     2/7 × 17500 gm
=12500 gm         :          =5000 gm
Now, When 1.250 kg (or 1250 gm) of zinc is mixed in 17 kg 500 gm of alloy.
Then, in the new mixed,
Amount of zinc = (5000 + 1250) gm
= 6250 gm
So, in new mixture, ratio of copper & zinc = 12500 gm : 6250 gm
= 2 : 1

4. (a);                                             Rs. 1               50 p                25 p
Respective values                         13         :         11         :          7
So, Respective no. of coins    (13 × 1)         (11 × 2)           (7 ×4)
Total no. of coins in 1 set = 13 + 22 + 28 = 63 coins
Total no. of sets of coins = 378/63 coins= 6 set
⇒ No. of 50 paise coins = 22 × 6 = 132 coins

5. (c);                         Rs. 500 Required Sum
Rate of Interest            12%               10%
S.I. after 4 years            480                 480
SI is same,
⇒ Rs. 500/ Required Sum = 10%/12%
⇒ Required sum = Rs. 500/5 × 6 = Rs. 600

6. (c); Decrease in rate of interest = 8% – 7 ¾ %
= 8% – 31/4% = ¼ %
Now, ATQ,
¼% of capital = Rs. 61.50
⇒ Total capital (i.e. 100%) = Rs. 61.50 × 4 × 100 = Rs. 24600

7. Total ages of 40 students = 40 × 15 = 600 years
Let the average age of 10 new students = x yr
(600 + 10x)/50 = 15.2
600 + 10x = 15.2 × 50
600 + 10x = 760
x = (760 – 600)/10
x = 160/10 = 16 yrs.

8. (b); T1 = 24/6 = 4 hrs, T2 = 24/8 = 3 hrs,
T3= 24/12 = 2 hrs
Average speed = (24 + 24 + 24)/(4 + 3 + 2) = 72/9 = 8 km/hr

9. (c); Let average of 8 innings = x
Total runs = 8x
(8x + 100)/9 = x + 9
x =  19
New average = x + 9 = 19 + 9 = 28

10. (b); Radius of the shot put ball = 7 cm
Height of the cylinder = 7/3 cm
Volume of the shot put = Volume of the cylinder
4/3 π × 73 = π × R2 × 7/3
R = √(4 × 72) = 2 × 7 = 14 cm

d = 2R = 2 × 14 = 28 cm

### Set B

1.  Simplify

(2 3/4 )/(1 5/6) ÷ 7/8 × (1/3 + 1/4) + 5/7 ÷ 3/4 of 3/4
(a) 56/77
(b) 49/80
(c)143/63
(d) 3 2/9

2. If x = (√3 + 1)/ (√3 – 1) and y = (√3 – 1)/ (√3 + 1), the value of x2 + y2 is
(a) 14
(b) 13
(c) 15
(d) 10

3.HCF and LCM of two numbers are 7 and 140 respectively. If the numbers are between 20 and 45, the sum of the numbers, is
(a) 70
(b) 77
(c) 63
(d) 56

4. The present age of a father is 3 years more than three times the age of his son. Three years hence, father’s age will be 10 years more than twice the age of the son. The father’s present age is
(a) 33 years
(b) 39 years
(c) 45 years
(d) 40 years

5.Three numbers which are co-prime to one another are such that the product of the first two is 551 and that of the last two is 1073. The sum of the three number is
(a) 75
(b) 81
(c) 85
(d) 89

6.A number of boys raised Rs. 400 for a famine relief fund, each boy giving as many 25 paise coins as there were boys. The number of boys was
(a) 40
(b) 16
(c) 20
(d) 100

7. A circular park has a path of uniform width around it. The difference between outer and inner circumference of the circular path is 132m. Its width is (Take π = 22/7)
(a) 22 m
(b) 20 m
(c) 21 m
(d) 24 m

8. The perimeter of a rhombus is 40 m and its height is 5 m. Its area is
(a) 60 sq. m.
(b) 50 sq. m.
(c) 45 sq. m.
(d) 55 sq. m.

9. The area of the biggest circle, which can be drawn inside a square of side 21 cm (Take π = 22/7) is
(a) 344.5 sq. cm.
(b)  364. 5 sq. m.
(c) 346.5 sq. cm.
(d) 366.5 sq. m.

10. If the length and the perimeter of a rectangle are in ratio 5 : 16, then its length and breadth will be in the ratio
(a) 5 : 11
(b) 5 : 8
(c) 5 : 4
(d) 5 : 3

Solution

1. (c); (2 3/4 )/(1 5/6) ÷ 7/8 × (1/3 + 1/4) + 5/7 ÷ 3/4 of 3/4
= 1 + 80/63 = 143/63

2. (a); x = (√3 + 1)/ (√3 – 1) = 2 + √3
y = (√3 + 1)2/ (3 – 1) = 2 – √3
x2 + y2 = (2 + √3)2 + (2 –√3)2
= 2[4 + 3] = 14

3. (c); H.C.F. × L.C.M. = 7 × 140 = 980
= 1st no. × 2nd no. (between 20 & 45)
= 28 × 35
So, sum of the number = 28 + 35 = 63

4. (a) Present age of son = x yrs.
Present age of father = 3x + 3 yrs
After 3 years, son = x + 3 yrs
Father = 3x + 3 + 3
= 3x + 6 yrs
ATQ,
3x + 6 = 2(x + 3) + 10
x = 10
Father’s present age = 3 × 10 + 3 = 33 yrs

5. (c); A B C — Let the three co-prime numbers
ATQ, A × B = 551 and B × C = 1073
And ∵ 19 × 29 = 551 and 29 × 37 = 1073
⇒  A = 19, B = 29 and C = 37

⇒ Sum of three numbers = 19 + 29 + 37 = 85

6. (a); No. of boys = x
Contribution of each = Rs. x/4
Total contribution = x × x/4
x × x/4 = 400
x = 40

7.(c); 2πr1 – 2πr2 = 132
(r1 – r2) = 132/2π
=132 × 7/(2×22) = 21 m
Width of the path = (r1 – r2) = 21 m

8. (b); Side of the rhombus = 10 m
Height = 5 m
Area = b × h = 105 = 50 sq. m.

9. (c); D. of the circle = 21 m
Area of the biggest circle = π × (21/2)2
22/7 × 21/2 × 21/2 = 346.5 sq. m .

10. (d); l/(2(l + b)) = 5/16
10l + 10b = 16l
10b = 6l
l/b = 10/6 = 5/3
l : b = 5 : 3

### Set C

1. n + (2/3) n + (1/2) n + (1/7) n = 97, then the value n is
(a) 40
(b) 42
(c) 44
(d) 46

2. The height of a cone is 30 cm. A small cone is cut off at the top by a plane parallel to its base. If its volume is 1/27 of the volume of the cone, at what height, above the base, is the section made?
(a) 6 cm
(b) 8 cm
(C) 10 cm
(d) 20 cm

3. A loss of 8.5% gets converted into a profit of 6.25% when the selling price of an article is increased by Rs. 590. The cost price of the article is
(a) Rs. 4000
(b) Rs. 5900
(C) Rs. 9440
(d) Rs. 26222.22

4. A tradesman marks his goods at 25% above its cost price and allows purchasers a discount of 12 ½% for cash payment. The profit, he thus makes, is
(a) 9 (3/8)%
(b) 9 (1/2)%
(C) 8 (1/2)%
(d) 8(3/8)%

5. A shopkeeper allows 4% discount on his marked price. If the cost price of an article is Rs. 100 and he has to make a profit of 20%, then his marked price must be
(a) 96
(b) 120
(C) 125
(d) 130

6. Diameter of a wheel is 3m. The wheel revolves 28 times in a minute. To cover             5.280km distance, the wheel will take (Take π = 22/7)
(a) 10 minutes
(b) 20 minutes
(C) 30 minutes
(d) 40 minutes

7. The sides of a triangle are 3 cm, 4 cm and 5 cm. The area (in cm2) of the triangle formed by joining the mid points of the sides of the triangle is
(a) 6
(b) 3
(C) 3/2
(d) 3/4

8. A cistern of capacity 8000 litres measures externally 3.3 m × 2.6 m × 1.1 m and its walls are 5 cm thick. The thickness of the bottom is
(a) l m
(b) 1.1m
(C) 1 dm
(d) 90 cm

9. Given √2 = 1.414. The value of √8 + 2√32 – 3√128 + 4√50 is
(a) 8.484
(b) 8.526
(C) 8.426
(d) 8.876

10. A cylindrical tube open at both ends is made of metal. The internal diameter of tube is 11.2 cm and its length is 21cm. The metal everywhere is 0.4 cm thick. The volume of the metal (Take π = 22/7) is
(a) 316 cm3
(b) 310 cm3
(C) 306.24cm3

(d) 280.52cm3

Solutions

1. (b); n + (2/3) n + (1/2) n + (1/7) n = 97
or, n ((42 + 28 + 21+ 6)/42) = 97
⇒ n = 42

2. (d); Here, [∵ rb/rs = hb/hs]

Volume of small cone = (value of bigger cone)/27
i.e. 1/3 π(rS)2(hS) = (1/3 π(rb)2(hb))/27
⇒ ((rb)2(hb))/((rS)2(hS)) = 27
or, ((rb × rb × hb)/(rS × rS × hS)) = ( 3× 3 × 3)/(1 × 1 ×1)
or, hb/hS = 3/1 ⇒ hS = hb/3 = 30/3 = 10 cm
⇒ The required height above the base = (30 – 10) cm
= 20 cm

3. (a); (8.5% + 6.25%) of C.P. = Rs. 590
i.e. 14.75% of C.P. = Rs. 590
So, C.P. (i.e. 100%) = Rs. 590 × 100/14.75 = Rs. 4000

4. (a); Let C.P. = x
So, M.P. = 25% of x
= (125/100) x
And S.P. = Price after discount of 12.5% on M.P.
= (87.5/100) × (125/100) x
= (35/32) x
So, % profit = ((35/32)x – x) × 100%/x
= 75/8 % = 9 3/8%

5. (c); C.P. (100)     +20%       S.P. (120)     –4% discount      M.P.
=  96% of MP
96% of M.P. = 120
So, M.P. (i.e. 100%) = 120 × 100/96 = 125

6. (b); Circumference of wheel = πd
= (22 × 3/7) m
= Distance covered in 1 minute = 28 × 22/7 × 3 m = 264 m
So, Time taken by wheel to cover a distance of 5.280 km (or, 5280 m)
= 5280/264 minute = 20 minutes

7. (c); Sides are 3, 4 and 5 cm
so, area of right angle triangle is=1/2*3*4=6 cm2
so the mid point of triangle divide area into 4 parts
then area covered by the mid point =6/4=  3/2 cm2

8. (c);
l                  b                 h                 Volume
Externally         3.3 m         2.6m         1.1 m
330 m      260 cm     110 cm         9438000cm3
Internally         320 m      250 cm                          8000000 cm3
⇒ Internal height = (8000000 cm3)/((320 × 250) cm2)
= 100 cm
⇒ thickness of the bottom
= (110 – 100) cm
= 10 cm
= 1 dm

9. (a); √8 + 2√32 – 3√128 + 4 √50
= √(2 × 4) + 2√(2 × 16) – 3√(2 × 64) + 4√(2 × 25)
= 6√2 = 6 × 1.414 = 8.484

10. (c); Volume of metal = External vol. of cylindrical tube – Internal vol. of cylindrical tube

= π(rex)2h – π(rin)2h = πh{rex)2 – (rin)2}
= πr {(12/2)2 – (11.2/2)2}
= 306.24 cm3